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Hindawi Publishing Corporation Mathematical Problems in Engineering Volume 2013, Article ID 273052, 14 pages http://dx.doi.org/10.1155/2013/273052 Research Article Analytical Solutions for Steady Heat Transfer in Longitudinal Fins with Temperature-Dependent Properties Partner L. Ndlovu and Raseelo J. Moitsheki Center for Differential Equations, Continuum Mechanics and Applications, School of Computational and Applied Mathematics, University of the Witwatersrand, Private Bag 3, Johannesburg 2050, South Africa Correspondence should be addressed to Raseelo J. Moitsheki; [email protected] Received 30 January 2013; Accepted 24 April 2013 Academic Editor: Ireneusz Zbicinski Copyright © 2013 P. L. Ndlovu and R. J. Moitsheki. is is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited. Explicit analytical expressions for the temperature profile, fin efficiency, and heat flux in a longitudinal fin are derived. Here, thermal conductivity and heat transfer coefficient depend on the temperature. e differential transform method (DTM) is employed to construct the analytical (series) solutions. ermal conductivity is considered to be given by the power law in one case and by the linear function of temperature in the other, whereas heat transfer coefficient is only given by the power law. e analytical solutions constructed by the DTM agree very well with the exact solutions even when both the thermal conductivity and the heat transfer coefficient are given by the power law. e analytical solutions are obtained for the problems which cannot be solved exactly. e effects of some physical parameters such as the thermogeometric fin parameter and thermal conductivity gradient on temperature distribution are illustrated and explained. 1. Introduction Fins are surfaces that extend from a hot object (body) to increase the rate of heat transfer to the surrounding fluid. In particular, fins are used extensively in various industrial applications such as the cooling of computer processors, air conditioning, and oil carrying pipe lines. A well-documented review of heat transfer in extended surfaces is presented by Kraus et al. [1]. e problems on heat transfer particularly in fins continue to be of scientific interest. ese problems are modeled by highly nonlinear differential equations which are difficult to solve exactly. However, Moitsheki et al. [24] have attempted to construct exact solutions for the steady state problems arising in heat flow through fins. A number of techniques, for example, Lie symmetry analysis [2], He’s variational iteration method [5], Adomain decomposition methods [6], homotopy perturbation methods [7], homotopy analysis methods [8], methods of successive approximations [9], and other approximation methods [10] have been used to determine solutions of the nonlinear differential equations describing heat transfer in fins. Recently, the solutions of the nonlinear ordinary differ- ential equations (ODEs) arising in extended surface heat transfer have been constructed using the DTM [1119]. e DTM is an analytical method based on the Taylor series expansion and was first introduced by Zhou [20] in 1986. e DTM approximates the exact solution by a polynomial, and previous studies have shown that it is an efficient means of solving nonlinear problems or systems with varying parame- ters [21]. Furthermore, DTM is a computational inexpensive tool for obtaining analytical solution, and it generalizes the Taylor method to problems involving procedures such as fractional derivative (see e.g., [2224]). Also, this method converges rapidly (see e.g., [25]). Models arising in heat transfer through fins may contain temperature-dependent properties such as thermal conduc- tivity and heat transfer coefficient. e dependency of ther- mal conductivity and heat transfer coefficient on temperature renders such problems highly nonlinear and difficult to solve, particularly exactly. ermal conductivity may be modeled for many engineering applications by the power law and by linear dependency on temperature. On the other hand, heat

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Page 1: Research Article Analytical Solutions for Steady Heat ...downloads.hindawi.com/journals/mpe/2013/273052.pdf · Research Article Analytical Solutions for Steady Heat Transfer in Longitudinal

Hindawi Publishing CorporationMathematical Problems in EngineeringVolume 2013 Article ID 273052 14 pageshttpdxdoiorg1011552013273052

Research ArticleAnalytical Solutions for Steady Heat Transfer inLongitudinal Fins with Temperature-Dependent Properties

Partner L Ndlovu and Raseelo J Moitsheki

Center for Differential Equations Continuum Mechanics and Applications School of Computational and Applied MathematicsUniversity of the Witwatersrand Private Bag 3 Johannesburg 2050 South Africa

Correspondence should be addressed to Raseelo J Moitsheki raseelomoitshekiwitsacza

Received 30 January 2013 Accepted 24 April 2013

Academic Editor Ireneusz Zbicinski

Copyright copy 2013 P L Ndlovu and R J Moitsheki This is an open access article distributed under the Creative CommonsAttribution License which permits unrestricted use distribution and reproduction in any medium provided the original work isproperly cited

Explicit analytical expressions for the temperature profile fin efficiency and heat flux in a longitudinal fin are derived Here thermalconductivity and heat transfer coefficient depend on the temperature The differential transform method (DTM) is employed toconstruct the analytical (series) solutions Thermal conductivity is considered to be given by the power law in one case and by thelinear function of temperature in the other whereas heat transfer coefficient is only given by the power lawThe analytical solutionsconstructed by the DTM agree very well with the exact solutions even when both the thermal conductivity and the heat transfercoefficient are given by the power law The analytical solutions are obtained for the problems which cannot be solved exactly Theeffects of some physical parameters such as the thermogeometric fin parameter and thermal conductivity gradient on temperaturedistribution are illustrated and explained

1 Introduction

Fins are surfaces that extend from a hot object (body) toincrease the rate of heat transfer to the surrounding fluidIn particular fins are used extensively in various industrialapplications such as the cooling of computer processors airconditioning and oil carrying pipe lines A well-documentedreview of heat transfer in extended surfaces is presented byKraus et al [1] The problems on heat transfer particularlyin fins continue to be of scientific interest These problemsare modeled by highly nonlinear differential equations whichare difficult to solve exactly However Moitsheki et al [2ndash4] have attempted to construct exact solutions for the steadystate problems arising in heat flow through fins A numberof techniques for example Lie symmetry analysis [2] Hersquosvariational iteration method [5] Adomain decompositionmethods [6] homotopy perturbationmethods [7] homotopyanalysis methods [8] methods of successive approximations[9] and other approximation methods [10] have been usedto determine solutions of the nonlinear differential equationsdescribing heat transfer in fins

Recently the solutions of the nonlinear ordinary differ-ential equations (ODEs) arising in extended surface heattransfer have been constructed using the DTM [11ndash19] TheDTM is an analytical method based on the Taylor seriesexpansion and was first introduced by Zhou [20] in 1986TheDTM approximates the exact solution by a polynomial andprevious studies have shown that it is an efficient means ofsolving nonlinear problems or systems with varying parame-ters [21] Furthermore DTM is a computational inexpensivetool for obtaining analytical solution and it generalizes theTaylor method to problems involving procedures such asfractional derivative (see eg [22ndash24]) Also this methodconverges rapidly (see eg [25])

Models arising in heat transfer through fins may containtemperature-dependent properties such as thermal conduc-tivity and heat transfer coefficient The dependency of ther-mal conductivity and heat transfer coefficient on temperaturerenders such problems highly nonlinear and difficult to solveparticularly exactly Thermal conductivity may be modeledfor many engineering applications by the power law and bylinear dependency on temperature On the other hand heat

2 Mathematical Problems in Engineering

transfer coefficient can be expressed as a power law for whichvalues of the exponent represent different phenomena (seeeg [26])

In this paper the DTM is employed to determine theanalytical solutions to the nonlinear boundary value problemdescribing heat transfer in longitudinal fins of rectangularexponential and convex parabolic profiles Both thermalconductivity and heat transfer coefficient are temperaturedependentWe adopt the terminology exact solutions to referto solutions given in terms of fundamental expressions suchas logarithmic trigonometric and exponential Howeveranalytical solutions will be series solutions and in partic-ular those constructed using the DTM The mathematicalmodelling of the problem under consideration is describedin Section 2 A brief discussion on the fundamentals of theDTM is be provided in Section 3 The comparison of theexact and analytical solutions constructed byDTM is given inSection 4 In Section 5 we provide analytical solutions for theheat transfer in longitudinal fins of various profilesHere heattransfer coefficient is given by the power law and we considertwo cases of the thermal conductivity namely the powerlaw and the linear function of temperature Furthermorewe describe the fin efficiency and the heat flux in Section 6Some exciting results are discussed in Section 7 Lastly theconcluding remarks are provided in Section 8

2 Mathematical Models

We consider a longitudinal one dimensional fin of cross-sectional area 119860

119888 The perimeter of the fin is denoted by

119875 and its length by 119871 The fin is attached to a fixed primesurface of temperature 119879

119887and extends to an ambient fluid

of temperature 119879119886 The fin thickness at the prime surface is

given by 120575119887and its profile is given by 119865(119883) Based on the one

dimensional heat conduction the energy balance equation isthen given by (see eg [1])

119860119888

119889

119889119883(120575119887

2119865 (119883)119870 (119879)

119889119879

119889119883) = 119875119867 (119879) (119879 minus 119879

119886)

0 le 119883 le 119871

(1)

where 119870 and 119867 are nonuniform temperature-dependentthermal conductivity and heat transfer coefficients respec-tively119879 is the temperature distribution119865(119883) is the fin profileand119883 is the space variable The length of the fin is measuredfrom the tip to the prime surface as shown in Figure 1Assuming that the fin tip is adiabatic (insulated) and the basetemperature is kept constant then the boundary conditionsare given by

119879 (119871) = 119879119887

119889119879

119889119883

10038161003816100381610038161003816100381610038161003816119883=0

= 0 (2)

Introducing the following dimensionless variables (see eg[1])

119909 =119883

119871 120579 =

119879 minus 119879119886

119879119887minus 119879119886

ℎ =119867

ℎ119887

119896 =119870

119896119886

1198722

=119875ℎ1198871198712

119860119888119896119886

119891 (119909) =120575119887

2119865 (119883)

(3)

119883 = 0 119883 = 119871

Fin tip

Fin profile

Prime surface

120575

119871

119882

Figure 1 Schematic representation of a longitudinal fin of anunspecified profile

with 119896119886being defined as the thermal conductivity at the

ambient temperature and ℎ119887as the heat transfer at the prime

surface (fin base) reduces (1) to

119889

119889119909[119891 (119909) 119896 (120579)

119889120579

119889119909] minus119872

2

120579ℎ (120579) = 0 0 le 119909 le 1 (4)

Here 120579 is the dimensionless temperature 119909 is the dimension-less space variable 119891(119909) is the dimensionless fin profile 119896 isthe dimensionless thermal conductivity ℎ is the dimension-less heat transfer coefficient and119872 is the thermogeometricfin parameter The dimensionless boundary conditions thenbecome

120579 (1) = 1 at the prime surface (5)

119889120579

119889119909

10038161003816100381610038161003816100381610038161003816119909=0

= 0 at the fin tip (6)

We assume that the heat transfer coefficient is given by thepower law

119867(119879) = ℎ119887(119879 minus 119879119886

119879119887minus 119879119886

)

119899

(7)

where the exponent 119899 is a real constant In fact the values of 119899may vary as between minus66 and 5 However in most practicalapplications it lies between minus3 and 3 [27] In dimensionlessvariables we have ℎ(120579) = 120579119899 We consider the two distinctcases of the thermal conductivity as follows

(a) the power law

119870 (119879) = 119896119886(119879 minus 119879119886

119879119887minus 119879119886

)

119898

(8)

with119898 being a constant and(b) the linear function

119870 (119879) = 119896119886[1 + 120574 (119879 minus 119879

119886)] (9)

Mathematical Problems in Engineering 3

The dimensionless thermal conductivity given by the powerlaw and the linear function of temperature is 119896(120579) = 120579119898 and119896(120579) = 1 + 120573120579 respectively Here the thermal conductivitygradient is120573 = 120574(119879

119887minus119879119886) Furthermore we consider a various

fin profiles including the longitudinal rectangular 119891(119909) =1 the longitudinal convex parabolic 119891(119909) = radic119909 and theexponential profile 119891(119909) = e119886119909 with 119886 being the constant (seealso [16])

3 Fundamentals of the DifferentialTransform Method

In this section the basic idea underlying the DTM is brieflyintroduced Let 119910(119905) be an analytic function in a domain DThe Taylor series expansion function of 119910(119905) with the centerlocated at 119905 = 119905

119895is given by [20]

119910 (119905) =

infin

sum

120581=0

(119905 minus 119905119895)120581

120581[119889120581

119910 (119905)

119889119905120581]

119905=119905119895

forall119905 isin D (10)

The particular case of (10) when 119905119895= 0 is referred to as

the Maclaurin series expansion of 119910(119905) and is expressed as

119910 (119905) =

infin

sum

120581=0

119905120581

120581[119889120581

119910 (119905)

119889119905120581]

119905=0

forall119905 isin D (11)

The differential transform of 119910(119905) is defined as follows

119884 (119905) =

infin

sum

120581=0

H120581

120581[119889120581

119910 (119905)

119889119905120581]

119905=0

(12)

where 119910(119905) is the original analytic function and 119884(119905) is thetransformed function The differential spectrum of 119884(119905) isconfined within the interval 119905 isin [0H] where H is aconstant From (11) and (12) the differential inverse transformof 119884(119905) is defined as follows

119910 (119905) =

infin

sum

120581=0

(119905

H)

120581

119884 (120581) (13)

and if 119910(119905) is expressed by a finite series then

119910 (119905) =

119903

sum

120581=0

(119905

H)

120581

119884 (120581) (14)

Some of the useful mathematical operations performedby the differential transform method are listed in Table 1

The delta function 120575(120581 minus 119904) is given by

120575 (120581 minus 119904) = 1 if 120581 = 1199040 if 120581 = 119904

(15)

4 Comparison of Exact andAnalytical Solutions

In this section we consider a model describing temperaturedistribution in a longitudinal rectangular fin with both ther-mal conductivity and heat transfer coefficient being functions

Table 1 Fundamental operations of the differential transformmethod

Original function Transformed function119910(119905) = 119909(119905) plusmn 119911(119905) 119884(119905) = 119883(119905) plusmn 119885(119905)

119910(119905) = 120572119909(119905) 119884(119905) = 120572119883(119905)

119910(119905) =119889119910(119905)

119889119905119884(119905) = (120581 + 1)119884(120581 + 1)

119910(119905) =1198892

119910(119905)

1198891199052119884(119905) = (120581 + 1)(120581 + 2)119884(120581 + 2)

119910(119905) =119889119904

119910(119905)

119889119905119904119884(119905) = (120581 + 1)(120581 + 2) sdot sdot sdot (120581 + 119904)119884(120581 + 119904)

119910(119905) = 119909(119905)119911(119905) 119884(119905) =

120581

sum

119894=0

119883(119894)119885(120581 minus 119894)

119910(119905) = 1 119884(119905) = 120575(120581)

119910(119905) = 119905 119884(119905) = 120575(120581 minus 1)

119910(119905) = 119905119904

119884(119905) = 120575(120581 minus 119904)

119910(119905) = exp(120582119905) 119884(119905) =120582120581

120581

119910(119905) = (1 + 119905)119904

119884(119905) =119904(119904 minus 1) sdot sdot sdot (119904 minus 120581 + 1)

120581

119910(119905) = sin(120596119905 + 120572) 119884(119905) =120596120581

120581sin(120587120581

2+ 120572)

119910(119905) = cos(120596119905 + 120572) 119884(119905) =120596120581

120581cos(120587120581

2+ 120572)

of temperature given by the power law (see eg [2])The exactsolution of (4) when both the power laws are given by thesame exponent is given by [2]

120579 (119909) = [

cosh (119872radic119899 + 1119909)

cosh (119872radic119899 + 1)]

1(119899+1)

(16)

We use this exact solution as a benchmark or validation ofthe DTM The effectiveness of the DTM is determined bycomparing the exact and the analytical solutionsWe comparethe results for the cases 119899 = 1 and 119899 = 2 with fixed values of119872

41 Case 119899=1 Applying the DTM to (4) with the power lawthermal conductivity 119891(119909) = 1 (rectangular profile) andgivenH = 1 one obtains the following recurrence relation

120581

sum

119894=0

[Θ (119894) (120581 minus 119894 + 1) (120581 minus 119894 + 2)Θ (120581 minus 119894 + 2)

+ (119894 + 1)Θ (119894 + 1) (120581 minus 119894 + 1)Θ (120581 minus 119894 + 1)

minus1198722

Θ (119894)Θ (120581 minus 119894)] = 0

(17)

Exerting the transformation to the boundary condition (6) ata point 119909 = 0

Θ (1) = 0 (18)

The other boundary conditions are considered as follows

Θ (0) = 119888 (19)

4 Mathematical Problems in Engineering

where 119888 is a constant Equation (17) is an iterative formula ofconstructing the power series solution as follows

Θ (2) =1198722

119888

2

(20)

Θ (3) = 0 (21)

Θ (4) = minus1198724

119888

24

(22)

Θ (5) = 0 (23)

Θ (6) =191198726

119888

720

(24)

Θ (7) = 0 (25)

Θ (8) = minus559119872

8

119888

40320

(26)

Θ (9) = 0 (27)

Θ (10) =29161119872

10

119888

3628800

(28)

Θ (11) = 0 (29)

Θ (12) = minus2368081119872

12

119888

479001600

(30)

Θ (13) = 0 (31)

Θ (14) =276580459119872

14

119888

87178291200

(32)

Θ (15) = 0

(33)

These terms may be taken as far as desired Substituting(18) to (32) into (13) we obtain the following analyticalsolution

120579 (119909) = 119888 +1198722

119888

21199092

minus1198724

119888

241199094

+191198726

119888

7201199096

minus559119872

8

119888

403201199098

+29161119872

10

119888

362880011990910

minus2368081119872

12

119888

47900160011990912

+276580459119872

14

119888

8717829120011990914

+ sdot sdot sdot

(34)

To obtain the value of 119888 we substitute the boundarycondition (5) into (34) at the point 119909 = 1 Thus we have

120579 (1) = 119888 +1198722

119888

2minus1198724

119888

24+191198726

119888

720minus559119872

8

119888

40320

+29161119872

10

119888

3628800minus2368081119872

12

119888

479001600

+276580459119872

14

119888

87178291200+ sdot sdot sdot = 1

(35)

Table 2 Results of the DTM and exact solutions for 119899 = 1119872 = 07

119909 DTM Exact Error0 0808093014 080809644 000000342601 0810072036 081007547 000000343402 0815999549 081600301 000000345903 0825847770 082585127 000000350004 0839573173 083957673 000000355905 0857120287 085712392 000000363306 0878426299 087843002 000000372207 0903426003 090342982 000000381408 0932056648 093206047 000000382009 0964262558 096426582 000000326310 1000000000 100000000 0000000000

Table 3 Results of theDTMand Exact Solutions for 119899 = 2119872 = 05

119909 DTM Exact Error0 0894109126 0894109793 000000066501 0895226066 0895226732 000000066602 0898568581 0898569249 000000066803 0904112232 0904112905 000000067204 0911817796 0911818474 000000067805 0921633352 0921634038 000000068506 0933496900 0933497594 000000069407 0947339244 0947339946 000000070208 0963086933 0963087627 000000069409 0980665073 0980665659 000000058610 1000000000 1000000000 0000000000

We omit presenting the tedious process of finding 119888 Howeverone may obtain the expression for 120579(119909) upon substituting theobtained value of 119888 into (34)

42 Case 119899=2 Following a similar approach given inSection 41 and given 119899 = 2 one obtains the analyticalsolution

120579 (119909) = 119888 +1198722

119888

21199092

minus1198724

119888

81199094

+231198726

119888

2401199096

minus1069119872

8

119888

134401199098

+9643119872

10

119888

13440011990910

minus1211729119872

12

119888

1774080011990912

+217994167119872

14

119888

322882560011990914

+ sdot sdot sdot

(36)

The constant 119888may be obtained using the boundary conditionat the fin base The comparison of the DTM and the exactsolutions are reflected in Tables 2 and 3 for different valuesof 119899 Furthermore the comparison of the exact and theanalytical solutions is depicted in Figures 2(a) and 2(b)

Mathematical Problems in Engineering 5

08

082

084

086

088

09

092

094

096

098

1

120579

0 02 04 06 08 1

119909

DTMExact

(a)

086

088

09

092

094

096

098

1

120579

0 02 04 06 08 1

119909

DTMExact

(b)

Figure 2 Comparison of analytical and exact solutions In (a) 119899 = 1119872 = 07 and (b) 119899 = 2119872 = 05

5 Analytical Solutions

It is well known that exact solutions forODEs such as (4) existonly when thermal conductivity and the term containingheat transfer coefficient are connected by differentiation(or simply if the ODE such as (4) is linearizable) [4] Inthis section we determine the analytical solutions for thenonlinearizable (4) firstlywhen thermal conductivity is givenby the power law and secondly as a linear function oftemperature In both cases and throughout this paper theheat transfer coefficient is assumed to be a power law functionof temperature These assumptions of the thermal propertiesare physical realistic We have noticed that DTM runs intodifficulty when the exponent of the power law of the thermalconductivity is given by fractional values and also when thefunction 119891(119909) is given in terms of fractional power law Onemay followMoradi and Ahmadikia [16] by introducing a newvariable to deal with fractional powers of 119891(119909) and on theother hand it is possible to remove the fractional exponent ofthe heat transfer coefficient by fundamental laws of exponentand binomial expansion

Proposition 1 A nonlinear ODE such as (4) may admit theDTM solution if 119891(119909) is a constant or exponential functionHowever if 119891(119909) is a power law then such an equation admitsa DTM solution if the product

119891 (119909) sdot 1198911015840

(119909) = 120572 (37)

holds Here 120572 is a real constant

Proof Introducing the new variable 120585 = 119891(119909) it follows fromchain rule that (4) becomes

120572119889120585

119889119909

119889

119889120585[119896 (120579)

119889120579

119889120585] minus119872

2

120579119899+1

= 0 (38)

The most general solution of condition (37) is

119891 (119909) = (2120572119909 + 120574)12

(39)

where 120574 is an integration constant

Example 2 (a) If 119891(119909) = 119909120590 then 120585 = 119909120590 transforms (4) into

119889

119889120585[119896 (120579)

119889120579

119889120585] minus 4119872

2

120585120579119899+1

= 0 (40)

only if 120590 = 12This example implies that the DTM may only be appli-

cable to problems describing heat transfer in fins withconvex parabolic profile In the next subsections we presentanalytical solutions for (4) with various functions 119891(119909) and119896(120579)

51 The Exponential Profile and Power Law Thermal Con-ductivity In this section we present solutions for equationdescribing heat transfer in a fin with exponential profile andpower law thermal conductivity and heat transfer coefficientThat is given (4) with 119891(119909) = e120590119909 and both heat transfercoefficient and thermal conductivity being power law func-tions of temperature we construct analytical solutions In ouranalysis we consider 119899 = 2 and 3 indicating the fin subject tonucleate boiling and radiation into free space at zero absolutetemperature respectively Firstly given 119899 = 3 and 119898 = 2

6 Mathematical Problems in Engineering

and applying the DTM one obtains the following recurrencerevelation

120581

sum

119894=0

120581minus119894

sum

119897=0

120581minus119894minus119897

sum

119901=0

[120590119901

119901Θ (119897) Θ (119894) (120581 minus 119894 minus 119897 minus 119901 + 1)

times (120581 minus 119894 minus 119897 minus 119901 + 2)Θ (120581 minus 119894 minus 119897 minus 119901 + 2)

+ 2120590119901

119901Θ (119897) (119894 + 1)Θ (119894 + 1)

times (120581 minus 119894 minus 119897 minus 119901 + 1)Θ (120581 minus 119894 minus 119897 minus 119901 + 1)

+ 2120590120590119901

119901Θ (119897) Θ (119894) (120581 minus 119894 minus 119897 minus 119901 + 1)

times Θ (120581 minus 119894 minus 119897 minus 119901 + 1)

minus1198722

Θ(119901)Θ (119897) Θ (119894) Θ (120581 minus 119894 minus 119897 minus 119901) ] = 0

(41)

One may recall the transformed prescribed boundaryconditions (18) and (19) Equation (41) is an iterative formulaof constructing the power series solution as follows

Θ (2) =1198722

1198882

2

Θ (3) = minus1205901198722

1198882

3

Θ (4) =

1198722

1198882

(31205902

minus 21198722

119888)

24

Θ (5) = minus

1198722

1198882

(1205903

minus 41205901198722

119888)

30

Θ (6) =

1198722

1198882

(51205904

minus 781205902

1198722

119888 + 581198724

1198882

)

720

(42)

The pervious process is continuous and onemay consideras many terms as desired (but bearing in mind that DTMconverges quite fast) Substituting (18) to (19) and (42) into(13) we obtain the following closed form of the solution

120579 (119909) = 119888 +1198722

1198882

21199092

minus1205901198722

1198882

31199093

+

1198722

1198882

(31205902

minus 21198722

119888)

241199094

minus

1198722

1198882

(1205903

minus 41205901198722

119888)

301199095

+

1198722

1198882

(51205904

minus 781205902

1198722

119888 + 581198724

1198882

)

7201199096

+ sdot sdot sdot

(43)

To obtain the value of 119888 we substitute the boundarycondition (5) into (43) at the point 119909 = 1 That is

120579 (1) = 119888 +1198722

1198882

2minus1205901198722

1198882

3+

1198722

1198882

(31205902

minus 21198722

119888)

24

minus

1198722

1198882

(1205903

minus 41205901198722

119888)

30

+

1198722

1198882

(51205904

minus 781205902

1198722

119888 + 581198724

1198882

)

720+ sdot sdot sdot = 1

(44)

Substituting this value of 119888 into (43) one finds theexpression for 120579(119909) On the other hand given (119899119898) = (2 3)one obtains the solution

120579 (119909) = 119888 +1198722

21199092

minus1205901198722

31199093

+

1198722

(1205902

119888 minus 21198722

)

81198881199094

minus

1205901198722

(21205902

119888 minus 211198722

)

601198881199095

+

1198722

(1801198724

minus 1861205902

1198722

119888 + 51205904

1198882

)

7201199096

+ sdot sdot sdot

(45)

Here the constant 119888 maybe obtained by evaluating theboundary condition 120579(1) = 1 The solutions (43) and (45) aredepicted in Figures 3(b) and 3(a) respectively

52 The Rectangular Profile and Power Law Thermal Con-ductivity In this section we provide a detailed constructionof analytical solutions for the heat transfer in a longitudinalrectangular finwith a power law thermal conductivity that iswe consider (4) with 119891(119909) = 1 and 119896(120579) = 120579119898 The analyticalsolutions are given in the following expressions

(a) Case (119899119898) = (2 3)

120579 (119909) = 119888 +1198722

21199092

minus1198724

41198881199094

+1198726

411988821199096

minus331198728

12211988831199098

+127119872

10

336119888411990910

+ sdot sdot sdot

(46)

(b) Case (119899119898) = (3 2)

120579 (119909) = 119888 +1198882

1198722

21199092

minus1198883

1198724

121199094

+291198884

1198726

3601199096

minus3071198885

1198728

50401199098

+23483119888

6

11987210

45360011990910

+ sdot sdot sdot

(47)

The constant 119888 is obtained by solving the appropriate 120579(119909)at the fin base boundary condition The analytical solutionsin (46) and (47) are depicted in Figures 4(a) and 4(b)respectively

Mathematical Problems in Engineering 7

0 02 04 06 08 1094

095

096

097

098

099

1

119909

120579

119872 = 02

119872 = 03

119872 = 04

119872 = 05

(a)

0 02 04 06 08 1093

094

095

096

097

098

099

1

119909

120579

119872 = 02

119872 = 03

119872 = 04

119872 = 05

(b)

Figure 3 Temperature profile in a longitudinal fin with exponential profile and power law thermal conductivity In (a) the exponents are(119899119898) = (2 3) and (b) (119899 119898) = (3 2)

119872 = 025

119872 = 04

119872 = 05

119872 = 065

082

084

086

088

09

092

094

096

098

1

120579

0 02 04 06 08 1

119909

(a)

084

086

088

09

092

094

096

098

1

120579

0 02 04 06 08 1

119909

119872 = 025

119872 = 04

119872 = 05

119872 = 065

(b)

Figure 4 Temperature profile in a longitudinal fin with rectangular profile and power law thermal conductivity In (a) the exponents are(119899119898) = (2 3) and (b) (119899 119898) = (3 2)

53 The Convex Parabolic Profile and Power Law ThermalConductivity In this section we present solutions for theequation describing the heat transfer in a fin with con-vex parabolic profile and power law thermal conductivityEquation (40) is considered Here we consider the values(119899119898) = (2 3) (3 2) The final analytical solution is givenby

(a) Case (119899119898) = (2 3)

120579 (119909) = 119888 +21198722

311990932

minus21198724

51198881199093

+231198726

45119888211990992

minus1909119872

8

247511988831199096

+329222119872

10

2598751198884119909152

+ sdot sdot sdot

(48)

8 Mathematical Problems in Engineering

075

08

085

09

095

1

119872 = 025

119872 = 04

119872 = 05

119872 = 065

120579

0 02 04 06 08 1119909

(a)

119872 = 025

119872 = 04

119872 = 05

119872 = 065

0 02 04 06 08 1119909

082

084

086

088

09

092

094

096

098

120579

1

(b)

Figure 5 Temperature profile in a longitudinal fin with convex parabolic profile and power law thermal conductivity In (a) the exponentsare (119899119898) = (2 3) and (b) (119899119898) = (3 2)

(b) Case (119899119898) = (3 2)

120579 (119909) = 119888 +21198882

1198722

311990932

minus41198883

1198724

451199093

+41198884

1198726

2711990992

minus9921198885

1198728

74251199096

+30064119888

6

11987210

212625119909152

+ sdot sdot sdot

(49)

The constant 119888 is obtained by evaluating the appropriate 120579(119909)at the fin base boundary condition The solutions in (48) and(49) are depicted in Figures 5(a) and 5(b) respectively

54 The Rectangular Profile and LinearThermal ConductivityIn this section we present solutions for the equation repre-senting the heat transfer in a fin with rectangular profile andthe thermal conductivity depending linearly on temperatureThat is we consider (4) with 119891(119909) = 1 and 119896(120579) = 1 + 120573120579 Theanalytical solutions for this problem for different values of 119899are given by

(a) Case 119899 = 0

120579 (119909)

= 119888 +

1198722

119888

2 (1 + 120573119888)

1199092

minus

1198724

119888 (minus1 + 2120573119888)

24(1 + 120573119888)31199094

+

1198726

119888 (1 minus 16120573119888 + 281205732

1198882

)

720(1 + 120573119888)51199096

minus

1198728

119888 (minus1 + 78120573119888 minus 6001205732

1198882

+ 8961205733

1198883

)

40320(1 + 120573119888)7

1199098

+

11987210

119888 (1 minus 332120573119888 + 78121205732

1198882

minus 398961205733

1198883

+ 511841205734

1198884

)

3628800(1 + 120573119888)9

11990910

+ sdot sdot sdot

(50)

(b) Case 119899 = 1

120579 (119909)

= 119888 +

11987221198882

2 (1 + 120573119888)

1199092minus

11987241198883(minus1 + 2120573119888)

24(1 + 120573119888)31199094

+

11987261198884(10 minus 16120573119888 + 19120573

21198882)

720(1 + 120573119888)5

1199096

minus

11987281198885(minus80 + 342120573119888 minus 594120573

21198882+ 559120573

31198883)

40320(1 + 120573119888)7

1199098

+

119872101198886(1000 minus 7820120573119888 + 24336120573

21198882minus 36908120573

31198883+ 29161120573

41198884)

3628800(1 + 120573119888)9

11990910

+ sdot sdot sdot

(51)

(c) Case 119899 = 2

120579 (119909)

= 119888 +

1198722

1198883

2 (1 + 120573119888)

1199092

+

1198724

1198885

8(1 + 120573119888)31199094

+

1198726

1198887

(3 + 21205732

1198882

)

80(1 + 120573119888)51199096

+

1198728

1198889

(49 minus 20120573119888 + 661205732

1198882

minus 401205733

1198883

)

4480(1 + 120573119888)7

1199098

+

11987210

11988811

(427 minus 440120573119888 + 11161205732

1198882

minus 10201205733

1198883

+ 6721205734

1198884

)

134400(1 + 120573119888)9

11990910

+ sdot sdot sdot

(52)

Mathematical Problems in Engineering 9

(d) Case 119899 = 3120579 (119909)

= 119888 +

11987221198884

2 (1 + 120573119888)

1199092+

11987241198887(4 + 120573119888)

24(1 + 120573119888)31199094

+

119872611988810(52 + 32120573119888 + 25120573

21198882)

720(1 + 120573119888)5

1199096

+

119872811988813(1288 + 1020120573119888 + 1212120573

21198882minus 95120573

31198883)

40320(1 + 120573119888)7

1199098

+

1198721011988816(52024+45688120573119888+77184120573

21198882minus680120573

31198883+ 15025120573

41198884)

3628800(1 + 120573119888)9

11990910

+

(53)

The constant 119888may be obtained from the boundary conditionon the appropriate solution The solutions in (50) (51) (52)and (53) are depicted in Figure 6

55The Convex Parabolic Profile and LinearThermal Conduc-tivity In this section we present solutions for the equationdescribing the heat transfer in a fin with convex parabolicprofile and the thermal conductivity depending linearly ontemperature That is we consider (40) with 119896(120579) = 1 + 120573120579The analytical solution for this problem for different values of119899 is given by

(a) Case 119899 = 0

120579 (119909)

= 119888 +21198722

119888

3 (1 + 120573119888)11990932

minus21198724

119888 (minus2 + 3120573119888)

45(1 + 120573119888)3

1199093

+

1198726

119888 (2 minus 25120573119888 + 331205732

1198882

)

405(1 + 120573119888)5

11990992

minus

1198728

119888 (minus10 + 599120573119888 minus 35821205732

1198882

+ 40591205733

1198883

)

66825(1 + 120573119888)7

1199096

+ sdot sdot sdot

(54)

(b) Case 119899 = 1

120579 (119909)

= 119888 +21198722

1198882

3 (1 + 120573119888)11990932

minus21198724

1198883

(minus4 + 120573119888)

45(1 + 120573119888)3

1199093

+

21198726

1198884

(9 minus 11120573119888 + 101205732

1198882

)

405(1 + 120573119888)5

11990992

minus

21198728

1198885

(minus330 + 1118120573119888 minus 15841205732

1198882

+ 10931205733

1198883

)

66825(1 + 120573119888)7

1199096

+ sdot sdot sdot

(55)

119899 = 0

119899 = 1

119899 = 2

119899 = 3

1

095

09

085

08

075

07

065

06

055

0 02 04 06 08 1

119909

120579

Figure 6 Temperature profile in a longitudinal rectangular fin withlinear thermal conductivity and varying values of 119899 Here 120573 = 05and119872 = 15 are fixed

(c) Case 119899 = 2

120579 (119909)

= 119888 +21198722

1198883

3 (1 + 120573119888)11990932

+21198724

1198885

(6 + 120573119888)

45(1 + 120573119888)31199093

+

1198726

1198887

(16 + 3120573119888 + 71205732

1198882

)

135(1 + 120573119888)5

11990992

+

1198728

1198889

(1160 minus 107120573119888 + 11161205732

1198882

minus 3671205733

1198883

)

22275(1 + 120573119888)7

1199096

+ sdot sdot sdot

(56)

(d) Case 119899 = 3

120579 (119909)

= 119888 +21198722

1198884

3 (1 + 120573119888)11990932

+21198724

1198887

(8 + 3120573119888)

45(1 + 120573119888)31199093

+

41198726

11988810

(23 + 17120573119888 + 91205732

1198882

)

405(1 + 120573119888)5

11990992

+

21198728

11988813

(5000+4868120573119888 + 43561205732

1198882

+ 3631205733

1198883

)

66825(1 + 120573119888)7

1199096

+ sdot sdot sdot

(57)

The constant 119888 may be obtained from the boundarycondition on the appropriate 120579(119909)The solutions in (54) (55)(56) and (57) are depicted in Figure 7

10 Mathematical Problems in Engineering

1

09

08

07

06

05

120579

119899 = 0

119899 = 1

119899 = 2

119899 = 3

0 02 04 06 08 1

119909

Figure 7 Temperature profile in a longitudinal convex parabolic finwith linear thermal conductivity and varying values of 119899 Here 120573 =05 and119872 = 15 are fixed

56 The Exponential Profile and LinearThermal ConductivityIn this section we present solutions for equation heat transferin a fin with convex parabolic profile and the thermalconductivity depending linearly on temperature That is weconsider (4) with 119896(120579) = 1 + 120573120579 and 119891(119909) = e120590119909 Here 120590is a constant The analytical solutions for this problem fordifferent values of 119899 are given by

(a) Case 119899 = 0

120579 (119909)

= 119888 +

1198722

119888

2 (1 + 120573119888)

1199092

minus

1205901198722

119888 (2 + 120573119888 + 1205732

119888)

6(1 + 120573119888)21199093

+

1198722

119888 (1198722

(1 minus 21205732

119888) + 1205902

(3 +3120573119888+1205733

1198882

+1205734

1198882

+1205732

119888 (3 + 119888)))

24(1 + 120573119888)4

1199094

+ sdot sdot sdot

(58)

(b) Case 119899 = 1

120579 (119909)

= 119888 +

11987221198882

2 (1 + 120573119888)

1199092minus

12059011987221198882(2 + 120573119888 + 120573

2119888)

6(1 + 120573119888)21199093

+

11987221198882(1198722119888 (2+120573119888minus2120573

2119888)+1205902(3+3120573119888+120573

31198882+12057341198882+1205732119888 (3+119888)))

24(1 + 120573119888)3

1199094

+ sdot sdot sdot

(59)

119899 = 0

119899 = 1

119899 = 2

119899 = 3

0 02 04 06 08 1

119909

1

095

09

085

08

075

07

120579

Figure 8 Temperature profile in a longitudinal fin of exponentialprofile with linear thermal conductivity and varying values of 119899Here 120573 = 05 and119872 = 15 are fixed

(c) Case 119899 = 2120579 (119909)

= 119888 +

11987221198883

2 (1 + 120573119888)

1199092minus

12059011987221198883(2 + 120573119888 + 120573

2119888)

6(1 + 120573119888)21199093

+

11987221198883(11987221198882(3+2120573119888minus2120573

2119888)+1205902(3+3120573119888+120573

31198882+12057341198882+1205732119888 (3+119888)))

24(1 + 120573119888)3

times1199094+ sdot sdot sdot

(60)

(d) Case 119899 = 3120579 (119909)

= 119888 +

11987221198884

2 (1 + 120573119888)

1199092minus

12059011987221198884(2 + 120573119888 + 120573

2119888)

6(1 + 120573119888)21199093

+

11987221198884(11987221198883(4+3120573119888minus2120573

2119888)+1205902(3+3120573119888+120573

31198882+12057341198882+1205732119888 (3+119888)))

24(1 + 120573119888)3

times1199094+ sdot sdot sdot

(61)

The constant 119888 may be obtained from the boundarycondition on the appropriate 120579(119909)The solutions in (58) (59)(60) and (61) are depicted in Figure 8

6 Fin Efficiency and Heat Flux

61 Fin Efficiency Theheat transfer rate from a fin is given byNewtonrsquos second law of cooling

119876 = int

119871

0

119875119867 (119879) (119879 minus 119879119886) 119889119909 (62)

Mathematical Problems in Engineering 11

Fin efficiency is defined as the ratio of the fin heat transferrate to the rate that would be if the entire fin were at the basetemperature and is given by (see eg [1])

120578=

119876

119876ideal=

int119871

0

119875119867 (119879) (119879 minus 119879119886) 119889119883

119875ℎ119887119871 (119879119887minus 119879119886)

(63)

In dimensionless variables we have

120578 = int

1

0

120579119899+1

119889119909 (64)

We consider the solutions (50) (51) (52) and (53) and depictthe fin efficiency (63) in Figure 9

62 Heat Flux The fin base heat flux is given by the Fourierrsquoslaw

119902119887= 119860119888119870 (119879)

119889119879

119889119909 (65)

The total heat flux of the fin is given by [1]

119902 =119902119887

119860119888119867(119879) (119879

119887minus 119879119886) (66)

Introducing the dimensionless variable as described inSection 2 implies

119902 =1

119861119894

119896 (120579)

ℎ (120579)

119889120579

119889119909 (67)

where the dimensionless parameter 119861119894 = ℎ119887119871119896119886is the Biot

number We consider a number of cases for the thermalconductivity and the heat transfer coefficient

621 Linear Thermal Conductivity and Power Law HeatTransfer Coefficient In this case (67) becomes

119902 =1

119861119894(1 + 120573120579) 120579

minus119899119889120579

119889119909 (68)

The heat flux in (68) at the base of the fin is plotted inFigure 10

622 Power LawThermal Conductivity and Heat TransferCoefficient In this case (67) becomes

119902 =1

119861119894120579119898minus119899

119889120579

119889119909 (69)

Not surprisingly heat flux in one-dimensional fins is highergiven values 119861119894 ≪ 1 The heat flux in (69) is plotted inFigures 11(a) 11(b) and 11(c)

7 Some Discussions

The DTM has resulted in some interesting observations andstudy We have observed in Figures 2(a) and 2(b) an excellentagreement between the analytical solutions generated byDTM and the exact solution obtained in [2] In particular we

119872

1

095

09

085

08

075

07

065

06

055

120578

2151050

119899 = 0

119899 = 1

119899 = 2

119899 = 3

Figure 9 Fin efficiency of a longitudinal rectangular fin Here 120573 =075

119872

25

20

15

10

5

0

119902

2151050

119899 = 0

119899 = 1

119899 = 2

119899 = 3

Figure 10 Base heat flux in a longitudinal rectangular finwith linearthermal conductivity Here 120573 = 01

considered a fin problem in which both thermal conductivityand heat transfer coefficient are given by the same power lawFurthermore we notice from Table 2 that an absolute errorof approximately 35119890 minus 005 is produced by DTM of order119874(15) In Table 3 an absolute error of approximately 65119890 minus006 is produced for the same order This confirms that theDTM converges faster and can provide accurate results witha minimum computation As such a tremendous confidencein the DTM in terms of the accuracy and effectiveness wasbuilt and thus we used this method to solve other problemsfor which exact solutions are harder to construct

In Figures 3(a) 3(b) 4(a) 4(b) 5(a) and 5(b) weobserve that the fin temperature increaseswith the decreasing

12 Mathematical Problems in Engineering

values of the thermogeometric fin parameter Here thevalues of the exponents are fixed Also we observe that fintemperature is higher when 119899 minus 119898 gt 0 that is when heattransfer coefficient is higher than the thermal conductivityWe observe in Figures 6 7 and 8 that the fin temperatureincreases with the increasing values of 119899 Furthermoreit appears that the fin with exponential profile performsthe least in transferring the heat from the base since thetemperature in such a fin is much higher than that ofthe rectangular and the convex parabolic profiles In otherwords heat dissipation to the fluid surrounding the extendedsurface is much faster in longitudinal fins of rectangular andconvex parabolic profiles In Figure 9 fin efficiency decreaseswith increasing thermogeometric fin parameter Also finefficiency increases with increasing values of 119899 It is easyto show that the thermogeometric fin parameter is directlyproportional to the aspect ratio (extension factor) with squareroot of the Biot number being the proportionality constantAs such shorter fins are more efficient than longer onesElse the increased Biot number results in less efficient finwhenever the space is confined that is where the length ofthe fin cannot be increased Figure 10 depicts the heat fluxat the fin base The amount of heat energy dissipated fromthe fin base is of immense interest in engineering [28] Weobserve in Figure 10 that the base heat flux increases withthe thermogeometric fin parameter for considered values ofthe exponent 119899 (see also [28]) Figures 11(a) 11(b) and 11(c)display the heat flux across the fin length We note that theheat flux across the fin length increases with increasing valuesof the thermogeometric fin parameter

8 Concluding Remarks

In this study we have successfully applied the DTM to highlynonlinear problems arising in heat transfer through longitu-dinal fins of various profiles Both thermal conductivity andheat transfer coefficient are given as functions of temperatureThe DTM agreed well with exact solutions when the thermalconductivity and heat transfer coefficient are given by thesame power law A rapid convergence to the exact solutionwas observed Following the confidence in DTM built by theresults mentioned we then solved various exciting problemsThe exotic results have been shown in tables and figures listedin this paper

The results obtained in this paper are significant improve-ments on the known results In particular both the heattransfer coefficient and the thermal conductivity are allowedto be given by the power law functions of temperature andalsowe considered a number of finprofilesWenote that exactsolutions are difficult if not impossible to construct when theexponents of these properties are distinct

Perhaps the notable advantage of the DTM is the general-ization of the Taylor method to problems involving unusualderivative procedures such as fractional fuzzy or q-derivative[22] Some generalizations have been made by Odibat et al[24] and they referred to their new method as the General-ized Differential Transform Method (GDTM) This showedgreat improvement compared to the Fractional Differential

0 02 04 06 08 1119909

45

40

35

30

25

20

15

10

5

0

minus5

119902

(a)

0 02 04 06 08 1119909

30

25

20

15

10

5

0

119902

(b)

35

30

25

20

15

10

5

0

119902

0 02 04 06 08 1119909

119872 = 025

119872 = 04

119872 = 05

119872 = 065

(c)

Figure 11 Heat flux across a longitudinal rectangular fin with linearthermal conductivity (a) 119899minus119898 lt 0 (b) 119899minus119898 gt 0 and (c) 119899minus119898 = 0Here 119861119894 = 001

Mathematical Problems in Engineering 13

Transform Method (FDTM) introduced by Arikoglu andOzkol [23]

We have shown with the help of an example that DTMmay only be applied to fin problems involving heat transferthrough fins with convex parabolic profile Note that given anODE such as (4) with a power law heat transfer coefficient ofa fractional exponent then one can easily remove the fractionby basic exponential rules and employment of the Binomialexpansion However using the DTM one runs into difficultyif the power law thermal conductivity in the same equation isgiven by the fractional exponent We do not know whetherthese observations call for the ldquomodifiedrdquo DTM to solveproblems arising in heat flow through fins with other profilessuch as longitudinal triangular and concave parabolic andalso with fractional power law thermal conductivity

The main results obtained in this paper give insightinto heat transfer in boiling liquids where the heat transfercoefficient is temperature dependent and may be given by apower law The thermal conductivity of some materials suchas gallium nitride (GaN) and Aluminium Nitride (AlN) maybe modeled by power law temperature dependency [29 30]Thus the solutions constructed here give a better comparisonof heat transfer in terms of material used since in many engi-neering applications thermal conductivity is given as a linearfunction of temperature Furthermore a good study in termsof performance and efficiency of fin with different profiles isundertaken These finding could help in the design of fins Itis claimed in [14] that DTM results are more accurate thanthose constructed by Variational Iteration Methods (VIM)and Homotopy Perturbation Methods (HPM) However itwould be risky to use the DTM approximate solutions asbenchmarks for the numerical schemes Nevertheless wehave also shown that DTM converges rapidly in just fifteenterms to the exact solution

Acknowledgments

The authors wish to thank the National Research Foundationof South Africa and the University of the Witwatersrandfor the generous financial support The authors are gratefulto Professor D P Mason and the anonymous reviewers fortheir invaluable comments which improved this paper signif-icantly

References

[1] A D Kraus A Aziz and J Welty Extended Surface Heat Trans-fer Wiley New York NY USA 2001

[2] R JMoitsheki T Hayat andM YMalik ldquoSome exact solutionsof the fin problem with a power law temperature-dependentthermal conductivityrdquo Nonlinear Analysis Real World Applica-tions vol 11 no 5 pp 3287ndash3294 2010

[3] R J Moitsheki ldquoSteady one-dimensional heat flow in a lon-gitudinal triangular and parabolic finrdquo Communications inNonlinear Science and Numerical Simulation vol 16 no 10 pp3971ndash3980 2011

[4] R J Moitsheki ldquoSteady heat transfer through a radial fin withrectangular and hyperbolic profilesrdquo Nonlinear Analysis RealWorld Applications vol 12 no 2 pp 867ndash874 2011

[5] Mo Miansari D D Ganji and Me Miansari ldquoApplication ofHersquos variational iteration method to nonlinear heat transferequationsrdquo Physics Letters A vol 372 no 6 pp 779ndash785 2008

[6] C H Chiu and C K Chen ldquoA decomposition method forsolving the convective longitudinal fins with variable thermalconductivityrdquo International Journal of Heat and Mass Transfervol 45 no 10 pp 2067ndash2075 2002

[7] A Rajabi ldquoHomotopy perturbation method for fin efficiencyof convective straight fins with temperature-dependent thermalconductivityrdquo Physics Letters A vol 364 no 1 pp 33ndash37 2007

[8] G Domairry and M Fazeli ldquoHomotopy analysis method todetermine the fin efficiency of convective straight fins withtemperature-dependent thermal conductivityrdquo Communica-tions in Nonlinear Science and Numerical Simulation vol 14 no2 pp 489ndash499 2009

[9] A Campo andR J Spaulding ldquoCoupling of themethods of suc-cessive approximations and undetermined coefficients for theprediction of the thermal behaviour of uniform circumferentialfinsrdquo Heat and Mass Transfer vol 34 no 6 pp 461ndash468 1999

[10] A Campo and F Rodrfguez ldquoApproximate analytic tempera-ture solution for uniform annular fins by adapting the powerseriesmethodrdquo International Communications inHeat andMassTransfer vol 25 no 6 pp 809ndash818 1998

[11] R Chiba ldquoApplication of differential transformmethod to ther-moelastic problem for annular disks of variable thickness withtemperature-dependent parametersrdquo International Journal ofThermophysics vol 33 pp 363ndash380 2012

[12] S Sadri M R Raveshi and S Amiri ldquoEfficiency analysis ofstraight fin with variable heat transfer coefficient and thermalconductivityrdquo Journal ofMechanical Science andTechnology vol26 no 4 pp 1283ndash1290 2012

[13] M Torabi H Yaghoori and A Aziz ldquoAnalytical solution forconvective-radiative continously moving fin with temperaturedependent thermal conductivityrdquo International Journal of Ther-mophysics vol 33 pp 924ndash941 2012

[14] M Torabi andH Yaghoobi ldquoTwo dominant analytical methodsfor thermal analysis of convective step fin with variable thermalconductivityrdquoThermal Science In press

[15] AMoradi ldquoAnalytical solutions for finwith temperature depen-dant heat transfer coefficientrdquo International Journal of Engineer-ing and Applied Sciences vol 3 no 2 pp 1ndash12 2011

[16] A Moradi and H Ahmadikia ldquoAnalytical solution for differentprofiles of fin with temperature-dependent thermal conductiv-ityrdquoMathematical Problems in Engineering vol 2010 Article ID568263 15 pages 2010

[17] H Yaghoobi and M Torabi ldquoThe application of differentialtransformation method to nonlinear equations arising in heattransferrdquo International Communications in Heat and MassTransfer vol 38 no 6 pp 815ndash820 2011

[18] S Ghafoori M Motevalli M G Nejad F Shakeri D D GanjiandM Jalaal ldquoEfficiency of differential transformation methodfor nonlinear oscillation Comparison with HPM and VIMrdquoCurrent Applied Physics vol 11 no 4 pp 965ndash971 2011

[19] A A Joneidi D D Ganji andM Babaelahi ldquoDifferential trans-formation method to determine fin efficiency of convectivestraight fins with temperature dependent thermal conductivityrdquoInternational Communications in Heat and Mass Transfer vol36 no 7 pp 757ndash762 2009

[20] J K Zhou Differential Transform Method and Its Applicationsfor Electric circuIts Huazhong University Press Wuhan China1986

14 Mathematical Problems in Engineering

[21] S Mukherjee ldquoReply to comment on lsquosolutions of the duffing-van der pol oscillator equation by the differential transformmethodrsquordquo Physica Scripta vol 84 Article ID 037003 2 pages2011

[22] C Bervillier ldquoStatus of the differential transformationmethodrdquoApplied Mathematics and Computation vol 218 no 20 pp10158ndash10170 2012

[23] A Arikoglu and I Ozkol ldquoSolution of fractional differentialequations by using differential transform methodrdquo ChaosSolitons and Fractals vol 34 no 5 pp 1473ndash1481 2007

[24] ZOdibat SMomani andV S Erturk ldquoGeneralized differentialtransform method application to differential equations offractional orderrdquo Applied Mathematics and Computation vol197 no 2 pp 467ndash477 2008

[25] CW Bert ldquoApplication of differential transformmethod to heatconduction in tapered finsrdquo Journal of Heat Transfer vol 124no 1 pp 208ndash209 2002

[26] F Khani and A Aziz ldquoThermal analysis of a longitudinal trape-zoidal fin with temperature-dependent thermal conductivityand heat transfer coefficientrdquo Communications in NonlinearScience and Numerical Simulation vol 15 no 3 pp 590ndash6012010

[27] H C Unal ldquoAn anlytical study of boiling heat transfer from afinrdquo International Journal of Heat and Mass Transfer vol 31 no7 pp 1483ndash1496 1988

[28] M H Chang ldquoA decomposition solution for fins with tempera-ture dependent surface heat fluxrdquo International Journal of Heatand Mass Transfer vol 48 no 9 pp 1819ndash1824 2005

[29] A Jezowski B A Danilchenko M Bockowski et al ldquoThermalconductivity of GaN crystals in 42ndash300 K rangerdquo Solid StateCommunications vol 128 no 2-3 pp 69ndash73 2003

[30] S Vitanov V Palankovski S Maroldt and R Quay ldquoHigh-temperature modeling of AlGaNGaN HEMTsrdquo Solid-StateElectronics vol 54 no 10 pp 1105ndash1112 2010

Submit your manuscripts athttpwwwhindawicom

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Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

Page 2: Research Article Analytical Solutions for Steady Heat ...downloads.hindawi.com/journals/mpe/2013/273052.pdf · Research Article Analytical Solutions for Steady Heat Transfer in Longitudinal

2 Mathematical Problems in Engineering

transfer coefficient can be expressed as a power law for whichvalues of the exponent represent different phenomena (seeeg [26])

In this paper the DTM is employed to determine theanalytical solutions to the nonlinear boundary value problemdescribing heat transfer in longitudinal fins of rectangularexponential and convex parabolic profiles Both thermalconductivity and heat transfer coefficient are temperaturedependentWe adopt the terminology exact solutions to referto solutions given in terms of fundamental expressions suchas logarithmic trigonometric and exponential Howeveranalytical solutions will be series solutions and in partic-ular those constructed using the DTM The mathematicalmodelling of the problem under consideration is describedin Section 2 A brief discussion on the fundamentals of theDTM is be provided in Section 3 The comparison of theexact and analytical solutions constructed byDTM is given inSection 4 In Section 5 we provide analytical solutions for theheat transfer in longitudinal fins of various profilesHere heattransfer coefficient is given by the power law and we considertwo cases of the thermal conductivity namely the powerlaw and the linear function of temperature Furthermorewe describe the fin efficiency and the heat flux in Section 6Some exciting results are discussed in Section 7 Lastly theconcluding remarks are provided in Section 8

2 Mathematical Models

We consider a longitudinal one dimensional fin of cross-sectional area 119860

119888 The perimeter of the fin is denoted by

119875 and its length by 119871 The fin is attached to a fixed primesurface of temperature 119879

119887and extends to an ambient fluid

of temperature 119879119886 The fin thickness at the prime surface is

given by 120575119887and its profile is given by 119865(119883) Based on the one

dimensional heat conduction the energy balance equation isthen given by (see eg [1])

119860119888

119889

119889119883(120575119887

2119865 (119883)119870 (119879)

119889119879

119889119883) = 119875119867 (119879) (119879 minus 119879

119886)

0 le 119883 le 119871

(1)

where 119870 and 119867 are nonuniform temperature-dependentthermal conductivity and heat transfer coefficients respec-tively119879 is the temperature distribution119865(119883) is the fin profileand119883 is the space variable The length of the fin is measuredfrom the tip to the prime surface as shown in Figure 1Assuming that the fin tip is adiabatic (insulated) and the basetemperature is kept constant then the boundary conditionsare given by

119879 (119871) = 119879119887

119889119879

119889119883

10038161003816100381610038161003816100381610038161003816119883=0

= 0 (2)

Introducing the following dimensionless variables (see eg[1])

119909 =119883

119871 120579 =

119879 minus 119879119886

119879119887minus 119879119886

ℎ =119867

ℎ119887

119896 =119870

119896119886

1198722

=119875ℎ1198871198712

119860119888119896119886

119891 (119909) =120575119887

2119865 (119883)

(3)

119883 = 0 119883 = 119871

Fin tip

Fin profile

Prime surface

120575

119871

119882

Figure 1 Schematic representation of a longitudinal fin of anunspecified profile

with 119896119886being defined as the thermal conductivity at the

ambient temperature and ℎ119887as the heat transfer at the prime

surface (fin base) reduces (1) to

119889

119889119909[119891 (119909) 119896 (120579)

119889120579

119889119909] minus119872

2

120579ℎ (120579) = 0 0 le 119909 le 1 (4)

Here 120579 is the dimensionless temperature 119909 is the dimension-less space variable 119891(119909) is the dimensionless fin profile 119896 isthe dimensionless thermal conductivity ℎ is the dimension-less heat transfer coefficient and119872 is the thermogeometricfin parameter The dimensionless boundary conditions thenbecome

120579 (1) = 1 at the prime surface (5)

119889120579

119889119909

10038161003816100381610038161003816100381610038161003816119909=0

= 0 at the fin tip (6)

We assume that the heat transfer coefficient is given by thepower law

119867(119879) = ℎ119887(119879 minus 119879119886

119879119887minus 119879119886

)

119899

(7)

where the exponent 119899 is a real constant In fact the values of 119899may vary as between minus66 and 5 However in most practicalapplications it lies between minus3 and 3 [27] In dimensionlessvariables we have ℎ(120579) = 120579119899 We consider the two distinctcases of the thermal conductivity as follows

(a) the power law

119870 (119879) = 119896119886(119879 minus 119879119886

119879119887minus 119879119886

)

119898

(8)

with119898 being a constant and(b) the linear function

119870 (119879) = 119896119886[1 + 120574 (119879 minus 119879

119886)] (9)

Mathematical Problems in Engineering 3

The dimensionless thermal conductivity given by the powerlaw and the linear function of temperature is 119896(120579) = 120579119898 and119896(120579) = 1 + 120573120579 respectively Here the thermal conductivitygradient is120573 = 120574(119879

119887minus119879119886) Furthermore we consider a various

fin profiles including the longitudinal rectangular 119891(119909) =1 the longitudinal convex parabolic 119891(119909) = radic119909 and theexponential profile 119891(119909) = e119886119909 with 119886 being the constant (seealso [16])

3 Fundamentals of the DifferentialTransform Method

In this section the basic idea underlying the DTM is brieflyintroduced Let 119910(119905) be an analytic function in a domain DThe Taylor series expansion function of 119910(119905) with the centerlocated at 119905 = 119905

119895is given by [20]

119910 (119905) =

infin

sum

120581=0

(119905 minus 119905119895)120581

120581[119889120581

119910 (119905)

119889119905120581]

119905=119905119895

forall119905 isin D (10)

The particular case of (10) when 119905119895= 0 is referred to as

the Maclaurin series expansion of 119910(119905) and is expressed as

119910 (119905) =

infin

sum

120581=0

119905120581

120581[119889120581

119910 (119905)

119889119905120581]

119905=0

forall119905 isin D (11)

The differential transform of 119910(119905) is defined as follows

119884 (119905) =

infin

sum

120581=0

H120581

120581[119889120581

119910 (119905)

119889119905120581]

119905=0

(12)

where 119910(119905) is the original analytic function and 119884(119905) is thetransformed function The differential spectrum of 119884(119905) isconfined within the interval 119905 isin [0H] where H is aconstant From (11) and (12) the differential inverse transformof 119884(119905) is defined as follows

119910 (119905) =

infin

sum

120581=0

(119905

H)

120581

119884 (120581) (13)

and if 119910(119905) is expressed by a finite series then

119910 (119905) =

119903

sum

120581=0

(119905

H)

120581

119884 (120581) (14)

Some of the useful mathematical operations performedby the differential transform method are listed in Table 1

The delta function 120575(120581 minus 119904) is given by

120575 (120581 minus 119904) = 1 if 120581 = 1199040 if 120581 = 119904

(15)

4 Comparison of Exact andAnalytical Solutions

In this section we consider a model describing temperaturedistribution in a longitudinal rectangular fin with both ther-mal conductivity and heat transfer coefficient being functions

Table 1 Fundamental operations of the differential transformmethod

Original function Transformed function119910(119905) = 119909(119905) plusmn 119911(119905) 119884(119905) = 119883(119905) plusmn 119885(119905)

119910(119905) = 120572119909(119905) 119884(119905) = 120572119883(119905)

119910(119905) =119889119910(119905)

119889119905119884(119905) = (120581 + 1)119884(120581 + 1)

119910(119905) =1198892

119910(119905)

1198891199052119884(119905) = (120581 + 1)(120581 + 2)119884(120581 + 2)

119910(119905) =119889119904

119910(119905)

119889119905119904119884(119905) = (120581 + 1)(120581 + 2) sdot sdot sdot (120581 + 119904)119884(120581 + 119904)

119910(119905) = 119909(119905)119911(119905) 119884(119905) =

120581

sum

119894=0

119883(119894)119885(120581 minus 119894)

119910(119905) = 1 119884(119905) = 120575(120581)

119910(119905) = 119905 119884(119905) = 120575(120581 minus 1)

119910(119905) = 119905119904

119884(119905) = 120575(120581 minus 119904)

119910(119905) = exp(120582119905) 119884(119905) =120582120581

120581

119910(119905) = (1 + 119905)119904

119884(119905) =119904(119904 minus 1) sdot sdot sdot (119904 minus 120581 + 1)

120581

119910(119905) = sin(120596119905 + 120572) 119884(119905) =120596120581

120581sin(120587120581

2+ 120572)

119910(119905) = cos(120596119905 + 120572) 119884(119905) =120596120581

120581cos(120587120581

2+ 120572)

of temperature given by the power law (see eg [2])The exactsolution of (4) when both the power laws are given by thesame exponent is given by [2]

120579 (119909) = [

cosh (119872radic119899 + 1119909)

cosh (119872radic119899 + 1)]

1(119899+1)

(16)

We use this exact solution as a benchmark or validation ofthe DTM The effectiveness of the DTM is determined bycomparing the exact and the analytical solutionsWe comparethe results for the cases 119899 = 1 and 119899 = 2 with fixed values of119872

41 Case 119899=1 Applying the DTM to (4) with the power lawthermal conductivity 119891(119909) = 1 (rectangular profile) andgivenH = 1 one obtains the following recurrence relation

120581

sum

119894=0

[Θ (119894) (120581 minus 119894 + 1) (120581 minus 119894 + 2)Θ (120581 minus 119894 + 2)

+ (119894 + 1)Θ (119894 + 1) (120581 minus 119894 + 1)Θ (120581 minus 119894 + 1)

minus1198722

Θ (119894)Θ (120581 minus 119894)] = 0

(17)

Exerting the transformation to the boundary condition (6) ata point 119909 = 0

Θ (1) = 0 (18)

The other boundary conditions are considered as follows

Θ (0) = 119888 (19)

4 Mathematical Problems in Engineering

where 119888 is a constant Equation (17) is an iterative formula ofconstructing the power series solution as follows

Θ (2) =1198722

119888

2

(20)

Θ (3) = 0 (21)

Θ (4) = minus1198724

119888

24

(22)

Θ (5) = 0 (23)

Θ (6) =191198726

119888

720

(24)

Θ (7) = 0 (25)

Θ (8) = minus559119872

8

119888

40320

(26)

Θ (9) = 0 (27)

Θ (10) =29161119872

10

119888

3628800

(28)

Θ (11) = 0 (29)

Θ (12) = minus2368081119872

12

119888

479001600

(30)

Θ (13) = 0 (31)

Θ (14) =276580459119872

14

119888

87178291200

(32)

Θ (15) = 0

(33)

These terms may be taken as far as desired Substituting(18) to (32) into (13) we obtain the following analyticalsolution

120579 (119909) = 119888 +1198722

119888

21199092

minus1198724

119888

241199094

+191198726

119888

7201199096

minus559119872

8

119888

403201199098

+29161119872

10

119888

362880011990910

minus2368081119872

12

119888

47900160011990912

+276580459119872

14

119888

8717829120011990914

+ sdot sdot sdot

(34)

To obtain the value of 119888 we substitute the boundarycondition (5) into (34) at the point 119909 = 1 Thus we have

120579 (1) = 119888 +1198722

119888

2minus1198724

119888

24+191198726

119888

720minus559119872

8

119888

40320

+29161119872

10

119888

3628800minus2368081119872

12

119888

479001600

+276580459119872

14

119888

87178291200+ sdot sdot sdot = 1

(35)

Table 2 Results of the DTM and exact solutions for 119899 = 1119872 = 07

119909 DTM Exact Error0 0808093014 080809644 000000342601 0810072036 081007547 000000343402 0815999549 081600301 000000345903 0825847770 082585127 000000350004 0839573173 083957673 000000355905 0857120287 085712392 000000363306 0878426299 087843002 000000372207 0903426003 090342982 000000381408 0932056648 093206047 000000382009 0964262558 096426582 000000326310 1000000000 100000000 0000000000

Table 3 Results of theDTMand Exact Solutions for 119899 = 2119872 = 05

119909 DTM Exact Error0 0894109126 0894109793 000000066501 0895226066 0895226732 000000066602 0898568581 0898569249 000000066803 0904112232 0904112905 000000067204 0911817796 0911818474 000000067805 0921633352 0921634038 000000068506 0933496900 0933497594 000000069407 0947339244 0947339946 000000070208 0963086933 0963087627 000000069409 0980665073 0980665659 000000058610 1000000000 1000000000 0000000000

We omit presenting the tedious process of finding 119888 Howeverone may obtain the expression for 120579(119909) upon substituting theobtained value of 119888 into (34)

42 Case 119899=2 Following a similar approach given inSection 41 and given 119899 = 2 one obtains the analyticalsolution

120579 (119909) = 119888 +1198722

119888

21199092

minus1198724

119888

81199094

+231198726

119888

2401199096

minus1069119872

8

119888

134401199098

+9643119872

10

119888

13440011990910

minus1211729119872

12

119888

1774080011990912

+217994167119872

14

119888

322882560011990914

+ sdot sdot sdot

(36)

The constant 119888may be obtained using the boundary conditionat the fin base The comparison of the DTM and the exactsolutions are reflected in Tables 2 and 3 for different valuesof 119899 Furthermore the comparison of the exact and theanalytical solutions is depicted in Figures 2(a) and 2(b)

Mathematical Problems in Engineering 5

08

082

084

086

088

09

092

094

096

098

1

120579

0 02 04 06 08 1

119909

DTMExact

(a)

086

088

09

092

094

096

098

1

120579

0 02 04 06 08 1

119909

DTMExact

(b)

Figure 2 Comparison of analytical and exact solutions In (a) 119899 = 1119872 = 07 and (b) 119899 = 2119872 = 05

5 Analytical Solutions

It is well known that exact solutions forODEs such as (4) existonly when thermal conductivity and the term containingheat transfer coefficient are connected by differentiation(or simply if the ODE such as (4) is linearizable) [4] Inthis section we determine the analytical solutions for thenonlinearizable (4) firstlywhen thermal conductivity is givenby the power law and secondly as a linear function oftemperature In both cases and throughout this paper theheat transfer coefficient is assumed to be a power law functionof temperature These assumptions of the thermal propertiesare physical realistic We have noticed that DTM runs intodifficulty when the exponent of the power law of the thermalconductivity is given by fractional values and also when thefunction 119891(119909) is given in terms of fractional power law Onemay followMoradi and Ahmadikia [16] by introducing a newvariable to deal with fractional powers of 119891(119909) and on theother hand it is possible to remove the fractional exponent ofthe heat transfer coefficient by fundamental laws of exponentand binomial expansion

Proposition 1 A nonlinear ODE such as (4) may admit theDTM solution if 119891(119909) is a constant or exponential functionHowever if 119891(119909) is a power law then such an equation admitsa DTM solution if the product

119891 (119909) sdot 1198911015840

(119909) = 120572 (37)

holds Here 120572 is a real constant

Proof Introducing the new variable 120585 = 119891(119909) it follows fromchain rule that (4) becomes

120572119889120585

119889119909

119889

119889120585[119896 (120579)

119889120579

119889120585] minus119872

2

120579119899+1

= 0 (38)

The most general solution of condition (37) is

119891 (119909) = (2120572119909 + 120574)12

(39)

where 120574 is an integration constant

Example 2 (a) If 119891(119909) = 119909120590 then 120585 = 119909120590 transforms (4) into

119889

119889120585[119896 (120579)

119889120579

119889120585] minus 4119872

2

120585120579119899+1

= 0 (40)

only if 120590 = 12This example implies that the DTM may only be appli-

cable to problems describing heat transfer in fins withconvex parabolic profile In the next subsections we presentanalytical solutions for (4) with various functions 119891(119909) and119896(120579)

51 The Exponential Profile and Power Law Thermal Con-ductivity In this section we present solutions for equationdescribing heat transfer in a fin with exponential profile andpower law thermal conductivity and heat transfer coefficientThat is given (4) with 119891(119909) = e120590119909 and both heat transfercoefficient and thermal conductivity being power law func-tions of temperature we construct analytical solutions In ouranalysis we consider 119899 = 2 and 3 indicating the fin subject tonucleate boiling and radiation into free space at zero absolutetemperature respectively Firstly given 119899 = 3 and 119898 = 2

6 Mathematical Problems in Engineering

and applying the DTM one obtains the following recurrencerevelation

120581

sum

119894=0

120581minus119894

sum

119897=0

120581minus119894minus119897

sum

119901=0

[120590119901

119901Θ (119897) Θ (119894) (120581 minus 119894 minus 119897 minus 119901 + 1)

times (120581 minus 119894 minus 119897 minus 119901 + 2)Θ (120581 minus 119894 minus 119897 minus 119901 + 2)

+ 2120590119901

119901Θ (119897) (119894 + 1)Θ (119894 + 1)

times (120581 minus 119894 minus 119897 minus 119901 + 1)Θ (120581 minus 119894 minus 119897 minus 119901 + 1)

+ 2120590120590119901

119901Θ (119897) Θ (119894) (120581 minus 119894 minus 119897 minus 119901 + 1)

times Θ (120581 minus 119894 minus 119897 minus 119901 + 1)

minus1198722

Θ(119901)Θ (119897) Θ (119894) Θ (120581 minus 119894 minus 119897 minus 119901) ] = 0

(41)

One may recall the transformed prescribed boundaryconditions (18) and (19) Equation (41) is an iterative formulaof constructing the power series solution as follows

Θ (2) =1198722

1198882

2

Θ (3) = minus1205901198722

1198882

3

Θ (4) =

1198722

1198882

(31205902

minus 21198722

119888)

24

Θ (5) = minus

1198722

1198882

(1205903

minus 41205901198722

119888)

30

Θ (6) =

1198722

1198882

(51205904

minus 781205902

1198722

119888 + 581198724

1198882

)

720

(42)

The pervious process is continuous and onemay consideras many terms as desired (but bearing in mind that DTMconverges quite fast) Substituting (18) to (19) and (42) into(13) we obtain the following closed form of the solution

120579 (119909) = 119888 +1198722

1198882

21199092

minus1205901198722

1198882

31199093

+

1198722

1198882

(31205902

minus 21198722

119888)

241199094

minus

1198722

1198882

(1205903

minus 41205901198722

119888)

301199095

+

1198722

1198882

(51205904

minus 781205902

1198722

119888 + 581198724

1198882

)

7201199096

+ sdot sdot sdot

(43)

To obtain the value of 119888 we substitute the boundarycondition (5) into (43) at the point 119909 = 1 That is

120579 (1) = 119888 +1198722

1198882

2minus1205901198722

1198882

3+

1198722

1198882

(31205902

minus 21198722

119888)

24

minus

1198722

1198882

(1205903

minus 41205901198722

119888)

30

+

1198722

1198882

(51205904

minus 781205902

1198722

119888 + 581198724

1198882

)

720+ sdot sdot sdot = 1

(44)

Substituting this value of 119888 into (43) one finds theexpression for 120579(119909) On the other hand given (119899119898) = (2 3)one obtains the solution

120579 (119909) = 119888 +1198722

21199092

minus1205901198722

31199093

+

1198722

(1205902

119888 minus 21198722

)

81198881199094

minus

1205901198722

(21205902

119888 minus 211198722

)

601198881199095

+

1198722

(1801198724

minus 1861205902

1198722

119888 + 51205904

1198882

)

7201199096

+ sdot sdot sdot

(45)

Here the constant 119888 maybe obtained by evaluating theboundary condition 120579(1) = 1 The solutions (43) and (45) aredepicted in Figures 3(b) and 3(a) respectively

52 The Rectangular Profile and Power Law Thermal Con-ductivity In this section we provide a detailed constructionof analytical solutions for the heat transfer in a longitudinalrectangular finwith a power law thermal conductivity that iswe consider (4) with 119891(119909) = 1 and 119896(120579) = 120579119898 The analyticalsolutions are given in the following expressions

(a) Case (119899119898) = (2 3)

120579 (119909) = 119888 +1198722

21199092

minus1198724

41198881199094

+1198726

411988821199096

minus331198728

12211988831199098

+127119872

10

336119888411990910

+ sdot sdot sdot

(46)

(b) Case (119899119898) = (3 2)

120579 (119909) = 119888 +1198882

1198722

21199092

minus1198883

1198724

121199094

+291198884

1198726

3601199096

minus3071198885

1198728

50401199098

+23483119888

6

11987210

45360011990910

+ sdot sdot sdot

(47)

The constant 119888 is obtained by solving the appropriate 120579(119909)at the fin base boundary condition The analytical solutionsin (46) and (47) are depicted in Figures 4(a) and 4(b)respectively

Mathematical Problems in Engineering 7

0 02 04 06 08 1094

095

096

097

098

099

1

119909

120579

119872 = 02

119872 = 03

119872 = 04

119872 = 05

(a)

0 02 04 06 08 1093

094

095

096

097

098

099

1

119909

120579

119872 = 02

119872 = 03

119872 = 04

119872 = 05

(b)

Figure 3 Temperature profile in a longitudinal fin with exponential profile and power law thermal conductivity In (a) the exponents are(119899119898) = (2 3) and (b) (119899 119898) = (3 2)

119872 = 025

119872 = 04

119872 = 05

119872 = 065

082

084

086

088

09

092

094

096

098

1

120579

0 02 04 06 08 1

119909

(a)

084

086

088

09

092

094

096

098

1

120579

0 02 04 06 08 1

119909

119872 = 025

119872 = 04

119872 = 05

119872 = 065

(b)

Figure 4 Temperature profile in a longitudinal fin with rectangular profile and power law thermal conductivity In (a) the exponents are(119899119898) = (2 3) and (b) (119899 119898) = (3 2)

53 The Convex Parabolic Profile and Power Law ThermalConductivity In this section we present solutions for theequation describing the heat transfer in a fin with con-vex parabolic profile and power law thermal conductivityEquation (40) is considered Here we consider the values(119899119898) = (2 3) (3 2) The final analytical solution is givenby

(a) Case (119899119898) = (2 3)

120579 (119909) = 119888 +21198722

311990932

minus21198724

51198881199093

+231198726

45119888211990992

minus1909119872

8

247511988831199096

+329222119872

10

2598751198884119909152

+ sdot sdot sdot

(48)

8 Mathematical Problems in Engineering

075

08

085

09

095

1

119872 = 025

119872 = 04

119872 = 05

119872 = 065

120579

0 02 04 06 08 1119909

(a)

119872 = 025

119872 = 04

119872 = 05

119872 = 065

0 02 04 06 08 1119909

082

084

086

088

09

092

094

096

098

120579

1

(b)

Figure 5 Temperature profile in a longitudinal fin with convex parabolic profile and power law thermal conductivity In (a) the exponentsare (119899119898) = (2 3) and (b) (119899119898) = (3 2)

(b) Case (119899119898) = (3 2)

120579 (119909) = 119888 +21198882

1198722

311990932

minus41198883

1198724

451199093

+41198884

1198726

2711990992

minus9921198885

1198728

74251199096

+30064119888

6

11987210

212625119909152

+ sdot sdot sdot

(49)

The constant 119888 is obtained by evaluating the appropriate 120579(119909)at the fin base boundary condition The solutions in (48) and(49) are depicted in Figures 5(a) and 5(b) respectively

54 The Rectangular Profile and LinearThermal ConductivityIn this section we present solutions for the equation repre-senting the heat transfer in a fin with rectangular profile andthe thermal conductivity depending linearly on temperatureThat is we consider (4) with 119891(119909) = 1 and 119896(120579) = 1 + 120573120579 Theanalytical solutions for this problem for different values of 119899are given by

(a) Case 119899 = 0

120579 (119909)

= 119888 +

1198722

119888

2 (1 + 120573119888)

1199092

minus

1198724

119888 (minus1 + 2120573119888)

24(1 + 120573119888)31199094

+

1198726

119888 (1 minus 16120573119888 + 281205732

1198882

)

720(1 + 120573119888)51199096

minus

1198728

119888 (minus1 + 78120573119888 minus 6001205732

1198882

+ 8961205733

1198883

)

40320(1 + 120573119888)7

1199098

+

11987210

119888 (1 minus 332120573119888 + 78121205732

1198882

minus 398961205733

1198883

+ 511841205734

1198884

)

3628800(1 + 120573119888)9

11990910

+ sdot sdot sdot

(50)

(b) Case 119899 = 1

120579 (119909)

= 119888 +

11987221198882

2 (1 + 120573119888)

1199092minus

11987241198883(minus1 + 2120573119888)

24(1 + 120573119888)31199094

+

11987261198884(10 minus 16120573119888 + 19120573

21198882)

720(1 + 120573119888)5

1199096

minus

11987281198885(minus80 + 342120573119888 minus 594120573

21198882+ 559120573

31198883)

40320(1 + 120573119888)7

1199098

+

119872101198886(1000 minus 7820120573119888 + 24336120573

21198882minus 36908120573

31198883+ 29161120573

41198884)

3628800(1 + 120573119888)9

11990910

+ sdot sdot sdot

(51)

(c) Case 119899 = 2

120579 (119909)

= 119888 +

1198722

1198883

2 (1 + 120573119888)

1199092

+

1198724

1198885

8(1 + 120573119888)31199094

+

1198726

1198887

(3 + 21205732

1198882

)

80(1 + 120573119888)51199096

+

1198728

1198889

(49 minus 20120573119888 + 661205732

1198882

minus 401205733

1198883

)

4480(1 + 120573119888)7

1199098

+

11987210

11988811

(427 minus 440120573119888 + 11161205732

1198882

minus 10201205733

1198883

+ 6721205734

1198884

)

134400(1 + 120573119888)9

11990910

+ sdot sdot sdot

(52)

Mathematical Problems in Engineering 9

(d) Case 119899 = 3120579 (119909)

= 119888 +

11987221198884

2 (1 + 120573119888)

1199092+

11987241198887(4 + 120573119888)

24(1 + 120573119888)31199094

+

119872611988810(52 + 32120573119888 + 25120573

21198882)

720(1 + 120573119888)5

1199096

+

119872811988813(1288 + 1020120573119888 + 1212120573

21198882minus 95120573

31198883)

40320(1 + 120573119888)7

1199098

+

1198721011988816(52024+45688120573119888+77184120573

21198882minus680120573

31198883+ 15025120573

41198884)

3628800(1 + 120573119888)9

11990910

+

(53)

The constant 119888may be obtained from the boundary conditionon the appropriate solution The solutions in (50) (51) (52)and (53) are depicted in Figure 6

55The Convex Parabolic Profile and LinearThermal Conduc-tivity In this section we present solutions for the equationdescribing the heat transfer in a fin with convex parabolicprofile and the thermal conductivity depending linearly ontemperature That is we consider (40) with 119896(120579) = 1 + 120573120579The analytical solution for this problem for different values of119899 is given by

(a) Case 119899 = 0

120579 (119909)

= 119888 +21198722

119888

3 (1 + 120573119888)11990932

minus21198724

119888 (minus2 + 3120573119888)

45(1 + 120573119888)3

1199093

+

1198726

119888 (2 minus 25120573119888 + 331205732

1198882

)

405(1 + 120573119888)5

11990992

minus

1198728

119888 (minus10 + 599120573119888 minus 35821205732

1198882

+ 40591205733

1198883

)

66825(1 + 120573119888)7

1199096

+ sdot sdot sdot

(54)

(b) Case 119899 = 1

120579 (119909)

= 119888 +21198722

1198882

3 (1 + 120573119888)11990932

minus21198724

1198883

(minus4 + 120573119888)

45(1 + 120573119888)3

1199093

+

21198726

1198884

(9 minus 11120573119888 + 101205732

1198882

)

405(1 + 120573119888)5

11990992

minus

21198728

1198885

(minus330 + 1118120573119888 minus 15841205732

1198882

+ 10931205733

1198883

)

66825(1 + 120573119888)7

1199096

+ sdot sdot sdot

(55)

119899 = 0

119899 = 1

119899 = 2

119899 = 3

1

095

09

085

08

075

07

065

06

055

0 02 04 06 08 1

119909

120579

Figure 6 Temperature profile in a longitudinal rectangular fin withlinear thermal conductivity and varying values of 119899 Here 120573 = 05and119872 = 15 are fixed

(c) Case 119899 = 2

120579 (119909)

= 119888 +21198722

1198883

3 (1 + 120573119888)11990932

+21198724

1198885

(6 + 120573119888)

45(1 + 120573119888)31199093

+

1198726

1198887

(16 + 3120573119888 + 71205732

1198882

)

135(1 + 120573119888)5

11990992

+

1198728

1198889

(1160 minus 107120573119888 + 11161205732

1198882

minus 3671205733

1198883

)

22275(1 + 120573119888)7

1199096

+ sdot sdot sdot

(56)

(d) Case 119899 = 3

120579 (119909)

= 119888 +21198722

1198884

3 (1 + 120573119888)11990932

+21198724

1198887

(8 + 3120573119888)

45(1 + 120573119888)31199093

+

41198726

11988810

(23 + 17120573119888 + 91205732

1198882

)

405(1 + 120573119888)5

11990992

+

21198728

11988813

(5000+4868120573119888 + 43561205732

1198882

+ 3631205733

1198883

)

66825(1 + 120573119888)7

1199096

+ sdot sdot sdot

(57)

The constant 119888 may be obtained from the boundarycondition on the appropriate 120579(119909)The solutions in (54) (55)(56) and (57) are depicted in Figure 7

10 Mathematical Problems in Engineering

1

09

08

07

06

05

120579

119899 = 0

119899 = 1

119899 = 2

119899 = 3

0 02 04 06 08 1

119909

Figure 7 Temperature profile in a longitudinal convex parabolic finwith linear thermal conductivity and varying values of 119899 Here 120573 =05 and119872 = 15 are fixed

56 The Exponential Profile and LinearThermal ConductivityIn this section we present solutions for equation heat transferin a fin with convex parabolic profile and the thermalconductivity depending linearly on temperature That is weconsider (4) with 119896(120579) = 1 + 120573120579 and 119891(119909) = e120590119909 Here 120590is a constant The analytical solutions for this problem fordifferent values of 119899 are given by

(a) Case 119899 = 0

120579 (119909)

= 119888 +

1198722

119888

2 (1 + 120573119888)

1199092

minus

1205901198722

119888 (2 + 120573119888 + 1205732

119888)

6(1 + 120573119888)21199093

+

1198722

119888 (1198722

(1 minus 21205732

119888) + 1205902

(3 +3120573119888+1205733

1198882

+1205734

1198882

+1205732

119888 (3 + 119888)))

24(1 + 120573119888)4

1199094

+ sdot sdot sdot

(58)

(b) Case 119899 = 1

120579 (119909)

= 119888 +

11987221198882

2 (1 + 120573119888)

1199092minus

12059011987221198882(2 + 120573119888 + 120573

2119888)

6(1 + 120573119888)21199093

+

11987221198882(1198722119888 (2+120573119888minus2120573

2119888)+1205902(3+3120573119888+120573

31198882+12057341198882+1205732119888 (3+119888)))

24(1 + 120573119888)3

1199094

+ sdot sdot sdot

(59)

119899 = 0

119899 = 1

119899 = 2

119899 = 3

0 02 04 06 08 1

119909

1

095

09

085

08

075

07

120579

Figure 8 Temperature profile in a longitudinal fin of exponentialprofile with linear thermal conductivity and varying values of 119899Here 120573 = 05 and119872 = 15 are fixed

(c) Case 119899 = 2120579 (119909)

= 119888 +

11987221198883

2 (1 + 120573119888)

1199092minus

12059011987221198883(2 + 120573119888 + 120573

2119888)

6(1 + 120573119888)21199093

+

11987221198883(11987221198882(3+2120573119888minus2120573

2119888)+1205902(3+3120573119888+120573

31198882+12057341198882+1205732119888 (3+119888)))

24(1 + 120573119888)3

times1199094+ sdot sdot sdot

(60)

(d) Case 119899 = 3120579 (119909)

= 119888 +

11987221198884

2 (1 + 120573119888)

1199092minus

12059011987221198884(2 + 120573119888 + 120573

2119888)

6(1 + 120573119888)21199093

+

11987221198884(11987221198883(4+3120573119888minus2120573

2119888)+1205902(3+3120573119888+120573

31198882+12057341198882+1205732119888 (3+119888)))

24(1 + 120573119888)3

times1199094+ sdot sdot sdot

(61)

The constant 119888 may be obtained from the boundarycondition on the appropriate 120579(119909)The solutions in (58) (59)(60) and (61) are depicted in Figure 8

6 Fin Efficiency and Heat Flux

61 Fin Efficiency Theheat transfer rate from a fin is given byNewtonrsquos second law of cooling

119876 = int

119871

0

119875119867 (119879) (119879 minus 119879119886) 119889119909 (62)

Mathematical Problems in Engineering 11

Fin efficiency is defined as the ratio of the fin heat transferrate to the rate that would be if the entire fin were at the basetemperature and is given by (see eg [1])

120578=

119876

119876ideal=

int119871

0

119875119867 (119879) (119879 minus 119879119886) 119889119883

119875ℎ119887119871 (119879119887minus 119879119886)

(63)

In dimensionless variables we have

120578 = int

1

0

120579119899+1

119889119909 (64)

We consider the solutions (50) (51) (52) and (53) and depictthe fin efficiency (63) in Figure 9

62 Heat Flux The fin base heat flux is given by the Fourierrsquoslaw

119902119887= 119860119888119870 (119879)

119889119879

119889119909 (65)

The total heat flux of the fin is given by [1]

119902 =119902119887

119860119888119867(119879) (119879

119887minus 119879119886) (66)

Introducing the dimensionless variable as described inSection 2 implies

119902 =1

119861119894

119896 (120579)

ℎ (120579)

119889120579

119889119909 (67)

where the dimensionless parameter 119861119894 = ℎ119887119871119896119886is the Biot

number We consider a number of cases for the thermalconductivity and the heat transfer coefficient

621 Linear Thermal Conductivity and Power Law HeatTransfer Coefficient In this case (67) becomes

119902 =1

119861119894(1 + 120573120579) 120579

minus119899119889120579

119889119909 (68)

The heat flux in (68) at the base of the fin is plotted inFigure 10

622 Power LawThermal Conductivity and Heat TransferCoefficient In this case (67) becomes

119902 =1

119861119894120579119898minus119899

119889120579

119889119909 (69)

Not surprisingly heat flux in one-dimensional fins is highergiven values 119861119894 ≪ 1 The heat flux in (69) is plotted inFigures 11(a) 11(b) and 11(c)

7 Some Discussions

The DTM has resulted in some interesting observations andstudy We have observed in Figures 2(a) and 2(b) an excellentagreement between the analytical solutions generated byDTM and the exact solution obtained in [2] In particular we

119872

1

095

09

085

08

075

07

065

06

055

120578

2151050

119899 = 0

119899 = 1

119899 = 2

119899 = 3

Figure 9 Fin efficiency of a longitudinal rectangular fin Here 120573 =075

119872

25

20

15

10

5

0

119902

2151050

119899 = 0

119899 = 1

119899 = 2

119899 = 3

Figure 10 Base heat flux in a longitudinal rectangular finwith linearthermal conductivity Here 120573 = 01

considered a fin problem in which both thermal conductivityand heat transfer coefficient are given by the same power lawFurthermore we notice from Table 2 that an absolute errorof approximately 35119890 minus 005 is produced by DTM of order119874(15) In Table 3 an absolute error of approximately 65119890 minus006 is produced for the same order This confirms that theDTM converges faster and can provide accurate results witha minimum computation As such a tremendous confidencein the DTM in terms of the accuracy and effectiveness wasbuilt and thus we used this method to solve other problemsfor which exact solutions are harder to construct

In Figures 3(a) 3(b) 4(a) 4(b) 5(a) and 5(b) weobserve that the fin temperature increaseswith the decreasing

12 Mathematical Problems in Engineering

values of the thermogeometric fin parameter Here thevalues of the exponents are fixed Also we observe that fintemperature is higher when 119899 minus 119898 gt 0 that is when heattransfer coefficient is higher than the thermal conductivityWe observe in Figures 6 7 and 8 that the fin temperatureincreases with the increasing values of 119899 Furthermoreit appears that the fin with exponential profile performsthe least in transferring the heat from the base since thetemperature in such a fin is much higher than that ofthe rectangular and the convex parabolic profiles In otherwords heat dissipation to the fluid surrounding the extendedsurface is much faster in longitudinal fins of rectangular andconvex parabolic profiles In Figure 9 fin efficiency decreaseswith increasing thermogeometric fin parameter Also finefficiency increases with increasing values of 119899 It is easyto show that the thermogeometric fin parameter is directlyproportional to the aspect ratio (extension factor) with squareroot of the Biot number being the proportionality constantAs such shorter fins are more efficient than longer onesElse the increased Biot number results in less efficient finwhenever the space is confined that is where the length ofthe fin cannot be increased Figure 10 depicts the heat fluxat the fin base The amount of heat energy dissipated fromthe fin base is of immense interest in engineering [28] Weobserve in Figure 10 that the base heat flux increases withthe thermogeometric fin parameter for considered values ofthe exponent 119899 (see also [28]) Figures 11(a) 11(b) and 11(c)display the heat flux across the fin length We note that theheat flux across the fin length increases with increasing valuesof the thermogeometric fin parameter

8 Concluding Remarks

In this study we have successfully applied the DTM to highlynonlinear problems arising in heat transfer through longitu-dinal fins of various profiles Both thermal conductivity andheat transfer coefficient are given as functions of temperatureThe DTM agreed well with exact solutions when the thermalconductivity and heat transfer coefficient are given by thesame power law A rapid convergence to the exact solutionwas observed Following the confidence in DTM built by theresults mentioned we then solved various exciting problemsThe exotic results have been shown in tables and figures listedin this paper

The results obtained in this paper are significant improve-ments on the known results In particular both the heattransfer coefficient and the thermal conductivity are allowedto be given by the power law functions of temperature andalsowe considered a number of finprofilesWenote that exactsolutions are difficult if not impossible to construct when theexponents of these properties are distinct

Perhaps the notable advantage of the DTM is the general-ization of the Taylor method to problems involving unusualderivative procedures such as fractional fuzzy or q-derivative[22] Some generalizations have been made by Odibat et al[24] and they referred to their new method as the General-ized Differential Transform Method (GDTM) This showedgreat improvement compared to the Fractional Differential

0 02 04 06 08 1119909

45

40

35

30

25

20

15

10

5

0

minus5

119902

(a)

0 02 04 06 08 1119909

30

25

20

15

10

5

0

119902

(b)

35

30

25

20

15

10

5

0

119902

0 02 04 06 08 1119909

119872 = 025

119872 = 04

119872 = 05

119872 = 065

(c)

Figure 11 Heat flux across a longitudinal rectangular fin with linearthermal conductivity (a) 119899minus119898 lt 0 (b) 119899minus119898 gt 0 and (c) 119899minus119898 = 0Here 119861119894 = 001

Mathematical Problems in Engineering 13

Transform Method (FDTM) introduced by Arikoglu andOzkol [23]

We have shown with the help of an example that DTMmay only be applied to fin problems involving heat transferthrough fins with convex parabolic profile Note that given anODE such as (4) with a power law heat transfer coefficient ofa fractional exponent then one can easily remove the fractionby basic exponential rules and employment of the Binomialexpansion However using the DTM one runs into difficultyif the power law thermal conductivity in the same equation isgiven by the fractional exponent We do not know whetherthese observations call for the ldquomodifiedrdquo DTM to solveproblems arising in heat flow through fins with other profilessuch as longitudinal triangular and concave parabolic andalso with fractional power law thermal conductivity

The main results obtained in this paper give insightinto heat transfer in boiling liquids where the heat transfercoefficient is temperature dependent and may be given by apower law The thermal conductivity of some materials suchas gallium nitride (GaN) and Aluminium Nitride (AlN) maybe modeled by power law temperature dependency [29 30]Thus the solutions constructed here give a better comparisonof heat transfer in terms of material used since in many engi-neering applications thermal conductivity is given as a linearfunction of temperature Furthermore a good study in termsof performance and efficiency of fin with different profiles isundertaken These finding could help in the design of fins Itis claimed in [14] that DTM results are more accurate thanthose constructed by Variational Iteration Methods (VIM)and Homotopy Perturbation Methods (HPM) However itwould be risky to use the DTM approximate solutions asbenchmarks for the numerical schemes Nevertheless wehave also shown that DTM converges rapidly in just fifteenterms to the exact solution

Acknowledgments

The authors wish to thank the National Research Foundationof South Africa and the University of the Witwatersrandfor the generous financial support The authors are gratefulto Professor D P Mason and the anonymous reviewers fortheir invaluable comments which improved this paper signif-icantly

References

[1] A D Kraus A Aziz and J Welty Extended Surface Heat Trans-fer Wiley New York NY USA 2001

[2] R JMoitsheki T Hayat andM YMalik ldquoSome exact solutionsof the fin problem with a power law temperature-dependentthermal conductivityrdquo Nonlinear Analysis Real World Applica-tions vol 11 no 5 pp 3287ndash3294 2010

[3] R J Moitsheki ldquoSteady one-dimensional heat flow in a lon-gitudinal triangular and parabolic finrdquo Communications inNonlinear Science and Numerical Simulation vol 16 no 10 pp3971ndash3980 2011

[4] R J Moitsheki ldquoSteady heat transfer through a radial fin withrectangular and hyperbolic profilesrdquo Nonlinear Analysis RealWorld Applications vol 12 no 2 pp 867ndash874 2011

[5] Mo Miansari D D Ganji and Me Miansari ldquoApplication ofHersquos variational iteration method to nonlinear heat transferequationsrdquo Physics Letters A vol 372 no 6 pp 779ndash785 2008

[6] C H Chiu and C K Chen ldquoA decomposition method forsolving the convective longitudinal fins with variable thermalconductivityrdquo International Journal of Heat and Mass Transfervol 45 no 10 pp 2067ndash2075 2002

[7] A Rajabi ldquoHomotopy perturbation method for fin efficiencyof convective straight fins with temperature-dependent thermalconductivityrdquo Physics Letters A vol 364 no 1 pp 33ndash37 2007

[8] G Domairry and M Fazeli ldquoHomotopy analysis method todetermine the fin efficiency of convective straight fins withtemperature-dependent thermal conductivityrdquo Communica-tions in Nonlinear Science and Numerical Simulation vol 14 no2 pp 489ndash499 2009

[9] A Campo andR J Spaulding ldquoCoupling of themethods of suc-cessive approximations and undetermined coefficients for theprediction of the thermal behaviour of uniform circumferentialfinsrdquo Heat and Mass Transfer vol 34 no 6 pp 461ndash468 1999

[10] A Campo and F Rodrfguez ldquoApproximate analytic tempera-ture solution for uniform annular fins by adapting the powerseriesmethodrdquo International Communications inHeat andMassTransfer vol 25 no 6 pp 809ndash818 1998

[11] R Chiba ldquoApplication of differential transformmethod to ther-moelastic problem for annular disks of variable thickness withtemperature-dependent parametersrdquo International Journal ofThermophysics vol 33 pp 363ndash380 2012

[12] S Sadri M R Raveshi and S Amiri ldquoEfficiency analysis ofstraight fin with variable heat transfer coefficient and thermalconductivityrdquo Journal ofMechanical Science andTechnology vol26 no 4 pp 1283ndash1290 2012

[13] M Torabi H Yaghoori and A Aziz ldquoAnalytical solution forconvective-radiative continously moving fin with temperaturedependent thermal conductivityrdquo International Journal of Ther-mophysics vol 33 pp 924ndash941 2012

[14] M Torabi andH Yaghoobi ldquoTwo dominant analytical methodsfor thermal analysis of convective step fin with variable thermalconductivityrdquoThermal Science In press

[15] AMoradi ldquoAnalytical solutions for finwith temperature depen-dant heat transfer coefficientrdquo International Journal of Engineer-ing and Applied Sciences vol 3 no 2 pp 1ndash12 2011

[16] A Moradi and H Ahmadikia ldquoAnalytical solution for differentprofiles of fin with temperature-dependent thermal conductiv-ityrdquoMathematical Problems in Engineering vol 2010 Article ID568263 15 pages 2010

[17] H Yaghoobi and M Torabi ldquoThe application of differentialtransformation method to nonlinear equations arising in heattransferrdquo International Communications in Heat and MassTransfer vol 38 no 6 pp 815ndash820 2011

[18] S Ghafoori M Motevalli M G Nejad F Shakeri D D GanjiandM Jalaal ldquoEfficiency of differential transformation methodfor nonlinear oscillation Comparison with HPM and VIMrdquoCurrent Applied Physics vol 11 no 4 pp 965ndash971 2011

[19] A A Joneidi D D Ganji andM Babaelahi ldquoDifferential trans-formation method to determine fin efficiency of convectivestraight fins with temperature dependent thermal conductivityrdquoInternational Communications in Heat and Mass Transfer vol36 no 7 pp 757ndash762 2009

[20] J K Zhou Differential Transform Method and Its Applicationsfor Electric circuIts Huazhong University Press Wuhan China1986

14 Mathematical Problems in Engineering

[21] S Mukherjee ldquoReply to comment on lsquosolutions of the duffing-van der pol oscillator equation by the differential transformmethodrsquordquo Physica Scripta vol 84 Article ID 037003 2 pages2011

[22] C Bervillier ldquoStatus of the differential transformationmethodrdquoApplied Mathematics and Computation vol 218 no 20 pp10158ndash10170 2012

[23] A Arikoglu and I Ozkol ldquoSolution of fractional differentialequations by using differential transform methodrdquo ChaosSolitons and Fractals vol 34 no 5 pp 1473ndash1481 2007

[24] ZOdibat SMomani andV S Erturk ldquoGeneralized differentialtransform method application to differential equations offractional orderrdquo Applied Mathematics and Computation vol197 no 2 pp 467ndash477 2008

[25] CW Bert ldquoApplication of differential transformmethod to heatconduction in tapered finsrdquo Journal of Heat Transfer vol 124no 1 pp 208ndash209 2002

[26] F Khani and A Aziz ldquoThermal analysis of a longitudinal trape-zoidal fin with temperature-dependent thermal conductivityand heat transfer coefficientrdquo Communications in NonlinearScience and Numerical Simulation vol 15 no 3 pp 590ndash6012010

[27] H C Unal ldquoAn anlytical study of boiling heat transfer from afinrdquo International Journal of Heat and Mass Transfer vol 31 no7 pp 1483ndash1496 1988

[28] M H Chang ldquoA decomposition solution for fins with tempera-ture dependent surface heat fluxrdquo International Journal of Heatand Mass Transfer vol 48 no 9 pp 1819ndash1824 2005

[29] A Jezowski B A Danilchenko M Bockowski et al ldquoThermalconductivity of GaN crystals in 42ndash300 K rangerdquo Solid StateCommunications vol 128 no 2-3 pp 69ndash73 2003

[30] S Vitanov V Palankovski S Maroldt and R Quay ldquoHigh-temperature modeling of AlGaNGaN HEMTsrdquo Solid-StateElectronics vol 54 no 10 pp 1105ndash1112 2010

Submit your manuscripts athttpwwwhindawicom

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Mathematical Problems in Engineering

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Differential EquationsInternational Journal of

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Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

Page 3: Research Article Analytical Solutions for Steady Heat ...downloads.hindawi.com/journals/mpe/2013/273052.pdf · Research Article Analytical Solutions for Steady Heat Transfer in Longitudinal

Mathematical Problems in Engineering 3

The dimensionless thermal conductivity given by the powerlaw and the linear function of temperature is 119896(120579) = 120579119898 and119896(120579) = 1 + 120573120579 respectively Here the thermal conductivitygradient is120573 = 120574(119879

119887minus119879119886) Furthermore we consider a various

fin profiles including the longitudinal rectangular 119891(119909) =1 the longitudinal convex parabolic 119891(119909) = radic119909 and theexponential profile 119891(119909) = e119886119909 with 119886 being the constant (seealso [16])

3 Fundamentals of the DifferentialTransform Method

In this section the basic idea underlying the DTM is brieflyintroduced Let 119910(119905) be an analytic function in a domain DThe Taylor series expansion function of 119910(119905) with the centerlocated at 119905 = 119905

119895is given by [20]

119910 (119905) =

infin

sum

120581=0

(119905 minus 119905119895)120581

120581[119889120581

119910 (119905)

119889119905120581]

119905=119905119895

forall119905 isin D (10)

The particular case of (10) when 119905119895= 0 is referred to as

the Maclaurin series expansion of 119910(119905) and is expressed as

119910 (119905) =

infin

sum

120581=0

119905120581

120581[119889120581

119910 (119905)

119889119905120581]

119905=0

forall119905 isin D (11)

The differential transform of 119910(119905) is defined as follows

119884 (119905) =

infin

sum

120581=0

H120581

120581[119889120581

119910 (119905)

119889119905120581]

119905=0

(12)

where 119910(119905) is the original analytic function and 119884(119905) is thetransformed function The differential spectrum of 119884(119905) isconfined within the interval 119905 isin [0H] where H is aconstant From (11) and (12) the differential inverse transformof 119884(119905) is defined as follows

119910 (119905) =

infin

sum

120581=0

(119905

H)

120581

119884 (120581) (13)

and if 119910(119905) is expressed by a finite series then

119910 (119905) =

119903

sum

120581=0

(119905

H)

120581

119884 (120581) (14)

Some of the useful mathematical operations performedby the differential transform method are listed in Table 1

The delta function 120575(120581 minus 119904) is given by

120575 (120581 minus 119904) = 1 if 120581 = 1199040 if 120581 = 119904

(15)

4 Comparison of Exact andAnalytical Solutions

In this section we consider a model describing temperaturedistribution in a longitudinal rectangular fin with both ther-mal conductivity and heat transfer coefficient being functions

Table 1 Fundamental operations of the differential transformmethod

Original function Transformed function119910(119905) = 119909(119905) plusmn 119911(119905) 119884(119905) = 119883(119905) plusmn 119885(119905)

119910(119905) = 120572119909(119905) 119884(119905) = 120572119883(119905)

119910(119905) =119889119910(119905)

119889119905119884(119905) = (120581 + 1)119884(120581 + 1)

119910(119905) =1198892

119910(119905)

1198891199052119884(119905) = (120581 + 1)(120581 + 2)119884(120581 + 2)

119910(119905) =119889119904

119910(119905)

119889119905119904119884(119905) = (120581 + 1)(120581 + 2) sdot sdot sdot (120581 + 119904)119884(120581 + 119904)

119910(119905) = 119909(119905)119911(119905) 119884(119905) =

120581

sum

119894=0

119883(119894)119885(120581 minus 119894)

119910(119905) = 1 119884(119905) = 120575(120581)

119910(119905) = 119905 119884(119905) = 120575(120581 minus 1)

119910(119905) = 119905119904

119884(119905) = 120575(120581 minus 119904)

119910(119905) = exp(120582119905) 119884(119905) =120582120581

120581

119910(119905) = (1 + 119905)119904

119884(119905) =119904(119904 minus 1) sdot sdot sdot (119904 minus 120581 + 1)

120581

119910(119905) = sin(120596119905 + 120572) 119884(119905) =120596120581

120581sin(120587120581

2+ 120572)

119910(119905) = cos(120596119905 + 120572) 119884(119905) =120596120581

120581cos(120587120581

2+ 120572)

of temperature given by the power law (see eg [2])The exactsolution of (4) when both the power laws are given by thesame exponent is given by [2]

120579 (119909) = [

cosh (119872radic119899 + 1119909)

cosh (119872radic119899 + 1)]

1(119899+1)

(16)

We use this exact solution as a benchmark or validation ofthe DTM The effectiveness of the DTM is determined bycomparing the exact and the analytical solutionsWe comparethe results for the cases 119899 = 1 and 119899 = 2 with fixed values of119872

41 Case 119899=1 Applying the DTM to (4) with the power lawthermal conductivity 119891(119909) = 1 (rectangular profile) andgivenH = 1 one obtains the following recurrence relation

120581

sum

119894=0

[Θ (119894) (120581 minus 119894 + 1) (120581 minus 119894 + 2)Θ (120581 minus 119894 + 2)

+ (119894 + 1)Θ (119894 + 1) (120581 minus 119894 + 1)Θ (120581 minus 119894 + 1)

minus1198722

Θ (119894)Θ (120581 minus 119894)] = 0

(17)

Exerting the transformation to the boundary condition (6) ata point 119909 = 0

Θ (1) = 0 (18)

The other boundary conditions are considered as follows

Θ (0) = 119888 (19)

4 Mathematical Problems in Engineering

where 119888 is a constant Equation (17) is an iterative formula ofconstructing the power series solution as follows

Θ (2) =1198722

119888

2

(20)

Θ (3) = 0 (21)

Θ (4) = minus1198724

119888

24

(22)

Θ (5) = 0 (23)

Θ (6) =191198726

119888

720

(24)

Θ (7) = 0 (25)

Θ (8) = minus559119872

8

119888

40320

(26)

Θ (9) = 0 (27)

Θ (10) =29161119872

10

119888

3628800

(28)

Θ (11) = 0 (29)

Θ (12) = minus2368081119872

12

119888

479001600

(30)

Θ (13) = 0 (31)

Θ (14) =276580459119872

14

119888

87178291200

(32)

Θ (15) = 0

(33)

These terms may be taken as far as desired Substituting(18) to (32) into (13) we obtain the following analyticalsolution

120579 (119909) = 119888 +1198722

119888

21199092

minus1198724

119888

241199094

+191198726

119888

7201199096

minus559119872

8

119888

403201199098

+29161119872

10

119888

362880011990910

minus2368081119872

12

119888

47900160011990912

+276580459119872

14

119888

8717829120011990914

+ sdot sdot sdot

(34)

To obtain the value of 119888 we substitute the boundarycondition (5) into (34) at the point 119909 = 1 Thus we have

120579 (1) = 119888 +1198722

119888

2minus1198724

119888

24+191198726

119888

720minus559119872

8

119888

40320

+29161119872

10

119888

3628800minus2368081119872

12

119888

479001600

+276580459119872

14

119888

87178291200+ sdot sdot sdot = 1

(35)

Table 2 Results of the DTM and exact solutions for 119899 = 1119872 = 07

119909 DTM Exact Error0 0808093014 080809644 000000342601 0810072036 081007547 000000343402 0815999549 081600301 000000345903 0825847770 082585127 000000350004 0839573173 083957673 000000355905 0857120287 085712392 000000363306 0878426299 087843002 000000372207 0903426003 090342982 000000381408 0932056648 093206047 000000382009 0964262558 096426582 000000326310 1000000000 100000000 0000000000

Table 3 Results of theDTMand Exact Solutions for 119899 = 2119872 = 05

119909 DTM Exact Error0 0894109126 0894109793 000000066501 0895226066 0895226732 000000066602 0898568581 0898569249 000000066803 0904112232 0904112905 000000067204 0911817796 0911818474 000000067805 0921633352 0921634038 000000068506 0933496900 0933497594 000000069407 0947339244 0947339946 000000070208 0963086933 0963087627 000000069409 0980665073 0980665659 000000058610 1000000000 1000000000 0000000000

We omit presenting the tedious process of finding 119888 Howeverone may obtain the expression for 120579(119909) upon substituting theobtained value of 119888 into (34)

42 Case 119899=2 Following a similar approach given inSection 41 and given 119899 = 2 one obtains the analyticalsolution

120579 (119909) = 119888 +1198722

119888

21199092

minus1198724

119888

81199094

+231198726

119888

2401199096

minus1069119872

8

119888

134401199098

+9643119872

10

119888

13440011990910

minus1211729119872

12

119888

1774080011990912

+217994167119872

14

119888

322882560011990914

+ sdot sdot sdot

(36)

The constant 119888may be obtained using the boundary conditionat the fin base The comparison of the DTM and the exactsolutions are reflected in Tables 2 and 3 for different valuesof 119899 Furthermore the comparison of the exact and theanalytical solutions is depicted in Figures 2(a) and 2(b)

Mathematical Problems in Engineering 5

08

082

084

086

088

09

092

094

096

098

1

120579

0 02 04 06 08 1

119909

DTMExact

(a)

086

088

09

092

094

096

098

1

120579

0 02 04 06 08 1

119909

DTMExact

(b)

Figure 2 Comparison of analytical and exact solutions In (a) 119899 = 1119872 = 07 and (b) 119899 = 2119872 = 05

5 Analytical Solutions

It is well known that exact solutions forODEs such as (4) existonly when thermal conductivity and the term containingheat transfer coefficient are connected by differentiation(or simply if the ODE such as (4) is linearizable) [4] Inthis section we determine the analytical solutions for thenonlinearizable (4) firstlywhen thermal conductivity is givenby the power law and secondly as a linear function oftemperature In both cases and throughout this paper theheat transfer coefficient is assumed to be a power law functionof temperature These assumptions of the thermal propertiesare physical realistic We have noticed that DTM runs intodifficulty when the exponent of the power law of the thermalconductivity is given by fractional values and also when thefunction 119891(119909) is given in terms of fractional power law Onemay followMoradi and Ahmadikia [16] by introducing a newvariable to deal with fractional powers of 119891(119909) and on theother hand it is possible to remove the fractional exponent ofthe heat transfer coefficient by fundamental laws of exponentand binomial expansion

Proposition 1 A nonlinear ODE such as (4) may admit theDTM solution if 119891(119909) is a constant or exponential functionHowever if 119891(119909) is a power law then such an equation admitsa DTM solution if the product

119891 (119909) sdot 1198911015840

(119909) = 120572 (37)

holds Here 120572 is a real constant

Proof Introducing the new variable 120585 = 119891(119909) it follows fromchain rule that (4) becomes

120572119889120585

119889119909

119889

119889120585[119896 (120579)

119889120579

119889120585] minus119872

2

120579119899+1

= 0 (38)

The most general solution of condition (37) is

119891 (119909) = (2120572119909 + 120574)12

(39)

where 120574 is an integration constant

Example 2 (a) If 119891(119909) = 119909120590 then 120585 = 119909120590 transforms (4) into

119889

119889120585[119896 (120579)

119889120579

119889120585] minus 4119872

2

120585120579119899+1

= 0 (40)

only if 120590 = 12This example implies that the DTM may only be appli-

cable to problems describing heat transfer in fins withconvex parabolic profile In the next subsections we presentanalytical solutions for (4) with various functions 119891(119909) and119896(120579)

51 The Exponential Profile and Power Law Thermal Con-ductivity In this section we present solutions for equationdescribing heat transfer in a fin with exponential profile andpower law thermal conductivity and heat transfer coefficientThat is given (4) with 119891(119909) = e120590119909 and both heat transfercoefficient and thermal conductivity being power law func-tions of temperature we construct analytical solutions In ouranalysis we consider 119899 = 2 and 3 indicating the fin subject tonucleate boiling and radiation into free space at zero absolutetemperature respectively Firstly given 119899 = 3 and 119898 = 2

6 Mathematical Problems in Engineering

and applying the DTM one obtains the following recurrencerevelation

120581

sum

119894=0

120581minus119894

sum

119897=0

120581minus119894minus119897

sum

119901=0

[120590119901

119901Θ (119897) Θ (119894) (120581 minus 119894 minus 119897 minus 119901 + 1)

times (120581 minus 119894 minus 119897 minus 119901 + 2)Θ (120581 minus 119894 minus 119897 minus 119901 + 2)

+ 2120590119901

119901Θ (119897) (119894 + 1)Θ (119894 + 1)

times (120581 minus 119894 minus 119897 minus 119901 + 1)Θ (120581 minus 119894 minus 119897 minus 119901 + 1)

+ 2120590120590119901

119901Θ (119897) Θ (119894) (120581 minus 119894 minus 119897 minus 119901 + 1)

times Θ (120581 minus 119894 minus 119897 minus 119901 + 1)

minus1198722

Θ(119901)Θ (119897) Θ (119894) Θ (120581 minus 119894 minus 119897 minus 119901) ] = 0

(41)

One may recall the transformed prescribed boundaryconditions (18) and (19) Equation (41) is an iterative formulaof constructing the power series solution as follows

Θ (2) =1198722

1198882

2

Θ (3) = minus1205901198722

1198882

3

Θ (4) =

1198722

1198882

(31205902

minus 21198722

119888)

24

Θ (5) = minus

1198722

1198882

(1205903

minus 41205901198722

119888)

30

Θ (6) =

1198722

1198882

(51205904

minus 781205902

1198722

119888 + 581198724

1198882

)

720

(42)

The pervious process is continuous and onemay consideras many terms as desired (but bearing in mind that DTMconverges quite fast) Substituting (18) to (19) and (42) into(13) we obtain the following closed form of the solution

120579 (119909) = 119888 +1198722

1198882

21199092

minus1205901198722

1198882

31199093

+

1198722

1198882

(31205902

minus 21198722

119888)

241199094

minus

1198722

1198882

(1205903

minus 41205901198722

119888)

301199095

+

1198722

1198882

(51205904

minus 781205902

1198722

119888 + 581198724

1198882

)

7201199096

+ sdot sdot sdot

(43)

To obtain the value of 119888 we substitute the boundarycondition (5) into (43) at the point 119909 = 1 That is

120579 (1) = 119888 +1198722

1198882

2minus1205901198722

1198882

3+

1198722

1198882

(31205902

minus 21198722

119888)

24

minus

1198722

1198882

(1205903

minus 41205901198722

119888)

30

+

1198722

1198882

(51205904

minus 781205902

1198722

119888 + 581198724

1198882

)

720+ sdot sdot sdot = 1

(44)

Substituting this value of 119888 into (43) one finds theexpression for 120579(119909) On the other hand given (119899119898) = (2 3)one obtains the solution

120579 (119909) = 119888 +1198722

21199092

minus1205901198722

31199093

+

1198722

(1205902

119888 minus 21198722

)

81198881199094

minus

1205901198722

(21205902

119888 minus 211198722

)

601198881199095

+

1198722

(1801198724

minus 1861205902

1198722

119888 + 51205904

1198882

)

7201199096

+ sdot sdot sdot

(45)

Here the constant 119888 maybe obtained by evaluating theboundary condition 120579(1) = 1 The solutions (43) and (45) aredepicted in Figures 3(b) and 3(a) respectively

52 The Rectangular Profile and Power Law Thermal Con-ductivity In this section we provide a detailed constructionof analytical solutions for the heat transfer in a longitudinalrectangular finwith a power law thermal conductivity that iswe consider (4) with 119891(119909) = 1 and 119896(120579) = 120579119898 The analyticalsolutions are given in the following expressions

(a) Case (119899119898) = (2 3)

120579 (119909) = 119888 +1198722

21199092

minus1198724

41198881199094

+1198726

411988821199096

minus331198728

12211988831199098

+127119872

10

336119888411990910

+ sdot sdot sdot

(46)

(b) Case (119899119898) = (3 2)

120579 (119909) = 119888 +1198882

1198722

21199092

minus1198883

1198724

121199094

+291198884

1198726

3601199096

minus3071198885

1198728

50401199098

+23483119888

6

11987210

45360011990910

+ sdot sdot sdot

(47)

The constant 119888 is obtained by solving the appropriate 120579(119909)at the fin base boundary condition The analytical solutionsin (46) and (47) are depicted in Figures 4(a) and 4(b)respectively

Mathematical Problems in Engineering 7

0 02 04 06 08 1094

095

096

097

098

099

1

119909

120579

119872 = 02

119872 = 03

119872 = 04

119872 = 05

(a)

0 02 04 06 08 1093

094

095

096

097

098

099

1

119909

120579

119872 = 02

119872 = 03

119872 = 04

119872 = 05

(b)

Figure 3 Temperature profile in a longitudinal fin with exponential profile and power law thermal conductivity In (a) the exponents are(119899119898) = (2 3) and (b) (119899 119898) = (3 2)

119872 = 025

119872 = 04

119872 = 05

119872 = 065

082

084

086

088

09

092

094

096

098

1

120579

0 02 04 06 08 1

119909

(a)

084

086

088

09

092

094

096

098

1

120579

0 02 04 06 08 1

119909

119872 = 025

119872 = 04

119872 = 05

119872 = 065

(b)

Figure 4 Temperature profile in a longitudinal fin with rectangular profile and power law thermal conductivity In (a) the exponents are(119899119898) = (2 3) and (b) (119899 119898) = (3 2)

53 The Convex Parabolic Profile and Power Law ThermalConductivity In this section we present solutions for theequation describing the heat transfer in a fin with con-vex parabolic profile and power law thermal conductivityEquation (40) is considered Here we consider the values(119899119898) = (2 3) (3 2) The final analytical solution is givenby

(a) Case (119899119898) = (2 3)

120579 (119909) = 119888 +21198722

311990932

minus21198724

51198881199093

+231198726

45119888211990992

minus1909119872

8

247511988831199096

+329222119872

10

2598751198884119909152

+ sdot sdot sdot

(48)

8 Mathematical Problems in Engineering

075

08

085

09

095

1

119872 = 025

119872 = 04

119872 = 05

119872 = 065

120579

0 02 04 06 08 1119909

(a)

119872 = 025

119872 = 04

119872 = 05

119872 = 065

0 02 04 06 08 1119909

082

084

086

088

09

092

094

096

098

120579

1

(b)

Figure 5 Temperature profile in a longitudinal fin with convex parabolic profile and power law thermal conductivity In (a) the exponentsare (119899119898) = (2 3) and (b) (119899119898) = (3 2)

(b) Case (119899119898) = (3 2)

120579 (119909) = 119888 +21198882

1198722

311990932

minus41198883

1198724

451199093

+41198884

1198726

2711990992

minus9921198885

1198728

74251199096

+30064119888

6

11987210

212625119909152

+ sdot sdot sdot

(49)

The constant 119888 is obtained by evaluating the appropriate 120579(119909)at the fin base boundary condition The solutions in (48) and(49) are depicted in Figures 5(a) and 5(b) respectively

54 The Rectangular Profile and LinearThermal ConductivityIn this section we present solutions for the equation repre-senting the heat transfer in a fin with rectangular profile andthe thermal conductivity depending linearly on temperatureThat is we consider (4) with 119891(119909) = 1 and 119896(120579) = 1 + 120573120579 Theanalytical solutions for this problem for different values of 119899are given by

(a) Case 119899 = 0

120579 (119909)

= 119888 +

1198722

119888

2 (1 + 120573119888)

1199092

minus

1198724

119888 (minus1 + 2120573119888)

24(1 + 120573119888)31199094

+

1198726

119888 (1 minus 16120573119888 + 281205732

1198882

)

720(1 + 120573119888)51199096

minus

1198728

119888 (minus1 + 78120573119888 minus 6001205732

1198882

+ 8961205733

1198883

)

40320(1 + 120573119888)7

1199098

+

11987210

119888 (1 minus 332120573119888 + 78121205732

1198882

minus 398961205733

1198883

+ 511841205734

1198884

)

3628800(1 + 120573119888)9

11990910

+ sdot sdot sdot

(50)

(b) Case 119899 = 1

120579 (119909)

= 119888 +

11987221198882

2 (1 + 120573119888)

1199092minus

11987241198883(minus1 + 2120573119888)

24(1 + 120573119888)31199094

+

11987261198884(10 minus 16120573119888 + 19120573

21198882)

720(1 + 120573119888)5

1199096

minus

11987281198885(minus80 + 342120573119888 minus 594120573

21198882+ 559120573

31198883)

40320(1 + 120573119888)7

1199098

+

119872101198886(1000 minus 7820120573119888 + 24336120573

21198882minus 36908120573

31198883+ 29161120573

41198884)

3628800(1 + 120573119888)9

11990910

+ sdot sdot sdot

(51)

(c) Case 119899 = 2

120579 (119909)

= 119888 +

1198722

1198883

2 (1 + 120573119888)

1199092

+

1198724

1198885

8(1 + 120573119888)31199094

+

1198726

1198887

(3 + 21205732

1198882

)

80(1 + 120573119888)51199096

+

1198728

1198889

(49 minus 20120573119888 + 661205732

1198882

minus 401205733

1198883

)

4480(1 + 120573119888)7

1199098

+

11987210

11988811

(427 minus 440120573119888 + 11161205732

1198882

minus 10201205733

1198883

+ 6721205734

1198884

)

134400(1 + 120573119888)9

11990910

+ sdot sdot sdot

(52)

Mathematical Problems in Engineering 9

(d) Case 119899 = 3120579 (119909)

= 119888 +

11987221198884

2 (1 + 120573119888)

1199092+

11987241198887(4 + 120573119888)

24(1 + 120573119888)31199094

+

119872611988810(52 + 32120573119888 + 25120573

21198882)

720(1 + 120573119888)5

1199096

+

119872811988813(1288 + 1020120573119888 + 1212120573

21198882minus 95120573

31198883)

40320(1 + 120573119888)7

1199098

+

1198721011988816(52024+45688120573119888+77184120573

21198882minus680120573

31198883+ 15025120573

41198884)

3628800(1 + 120573119888)9

11990910

+

(53)

The constant 119888may be obtained from the boundary conditionon the appropriate solution The solutions in (50) (51) (52)and (53) are depicted in Figure 6

55The Convex Parabolic Profile and LinearThermal Conduc-tivity In this section we present solutions for the equationdescribing the heat transfer in a fin with convex parabolicprofile and the thermal conductivity depending linearly ontemperature That is we consider (40) with 119896(120579) = 1 + 120573120579The analytical solution for this problem for different values of119899 is given by

(a) Case 119899 = 0

120579 (119909)

= 119888 +21198722

119888

3 (1 + 120573119888)11990932

minus21198724

119888 (minus2 + 3120573119888)

45(1 + 120573119888)3

1199093

+

1198726

119888 (2 minus 25120573119888 + 331205732

1198882

)

405(1 + 120573119888)5

11990992

minus

1198728

119888 (minus10 + 599120573119888 minus 35821205732

1198882

+ 40591205733

1198883

)

66825(1 + 120573119888)7

1199096

+ sdot sdot sdot

(54)

(b) Case 119899 = 1

120579 (119909)

= 119888 +21198722

1198882

3 (1 + 120573119888)11990932

minus21198724

1198883

(minus4 + 120573119888)

45(1 + 120573119888)3

1199093

+

21198726

1198884

(9 minus 11120573119888 + 101205732

1198882

)

405(1 + 120573119888)5

11990992

minus

21198728

1198885

(minus330 + 1118120573119888 minus 15841205732

1198882

+ 10931205733

1198883

)

66825(1 + 120573119888)7

1199096

+ sdot sdot sdot

(55)

119899 = 0

119899 = 1

119899 = 2

119899 = 3

1

095

09

085

08

075

07

065

06

055

0 02 04 06 08 1

119909

120579

Figure 6 Temperature profile in a longitudinal rectangular fin withlinear thermal conductivity and varying values of 119899 Here 120573 = 05and119872 = 15 are fixed

(c) Case 119899 = 2

120579 (119909)

= 119888 +21198722

1198883

3 (1 + 120573119888)11990932

+21198724

1198885

(6 + 120573119888)

45(1 + 120573119888)31199093

+

1198726

1198887

(16 + 3120573119888 + 71205732

1198882

)

135(1 + 120573119888)5

11990992

+

1198728

1198889

(1160 minus 107120573119888 + 11161205732

1198882

minus 3671205733

1198883

)

22275(1 + 120573119888)7

1199096

+ sdot sdot sdot

(56)

(d) Case 119899 = 3

120579 (119909)

= 119888 +21198722

1198884

3 (1 + 120573119888)11990932

+21198724

1198887

(8 + 3120573119888)

45(1 + 120573119888)31199093

+

41198726

11988810

(23 + 17120573119888 + 91205732

1198882

)

405(1 + 120573119888)5

11990992

+

21198728

11988813

(5000+4868120573119888 + 43561205732

1198882

+ 3631205733

1198883

)

66825(1 + 120573119888)7

1199096

+ sdot sdot sdot

(57)

The constant 119888 may be obtained from the boundarycondition on the appropriate 120579(119909)The solutions in (54) (55)(56) and (57) are depicted in Figure 7

10 Mathematical Problems in Engineering

1

09

08

07

06

05

120579

119899 = 0

119899 = 1

119899 = 2

119899 = 3

0 02 04 06 08 1

119909

Figure 7 Temperature profile in a longitudinal convex parabolic finwith linear thermal conductivity and varying values of 119899 Here 120573 =05 and119872 = 15 are fixed

56 The Exponential Profile and LinearThermal ConductivityIn this section we present solutions for equation heat transferin a fin with convex parabolic profile and the thermalconductivity depending linearly on temperature That is weconsider (4) with 119896(120579) = 1 + 120573120579 and 119891(119909) = e120590119909 Here 120590is a constant The analytical solutions for this problem fordifferent values of 119899 are given by

(a) Case 119899 = 0

120579 (119909)

= 119888 +

1198722

119888

2 (1 + 120573119888)

1199092

minus

1205901198722

119888 (2 + 120573119888 + 1205732

119888)

6(1 + 120573119888)21199093

+

1198722

119888 (1198722

(1 minus 21205732

119888) + 1205902

(3 +3120573119888+1205733

1198882

+1205734

1198882

+1205732

119888 (3 + 119888)))

24(1 + 120573119888)4

1199094

+ sdot sdot sdot

(58)

(b) Case 119899 = 1

120579 (119909)

= 119888 +

11987221198882

2 (1 + 120573119888)

1199092minus

12059011987221198882(2 + 120573119888 + 120573

2119888)

6(1 + 120573119888)21199093

+

11987221198882(1198722119888 (2+120573119888minus2120573

2119888)+1205902(3+3120573119888+120573

31198882+12057341198882+1205732119888 (3+119888)))

24(1 + 120573119888)3

1199094

+ sdot sdot sdot

(59)

119899 = 0

119899 = 1

119899 = 2

119899 = 3

0 02 04 06 08 1

119909

1

095

09

085

08

075

07

120579

Figure 8 Temperature profile in a longitudinal fin of exponentialprofile with linear thermal conductivity and varying values of 119899Here 120573 = 05 and119872 = 15 are fixed

(c) Case 119899 = 2120579 (119909)

= 119888 +

11987221198883

2 (1 + 120573119888)

1199092minus

12059011987221198883(2 + 120573119888 + 120573

2119888)

6(1 + 120573119888)21199093

+

11987221198883(11987221198882(3+2120573119888minus2120573

2119888)+1205902(3+3120573119888+120573

31198882+12057341198882+1205732119888 (3+119888)))

24(1 + 120573119888)3

times1199094+ sdot sdot sdot

(60)

(d) Case 119899 = 3120579 (119909)

= 119888 +

11987221198884

2 (1 + 120573119888)

1199092minus

12059011987221198884(2 + 120573119888 + 120573

2119888)

6(1 + 120573119888)21199093

+

11987221198884(11987221198883(4+3120573119888minus2120573

2119888)+1205902(3+3120573119888+120573

31198882+12057341198882+1205732119888 (3+119888)))

24(1 + 120573119888)3

times1199094+ sdot sdot sdot

(61)

The constant 119888 may be obtained from the boundarycondition on the appropriate 120579(119909)The solutions in (58) (59)(60) and (61) are depicted in Figure 8

6 Fin Efficiency and Heat Flux

61 Fin Efficiency Theheat transfer rate from a fin is given byNewtonrsquos second law of cooling

119876 = int

119871

0

119875119867 (119879) (119879 minus 119879119886) 119889119909 (62)

Mathematical Problems in Engineering 11

Fin efficiency is defined as the ratio of the fin heat transferrate to the rate that would be if the entire fin were at the basetemperature and is given by (see eg [1])

120578=

119876

119876ideal=

int119871

0

119875119867 (119879) (119879 minus 119879119886) 119889119883

119875ℎ119887119871 (119879119887minus 119879119886)

(63)

In dimensionless variables we have

120578 = int

1

0

120579119899+1

119889119909 (64)

We consider the solutions (50) (51) (52) and (53) and depictthe fin efficiency (63) in Figure 9

62 Heat Flux The fin base heat flux is given by the Fourierrsquoslaw

119902119887= 119860119888119870 (119879)

119889119879

119889119909 (65)

The total heat flux of the fin is given by [1]

119902 =119902119887

119860119888119867(119879) (119879

119887minus 119879119886) (66)

Introducing the dimensionless variable as described inSection 2 implies

119902 =1

119861119894

119896 (120579)

ℎ (120579)

119889120579

119889119909 (67)

where the dimensionless parameter 119861119894 = ℎ119887119871119896119886is the Biot

number We consider a number of cases for the thermalconductivity and the heat transfer coefficient

621 Linear Thermal Conductivity and Power Law HeatTransfer Coefficient In this case (67) becomes

119902 =1

119861119894(1 + 120573120579) 120579

minus119899119889120579

119889119909 (68)

The heat flux in (68) at the base of the fin is plotted inFigure 10

622 Power LawThermal Conductivity and Heat TransferCoefficient In this case (67) becomes

119902 =1

119861119894120579119898minus119899

119889120579

119889119909 (69)

Not surprisingly heat flux in one-dimensional fins is highergiven values 119861119894 ≪ 1 The heat flux in (69) is plotted inFigures 11(a) 11(b) and 11(c)

7 Some Discussions

The DTM has resulted in some interesting observations andstudy We have observed in Figures 2(a) and 2(b) an excellentagreement between the analytical solutions generated byDTM and the exact solution obtained in [2] In particular we

119872

1

095

09

085

08

075

07

065

06

055

120578

2151050

119899 = 0

119899 = 1

119899 = 2

119899 = 3

Figure 9 Fin efficiency of a longitudinal rectangular fin Here 120573 =075

119872

25

20

15

10

5

0

119902

2151050

119899 = 0

119899 = 1

119899 = 2

119899 = 3

Figure 10 Base heat flux in a longitudinal rectangular finwith linearthermal conductivity Here 120573 = 01

considered a fin problem in which both thermal conductivityand heat transfer coefficient are given by the same power lawFurthermore we notice from Table 2 that an absolute errorof approximately 35119890 minus 005 is produced by DTM of order119874(15) In Table 3 an absolute error of approximately 65119890 minus006 is produced for the same order This confirms that theDTM converges faster and can provide accurate results witha minimum computation As such a tremendous confidencein the DTM in terms of the accuracy and effectiveness wasbuilt and thus we used this method to solve other problemsfor which exact solutions are harder to construct

In Figures 3(a) 3(b) 4(a) 4(b) 5(a) and 5(b) weobserve that the fin temperature increaseswith the decreasing

12 Mathematical Problems in Engineering

values of the thermogeometric fin parameter Here thevalues of the exponents are fixed Also we observe that fintemperature is higher when 119899 minus 119898 gt 0 that is when heattransfer coefficient is higher than the thermal conductivityWe observe in Figures 6 7 and 8 that the fin temperatureincreases with the increasing values of 119899 Furthermoreit appears that the fin with exponential profile performsthe least in transferring the heat from the base since thetemperature in such a fin is much higher than that ofthe rectangular and the convex parabolic profiles In otherwords heat dissipation to the fluid surrounding the extendedsurface is much faster in longitudinal fins of rectangular andconvex parabolic profiles In Figure 9 fin efficiency decreaseswith increasing thermogeometric fin parameter Also finefficiency increases with increasing values of 119899 It is easyto show that the thermogeometric fin parameter is directlyproportional to the aspect ratio (extension factor) with squareroot of the Biot number being the proportionality constantAs such shorter fins are more efficient than longer onesElse the increased Biot number results in less efficient finwhenever the space is confined that is where the length ofthe fin cannot be increased Figure 10 depicts the heat fluxat the fin base The amount of heat energy dissipated fromthe fin base is of immense interest in engineering [28] Weobserve in Figure 10 that the base heat flux increases withthe thermogeometric fin parameter for considered values ofthe exponent 119899 (see also [28]) Figures 11(a) 11(b) and 11(c)display the heat flux across the fin length We note that theheat flux across the fin length increases with increasing valuesof the thermogeometric fin parameter

8 Concluding Remarks

In this study we have successfully applied the DTM to highlynonlinear problems arising in heat transfer through longitu-dinal fins of various profiles Both thermal conductivity andheat transfer coefficient are given as functions of temperatureThe DTM agreed well with exact solutions when the thermalconductivity and heat transfer coefficient are given by thesame power law A rapid convergence to the exact solutionwas observed Following the confidence in DTM built by theresults mentioned we then solved various exciting problemsThe exotic results have been shown in tables and figures listedin this paper

The results obtained in this paper are significant improve-ments on the known results In particular both the heattransfer coefficient and the thermal conductivity are allowedto be given by the power law functions of temperature andalsowe considered a number of finprofilesWenote that exactsolutions are difficult if not impossible to construct when theexponents of these properties are distinct

Perhaps the notable advantage of the DTM is the general-ization of the Taylor method to problems involving unusualderivative procedures such as fractional fuzzy or q-derivative[22] Some generalizations have been made by Odibat et al[24] and they referred to their new method as the General-ized Differential Transform Method (GDTM) This showedgreat improvement compared to the Fractional Differential

0 02 04 06 08 1119909

45

40

35

30

25

20

15

10

5

0

minus5

119902

(a)

0 02 04 06 08 1119909

30

25

20

15

10

5

0

119902

(b)

35

30

25

20

15

10

5

0

119902

0 02 04 06 08 1119909

119872 = 025

119872 = 04

119872 = 05

119872 = 065

(c)

Figure 11 Heat flux across a longitudinal rectangular fin with linearthermal conductivity (a) 119899minus119898 lt 0 (b) 119899minus119898 gt 0 and (c) 119899minus119898 = 0Here 119861119894 = 001

Mathematical Problems in Engineering 13

Transform Method (FDTM) introduced by Arikoglu andOzkol [23]

We have shown with the help of an example that DTMmay only be applied to fin problems involving heat transferthrough fins with convex parabolic profile Note that given anODE such as (4) with a power law heat transfer coefficient ofa fractional exponent then one can easily remove the fractionby basic exponential rules and employment of the Binomialexpansion However using the DTM one runs into difficultyif the power law thermal conductivity in the same equation isgiven by the fractional exponent We do not know whetherthese observations call for the ldquomodifiedrdquo DTM to solveproblems arising in heat flow through fins with other profilessuch as longitudinal triangular and concave parabolic andalso with fractional power law thermal conductivity

The main results obtained in this paper give insightinto heat transfer in boiling liquids where the heat transfercoefficient is temperature dependent and may be given by apower law The thermal conductivity of some materials suchas gallium nitride (GaN) and Aluminium Nitride (AlN) maybe modeled by power law temperature dependency [29 30]Thus the solutions constructed here give a better comparisonof heat transfer in terms of material used since in many engi-neering applications thermal conductivity is given as a linearfunction of temperature Furthermore a good study in termsof performance and efficiency of fin with different profiles isundertaken These finding could help in the design of fins Itis claimed in [14] that DTM results are more accurate thanthose constructed by Variational Iteration Methods (VIM)and Homotopy Perturbation Methods (HPM) However itwould be risky to use the DTM approximate solutions asbenchmarks for the numerical schemes Nevertheless wehave also shown that DTM converges rapidly in just fifteenterms to the exact solution

Acknowledgments

The authors wish to thank the National Research Foundationof South Africa and the University of the Witwatersrandfor the generous financial support The authors are gratefulto Professor D P Mason and the anonymous reviewers fortheir invaluable comments which improved this paper signif-icantly

References

[1] A D Kraus A Aziz and J Welty Extended Surface Heat Trans-fer Wiley New York NY USA 2001

[2] R JMoitsheki T Hayat andM YMalik ldquoSome exact solutionsof the fin problem with a power law temperature-dependentthermal conductivityrdquo Nonlinear Analysis Real World Applica-tions vol 11 no 5 pp 3287ndash3294 2010

[3] R J Moitsheki ldquoSteady one-dimensional heat flow in a lon-gitudinal triangular and parabolic finrdquo Communications inNonlinear Science and Numerical Simulation vol 16 no 10 pp3971ndash3980 2011

[4] R J Moitsheki ldquoSteady heat transfer through a radial fin withrectangular and hyperbolic profilesrdquo Nonlinear Analysis RealWorld Applications vol 12 no 2 pp 867ndash874 2011

[5] Mo Miansari D D Ganji and Me Miansari ldquoApplication ofHersquos variational iteration method to nonlinear heat transferequationsrdquo Physics Letters A vol 372 no 6 pp 779ndash785 2008

[6] C H Chiu and C K Chen ldquoA decomposition method forsolving the convective longitudinal fins with variable thermalconductivityrdquo International Journal of Heat and Mass Transfervol 45 no 10 pp 2067ndash2075 2002

[7] A Rajabi ldquoHomotopy perturbation method for fin efficiencyof convective straight fins with temperature-dependent thermalconductivityrdquo Physics Letters A vol 364 no 1 pp 33ndash37 2007

[8] G Domairry and M Fazeli ldquoHomotopy analysis method todetermine the fin efficiency of convective straight fins withtemperature-dependent thermal conductivityrdquo Communica-tions in Nonlinear Science and Numerical Simulation vol 14 no2 pp 489ndash499 2009

[9] A Campo andR J Spaulding ldquoCoupling of themethods of suc-cessive approximations and undetermined coefficients for theprediction of the thermal behaviour of uniform circumferentialfinsrdquo Heat and Mass Transfer vol 34 no 6 pp 461ndash468 1999

[10] A Campo and F Rodrfguez ldquoApproximate analytic tempera-ture solution for uniform annular fins by adapting the powerseriesmethodrdquo International Communications inHeat andMassTransfer vol 25 no 6 pp 809ndash818 1998

[11] R Chiba ldquoApplication of differential transformmethod to ther-moelastic problem for annular disks of variable thickness withtemperature-dependent parametersrdquo International Journal ofThermophysics vol 33 pp 363ndash380 2012

[12] S Sadri M R Raveshi and S Amiri ldquoEfficiency analysis ofstraight fin with variable heat transfer coefficient and thermalconductivityrdquo Journal ofMechanical Science andTechnology vol26 no 4 pp 1283ndash1290 2012

[13] M Torabi H Yaghoori and A Aziz ldquoAnalytical solution forconvective-radiative continously moving fin with temperaturedependent thermal conductivityrdquo International Journal of Ther-mophysics vol 33 pp 924ndash941 2012

[14] M Torabi andH Yaghoobi ldquoTwo dominant analytical methodsfor thermal analysis of convective step fin with variable thermalconductivityrdquoThermal Science In press

[15] AMoradi ldquoAnalytical solutions for finwith temperature depen-dant heat transfer coefficientrdquo International Journal of Engineer-ing and Applied Sciences vol 3 no 2 pp 1ndash12 2011

[16] A Moradi and H Ahmadikia ldquoAnalytical solution for differentprofiles of fin with temperature-dependent thermal conductiv-ityrdquoMathematical Problems in Engineering vol 2010 Article ID568263 15 pages 2010

[17] H Yaghoobi and M Torabi ldquoThe application of differentialtransformation method to nonlinear equations arising in heattransferrdquo International Communications in Heat and MassTransfer vol 38 no 6 pp 815ndash820 2011

[18] S Ghafoori M Motevalli M G Nejad F Shakeri D D GanjiandM Jalaal ldquoEfficiency of differential transformation methodfor nonlinear oscillation Comparison with HPM and VIMrdquoCurrent Applied Physics vol 11 no 4 pp 965ndash971 2011

[19] A A Joneidi D D Ganji andM Babaelahi ldquoDifferential trans-formation method to determine fin efficiency of convectivestraight fins with temperature dependent thermal conductivityrdquoInternational Communications in Heat and Mass Transfer vol36 no 7 pp 757ndash762 2009

[20] J K Zhou Differential Transform Method and Its Applicationsfor Electric circuIts Huazhong University Press Wuhan China1986

14 Mathematical Problems in Engineering

[21] S Mukherjee ldquoReply to comment on lsquosolutions of the duffing-van der pol oscillator equation by the differential transformmethodrsquordquo Physica Scripta vol 84 Article ID 037003 2 pages2011

[22] C Bervillier ldquoStatus of the differential transformationmethodrdquoApplied Mathematics and Computation vol 218 no 20 pp10158ndash10170 2012

[23] A Arikoglu and I Ozkol ldquoSolution of fractional differentialequations by using differential transform methodrdquo ChaosSolitons and Fractals vol 34 no 5 pp 1473ndash1481 2007

[24] ZOdibat SMomani andV S Erturk ldquoGeneralized differentialtransform method application to differential equations offractional orderrdquo Applied Mathematics and Computation vol197 no 2 pp 467ndash477 2008

[25] CW Bert ldquoApplication of differential transformmethod to heatconduction in tapered finsrdquo Journal of Heat Transfer vol 124no 1 pp 208ndash209 2002

[26] F Khani and A Aziz ldquoThermal analysis of a longitudinal trape-zoidal fin with temperature-dependent thermal conductivityand heat transfer coefficientrdquo Communications in NonlinearScience and Numerical Simulation vol 15 no 3 pp 590ndash6012010

[27] H C Unal ldquoAn anlytical study of boiling heat transfer from afinrdquo International Journal of Heat and Mass Transfer vol 31 no7 pp 1483ndash1496 1988

[28] M H Chang ldquoA decomposition solution for fins with tempera-ture dependent surface heat fluxrdquo International Journal of Heatand Mass Transfer vol 48 no 9 pp 1819ndash1824 2005

[29] A Jezowski B A Danilchenko M Bockowski et al ldquoThermalconductivity of GaN crystals in 42ndash300 K rangerdquo Solid StateCommunications vol 128 no 2-3 pp 69ndash73 2003

[30] S Vitanov V Palankovski S Maroldt and R Quay ldquoHigh-temperature modeling of AlGaNGaN HEMTsrdquo Solid-StateElectronics vol 54 no 10 pp 1105ndash1112 2010

Submit your manuscripts athttpwwwhindawicom

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Mathematical Problems in Engineering

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Differential EquationsInternational Journal of

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Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

Page 4: Research Article Analytical Solutions for Steady Heat ...downloads.hindawi.com/journals/mpe/2013/273052.pdf · Research Article Analytical Solutions for Steady Heat Transfer in Longitudinal

4 Mathematical Problems in Engineering

where 119888 is a constant Equation (17) is an iterative formula ofconstructing the power series solution as follows

Θ (2) =1198722

119888

2

(20)

Θ (3) = 0 (21)

Θ (4) = minus1198724

119888

24

(22)

Θ (5) = 0 (23)

Θ (6) =191198726

119888

720

(24)

Θ (7) = 0 (25)

Θ (8) = minus559119872

8

119888

40320

(26)

Θ (9) = 0 (27)

Θ (10) =29161119872

10

119888

3628800

(28)

Θ (11) = 0 (29)

Θ (12) = minus2368081119872

12

119888

479001600

(30)

Θ (13) = 0 (31)

Θ (14) =276580459119872

14

119888

87178291200

(32)

Θ (15) = 0

(33)

These terms may be taken as far as desired Substituting(18) to (32) into (13) we obtain the following analyticalsolution

120579 (119909) = 119888 +1198722

119888

21199092

minus1198724

119888

241199094

+191198726

119888

7201199096

minus559119872

8

119888

403201199098

+29161119872

10

119888

362880011990910

minus2368081119872

12

119888

47900160011990912

+276580459119872

14

119888

8717829120011990914

+ sdot sdot sdot

(34)

To obtain the value of 119888 we substitute the boundarycondition (5) into (34) at the point 119909 = 1 Thus we have

120579 (1) = 119888 +1198722

119888

2minus1198724

119888

24+191198726

119888

720minus559119872

8

119888

40320

+29161119872

10

119888

3628800minus2368081119872

12

119888

479001600

+276580459119872

14

119888

87178291200+ sdot sdot sdot = 1

(35)

Table 2 Results of the DTM and exact solutions for 119899 = 1119872 = 07

119909 DTM Exact Error0 0808093014 080809644 000000342601 0810072036 081007547 000000343402 0815999549 081600301 000000345903 0825847770 082585127 000000350004 0839573173 083957673 000000355905 0857120287 085712392 000000363306 0878426299 087843002 000000372207 0903426003 090342982 000000381408 0932056648 093206047 000000382009 0964262558 096426582 000000326310 1000000000 100000000 0000000000

Table 3 Results of theDTMand Exact Solutions for 119899 = 2119872 = 05

119909 DTM Exact Error0 0894109126 0894109793 000000066501 0895226066 0895226732 000000066602 0898568581 0898569249 000000066803 0904112232 0904112905 000000067204 0911817796 0911818474 000000067805 0921633352 0921634038 000000068506 0933496900 0933497594 000000069407 0947339244 0947339946 000000070208 0963086933 0963087627 000000069409 0980665073 0980665659 000000058610 1000000000 1000000000 0000000000

We omit presenting the tedious process of finding 119888 Howeverone may obtain the expression for 120579(119909) upon substituting theobtained value of 119888 into (34)

42 Case 119899=2 Following a similar approach given inSection 41 and given 119899 = 2 one obtains the analyticalsolution

120579 (119909) = 119888 +1198722

119888

21199092

minus1198724

119888

81199094

+231198726

119888

2401199096

minus1069119872

8

119888

134401199098

+9643119872

10

119888

13440011990910

minus1211729119872

12

119888

1774080011990912

+217994167119872

14

119888

322882560011990914

+ sdot sdot sdot

(36)

The constant 119888may be obtained using the boundary conditionat the fin base The comparison of the DTM and the exactsolutions are reflected in Tables 2 and 3 for different valuesof 119899 Furthermore the comparison of the exact and theanalytical solutions is depicted in Figures 2(a) and 2(b)

Mathematical Problems in Engineering 5

08

082

084

086

088

09

092

094

096

098

1

120579

0 02 04 06 08 1

119909

DTMExact

(a)

086

088

09

092

094

096

098

1

120579

0 02 04 06 08 1

119909

DTMExact

(b)

Figure 2 Comparison of analytical and exact solutions In (a) 119899 = 1119872 = 07 and (b) 119899 = 2119872 = 05

5 Analytical Solutions

It is well known that exact solutions forODEs such as (4) existonly when thermal conductivity and the term containingheat transfer coefficient are connected by differentiation(or simply if the ODE such as (4) is linearizable) [4] Inthis section we determine the analytical solutions for thenonlinearizable (4) firstlywhen thermal conductivity is givenby the power law and secondly as a linear function oftemperature In both cases and throughout this paper theheat transfer coefficient is assumed to be a power law functionof temperature These assumptions of the thermal propertiesare physical realistic We have noticed that DTM runs intodifficulty when the exponent of the power law of the thermalconductivity is given by fractional values and also when thefunction 119891(119909) is given in terms of fractional power law Onemay followMoradi and Ahmadikia [16] by introducing a newvariable to deal with fractional powers of 119891(119909) and on theother hand it is possible to remove the fractional exponent ofthe heat transfer coefficient by fundamental laws of exponentand binomial expansion

Proposition 1 A nonlinear ODE such as (4) may admit theDTM solution if 119891(119909) is a constant or exponential functionHowever if 119891(119909) is a power law then such an equation admitsa DTM solution if the product

119891 (119909) sdot 1198911015840

(119909) = 120572 (37)

holds Here 120572 is a real constant

Proof Introducing the new variable 120585 = 119891(119909) it follows fromchain rule that (4) becomes

120572119889120585

119889119909

119889

119889120585[119896 (120579)

119889120579

119889120585] minus119872

2

120579119899+1

= 0 (38)

The most general solution of condition (37) is

119891 (119909) = (2120572119909 + 120574)12

(39)

where 120574 is an integration constant

Example 2 (a) If 119891(119909) = 119909120590 then 120585 = 119909120590 transforms (4) into

119889

119889120585[119896 (120579)

119889120579

119889120585] minus 4119872

2

120585120579119899+1

= 0 (40)

only if 120590 = 12This example implies that the DTM may only be appli-

cable to problems describing heat transfer in fins withconvex parabolic profile In the next subsections we presentanalytical solutions for (4) with various functions 119891(119909) and119896(120579)

51 The Exponential Profile and Power Law Thermal Con-ductivity In this section we present solutions for equationdescribing heat transfer in a fin with exponential profile andpower law thermal conductivity and heat transfer coefficientThat is given (4) with 119891(119909) = e120590119909 and both heat transfercoefficient and thermal conductivity being power law func-tions of temperature we construct analytical solutions In ouranalysis we consider 119899 = 2 and 3 indicating the fin subject tonucleate boiling and radiation into free space at zero absolutetemperature respectively Firstly given 119899 = 3 and 119898 = 2

6 Mathematical Problems in Engineering

and applying the DTM one obtains the following recurrencerevelation

120581

sum

119894=0

120581minus119894

sum

119897=0

120581minus119894minus119897

sum

119901=0

[120590119901

119901Θ (119897) Θ (119894) (120581 minus 119894 minus 119897 minus 119901 + 1)

times (120581 minus 119894 minus 119897 minus 119901 + 2)Θ (120581 minus 119894 minus 119897 minus 119901 + 2)

+ 2120590119901

119901Θ (119897) (119894 + 1)Θ (119894 + 1)

times (120581 minus 119894 minus 119897 minus 119901 + 1)Θ (120581 minus 119894 minus 119897 minus 119901 + 1)

+ 2120590120590119901

119901Θ (119897) Θ (119894) (120581 minus 119894 minus 119897 minus 119901 + 1)

times Θ (120581 minus 119894 minus 119897 minus 119901 + 1)

minus1198722

Θ(119901)Θ (119897) Θ (119894) Θ (120581 minus 119894 minus 119897 minus 119901) ] = 0

(41)

One may recall the transformed prescribed boundaryconditions (18) and (19) Equation (41) is an iterative formulaof constructing the power series solution as follows

Θ (2) =1198722

1198882

2

Θ (3) = minus1205901198722

1198882

3

Θ (4) =

1198722

1198882

(31205902

minus 21198722

119888)

24

Θ (5) = minus

1198722

1198882

(1205903

minus 41205901198722

119888)

30

Θ (6) =

1198722

1198882

(51205904

minus 781205902

1198722

119888 + 581198724

1198882

)

720

(42)

The pervious process is continuous and onemay consideras many terms as desired (but bearing in mind that DTMconverges quite fast) Substituting (18) to (19) and (42) into(13) we obtain the following closed form of the solution

120579 (119909) = 119888 +1198722

1198882

21199092

minus1205901198722

1198882

31199093

+

1198722

1198882

(31205902

minus 21198722

119888)

241199094

minus

1198722

1198882

(1205903

minus 41205901198722

119888)

301199095

+

1198722

1198882

(51205904

minus 781205902

1198722

119888 + 581198724

1198882

)

7201199096

+ sdot sdot sdot

(43)

To obtain the value of 119888 we substitute the boundarycondition (5) into (43) at the point 119909 = 1 That is

120579 (1) = 119888 +1198722

1198882

2minus1205901198722

1198882

3+

1198722

1198882

(31205902

minus 21198722

119888)

24

minus

1198722

1198882

(1205903

minus 41205901198722

119888)

30

+

1198722

1198882

(51205904

minus 781205902

1198722

119888 + 581198724

1198882

)

720+ sdot sdot sdot = 1

(44)

Substituting this value of 119888 into (43) one finds theexpression for 120579(119909) On the other hand given (119899119898) = (2 3)one obtains the solution

120579 (119909) = 119888 +1198722

21199092

minus1205901198722

31199093

+

1198722

(1205902

119888 minus 21198722

)

81198881199094

minus

1205901198722

(21205902

119888 minus 211198722

)

601198881199095

+

1198722

(1801198724

minus 1861205902

1198722

119888 + 51205904

1198882

)

7201199096

+ sdot sdot sdot

(45)

Here the constant 119888 maybe obtained by evaluating theboundary condition 120579(1) = 1 The solutions (43) and (45) aredepicted in Figures 3(b) and 3(a) respectively

52 The Rectangular Profile and Power Law Thermal Con-ductivity In this section we provide a detailed constructionof analytical solutions for the heat transfer in a longitudinalrectangular finwith a power law thermal conductivity that iswe consider (4) with 119891(119909) = 1 and 119896(120579) = 120579119898 The analyticalsolutions are given in the following expressions

(a) Case (119899119898) = (2 3)

120579 (119909) = 119888 +1198722

21199092

minus1198724

41198881199094

+1198726

411988821199096

minus331198728

12211988831199098

+127119872

10

336119888411990910

+ sdot sdot sdot

(46)

(b) Case (119899119898) = (3 2)

120579 (119909) = 119888 +1198882

1198722

21199092

minus1198883

1198724

121199094

+291198884

1198726

3601199096

minus3071198885

1198728

50401199098

+23483119888

6

11987210

45360011990910

+ sdot sdot sdot

(47)

The constant 119888 is obtained by solving the appropriate 120579(119909)at the fin base boundary condition The analytical solutionsin (46) and (47) are depicted in Figures 4(a) and 4(b)respectively

Mathematical Problems in Engineering 7

0 02 04 06 08 1094

095

096

097

098

099

1

119909

120579

119872 = 02

119872 = 03

119872 = 04

119872 = 05

(a)

0 02 04 06 08 1093

094

095

096

097

098

099

1

119909

120579

119872 = 02

119872 = 03

119872 = 04

119872 = 05

(b)

Figure 3 Temperature profile in a longitudinal fin with exponential profile and power law thermal conductivity In (a) the exponents are(119899119898) = (2 3) and (b) (119899 119898) = (3 2)

119872 = 025

119872 = 04

119872 = 05

119872 = 065

082

084

086

088

09

092

094

096

098

1

120579

0 02 04 06 08 1

119909

(a)

084

086

088

09

092

094

096

098

1

120579

0 02 04 06 08 1

119909

119872 = 025

119872 = 04

119872 = 05

119872 = 065

(b)

Figure 4 Temperature profile in a longitudinal fin with rectangular profile and power law thermal conductivity In (a) the exponents are(119899119898) = (2 3) and (b) (119899 119898) = (3 2)

53 The Convex Parabolic Profile and Power Law ThermalConductivity In this section we present solutions for theequation describing the heat transfer in a fin with con-vex parabolic profile and power law thermal conductivityEquation (40) is considered Here we consider the values(119899119898) = (2 3) (3 2) The final analytical solution is givenby

(a) Case (119899119898) = (2 3)

120579 (119909) = 119888 +21198722

311990932

minus21198724

51198881199093

+231198726

45119888211990992

minus1909119872

8

247511988831199096

+329222119872

10

2598751198884119909152

+ sdot sdot sdot

(48)

8 Mathematical Problems in Engineering

075

08

085

09

095

1

119872 = 025

119872 = 04

119872 = 05

119872 = 065

120579

0 02 04 06 08 1119909

(a)

119872 = 025

119872 = 04

119872 = 05

119872 = 065

0 02 04 06 08 1119909

082

084

086

088

09

092

094

096

098

120579

1

(b)

Figure 5 Temperature profile in a longitudinal fin with convex parabolic profile and power law thermal conductivity In (a) the exponentsare (119899119898) = (2 3) and (b) (119899119898) = (3 2)

(b) Case (119899119898) = (3 2)

120579 (119909) = 119888 +21198882

1198722

311990932

minus41198883

1198724

451199093

+41198884

1198726

2711990992

minus9921198885

1198728

74251199096

+30064119888

6

11987210

212625119909152

+ sdot sdot sdot

(49)

The constant 119888 is obtained by evaluating the appropriate 120579(119909)at the fin base boundary condition The solutions in (48) and(49) are depicted in Figures 5(a) and 5(b) respectively

54 The Rectangular Profile and LinearThermal ConductivityIn this section we present solutions for the equation repre-senting the heat transfer in a fin with rectangular profile andthe thermal conductivity depending linearly on temperatureThat is we consider (4) with 119891(119909) = 1 and 119896(120579) = 1 + 120573120579 Theanalytical solutions for this problem for different values of 119899are given by

(a) Case 119899 = 0

120579 (119909)

= 119888 +

1198722

119888

2 (1 + 120573119888)

1199092

minus

1198724

119888 (minus1 + 2120573119888)

24(1 + 120573119888)31199094

+

1198726

119888 (1 minus 16120573119888 + 281205732

1198882

)

720(1 + 120573119888)51199096

minus

1198728

119888 (minus1 + 78120573119888 minus 6001205732

1198882

+ 8961205733

1198883

)

40320(1 + 120573119888)7

1199098

+

11987210

119888 (1 minus 332120573119888 + 78121205732

1198882

minus 398961205733

1198883

+ 511841205734

1198884

)

3628800(1 + 120573119888)9

11990910

+ sdot sdot sdot

(50)

(b) Case 119899 = 1

120579 (119909)

= 119888 +

11987221198882

2 (1 + 120573119888)

1199092minus

11987241198883(minus1 + 2120573119888)

24(1 + 120573119888)31199094

+

11987261198884(10 minus 16120573119888 + 19120573

21198882)

720(1 + 120573119888)5

1199096

minus

11987281198885(minus80 + 342120573119888 minus 594120573

21198882+ 559120573

31198883)

40320(1 + 120573119888)7

1199098

+

119872101198886(1000 minus 7820120573119888 + 24336120573

21198882minus 36908120573

31198883+ 29161120573

41198884)

3628800(1 + 120573119888)9

11990910

+ sdot sdot sdot

(51)

(c) Case 119899 = 2

120579 (119909)

= 119888 +

1198722

1198883

2 (1 + 120573119888)

1199092

+

1198724

1198885

8(1 + 120573119888)31199094

+

1198726

1198887

(3 + 21205732

1198882

)

80(1 + 120573119888)51199096

+

1198728

1198889

(49 minus 20120573119888 + 661205732

1198882

minus 401205733

1198883

)

4480(1 + 120573119888)7

1199098

+

11987210

11988811

(427 minus 440120573119888 + 11161205732

1198882

minus 10201205733

1198883

+ 6721205734

1198884

)

134400(1 + 120573119888)9

11990910

+ sdot sdot sdot

(52)

Mathematical Problems in Engineering 9

(d) Case 119899 = 3120579 (119909)

= 119888 +

11987221198884

2 (1 + 120573119888)

1199092+

11987241198887(4 + 120573119888)

24(1 + 120573119888)31199094

+

119872611988810(52 + 32120573119888 + 25120573

21198882)

720(1 + 120573119888)5

1199096

+

119872811988813(1288 + 1020120573119888 + 1212120573

21198882minus 95120573

31198883)

40320(1 + 120573119888)7

1199098

+

1198721011988816(52024+45688120573119888+77184120573

21198882minus680120573

31198883+ 15025120573

41198884)

3628800(1 + 120573119888)9

11990910

+

(53)

The constant 119888may be obtained from the boundary conditionon the appropriate solution The solutions in (50) (51) (52)and (53) are depicted in Figure 6

55The Convex Parabolic Profile and LinearThermal Conduc-tivity In this section we present solutions for the equationdescribing the heat transfer in a fin with convex parabolicprofile and the thermal conductivity depending linearly ontemperature That is we consider (40) with 119896(120579) = 1 + 120573120579The analytical solution for this problem for different values of119899 is given by

(a) Case 119899 = 0

120579 (119909)

= 119888 +21198722

119888

3 (1 + 120573119888)11990932

minus21198724

119888 (minus2 + 3120573119888)

45(1 + 120573119888)3

1199093

+

1198726

119888 (2 minus 25120573119888 + 331205732

1198882

)

405(1 + 120573119888)5

11990992

minus

1198728

119888 (minus10 + 599120573119888 minus 35821205732

1198882

+ 40591205733

1198883

)

66825(1 + 120573119888)7

1199096

+ sdot sdot sdot

(54)

(b) Case 119899 = 1

120579 (119909)

= 119888 +21198722

1198882

3 (1 + 120573119888)11990932

minus21198724

1198883

(minus4 + 120573119888)

45(1 + 120573119888)3

1199093

+

21198726

1198884

(9 minus 11120573119888 + 101205732

1198882

)

405(1 + 120573119888)5

11990992

minus

21198728

1198885

(minus330 + 1118120573119888 minus 15841205732

1198882

+ 10931205733

1198883

)

66825(1 + 120573119888)7

1199096

+ sdot sdot sdot

(55)

119899 = 0

119899 = 1

119899 = 2

119899 = 3

1

095

09

085

08

075

07

065

06

055

0 02 04 06 08 1

119909

120579

Figure 6 Temperature profile in a longitudinal rectangular fin withlinear thermal conductivity and varying values of 119899 Here 120573 = 05and119872 = 15 are fixed

(c) Case 119899 = 2

120579 (119909)

= 119888 +21198722

1198883

3 (1 + 120573119888)11990932

+21198724

1198885

(6 + 120573119888)

45(1 + 120573119888)31199093

+

1198726

1198887

(16 + 3120573119888 + 71205732

1198882

)

135(1 + 120573119888)5

11990992

+

1198728

1198889

(1160 minus 107120573119888 + 11161205732

1198882

minus 3671205733

1198883

)

22275(1 + 120573119888)7

1199096

+ sdot sdot sdot

(56)

(d) Case 119899 = 3

120579 (119909)

= 119888 +21198722

1198884

3 (1 + 120573119888)11990932

+21198724

1198887

(8 + 3120573119888)

45(1 + 120573119888)31199093

+

41198726

11988810

(23 + 17120573119888 + 91205732

1198882

)

405(1 + 120573119888)5

11990992

+

21198728

11988813

(5000+4868120573119888 + 43561205732

1198882

+ 3631205733

1198883

)

66825(1 + 120573119888)7

1199096

+ sdot sdot sdot

(57)

The constant 119888 may be obtained from the boundarycondition on the appropriate 120579(119909)The solutions in (54) (55)(56) and (57) are depicted in Figure 7

10 Mathematical Problems in Engineering

1

09

08

07

06

05

120579

119899 = 0

119899 = 1

119899 = 2

119899 = 3

0 02 04 06 08 1

119909

Figure 7 Temperature profile in a longitudinal convex parabolic finwith linear thermal conductivity and varying values of 119899 Here 120573 =05 and119872 = 15 are fixed

56 The Exponential Profile and LinearThermal ConductivityIn this section we present solutions for equation heat transferin a fin with convex parabolic profile and the thermalconductivity depending linearly on temperature That is weconsider (4) with 119896(120579) = 1 + 120573120579 and 119891(119909) = e120590119909 Here 120590is a constant The analytical solutions for this problem fordifferent values of 119899 are given by

(a) Case 119899 = 0

120579 (119909)

= 119888 +

1198722

119888

2 (1 + 120573119888)

1199092

minus

1205901198722

119888 (2 + 120573119888 + 1205732

119888)

6(1 + 120573119888)21199093

+

1198722

119888 (1198722

(1 minus 21205732

119888) + 1205902

(3 +3120573119888+1205733

1198882

+1205734

1198882

+1205732

119888 (3 + 119888)))

24(1 + 120573119888)4

1199094

+ sdot sdot sdot

(58)

(b) Case 119899 = 1

120579 (119909)

= 119888 +

11987221198882

2 (1 + 120573119888)

1199092minus

12059011987221198882(2 + 120573119888 + 120573

2119888)

6(1 + 120573119888)21199093

+

11987221198882(1198722119888 (2+120573119888minus2120573

2119888)+1205902(3+3120573119888+120573

31198882+12057341198882+1205732119888 (3+119888)))

24(1 + 120573119888)3

1199094

+ sdot sdot sdot

(59)

119899 = 0

119899 = 1

119899 = 2

119899 = 3

0 02 04 06 08 1

119909

1

095

09

085

08

075

07

120579

Figure 8 Temperature profile in a longitudinal fin of exponentialprofile with linear thermal conductivity and varying values of 119899Here 120573 = 05 and119872 = 15 are fixed

(c) Case 119899 = 2120579 (119909)

= 119888 +

11987221198883

2 (1 + 120573119888)

1199092minus

12059011987221198883(2 + 120573119888 + 120573

2119888)

6(1 + 120573119888)21199093

+

11987221198883(11987221198882(3+2120573119888minus2120573

2119888)+1205902(3+3120573119888+120573

31198882+12057341198882+1205732119888 (3+119888)))

24(1 + 120573119888)3

times1199094+ sdot sdot sdot

(60)

(d) Case 119899 = 3120579 (119909)

= 119888 +

11987221198884

2 (1 + 120573119888)

1199092minus

12059011987221198884(2 + 120573119888 + 120573

2119888)

6(1 + 120573119888)21199093

+

11987221198884(11987221198883(4+3120573119888minus2120573

2119888)+1205902(3+3120573119888+120573

31198882+12057341198882+1205732119888 (3+119888)))

24(1 + 120573119888)3

times1199094+ sdot sdot sdot

(61)

The constant 119888 may be obtained from the boundarycondition on the appropriate 120579(119909)The solutions in (58) (59)(60) and (61) are depicted in Figure 8

6 Fin Efficiency and Heat Flux

61 Fin Efficiency Theheat transfer rate from a fin is given byNewtonrsquos second law of cooling

119876 = int

119871

0

119875119867 (119879) (119879 minus 119879119886) 119889119909 (62)

Mathematical Problems in Engineering 11

Fin efficiency is defined as the ratio of the fin heat transferrate to the rate that would be if the entire fin were at the basetemperature and is given by (see eg [1])

120578=

119876

119876ideal=

int119871

0

119875119867 (119879) (119879 minus 119879119886) 119889119883

119875ℎ119887119871 (119879119887minus 119879119886)

(63)

In dimensionless variables we have

120578 = int

1

0

120579119899+1

119889119909 (64)

We consider the solutions (50) (51) (52) and (53) and depictthe fin efficiency (63) in Figure 9

62 Heat Flux The fin base heat flux is given by the Fourierrsquoslaw

119902119887= 119860119888119870 (119879)

119889119879

119889119909 (65)

The total heat flux of the fin is given by [1]

119902 =119902119887

119860119888119867(119879) (119879

119887minus 119879119886) (66)

Introducing the dimensionless variable as described inSection 2 implies

119902 =1

119861119894

119896 (120579)

ℎ (120579)

119889120579

119889119909 (67)

where the dimensionless parameter 119861119894 = ℎ119887119871119896119886is the Biot

number We consider a number of cases for the thermalconductivity and the heat transfer coefficient

621 Linear Thermal Conductivity and Power Law HeatTransfer Coefficient In this case (67) becomes

119902 =1

119861119894(1 + 120573120579) 120579

minus119899119889120579

119889119909 (68)

The heat flux in (68) at the base of the fin is plotted inFigure 10

622 Power LawThermal Conductivity and Heat TransferCoefficient In this case (67) becomes

119902 =1

119861119894120579119898minus119899

119889120579

119889119909 (69)

Not surprisingly heat flux in one-dimensional fins is highergiven values 119861119894 ≪ 1 The heat flux in (69) is plotted inFigures 11(a) 11(b) and 11(c)

7 Some Discussions

The DTM has resulted in some interesting observations andstudy We have observed in Figures 2(a) and 2(b) an excellentagreement between the analytical solutions generated byDTM and the exact solution obtained in [2] In particular we

119872

1

095

09

085

08

075

07

065

06

055

120578

2151050

119899 = 0

119899 = 1

119899 = 2

119899 = 3

Figure 9 Fin efficiency of a longitudinal rectangular fin Here 120573 =075

119872

25

20

15

10

5

0

119902

2151050

119899 = 0

119899 = 1

119899 = 2

119899 = 3

Figure 10 Base heat flux in a longitudinal rectangular finwith linearthermal conductivity Here 120573 = 01

considered a fin problem in which both thermal conductivityand heat transfer coefficient are given by the same power lawFurthermore we notice from Table 2 that an absolute errorof approximately 35119890 minus 005 is produced by DTM of order119874(15) In Table 3 an absolute error of approximately 65119890 minus006 is produced for the same order This confirms that theDTM converges faster and can provide accurate results witha minimum computation As such a tremendous confidencein the DTM in terms of the accuracy and effectiveness wasbuilt and thus we used this method to solve other problemsfor which exact solutions are harder to construct

In Figures 3(a) 3(b) 4(a) 4(b) 5(a) and 5(b) weobserve that the fin temperature increaseswith the decreasing

12 Mathematical Problems in Engineering

values of the thermogeometric fin parameter Here thevalues of the exponents are fixed Also we observe that fintemperature is higher when 119899 minus 119898 gt 0 that is when heattransfer coefficient is higher than the thermal conductivityWe observe in Figures 6 7 and 8 that the fin temperatureincreases with the increasing values of 119899 Furthermoreit appears that the fin with exponential profile performsthe least in transferring the heat from the base since thetemperature in such a fin is much higher than that ofthe rectangular and the convex parabolic profiles In otherwords heat dissipation to the fluid surrounding the extendedsurface is much faster in longitudinal fins of rectangular andconvex parabolic profiles In Figure 9 fin efficiency decreaseswith increasing thermogeometric fin parameter Also finefficiency increases with increasing values of 119899 It is easyto show that the thermogeometric fin parameter is directlyproportional to the aspect ratio (extension factor) with squareroot of the Biot number being the proportionality constantAs such shorter fins are more efficient than longer onesElse the increased Biot number results in less efficient finwhenever the space is confined that is where the length ofthe fin cannot be increased Figure 10 depicts the heat fluxat the fin base The amount of heat energy dissipated fromthe fin base is of immense interest in engineering [28] Weobserve in Figure 10 that the base heat flux increases withthe thermogeometric fin parameter for considered values ofthe exponent 119899 (see also [28]) Figures 11(a) 11(b) and 11(c)display the heat flux across the fin length We note that theheat flux across the fin length increases with increasing valuesof the thermogeometric fin parameter

8 Concluding Remarks

In this study we have successfully applied the DTM to highlynonlinear problems arising in heat transfer through longitu-dinal fins of various profiles Both thermal conductivity andheat transfer coefficient are given as functions of temperatureThe DTM agreed well with exact solutions when the thermalconductivity and heat transfer coefficient are given by thesame power law A rapid convergence to the exact solutionwas observed Following the confidence in DTM built by theresults mentioned we then solved various exciting problemsThe exotic results have been shown in tables and figures listedin this paper

The results obtained in this paper are significant improve-ments on the known results In particular both the heattransfer coefficient and the thermal conductivity are allowedto be given by the power law functions of temperature andalsowe considered a number of finprofilesWenote that exactsolutions are difficult if not impossible to construct when theexponents of these properties are distinct

Perhaps the notable advantage of the DTM is the general-ization of the Taylor method to problems involving unusualderivative procedures such as fractional fuzzy or q-derivative[22] Some generalizations have been made by Odibat et al[24] and they referred to their new method as the General-ized Differential Transform Method (GDTM) This showedgreat improvement compared to the Fractional Differential

0 02 04 06 08 1119909

45

40

35

30

25

20

15

10

5

0

minus5

119902

(a)

0 02 04 06 08 1119909

30

25

20

15

10

5

0

119902

(b)

35

30

25

20

15

10

5

0

119902

0 02 04 06 08 1119909

119872 = 025

119872 = 04

119872 = 05

119872 = 065

(c)

Figure 11 Heat flux across a longitudinal rectangular fin with linearthermal conductivity (a) 119899minus119898 lt 0 (b) 119899minus119898 gt 0 and (c) 119899minus119898 = 0Here 119861119894 = 001

Mathematical Problems in Engineering 13

Transform Method (FDTM) introduced by Arikoglu andOzkol [23]

We have shown with the help of an example that DTMmay only be applied to fin problems involving heat transferthrough fins with convex parabolic profile Note that given anODE such as (4) with a power law heat transfer coefficient ofa fractional exponent then one can easily remove the fractionby basic exponential rules and employment of the Binomialexpansion However using the DTM one runs into difficultyif the power law thermal conductivity in the same equation isgiven by the fractional exponent We do not know whetherthese observations call for the ldquomodifiedrdquo DTM to solveproblems arising in heat flow through fins with other profilessuch as longitudinal triangular and concave parabolic andalso with fractional power law thermal conductivity

The main results obtained in this paper give insightinto heat transfer in boiling liquids where the heat transfercoefficient is temperature dependent and may be given by apower law The thermal conductivity of some materials suchas gallium nitride (GaN) and Aluminium Nitride (AlN) maybe modeled by power law temperature dependency [29 30]Thus the solutions constructed here give a better comparisonof heat transfer in terms of material used since in many engi-neering applications thermal conductivity is given as a linearfunction of temperature Furthermore a good study in termsof performance and efficiency of fin with different profiles isundertaken These finding could help in the design of fins Itis claimed in [14] that DTM results are more accurate thanthose constructed by Variational Iteration Methods (VIM)and Homotopy Perturbation Methods (HPM) However itwould be risky to use the DTM approximate solutions asbenchmarks for the numerical schemes Nevertheless wehave also shown that DTM converges rapidly in just fifteenterms to the exact solution

Acknowledgments

The authors wish to thank the National Research Foundationof South Africa and the University of the Witwatersrandfor the generous financial support The authors are gratefulto Professor D P Mason and the anonymous reviewers fortheir invaluable comments which improved this paper signif-icantly

References

[1] A D Kraus A Aziz and J Welty Extended Surface Heat Trans-fer Wiley New York NY USA 2001

[2] R JMoitsheki T Hayat andM YMalik ldquoSome exact solutionsof the fin problem with a power law temperature-dependentthermal conductivityrdquo Nonlinear Analysis Real World Applica-tions vol 11 no 5 pp 3287ndash3294 2010

[3] R J Moitsheki ldquoSteady one-dimensional heat flow in a lon-gitudinal triangular and parabolic finrdquo Communications inNonlinear Science and Numerical Simulation vol 16 no 10 pp3971ndash3980 2011

[4] R J Moitsheki ldquoSteady heat transfer through a radial fin withrectangular and hyperbolic profilesrdquo Nonlinear Analysis RealWorld Applications vol 12 no 2 pp 867ndash874 2011

[5] Mo Miansari D D Ganji and Me Miansari ldquoApplication ofHersquos variational iteration method to nonlinear heat transferequationsrdquo Physics Letters A vol 372 no 6 pp 779ndash785 2008

[6] C H Chiu and C K Chen ldquoA decomposition method forsolving the convective longitudinal fins with variable thermalconductivityrdquo International Journal of Heat and Mass Transfervol 45 no 10 pp 2067ndash2075 2002

[7] A Rajabi ldquoHomotopy perturbation method for fin efficiencyof convective straight fins with temperature-dependent thermalconductivityrdquo Physics Letters A vol 364 no 1 pp 33ndash37 2007

[8] G Domairry and M Fazeli ldquoHomotopy analysis method todetermine the fin efficiency of convective straight fins withtemperature-dependent thermal conductivityrdquo Communica-tions in Nonlinear Science and Numerical Simulation vol 14 no2 pp 489ndash499 2009

[9] A Campo andR J Spaulding ldquoCoupling of themethods of suc-cessive approximations and undetermined coefficients for theprediction of the thermal behaviour of uniform circumferentialfinsrdquo Heat and Mass Transfer vol 34 no 6 pp 461ndash468 1999

[10] A Campo and F Rodrfguez ldquoApproximate analytic tempera-ture solution for uniform annular fins by adapting the powerseriesmethodrdquo International Communications inHeat andMassTransfer vol 25 no 6 pp 809ndash818 1998

[11] R Chiba ldquoApplication of differential transformmethod to ther-moelastic problem for annular disks of variable thickness withtemperature-dependent parametersrdquo International Journal ofThermophysics vol 33 pp 363ndash380 2012

[12] S Sadri M R Raveshi and S Amiri ldquoEfficiency analysis ofstraight fin with variable heat transfer coefficient and thermalconductivityrdquo Journal ofMechanical Science andTechnology vol26 no 4 pp 1283ndash1290 2012

[13] M Torabi H Yaghoori and A Aziz ldquoAnalytical solution forconvective-radiative continously moving fin with temperaturedependent thermal conductivityrdquo International Journal of Ther-mophysics vol 33 pp 924ndash941 2012

[14] M Torabi andH Yaghoobi ldquoTwo dominant analytical methodsfor thermal analysis of convective step fin with variable thermalconductivityrdquoThermal Science In press

[15] AMoradi ldquoAnalytical solutions for finwith temperature depen-dant heat transfer coefficientrdquo International Journal of Engineer-ing and Applied Sciences vol 3 no 2 pp 1ndash12 2011

[16] A Moradi and H Ahmadikia ldquoAnalytical solution for differentprofiles of fin with temperature-dependent thermal conductiv-ityrdquoMathematical Problems in Engineering vol 2010 Article ID568263 15 pages 2010

[17] H Yaghoobi and M Torabi ldquoThe application of differentialtransformation method to nonlinear equations arising in heattransferrdquo International Communications in Heat and MassTransfer vol 38 no 6 pp 815ndash820 2011

[18] S Ghafoori M Motevalli M G Nejad F Shakeri D D GanjiandM Jalaal ldquoEfficiency of differential transformation methodfor nonlinear oscillation Comparison with HPM and VIMrdquoCurrent Applied Physics vol 11 no 4 pp 965ndash971 2011

[19] A A Joneidi D D Ganji andM Babaelahi ldquoDifferential trans-formation method to determine fin efficiency of convectivestraight fins with temperature dependent thermal conductivityrdquoInternational Communications in Heat and Mass Transfer vol36 no 7 pp 757ndash762 2009

[20] J K Zhou Differential Transform Method and Its Applicationsfor Electric circuIts Huazhong University Press Wuhan China1986

14 Mathematical Problems in Engineering

[21] S Mukherjee ldquoReply to comment on lsquosolutions of the duffing-van der pol oscillator equation by the differential transformmethodrsquordquo Physica Scripta vol 84 Article ID 037003 2 pages2011

[22] C Bervillier ldquoStatus of the differential transformationmethodrdquoApplied Mathematics and Computation vol 218 no 20 pp10158ndash10170 2012

[23] A Arikoglu and I Ozkol ldquoSolution of fractional differentialequations by using differential transform methodrdquo ChaosSolitons and Fractals vol 34 no 5 pp 1473ndash1481 2007

[24] ZOdibat SMomani andV S Erturk ldquoGeneralized differentialtransform method application to differential equations offractional orderrdquo Applied Mathematics and Computation vol197 no 2 pp 467ndash477 2008

[25] CW Bert ldquoApplication of differential transformmethod to heatconduction in tapered finsrdquo Journal of Heat Transfer vol 124no 1 pp 208ndash209 2002

[26] F Khani and A Aziz ldquoThermal analysis of a longitudinal trape-zoidal fin with temperature-dependent thermal conductivityand heat transfer coefficientrdquo Communications in NonlinearScience and Numerical Simulation vol 15 no 3 pp 590ndash6012010

[27] H C Unal ldquoAn anlytical study of boiling heat transfer from afinrdquo International Journal of Heat and Mass Transfer vol 31 no7 pp 1483ndash1496 1988

[28] M H Chang ldquoA decomposition solution for fins with tempera-ture dependent surface heat fluxrdquo International Journal of Heatand Mass Transfer vol 48 no 9 pp 1819ndash1824 2005

[29] A Jezowski B A Danilchenko M Bockowski et al ldquoThermalconductivity of GaN crystals in 42ndash300 K rangerdquo Solid StateCommunications vol 128 no 2-3 pp 69ndash73 2003

[30] S Vitanov V Palankovski S Maroldt and R Quay ldquoHigh-temperature modeling of AlGaNGaN HEMTsrdquo Solid-StateElectronics vol 54 no 10 pp 1105ndash1112 2010

Submit your manuscripts athttpwwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

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Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical Problems in Engineering

Hindawi Publishing Corporationhttpwwwhindawicom

Differential EquationsInternational Journal of

Volume 2014

Applied MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

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Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

OptimizationJournal of

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CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of

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Journal of

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Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

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The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014

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Algebra

Discrete Dynamics in Nature and Society

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Decision SciencesAdvances in

Discrete MathematicsJournal of

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Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

Page 5: Research Article Analytical Solutions for Steady Heat ...downloads.hindawi.com/journals/mpe/2013/273052.pdf · Research Article Analytical Solutions for Steady Heat Transfer in Longitudinal

Mathematical Problems in Engineering 5

08

082

084

086

088

09

092

094

096

098

1

120579

0 02 04 06 08 1

119909

DTMExact

(a)

086

088

09

092

094

096

098

1

120579

0 02 04 06 08 1

119909

DTMExact

(b)

Figure 2 Comparison of analytical and exact solutions In (a) 119899 = 1119872 = 07 and (b) 119899 = 2119872 = 05

5 Analytical Solutions

It is well known that exact solutions forODEs such as (4) existonly when thermal conductivity and the term containingheat transfer coefficient are connected by differentiation(or simply if the ODE such as (4) is linearizable) [4] Inthis section we determine the analytical solutions for thenonlinearizable (4) firstlywhen thermal conductivity is givenby the power law and secondly as a linear function oftemperature In both cases and throughout this paper theheat transfer coefficient is assumed to be a power law functionof temperature These assumptions of the thermal propertiesare physical realistic We have noticed that DTM runs intodifficulty when the exponent of the power law of the thermalconductivity is given by fractional values and also when thefunction 119891(119909) is given in terms of fractional power law Onemay followMoradi and Ahmadikia [16] by introducing a newvariable to deal with fractional powers of 119891(119909) and on theother hand it is possible to remove the fractional exponent ofthe heat transfer coefficient by fundamental laws of exponentand binomial expansion

Proposition 1 A nonlinear ODE such as (4) may admit theDTM solution if 119891(119909) is a constant or exponential functionHowever if 119891(119909) is a power law then such an equation admitsa DTM solution if the product

119891 (119909) sdot 1198911015840

(119909) = 120572 (37)

holds Here 120572 is a real constant

Proof Introducing the new variable 120585 = 119891(119909) it follows fromchain rule that (4) becomes

120572119889120585

119889119909

119889

119889120585[119896 (120579)

119889120579

119889120585] minus119872

2

120579119899+1

= 0 (38)

The most general solution of condition (37) is

119891 (119909) = (2120572119909 + 120574)12

(39)

where 120574 is an integration constant

Example 2 (a) If 119891(119909) = 119909120590 then 120585 = 119909120590 transforms (4) into

119889

119889120585[119896 (120579)

119889120579

119889120585] minus 4119872

2

120585120579119899+1

= 0 (40)

only if 120590 = 12This example implies that the DTM may only be appli-

cable to problems describing heat transfer in fins withconvex parabolic profile In the next subsections we presentanalytical solutions for (4) with various functions 119891(119909) and119896(120579)

51 The Exponential Profile and Power Law Thermal Con-ductivity In this section we present solutions for equationdescribing heat transfer in a fin with exponential profile andpower law thermal conductivity and heat transfer coefficientThat is given (4) with 119891(119909) = e120590119909 and both heat transfercoefficient and thermal conductivity being power law func-tions of temperature we construct analytical solutions In ouranalysis we consider 119899 = 2 and 3 indicating the fin subject tonucleate boiling and radiation into free space at zero absolutetemperature respectively Firstly given 119899 = 3 and 119898 = 2

6 Mathematical Problems in Engineering

and applying the DTM one obtains the following recurrencerevelation

120581

sum

119894=0

120581minus119894

sum

119897=0

120581minus119894minus119897

sum

119901=0

[120590119901

119901Θ (119897) Θ (119894) (120581 minus 119894 minus 119897 minus 119901 + 1)

times (120581 minus 119894 minus 119897 minus 119901 + 2)Θ (120581 minus 119894 minus 119897 minus 119901 + 2)

+ 2120590119901

119901Θ (119897) (119894 + 1)Θ (119894 + 1)

times (120581 minus 119894 minus 119897 minus 119901 + 1)Θ (120581 minus 119894 minus 119897 minus 119901 + 1)

+ 2120590120590119901

119901Θ (119897) Θ (119894) (120581 minus 119894 minus 119897 minus 119901 + 1)

times Θ (120581 minus 119894 minus 119897 minus 119901 + 1)

minus1198722

Θ(119901)Θ (119897) Θ (119894) Θ (120581 minus 119894 minus 119897 minus 119901) ] = 0

(41)

One may recall the transformed prescribed boundaryconditions (18) and (19) Equation (41) is an iterative formulaof constructing the power series solution as follows

Θ (2) =1198722

1198882

2

Θ (3) = minus1205901198722

1198882

3

Θ (4) =

1198722

1198882

(31205902

minus 21198722

119888)

24

Θ (5) = minus

1198722

1198882

(1205903

minus 41205901198722

119888)

30

Θ (6) =

1198722

1198882

(51205904

minus 781205902

1198722

119888 + 581198724

1198882

)

720

(42)

The pervious process is continuous and onemay consideras many terms as desired (but bearing in mind that DTMconverges quite fast) Substituting (18) to (19) and (42) into(13) we obtain the following closed form of the solution

120579 (119909) = 119888 +1198722

1198882

21199092

minus1205901198722

1198882

31199093

+

1198722

1198882

(31205902

minus 21198722

119888)

241199094

minus

1198722

1198882

(1205903

minus 41205901198722

119888)

301199095

+

1198722

1198882

(51205904

minus 781205902

1198722

119888 + 581198724

1198882

)

7201199096

+ sdot sdot sdot

(43)

To obtain the value of 119888 we substitute the boundarycondition (5) into (43) at the point 119909 = 1 That is

120579 (1) = 119888 +1198722

1198882

2minus1205901198722

1198882

3+

1198722

1198882

(31205902

minus 21198722

119888)

24

minus

1198722

1198882

(1205903

minus 41205901198722

119888)

30

+

1198722

1198882

(51205904

minus 781205902

1198722

119888 + 581198724

1198882

)

720+ sdot sdot sdot = 1

(44)

Substituting this value of 119888 into (43) one finds theexpression for 120579(119909) On the other hand given (119899119898) = (2 3)one obtains the solution

120579 (119909) = 119888 +1198722

21199092

minus1205901198722

31199093

+

1198722

(1205902

119888 minus 21198722

)

81198881199094

minus

1205901198722

(21205902

119888 minus 211198722

)

601198881199095

+

1198722

(1801198724

minus 1861205902

1198722

119888 + 51205904

1198882

)

7201199096

+ sdot sdot sdot

(45)

Here the constant 119888 maybe obtained by evaluating theboundary condition 120579(1) = 1 The solutions (43) and (45) aredepicted in Figures 3(b) and 3(a) respectively

52 The Rectangular Profile and Power Law Thermal Con-ductivity In this section we provide a detailed constructionof analytical solutions for the heat transfer in a longitudinalrectangular finwith a power law thermal conductivity that iswe consider (4) with 119891(119909) = 1 and 119896(120579) = 120579119898 The analyticalsolutions are given in the following expressions

(a) Case (119899119898) = (2 3)

120579 (119909) = 119888 +1198722

21199092

minus1198724

41198881199094

+1198726

411988821199096

minus331198728

12211988831199098

+127119872

10

336119888411990910

+ sdot sdot sdot

(46)

(b) Case (119899119898) = (3 2)

120579 (119909) = 119888 +1198882

1198722

21199092

minus1198883

1198724

121199094

+291198884

1198726

3601199096

minus3071198885

1198728

50401199098

+23483119888

6

11987210

45360011990910

+ sdot sdot sdot

(47)

The constant 119888 is obtained by solving the appropriate 120579(119909)at the fin base boundary condition The analytical solutionsin (46) and (47) are depicted in Figures 4(a) and 4(b)respectively

Mathematical Problems in Engineering 7

0 02 04 06 08 1094

095

096

097

098

099

1

119909

120579

119872 = 02

119872 = 03

119872 = 04

119872 = 05

(a)

0 02 04 06 08 1093

094

095

096

097

098

099

1

119909

120579

119872 = 02

119872 = 03

119872 = 04

119872 = 05

(b)

Figure 3 Temperature profile in a longitudinal fin with exponential profile and power law thermal conductivity In (a) the exponents are(119899119898) = (2 3) and (b) (119899 119898) = (3 2)

119872 = 025

119872 = 04

119872 = 05

119872 = 065

082

084

086

088

09

092

094

096

098

1

120579

0 02 04 06 08 1

119909

(a)

084

086

088

09

092

094

096

098

1

120579

0 02 04 06 08 1

119909

119872 = 025

119872 = 04

119872 = 05

119872 = 065

(b)

Figure 4 Temperature profile in a longitudinal fin with rectangular profile and power law thermal conductivity In (a) the exponents are(119899119898) = (2 3) and (b) (119899 119898) = (3 2)

53 The Convex Parabolic Profile and Power Law ThermalConductivity In this section we present solutions for theequation describing the heat transfer in a fin with con-vex parabolic profile and power law thermal conductivityEquation (40) is considered Here we consider the values(119899119898) = (2 3) (3 2) The final analytical solution is givenby

(a) Case (119899119898) = (2 3)

120579 (119909) = 119888 +21198722

311990932

minus21198724

51198881199093

+231198726

45119888211990992

minus1909119872

8

247511988831199096

+329222119872

10

2598751198884119909152

+ sdot sdot sdot

(48)

8 Mathematical Problems in Engineering

075

08

085

09

095

1

119872 = 025

119872 = 04

119872 = 05

119872 = 065

120579

0 02 04 06 08 1119909

(a)

119872 = 025

119872 = 04

119872 = 05

119872 = 065

0 02 04 06 08 1119909

082

084

086

088

09

092

094

096

098

120579

1

(b)

Figure 5 Temperature profile in a longitudinal fin with convex parabolic profile and power law thermal conductivity In (a) the exponentsare (119899119898) = (2 3) and (b) (119899119898) = (3 2)

(b) Case (119899119898) = (3 2)

120579 (119909) = 119888 +21198882

1198722

311990932

minus41198883

1198724

451199093

+41198884

1198726

2711990992

minus9921198885

1198728

74251199096

+30064119888

6

11987210

212625119909152

+ sdot sdot sdot

(49)

The constant 119888 is obtained by evaluating the appropriate 120579(119909)at the fin base boundary condition The solutions in (48) and(49) are depicted in Figures 5(a) and 5(b) respectively

54 The Rectangular Profile and LinearThermal ConductivityIn this section we present solutions for the equation repre-senting the heat transfer in a fin with rectangular profile andthe thermal conductivity depending linearly on temperatureThat is we consider (4) with 119891(119909) = 1 and 119896(120579) = 1 + 120573120579 Theanalytical solutions for this problem for different values of 119899are given by

(a) Case 119899 = 0

120579 (119909)

= 119888 +

1198722

119888

2 (1 + 120573119888)

1199092

minus

1198724

119888 (minus1 + 2120573119888)

24(1 + 120573119888)31199094

+

1198726

119888 (1 minus 16120573119888 + 281205732

1198882

)

720(1 + 120573119888)51199096

minus

1198728

119888 (minus1 + 78120573119888 minus 6001205732

1198882

+ 8961205733

1198883

)

40320(1 + 120573119888)7

1199098

+

11987210

119888 (1 minus 332120573119888 + 78121205732

1198882

minus 398961205733

1198883

+ 511841205734

1198884

)

3628800(1 + 120573119888)9

11990910

+ sdot sdot sdot

(50)

(b) Case 119899 = 1

120579 (119909)

= 119888 +

11987221198882

2 (1 + 120573119888)

1199092minus

11987241198883(minus1 + 2120573119888)

24(1 + 120573119888)31199094

+

11987261198884(10 minus 16120573119888 + 19120573

21198882)

720(1 + 120573119888)5

1199096

minus

11987281198885(minus80 + 342120573119888 minus 594120573

21198882+ 559120573

31198883)

40320(1 + 120573119888)7

1199098

+

119872101198886(1000 minus 7820120573119888 + 24336120573

21198882minus 36908120573

31198883+ 29161120573

41198884)

3628800(1 + 120573119888)9

11990910

+ sdot sdot sdot

(51)

(c) Case 119899 = 2

120579 (119909)

= 119888 +

1198722

1198883

2 (1 + 120573119888)

1199092

+

1198724

1198885

8(1 + 120573119888)31199094

+

1198726

1198887

(3 + 21205732

1198882

)

80(1 + 120573119888)51199096

+

1198728

1198889

(49 minus 20120573119888 + 661205732

1198882

minus 401205733

1198883

)

4480(1 + 120573119888)7

1199098

+

11987210

11988811

(427 minus 440120573119888 + 11161205732

1198882

minus 10201205733

1198883

+ 6721205734

1198884

)

134400(1 + 120573119888)9

11990910

+ sdot sdot sdot

(52)

Mathematical Problems in Engineering 9

(d) Case 119899 = 3120579 (119909)

= 119888 +

11987221198884

2 (1 + 120573119888)

1199092+

11987241198887(4 + 120573119888)

24(1 + 120573119888)31199094

+

119872611988810(52 + 32120573119888 + 25120573

21198882)

720(1 + 120573119888)5

1199096

+

119872811988813(1288 + 1020120573119888 + 1212120573

21198882minus 95120573

31198883)

40320(1 + 120573119888)7

1199098

+

1198721011988816(52024+45688120573119888+77184120573

21198882minus680120573

31198883+ 15025120573

41198884)

3628800(1 + 120573119888)9

11990910

+

(53)

The constant 119888may be obtained from the boundary conditionon the appropriate solution The solutions in (50) (51) (52)and (53) are depicted in Figure 6

55The Convex Parabolic Profile and LinearThermal Conduc-tivity In this section we present solutions for the equationdescribing the heat transfer in a fin with convex parabolicprofile and the thermal conductivity depending linearly ontemperature That is we consider (40) with 119896(120579) = 1 + 120573120579The analytical solution for this problem for different values of119899 is given by

(a) Case 119899 = 0

120579 (119909)

= 119888 +21198722

119888

3 (1 + 120573119888)11990932

minus21198724

119888 (minus2 + 3120573119888)

45(1 + 120573119888)3

1199093

+

1198726

119888 (2 minus 25120573119888 + 331205732

1198882

)

405(1 + 120573119888)5

11990992

minus

1198728

119888 (minus10 + 599120573119888 minus 35821205732

1198882

+ 40591205733

1198883

)

66825(1 + 120573119888)7

1199096

+ sdot sdot sdot

(54)

(b) Case 119899 = 1

120579 (119909)

= 119888 +21198722

1198882

3 (1 + 120573119888)11990932

minus21198724

1198883

(minus4 + 120573119888)

45(1 + 120573119888)3

1199093

+

21198726

1198884

(9 minus 11120573119888 + 101205732

1198882

)

405(1 + 120573119888)5

11990992

minus

21198728

1198885

(minus330 + 1118120573119888 minus 15841205732

1198882

+ 10931205733

1198883

)

66825(1 + 120573119888)7

1199096

+ sdot sdot sdot

(55)

119899 = 0

119899 = 1

119899 = 2

119899 = 3

1

095

09

085

08

075

07

065

06

055

0 02 04 06 08 1

119909

120579

Figure 6 Temperature profile in a longitudinal rectangular fin withlinear thermal conductivity and varying values of 119899 Here 120573 = 05and119872 = 15 are fixed

(c) Case 119899 = 2

120579 (119909)

= 119888 +21198722

1198883

3 (1 + 120573119888)11990932

+21198724

1198885

(6 + 120573119888)

45(1 + 120573119888)31199093

+

1198726

1198887

(16 + 3120573119888 + 71205732

1198882

)

135(1 + 120573119888)5

11990992

+

1198728

1198889

(1160 minus 107120573119888 + 11161205732

1198882

minus 3671205733

1198883

)

22275(1 + 120573119888)7

1199096

+ sdot sdot sdot

(56)

(d) Case 119899 = 3

120579 (119909)

= 119888 +21198722

1198884

3 (1 + 120573119888)11990932

+21198724

1198887

(8 + 3120573119888)

45(1 + 120573119888)31199093

+

41198726

11988810

(23 + 17120573119888 + 91205732

1198882

)

405(1 + 120573119888)5

11990992

+

21198728

11988813

(5000+4868120573119888 + 43561205732

1198882

+ 3631205733

1198883

)

66825(1 + 120573119888)7

1199096

+ sdot sdot sdot

(57)

The constant 119888 may be obtained from the boundarycondition on the appropriate 120579(119909)The solutions in (54) (55)(56) and (57) are depicted in Figure 7

10 Mathematical Problems in Engineering

1

09

08

07

06

05

120579

119899 = 0

119899 = 1

119899 = 2

119899 = 3

0 02 04 06 08 1

119909

Figure 7 Temperature profile in a longitudinal convex parabolic finwith linear thermal conductivity and varying values of 119899 Here 120573 =05 and119872 = 15 are fixed

56 The Exponential Profile and LinearThermal ConductivityIn this section we present solutions for equation heat transferin a fin with convex parabolic profile and the thermalconductivity depending linearly on temperature That is weconsider (4) with 119896(120579) = 1 + 120573120579 and 119891(119909) = e120590119909 Here 120590is a constant The analytical solutions for this problem fordifferent values of 119899 are given by

(a) Case 119899 = 0

120579 (119909)

= 119888 +

1198722

119888

2 (1 + 120573119888)

1199092

minus

1205901198722

119888 (2 + 120573119888 + 1205732

119888)

6(1 + 120573119888)21199093

+

1198722

119888 (1198722

(1 minus 21205732

119888) + 1205902

(3 +3120573119888+1205733

1198882

+1205734

1198882

+1205732

119888 (3 + 119888)))

24(1 + 120573119888)4

1199094

+ sdot sdot sdot

(58)

(b) Case 119899 = 1

120579 (119909)

= 119888 +

11987221198882

2 (1 + 120573119888)

1199092minus

12059011987221198882(2 + 120573119888 + 120573

2119888)

6(1 + 120573119888)21199093

+

11987221198882(1198722119888 (2+120573119888minus2120573

2119888)+1205902(3+3120573119888+120573

31198882+12057341198882+1205732119888 (3+119888)))

24(1 + 120573119888)3

1199094

+ sdot sdot sdot

(59)

119899 = 0

119899 = 1

119899 = 2

119899 = 3

0 02 04 06 08 1

119909

1

095

09

085

08

075

07

120579

Figure 8 Temperature profile in a longitudinal fin of exponentialprofile with linear thermal conductivity and varying values of 119899Here 120573 = 05 and119872 = 15 are fixed

(c) Case 119899 = 2120579 (119909)

= 119888 +

11987221198883

2 (1 + 120573119888)

1199092minus

12059011987221198883(2 + 120573119888 + 120573

2119888)

6(1 + 120573119888)21199093

+

11987221198883(11987221198882(3+2120573119888minus2120573

2119888)+1205902(3+3120573119888+120573

31198882+12057341198882+1205732119888 (3+119888)))

24(1 + 120573119888)3

times1199094+ sdot sdot sdot

(60)

(d) Case 119899 = 3120579 (119909)

= 119888 +

11987221198884

2 (1 + 120573119888)

1199092minus

12059011987221198884(2 + 120573119888 + 120573

2119888)

6(1 + 120573119888)21199093

+

11987221198884(11987221198883(4+3120573119888minus2120573

2119888)+1205902(3+3120573119888+120573

31198882+12057341198882+1205732119888 (3+119888)))

24(1 + 120573119888)3

times1199094+ sdot sdot sdot

(61)

The constant 119888 may be obtained from the boundarycondition on the appropriate 120579(119909)The solutions in (58) (59)(60) and (61) are depicted in Figure 8

6 Fin Efficiency and Heat Flux

61 Fin Efficiency Theheat transfer rate from a fin is given byNewtonrsquos second law of cooling

119876 = int

119871

0

119875119867 (119879) (119879 minus 119879119886) 119889119909 (62)

Mathematical Problems in Engineering 11

Fin efficiency is defined as the ratio of the fin heat transferrate to the rate that would be if the entire fin were at the basetemperature and is given by (see eg [1])

120578=

119876

119876ideal=

int119871

0

119875119867 (119879) (119879 minus 119879119886) 119889119883

119875ℎ119887119871 (119879119887minus 119879119886)

(63)

In dimensionless variables we have

120578 = int

1

0

120579119899+1

119889119909 (64)

We consider the solutions (50) (51) (52) and (53) and depictthe fin efficiency (63) in Figure 9

62 Heat Flux The fin base heat flux is given by the Fourierrsquoslaw

119902119887= 119860119888119870 (119879)

119889119879

119889119909 (65)

The total heat flux of the fin is given by [1]

119902 =119902119887

119860119888119867(119879) (119879

119887minus 119879119886) (66)

Introducing the dimensionless variable as described inSection 2 implies

119902 =1

119861119894

119896 (120579)

ℎ (120579)

119889120579

119889119909 (67)

where the dimensionless parameter 119861119894 = ℎ119887119871119896119886is the Biot

number We consider a number of cases for the thermalconductivity and the heat transfer coefficient

621 Linear Thermal Conductivity and Power Law HeatTransfer Coefficient In this case (67) becomes

119902 =1

119861119894(1 + 120573120579) 120579

minus119899119889120579

119889119909 (68)

The heat flux in (68) at the base of the fin is plotted inFigure 10

622 Power LawThermal Conductivity and Heat TransferCoefficient In this case (67) becomes

119902 =1

119861119894120579119898minus119899

119889120579

119889119909 (69)

Not surprisingly heat flux in one-dimensional fins is highergiven values 119861119894 ≪ 1 The heat flux in (69) is plotted inFigures 11(a) 11(b) and 11(c)

7 Some Discussions

The DTM has resulted in some interesting observations andstudy We have observed in Figures 2(a) and 2(b) an excellentagreement between the analytical solutions generated byDTM and the exact solution obtained in [2] In particular we

119872

1

095

09

085

08

075

07

065

06

055

120578

2151050

119899 = 0

119899 = 1

119899 = 2

119899 = 3

Figure 9 Fin efficiency of a longitudinal rectangular fin Here 120573 =075

119872

25

20

15

10

5

0

119902

2151050

119899 = 0

119899 = 1

119899 = 2

119899 = 3

Figure 10 Base heat flux in a longitudinal rectangular finwith linearthermal conductivity Here 120573 = 01

considered a fin problem in which both thermal conductivityand heat transfer coefficient are given by the same power lawFurthermore we notice from Table 2 that an absolute errorof approximately 35119890 minus 005 is produced by DTM of order119874(15) In Table 3 an absolute error of approximately 65119890 minus006 is produced for the same order This confirms that theDTM converges faster and can provide accurate results witha minimum computation As such a tremendous confidencein the DTM in terms of the accuracy and effectiveness wasbuilt and thus we used this method to solve other problemsfor which exact solutions are harder to construct

In Figures 3(a) 3(b) 4(a) 4(b) 5(a) and 5(b) weobserve that the fin temperature increaseswith the decreasing

12 Mathematical Problems in Engineering

values of the thermogeometric fin parameter Here thevalues of the exponents are fixed Also we observe that fintemperature is higher when 119899 minus 119898 gt 0 that is when heattransfer coefficient is higher than the thermal conductivityWe observe in Figures 6 7 and 8 that the fin temperatureincreases with the increasing values of 119899 Furthermoreit appears that the fin with exponential profile performsthe least in transferring the heat from the base since thetemperature in such a fin is much higher than that ofthe rectangular and the convex parabolic profiles In otherwords heat dissipation to the fluid surrounding the extendedsurface is much faster in longitudinal fins of rectangular andconvex parabolic profiles In Figure 9 fin efficiency decreaseswith increasing thermogeometric fin parameter Also finefficiency increases with increasing values of 119899 It is easyto show that the thermogeometric fin parameter is directlyproportional to the aspect ratio (extension factor) with squareroot of the Biot number being the proportionality constantAs such shorter fins are more efficient than longer onesElse the increased Biot number results in less efficient finwhenever the space is confined that is where the length ofthe fin cannot be increased Figure 10 depicts the heat fluxat the fin base The amount of heat energy dissipated fromthe fin base is of immense interest in engineering [28] Weobserve in Figure 10 that the base heat flux increases withthe thermogeometric fin parameter for considered values ofthe exponent 119899 (see also [28]) Figures 11(a) 11(b) and 11(c)display the heat flux across the fin length We note that theheat flux across the fin length increases with increasing valuesof the thermogeometric fin parameter

8 Concluding Remarks

In this study we have successfully applied the DTM to highlynonlinear problems arising in heat transfer through longitu-dinal fins of various profiles Both thermal conductivity andheat transfer coefficient are given as functions of temperatureThe DTM agreed well with exact solutions when the thermalconductivity and heat transfer coefficient are given by thesame power law A rapid convergence to the exact solutionwas observed Following the confidence in DTM built by theresults mentioned we then solved various exciting problemsThe exotic results have been shown in tables and figures listedin this paper

The results obtained in this paper are significant improve-ments on the known results In particular both the heattransfer coefficient and the thermal conductivity are allowedto be given by the power law functions of temperature andalsowe considered a number of finprofilesWenote that exactsolutions are difficult if not impossible to construct when theexponents of these properties are distinct

Perhaps the notable advantage of the DTM is the general-ization of the Taylor method to problems involving unusualderivative procedures such as fractional fuzzy or q-derivative[22] Some generalizations have been made by Odibat et al[24] and they referred to their new method as the General-ized Differential Transform Method (GDTM) This showedgreat improvement compared to the Fractional Differential

0 02 04 06 08 1119909

45

40

35

30

25

20

15

10

5

0

minus5

119902

(a)

0 02 04 06 08 1119909

30

25

20

15

10

5

0

119902

(b)

35

30

25

20

15

10

5

0

119902

0 02 04 06 08 1119909

119872 = 025

119872 = 04

119872 = 05

119872 = 065

(c)

Figure 11 Heat flux across a longitudinal rectangular fin with linearthermal conductivity (a) 119899minus119898 lt 0 (b) 119899minus119898 gt 0 and (c) 119899minus119898 = 0Here 119861119894 = 001

Mathematical Problems in Engineering 13

Transform Method (FDTM) introduced by Arikoglu andOzkol [23]

We have shown with the help of an example that DTMmay only be applied to fin problems involving heat transferthrough fins with convex parabolic profile Note that given anODE such as (4) with a power law heat transfer coefficient ofa fractional exponent then one can easily remove the fractionby basic exponential rules and employment of the Binomialexpansion However using the DTM one runs into difficultyif the power law thermal conductivity in the same equation isgiven by the fractional exponent We do not know whetherthese observations call for the ldquomodifiedrdquo DTM to solveproblems arising in heat flow through fins with other profilessuch as longitudinal triangular and concave parabolic andalso with fractional power law thermal conductivity

The main results obtained in this paper give insightinto heat transfer in boiling liquids where the heat transfercoefficient is temperature dependent and may be given by apower law The thermal conductivity of some materials suchas gallium nitride (GaN) and Aluminium Nitride (AlN) maybe modeled by power law temperature dependency [29 30]Thus the solutions constructed here give a better comparisonof heat transfer in terms of material used since in many engi-neering applications thermal conductivity is given as a linearfunction of temperature Furthermore a good study in termsof performance and efficiency of fin with different profiles isundertaken These finding could help in the design of fins Itis claimed in [14] that DTM results are more accurate thanthose constructed by Variational Iteration Methods (VIM)and Homotopy Perturbation Methods (HPM) However itwould be risky to use the DTM approximate solutions asbenchmarks for the numerical schemes Nevertheless wehave also shown that DTM converges rapidly in just fifteenterms to the exact solution

Acknowledgments

The authors wish to thank the National Research Foundationof South Africa and the University of the Witwatersrandfor the generous financial support The authors are gratefulto Professor D P Mason and the anonymous reviewers fortheir invaluable comments which improved this paper signif-icantly

References

[1] A D Kraus A Aziz and J Welty Extended Surface Heat Trans-fer Wiley New York NY USA 2001

[2] R JMoitsheki T Hayat andM YMalik ldquoSome exact solutionsof the fin problem with a power law temperature-dependentthermal conductivityrdquo Nonlinear Analysis Real World Applica-tions vol 11 no 5 pp 3287ndash3294 2010

[3] R J Moitsheki ldquoSteady one-dimensional heat flow in a lon-gitudinal triangular and parabolic finrdquo Communications inNonlinear Science and Numerical Simulation vol 16 no 10 pp3971ndash3980 2011

[4] R J Moitsheki ldquoSteady heat transfer through a radial fin withrectangular and hyperbolic profilesrdquo Nonlinear Analysis RealWorld Applications vol 12 no 2 pp 867ndash874 2011

[5] Mo Miansari D D Ganji and Me Miansari ldquoApplication ofHersquos variational iteration method to nonlinear heat transferequationsrdquo Physics Letters A vol 372 no 6 pp 779ndash785 2008

[6] C H Chiu and C K Chen ldquoA decomposition method forsolving the convective longitudinal fins with variable thermalconductivityrdquo International Journal of Heat and Mass Transfervol 45 no 10 pp 2067ndash2075 2002

[7] A Rajabi ldquoHomotopy perturbation method for fin efficiencyof convective straight fins with temperature-dependent thermalconductivityrdquo Physics Letters A vol 364 no 1 pp 33ndash37 2007

[8] G Domairry and M Fazeli ldquoHomotopy analysis method todetermine the fin efficiency of convective straight fins withtemperature-dependent thermal conductivityrdquo Communica-tions in Nonlinear Science and Numerical Simulation vol 14 no2 pp 489ndash499 2009

[9] A Campo andR J Spaulding ldquoCoupling of themethods of suc-cessive approximations and undetermined coefficients for theprediction of the thermal behaviour of uniform circumferentialfinsrdquo Heat and Mass Transfer vol 34 no 6 pp 461ndash468 1999

[10] A Campo and F Rodrfguez ldquoApproximate analytic tempera-ture solution for uniform annular fins by adapting the powerseriesmethodrdquo International Communications inHeat andMassTransfer vol 25 no 6 pp 809ndash818 1998

[11] R Chiba ldquoApplication of differential transformmethod to ther-moelastic problem for annular disks of variable thickness withtemperature-dependent parametersrdquo International Journal ofThermophysics vol 33 pp 363ndash380 2012

[12] S Sadri M R Raveshi and S Amiri ldquoEfficiency analysis ofstraight fin with variable heat transfer coefficient and thermalconductivityrdquo Journal ofMechanical Science andTechnology vol26 no 4 pp 1283ndash1290 2012

[13] M Torabi H Yaghoori and A Aziz ldquoAnalytical solution forconvective-radiative continously moving fin with temperaturedependent thermal conductivityrdquo International Journal of Ther-mophysics vol 33 pp 924ndash941 2012

[14] M Torabi andH Yaghoobi ldquoTwo dominant analytical methodsfor thermal analysis of convective step fin with variable thermalconductivityrdquoThermal Science In press

[15] AMoradi ldquoAnalytical solutions for finwith temperature depen-dant heat transfer coefficientrdquo International Journal of Engineer-ing and Applied Sciences vol 3 no 2 pp 1ndash12 2011

[16] A Moradi and H Ahmadikia ldquoAnalytical solution for differentprofiles of fin with temperature-dependent thermal conductiv-ityrdquoMathematical Problems in Engineering vol 2010 Article ID568263 15 pages 2010

[17] H Yaghoobi and M Torabi ldquoThe application of differentialtransformation method to nonlinear equations arising in heattransferrdquo International Communications in Heat and MassTransfer vol 38 no 6 pp 815ndash820 2011

[18] S Ghafoori M Motevalli M G Nejad F Shakeri D D GanjiandM Jalaal ldquoEfficiency of differential transformation methodfor nonlinear oscillation Comparison with HPM and VIMrdquoCurrent Applied Physics vol 11 no 4 pp 965ndash971 2011

[19] A A Joneidi D D Ganji andM Babaelahi ldquoDifferential trans-formation method to determine fin efficiency of convectivestraight fins with temperature dependent thermal conductivityrdquoInternational Communications in Heat and Mass Transfer vol36 no 7 pp 757ndash762 2009

[20] J K Zhou Differential Transform Method and Its Applicationsfor Electric circuIts Huazhong University Press Wuhan China1986

14 Mathematical Problems in Engineering

[21] S Mukherjee ldquoReply to comment on lsquosolutions of the duffing-van der pol oscillator equation by the differential transformmethodrsquordquo Physica Scripta vol 84 Article ID 037003 2 pages2011

[22] C Bervillier ldquoStatus of the differential transformationmethodrdquoApplied Mathematics and Computation vol 218 no 20 pp10158ndash10170 2012

[23] A Arikoglu and I Ozkol ldquoSolution of fractional differentialequations by using differential transform methodrdquo ChaosSolitons and Fractals vol 34 no 5 pp 1473ndash1481 2007

[24] ZOdibat SMomani andV S Erturk ldquoGeneralized differentialtransform method application to differential equations offractional orderrdquo Applied Mathematics and Computation vol197 no 2 pp 467ndash477 2008

[25] CW Bert ldquoApplication of differential transformmethod to heatconduction in tapered finsrdquo Journal of Heat Transfer vol 124no 1 pp 208ndash209 2002

[26] F Khani and A Aziz ldquoThermal analysis of a longitudinal trape-zoidal fin with temperature-dependent thermal conductivityand heat transfer coefficientrdquo Communications in NonlinearScience and Numerical Simulation vol 15 no 3 pp 590ndash6012010

[27] H C Unal ldquoAn anlytical study of boiling heat transfer from afinrdquo International Journal of Heat and Mass Transfer vol 31 no7 pp 1483ndash1496 1988

[28] M H Chang ldquoA decomposition solution for fins with tempera-ture dependent surface heat fluxrdquo International Journal of Heatand Mass Transfer vol 48 no 9 pp 1819ndash1824 2005

[29] A Jezowski B A Danilchenko M Bockowski et al ldquoThermalconductivity of GaN crystals in 42ndash300 K rangerdquo Solid StateCommunications vol 128 no 2-3 pp 69ndash73 2003

[30] S Vitanov V Palankovski S Maroldt and R Quay ldquoHigh-temperature modeling of AlGaNGaN HEMTsrdquo Solid-StateElectronics vol 54 no 10 pp 1105ndash1112 2010

Submit your manuscripts athttpwwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

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Mathematical Problems in Engineering

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Differential EquationsInternational Journal of

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Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

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Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

Page 6: Research Article Analytical Solutions for Steady Heat ...downloads.hindawi.com/journals/mpe/2013/273052.pdf · Research Article Analytical Solutions for Steady Heat Transfer in Longitudinal

6 Mathematical Problems in Engineering

and applying the DTM one obtains the following recurrencerevelation

120581

sum

119894=0

120581minus119894

sum

119897=0

120581minus119894minus119897

sum

119901=0

[120590119901

119901Θ (119897) Θ (119894) (120581 minus 119894 minus 119897 minus 119901 + 1)

times (120581 minus 119894 minus 119897 minus 119901 + 2)Θ (120581 minus 119894 minus 119897 minus 119901 + 2)

+ 2120590119901

119901Θ (119897) (119894 + 1)Θ (119894 + 1)

times (120581 minus 119894 minus 119897 minus 119901 + 1)Θ (120581 minus 119894 minus 119897 minus 119901 + 1)

+ 2120590120590119901

119901Θ (119897) Θ (119894) (120581 minus 119894 minus 119897 minus 119901 + 1)

times Θ (120581 minus 119894 minus 119897 minus 119901 + 1)

minus1198722

Θ(119901)Θ (119897) Θ (119894) Θ (120581 minus 119894 minus 119897 minus 119901) ] = 0

(41)

One may recall the transformed prescribed boundaryconditions (18) and (19) Equation (41) is an iterative formulaof constructing the power series solution as follows

Θ (2) =1198722

1198882

2

Θ (3) = minus1205901198722

1198882

3

Θ (4) =

1198722

1198882

(31205902

minus 21198722

119888)

24

Θ (5) = minus

1198722

1198882

(1205903

minus 41205901198722

119888)

30

Θ (6) =

1198722

1198882

(51205904

minus 781205902

1198722

119888 + 581198724

1198882

)

720

(42)

The pervious process is continuous and onemay consideras many terms as desired (but bearing in mind that DTMconverges quite fast) Substituting (18) to (19) and (42) into(13) we obtain the following closed form of the solution

120579 (119909) = 119888 +1198722

1198882

21199092

minus1205901198722

1198882

31199093

+

1198722

1198882

(31205902

minus 21198722

119888)

241199094

minus

1198722

1198882

(1205903

minus 41205901198722

119888)

301199095

+

1198722

1198882

(51205904

minus 781205902

1198722

119888 + 581198724

1198882

)

7201199096

+ sdot sdot sdot

(43)

To obtain the value of 119888 we substitute the boundarycondition (5) into (43) at the point 119909 = 1 That is

120579 (1) = 119888 +1198722

1198882

2minus1205901198722

1198882

3+

1198722

1198882

(31205902

minus 21198722

119888)

24

minus

1198722

1198882

(1205903

minus 41205901198722

119888)

30

+

1198722

1198882

(51205904

minus 781205902

1198722

119888 + 581198724

1198882

)

720+ sdot sdot sdot = 1

(44)

Substituting this value of 119888 into (43) one finds theexpression for 120579(119909) On the other hand given (119899119898) = (2 3)one obtains the solution

120579 (119909) = 119888 +1198722

21199092

minus1205901198722

31199093

+

1198722

(1205902

119888 minus 21198722

)

81198881199094

minus

1205901198722

(21205902

119888 minus 211198722

)

601198881199095

+

1198722

(1801198724

minus 1861205902

1198722

119888 + 51205904

1198882

)

7201199096

+ sdot sdot sdot

(45)

Here the constant 119888 maybe obtained by evaluating theboundary condition 120579(1) = 1 The solutions (43) and (45) aredepicted in Figures 3(b) and 3(a) respectively

52 The Rectangular Profile and Power Law Thermal Con-ductivity In this section we provide a detailed constructionof analytical solutions for the heat transfer in a longitudinalrectangular finwith a power law thermal conductivity that iswe consider (4) with 119891(119909) = 1 and 119896(120579) = 120579119898 The analyticalsolutions are given in the following expressions

(a) Case (119899119898) = (2 3)

120579 (119909) = 119888 +1198722

21199092

minus1198724

41198881199094

+1198726

411988821199096

minus331198728

12211988831199098

+127119872

10

336119888411990910

+ sdot sdot sdot

(46)

(b) Case (119899119898) = (3 2)

120579 (119909) = 119888 +1198882

1198722

21199092

minus1198883

1198724

121199094

+291198884

1198726

3601199096

minus3071198885

1198728

50401199098

+23483119888

6

11987210

45360011990910

+ sdot sdot sdot

(47)

The constant 119888 is obtained by solving the appropriate 120579(119909)at the fin base boundary condition The analytical solutionsin (46) and (47) are depicted in Figures 4(a) and 4(b)respectively

Mathematical Problems in Engineering 7

0 02 04 06 08 1094

095

096

097

098

099

1

119909

120579

119872 = 02

119872 = 03

119872 = 04

119872 = 05

(a)

0 02 04 06 08 1093

094

095

096

097

098

099

1

119909

120579

119872 = 02

119872 = 03

119872 = 04

119872 = 05

(b)

Figure 3 Temperature profile in a longitudinal fin with exponential profile and power law thermal conductivity In (a) the exponents are(119899119898) = (2 3) and (b) (119899 119898) = (3 2)

119872 = 025

119872 = 04

119872 = 05

119872 = 065

082

084

086

088

09

092

094

096

098

1

120579

0 02 04 06 08 1

119909

(a)

084

086

088

09

092

094

096

098

1

120579

0 02 04 06 08 1

119909

119872 = 025

119872 = 04

119872 = 05

119872 = 065

(b)

Figure 4 Temperature profile in a longitudinal fin with rectangular profile and power law thermal conductivity In (a) the exponents are(119899119898) = (2 3) and (b) (119899 119898) = (3 2)

53 The Convex Parabolic Profile and Power Law ThermalConductivity In this section we present solutions for theequation describing the heat transfer in a fin with con-vex parabolic profile and power law thermal conductivityEquation (40) is considered Here we consider the values(119899119898) = (2 3) (3 2) The final analytical solution is givenby

(a) Case (119899119898) = (2 3)

120579 (119909) = 119888 +21198722

311990932

minus21198724

51198881199093

+231198726

45119888211990992

minus1909119872

8

247511988831199096

+329222119872

10

2598751198884119909152

+ sdot sdot sdot

(48)

8 Mathematical Problems in Engineering

075

08

085

09

095

1

119872 = 025

119872 = 04

119872 = 05

119872 = 065

120579

0 02 04 06 08 1119909

(a)

119872 = 025

119872 = 04

119872 = 05

119872 = 065

0 02 04 06 08 1119909

082

084

086

088

09

092

094

096

098

120579

1

(b)

Figure 5 Temperature profile in a longitudinal fin with convex parabolic profile and power law thermal conductivity In (a) the exponentsare (119899119898) = (2 3) and (b) (119899119898) = (3 2)

(b) Case (119899119898) = (3 2)

120579 (119909) = 119888 +21198882

1198722

311990932

minus41198883

1198724

451199093

+41198884

1198726

2711990992

minus9921198885

1198728

74251199096

+30064119888

6

11987210

212625119909152

+ sdot sdot sdot

(49)

The constant 119888 is obtained by evaluating the appropriate 120579(119909)at the fin base boundary condition The solutions in (48) and(49) are depicted in Figures 5(a) and 5(b) respectively

54 The Rectangular Profile and LinearThermal ConductivityIn this section we present solutions for the equation repre-senting the heat transfer in a fin with rectangular profile andthe thermal conductivity depending linearly on temperatureThat is we consider (4) with 119891(119909) = 1 and 119896(120579) = 1 + 120573120579 Theanalytical solutions for this problem for different values of 119899are given by

(a) Case 119899 = 0

120579 (119909)

= 119888 +

1198722

119888

2 (1 + 120573119888)

1199092

minus

1198724

119888 (minus1 + 2120573119888)

24(1 + 120573119888)31199094

+

1198726

119888 (1 minus 16120573119888 + 281205732

1198882

)

720(1 + 120573119888)51199096

minus

1198728

119888 (minus1 + 78120573119888 minus 6001205732

1198882

+ 8961205733

1198883

)

40320(1 + 120573119888)7

1199098

+

11987210

119888 (1 minus 332120573119888 + 78121205732

1198882

minus 398961205733

1198883

+ 511841205734

1198884

)

3628800(1 + 120573119888)9

11990910

+ sdot sdot sdot

(50)

(b) Case 119899 = 1

120579 (119909)

= 119888 +

11987221198882

2 (1 + 120573119888)

1199092minus

11987241198883(minus1 + 2120573119888)

24(1 + 120573119888)31199094

+

11987261198884(10 minus 16120573119888 + 19120573

21198882)

720(1 + 120573119888)5

1199096

minus

11987281198885(minus80 + 342120573119888 minus 594120573

21198882+ 559120573

31198883)

40320(1 + 120573119888)7

1199098

+

119872101198886(1000 minus 7820120573119888 + 24336120573

21198882minus 36908120573

31198883+ 29161120573

41198884)

3628800(1 + 120573119888)9

11990910

+ sdot sdot sdot

(51)

(c) Case 119899 = 2

120579 (119909)

= 119888 +

1198722

1198883

2 (1 + 120573119888)

1199092

+

1198724

1198885

8(1 + 120573119888)31199094

+

1198726

1198887

(3 + 21205732

1198882

)

80(1 + 120573119888)51199096

+

1198728

1198889

(49 minus 20120573119888 + 661205732

1198882

minus 401205733

1198883

)

4480(1 + 120573119888)7

1199098

+

11987210

11988811

(427 minus 440120573119888 + 11161205732

1198882

minus 10201205733

1198883

+ 6721205734

1198884

)

134400(1 + 120573119888)9

11990910

+ sdot sdot sdot

(52)

Mathematical Problems in Engineering 9

(d) Case 119899 = 3120579 (119909)

= 119888 +

11987221198884

2 (1 + 120573119888)

1199092+

11987241198887(4 + 120573119888)

24(1 + 120573119888)31199094

+

119872611988810(52 + 32120573119888 + 25120573

21198882)

720(1 + 120573119888)5

1199096

+

119872811988813(1288 + 1020120573119888 + 1212120573

21198882minus 95120573

31198883)

40320(1 + 120573119888)7

1199098

+

1198721011988816(52024+45688120573119888+77184120573

21198882minus680120573

31198883+ 15025120573

41198884)

3628800(1 + 120573119888)9

11990910

+

(53)

The constant 119888may be obtained from the boundary conditionon the appropriate solution The solutions in (50) (51) (52)and (53) are depicted in Figure 6

55The Convex Parabolic Profile and LinearThermal Conduc-tivity In this section we present solutions for the equationdescribing the heat transfer in a fin with convex parabolicprofile and the thermal conductivity depending linearly ontemperature That is we consider (40) with 119896(120579) = 1 + 120573120579The analytical solution for this problem for different values of119899 is given by

(a) Case 119899 = 0

120579 (119909)

= 119888 +21198722

119888

3 (1 + 120573119888)11990932

minus21198724

119888 (minus2 + 3120573119888)

45(1 + 120573119888)3

1199093

+

1198726

119888 (2 minus 25120573119888 + 331205732

1198882

)

405(1 + 120573119888)5

11990992

minus

1198728

119888 (minus10 + 599120573119888 minus 35821205732

1198882

+ 40591205733

1198883

)

66825(1 + 120573119888)7

1199096

+ sdot sdot sdot

(54)

(b) Case 119899 = 1

120579 (119909)

= 119888 +21198722

1198882

3 (1 + 120573119888)11990932

minus21198724

1198883

(minus4 + 120573119888)

45(1 + 120573119888)3

1199093

+

21198726

1198884

(9 minus 11120573119888 + 101205732

1198882

)

405(1 + 120573119888)5

11990992

minus

21198728

1198885

(minus330 + 1118120573119888 minus 15841205732

1198882

+ 10931205733

1198883

)

66825(1 + 120573119888)7

1199096

+ sdot sdot sdot

(55)

119899 = 0

119899 = 1

119899 = 2

119899 = 3

1

095

09

085

08

075

07

065

06

055

0 02 04 06 08 1

119909

120579

Figure 6 Temperature profile in a longitudinal rectangular fin withlinear thermal conductivity and varying values of 119899 Here 120573 = 05and119872 = 15 are fixed

(c) Case 119899 = 2

120579 (119909)

= 119888 +21198722

1198883

3 (1 + 120573119888)11990932

+21198724

1198885

(6 + 120573119888)

45(1 + 120573119888)31199093

+

1198726

1198887

(16 + 3120573119888 + 71205732

1198882

)

135(1 + 120573119888)5

11990992

+

1198728

1198889

(1160 minus 107120573119888 + 11161205732

1198882

minus 3671205733

1198883

)

22275(1 + 120573119888)7

1199096

+ sdot sdot sdot

(56)

(d) Case 119899 = 3

120579 (119909)

= 119888 +21198722

1198884

3 (1 + 120573119888)11990932

+21198724

1198887

(8 + 3120573119888)

45(1 + 120573119888)31199093

+

41198726

11988810

(23 + 17120573119888 + 91205732

1198882

)

405(1 + 120573119888)5

11990992

+

21198728

11988813

(5000+4868120573119888 + 43561205732

1198882

+ 3631205733

1198883

)

66825(1 + 120573119888)7

1199096

+ sdot sdot sdot

(57)

The constant 119888 may be obtained from the boundarycondition on the appropriate 120579(119909)The solutions in (54) (55)(56) and (57) are depicted in Figure 7

10 Mathematical Problems in Engineering

1

09

08

07

06

05

120579

119899 = 0

119899 = 1

119899 = 2

119899 = 3

0 02 04 06 08 1

119909

Figure 7 Temperature profile in a longitudinal convex parabolic finwith linear thermal conductivity and varying values of 119899 Here 120573 =05 and119872 = 15 are fixed

56 The Exponential Profile and LinearThermal ConductivityIn this section we present solutions for equation heat transferin a fin with convex parabolic profile and the thermalconductivity depending linearly on temperature That is weconsider (4) with 119896(120579) = 1 + 120573120579 and 119891(119909) = e120590119909 Here 120590is a constant The analytical solutions for this problem fordifferent values of 119899 are given by

(a) Case 119899 = 0

120579 (119909)

= 119888 +

1198722

119888

2 (1 + 120573119888)

1199092

minus

1205901198722

119888 (2 + 120573119888 + 1205732

119888)

6(1 + 120573119888)21199093

+

1198722

119888 (1198722

(1 minus 21205732

119888) + 1205902

(3 +3120573119888+1205733

1198882

+1205734

1198882

+1205732

119888 (3 + 119888)))

24(1 + 120573119888)4

1199094

+ sdot sdot sdot

(58)

(b) Case 119899 = 1

120579 (119909)

= 119888 +

11987221198882

2 (1 + 120573119888)

1199092minus

12059011987221198882(2 + 120573119888 + 120573

2119888)

6(1 + 120573119888)21199093

+

11987221198882(1198722119888 (2+120573119888minus2120573

2119888)+1205902(3+3120573119888+120573

31198882+12057341198882+1205732119888 (3+119888)))

24(1 + 120573119888)3

1199094

+ sdot sdot sdot

(59)

119899 = 0

119899 = 1

119899 = 2

119899 = 3

0 02 04 06 08 1

119909

1

095

09

085

08

075

07

120579

Figure 8 Temperature profile in a longitudinal fin of exponentialprofile with linear thermal conductivity and varying values of 119899Here 120573 = 05 and119872 = 15 are fixed

(c) Case 119899 = 2120579 (119909)

= 119888 +

11987221198883

2 (1 + 120573119888)

1199092minus

12059011987221198883(2 + 120573119888 + 120573

2119888)

6(1 + 120573119888)21199093

+

11987221198883(11987221198882(3+2120573119888minus2120573

2119888)+1205902(3+3120573119888+120573

31198882+12057341198882+1205732119888 (3+119888)))

24(1 + 120573119888)3

times1199094+ sdot sdot sdot

(60)

(d) Case 119899 = 3120579 (119909)

= 119888 +

11987221198884

2 (1 + 120573119888)

1199092minus

12059011987221198884(2 + 120573119888 + 120573

2119888)

6(1 + 120573119888)21199093

+

11987221198884(11987221198883(4+3120573119888minus2120573

2119888)+1205902(3+3120573119888+120573

31198882+12057341198882+1205732119888 (3+119888)))

24(1 + 120573119888)3

times1199094+ sdot sdot sdot

(61)

The constant 119888 may be obtained from the boundarycondition on the appropriate 120579(119909)The solutions in (58) (59)(60) and (61) are depicted in Figure 8

6 Fin Efficiency and Heat Flux

61 Fin Efficiency Theheat transfer rate from a fin is given byNewtonrsquos second law of cooling

119876 = int

119871

0

119875119867 (119879) (119879 minus 119879119886) 119889119909 (62)

Mathematical Problems in Engineering 11

Fin efficiency is defined as the ratio of the fin heat transferrate to the rate that would be if the entire fin were at the basetemperature and is given by (see eg [1])

120578=

119876

119876ideal=

int119871

0

119875119867 (119879) (119879 minus 119879119886) 119889119883

119875ℎ119887119871 (119879119887minus 119879119886)

(63)

In dimensionless variables we have

120578 = int

1

0

120579119899+1

119889119909 (64)

We consider the solutions (50) (51) (52) and (53) and depictthe fin efficiency (63) in Figure 9

62 Heat Flux The fin base heat flux is given by the Fourierrsquoslaw

119902119887= 119860119888119870 (119879)

119889119879

119889119909 (65)

The total heat flux of the fin is given by [1]

119902 =119902119887

119860119888119867(119879) (119879

119887minus 119879119886) (66)

Introducing the dimensionless variable as described inSection 2 implies

119902 =1

119861119894

119896 (120579)

ℎ (120579)

119889120579

119889119909 (67)

where the dimensionless parameter 119861119894 = ℎ119887119871119896119886is the Biot

number We consider a number of cases for the thermalconductivity and the heat transfer coefficient

621 Linear Thermal Conductivity and Power Law HeatTransfer Coefficient In this case (67) becomes

119902 =1

119861119894(1 + 120573120579) 120579

minus119899119889120579

119889119909 (68)

The heat flux in (68) at the base of the fin is plotted inFigure 10

622 Power LawThermal Conductivity and Heat TransferCoefficient In this case (67) becomes

119902 =1

119861119894120579119898minus119899

119889120579

119889119909 (69)

Not surprisingly heat flux in one-dimensional fins is highergiven values 119861119894 ≪ 1 The heat flux in (69) is plotted inFigures 11(a) 11(b) and 11(c)

7 Some Discussions

The DTM has resulted in some interesting observations andstudy We have observed in Figures 2(a) and 2(b) an excellentagreement between the analytical solutions generated byDTM and the exact solution obtained in [2] In particular we

119872

1

095

09

085

08

075

07

065

06

055

120578

2151050

119899 = 0

119899 = 1

119899 = 2

119899 = 3

Figure 9 Fin efficiency of a longitudinal rectangular fin Here 120573 =075

119872

25

20

15

10

5

0

119902

2151050

119899 = 0

119899 = 1

119899 = 2

119899 = 3

Figure 10 Base heat flux in a longitudinal rectangular finwith linearthermal conductivity Here 120573 = 01

considered a fin problem in which both thermal conductivityand heat transfer coefficient are given by the same power lawFurthermore we notice from Table 2 that an absolute errorof approximately 35119890 minus 005 is produced by DTM of order119874(15) In Table 3 an absolute error of approximately 65119890 minus006 is produced for the same order This confirms that theDTM converges faster and can provide accurate results witha minimum computation As such a tremendous confidencein the DTM in terms of the accuracy and effectiveness wasbuilt and thus we used this method to solve other problemsfor which exact solutions are harder to construct

In Figures 3(a) 3(b) 4(a) 4(b) 5(a) and 5(b) weobserve that the fin temperature increaseswith the decreasing

12 Mathematical Problems in Engineering

values of the thermogeometric fin parameter Here thevalues of the exponents are fixed Also we observe that fintemperature is higher when 119899 minus 119898 gt 0 that is when heattransfer coefficient is higher than the thermal conductivityWe observe in Figures 6 7 and 8 that the fin temperatureincreases with the increasing values of 119899 Furthermoreit appears that the fin with exponential profile performsthe least in transferring the heat from the base since thetemperature in such a fin is much higher than that ofthe rectangular and the convex parabolic profiles In otherwords heat dissipation to the fluid surrounding the extendedsurface is much faster in longitudinal fins of rectangular andconvex parabolic profiles In Figure 9 fin efficiency decreaseswith increasing thermogeometric fin parameter Also finefficiency increases with increasing values of 119899 It is easyto show that the thermogeometric fin parameter is directlyproportional to the aspect ratio (extension factor) with squareroot of the Biot number being the proportionality constantAs such shorter fins are more efficient than longer onesElse the increased Biot number results in less efficient finwhenever the space is confined that is where the length ofthe fin cannot be increased Figure 10 depicts the heat fluxat the fin base The amount of heat energy dissipated fromthe fin base is of immense interest in engineering [28] Weobserve in Figure 10 that the base heat flux increases withthe thermogeometric fin parameter for considered values ofthe exponent 119899 (see also [28]) Figures 11(a) 11(b) and 11(c)display the heat flux across the fin length We note that theheat flux across the fin length increases with increasing valuesof the thermogeometric fin parameter

8 Concluding Remarks

In this study we have successfully applied the DTM to highlynonlinear problems arising in heat transfer through longitu-dinal fins of various profiles Both thermal conductivity andheat transfer coefficient are given as functions of temperatureThe DTM agreed well with exact solutions when the thermalconductivity and heat transfer coefficient are given by thesame power law A rapid convergence to the exact solutionwas observed Following the confidence in DTM built by theresults mentioned we then solved various exciting problemsThe exotic results have been shown in tables and figures listedin this paper

The results obtained in this paper are significant improve-ments on the known results In particular both the heattransfer coefficient and the thermal conductivity are allowedto be given by the power law functions of temperature andalsowe considered a number of finprofilesWenote that exactsolutions are difficult if not impossible to construct when theexponents of these properties are distinct

Perhaps the notable advantage of the DTM is the general-ization of the Taylor method to problems involving unusualderivative procedures such as fractional fuzzy or q-derivative[22] Some generalizations have been made by Odibat et al[24] and they referred to their new method as the General-ized Differential Transform Method (GDTM) This showedgreat improvement compared to the Fractional Differential

0 02 04 06 08 1119909

45

40

35

30

25

20

15

10

5

0

minus5

119902

(a)

0 02 04 06 08 1119909

30

25

20

15

10

5

0

119902

(b)

35

30

25

20

15

10

5

0

119902

0 02 04 06 08 1119909

119872 = 025

119872 = 04

119872 = 05

119872 = 065

(c)

Figure 11 Heat flux across a longitudinal rectangular fin with linearthermal conductivity (a) 119899minus119898 lt 0 (b) 119899minus119898 gt 0 and (c) 119899minus119898 = 0Here 119861119894 = 001

Mathematical Problems in Engineering 13

Transform Method (FDTM) introduced by Arikoglu andOzkol [23]

We have shown with the help of an example that DTMmay only be applied to fin problems involving heat transferthrough fins with convex parabolic profile Note that given anODE such as (4) with a power law heat transfer coefficient ofa fractional exponent then one can easily remove the fractionby basic exponential rules and employment of the Binomialexpansion However using the DTM one runs into difficultyif the power law thermal conductivity in the same equation isgiven by the fractional exponent We do not know whetherthese observations call for the ldquomodifiedrdquo DTM to solveproblems arising in heat flow through fins with other profilessuch as longitudinal triangular and concave parabolic andalso with fractional power law thermal conductivity

The main results obtained in this paper give insightinto heat transfer in boiling liquids where the heat transfercoefficient is temperature dependent and may be given by apower law The thermal conductivity of some materials suchas gallium nitride (GaN) and Aluminium Nitride (AlN) maybe modeled by power law temperature dependency [29 30]Thus the solutions constructed here give a better comparisonof heat transfer in terms of material used since in many engi-neering applications thermal conductivity is given as a linearfunction of temperature Furthermore a good study in termsof performance and efficiency of fin with different profiles isundertaken These finding could help in the design of fins Itis claimed in [14] that DTM results are more accurate thanthose constructed by Variational Iteration Methods (VIM)and Homotopy Perturbation Methods (HPM) However itwould be risky to use the DTM approximate solutions asbenchmarks for the numerical schemes Nevertheless wehave also shown that DTM converges rapidly in just fifteenterms to the exact solution

Acknowledgments

The authors wish to thank the National Research Foundationof South Africa and the University of the Witwatersrandfor the generous financial support The authors are gratefulto Professor D P Mason and the anonymous reviewers fortheir invaluable comments which improved this paper signif-icantly

References

[1] A D Kraus A Aziz and J Welty Extended Surface Heat Trans-fer Wiley New York NY USA 2001

[2] R JMoitsheki T Hayat andM YMalik ldquoSome exact solutionsof the fin problem with a power law temperature-dependentthermal conductivityrdquo Nonlinear Analysis Real World Applica-tions vol 11 no 5 pp 3287ndash3294 2010

[3] R J Moitsheki ldquoSteady one-dimensional heat flow in a lon-gitudinal triangular and parabolic finrdquo Communications inNonlinear Science and Numerical Simulation vol 16 no 10 pp3971ndash3980 2011

[4] R J Moitsheki ldquoSteady heat transfer through a radial fin withrectangular and hyperbolic profilesrdquo Nonlinear Analysis RealWorld Applications vol 12 no 2 pp 867ndash874 2011

[5] Mo Miansari D D Ganji and Me Miansari ldquoApplication ofHersquos variational iteration method to nonlinear heat transferequationsrdquo Physics Letters A vol 372 no 6 pp 779ndash785 2008

[6] C H Chiu and C K Chen ldquoA decomposition method forsolving the convective longitudinal fins with variable thermalconductivityrdquo International Journal of Heat and Mass Transfervol 45 no 10 pp 2067ndash2075 2002

[7] A Rajabi ldquoHomotopy perturbation method for fin efficiencyof convective straight fins with temperature-dependent thermalconductivityrdquo Physics Letters A vol 364 no 1 pp 33ndash37 2007

[8] G Domairry and M Fazeli ldquoHomotopy analysis method todetermine the fin efficiency of convective straight fins withtemperature-dependent thermal conductivityrdquo Communica-tions in Nonlinear Science and Numerical Simulation vol 14 no2 pp 489ndash499 2009

[9] A Campo andR J Spaulding ldquoCoupling of themethods of suc-cessive approximations and undetermined coefficients for theprediction of the thermal behaviour of uniform circumferentialfinsrdquo Heat and Mass Transfer vol 34 no 6 pp 461ndash468 1999

[10] A Campo and F Rodrfguez ldquoApproximate analytic tempera-ture solution for uniform annular fins by adapting the powerseriesmethodrdquo International Communications inHeat andMassTransfer vol 25 no 6 pp 809ndash818 1998

[11] R Chiba ldquoApplication of differential transformmethod to ther-moelastic problem for annular disks of variable thickness withtemperature-dependent parametersrdquo International Journal ofThermophysics vol 33 pp 363ndash380 2012

[12] S Sadri M R Raveshi and S Amiri ldquoEfficiency analysis ofstraight fin with variable heat transfer coefficient and thermalconductivityrdquo Journal ofMechanical Science andTechnology vol26 no 4 pp 1283ndash1290 2012

[13] M Torabi H Yaghoori and A Aziz ldquoAnalytical solution forconvective-radiative continously moving fin with temperaturedependent thermal conductivityrdquo International Journal of Ther-mophysics vol 33 pp 924ndash941 2012

[14] M Torabi andH Yaghoobi ldquoTwo dominant analytical methodsfor thermal analysis of convective step fin with variable thermalconductivityrdquoThermal Science In press

[15] AMoradi ldquoAnalytical solutions for finwith temperature depen-dant heat transfer coefficientrdquo International Journal of Engineer-ing and Applied Sciences vol 3 no 2 pp 1ndash12 2011

[16] A Moradi and H Ahmadikia ldquoAnalytical solution for differentprofiles of fin with temperature-dependent thermal conductiv-ityrdquoMathematical Problems in Engineering vol 2010 Article ID568263 15 pages 2010

[17] H Yaghoobi and M Torabi ldquoThe application of differentialtransformation method to nonlinear equations arising in heattransferrdquo International Communications in Heat and MassTransfer vol 38 no 6 pp 815ndash820 2011

[18] S Ghafoori M Motevalli M G Nejad F Shakeri D D GanjiandM Jalaal ldquoEfficiency of differential transformation methodfor nonlinear oscillation Comparison with HPM and VIMrdquoCurrent Applied Physics vol 11 no 4 pp 965ndash971 2011

[19] A A Joneidi D D Ganji andM Babaelahi ldquoDifferential trans-formation method to determine fin efficiency of convectivestraight fins with temperature dependent thermal conductivityrdquoInternational Communications in Heat and Mass Transfer vol36 no 7 pp 757ndash762 2009

[20] J K Zhou Differential Transform Method and Its Applicationsfor Electric circuIts Huazhong University Press Wuhan China1986

14 Mathematical Problems in Engineering

[21] S Mukherjee ldquoReply to comment on lsquosolutions of the duffing-van der pol oscillator equation by the differential transformmethodrsquordquo Physica Scripta vol 84 Article ID 037003 2 pages2011

[22] C Bervillier ldquoStatus of the differential transformationmethodrdquoApplied Mathematics and Computation vol 218 no 20 pp10158ndash10170 2012

[23] A Arikoglu and I Ozkol ldquoSolution of fractional differentialequations by using differential transform methodrdquo ChaosSolitons and Fractals vol 34 no 5 pp 1473ndash1481 2007

[24] ZOdibat SMomani andV S Erturk ldquoGeneralized differentialtransform method application to differential equations offractional orderrdquo Applied Mathematics and Computation vol197 no 2 pp 467ndash477 2008

[25] CW Bert ldquoApplication of differential transformmethod to heatconduction in tapered finsrdquo Journal of Heat Transfer vol 124no 1 pp 208ndash209 2002

[26] F Khani and A Aziz ldquoThermal analysis of a longitudinal trape-zoidal fin with temperature-dependent thermal conductivityand heat transfer coefficientrdquo Communications in NonlinearScience and Numerical Simulation vol 15 no 3 pp 590ndash6012010

[27] H C Unal ldquoAn anlytical study of boiling heat transfer from afinrdquo International Journal of Heat and Mass Transfer vol 31 no7 pp 1483ndash1496 1988

[28] M H Chang ldquoA decomposition solution for fins with tempera-ture dependent surface heat fluxrdquo International Journal of Heatand Mass Transfer vol 48 no 9 pp 1819ndash1824 2005

[29] A Jezowski B A Danilchenko M Bockowski et al ldquoThermalconductivity of GaN crystals in 42ndash300 K rangerdquo Solid StateCommunications vol 128 no 2-3 pp 69ndash73 2003

[30] S Vitanov V Palankovski S Maroldt and R Quay ldquoHigh-temperature modeling of AlGaNGaN HEMTsrdquo Solid-StateElectronics vol 54 no 10 pp 1105ndash1112 2010

Submit your manuscripts athttpwwwhindawicom

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Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical Problems in Engineering

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Differential EquationsInternational Journal of

Volume 2014

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Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

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OptimizationJournal of

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CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of

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Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

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Discrete Dynamics in Nature and Society

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Decision SciencesAdvances in

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Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

Page 7: Research Article Analytical Solutions for Steady Heat ...downloads.hindawi.com/journals/mpe/2013/273052.pdf · Research Article Analytical Solutions for Steady Heat Transfer in Longitudinal

Mathematical Problems in Engineering 7

0 02 04 06 08 1094

095

096

097

098

099

1

119909

120579

119872 = 02

119872 = 03

119872 = 04

119872 = 05

(a)

0 02 04 06 08 1093

094

095

096

097

098

099

1

119909

120579

119872 = 02

119872 = 03

119872 = 04

119872 = 05

(b)

Figure 3 Temperature profile in a longitudinal fin with exponential profile and power law thermal conductivity In (a) the exponents are(119899119898) = (2 3) and (b) (119899 119898) = (3 2)

119872 = 025

119872 = 04

119872 = 05

119872 = 065

082

084

086

088

09

092

094

096

098

1

120579

0 02 04 06 08 1

119909

(a)

084

086

088

09

092

094

096

098

1

120579

0 02 04 06 08 1

119909

119872 = 025

119872 = 04

119872 = 05

119872 = 065

(b)

Figure 4 Temperature profile in a longitudinal fin with rectangular profile and power law thermal conductivity In (a) the exponents are(119899119898) = (2 3) and (b) (119899 119898) = (3 2)

53 The Convex Parabolic Profile and Power Law ThermalConductivity In this section we present solutions for theequation describing the heat transfer in a fin with con-vex parabolic profile and power law thermal conductivityEquation (40) is considered Here we consider the values(119899119898) = (2 3) (3 2) The final analytical solution is givenby

(a) Case (119899119898) = (2 3)

120579 (119909) = 119888 +21198722

311990932

minus21198724

51198881199093

+231198726

45119888211990992

minus1909119872

8

247511988831199096

+329222119872

10

2598751198884119909152

+ sdot sdot sdot

(48)

8 Mathematical Problems in Engineering

075

08

085

09

095

1

119872 = 025

119872 = 04

119872 = 05

119872 = 065

120579

0 02 04 06 08 1119909

(a)

119872 = 025

119872 = 04

119872 = 05

119872 = 065

0 02 04 06 08 1119909

082

084

086

088

09

092

094

096

098

120579

1

(b)

Figure 5 Temperature profile in a longitudinal fin with convex parabolic profile and power law thermal conductivity In (a) the exponentsare (119899119898) = (2 3) and (b) (119899119898) = (3 2)

(b) Case (119899119898) = (3 2)

120579 (119909) = 119888 +21198882

1198722

311990932

minus41198883

1198724

451199093

+41198884

1198726

2711990992

minus9921198885

1198728

74251199096

+30064119888

6

11987210

212625119909152

+ sdot sdot sdot

(49)

The constant 119888 is obtained by evaluating the appropriate 120579(119909)at the fin base boundary condition The solutions in (48) and(49) are depicted in Figures 5(a) and 5(b) respectively

54 The Rectangular Profile and LinearThermal ConductivityIn this section we present solutions for the equation repre-senting the heat transfer in a fin with rectangular profile andthe thermal conductivity depending linearly on temperatureThat is we consider (4) with 119891(119909) = 1 and 119896(120579) = 1 + 120573120579 Theanalytical solutions for this problem for different values of 119899are given by

(a) Case 119899 = 0

120579 (119909)

= 119888 +

1198722

119888

2 (1 + 120573119888)

1199092

minus

1198724

119888 (minus1 + 2120573119888)

24(1 + 120573119888)31199094

+

1198726

119888 (1 minus 16120573119888 + 281205732

1198882

)

720(1 + 120573119888)51199096

minus

1198728

119888 (minus1 + 78120573119888 minus 6001205732

1198882

+ 8961205733

1198883

)

40320(1 + 120573119888)7

1199098

+

11987210

119888 (1 minus 332120573119888 + 78121205732

1198882

minus 398961205733

1198883

+ 511841205734

1198884

)

3628800(1 + 120573119888)9

11990910

+ sdot sdot sdot

(50)

(b) Case 119899 = 1

120579 (119909)

= 119888 +

11987221198882

2 (1 + 120573119888)

1199092minus

11987241198883(minus1 + 2120573119888)

24(1 + 120573119888)31199094

+

11987261198884(10 minus 16120573119888 + 19120573

21198882)

720(1 + 120573119888)5

1199096

minus

11987281198885(minus80 + 342120573119888 minus 594120573

21198882+ 559120573

31198883)

40320(1 + 120573119888)7

1199098

+

119872101198886(1000 minus 7820120573119888 + 24336120573

21198882minus 36908120573

31198883+ 29161120573

41198884)

3628800(1 + 120573119888)9

11990910

+ sdot sdot sdot

(51)

(c) Case 119899 = 2

120579 (119909)

= 119888 +

1198722

1198883

2 (1 + 120573119888)

1199092

+

1198724

1198885

8(1 + 120573119888)31199094

+

1198726

1198887

(3 + 21205732

1198882

)

80(1 + 120573119888)51199096

+

1198728

1198889

(49 minus 20120573119888 + 661205732

1198882

minus 401205733

1198883

)

4480(1 + 120573119888)7

1199098

+

11987210

11988811

(427 minus 440120573119888 + 11161205732

1198882

minus 10201205733

1198883

+ 6721205734

1198884

)

134400(1 + 120573119888)9

11990910

+ sdot sdot sdot

(52)

Mathematical Problems in Engineering 9

(d) Case 119899 = 3120579 (119909)

= 119888 +

11987221198884

2 (1 + 120573119888)

1199092+

11987241198887(4 + 120573119888)

24(1 + 120573119888)31199094

+

119872611988810(52 + 32120573119888 + 25120573

21198882)

720(1 + 120573119888)5

1199096

+

119872811988813(1288 + 1020120573119888 + 1212120573

21198882minus 95120573

31198883)

40320(1 + 120573119888)7

1199098

+

1198721011988816(52024+45688120573119888+77184120573

21198882minus680120573

31198883+ 15025120573

41198884)

3628800(1 + 120573119888)9

11990910

+

(53)

The constant 119888may be obtained from the boundary conditionon the appropriate solution The solutions in (50) (51) (52)and (53) are depicted in Figure 6

55The Convex Parabolic Profile and LinearThermal Conduc-tivity In this section we present solutions for the equationdescribing the heat transfer in a fin with convex parabolicprofile and the thermal conductivity depending linearly ontemperature That is we consider (40) with 119896(120579) = 1 + 120573120579The analytical solution for this problem for different values of119899 is given by

(a) Case 119899 = 0

120579 (119909)

= 119888 +21198722

119888

3 (1 + 120573119888)11990932

minus21198724

119888 (minus2 + 3120573119888)

45(1 + 120573119888)3

1199093

+

1198726

119888 (2 minus 25120573119888 + 331205732

1198882

)

405(1 + 120573119888)5

11990992

minus

1198728

119888 (minus10 + 599120573119888 minus 35821205732

1198882

+ 40591205733

1198883

)

66825(1 + 120573119888)7

1199096

+ sdot sdot sdot

(54)

(b) Case 119899 = 1

120579 (119909)

= 119888 +21198722

1198882

3 (1 + 120573119888)11990932

minus21198724

1198883

(minus4 + 120573119888)

45(1 + 120573119888)3

1199093

+

21198726

1198884

(9 minus 11120573119888 + 101205732

1198882

)

405(1 + 120573119888)5

11990992

minus

21198728

1198885

(minus330 + 1118120573119888 minus 15841205732

1198882

+ 10931205733

1198883

)

66825(1 + 120573119888)7

1199096

+ sdot sdot sdot

(55)

119899 = 0

119899 = 1

119899 = 2

119899 = 3

1

095

09

085

08

075

07

065

06

055

0 02 04 06 08 1

119909

120579

Figure 6 Temperature profile in a longitudinal rectangular fin withlinear thermal conductivity and varying values of 119899 Here 120573 = 05and119872 = 15 are fixed

(c) Case 119899 = 2

120579 (119909)

= 119888 +21198722

1198883

3 (1 + 120573119888)11990932

+21198724

1198885

(6 + 120573119888)

45(1 + 120573119888)31199093

+

1198726

1198887

(16 + 3120573119888 + 71205732

1198882

)

135(1 + 120573119888)5

11990992

+

1198728

1198889

(1160 minus 107120573119888 + 11161205732

1198882

minus 3671205733

1198883

)

22275(1 + 120573119888)7

1199096

+ sdot sdot sdot

(56)

(d) Case 119899 = 3

120579 (119909)

= 119888 +21198722

1198884

3 (1 + 120573119888)11990932

+21198724

1198887

(8 + 3120573119888)

45(1 + 120573119888)31199093

+

41198726

11988810

(23 + 17120573119888 + 91205732

1198882

)

405(1 + 120573119888)5

11990992

+

21198728

11988813

(5000+4868120573119888 + 43561205732

1198882

+ 3631205733

1198883

)

66825(1 + 120573119888)7

1199096

+ sdot sdot sdot

(57)

The constant 119888 may be obtained from the boundarycondition on the appropriate 120579(119909)The solutions in (54) (55)(56) and (57) are depicted in Figure 7

10 Mathematical Problems in Engineering

1

09

08

07

06

05

120579

119899 = 0

119899 = 1

119899 = 2

119899 = 3

0 02 04 06 08 1

119909

Figure 7 Temperature profile in a longitudinal convex parabolic finwith linear thermal conductivity and varying values of 119899 Here 120573 =05 and119872 = 15 are fixed

56 The Exponential Profile and LinearThermal ConductivityIn this section we present solutions for equation heat transferin a fin with convex parabolic profile and the thermalconductivity depending linearly on temperature That is weconsider (4) with 119896(120579) = 1 + 120573120579 and 119891(119909) = e120590119909 Here 120590is a constant The analytical solutions for this problem fordifferent values of 119899 are given by

(a) Case 119899 = 0

120579 (119909)

= 119888 +

1198722

119888

2 (1 + 120573119888)

1199092

minus

1205901198722

119888 (2 + 120573119888 + 1205732

119888)

6(1 + 120573119888)21199093

+

1198722

119888 (1198722

(1 minus 21205732

119888) + 1205902

(3 +3120573119888+1205733

1198882

+1205734

1198882

+1205732

119888 (3 + 119888)))

24(1 + 120573119888)4

1199094

+ sdot sdot sdot

(58)

(b) Case 119899 = 1

120579 (119909)

= 119888 +

11987221198882

2 (1 + 120573119888)

1199092minus

12059011987221198882(2 + 120573119888 + 120573

2119888)

6(1 + 120573119888)21199093

+

11987221198882(1198722119888 (2+120573119888minus2120573

2119888)+1205902(3+3120573119888+120573

31198882+12057341198882+1205732119888 (3+119888)))

24(1 + 120573119888)3

1199094

+ sdot sdot sdot

(59)

119899 = 0

119899 = 1

119899 = 2

119899 = 3

0 02 04 06 08 1

119909

1

095

09

085

08

075

07

120579

Figure 8 Temperature profile in a longitudinal fin of exponentialprofile with linear thermal conductivity and varying values of 119899Here 120573 = 05 and119872 = 15 are fixed

(c) Case 119899 = 2120579 (119909)

= 119888 +

11987221198883

2 (1 + 120573119888)

1199092minus

12059011987221198883(2 + 120573119888 + 120573

2119888)

6(1 + 120573119888)21199093

+

11987221198883(11987221198882(3+2120573119888minus2120573

2119888)+1205902(3+3120573119888+120573

31198882+12057341198882+1205732119888 (3+119888)))

24(1 + 120573119888)3

times1199094+ sdot sdot sdot

(60)

(d) Case 119899 = 3120579 (119909)

= 119888 +

11987221198884

2 (1 + 120573119888)

1199092minus

12059011987221198884(2 + 120573119888 + 120573

2119888)

6(1 + 120573119888)21199093

+

11987221198884(11987221198883(4+3120573119888minus2120573

2119888)+1205902(3+3120573119888+120573

31198882+12057341198882+1205732119888 (3+119888)))

24(1 + 120573119888)3

times1199094+ sdot sdot sdot

(61)

The constant 119888 may be obtained from the boundarycondition on the appropriate 120579(119909)The solutions in (58) (59)(60) and (61) are depicted in Figure 8

6 Fin Efficiency and Heat Flux

61 Fin Efficiency Theheat transfer rate from a fin is given byNewtonrsquos second law of cooling

119876 = int

119871

0

119875119867 (119879) (119879 minus 119879119886) 119889119909 (62)

Mathematical Problems in Engineering 11

Fin efficiency is defined as the ratio of the fin heat transferrate to the rate that would be if the entire fin were at the basetemperature and is given by (see eg [1])

120578=

119876

119876ideal=

int119871

0

119875119867 (119879) (119879 minus 119879119886) 119889119883

119875ℎ119887119871 (119879119887minus 119879119886)

(63)

In dimensionless variables we have

120578 = int

1

0

120579119899+1

119889119909 (64)

We consider the solutions (50) (51) (52) and (53) and depictthe fin efficiency (63) in Figure 9

62 Heat Flux The fin base heat flux is given by the Fourierrsquoslaw

119902119887= 119860119888119870 (119879)

119889119879

119889119909 (65)

The total heat flux of the fin is given by [1]

119902 =119902119887

119860119888119867(119879) (119879

119887minus 119879119886) (66)

Introducing the dimensionless variable as described inSection 2 implies

119902 =1

119861119894

119896 (120579)

ℎ (120579)

119889120579

119889119909 (67)

where the dimensionless parameter 119861119894 = ℎ119887119871119896119886is the Biot

number We consider a number of cases for the thermalconductivity and the heat transfer coefficient

621 Linear Thermal Conductivity and Power Law HeatTransfer Coefficient In this case (67) becomes

119902 =1

119861119894(1 + 120573120579) 120579

minus119899119889120579

119889119909 (68)

The heat flux in (68) at the base of the fin is plotted inFigure 10

622 Power LawThermal Conductivity and Heat TransferCoefficient In this case (67) becomes

119902 =1

119861119894120579119898minus119899

119889120579

119889119909 (69)

Not surprisingly heat flux in one-dimensional fins is highergiven values 119861119894 ≪ 1 The heat flux in (69) is plotted inFigures 11(a) 11(b) and 11(c)

7 Some Discussions

The DTM has resulted in some interesting observations andstudy We have observed in Figures 2(a) and 2(b) an excellentagreement between the analytical solutions generated byDTM and the exact solution obtained in [2] In particular we

119872

1

095

09

085

08

075

07

065

06

055

120578

2151050

119899 = 0

119899 = 1

119899 = 2

119899 = 3

Figure 9 Fin efficiency of a longitudinal rectangular fin Here 120573 =075

119872

25

20

15

10

5

0

119902

2151050

119899 = 0

119899 = 1

119899 = 2

119899 = 3

Figure 10 Base heat flux in a longitudinal rectangular finwith linearthermal conductivity Here 120573 = 01

considered a fin problem in which both thermal conductivityand heat transfer coefficient are given by the same power lawFurthermore we notice from Table 2 that an absolute errorof approximately 35119890 minus 005 is produced by DTM of order119874(15) In Table 3 an absolute error of approximately 65119890 minus006 is produced for the same order This confirms that theDTM converges faster and can provide accurate results witha minimum computation As such a tremendous confidencein the DTM in terms of the accuracy and effectiveness wasbuilt and thus we used this method to solve other problemsfor which exact solutions are harder to construct

In Figures 3(a) 3(b) 4(a) 4(b) 5(a) and 5(b) weobserve that the fin temperature increaseswith the decreasing

12 Mathematical Problems in Engineering

values of the thermogeometric fin parameter Here thevalues of the exponents are fixed Also we observe that fintemperature is higher when 119899 minus 119898 gt 0 that is when heattransfer coefficient is higher than the thermal conductivityWe observe in Figures 6 7 and 8 that the fin temperatureincreases with the increasing values of 119899 Furthermoreit appears that the fin with exponential profile performsthe least in transferring the heat from the base since thetemperature in such a fin is much higher than that ofthe rectangular and the convex parabolic profiles In otherwords heat dissipation to the fluid surrounding the extendedsurface is much faster in longitudinal fins of rectangular andconvex parabolic profiles In Figure 9 fin efficiency decreaseswith increasing thermogeometric fin parameter Also finefficiency increases with increasing values of 119899 It is easyto show that the thermogeometric fin parameter is directlyproportional to the aspect ratio (extension factor) with squareroot of the Biot number being the proportionality constantAs such shorter fins are more efficient than longer onesElse the increased Biot number results in less efficient finwhenever the space is confined that is where the length ofthe fin cannot be increased Figure 10 depicts the heat fluxat the fin base The amount of heat energy dissipated fromthe fin base is of immense interest in engineering [28] Weobserve in Figure 10 that the base heat flux increases withthe thermogeometric fin parameter for considered values ofthe exponent 119899 (see also [28]) Figures 11(a) 11(b) and 11(c)display the heat flux across the fin length We note that theheat flux across the fin length increases with increasing valuesof the thermogeometric fin parameter

8 Concluding Remarks

In this study we have successfully applied the DTM to highlynonlinear problems arising in heat transfer through longitu-dinal fins of various profiles Both thermal conductivity andheat transfer coefficient are given as functions of temperatureThe DTM agreed well with exact solutions when the thermalconductivity and heat transfer coefficient are given by thesame power law A rapid convergence to the exact solutionwas observed Following the confidence in DTM built by theresults mentioned we then solved various exciting problemsThe exotic results have been shown in tables and figures listedin this paper

The results obtained in this paper are significant improve-ments on the known results In particular both the heattransfer coefficient and the thermal conductivity are allowedto be given by the power law functions of temperature andalsowe considered a number of finprofilesWenote that exactsolutions are difficult if not impossible to construct when theexponents of these properties are distinct

Perhaps the notable advantage of the DTM is the general-ization of the Taylor method to problems involving unusualderivative procedures such as fractional fuzzy or q-derivative[22] Some generalizations have been made by Odibat et al[24] and they referred to their new method as the General-ized Differential Transform Method (GDTM) This showedgreat improvement compared to the Fractional Differential

0 02 04 06 08 1119909

45

40

35

30

25

20

15

10

5

0

minus5

119902

(a)

0 02 04 06 08 1119909

30

25

20

15

10

5

0

119902

(b)

35

30

25

20

15

10

5

0

119902

0 02 04 06 08 1119909

119872 = 025

119872 = 04

119872 = 05

119872 = 065

(c)

Figure 11 Heat flux across a longitudinal rectangular fin with linearthermal conductivity (a) 119899minus119898 lt 0 (b) 119899minus119898 gt 0 and (c) 119899minus119898 = 0Here 119861119894 = 001

Mathematical Problems in Engineering 13

Transform Method (FDTM) introduced by Arikoglu andOzkol [23]

We have shown with the help of an example that DTMmay only be applied to fin problems involving heat transferthrough fins with convex parabolic profile Note that given anODE such as (4) with a power law heat transfer coefficient ofa fractional exponent then one can easily remove the fractionby basic exponential rules and employment of the Binomialexpansion However using the DTM one runs into difficultyif the power law thermal conductivity in the same equation isgiven by the fractional exponent We do not know whetherthese observations call for the ldquomodifiedrdquo DTM to solveproblems arising in heat flow through fins with other profilessuch as longitudinal triangular and concave parabolic andalso with fractional power law thermal conductivity

The main results obtained in this paper give insightinto heat transfer in boiling liquids where the heat transfercoefficient is temperature dependent and may be given by apower law The thermal conductivity of some materials suchas gallium nitride (GaN) and Aluminium Nitride (AlN) maybe modeled by power law temperature dependency [29 30]Thus the solutions constructed here give a better comparisonof heat transfer in terms of material used since in many engi-neering applications thermal conductivity is given as a linearfunction of temperature Furthermore a good study in termsof performance and efficiency of fin with different profiles isundertaken These finding could help in the design of fins Itis claimed in [14] that DTM results are more accurate thanthose constructed by Variational Iteration Methods (VIM)and Homotopy Perturbation Methods (HPM) However itwould be risky to use the DTM approximate solutions asbenchmarks for the numerical schemes Nevertheless wehave also shown that DTM converges rapidly in just fifteenterms to the exact solution

Acknowledgments

The authors wish to thank the National Research Foundationof South Africa and the University of the Witwatersrandfor the generous financial support The authors are gratefulto Professor D P Mason and the anonymous reviewers fortheir invaluable comments which improved this paper signif-icantly

References

[1] A D Kraus A Aziz and J Welty Extended Surface Heat Trans-fer Wiley New York NY USA 2001

[2] R JMoitsheki T Hayat andM YMalik ldquoSome exact solutionsof the fin problem with a power law temperature-dependentthermal conductivityrdquo Nonlinear Analysis Real World Applica-tions vol 11 no 5 pp 3287ndash3294 2010

[3] R J Moitsheki ldquoSteady one-dimensional heat flow in a lon-gitudinal triangular and parabolic finrdquo Communications inNonlinear Science and Numerical Simulation vol 16 no 10 pp3971ndash3980 2011

[4] R J Moitsheki ldquoSteady heat transfer through a radial fin withrectangular and hyperbolic profilesrdquo Nonlinear Analysis RealWorld Applications vol 12 no 2 pp 867ndash874 2011

[5] Mo Miansari D D Ganji and Me Miansari ldquoApplication ofHersquos variational iteration method to nonlinear heat transferequationsrdquo Physics Letters A vol 372 no 6 pp 779ndash785 2008

[6] C H Chiu and C K Chen ldquoA decomposition method forsolving the convective longitudinal fins with variable thermalconductivityrdquo International Journal of Heat and Mass Transfervol 45 no 10 pp 2067ndash2075 2002

[7] A Rajabi ldquoHomotopy perturbation method for fin efficiencyof convective straight fins with temperature-dependent thermalconductivityrdquo Physics Letters A vol 364 no 1 pp 33ndash37 2007

[8] G Domairry and M Fazeli ldquoHomotopy analysis method todetermine the fin efficiency of convective straight fins withtemperature-dependent thermal conductivityrdquo Communica-tions in Nonlinear Science and Numerical Simulation vol 14 no2 pp 489ndash499 2009

[9] A Campo andR J Spaulding ldquoCoupling of themethods of suc-cessive approximations and undetermined coefficients for theprediction of the thermal behaviour of uniform circumferentialfinsrdquo Heat and Mass Transfer vol 34 no 6 pp 461ndash468 1999

[10] A Campo and F Rodrfguez ldquoApproximate analytic tempera-ture solution for uniform annular fins by adapting the powerseriesmethodrdquo International Communications inHeat andMassTransfer vol 25 no 6 pp 809ndash818 1998

[11] R Chiba ldquoApplication of differential transformmethod to ther-moelastic problem for annular disks of variable thickness withtemperature-dependent parametersrdquo International Journal ofThermophysics vol 33 pp 363ndash380 2012

[12] S Sadri M R Raveshi and S Amiri ldquoEfficiency analysis ofstraight fin with variable heat transfer coefficient and thermalconductivityrdquo Journal ofMechanical Science andTechnology vol26 no 4 pp 1283ndash1290 2012

[13] M Torabi H Yaghoori and A Aziz ldquoAnalytical solution forconvective-radiative continously moving fin with temperaturedependent thermal conductivityrdquo International Journal of Ther-mophysics vol 33 pp 924ndash941 2012

[14] M Torabi andH Yaghoobi ldquoTwo dominant analytical methodsfor thermal analysis of convective step fin with variable thermalconductivityrdquoThermal Science In press

[15] AMoradi ldquoAnalytical solutions for finwith temperature depen-dant heat transfer coefficientrdquo International Journal of Engineer-ing and Applied Sciences vol 3 no 2 pp 1ndash12 2011

[16] A Moradi and H Ahmadikia ldquoAnalytical solution for differentprofiles of fin with temperature-dependent thermal conductiv-ityrdquoMathematical Problems in Engineering vol 2010 Article ID568263 15 pages 2010

[17] H Yaghoobi and M Torabi ldquoThe application of differentialtransformation method to nonlinear equations arising in heattransferrdquo International Communications in Heat and MassTransfer vol 38 no 6 pp 815ndash820 2011

[18] S Ghafoori M Motevalli M G Nejad F Shakeri D D GanjiandM Jalaal ldquoEfficiency of differential transformation methodfor nonlinear oscillation Comparison with HPM and VIMrdquoCurrent Applied Physics vol 11 no 4 pp 965ndash971 2011

[19] A A Joneidi D D Ganji andM Babaelahi ldquoDifferential trans-formation method to determine fin efficiency of convectivestraight fins with temperature dependent thermal conductivityrdquoInternational Communications in Heat and Mass Transfer vol36 no 7 pp 757ndash762 2009

[20] J K Zhou Differential Transform Method and Its Applicationsfor Electric circuIts Huazhong University Press Wuhan China1986

14 Mathematical Problems in Engineering

[21] S Mukherjee ldquoReply to comment on lsquosolutions of the duffing-van der pol oscillator equation by the differential transformmethodrsquordquo Physica Scripta vol 84 Article ID 037003 2 pages2011

[22] C Bervillier ldquoStatus of the differential transformationmethodrdquoApplied Mathematics and Computation vol 218 no 20 pp10158ndash10170 2012

[23] A Arikoglu and I Ozkol ldquoSolution of fractional differentialequations by using differential transform methodrdquo ChaosSolitons and Fractals vol 34 no 5 pp 1473ndash1481 2007

[24] ZOdibat SMomani andV S Erturk ldquoGeneralized differentialtransform method application to differential equations offractional orderrdquo Applied Mathematics and Computation vol197 no 2 pp 467ndash477 2008

[25] CW Bert ldquoApplication of differential transformmethod to heatconduction in tapered finsrdquo Journal of Heat Transfer vol 124no 1 pp 208ndash209 2002

[26] F Khani and A Aziz ldquoThermal analysis of a longitudinal trape-zoidal fin with temperature-dependent thermal conductivityand heat transfer coefficientrdquo Communications in NonlinearScience and Numerical Simulation vol 15 no 3 pp 590ndash6012010

[27] H C Unal ldquoAn anlytical study of boiling heat transfer from afinrdquo International Journal of Heat and Mass Transfer vol 31 no7 pp 1483ndash1496 1988

[28] M H Chang ldquoA decomposition solution for fins with tempera-ture dependent surface heat fluxrdquo International Journal of Heatand Mass Transfer vol 48 no 9 pp 1819ndash1824 2005

[29] A Jezowski B A Danilchenko M Bockowski et al ldquoThermalconductivity of GaN crystals in 42ndash300 K rangerdquo Solid StateCommunications vol 128 no 2-3 pp 69ndash73 2003

[30] S Vitanov V Palankovski S Maroldt and R Quay ldquoHigh-temperature modeling of AlGaNGaN HEMTsrdquo Solid-StateElectronics vol 54 no 10 pp 1105ndash1112 2010

Submit your manuscripts athttpwwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

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Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical Problems in Engineering

Hindawi Publishing Corporationhttpwwwhindawicom

Differential EquationsInternational Journal of

Volume 2014

Applied MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical PhysicsAdvances in

Complex AnalysisJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

OptimizationJournal of

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CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of

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Operations ResearchAdvances in

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Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of Mathematics and Mathematical Sciences

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The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Algebra

Discrete Dynamics in Nature and Society

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Decision SciencesAdvances in

Discrete MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom

Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

Page 8: Research Article Analytical Solutions for Steady Heat ...downloads.hindawi.com/journals/mpe/2013/273052.pdf · Research Article Analytical Solutions for Steady Heat Transfer in Longitudinal

8 Mathematical Problems in Engineering

075

08

085

09

095

1

119872 = 025

119872 = 04

119872 = 05

119872 = 065

120579

0 02 04 06 08 1119909

(a)

119872 = 025

119872 = 04

119872 = 05

119872 = 065

0 02 04 06 08 1119909

082

084

086

088

09

092

094

096

098

120579

1

(b)

Figure 5 Temperature profile in a longitudinal fin with convex parabolic profile and power law thermal conductivity In (a) the exponentsare (119899119898) = (2 3) and (b) (119899119898) = (3 2)

(b) Case (119899119898) = (3 2)

120579 (119909) = 119888 +21198882

1198722

311990932

minus41198883

1198724

451199093

+41198884

1198726

2711990992

minus9921198885

1198728

74251199096

+30064119888

6

11987210

212625119909152

+ sdot sdot sdot

(49)

The constant 119888 is obtained by evaluating the appropriate 120579(119909)at the fin base boundary condition The solutions in (48) and(49) are depicted in Figures 5(a) and 5(b) respectively

54 The Rectangular Profile and LinearThermal ConductivityIn this section we present solutions for the equation repre-senting the heat transfer in a fin with rectangular profile andthe thermal conductivity depending linearly on temperatureThat is we consider (4) with 119891(119909) = 1 and 119896(120579) = 1 + 120573120579 Theanalytical solutions for this problem for different values of 119899are given by

(a) Case 119899 = 0

120579 (119909)

= 119888 +

1198722

119888

2 (1 + 120573119888)

1199092

minus

1198724

119888 (minus1 + 2120573119888)

24(1 + 120573119888)31199094

+

1198726

119888 (1 minus 16120573119888 + 281205732

1198882

)

720(1 + 120573119888)51199096

minus

1198728

119888 (minus1 + 78120573119888 minus 6001205732

1198882

+ 8961205733

1198883

)

40320(1 + 120573119888)7

1199098

+

11987210

119888 (1 minus 332120573119888 + 78121205732

1198882

minus 398961205733

1198883

+ 511841205734

1198884

)

3628800(1 + 120573119888)9

11990910

+ sdot sdot sdot

(50)

(b) Case 119899 = 1

120579 (119909)

= 119888 +

11987221198882

2 (1 + 120573119888)

1199092minus

11987241198883(minus1 + 2120573119888)

24(1 + 120573119888)31199094

+

11987261198884(10 minus 16120573119888 + 19120573

21198882)

720(1 + 120573119888)5

1199096

minus

11987281198885(minus80 + 342120573119888 minus 594120573

21198882+ 559120573

31198883)

40320(1 + 120573119888)7

1199098

+

119872101198886(1000 minus 7820120573119888 + 24336120573

21198882minus 36908120573

31198883+ 29161120573

41198884)

3628800(1 + 120573119888)9

11990910

+ sdot sdot sdot

(51)

(c) Case 119899 = 2

120579 (119909)

= 119888 +

1198722

1198883

2 (1 + 120573119888)

1199092

+

1198724

1198885

8(1 + 120573119888)31199094

+

1198726

1198887

(3 + 21205732

1198882

)

80(1 + 120573119888)51199096

+

1198728

1198889

(49 minus 20120573119888 + 661205732

1198882

minus 401205733

1198883

)

4480(1 + 120573119888)7

1199098

+

11987210

11988811

(427 minus 440120573119888 + 11161205732

1198882

minus 10201205733

1198883

+ 6721205734

1198884

)

134400(1 + 120573119888)9

11990910

+ sdot sdot sdot

(52)

Mathematical Problems in Engineering 9

(d) Case 119899 = 3120579 (119909)

= 119888 +

11987221198884

2 (1 + 120573119888)

1199092+

11987241198887(4 + 120573119888)

24(1 + 120573119888)31199094

+

119872611988810(52 + 32120573119888 + 25120573

21198882)

720(1 + 120573119888)5

1199096

+

119872811988813(1288 + 1020120573119888 + 1212120573

21198882minus 95120573

31198883)

40320(1 + 120573119888)7

1199098

+

1198721011988816(52024+45688120573119888+77184120573

21198882minus680120573

31198883+ 15025120573

41198884)

3628800(1 + 120573119888)9

11990910

+

(53)

The constant 119888may be obtained from the boundary conditionon the appropriate solution The solutions in (50) (51) (52)and (53) are depicted in Figure 6

55The Convex Parabolic Profile and LinearThermal Conduc-tivity In this section we present solutions for the equationdescribing the heat transfer in a fin with convex parabolicprofile and the thermal conductivity depending linearly ontemperature That is we consider (40) with 119896(120579) = 1 + 120573120579The analytical solution for this problem for different values of119899 is given by

(a) Case 119899 = 0

120579 (119909)

= 119888 +21198722

119888

3 (1 + 120573119888)11990932

minus21198724

119888 (minus2 + 3120573119888)

45(1 + 120573119888)3

1199093

+

1198726

119888 (2 minus 25120573119888 + 331205732

1198882

)

405(1 + 120573119888)5

11990992

minus

1198728

119888 (minus10 + 599120573119888 minus 35821205732

1198882

+ 40591205733

1198883

)

66825(1 + 120573119888)7

1199096

+ sdot sdot sdot

(54)

(b) Case 119899 = 1

120579 (119909)

= 119888 +21198722

1198882

3 (1 + 120573119888)11990932

minus21198724

1198883

(minus4 + 120573119888)

45(1 + 120573119888)3

1199093

+

21198726

1198884

(9 minus 11120573119888 + 101205732

1198882

)

405(1 + 120573119888)5

11990992

minus

21198728

1198885

(minus330 + 1118120573119888 minus 15841205732

1198882

+ 10931205733

1198883

)

66825(1 + 120573119888)7

1199096

+ sdot sdot sdot

(55)

119899 = 0

119899 = 1

119899 = 2

119899 = 3

1

095

09

085

08

075

07

065

06

055

0 02 04 06 08 1

119909

120579

Figure 6 Temperature profile in a longitudinal rectangular fin withlinear thermal conductivity and varying values of 119899 Here 120573 = 05and119872 = 15 are fixed

(c) Case 119899 = 2

120579 (119909)

= 119888 +21198722

1198883

3 (1 + 120573119888)11990932

+21198724

1198885

(6 + 120573119888)

45(1 + 120573119888)31199093

+

1198726

1198887

(16 + 3120573119888 + 71205732

1198882

)

135(1 + 120573119888)5

11990992

+

1198728

1198889

(1160 minus 107120573119888 + 11161205732

1198882

minus 3671205733

1198883

)

22275(1 + 120573119888)7

1199096

+ sdot sdot sdot

(56)

(d) Case 119899 = 3

120579 (119909)

= 119888 +21198722

1198884

3 (1 + 120573119888)11990932

+21198724

1198887

(8 + 3120573119888)

45(1 + 120573119888)31199093

+

41198726

11988810

(23 + 17120573119888 + 91205732

1198882

)

405(1 + 120573119888)5

11990992

+

21198728

11988813

(5000+4868120573119888 + 43561205732

1198882

+ 3631205733

1198883

)

66825(1 + 120573119888)7

1199096

+ sdot sdot sdot

(57)

The constant 119888 may be obtained from the boundarycondition on the appropriate 120579(119909)The solutions in (54) (55)(56) and (57) are depicted in Figure 7

10 Mathematical Problems in Engineering

1

09

08

07

06

05

120579

119899 = 0

119899 = 1

119899 = 2

119899 = 3

0 02 04 06 08 1

119909

Figure 7 Temperature profile in a longitudinal convex parabolic finwith linear thermal conductivity and varying values of 119899 Here 120573 =05 and119872 = 15 are fixed

56 The Exponential Profile and LinearThermal ConductivityIn this section we present solutions for equation heat transferin a fin with convex parabolic profile and the thermalconductivity depending linearly on temperature That is weconsider (4) with 119896(120579) = 1 + 120573120579 and 119891(119909) = e120590119909 Here 120590is a constant The analytical solutions for this problem fordifferent values of 119899 are given by

(a) Case 119899 = 0

120579 (119909)

= 119888 +

1198722

119888

2 (1 + 120573119888)

1199092

minus

1205901198722

119888 (2 + 120573119888 + 1205732

119888)

6(1 + 120573119888)21199093

+

1198722

119888 (1198722

(1 minus 21205732

119888) + 1205902

(3 +3120573119888+1205733

1198882

+1205734

1198882

+1205732

119888 (3 + 119888)))

24(1 + 120573119888)4

1199094

+ sdot sdot sdot

(58)

(b) Case 119899 = 1

120579 (119909)

= 119888 +

11987221198882

2 (1 + 120573119888)

1199092minus

12059011987221198882(2 + 120573119888 + 120573

2119888)

6(1 + 120573119888)21199093

+

11987221198882(1198722119888 (2+120573119888minus2120573

2119888)+1205902(3+3120573119888+120573

31198882+12057341198882+1205732119888 (3+119888)))

24(1 + 120573119888)3

1199094

+ sdot sdot sdot

(59)

119899 = 0

119899 = 1

119899 = 2

119899 = 3

0 02 04 06 08 1

119909

1

095

09

085

08

075

07

120579

Figure 8 Temperature profile in a longitudinal fin of exponentialprofile with linear thermal conductivity and varying values of 119899Here 120573 = 05 and119872 = 15 are fixed

(c) Case 119899 = 2120579 (119909)

= 119888 +

11987221198883

2 (1 + 120573119888)

1199092minus

12059011987221198883(2 + 120573119888 + 120573

2119888)

6(1 + 120573119888)21199093

+

11987221198883(11987221198882(3+2120573119888minus2120573

2119888)+1205902(3+3120573119888+120573

31198882+12057341198882+1205732119888 (3+119888)))

24(1 + 120573119888)3

times1199094+ sdot sdot sdot

(60)

(d) Case 119899 = 3120579 (119909)

= 119888 +

11987221198884

2 (1 + 120573119888)

1199092minus

12059011987221198884(2 + 120573119888 + 120573

2119888)

6(1 + 120573119888)21199093

+

11987221198884(11987221198883(4+3120573119888minus2120573

2119888)+1205902(3+3120573119888+120573

31198882+12057341198882+1205732119888 (3+119888)))

24(1 + 120573119888)3

times1199094+ sdot sdot sdot

(61)

The constant 119888 may be obtained from the boundarycondition on the appropriate 120579(119909)The solutions in (58) (59)(60) and (61) are depicted in Figure 8

6 Fin Efficiency and Heat Flux

61 Fin Efficiency Theheat transfer rate from a fin is given byNewtonrsquos second law of cooling

119876 = int

119871

0

119875119867 (119879) (119879 minus 119879119886) 119889119909 (62)

Mathematical Problems in Engineering 11

Fin efficiency is defined as the ratio of the fin heat transferrate to the rate that would be if the entire fin were at the basetemperature and is given by (see eg [1])

120578=

119876

119876ideal=

int119871

0

119875119867 (119879) (119879 minus 119879119886) 119889119883

119875ℎ119887119871 (119879119887minus 119879119886)

(63)

In dimensionless variables we have

120578 = int

1

0

120579119899+1

119889119909 (64)

We consider the solutions (50) (51) (52) and (53) and depictthe fin efficiency (63) in Figure 9

62 Heat Flux The fin base heat flux is given by the Fourierrsquoslaw

119902119887= 119860119888119870 (119879)

119889119879

119889119909 (65)

The total heat flux of the fin is given by [1]

119902 =119902119887

119860119888119867(119879) (119879

119887minus 119879119886) (66)

Introducing the dimensionless variable as described inSection 2 implies

119902 =1

119861119894

119896 (120579)

ℎ (120579)

119889120579

119889119909 (67)

where the dimensionless parameter 119861119894 = ℎ119887119871119896119886is the Biot

number We consider a number of cases for the thermalconductivity and the heat transfer coefficient

621 Linear Thermal Conductivity and Power Law HeatTransfer Coefficient In this case (67) becomes

119902 =1

119861119894(1 + 120573120579) 120579

minus119899119889120579

119889119909 (68)

The heat flux in (68) at the base of the fin is plotted inFigure 10

622 Power LawThermal Conductivity and Heat TransferCoefficient In this case (67) becomes

119902 =1

119861119894120579119898minus119899

119889120579

119889119909 (69)

Not surprisingly heat flux in one-dimensional fins is highergiven values 119861119894 ≪ 1 The heat flux in (69) is plotted inFigures 11(a) 11(b) and 11(c)

7 Some Discussions

The DTM has resulted in some interesting observations andstudy We have observed in Figures 2(a) and 2(b) an excellentagreement between the analytical solutions generated byDTM and the exact solution obtained in [2] In particular we

119872

1

095

09

085

08

075

07

065

06

055

120578

2151050

119899 = 0

119899 = 1

119899 = 2

119899 = 3

Figure 9 Fin efficiency of a longitudinal rectangular fin Here 120573 =075

119872

25

20

15

10

5

0

119902

2151050

119899 = 0

119899 = 1

119899 = 2

119899 = 3

Figure 10 Base heat flux in a longitudinal rectangular finwith linearthermal conductivity Here 120573 = 01

considered a fin problem in which both thermal conductivityand heat transfer coefficient are given by the same power lawFurthermore we notice from Table 2 that an absolute errorof approximately 35119890 minus 005 is produced by DTM of order119874(15) In Table 3 an absolute error of approximately 65119890 minus006 is produced for the same order This confirms that theDTM converges faster and can provide accurate results witha minimum computation As such a tremendous confidencein the DTM in terms of the accuracy and effectiveness wasbuilt and thus we used this method to solve other problemsfor which exact solutions are harder to construct

In Figures 3(a) 3(b) 4(a) 4(b) 5(a) and 5(b) weobserve that the fin temperature increaseswith the decreasing

12 Mathematical Problems in Engineering

values of the thermogeometric fin parameter Here thevalues of the exponents are fixed Also we observe that fintemperature is higher when 119899 minus 119898 gt 0 that is when heattransfer coefficient is higher than the thermal conductivityWe observe in Figures 6 7 and 8 that the fin temperatureincreases with the increasing values of 119899 Furthermoreit appears that the fin with exponential profile performsthe least in transferring the heat from the base since thetemperature in such a fin is much higher than that ofthe rectangular and the convex parabolic profiles In otherwords heat dissipation to the fluid surrounding the extendedsurface is much faster in longitudinal fins of rectangular andconvex parabolic profiles In Figure 9 fin efficiency decreaseswith increasing thermogeometric fin parameter Also finefficiency increases with increasing values of 119899 It is easyto show that the thermogeometric fin parameter is directlyproportional to the aspect ratio (extension factor) with squareroot of the Biot number being the proportionality constantAs such shorter fins are more efficient than longer onesElse the increased Biot number results in less efficient finwhenever the space is confined that is where the length ofthe fin cannot be increased Figure 10 depicts the heat fluxat the fin base The amount of heat energy dissipated fromthe fin base is of immense interest in engineering [28] Weobserve in Figure 10 that the base heat flux increases withthe thermogeometric fin parameter for considered values ofthe exponent 119899 (see also [28]) Figures 11(a) 11(b) and 11(c)display the heat flux across the fin length We note that theheat flux across the fin length increases with increasing valuesof the thermogeometric fin parameter

8 Concluding Remarks

In this study we have successfully applied the DTM to highlynonlinear problems arising in heat transfer through longitu-dinal fins of various profiles Both thermal conductivity andheat transfer coefficient are given as functions of temperatureThe DTM agreed well with exact solutions when the thermalconductivity and heat transfer coefficient are given by thesame power law A rapid convergence to the exact solutionwas observed Following the confidence in DTM built by theresults mentioned we then solved various exciting problemsThe exotic results have been shown in tables and figures listedin this paper

The results obtained in this paper are significant improve-ments on the known results In particular both the heattransfer coefficient and the thermal conductivity are allowedto be given by the power law functions of temperature andalsowe considered a number of finprofilesWenote that exactsolutions are difficult if not impossible to construct when theexponents of these properties are distinct

Perhaps the notable advantage of the DTM is the general-ization of the Taylor method to problems involving unusualderivative procedures such as fractional fuzzy or q-derivative[22] Some generalizations have been made by Odibat et al[24] and they referred to their new method as the General-ized Differential Transform Method (GDTM) This showedgreat improvement compared to the Fractional Differential

0 02 04 06 08 1119909

45

40

35

30

25

20

15

10

5

0

minus5

119902

(a)

0 02 04 06 08 1119909

30

25

20

15

10

5

0

119902

(b)

35

30

25

20

15

10

5

0

119902

0 02 04 06 08 1119909

119872 = 025

119872 = 04

119872 = 05

119872 = 065

(c)

Figure 11 Heat flux across a longitudinal rectangular fin with linearthermal conductivity (a) 119899minus119898 lt 0 (b) 119899minus119898 gt 0 and (c) 119899minus119898 = 0Here 119861119894 = 001

Mathematical Problems in Engineering 13

Transform Method (FDTM) introduced by Arikoglu andOzkol [23]

We have shown with the help of an example that DTMmay only be applied to fin problems involving heat transferthrough fins with convex parabolic profile Note that given anODE such as (4) with a power law heat transfer coefficient ofa fractional exponent then one can easily remove the fractionby basic exponential rules and employment of the Binomialexpansion However using the DTM one runs into difficultyif the power law thermal conductivity in the same equation isgiven by the fractional exponent We do not know whetherthese observations call for the ldquomodifiedrdquo DTM to solveproblems arising in heat flow through fins with other profilessuch as longitudinal triangular and concave parabolic andalso with fractional power law thermal conductivity

The main results obtained in this paper give insightinto heat transfer in boiling liquids where the heat transfercoefficient is temperature dependent and may be given by apower law The thermal conductivity of some materials suchas gallium nitride (GaN) and Aluminium Nitride (AlN) maybe modeled by power law temperature dependency [29 30]Thus the solutions constructed here give a better comparisonof heat transfer in terms of material used since in many engi-neering applications thermal conductivity is given as a linearfunction of temperature Furthermore a good study in termsof performance and efficiency of fin with different profiles isundertaken These finding could help in the design of fins Itis claimed in [14] that DTM results are more accurate thanthose constructed by Variational Iteration Methods (VIM)and Homotopy Perturbation Methods (HPM) However itwould be risky to use the DTM approximate solutions asbenchmarks for the numerical schemes Nevertheless wehave also shown that DTM converges rapidly in just fifteenterms to the exact solution

Acknowledgments

The authors wish to thank the National Research Foundationof South Africa and the University of the Witwatersrandfor the generous financial support The authors are gratefulto Professor D P Mason and the anonymous reviewers fortheir invaluable comments which improved this paper signif-icantly

References

[1] A D Kraus A Aziz and J Welty Extended Surface Heat Trans-fer Wiley New York NY USA 2001

[2] R JMoitsheki T Hayat andM YMalik ldquoSome exact solutionsof the fin problem with a power law temperature-dependentthermal conductivityrdquo Nonlinear Analysis Real World Applica-tions vol 11 no 5 pp 3287ndash3294 2010

[3] R J Moitsheki ldquoSteady one-dimensional heat flow in a lon-gitudinal triangular and parabolic finrdquo Communications inNonlinear Science and Numerical Simulation vol 16 no 10 pp3971ndash3980 2011

[4] R J Moitsheki ldquoSteady heat transfer through a radial fin withrectangular and hyperbolic profilesrdquo Nonlinear Analysis RealWorld Applications vol 12 no 2 pp 867ndash874 2011

[5] Mo Miansari D D Ganji and Me Miansari ldquoApplication ofHersquos variational iteration method to nonlinear heat transferequationsrdquo Physics Letters A vol 372 no 6 pp 779ndash785 2008

[6] C H Chiu and C K Chen ldquoA decomposition method forsolving the convective longitudinal fins with variable thermalconductivityrdquo International Journal of Heat and Mass Transfervol 45 no 10 pp 2067ndash2075 2002

[7] A Rajabi ldquoHomotopy perturbation method for fin efficiencyof convective straight fins with temperature-dependent thermalconductivityrdquo Physics Letters A vol 364 no 1 pp 33ndash37 2007

[8] G Domairry and M Fazeli ldquoHomotopy analysis method todetermine the fin efficiency of convective straight fins withtemperature-dependent thermal conductivityrdquo Communica-tions in Nonlinear Science and Numerical Simulation vol 14 no2 pp 489ndash499 2009

[9] A Campo andR J Spaulding ldquoCoupling of themethods of suc-cessive approximations and undetermined coefficients for theprediction of the thermal behaviour of uniform circumferentialfinsrdquo Heat and Mass Transfer vol 34 no 6 pp 461ndash468 1999

[10] A Campo and F Rodrfguez ldquoApproximate analytic tempera-ture solution for uniform annular fins by adapting the powerseriesmethodrdquo International Communications inHeat andMassTransfer vol 25 no 6 pp 809ndash818 1998

[11] R Chiba ldquoApplication of differential transformmethod to ther-moelastic problem for annular disks of variable thickness withtemperature-dependent parametersrdquo International Journal ofThermophysics vol 33 pp 363ndash380 2012

[12] S Sadri M R Raveshi and S Amiri ldquoEfficiency analysis ofstraight fin with variable heat transfer coefficient and thermalconductivityrdquo Journal ofMechanical Science andTechnology vol26 no 4 pp 1283ndash1290 2012

[13] M Torabi H Yaghoori and A Aziz ldquoAnalytical solution forconvective-radiative continously moving fin with temperaturedependent thermal conductivityrdquo International Journal of Ther-mophysics vol 33 pp 924ndash941 2012

[14] M Torabi andH Yaghoobi ldquoTwo dominant analytical methodsfor thermal analysis of convective step fin with variable thermalconductivityrdquoThermal Science In press

[15] AMoradi ldquoAnalytical solutions for finwith temperature depen-dant heat transfer coefficientrdquo International Journal of Engineer-ing and Applied Sciences vol 3 no 2 pp 1ndash12 2011

[16] A Moradi and H Ahmadikia ldquoAnalytical solution for differentprofiles of fin with temperature-dependent thermal conductiv-ityrdquoMathematical Problems in Engineering vol 2010 Article ID568263 15 pages 2010

[17] H Yaghoobi and M Torabi ldquoThe application of differentialtransformation method to nonlinear equations arising in heattransferrdquo International Communications in Heat and MassTransfer vol 38 no 6 pp 815ndash820 2011

[18] S Ghafoori M Motevalli M G Nejad F Shakeri D D GanjiandM Jalaal ldquoEfficiency of differential transformation methodfor nonlinear oscillation Comparison with HPM and VIMrdquoCurrent Applied Physics vol 11 no 4 pp 965ndash971 2011

[19] A A Joneidi D D Ganji andM Babaelahi ldquoDifferential trans-formation method to determine fin efficiency of convectivestraight fins with temperature dependent thermal conductivityrdquoInternational Communications in Heat and Mass Transfer vol36 no 7 pp 757ndash762 2009

[20] J K Zhou Differential Transform Method and Its Applicationsfor Electric circuIts Huazhong University Press Wuhan China1986

14 Mathematical Problems in Engineering

[21] S Mukherjee ldquoReply to comment on lsquosolutions of the duffing-van der pol oscillator equation by the differential transformmethodrsquordquo Physica Scripta vol 84 Article ID 037003 2 pages2011

[22] C Bervillier ldquoStatus of the differential transformationmethodrdquoApplied Mathematics and Computation vol 218 no 20 pp10158ndash10170 2012

[23] A Arikoglu and I Ozkol ldquoSolution of fractional differentialequations by using differential transform methodrdquo ChaosSolitons and Fractals vol 34 no 5 pp 1473ndash1481 2007

[24] ZOdibat SMomani andV S Erturk ldquoGeneralized differentialtransform method application to differential equations offractional orderrdquo Applied Mathematics and Computation vol197 no 2 pp 467ndash477 2008

[25] CW Bert ldquoApplication of differential transformmethod to heatconduction in tapered finsrdquo Journal of Heat Transfer vol 124no 1 pp 208ndash209 2002

[26] F Khani and A Aziz ldquoThermal analysis of a longitudinal trape-zoidal fin with temperature-dependent thermal conductivityand heat transfer coefficientrdquo Communications in NonlinearScience and Numerical Simulation vol 15 no 3 pp 590ndash6012010

[27] H C Unal ldquoAn anlytical study of boiling heat transfer from afinrdquo International Journal of Heat and Mass Transfer vol 31 no7 pp 1483ndash1496 1988

[28] M H Chang ldquoA decomposition solution for fins with tempera-ture dependent surface heat fluxrdquo International Journal of Heatand Mass Transfer vol 48 no 9 pp 1819ndash1824 2005

[29] A Jezowski B A Danilchenko M Bockowski et al ldquoThermalconductivity of GaN crystals in 42ndash300 K rangerdquo Solid StateCommunications vol 128 no 2-3 pp 69ndash73 2003

[30] S Vitanov V Palankovski S Maroldt and R Quay ldquoHigh-temperature modeling of AlGaNGaN HEMTsrdquo Solid-StateElectronics vol 54 no 10 pp 1105ndash1112 2010

Submit your manuscripts athttpwwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical Problems in Engineering

Hindawi Publishing Corporationhttpwwwhindawicom

Differential EquationsInternational Journal of

Volume 2014

Applied MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical PhysicsAdvances in

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Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

OptimizationJournal of

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CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of

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Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

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The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Algebra

Discrete Dynamics in Nature and Society

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

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Decision SciencesAdvances in

Discrete MathematicsJournal of

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Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

Page 9: Research Article Analytical Solutions for Steady Heat ...downloads.hindawi.com/journals/mpe/2013/273052.pdf · Research Article Analytical Solutions for Steady Heat Transfer in Longitudinal

Mathematical Problems in Engineering 9

(d) Case 119899 = 3120579 (119909)

= 119888 +

11987221198884

2 (1 + 120573119888)

1199092+

11987241198887(4 + 120573119888)

24(1 + 120573119888)31199094

+

119872611988810(52 + 32120573119888 + 25120573

21198882)

720(1 + 120573119888)5

1199096

+

119872811988813(1288 + 1020120573119888 + 1212120573

21198882minus 95120573

31198883)

40320(1 + 120573119888)7

1199098

+

1198721011988816(52024+45688120573119888+77184120573

21198882minus680120573

31198883+ 15025120573

41198884)

3628800(1 + 120573119888)9

11990910

+

(53)

The constant 119888may be obtained from the boundary conditionon the appropriate solution The solutions in (50) (51) (52)and (53) are depicted in Figure 6

55The Convex Parabolic Profile and LinearThermal Conduc-tivity In this section we present solutions for the equationdescribing the heat transfer in a fin with convex parabolicprofile and the thermal conductivity depending linearly ontemperature That is we consider (40) with 119896(120579) = 1 + 120573120579The analytical solution for this problem for different values of119899 is given by

(a) Case 119899 = 0

120579 (119909)

= 119888 +21198722

119888

3 (1 + 120573119888)11990932

minus21198724

119888 (minus2 + 3120573119888)

45(1 + 120573119888)3

1199093

+

1198726

119888 (2 minus 25120573119888 + 331205732

1198882

)

405(1 + 120573119888)5

11990992

minus

1198728

119888 (minus10 + 599120573119888 minus 35821205732

1198882

+ 40591205733

1198883

)

66825(1 + 120573119888)7

1199096

+ sdot sdot sdot

(54)

(b) Case 119899 = 1

120579 (119909)

= 119888 +21198722

1198882

3 (1 + 120573119888)11990932

minus21198724

1198883

(minus4 + 120573119888)

45(1 + 120573119888)3

1199093

+

21198726

1198884

(9 minus 11120573119888 + 101205732

1198882

)

405(1 + 120573119888)5

11990992

minus

21198728

1198885

(minus330 + 1118120573119888 minus 15841205732

1198882

+ 10931205733

1198883

)

66825(1 + 120573119888)7

1199096

+ sdot sdot sdot

(55)

119899 = 0

119899 = 1

119899 = 2

119899 = 3

1

095

09

085

08

075

07

065

06

055

0 02 04 06 08 1

119909

120579

Figure 6 Temperature profile in a longitudinal rectangular fin withlinear thermal conductivity and varying values of 119899 Here 120573 = 05and119872 = 15 are fixed

(c) Case 119899 = 2

120579 (119909)

= 119888 +21198722

1198883

3 (1 + 120573119888)11990932

+21198724

1198885

(6 + 120573119888)

45(1 + 120573119888)31199093

+

1198726

1198887

(16 + 3120573119888 + 71205732

1198882

)

135(1 + 120573119888)5

11990992

+

1198728

1198889

(1160 minus 107120573119888 + 11161205732

1198882

minus 3671205733

1198883

)

22275(1 + 120573119888)7

1199096

+ sdot sdot sdot

(56)

(d) Case 119899 = 3

120579 (119909)

= 119888 +21198722

1198884

3 (1 + 120573119888)11990932

+21198724

1198887

(8 + 3120573119888)

45(1 + 120573119888)31199093

+

41198726

11988810

(23 + 17120573119888 + 91205732

1198882

)

405(1 + 120573119888)5

11990992

+

21198728

11988813

(5000+4868120573119888 + 43561205732

1198882

+ 3631205733

1198883

)

66825(1 + 120573119888)7

1199096

+ sdot sdot sdot

(57)

The constant 119888 may be obtained from the boundarycondition on the appropriate 120579(119909)The solutions in (54) (55)(56) and (57) are depicted in Figure 7

10 Mathematical Problems in Engineering

1

09

08

07

06

05

120579

119899 = 0

119899 = 1

119899 = 2

119899 = 3

0 02 04 06 08 1

119909

Figure 7 Temperature profile in a longitudinal convex parabolic finwith linear thermal conductivity and varying values of 119899 Here 120573 =05 and119872 = 15 are fixed

56 The Exponential Profile and LinearThermal ConductivityIn this section we present solutions for equation heat transferin a fin with convex parabolic profile and the thermalconductivity depending linearly on temperature That is weconsider (4) with 119896(120579) = 1 + 120573120579 and 119891(119909) = e120590119909 Here 120590is a constant The analytical solutions for this problem fordifferent values of 119899 are given by

(a) Case 119899 = 0

120579 (119909)

= 119888 +

1198722

119888

2 (1 + 120573119888)

1199092

minus

1205901198722

119888 (2 + 120573119888 + 1205732

119888)

6(1 + 120573119888)21199093

+

1198722

119888 (1198722

(1 minus 21205732

119888) + 1205902

(3 +3120573119888+1205733

1198882

+1205734

1198882

+1205732

119888 (3 + 119888)))

24(1 + 120573119888)4

1199094

+ sdot sdot sdot

(58)

(b) Case 119899 = 1

120579 (119909)

= 119888 +

11987221198882

2 (1 + 120573119888)

1199092minus

12059011987221198882(2 + 120573119888 + 120573

2119888)

6(1 + 120573119888)21199093

+

11987221198882(1198722119888 (2+120573119888minus2120573

2119888)+1205902(3+3120573119888+120573

31198882+12057341198882+1205732119888 (3+119888)))

24(1 + 120573119888)3

1199094

+ sdot sdot sdot

(59)

119899 = 0

119899 = 1

119899 = 2

119899 = 3

0 02 04 06 08 1

119909

1

095

09

085

08

075

07

120579

Figure 8 Temperature profile in a longitudinal fin of exponentialprofile with linear thermal conductivity and varying values of 119899Here 120573 = 05 and119872 = 15 are fixed

(c) Case 119899 = 2120579 (119909)

= 119888 +

11987221198883

2 (1 + 120573119888)

1199092minus

12059011987221198883(2 + 120573119888 + 120573

2119888)

6(1 + 120573119888)21199093

+

11987221198883(11987221198882(3+2120573119888minus2120573

2119888)+1205902(3+3120573119888+120573

31198882+12057341198882+1205732119888 (3+119888)))

24(1 + 120573119888)3

times1199094+ sdot sdot sdot

(60)

(d) Case 119899 = 3120579 (119909)

= 119888 +

11987221198884

2 (1 + 120573119888)

1199092minus

12059011987221198884(2 + 120573119888 + 120573

2119888)

6(1 + 120573119888)21199093

+

11987221198884(11987221198883(4+3120573119888minus2120573

2119888)+1205902(3+3120573119888+120573

31198882+12057341198882+1205732119888 (3+119888)))

24(1 + 120573119888)3

times1199094+ sdot sdot sdot

(61)

The constant 119888 may be obtained from the boundarycondition on the appropriate 120579(119909)The solutions in (58) (59)(60) and (61) are depicted in Figure 8

6 Fin Efficiency and Heat Flux

61 Fin Efficiency Theheat transfer rate from a fin is given byNewtonrsquos second law of cooling

119876 = int

119871

0

119875119867 (119879) (119879 minus 119879119886) 119889119909 (62)

Mathematical Problems in Engineering 11

Fin efficiency is defined as the ratio of the fin heat transferrate to the rate that would be if the entire fin were at the basetemperature and is given by (see eg [1])

120578=

119876

119876ideal=

int119871

0

119875119867 (119879) (119879 minus 119879119886) 119889119883

119875ℎ119887119871 (119879119887minus 119879119886)

(63)

In dimensionless variables we have

120578 = int

1

0

120579119899+1

119889119909 (64)

We consider the solutions (50) (51) (52) and (53) and depictthe fin efficiency (63) in Figure 9

62 Heat Flux The fin base heat flux is given by the Fourierrsquoslaw

119902119887= 119860119888119870 (119879)

119889119879

119889119909 (65)

The total heat flux of the fin is given by [1]

119902 =119902119887

119860119888119867(119879) (119879

119887minus 119879119886) (66)

Introducing the dimensionless variable as described inSection 2 implies

119902 =1

119861119894

119896 (120579)

ℎ (120579)

119889120579

119889119909 (67)

where the dimensionless parameter 119861119894 = ℎ119887119871119896119886is the Biot

number We consider a number of cases for the thermalconductivity and the heat transfer coefficient

621 Linear Thermal Conductivity and Power Law HeatTransfer Coefficient In this case (67) becomes

119902 =1

119861119894(1 + 120573120579) 120579

minus119899119889120579

119889119909 (68)

The heat flux in (68) at the base of the fin is plotted inFigure 10

622 Power LawThermal Conductivity and Heat TransferCoefficient In this case (67) becomes

119902 =1

119861119894120579119898minus119899

119889120579

119889119909 (69)

Not surprisingly heat flux in one-dimensional fins is highergiven values 119861119894 ≪ 1 The heat flux in (69) is plotted inFigures 11(a) 11(b) and 11(c)

7 Some Discussions

The DTM has resulted in some interesting observations andstudy We have observed in Figures 2(a) and 2(b) an excellentagreement between the analytical solutions generated byDTM and the exact solution obtained in [2] In particular we

119872

1

095

09

085

08

075

07

065

06

055

120578

2151050

119899 = 0

119899 = 1

119899 = 2

119899 = 3

Figure 9 Fin efficiency of a longitudinal rectangular fin Here 120573 =075

119872

25

20

15

10

5

0

119902

2151050

119899 = 0

119899 = 1

119899 = 2

119899 = 3

Figure 10 Base heat flux in a longitudinal rectangular finwith linearthermal conductivity Here 120573 = 01

considered a fin problem in which both thermal conductivityand heat transfer coefficient are given by the same power lawFurthermore we notice from Table 2 that an absolute errorof approximately 35119890 minus 005 is produced by DTM of order119874(15) In Table 3 an absolute error of approximately 65119890 minus006 is produced for the same order This confirms that theDTM converges faster and can provide accurate results witha minimum computation As such a tremendous confidencein the DTM in terms of the accuracy and effectiveness wasbuilt and thus we used this method to solve other problemsfor which exact solutions are harder to construct

In Figures 3(a) 3(b) 4(a) 4(b) 5(a) and 5(b) weobserve that the fin temperature increaseswith the decreasing

12 Mathematical Problems in Engineering

values of the thermogeometric fin parameter Here thevalues of the exponents are fixed Also we observe that fintemperature is higher when 119899 minus 119898 gt 0 that is when heattransfer coefficient is higher than the thermal conductivityWe observe in Figures 6 7 and 8 that the fin temperatureincreases with the increasing values of 119899 Furthermoreit appears that the fin with exponential profile performsthe least in transferring the heat from the base since thetemperature in such a fin is much higher than that ofthe rectangular and the convex parabolic profiles In otherwords heat dissipation to the fluid surrounding the extendedsurface is much faster in longitudinal fins of rectangular andconvex parabolic profiles In Figure 9 fin efficiency decreaseswith increasing thermogeometric fin parameter Also finefficiency increases with increasing values of 119899 It is easyto show that the thermogeometric fin parameter is directlyproportional to the aspect ratio (extension factor) with squareroot of the Biot number being the proportionality constantAs such shorter fins are more efficient than longer onesElse the increased Biot number results in less efficient finwhenever the space is confined that is where the length ofthe fin cannot be increased Figure 10 depicts the heat fluxat the fin base The amount of heat energy dissipated fromthe fin base is of immense interest in engineering [28] Weobserve in Figure 10 that the base heat flux increases withthe thermogeometric fin parameter for considered values ofthe exponent 119899 (see also [28]) Figures 11(a) 11(b) and 11(c)display the heat flux across the fin length We note that theheat flux across the fin length increases with increasing valuesof the thermogeometric fin parameter

8 Concluding Remarks

In this study we have successfully applied the DTM to highlynonlinear problems arising in heat transfer through longitu-dinal fins of various profiles Both thermal conductivity andheat transfer coefficient are given as functions of temperatureThe DTM agreed well with exact solutions when the thermalconductivity and heat transfer coefficient are given by thesame power law A rapid convergence to the exact solutionwas observed Following the confidence in DTM built by theresults mentioned we then solved various exciting problemsThe exotic results have been shown in tables and figures listedin this paper

The results obtained in this paper are significant improve-ments on the known results In particular both the heattransfer coefficient and the thermal conductivity are allowedto be given by the power law functions of temperature andalsowe considered a number of finprofilesWenote that exactsolutions are difficult if not impossible to construct when theexponents of these properties are distinct

Perhaps the notable advantage of the DTM is the general-ization of the Taylor method to problems involving unusualderivative procedures such as fractional fuzzy or q-derivative[22] Some generalizations have been made by Odibat et al[24] and they referred to their new method as the General-ized Differential Transform Method (GDTM) This showedgreat improvement compared to the Fractional Differential

0 02 04 06 08 1119909

45

40

35

30

25

20

15

10

5

0

minus5

119902

(a)

0 02 04 06 08 1119909

30

25

20

15

10

5

0

119902

(b)

35

30

25

20

15

10

5

0

119902

0 02 04 06 08 1119909

119872 = 025

119872 = 04

119872 = 05

119872 = 065

(c)

Figure 11 Heat flux across a longitudinal rectangular fin with linearthermal conductivity (a) 119899minus119898 lt 0 (b) 119899minus119898 gt 0 and (c) 119899minus119898 = 0Here 119861119894 = 001

Mathematical Problems in Engineering 13

Transform Method (FDTM) introduced by Arikoglu andOzkol [23]

We have shown with the help of an example that DTMmay only be applied to fin problems involving heat transferthrough fins with convex parabolic profile Note that given anODE such as (4) with a power law heat transfer coefficient ofa fractional exponent then one can easily remove the fractionby basic exponential rules and employment of the Binomialexpansion However using the DTM one runs into difficultyif the power law thermal conductivity in the same equation isgiven by the fractional exponent We do not know whetherthese observations call for the ldquomodifiedrdquo DTM to solveproblems arising in heat flow through fins with other profilessuch as longitudinal triangular and concave parabolic andalso with fractional power law thermal conductivity

The main results obtained in this paper give insightinto heat transfer in boiling liquids where the heat transfercoefficient is temperature dependent and may be given by apower law The thermal conductivity of some materials suchas gallium nitride (GaN) and Aluminium Nitride (AlN) maybe modeled by power law temperature dependency [29 30]Thus the solutions constructed here give a better comparisonof heat transfer in terms of material used since in many engi-neering applications thermal conductivity is given as a linearfunction of temperature Furthermore a good study in termsof performance and efficiency of fin with different profiles isundertaken These finding could help in the design of fins Itis claimed in [14] that DTM results are more accurate thanthose constructed by Variational Iteration Methods (VIM)and Homotopy Perturbation Methods (HPM) However itwould be risky to use the DTM approximate solutions asbenchmarks for the numerical schemes Nevertheless wehave also shown that DTM converges rapidly in just fifteenterms to the exact solution

Acknowledgments

The authors wish to thank the National Research Foundationof South Africa and the University of the Witwatersrandfor the generous financial support The authors are gratefulto Professor D P Mason and the anonymous reviewers fortheir invaluable comments which improved this paper signif-icantly

References

[1] A D Kraus A Aziz and J Welty Extended Surface Heat Trans-fer Wiley New York NY USA 2001

[2] R JMoitsheki T Hayat andM YMalik ldquoSome exact solutionsof the fin problem with a power law temperature-dependentthermal conductivityrdquo Nonlinear Analysis Real World Applica-tions vol 11 no 5 pp 3287ndash3294 2010

[3] R J Moitsheki ldquoSteady one-dimensional heat flow in a lon-gitudinal triangular and parabolic finrdquo Communications inNonlinear Science and Numerical Simulation vol 16 no 10 pp3971ndash3980 2011

[4] R J Moitsheki ldquoSteady heat transfer through a radial fin withrectangular and hyperbolic profilesrdquo Nonlinear Analysis RealWorld Applications vol 12 no 2 pp 867ndash874 2011

[5] Mo Miansari D D Ganji and Me Miansari ldquoApplication ofHersquos variational iteration method to nonlinear heat transferequationsrdquo Physics Letters A vol 372 no 6 pp 779ndash785 2008

[6] C H Chiu and C K Chen ldquoA decomposition method forsolving the convective longitudinal fins with variable thermalconductivityrdquo International Journal of Heat and Mass Transfervol 45 no 10 pp 2067ndash2075 2002

[7] A Rajabi ldquoHomotopy perturbation method for fin efficiencyof convective straight fins with temperature-dependent thermalconductivityrdquo Physics Letters A vol 364 no 1 pp 33ndash37 2007

[8] G Domairry and M Fazeli ldquoHomotopy analysis method todetermine the fin efficiency of convective straight fins withtemperature-dependent thermal conductivityrdquo Communica-tions in Nonlinear Science and Numerical Simulation vol 14 no2 pp 489ndash499 2009

[9] A Campo andR J Spaulding ldquoCoupling of themethods of suc-cessive approximations and undetermined coefficients for theprediction of the thermal behaviour of uniform circumferentialfinsrdquo Heat and Mass Transfer vol 34 no 6 pp 461ndash468 1999

[10] A Campo and F Rodrfguez ldquoApproximate analytic tempera-ture solution for uniform annular fins by adapting the powerseriesmethodrdquo International Communications inHeat andMassTransfer vol 25 no 6 pp 809ndash818 1998

[11] R Chiba ldquoApplication of differential transformmethod to ther-moelastic problem for annular disks of variable thickness withtemperature-dependent parametersrdquo International Journal ofThermophysics vol 33 pp 363ndash380 2012

[12] S Sadri M R Raveshi and S Amiri ldquoEfficiency analysis ofstraight fin with variable heat transfer coefficient and thermalconductivityrdquo Journal ofMechanical Science andTechnology vol26 no 4 pp 1283ndash1290 2012

[13] M Torabi H Yaghoori and A Aziz ldquoAnalytical solution forconvective-radiative continously moving fin with temperaturedependent thermal conductivityrdquo International Journal of Ther-mophysics vol 33 pp 924ndash941 2012

[14] M Torabi andH Yaghoobi ldquoTwo dominant analytical methodsfor thermal analysis of convective step fin with variable thermalconductivityrdquoThermal Science In press

[15] AMoradi ldquoAnalytical solutions for finwith temperature depen-dant heat transfer coefficientrdquo International Journal of Engineer-ing and Applied Sciences vol 3 no 2 pp 1ndash12 2011

[16] A Moradi and H Ahmadikia ldquoAnalytical solution for differentprofiles of fin with temperature-dependent thermal conductiv-ityrdquoMathematical Problems in Engineering vol 2010 Article ID568263 15 pages 2010

[17] H Yaghoobi and M Torabi ldquoThe application of differentialtransformation method to nonlinear equations arising in heattransferrdquo International Communications in Heat and MassTransfer vol 38 no 6 pp 815ndash820 2011

[18] S Ghafoori M Motevalli M G Nejad F Shakeri D D GanjiandM Jalaal ldquoEfficiency of differential transformation methodfor nonlinear oscillation Comparison with HPM and VIMrdquoCurrent Applied Physics vol 11 no 4 pp 965ndash971 2011

[19] A A Joneidi D D Ganji andM Babaelahi ldquoDifferential trans-formation method to determine fin efficiency of convectivestraight fins with temperature dependent thermal conductivityrdquoInternational Communications in Heat and Mass Transfer vol36 no 7 pp 757ndash762 2009

[20] J K Zhou Differential Transform Method and Its Applicationsfor Electric circuIts Huazhong University Press Wuhan China1986

14 Mathematical Problems in Engineering

[21] S Mukherjee ldquoReply to comment on lsquosolutions of the duffing-van der pol oscillator equation by the differential transformmethodrsquordquo Physica Scripta vol 84 Article ID 037003 2 pages2011

[22] C Bervillier ldquoStatus of the differential transformationmethodrdquoApplied Mathematics and Computation vol 218 no 20 pp10158ndash10170 2012

[23] A Arikoglu and I Ozkol ldquoSolution of fractional differentialequations by using differential transform methodrdquo ChaosSolitons and Fractals vol 34 no 5 pp 1473ndash1481 2007

[24] ZOdibat SMomani andV S Erturk ldquoGeneralized differentialtransform method application to differential equations offractional orderrdquo Applied Mathematics and Computation vol197 no 2 pp 467ndash477 2008

[25] CW Bert ldquoApplication of differential transformmethod to heatconduction in tapered finsrdquo Journal of Heat Transfer vol 124no 1 pp 208ndash209 2002

[26] F Khani and A Aziz ldquoThermal analysis of a longitudinal trape-zoidal fin with temperature-dependent thermal conductivityand heat transfer coefficientrdquo Communications in NonlinearScience and Numerical Simulation vol 15 no 3 pp 590ndash6012010

[27] H C Unal ldquoAn anlytical study of boiling heat transfer from afinrdquo International Journal of Heat and Mass Transfer vol 31 no7 pp 1483ndash1496 1988

[28] M H Chang ldquoA decomposition solution for fins with tempera-ture dependent surface heat fluxrdquo International Journal of Heatand Mass Transfer vol 48 no 9 pp 1819ndash1824 2005

[29] A Jezowski B A Danilchenko M Bockowski et al ldquoThermalconductivity of GaN crystals in 42ndash300 K rangerdquo Solid StateCommunications vol 128 no 2-3 pp 69ndash73 2003

[30] S Vitanov V Palankovski S Maroldt and R Quay ldquoHigh-temperature modeling of AlGaNGaN HEMTsrdquo Solid-StateElectronics vol 54 no 10 pp 1105ndash1112 2010

Submit your manuscripts athttpwwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical Problems in Engineering

Hindawi Publishing Corporationhttpwwwhindawicom

Differential EquationsInternational Journal of

Volume 2014

Applied MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical PhysicsAdvances in

Complex AnalysisJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

OptimizationJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Operations ResearchAdvances in

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of Mathematics and Mathematical Sciences

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Algebra

Discrete Dynamics in Nature and Society

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Decision SciencesAdvances in

Discrete MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom

Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

Page 10: Research Article Analytical Solutions for Steady Heat ...downloads.hindawi.com/journals/mpe/2013/273052.pdf · Research Article Analytical Solutions for Steady Heat Transfer in Longitudinal

10 Mathematical Problems in Engineering

1

09

08

07

06

05

120579

119899 = 0

119899 = 1

119899 = 2

119899 = 3

0 02 04 06 08 1

119909

Figure 7 Temperature profile in a longitudinal convex parabolic finwith linear thermal conductivity and varying values of 119899 Here 120573 =05 and119872 = 15 are fixed

56 The Exponential Profile and LinearThermal ConductivityIn this section we present solutions for equation heat transferin a fin with convex parabolic profile and the thermalconductivity depending linearly on temperature That is weconsider (4) with 119896(120579) = 1 + 120573120579 and 119891(119909) = e120590119909 Here 120590is a constant The analytical solutions for this problem fordifferent values of 119899 are given by

(a) Case 119899 = 0

120579 (119909)

= 119888 +

1198722

119888

2 (1 + 120573119888)

1199092

minus

1205901198722

119888 (2 + 120573119888 + 1205732

119888)

6(1 + 120573119888)21199093

+

1198722

119888 (1198722

(1 minus 21205732

119888) + 1205902

(3 +3120573119888+1205733

1198882

+1205734

1198882

+1205732

119888 (3 + 119888)))

24(1 + 120573119888)4

1199094

+ sdot sdot sdot

(58)

(b) Case 119899 = 1

120579 (119909)

= 119888 +

11987221198882

2 (1 + 120573119888)

1199092minus

12059011987221198882(2 + 120573119888 + 120573

2119888)

6(1 + 120573119888)21199093

+

11987221198882(1198722119888 (2+120573119888minus2120573

2119888)+1205902(3+3120573119888+120573

31198882+12057341198882+1205732119888 (3+119888)))

24(1 + 120573119888)3

1199094

+ sdot sdot sdot

(59)

119899 = 0

119899 = 1

119899 = 2

119899 = 3

0 02 04 06 08 1

119909

1

095

09

085

08

075

07

120579

Figure 8 Temperature profile in a longitudinal fin of exponentialprofile with linear thermal conductivity and varying values of 119899Here 120573 = 05 and119872 = 15 are fixed

(c) Case 119899 = 2120579 (119909)

= 119888 +

11987221198883

2 (1 + 120573119888)

1199092minus

12059011987221198883(2 + 120573119888 + 120573

2119888)

6(1 + 120573119888)21199093

+

11987221198883(11987221198882(3+2120573119888minus2120573

2119888)+1205902(3+3120573119888+120573

31198882+12057341198882+1205732119888 (3+119888)))

24(1 + 120573119888)3

times1199094+ sdot sdot sdot

(60)

(d) Case 119899 = 3120579 (119909)

= 119888 +

11987221198884

2 (1 + 120573119888)

1199092minus

12059011987221198884(2 + 120573119888 + 120573

2119888)

6(1 + 120573119888)21199093

+

11987221198884(11987221198883(4+3120573119888minus2120573

2119888)+1205902(3+3120573119888+120573

31198882+12057341198882+1205732119888 (3+119888)))

24(1 + 120573119888)3

times1199094+ sdot sdot sdot

(61)

The constant 119888 may be obtained from the boundarycondition on the appropriate 120579(119909)The solutions in (58) (59)(60) and (61) are depicted in Figure 8

6 Fin Efficiency and Heat Flux

61 Fin Efficiency Theheat transfer rate from a fin is given byNewtonrsquos second law of cooling

119876 = int

119871

0

119875119867 (119879) (119879 minus 119879119886) 119889119909 (62)

Mathematical Problems in Engineering 11

Fin efficiency is defined as the ratio of the fin heat transferrate to the rate that would be if the entire fin were at the basetemperature and is given by (see eg [1])

120578=

119876

119876ideal=

int119871

0

119875119867 (119879) (119879 minus 119879119886) 119889119883

119875ℎ119887119871 (119879119887minus 119879119886)

(63)

In dimensionless variables we have

120578 = int

1

0

120579119899+1

119889119909 (64)

We consider the solutions (50) (51) (52) and (53) and depictthe fin efficiency (63) in Figure 9

62 Heat Flux The fin base heat flux is given by the Fourierrsquoslaw

119902119887= 119860119888119870 (119879)

119889119879

119889119909 (65)

The total heat flux of the fin is given by [1]

119902 =119902119887

119860119888119867(119879) (119879

119887minus 119879119886) (66)

Introducing the dimensionless variable as described inSection 2 implies

119902 =1

119861119894

119896 (120579)

ℎ (120579)

119889120579

119889119909 (67)

where the dimensionless parameter 119861119894 = ℎ119887119871119896119886is the Biot

number We consider a number of cases for the thermalconductivity and the heat transfer coefficient

621 Linear Thermal Conductivity and Power Law HeatTransfer Coefficient In this case (67) becomes

119902 =1

119861119894(1 + 120573120579) 120579

minus119899119889120579

119889119909 (68)

The heat flux in (68) at the base of the fin is plotted inFigure 10

622 Power LawThermal Conductivity and Heat TransferCoefficient In this case (67) becomes

119902 =1

119861119894120579119898minus119899

119889120579

119889119909 (69)

Not surprisingly heat flux in one-dimensional fins is highergiven values 119861119894 ≪ 1 The heat flux in (69) is plotted inFigures 11(a) 11(b) and 11(c)

7 Some Discussions

The DTM has resulted in some interesting observations andstudy We have observed in Figures 2(a) and 2(b) an excellentagreement between the analytical solutions generated byDTM and the exact solution obtained in [2] In particular we

119872

1

095

09

085

08

075

07

065

06

055

120578

2151050

119899 = 0

119899 = 1

119899 = 2

119899 = 3

Figure 9 Fin efficiency of a longitudinal rectangular fin Here 120573 =075

119872

25

20

15

10

5

0

119902

2151050

119899 = 0

119899 = 1

119899 = 2

119899 = 3

Figure 10 Base heat flux in a longitudinal rectangular finwith linearthermal conductivity Here 120573 = 01

considered a fin problem in which both thermal conductivityand heat transfer coefficient are given by the same power lawFurthermore we notice from Table 2 that an absolute errorof approximately 35119890 minus 005 is produced by DTM of order119874(15) In Table 3 an absolute error of approximately 65119890 minus006 is produced for the same order This confirms that theDTM converges faster and can provide accurate results witha minimum computation As such a tremendous confidencein the DTM in terms of the accuracy and effectiveness wasbuilt and thus we used this method to solve other problemsfor which exact solutions are harder to construct

In Figures 3(a) 3(b) 4(a) 4(b) 5(a) and 5(b) weobserve that the fin temperature increaseswith the decreasing

12 Mathematical Problems in Engineering

values of the thermogeometric fin parameter Here thevalues of the exponents are fixed Also we observe that fintemperature is higher when 119899 minus 119898 gt 0 that is when heattransfer coefficient is higher than the thermal conductivityWe observe in Figures 6 7 and 8 that the fin temperatureincreases with the increasing values of 119899 Furthermoreit appears that the fin with exponential profile performsthe least in transferring the heat from the base since thetemperature in such a fin is much higher than that ofthe rectangular and the convex parabolic profiles In otherwords heat dissipation to the fluid surrounding the extendedsurface is much faster in longitudinal fins of rectangular andconvex parabolic profiles In Figure 9 fin efficiency decreaseswith increasing thermogeometric fin parameter Also finefficiency increases with increasing values of 119899 It is easyto show that the thermogeometric fin parameter is directlyproportional to the aspect ratio (extension factor) with squareroot of the Biot number being the proportionality constantAs such shorter fins are more efficient than longer onesElse the increased Biot number results in less efficient finwhenever the space is confined that is where the length ofthe fin cannot be increased Figure 10 depicts the heat fluxat the fin base The amount of heat energy dissipated fromthe fin base is of immense interest in engineering [28] Weobserve in Figure 10 that the base heat flux increases withthe thermogeometric fin parameter for considered values ofthe exponent 119899 (see also [28]) Figures 11(a) 11(b) and 11(c)display the heat flux across the fin length We note that theheat flux across the fin length increases with increasing valuesof the thermogeometric fin parameter

8 Concluding Remarks

In this study we have successfully applied the DTM to highlynonlinear problems arising in heat transfer through longitu-dinal fins of various profiles Both thermal conductivity andheat transfer coefficient are given as functions of temperatureThe DTM agreed well with exact solutions when the thermalconductivity and heat transfer coefficient are given by thesame power law A rapid convergence to the exact solutionwas observed Following the confidence in DTM built by theresults mentioned we then solved various exciting problemsThe exotic results have been shown in tables and figures listedin this paper

The results obtained in this paper are significant improve-ments on the known results In particular both the heattransfer coefficient and the thermal conductivity are allowedto be given by the power law functions of temperature andalsowe considered a number of finprofilesWenote that exactsolutions are difficult if not impossible to construct when theexponents of these properties are distinct

Perhaps the notable advantage of the DTM is the general-ization of the Taylor method to problems involving unusualderivative procedures such as fractional fuzzy or q-derivative[22] Some generalizations have been made by Odibat et al[24] and they referred to their new method as the General-ized Differential Transform Method (GDTM) This showedgreat improvement compared to the Fractional Differential

0 02 04 06 08 1119909

45

40

35

30

25

20

15

10

5

0

minus5

119902

(a)

0 02 04 06 08 1119909

30

25

20

15

10

5

0

119902

(b)

35

30

25

20

15

10

5

0

119902

0 02 04 06 08 1119909

119872 = 025

119872 = 04

119872 = 05

119872 = 065

(c)

Figure 11 Heat flux across a longitudinal rectangular fin with linearthermal conductivity (a) 119899minus119898 lt 0 (b) 119899minus119898 gt 0 and (c) 119899minus119898 = 0Here 119861119894 = 001

Mathematical Problems in Engineering 13

Transform Method (FDTM) introduced by Arikoglu andOzkol [23]

We have shown with the help of an example that DTMmay only be applied to fin problems involving heat transferthrough fins with convex parabolic profile Note that given anODE such as (4) with a power law heat transfer coefficient ofa fractional exponent then one can easily remove the fractionby basic exponential rules and employment of the Binomialexpansion However using the DTM one runs into difficultyif the power law thermal conductivity in the same equation isgiven by the fractional exponent We do not know whetherthese observations call for the ldquomodifiedrdquo DTM to solveproblems arising in heat flow through fins with other profilessuch as longitudinal triangular and concave parabolic andalso with fractional power law thermal conductivity

The main results obtained in this paper give insightinto heat transfer in boiling liquids where the heat transfercoefficient is temperature dependent and may be given by apower law The thermal conductivity of some materials suchas gallium nitride (GaN) and Aluminium Nitride (AlN) maybe modeled by power law temperature dependency [29 30]Thus the solutions constructed here give a better comparisonof heat transfer in terms of material used since in many engi-neering applications thermal conductivity is given as a linearfunction of temperature Furthermore a good study in termsof performance and efficiency of fin with different profiles isundertaken These finding could help in the design of fins Itis claimed in [14] that DTM results are more accurate thanthose constructed by Variational Iteration Methods (VIM)and Homotopy Perturbation Methods (HPM) However itwould be risky to use the DTM approximate solutions asbenchmarks for the numerical schemes Nevertheless wehave also shown that DTM converges rapidly in just fifteenterms to the exact solution

Acknowledgments

The authors wish to thank the National Research Foundationof South Africa and the University of the Witwatersrandfor the generous financial support The authors are gratefulto Professor D P Mason and the anonymous reviewers fortheir invaluable comments which improved this paper signif-icantly

References

[1] A D Kraus A Aziz and J Welty Extended Surface Heat Trans-fer Wiley New York NY USA 2001

[2] R JMoitsheki T Hayat andM YMalik ldquoSome exact solutionsof the fin problem with a power law temperature-dependentthermal conductivityrdquo Nonlinear Analysis Real World Applica-tions vol 11 no 5 pp 3287ndash3294 2010

[3] R J Moitsheki ldquoSteady one-dimensional heat flow in a lon-gitudinal triangular and parabolic finrdquo Communications inNonlinear Science and Numerical Simulation vol 16 no 10 pp3971ndash3980 2011

[4] R J Moitsheki ldquoSteady heat transfer through a radial fin withrectangular and hyperbolic profilesrdquo Nonlinear Analysis RealWorld Applications vol 12 no 2 pp 867ndash874 2011

[5] Mo Miansari D D Ganji and Me Miansari ldquoApplication ofHersquos variational iteration method to nonlinear heat transferequationsrdquo Physics Letters A vol 372 no 6 pp 779ndash785 2008

[6] C H Chiu and C K Chen ldquoA decomposition method forsolving the convective longitudinal fins with variable thermalconductivityrdquo International Journal of Heat and Mass Transfervol 45 no 10 pp 2067ndash2075 2002

[7] A Rajabi ldquoHomotopy perturbation method for fin efficiencyof convective straight fins with temperature-dependent thermalconductivityrdquo Physics Letters A vol 364 no 1 pp 33ndash37 2007

[8] G Domairry and M Fazeli ldquoHomotopy analysis method todetermine the fin efficiency of convective straight fins withtemperature-dependent thermal conductivityrdquo Communica-tions in Nonlinear Science and Numerical Simulation vol 14 no2 pp 489ndash499 2009

[9] A Campo andR J Spaulding ldquoCoupling of themethods of suc-cessive approximations and undetermined coefficients for theprediction of the thermal behaviour of uniform circumferentialfinsrdquo Heat and Mass Transfer vol 34 no 6 pp 461ndash468 1999

[10] A Campo and F Rodrfguez ldquoApproximate analytic tempera-ture solution for uniform annular fins by adapting the powerseriesmethodrdquo International Communications inHeat andMassTransfer vol 25 no 6 pp 809ndash818 1998

[11] R Chiba ldquoApplication of differential transformmethod to ther-moelastic problem for annular disks of variable thickness withtemperature-dependent parametersrdquo International Journal ofThermophysics vol 33 pp 363ndash380 2012

[12] S Sadri M R Raveshi and S Amiri ldquoEfficiency analysis ofstraight fin with variable heat transfer coefficient and thermalconductivityrdquo Journal ofMechanical Science andTechnology vol26 no 4 pp 1283ndash1290 2012

[13] M Torabi H Yaghoori and A Aziz ldquoAnalytical solution forconvective-radiative continously moving fin with temperaturedependent thermal conductivityrdquo International Journal of Ther-mophysics vol 33 pp 924ndash941 2012

[14] M Torabi andH Yaghoobi ldquoTwo dominant analytical methodsfor thermal analysis of convective step fin with variable thermalconductivityrdquoThermal Science In press

[15] AMoradi ldquoAnalytical solutions for finwith temperature depen-dant heat transfer coefficientrdquo International Journal of Engineer-ing and Applied Sciences vol 3 no 2 pp 1ndash12 2011

[16] A Moradi and H Ahmadikia ldquoAnalytical solution for differentprofiles of fin with temperature-dependent thermal conductiv-ityrdquoMathematical Problems in Engineering vol 2010 Article ID568263 15 pages 2010

[17] H Yaghoobi and M Torabi ldquoThe application of differentialtransformation method to nonlinear equations arising in heattransferrdquo International Communications in Heat and MassTransfer vol 38 no 6 pp 815ndash820 2011

[18] S Ghafoori M Motevalli M G Nejad F Shakeri D D GanjiandM Jalaal ldquoEfficiency of differential transformation methodfor nonlinear oscillation Comparison with HPM and VIMrdquoCurrent Applied Physics vol 11 no 4 pp 965ndash971 2011

[19] A A Joneidi D D Ganji andM Babaelahi ldquoDifferential trans-formation method to determine fin efficiency of convectivestraight fins with temperature dependent thermal conductivityrdquoInternational Communications in Heat and Mass Transfer vol36 no 7 pp 757ndash762 2009

[20] J K Zhou Differential Transform Method and Its Applicationsfor Electric circuIts Huazhong University Press Wuhan China1986

14 Mathematical Problems in Engineering

[21] S Mukherjee ldquoReply to comment on lsquosolutions of the duffing-van der pol oscillator equation by the differential transformmethodrsquordquo Physica Scripta vol 84 Article ID 037003 2 pages2011

[22] C Bervillier ldquoStatus of the differential transformationmethodrdquoApplied Mathematics and Computation vol 218 no 20 pp10158ndash10170 2012

[23] A Arikoglu and I Ozkol ldquoSolution of fractional differentialequations by using differential transform methodrdquo ChaosSolitons and Fractals vol 34 no 5 pp 1473ndash1481 2007

[24] ZOdibat SMomani andV S Erturk ldquoGeneralized differentialtransform method application to differential equations offractional orderrdquo Applied Mathematics and Computation vol197 no 2 pp 467ndash477 2008

[25] CW Bert ldquoApplication of differential transformmethod to heatconduction in tapered finsrdquo Journal of Heat Transfer vol 124no 1 pp 208ndash209 2002

[26] F Khani and A Aziz ldquoThermal analysis of a longitudinal trape-zoidal fin with temperature-dependent thermal conductivityand heat transfer coefficientrdquo Communications in NonlinearScience and Numerical Simulation vol 15 no 3 pp 590ndash6012010

[27] H C Unal ldquoAn anlytical study of boiling heat transfer from afinrdquo International Journal of Heat and Mass Transfer vol 31 no7 pp 1483ndash1496 1988

[28] M H Chang ldquoA decomposition solution for fins with tempera-ture dependent surface heat fluxrdquo International Journal of Heatand Mass Transfer vol 48 no 9 pp 1819ndash1824 2005

[29] A Jezowski B A Danilchenko M Bockowski et al ldquoThermalconductivity of GaN crystals in 42ndash300 K rangerdquo Solid StateCommunications vol 128 no 2-3 pp 69ndash73 2003

[30] S Vitanov V Palankovski S Maroldt and R Quay ldquoHigh-temperature modeling of AlGaNGaN HEMTsrdquo Solid-StateElectronics vol 54 no 10 pp 1105ndash1112 2010

Submit your manuscripts athttpwwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical Problems in Engineering

Hindawi Publishing Corporationhttpwwwhindawicom

Differential EquationsInternational Journal of

Volume 2014

Applied MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical PhysicsAdvances in

Complex AnalysisJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

OptimizationJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Operations ResearchAdvances in

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of Mathematics and Mathematical Sciences

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Algebra

Discrete Dynamics in Nature and Society

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Decision SciencesAdvances in

Discrete MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom

Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

Page 11: Research Article Analytical Solutions for Steady Heat ...downloads.hindawi.com/journals/mpe/2013/273052.pdf · Research Article Analytical Solutions for Steady Heat Transfer in Longitudinal

Mathematical Problems in Engineering 11

Fin efficiency is defined as the ratio of the fin heat transferrate to the rate that would be if the entire fin were at the basetemperature and is given by (see eg [1])

120578=

119876

119876ideal=

int119871

0

119875119867 (119879) (119879 minus 119879119886) 119889119883

119875ℎ119887119871 (119879119887minus 119879119886)

(63)

In dimensionless variables we have

120578 = int

1

0

120579119899+1

119889119909 (64)

We consider the solutions (50) (51) (52) and (53) and depictthe fin efficiency (63) in Figure 9

62 Heat Flux The fin base heat flux is given by the Fourierrsquoslaw

119902119887= 119860119888119870 (119879)

119889119879

119889119909 (65)

The total heat flux of the fin is given by [1]

119902 =119902119887

119860119888119867(119879) (119879

119887minus 119879119886) (66)

Introducing the dimensionless variable as described inSection 2 implies

119902 =1

119861119894

119896 (120579)

ℎ (120579)

119889120579

119889119909 (67)

where the dimensionless parameter 119861119894 = ℎ119887119871119896119886is the Biot

number We consider a number of cases for the thermalconductivity and the heat transfer coefficient

621 Linear Thermal Conductivity and Power Law HeatTransfer Coefficient In this case (67) becomes

119902 =1

119861119894(1 + 120573120579) 120579

minus119899119889120579

119889119909 (68)

The heat flux in (68) at the base of the fin is plotted inFigure 10

622 Power LawThermal Conductivity and Heat TransferCoefficient In this case (67) becomes

119902 =1

119861119894120579119898minus119899

119889120579

119889119909 (69)

Not surprisingly heat flux in one-dimensional fins is highergiven values 119861119894 ≪ 1 The heat flux in (69) is plotted inFigures 11(a) 11(b) and 11(c)

7 Some Discussions

The DTM has resulted in some interesting observations andstudy We have observed in Figures 2(a) and 2(b) an excellentagreement between the analytical solutions generated byDTM and the exact solution obtained in [2] In particular we

119872

1

095

09

085

08

075

07

065

06

055

120578

2151050

119899 = 0

119899 = 1

119899 = 2

119899 = 3

Figure 9 Fin efficiency of a longitudinal rectangular fin Here 120573 =075

119872

25

20

15

10

5

0

119902

2151050

119899 = 0

119899 = 1

119899 = 2

119899 = 3

Figure 10 Base heat flux in a longitudinal rectangular finwith linearthermal conductivity Here 120573 = 01

considered a fin problem in which both thermal conductivityand heat transfer coefficient are given by the same power lawFurthermore we notice from Table 2 that an absolute errorof approximately 35119890 minus 005 is produced by DTM of order119874(15) In Table 3 an absolute error of approximately 65119890 minus006 is produced for the same order This confirms that theDTM converges faster and can provide accurate results witha minimum computation As such a tremendous confidencein the DTM in terms of the accuracy and effectiveness wasbuilt and thus we used this method to solve other problemsfor which exact solutions are harder to construct

In Figures 3(a) 3(b) 4(a) 4(b) 5(a) and 5(b) weobserve that the fin temperature increaseswith the decreasing

12 Mathematical Problems in Engineering

values of the thermogeometric fin parameter Here thevalues of the exponents are fixed Also we observe that fintemperature is higher when 119899 minus 119898 gt 0 that is when heattransfer coefficient is higher than the thermal conductivityWe observe in Figures 6 7 and 8 that the fin temperatureincreases with the increasing values of 119899 Furthermoreit appears that the fin with exponential profile performsthe least in transferring the heat from the base since thetemperature in such a fin is much higher than that ofthe rectangular and the convex parabolic profiles In otherwords heat dissipation to the fluid surrounding the extendedsurface is much faster in longitudinal fins of rectangular andconvex parabolic profiles In Figure 9 fin efficiency decreaseswith increasing thermogeometric fin parameter Also finefficiency increases with increasing values of 119899 It is easyto show that the thermogeometric fin parameter is directlyproportional to the aspect ratio (extension factor) with squareroot of the Biot number being the proportionality constantAs such shorter fins are more efficient than longer onesElse the increased Biot number results in less efficient finwhenever the space is confined that is where the length ofthe fin cannot be increased Figure 10 depicts the heat fluxat the fin base The amount of heat energy dissipated fromthe fin base is of immense interest in engineering [28] Weobserve in Figure 10 that the base heat flux increases withthe thermogeometric fin parameter for considered values ofthe exponent 119899 (see also [28]) Figures 11(a) 11(b) and 11(c)display the heat flux across the fin length We note that theheat flux across the fin length increases with increasing valuesof the thermogeometric fin parameter

8 Concluding Remarks

In this study we have successfully applied the DTM to highlynonlinear problems arising in heat transfer through longitu-dinal fins of various profiles Both thermal conductivity andheat transfer coefficient are given as functions of temperatureThe DTM agreed well with exact solutions when the thermalconductivity and heat transfer coefficient are given by thesame power law A rapid convergence to the exact solutionwas observed Following the confidence in DTM built by theresults mentioned we then solved various exciting problemsThe exotic results have been shown in tables and figures listedin this paper

The results obtained in this paper are significant improve-ments on the known results In particular both the heattransfer coefficient and the thermal conductivity are allowedto be given by the power law functions of temperature andalsowe considered a number of finprofilesWenote that exactsolutions are difficult if not impossible to construct when theexponents of these properties are distinct

Perhaps the notable advantage of the DTM is the general-ization of the Taylor method to problems involving unusualderivative procedures such as fractional fuzzy or q-derivative[22] Some generalizations have been made by Odibat et al[24] and they referred to their new method as the General-ized Differential Transform Method (GDTM) This showedgreat improvement compared to the Fractional Differential

0 02 04 06 08 1119909

45

40

35

30

25

20

15

10

5

0

minus5

119902

(a)

0 02 04 06 08 1119909

30

25

20

15

10

5

0

119902

(b)

35

30

25

20

15

10

5

0

119902

0 02 04 06 08 1119909

119872 = 025

119872 = 04

119872 = 05

119872 = 065

(c)

Figure 11 Heat flux across a longitudinal rectangular fin with linearthermal conductivity (a) 119899minus119898 lt 0 (b) 119899minus119898 gt 0 and (c) 119899minus119898 = 0Here 119861119894 = 001

Mathematical Problems in Engineering 13

Transform Method (FDTM) introduced by Arikoglu andOzkol [23]

We have shown with the help of an example that DTMmay only be applied to fin problems involving heat transferthrough fins with convex parabolic profile Note that given anODE such as (4) with a power law heat transfer coefficient ofa fractional exponent then one can easily remove the fractionby basic exponential rules and employment of the Binomialexpansion However using the DTM one runs into difficultyif the power law thermal conductivity in the same equation isgiven by the fractional exponent We do not know whetherthese observations call for the ldquomodifiedrdquo DTM to solveproblems arising in heat flow through fins with other profilessuch as longitudinal triangular and concave parabolic andalso with fractional power law thermal conductivity

The main results obtained in this paper give insightinto heat transfer in boiling liquids where the heat transfercoefficient is temperature dependent and may be given by apower law The thermal conductivity of some materials suchas gallium nitride (GaN) and Aluminium Nitride (AlN) maybe modeled by power law temperature dependency [29 30]Thus the solutions constructed here give a better comparisonof heat transfer in terms of material used since in many engi-neering applications thermal conductivity is given as a linearfunction of temperature Furthermore a good study in termsof performance and efficiency of fin with different profiles isundertaken These finding could help in the design of fins Itis claimed in [14] that DTM results are more accurate thanthose constructed by Variational Iteration Methods (VIM)and Homotopy Perturbation Methods (HPM) However itwould be risky to use the DTM approximate solutions asbenchmarks for the numerical schemes Nevertheless wehave also shown that DTM converges rapidly in just fifteenterms to the exact solution

Acknowledgments

The authors wish to thank the National Research Foundationof South Africa and the University of the Witwatersrandfor the generous financial support The authors are gratefulto Professor D P Mason and the anonymous reviewers fortheir invaluable comments which improved this paper signif-icantly

References

[1] A D Kraus A Aziz and J Welty Extended Surface Heat Trans-fer Wiley New York NY USA 2001

[2] R JMoitsheki T Hayat andM YMalik ldquoSome exact solutionsof the fin problem with a power law temperature-dependentthermal conductivityrdquo Nonlinear Analysis Real World Applica-tions vol 11 no 5 pp 3287ndash3294 2010

[3] R J Moitsheki ldquoSteady one-dimensional heat flow in a lon-gitudinal triangular and parabolic finrdquo Communications inNonlinear Science and Numerical Simulation vol 16 no 10 pp3971ndash3980 2011

[4] R J Moitsheki ldquoSteady heat transfer through a radial fin withrectangular and hyperbolic profilesrdquo Nonlinear Analysis RealWorld Applications vol 12 no 2 pp 867ndash874 2011

[5] Mo Miansari D D Ganji and Me Miansari ldquoApplication ofHersquos variational iteration method to nonlinear heat transferequationsrdquo Physics Letters A vol 372 no 6 pp 779ndash785 2008

[6] C H Chiu and C K Chen ldquoA decomposition method forsolving the convective longitudinal fins with variable thermalconductivityrdquo International Journal of Heat and Mass Transfervol 45 no 10 pp 2067ndash2075 2002

[7] A Rajabi ldquoHomotopy perturbation method for fin efficiencyof convective straight fins with temperature-dependent thermalconductivityrdquo Physics Letters A vol 364 no 1 pp 33ndash37 2007

[8] G Domairry and M Fazeli ldquoHomotopy analysis method todetermine the fin efficiency of convective straight fins withtemperature-dependent thermal conductivityrdquo Communica-tions in Nonlinear Science and Numerical Simulation vol 14 no2 pp 489ndash499 2009

[9] A Campo andR J Spaulding ldquoCoupling of themethods of suc-cessive approximations and undetermined coefficients for theprediction of the thermal behaviour of uniform circumferentialfinsrdquo Heat and Mass Transfer vol 34 no 6 pp 461ndash468 1999

[10] A Campo and F Rodrfguez ldquoApproximate analytic tempera-ture solution for uniform annular fins by adapting the powerseriesmethodrdquo International Communications inHeat andMassTransfer vol 25 no 6 pp 809ndash818 1998

[11] R Chiba ldquoApplication of differential transformmethod to ther-moelastic problem for annular disks of variable thickness withtemperature-dependent parametersrdquo International Journal ofThermophysics vol 33 pp 363ndash380 2012

[12] S Sadri M R Raveshi and S Amiri ldquoEfficiency analysis ofstraight fin with variable heat transfer coefficient and thermalconductivityrdquo Journal ofMechanical Science andTechnology vol26 no 4 pp 1283ndash1290 2012

[13] M Torabi H Yaghoori and A Aziz ldquoAnalytical solution forconvective-radiative continously moving fin with temperaturedependent thermal conductivityrdquo International Journal of Ther-mophysics vol 33 pp 924ndash941 2012

[14] M Torabi andH Yaghoobi ldquoTwo dominant analytical methodsfor thermal analysis of convective step fin with variable thermalconductivityrdquoThermal Science In press

[15] AMoradi ldquoAnalytical solutions for finwith temperature depen-dant heat transfer coefficientrdquo International Journal of Engineer-ing and Applied Sciences vol 3 no 2 pp 1ndash12 2011

[16] A Moradi and H Ahmadikia ldquoAnalytical solution for differentprofiles of fin with temperature-dependent thermal conductiv-ityrdquoMathematical Problems in Engineering vol 2010 Article ID568263 15 pages 2010

[17] H Yaghoobi and M Torabi ldquoThe application of differentialtransformation method to nonlinear equations arising in heattransferrdquo International Communications in Heat and MassTransfer vol 38 no 6 pp 815ndash820 2011

[18] S Ghafoori M Motevalli M G Nejad F Shakeri D D GanjiandM Jalaal ldquoEfficiency of differential transformation methodfor nonlinear oscillation Comparison with HPM and VIMrdquoCurrent Applied Physics vol 11 no 4 pp 965ndash971 2011

[19] A A Joneidi D D Ganji andM Babaelahi ldquoDifferential trans-formation method to determine fin efficiency of convectivestraight fins with temperature dependent thermal conductivityrdquoInternational Communications in Heat and Mass Transfer vol36 no 7 pp 757ndash762 2009

[20] J K Zhou Differential Transform Method and Its Applicationsfor Electric circuIts Huazhong University Press Wuhan China1986

14 Mathematical Problems in Engineering

[21] S Mukherjee ldquoReply to comment on lsquosolutions of the duffing-van der pol oscillator equation by the differential transformmethodrsquordquo Physica Scripta vol 84 Article ID 037003 2 pages2011

[22] C Bervillier ldquoStatus of the differential transformationmethodrdquoApplied Mathematics and Computation vol 218 no 20 pp10158ndash10170 2012

[23] A Arikoglu and I Ozkol ldquoSolution of fractional differentialequations by using differential transform methodrdquo ChaosSolitons and Fractals vol 34 no 5 pp 1473ndash1481 2007

[24] ZOdibat SMomani andV S Erturk ldquoGeneralized differentialtransform method application to differential equations offractional orderrdquo Applied Mathematics and Computation vol197 no 2 pp 467ndash477 2008

[25] CW Bert ldquoApplication of differential transformmethod to heatconduction in tapered finsrdquo Journal of Heat Transfer vol 124no 1 pp 208ndash209 2002

[26] F Khani and A Aziz ldquoThermal analysis of a longitudinal trape-zoidal fin with temperature-dependent thermal conductivityand heat transfer coefficientrdquo Communications in NonlinearScience and Numerical Simulation vol 15 no 3 pp 590ndash6012010

[27] H C Unal ldquoAn anlytical study of boiling heat transfer from afinrdquo International Journal of Heat and Mass Transfer vol 31 no7 pp 1483ndash1496 1988

[28] M H Chang ldquoA decomposition solution for fins with tempera-ture dependent surface heat fluxrdquo International Journal of Heatand Mass Transfer vol 48 no 9 pp 1819ndash1824 2005

[29] A Jezowski B A Danilchenko M Bockowski et al ldquoThermalconductivity of GaN crystals in 42ndash300 K rangerdquo Solid StateCommunications vol 128 no 2-3 pp 69ndash73 2003

[30] S Vitanov V Palankovski S Maroldt and R Quay ldquoHigh-temperature modeling of AlGaNGaN HEMTsrdquo Solid-StateElectronics vol 54 no 10 pp 1105ndash1112 2010

Submit your manuscripts athttpwwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical Problems in Engineering

Hindawi Publishing Corporationhttpwwwhindawicom

Differential EquationsInternational Journal of

Volume 2014

Applied MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical PhysicsAdvances in

Complex AnalysisJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

OptimizationJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Operations ResearchAdvances in

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of Mathematics and Mathematical Sciences

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Algebra

Discrete Dynamics in Nature and Society

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Decision SciencesAdvances in

Discrete MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom

Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

Page 12: Research Article Analytical Solutions for Steady Heat ...downloads.hindawi.com/journals/mpe/2013/273052.pdf · Research Article Analytical Solutions for Steady Heat Transfer in Longitudinal

12 Mathematical Problems in Engineering

values of the thermogeometric fin parameter Here thevalues of the exponents are fixed Also we observe that fintemperature is higher when 119899 minus 119898 gt 0 that is when heattransfer coefficient is higher than the thermal conductivityWe observe in Figures 6 7 and 8 that the fin temperatureincreases with the increasing values of 119899 Furthermoreit appears that the fin with exponential profile performsthe least in transferring the heat from the base since thetemperature in such a fin is much higher than that ofthe rectangular and the convex parabolic profiles In otherwords heat dissipation to the fluid surrounding the extendedsurface is much faster in longitudinal fins of rectangular andconvex parabolic profiles In Figure 9 fin efficiency decreaseswith increasing thermogeometric fin parameter Also finefficiency increases with increasing values of 119899 It is easyto show that the thermogeometric fin parameter is directlyproportional to the aspect ratio (extension factor) with squareroot of the Biot number being the proportionality constantAs such shorter fins are more efficient than longer onesElse the increased Biot number results in less efficient finwhenever the space is confined that is where the length ofthe fin cannot be increased Figure 10 depicts the heat fluxat the fin base The amount of heat energy dissipated fromthe fin base is of immense interest in engineering [28] Weobserve in Figure 10 that the base heat flux increases withthe thermogeometric fin parameter for considered values ofthe exponent 119899 (see also [28]) Figures 11(a) 11(b) and 11(c)display the heat flux across the fin length We note that theheat flux across the fin length increases with increasing valuesof the thermogeometric fin parameter

8 Concluding Remarks

In this study we have successfully applied the DTM to highlynonlinear problems arising in heat transfer through longitu-dinal fins of various profiles Both thermal conductivity andheat transfer coefficient are given as functions of temperatureThe DTM agreed well with exact solutions when the thermalconductivity and heat transfer coefficient are given by thesame power law A rapid convergence to the exact solutionwas observed Following the confidence in DTM built by theresults mentioned we then solved various exciting problemsThe exotic results have been shown in tables and figures listedin this paper

The results obtained in this paper are significant improve-ments on the known results In particular both the heattransfer coefficient and the thermal conductivity are allowedto be given by the power law functions of temperature andalsowe considered a number of finprofilesWenote that exactsolutions are difficult if not impossible to construct when theexponents of these properties are distinct

Perhaps the notable advantage of the DTM is the general-ization of the Taylor method to problems involving unusualderivative procedures such as fractional fuzzy or q-derivative[22] Some generalizations have been made by Odibat et al[24] and they referred to their new method as the General-ized Differential Transform Method (GDTM) This showedgreat improvement compared to the Fractional Differential

0 02 04 06 08 1119909

45

40

35

30

25

20

15

10

5

0

minus5

119902

(a)

0 02 04 06 08 1119909

30

25

20

15

10

5

0

119902

(b)

35

30

25

20

15

10

5

0

119902

0 02 04 06 08 1119909

119872 = 025

119872 = 04

119872 = 05

119872 = 065

(c)

Figure 11 Heat flux across a longitudinal rectangular fin with linearthermal conductivity (a) 119899minus119898 lt 0 (b) 119899minus119898 gt 0 and (c) 119899minus119898 = 0Here 119861119894 = 001

Mathematical Problems in Engineering 13

Transform Method (FDTM) introduced by Arikoglu andOzkol [23]

We have shown with the help of an example that DTMmay only be applied to fin problems involving heat transferthrough fins with convex parabolic profile Note that given anODE such as (4) with a power law heat transfer coefficient ofa fractional exponent then one can easily remove the fractionby basic exponential rules and employment of the Binomialexpansion However using the DTM one runs into difficultyif the power law thermal conductivity in the same equation isgiven by the fractional exponent We do not know whetherthese observations call for the ldquomodifiedrdquo DTM to solveproblems arising in heat flow through fins with other profilessuch as longitudinal triangular and concave parabolic andalso with fractional power law thermal conductivity

The main results obtained in this paper give insightinto heat transfer in boiling liquids where the heat transfercoefficient is temperature dependent and may be given by apower law The thermal conductivity of some materials suchas gallium nitride (GaN) and Aluminium Nitride (AlN) maybe modeled by power law temperature dependency [29 30]Thus the solutions constructed here give a better comparisonof heat transfer in terms of material used since in many engi-neering applications thermal conductivity is given as a linearfunction of temperature Furthermore a good study in termsof performance and efficiency of fin with different profiles isundertaken These finding could help in the design of fins Itis claimed in [14] that DTM results are more accurate thanthose constructed by Variational Iteration Methods (VIM)and Homotopy Perturbation Methods (HPM) However itwould be risky to use the DTM approximate solutions asbenchmarks for the numerical schemes Nevertheless wehave also shown that DTM converges rapidly in just fifteenterms to the exact solution

Acknowledgments

The authors wish to thank the National Research Foundationof South Africa and the University of the Witwatersrandfor the generous financial support The authors are gratefulto Professor D P Mason and the anonymous reviewers fortheir invaluable comments which improved this paper signif-icantly

References

[1] A D Kraus A Aziz and J Welty Extended Surface Heat Trans-fer Wiley New York NY USA 2001

[2] R JMoitsheki T Hayat andM YMalik ldquoSome exact solutionsof the fin problem with a power law temperature-dependentthermal conductivityrdquo Nonlinear Analysis Real World Applica-tions vol 11 no 5 pp 3287ndash3294 2010

[3] R J Moitsheki ldquoSteady one-dimensional heat flow in a lon-gitudinal triangular and parabolic finrdquo Communications inNonlinear Science and Numerical Simulation vol 16 no 10 pp3971ndash3980 2011

[4] R J Moitsheki ldquoSteady heat transfer through a radial fin withrectangular and hyperbolic profilesrdquo Nonlinear Analysis RealWorld Applications vol 12 no 2 pp 867ndash874 2011

[5] Mo Miansari D D Ganji and Me Miansari ldquoApplication ofHersquos variational iteration method to nonlinear heat transferequationsrdquo Physics Letters A vol 372 no 6 pp 779ndash785 2008

[6] C H Chiu and C K Chen ldquoA decomposition method forsolving the convective longitudinal fins with variable thermalconductivityrdquo International Journal of Heat and Mass Transfervol 45 no 10 pp 2067ndash2075 2002

[7] A Rajabi ldquoHomotopy perturbation method for fin efficiencyof convective straight fins with temperature-dependent thermalconductivityrdquo Physics Letters A vol 364 no 1 pp 33ndash37 2007

[8] G Domairry and M Fazeli ldquoHomotopy analysis method todetermine the fin efficiency of convective straight fins withtemperature-dependent thermal conductivityrdquo Communica-tions in Nonlinear Science and Numerical Simulation vol 14 no2 pp 489ndash499 2009

[9] A Campo andR J Spaulding ldquoCoupling of themethods of suc-cessive approximations and undetermined coefficients for theprediction of the thermal behaviour of uniform circumferentialfinsrdquo Heat and Mass Transfer vol 34 no 6 pp 461ndash468 1999

[10] A Campo and F Rodrfguez ldquoApproximate analytic tempera-ture solution for uniform annular fins by adapting the powerseriesmethodrdquo International Communications inHeat andMassTransfer vol 25 no 6 pp 809ndash818 1998

[11] R Chiba ldquoApplication of differential transformmethod to ther-moelastic problem for annular disks of variable thickness withtemperature-dependent parametersrdquo International Journal ofThermophysics vol 33 pp 363ndash380 2012

[12] S Sadri M R Raveshi and S Amiri ldquoEfficiency analysis ofstraight fin with variable heat transfer coefficient and thermalconductivityrdquo Journal ofMechanical Science andTechnology vol26 no 4 pp 1283ndash1290 2012

[13] M Torabi H Yaghoori and A Aziz ldquoAnalytical solution forconvective-radiative continously moving fin with temperaturedependent thermal conductivityrdquo International Journal of Ther-mophysics vol 33 pp 924ndash941 2012

[14] M Torabi andH Yaghoobi ldquoTwo dominant analytical methodsfor thermal analysis of convective step fin with variable thermalconductivityrdquoThermal Science In press

[15] AMoradi ldquoAnalytical solutions for finwith temperature depen-dant heat transfer coefficientrdquo International Journal of Engineer-ing and Applied Sciences vol 3 no 2 pp 1ndash12 2011

[16] A Moradi and H Ahmadikia ldquoAnalytical solution for differentprofiles of fin with temperature-dependent thermal conductiv-ityrdquoMathematical Problems in Engineering vol 2010 Article ID568263 15 pages 2010

[17] H Yaghoobi and M Torabi ldquoThe application of differentialtransformation method to nonlinear equations arising in heattransferrdquo International Communications in Heat and MassTransfer vol 38 no 6 pp 815ndash820 2011

[18] S Ghafoori M Motevalli M G Nejad F Shakeri D D GanjiandM Jalaal ldquoEfficiency of differential transformation methodfor nonlinear oscillation Comparison with HPM and VIMrdquoCurrent Applied Physics vol 11 no 4 pp 965ndash971 2011

[19] A A Joneidi D D Ganji andM Babaelahi ldquoDifferential trans-formation method to determine fin efficiency of convectivestraight fins with temperature dependent thermal conductivityrdquoInternational Communications in Heat and Mass Transfer vol36 no 7 pp 757ndash762 2009

[20] J K Zhou Differential Transform Method and Its Applicationsfor Electric circuIts Huazhong University Press Wuhan China1986

14 Mathematical Problems in Engineering

[21] S Mukherjee ldquoReply to comment on lsquosolutions of the duffing-van der pol oscillator equation by the differential transformmethodrsquordquo Physica Scripta vol 84 Article ID 037003 2 pages2011

[22] C Bervillier ldquoStatus of the differential transformationmethodrdquoApplied Mathematics and Computation vol 218 no 20 pp10158ndash10170 2012

[23] A Arikoglu and I Ozkol ldquoSolution of fractional differentialequations by using differential transform methodrdquo ChaosSolitons and Fractals vol 34 no 5 pp 1473ndash1481 2007

[24] ZOdibat SMomani andV S Erturk ldquoGeneralized differentialtransform method application to differential equations offractional orderrdquo Applied Mathematics and Computation vol197 no 2 pp 467ndash477 2008

[25] CW Bert ldquoApplication of differential transformmethod to heatconduction in tapered finsrdquo Journal of Heat Transfer vol 124no 1 pp 208ndash209 2002

[26] F Khani and A Aziz ldquoThermal analysis of a longitudinal trape-zoidal fin with temperature-dependent thermal conductivityand heat transfer coefficientrdquo Communications in NonlinearScience and Numerical Simulation vol 15 no 3 pp 590ndash6012010

[27] H C Unal ldquoAn anlytical study of boiling heat transfer from afinrdquo International Journal of Heat and Mass Transfer vol 31 no7 pp 1483ndash1496 1988

[28] M H Chang ldquoA decomposition solution for fins with tempera-ture dependent surface heat fluxrdquo International Journal of Heatand Mass Transfer vol 48 no 9 pp 1819ndash1824 2005

[29] A Jezowski B A Danilchenko M Bockowski et al ldquoThermalconductivity of GaN crystals in 42ndash300 K rangerdquo Solid StateCommunications vol 128 no 2-3 pp 69ndash73 2003

[30] S Vitanov V Palankovski S Maroldt and R Quay ldquoHigh-temperature modeling of AlGaNGaN HEMTsrdquo Solid-StateElectronics vol 54 no 10 pp 1105ndash1112 2010

Submit your manuscripts athttpwwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical Problems in Engineering

Hindawi Publishing Corporationhttpwwwhindawicom

Differential EquationsInternational Journal of

Volume 2014

Applied MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical PhysicsAdvances in

Complex AnalysisJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

OptimizationJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Operations ResearchAdvances in

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of Mathematics and Mathematical Sciences

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Algebra

Discrete Dynamics in Nature and Society

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Decision SciencesAdvances in

Discrete MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom

Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

Page 13: Research Article Analytical Solutions for Steady Heat ...downloads.hindawi.com/journals/mpe/2013/273052.pdf · Research Article Analytical Solutions for Steady Heat Transfer in Longitudinal

Mathematical Problems in Engineering 13

Transform Method (FDTM) introduced by Arikoglu andOzkol [23]

We have shown with the help of an example that DTMmay only be applied to fin problems involving heat transferthrough fins with convex parabolic profile Note that given anODE such as (4) with a power law heat transfer coefficient ofa fractional exponent then one can easily remove the fractionby basic exponential rules and employment of the Binomialexpansion However using the DTM one runs into difficultyif the power law thermal conductivity in the same equation isgiven by the fractional exponent We do not know whetherthese observations call for the ldquomodifiedrdquo DTM to solveproblems arising in heat flow through fins with other profilessuch as longitudinal triangular and concave parabolic andalso with fractional power law thermal conductivity

The main results obtained in this paper give insightinto heat transfer in boiling liquids where the heat transfercoefficient is temperature dependent and may be given by apower law The thermal conductivity of some materials suchas gallium nitride (GaN) and Aluminium Nitride (AlN) maybe modeled by power law temperature dependency [29 30]Thus the solutions constructed here give a better comparisonof heat transfer in terms of material used since in many engi-neering applications thermal conductivity is given as a linearfunction of temperature Furthermore a good study in termsof performance and efficiency of fin with different profiles isundertaken These finding could help in the design of fins Itis claimed in [14] that DTM results are more accurate thanthose constructed by Variational Iteration Methods (VIM)and Homotopy Perturbation Methods (HPM) However itwould be risky to use the DTM approximate solutions asbenchmarks for the numerical schemes Nevertheless wehave also shown that DTM converges rapidly in just fifteenterms to the exact solution

Acknowledgments

The authors wish to thank the National Research Foundationof South Africa and the University of the Witwatersrandfor the generous financial support The authors are gratefulto Professor D P Mason and the anonymous reviewers fortheir invaluable comments which improved this paper signif-icantly

References

[1] A D Kraus A Aziz and J Welty Extended Surface Heat Trans-fer Wiley New York NY USA 2001

[2] R JMoitsheki T Hayat andM YMalik ldquoSome exact solutionsof the fin problem with a power law temperature-dependentthermal conductivityrdquo Nonlinear Analysis Real World Applica-tions vol 11 no 5 pp 3287ndash3294 2010

[3] R J Moitsheki ldquoSteady one-dimensional heat flow in a lon-gitudinal triangular and parabolic finrdquo Communications inNonlinear Science and Numerical Simulation vol 16 no 10 pp3971ndash3980 2011

[4] R J Moitsheki ldquoSteady heat transfer through a radial fin withrectangular and hyperbolic profilesrdquo Nonlinear Analysis RealWorld Applications vol 12 no 2 pp 867ndash874 2011

[5] Mo Miansari D D Ganji and Me Miansari ldquoApplication ofHersquos variational iteration method to nonlinear heat transferequationsrdquo Physics Letters A vol 372 no 6 pp 779ndash785 2008

[6] C H Chiu and C K Chen ldquoA decomposition method forsolving the convective longitudinal fins with variable thermalconductivityrdquo International Journal of Heat and Mass Transfervol 45 no 10 pp 2067ndash2075 2002

[7] A Rajabi ldquoHomotopy perturbation method for fin efficiencyof convective straight fins with temperature-dependent thermalconductivityrdquo Physics Letters A vol 364 no 1 pp 33ndash37 2007

[8] G Domairry and M Fazeli ldquoHomotopy analysis method todetermine the fin efficiency of convective straight fins withtemperature-dependent thermal conductivityrdquo Communica-tions in Nonlinear Science and Numerical Simulation vol 14 no2 pp 489ndash499 2009

[9] A Campo andR J Spaulding ldquoCoupling of themethods of suc-cessive approximations and undetermined coefficients for theprediction of the thermal behaviour of uniform circumferentialfinsrdquo Heat and Mass Transfer vol 34 no 6 pp 461ndash468 1999

[10] A Campo and F Rodrfguez ldquoApproximate analytic tempera-ture solution for uniform annular fins by adapting the powerseriesmethodrdquo International Communications inHeat andMassTransfer vol 25 no 6 pp 809ndash818 1998

[11] R Chiba ldquoApplication of differential transformmethod to ther-moelastic problem for annular disks of variable thickness withtemperature-dependent parametersrdquo International Journal ofThermophysics vol 33 pp 363ndash380 2012

[12] S Sadri M R Raveshi and S Amiri ldquoEfficiency analysis ofstraight fin with variable heat transfer coefficient and thermalconductivityrdquo Journal ofMechanical Science andTechnology vol26 no 4 pp 1283ndash1290 2012

[13] M Torabi H Yaghoori and A Aziz ldquoAnalytical solution forconvective-radiative continously moving fin with temperaturedependent thermal conductivityrdquo International Journal of Ther-mophysics vol 33 pp 924ndash941 2012

[14] M Torabi andH Yaghoobi ldquoTwo dominant analytical methodsfor thermal analysis of convective step fin with variable thermalconductivityrdquoThermal Science In press

[15] AMoradi ldquoAnalytical solutions for finwith temperature depen-dant heat transfer coefficientrdquo International Journal of Engineer-ing and Applied Sciences vol 3 no 2 pp 1ndash12 2011

[16] A Moradi and H Ahmadikia ldquoAnalytical solution for differentprofiles of fin with temperature-dependent thermal conductiv-ityrdquoMathematical Problems in Engineering vol 2010 Article ID568263 15 pages 2010

[17] H Yaghoobi and M Torabi ldquoThe application of differentialtransformation method to nonlinear equations arising in heattransferrdquo International Communications in Heat and MassTransfer vol 38 no 6 pp 815ndash820 2011

[18] S Ghafoori M Motevalli M G Nejad F Shakeri D D GanjiandM Jalaal ldquoEfficiency of differential transformation methodfor nonlinear oscillation Comparison with HPM and VIMrdquoCurrent Applied Physics vol 11 no 4 pp 965ndash971 2011

[19] A A Joneidi D D Ganji andM Babaelahi ldquoDifferential trans-formation method to determine fin efficiency of convectivestraight fins with temperature dependent thermal conductivityrdquoInternational Communications in Heat and Mass Transfer vol36 no 7 pp 757ndash762 2009

[20] J K Zhou Differential Transform Method and Its Applicationsfor Electric circuIts Huazhong University Press Wuhan China1986

14 Mathematical Problems in Engineering

[21] S Mukherjee ldquoReply to comment on lsquosolutions of the duffing-van der pol oscillator equation by the differential transformmethodrsquordquo Physica Scripta vol 84 Article ID 037003 2 pages2011

[22] C Bervillier ldquoStatus of the differential transformationmethodrdquoApplied Mathematics and Computation vol 218 no 20 pp10158ndash10170 2012

[23] A Arikoglu and I Ozkol ldquoSolution of fractional differentialequations by using differential transform methodrdquo ChaosSolitons and Fractals vol 34 no 5 pp 1473ndash1481 2007

[24] ZOdibat SMomani andV S Erturk ldquoGeneralized differentialtransform method application to differential equations offractional orderrdquo Applied Mathematics and Computation vol197 no 2 pp 467ndash477 2008

[25] CW Bert ldquoApplication of differential transformmethod to heatconduction in tapered finsrdquo Journal of Heat Transfer vol 124no 1 pp 208ndash209 2002

[26] F Khani and A Aziz ldquoThermal analysis of a longitudinal trape-zoidal fin with temperature-dependent thermal conductivityand heat transfer coefficientrdquo Communications in NonlinearScience and Numerical Simulation vol 15 no 3 pp 590ndash6012010

[27] H C Unal ldquoAn anlytical study of boiling heat transfer from afinrdquo International Journal of Heat and Mass Transfer vol 31 no7 pp 1483ndash1496 1988

[28] M H Chang ldquoA decomposition solution for fins with tempera-ture dependent surface heat fluxrdquo International Journal of Heatand Mass Transfer vol 48 no 9 pp 1819ndash1824 2005

[29] A Jezowski B A Danilchenko M Bockowski et al ldquoThermalconductivity of GaN crystals in 42ndash300 K rangerdquo Solid StateCommunications vol 128 no 2-3 pp 69ndash73 2003

[30] S Vitanov V Palankovski S Maroldt and R Quay ldquoHigh-temperature modeling of AlGaNGaN HEMTsrdquo Solid-StateElectronics vol 54 no 10 pp 1105ndash1112 2010

Submit your manuscripts athttpwwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical Problems in Engineering

Hindawi Publishing Corporationhttpwwwhindawicom

Differential EquationsInternational Journal of

Volume 2014

Applied MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical PhysicsAdvances in

Complex AnalysisJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

OptimizationJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Operations ResearchAdvances in

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of Mathematics and Mathematical Sciences

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Algebra

Discrete Dynamics in Nature and Society

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Decision SciencesAdvances in

Discrete MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom

Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

Page 14: Research Article Analytical Solutions for Steady Heat ...downloads.hindawi.com/journals/mpe/2013/273052.pdf · Research Article Analytical Solutions for Steady Heat Transfer in Longitudinal

14 Mathematical Problems in Engineering

[21] S Mukherjee ldquoReply to comment on lsquosolutions of the duffing-van der pol oscillator equation by the differential transformmethodrsquordquo Physica Scripta vol 84 Article ID 037003 2 pages2011

[22] C Bervillier ldquoStatus of the differential transformationmethodrdquoApplied Mathematics and Computation vol 218 no 20 pp10158ndash10170 2012

[23] A Arikoglu and I Ozkol ldquoSolution of fractional differentialequations by using differential transform methodrdquo ChaosSolitons and Fractals vol 34 no 5 pp 1473ndash1481 2007

[24] ZOdibat SMomani andV S Erturk ldquoGeneralized differentialtransform method application to differential equations offractional orderrdquo Applied Mathematics and Computation vol197 no 2 pp 467ndash477 2008

[25] CW Bert ldquoApplication of differential transformmethod to heatconduction in tapered finsrdquo Journal of Heat Transfer vol 124no 1 pp 208ndash209 2002

[26] F Khani and A Aziz ldquoThermal analysis of a longitudinal trape-zoidal fin with temperature-dependent thermal conductivityand heat transfer coefficientrdquo Communications in NonlinearScience and Numerical Simulation vol 15 no 3 pp 590ndash6012010

[27] H C Unal ldquoAn anlytical study of boiling heat transfer from afinrdquo International Journal of Heat and Mass Transfer vol 31 no7 pp 1483ndash1496 1988

[28] M H Chang ldquoA decomposition solution for fins with tempera-ture dependent surface heat fluxrdquo International Journal of Heatand Mass Transfer vol 48 no 9 pp 1819ndash1824 2005

[29] A Jezowski B A Danilchenko M Bockowski et al ldquoThermalconductivity of GaN crystals in 42ndash300 K rangerdquo Solid StateCommunications vol 128 no 2-3 pp 69ndash73 2003

[30] S Vitanov V Palankovski S Maroldt and R Quay ldquoHigh-temperature modeling of AlGaNGaN HEMTsrdquo Solid-StateElectronics vol 54 no 10 pp 1105ndash1112 2010

Submit your manuscripts athttpwwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical Problems in Engineering

Hindawi Publishing Corporationhttpwwwhindawicom

Differential EquationsInternational Journal of

Volume 2014

Applied MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical PhysicsAdvances in

Complex AnalysisJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

OptimizationJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Operations ResearchAdvances in

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of Mathematics and Mathematical Sciences

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Algebra

Discrete Dynamics in Nature and Society

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Decision SciencesAdvances in

Discrete MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom

Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

Page 15: Research Article Analytical Solutions for Steady Heat ...downloads.hindawi.com/journals/mpe/2013/273052.pdf · Research Article Analytical Solutions for Steady Heat Transfer in Longitudinal

Submit your manuscripts athttpwwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical Problems in Engineering

Hindawi Publishing Corporationhttpwwwhindawicom

Differential EquationsInternational Journal of

Volume 2014

Applied MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical PhysicsAdvances in

Complex AnalysisJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

OptimizationJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Operations ResearchAdvances in

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of Mathematics and Mathematical Sciences

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Algebra

Discrete Dynamics in Nature and Society

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Decision SciencesAdvances in

Discrete MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom

Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of