AHT Assignment 2

Submitted by

Aravind G P

SC15D024

Phd

IIST

Consider the radial fin of triangular profile as shown above.

The generalized equation for any arbitrary shape radial fin is given by,

0)()(

)( 22

2

2

2

k

h

dr

d

dr

rdf

dr

d

r

rf

dr

drf

In case of a triangular fin, )(2

)(2 rrb

rf ab

rdr

rdf b

2

)(2

Fin height, b = ra-rb

So the governing equation will become

0)2()( 22

2

rbmdr

drr

dr

drrr aa

2

1

2

bk

hm

Let rrv a

drdv

Differential equation is transformed into

0)()2()( 22

2

vrbmdv

dvra

dv

dvrv aa

This equation can be solved by the method of Frobenius

Assume that

...)1262(

...)432(2.......)()1(

)432(.......)(

.......)(

2

432

3

4

2

321

13

3

2

210

2

2

2

3

4

2

321

3

3

2

210

1

3

3

2

210

vavaav

vavavaapvvavavaavppdv

d

and

vavavaavvavavaapvdv

d

So

vavavaav

p

pp

pp

p

Substituting the assumed values of and its derivatives in to the transformed differential

equation, we will get a series involving v to various powers of p results:

02122223

0

22

1

2

2

0

2

1

321

2

4

1

3

21

1

)23()44(

)()12(

,,

0......

abmarbmpparppA

arbmpparppA

arpA

areAAA

vAvAvAvA

aa

aa

a

pppp

The indical equation derives from the sum of the entities in the column headed by the lowest

power of v. it is observed that 02 arp a =0 and because ra is a physical dimension that cannot

equal to zero, a trivial solution would result if a0=0. Thus the only alternative is that p=0. The

theory then says that the solution for will be

0 0

2

0

1 lnk k

k

k

k

k

k

k

k vbvavCvaC

But to keep finite at r=ra, where v = ra-ra =0, C2 must be zero, and there after application of

the boundary condition, the solution will become

0

1

k

k

kvaC

which is the required temperature distribution.

To Find:

(a) Temperature distribution within the plate using prescribed grid spacing,

(b) Sketch isotherms to illustrate temperature distribution,

(c) Heat loss by convection from exposed surface (compare with element dissipation rate)

Assumptions:

(1) Steady-state, two-dimensional conduction in ceramic plate

(2) Constant properties

(3) No internal generation, except for Node 7

(4) Heating element approximates a line source of negligible wire diameter.

Ans:

(a) The prescribed grid for the symmetry element shown above consists of 12 nodal points.

Nodes 1-3 are points on a surface experiencing convection; nodes 4-6 and 8-12 are interior

nodes. Node 7 is a special case of the interior node having a generation term; because of

symmetry, qht = 25 W/m.

The finite-difference equations are derived as follows:

From an energy balance on Node 2 with x/y = 3,

Ein Eout = qa + qb + qc + qd = 0;

022

)(2

25232

21

y

TTyxk

x

TTykTTxh

x

TTyk

T

y

xNT

y

xTT

y

x

y

xNT 222121 5

2

31

2

2

where

N = hx/k = K30.02

006.0100

Hence, with T = 30C,

T2 = 0.04587T1 + 0.04587T3 + 0.82569T5 + 2.4771

From this FDE, the forms for nodes 1 and 3 can also be deduced.

From an energy balance for node 7, with x/y = 3,

Ein Eg = 0

where Eg = 2 qht and Ein represents the conduction terms.

Hence,

qa + qb + qc + qd + 2qht = 0

02710787478

htq

y

TTxk

x

TTyk

y

TTxk

x

TTyk

Regrouping,

y

x

k

qT

y

xTT

y

xT

y

x

y

xT ht

211 10

2

87

2

8

22

7

Recognizing that x/y = 3, qht = 25 W/m and k = 2 W/mK, the FDE is

T7 = 0.0500T8 + 0.4500T4 + 0.0500T8 + 0.4500T10 + 3.7500

The FDEs for the remaining nodes may be deduced from this form. Following the procedure

described in Section 4.6.2 of Fundamentals of Heat and Mass Transfer by Incropera (Wiley

India Edition) for the Gauss-Seidel method, the system of FDEs has the form:

kkk

kkkk

kkk

kkkk

kkkkk

kkkk

kkkk

kkkkk

kkkk

kkk

kkkk

kkk

TTT

TTTT

TTT

TTTT

TTTTT

TTTT

TTTT

TTTTT

TTTT

TTT

TTTT

TTT

11912

1

12

1

10811

1

11710

1

12869

1

11

1

9758

1

10

1

847

1

9546

1

8

1

6425

1

7

1

514

1

623

1

5

1

312

1

4

1

21

10.09.0

05.005.09.0

10.09.0

45.010.045.0

45.005.005.045.0

75.345.010.045.0

45.010.045.0

45.005.005.045.0

45.01.045.0

45771.28257.009174.0

4771.28257.004587.004587.0

45771.28257.009174.0

Note the use of the superscript k to denote the level of iteration. Begin the iteration

procedure with rational estimates for Ti (k = 0)

k/Ti 1 2 3 4 5 6 7 8 9 10 11 12

0 55.0 50.0 45.0 61.0 54.0 47.0 65.0 56.0 49.0 60.0 55.0 50.0

1 57.4 51.7 46.0 60.4 53.8 48.1 63.5 54.6 49.6 62.7 54.8 50.1

2 57.1 51.6 46.9 59.7 53.2 48.7 64.3 54.3 49.9 63.4 54.5 50.4

55.80 49.93 47.67 59.03 51.72 49.19 63.89 52.98 50.14 62.84 53.35 50.46

The last row with k = corresponds to the solution obtained by matrix inversion. It appears

that at least 20 iterations would be required to satisfy the convergence criterion using the

Gauss-Seidel method.

(b) Selected isotherms are shown in the sketch of the nodal network.

(c) The heat loss by convection can be expressed as

)(

2

1)()(

2

1321 TTxTTxTTxhqconv

)3067.47(006.0

2

1)3093.49(006.0)3080.55(006.0

2

1100convq 25W/m

The heat loss by convection is equal to the heater element dissipation.

References

1. Extended Surface Heat Transfer by Allan D Kraus, Abdul Aziz, James Welty

2. Fundamentals of Heat and Mass Transfer by Frank P Incropera, David P Dewitt,

Theodere L Bergman, Adrienne S Lavine

3. Heat Transfer by A F Mills and V Ganesan