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CHAPTER I
INTRODUCTION
Heat transfer is energy in transit, which occurs as a result of a temperature gradient or
difference. This temperature difference is thought of as a driving force that causes heat to
flow. Heat transfer occurs by three basic mechanisms or modes: conduction, convection,
and radiation.
Conduction
Conduction heat transfer is defined as heat transfer in solids and fluids without bulk
motion. Heat conduction generally takes place in solids, through it may occur in fluids
without bulk motion or with rigid body motion. In fluids, conduction is due to the
collusions of the molecules during their random motion. In solids, it is due to the
combination of vibrations of molecules in a lattice and the energy transport by free
electrons.
heat (q)
cold hot
T1 T2
L
concrete wall
Figure 1.1 Heat conduction through a concrete wall.
It is observed that the rate of heat conduction through a wall (qx) with constant
thickness is proportional to the temperature difference (T2 - T1) between the surfaces and
the area normal to the heat flow direction (A) and is inversely proportional to the thickness
of the wall (L).
LTT
Akq 12x
−⋅⋅−= (1.1)
where k is thermal conductivity of the wall’s material (W/m.K) A is cross-sectional area of the wall (m2) L is the thickness of the wall (m) T1, T2 is surface temperatures of the wall (°C) Table 1.1 Thermal conductivity of some material at 300 K. ----------------------------------------------------------------------------------------------------------- Material k (W/m.K) ----------------------------------------------------------------------------------------------------------- Silver (pure) 427 Copper (pure) 399 Gold 316 Stainless steel (316) 14.4 Glass 0.81 Concrete 0.128 Fiberglass wool 0.040 Air 0.0262 Water 0.540 Wood 0.17 ----------------------------------------------------------------------------------------------------------- Convection
Convection is the mode of heat transfer between a solid surface and the adjacent fluid that
is in motion, and it involves the combined effects of conduction and fluid motion. The
faster the fluid motion, the greater the convection heat transfer. In the absence of any bulk
fluid motion, heat transfer between a solid surface and the adjacent fluid is by pure
conduction. The presence of bulk motion of the fluid enhances the heat transfer between
the solid surface and the fluid.
Convection is called forced convection if the fluid is forced to flow over the surface
by external means such as a fan, pump, or the wind. In contrast, convection is called
natural or free convection if the fluid motion is caused by buoyancy force that is induced
by density difference due to the variation of temperature in the fluid. Heat transfer from a
solid surface can be obtained from:
)TT(Ahq wc ∞−⋅⋅= (1.2)
where ch is convective heat transfer coefficient (W/m2.K) A is heat transfer area (m2) Tw is surface temperatures of the wall (°C) T∞ is the bulk fluid temperature (°C) Table 1.2 Typical values of convective heat transfer coefficients ----------------------------------------------------------------------------------------------------------- Type of convection h (W/m2.K) ----------------------------------------------------------------------------------------------------------- Natural convection (air) 5 - 15 Natural convection (water) 500 - 1000 Force convection (air) 10 - 200 Force convection (oil) 20 - 2000 Force convection (water) 300 - 20000 Water boiling 3000 - 100000 Water condensing 5000- 10000 ----------------------------------------------------------------------------------------------------------- Radiation
Radiation is the energy emitted by matter in the form of electromagnetic waves as a result
of the changes in the electronic configurations of the atoms or molecules. Unlike
conduction and convection, the transfer of energy by radiation does not require the
presence of an intervening medium. In fact, energy transfer by radiation is the fastest (at
the speed of light) and it suffers no attenuation in a vacuum. This is exactly how the energy
of the sun reaches the earth.
The maximum rate of radiation which can be emitted from a surface at an absolute
temperature (Ts) is given by the Stefan-Boltzmann law as:
4semit TAq ⋅⋅σ⋅ε= (1.3)
Where = 5.67 × 10σ -8 W/m2.K4 is the Stefan-Boltzmann constant and ε is the emissivity of
the surface. An idealized surface, which emits radiation at a maximum rate has ε = 1, is
known as a blackbody. The radiation emitted by actual surfaces is less than that emitted by
the blackbody. The value of ε is in the range 0 ≤ ε ≤ 1, is a measure of how closely a
surface approximates a blackbody.
Table 1.3 Emissivity of some materials at 300 K ---------------------------------------------------------------------------------------------------------------------------- material ε material ε ---------------------------------------------------------------------------------------------------------------------------- aluminum foil 0.07 Black paint 0.98 anodized aluminum 0.82 White paint 0.90 polished copper 0.03 White paper 0.92-0.97 polished gold 0.03 Asphalt pavement 0.85-0.93 polished silver 0.02 Human skin 0.95 polished stainless-steel 0.17 Wood 0.82-0.92 black paint 0.98 Soil 0.93-0.96 Water 0.96 ----------------------------------------------------------------------------------------------------------------------------
Another important radiation property of a surface is its absorptivity (α), which is
the fraction of the radiation energy incident on a surface which is absorbed by the surface.
Like emissivity, its value is in the range 0 ≤ α ≤ 1. A blackbody absorbs the entire
radiation incident on it. That is, a black is a perfect absorber (α = 1) as well as a perfect
emitter. In practice, α and ε are assumed to independent from temperature and wavelength
of the radiation. The average absorptivity of a surface is taken to be equal to its average
emissivity. The rate at which a surface absorbs radiation is determined from:
incabs qq ⋅α= (1.4)
Where qinc is the rate at which radiation is incident on the surface. For nontransparent
surfaces, the portion of incident radiation not absorbed by the surface
qref = (1 - α)qinc
qinc
surface qabs = α qinc
Figure 1.2 The absorption of radiation incident on an opaque surface of absorptivity ∝.
The difference between the rates of radiation emitted by the surface and the
radiation absorbed is the net radiation heat transfer. If the rate of radiation absorption is
greater than the rate of radiation emission, the surface is said to be gaining energy by
radiation. Otherwise, the surface is said to be losing energy by radiation.
CHAPTER II
STEADY-STATE CONDUCTION IN ONE DIMENSION
2.1 Fourier’s law of heat conduction
Consider a wall as shown, we know that the rate of heat transfer through the wall increases
when:
• The temperatures difference between the left and right surfaces increase,
• The wall surface area increases,
• The wall thickness reduces,
• The wall is change from brick to aluminum.
If we measure temperatures of the wall from left to right and plot the temperature variation
with the wall thickness, we get:
Tcold
L
q dX
dTslope = Thot
Tcold
Thot
0 L
Figure 2.1 Heat conduction through a wall.
Relative to the heat flow direction, the slope of the temperature line is negative as the
temperature decrease with the heat flow direction.
Therefore, the relation can be written as:
x
TAkqx∂
∂⋅⋅−= (2.1)
This relation is known a Fourier’s law of heat conduction.
2.2 General conduction equation based on Cartesian Coordinates
qx
qz
qy
qy+d
qz+d
qx+d
Figure 2.2 A control volume for deriving the three-dimensional conduction equation in Cartesian Coordinates.
Apply the first law of thermodynamics to the element, we get:
Rate of energy stored inside the control volume
+Rate of energy conducted out off the control
+Rate of energy generated inside the control volume
Rate of energy conducted into the control volume
Rate of energy conducted into the system:
x
Tdzdykq x∂
∂⋅⋅⋅−= ,
y
Tdzdxky∂
∂⋅⋅⋅−=q ,
z
Tdxdykz∂
∂⋅⋅⋅−=q (2.2)
Rate of energy conducted out off the system:
dxx
qqq xxdxx ⋅
∂
∂+=+ , dy
y
qqq y
ydyy ⋅∂
∂+=+ , dz
z
qqq z
zdzz ⋅∂
∂+=+ (2 .3)
Rate of energy generated inside the system:
dzdydxqqgen ⋅⋅⋅′′′= (2.4)
Rate of energy stored inside the system
tTdzdydxCq store ∂
∂⋅⋅⋅⋅ρ⋅= (2.5)
By combining equations 2.2 to 2.5, then:
t
T1
z
T
y
T
x
Tkq2
2
2
2
2
2
∂
∂⋅
α+
∂
∂+
∂
∂+
∂
∂−=′′′ (2.6)
Where α is called thermal diffusivity of material (m2/sec):
Ck⋅ρ
=α (2.7)
Note that the thermal conductivity (k) represented how well a material conducts heat, and
the heat capacity represents how much energy a material stores per unit volume. The
larger the diffusivity, the faster the propagation of heat into the medium. A small value of
thermal diffusivity means that heat is mostly abs0rbed by the material and a small amount
of heat will be conducted further.
)C( ⋅ρ
2.2.1 Heat transfer through a wall
x
T2
T
For this case, the process is steady-state, no internal heat
generated, and one dimensional heat flow, therefore
equation 2.6 can be simplified as:
T1
0x
T2
2=
∂
∂ (2.8)
Figure 2.3 Steady-state temperature
L
distribution within a plane wall.
By integrating equation 2.8:
1Cdx
dT= (2.9)
21 CxCT +⋅= (2.10)
and we know at x = 0, T = T1, and at x = L, T = T2, therefore:
at x = 0
21 CT =
at x = L
112 TLCT +⋅=
then we have
112x TL
x)TT(T +⋅−= (2.11)
by differentiating equation 2.11 and applying to a Fourier’s law of heat conduction, the
heat transfer rate through the wall is then obtained from:
L
TTAkq 12x
−⋅⋅−= (2.12)
If we define a conductive thermal resistance as:
Ak
LRk⋅
= (2.13)
Then equation 2.12 may be written as:
k
12x
R
TTq −= (2.14)
This equation is analogous to the relation for electric current flow:
e
12
R
VVI −= (2.15)
Based on this analogy, the system can be drawn schematically as:
T2 T1
Rk
Figure 2.4 Conductive thermal resistance.
2.2.2 Composite wall (materials in series)
Rk-2
T2 Rk-1
T1 T0 Rk-3
T3
∆ x3∆ x1 ∆ x2
T0
T3T2
T1
Figure 2.5 Composite wall with material in series.
Heat transfer rate through material 1
1k
1010
1
11
R
)TT()TT(
x
Akq−
−=−⋅
∆
⋅= (2.13)
Heat transfer rate through material 2
2k
2121
2
22
R
)TT()TT(x
Akq−
−=−⋅
∆
⋅= (2.14)
Heat transfer rate through material 3
3k
3232
3
31 R
)TT()TT(
xAk
q−
−=−⋅
∆⋅
= (2.15)
As the system is steady-state and no internal heat generated, the heat flows enter and exit
each layer are equal. Therefore:
x321 qqqq === (2.16)
Then, by combining equations 2.13 to 2.14:
n
30
3k2k1k
30x
R
TT
RRR
TTq
Σ
−=
++
−=
−−−
(2.17)
2.2.3 Composite wall (material in parallel)
Rk-3
Rk-2b
Rk-2a
T3 T2
Rk-1
T1
T0
3
2b
2a
1
Figure 2.6 Composite wall in a series/parallel arrangement.
If it is assumed that each layer has a uniform temperature:
x3b2a21 qqqqq ==+= (2.18)
3k
32
b2k
21
a2k
21
1k
10
R
TT
R
TT
R
TT
R
TT
−−−−
−=
−+
−=
− (2.19)
The overall thermal resistance of layer 2 is:
b2ka2k2k R
1
R
1
R
1
−−−
+= (2.20)
Therefore, the heat transfer through this wall is:
3k2k1k
30x
RRR
TTq
−−− ++
−= (2.21)
3
3
b2a2
2
1
1
30
)Ak(
x
)Ak()Ak(
x
)Ak(
xTT
⋅
∆+
⋅+⋅
∆+
⋅
∆−
= (2.22)
2.2.4 Plane wall with internal heat generated
A wall in which there is internal heat
generation per unit volume, q . The heat
source is at the center plane, thus we can
expect a temperature profile that is symmetric
about the center. For this case, the process is
steady-state and one-dimensional heat flow,
therefore, equation 2.6 can be simplified as:
′′′
heat source
+X
L L
T = Tmax
T = Tw
T
- X
Figure 2.7 Plane wall with internal heat generation.
2
2
xd
Td
k
q=
′′′− (2.23)
By integrating this equation:
1Ck
xqdx
dT+′′′−= (2.24)
21
2CxC
k2
xqT +⋅+⋅
′′′−= (2.25)
We know that at the surface T = Tw, thus from equation 2.25:
at x = L
21
2
w CLCk2
LqT +⋅+⋅
′′′−= (2.26)
at x = -L
21
2
w CLCk2
LqT +⋅−⋅
′′′−= (2.27)
By combing equation 2.26 and 2.27:
2
2
w2 Ck2
LqTC +⋅
′′′+= (2.28)
At the x = 0 (center plane), T = Tmax and 0dx
dT= , thus from equation 2.24:
0C1 = (2.27)
As C1 and C2 are already known, the solution of equation 2.23 is:
−
⋅′′′+=
2
22
wxL
x1k2
LqTT (2.28)
2.2.5 Overall heat transfer coefficient
Heat transfer by convection between the wall’s surface and fluid can be obtained from:
)TT(Ahq wc ∞−⋅⋅= (2.29)
and the convective thermal resistance is:
Ah
1Rc
c⋅
= (2.30)
Rc-2 T2
Rc-1 T1 T∞ - 1
Rk T∞ - 2
T2
T1 T∞ -1
T∞ - 2
Figure 2.8 Combine convection-conduction heat transfer through a wall.
The heat transfer rate through the wall in term of fluid temperatures is:
Ah1
Akx
Ah1
TTq
2c1c
21
⋅+
⋅∆
+⋅
−=
−−
−∞−∞ (2.31)
2ck1c
21
RRRTT
−−
−∞−∞
++−
= (2.32)
It is convenient to express the heat transfer in term of a single value that accounts for both
conduction and convection resistances, thus:
)TT(AUq 21 −∞−∞ −⋅⋅= (2.32)
where U is defined as an over heat transfer coefficient and:
)RRR(A
1
h
1
k
x
h
11U
2ck1c
2c1c
−−
−−
++=
+∆
+= (2.33)
Then
kc RR
1AUΣ+Σ
=⋅ (2.34)
2.3 General conduction equation based on Polar Cylindrical Coordinates
qθ+dθ
qθ
qz+dz
qz
dr
dz
r⋅dθ
qr+dr
qr
Figure 2.9 Volume element in a cylindrical coordinates.
Similar to the case for Cartesian Coordinates, energy equation can be obtained as:
tTdzdr)dr(Cdz
zq
qdq
qdrr
qqdzdr)dr(qqqq
z
zr
rzr
∂∂
⋅θ⋅ρ⋅+∂
∂+
+θθ∂
∂++
∂∂
+=⋅θ⋅′′′+++ θθθ
(2.35)
Equation 2.35 can be rewritten as:
t
T1
k
q
z
TT
r
1
r
T
r
1
r
T2
2
2
2
22
2
∂
∂
α=
′′′+
∂
∂+
θ∂
∂+
∂
∂+
∂
∂ (2.36)
2.3.1 Heat transfer to/from a circular duct
LT
T1
T2
R1
R2
r
Figure 2.10 Heat conduction through a cylindrical wall.
Assuming that, the system is steady-state, there is no internal heat generation, temperature
varies only with r. Thus equation 2.36 is simplified to:
0rd
Tdrrd
d
r
1
r
T
r
1
r
T2
2=
⋅=
∂
∂+
∂
∂ (2.37)
0rd
Tdrdr
d=
(2.38)
By integrating equation 2.36:
1Crd
Tdr = r
C
rd
Td 1= (2.39)
By integrating again:
21 C)r(lnCT += (2.40)
The boundary condition are: at r = R1, T = T1 and at r = R2, T = T2.
at R1
(2.41) 2111 C)R(lnCT +=
at R2
2212 C)R(lnCT += (2.42)
then we can solve for C1 and C2, thus:
)R/R(ln
)R/r(ln
TT
TT
21
1
12
1 =−
− (2.43)
equation 2.41 can be used to calculate temperature at r when R1 < r < R2. The heat transfer
through the pipe wall can be obtained by applying Fourier’s law of heat conduction, thus:
r
T)Lr2(kqr∂
∂⋅⋅π⋅⋅−= (2.44)
by differentiating equation 2.43 and combining with equation 2.44, thus:
−⋅⋅π⋅⋅−=
)R/R(ln
TT
r
1)Lr2(kq21
21r (2.45)
)TT()R/R(ln
Lk221
12
−⋅⋅π⋅
= (2.46)
k
21
R
TT −= (2.47)
Thus the thermal resistance is:
Lk2
)R/R(lnR 12k
⋅⋅π⋅= (2.48)
2.3.2 Overall heat transfer coefficient
Rk-1 Rc-1 Rc-2 Rk-2
T∞1
T3T2
T1
T∞2
T∞1
R1
R3
R2
insulation
pipe
T∞2 T1 T2 T3
Figure 2.11 Heat flow through a cylinder with convection.
The heat transfer between the fluids can be obtained similar to the case for composites wall
(series), thus:
2c2k1k1c
21r RRRR
TTq
−−−−
−∞−∞
+++−
= (2.49)
and the thermal resistances are:
LR2h
1R11c
1c⋅⋅π⋅⋅
=
−
−
LR2h
1R22c
2c⋅⋅π⋅⋅
=
−
−
Lk2
)R/R(lnR1
121k
⋅⋅π⋅=−
Lk2
)R/R(lnR
2
232k
⋅⋅π⋅=−
Based on the inner surface area, the heat transfer rate may be obtained as:
)TT)(LR2(Uq 2111r −∞−∞ −⋅⋅π⋅⋅= (2.50)
The overall heat transfer coefficient based on the inner surface area is:
2c3
1
2
231
1
121
1c
1
hR
R
k
)R/R(lnR
k
)R/R(lnR
h
11U
−− ⋅+++
= (2.51)
Based on the outer surface area, the heat transfer rate may be obtained as:
)TT)(LR2(Uq 2132r −∞−∞ −⋅⋅π⋅⋅= (2.52)
The overall heat transfer coefficient based on the outer surface area is:
2c2
233
1
123
1c1
32
h
1
k
)R/R(lnR
k
)R/R(lnR
hR
R1U
−−
+++⋅
= (2.53)
2.3.3 Critical thickness of insulation
In order to reduce heat loss from the hot fluid flowing inside a pipe, insulating
material is normally used. When we add the insulation, the conductive thermal resistance is
definitely increased. This causes the heat loss to reduce. However, at the same time, adding
the insulation also increases the outer surface area. This causes the heat loss to increase.
Therefore, in some case, adding insulation may increase the heat loss. Particularly for a
large pipe and when the outer convective coefficient and thermal conductivity of the
insulation used are high.
Consider a 1 inch pipe (OD = 3.34 cm) covered with kapok insulation (k = 0.035
W/m.K). Assume that the outside-pipe-wall temperature is 200°C and the ambient
temperature is 20°C. The outer convective coefficient is 1.7 W/m2.K. The heat loss from
the pipe is:
)R2(h1
k2)R/R(ln
TTL/q
32c2
23
22
⋅π⋅+
⋅π⋅
−=
−
−∞ (2.54)
3
3
R7.11
035.0)0167.0/R(ln
)20200(2L/q
⋅+
−⋅π⋅= (2.55)
The table below shows calculated heat loss when the insulation thickness (R3 - R2)
increases from 0 to 2.5 cm.
---------------------------------------------------- R3 –R2 (cm) heat loss (W/m) ----------------------------------------------------
0 32.1 0.5 32.7 1.0 31.9 1.5 30.7 2.0 29.4 2.5 28.1
----------------------------------------------------
Figure 2.12 Heat loss from an insulated pipe as a function of insulation thickness.
Rcri – R1
q =qmax when dq/dR2 = 0 q / L
R2 - R1
from the graph, it can be seen that when the insulation thickness is greater than the critical
value, the heat loss reduce when the thickness increases. The critical radius is obtained as
the heat loss is at maximum value, thus:
by differentiating equation 2.54:
2
32c2
23
232c322
3
Rh1
k)R/R(ln
Rh/1Rk/1)(TT(20
Rd)L/q(d
⋅+
−−π⋅−==
−
−−∞ (2.56)
solving for R3-cri gives:
2c
2cri3
h
kR−
− = (2.57)
2.3.4 Cylinder with internal heat generation
Tw
T
R
Figure 2.13 Solid cylinder with uniform internal heat generation.
Examples of this system are a tungsten wire in an electric light bulb, an electric heating
element. For this case, the system is steady state and temperature varies only r, thus
equation 2.36 may be simplified as:
0k
qr
rd
Tdrdr
d=
′′′⋅+
(2.58)
rearranging and integrating once:
1
2C
k2
qr
rd
Tdr +⋅
′′′−= (2.59)
dividing by r and integrating again:
21
2C)r(lnC
k4
qrT ++⋅
′′′−= (2.60)
The boundary conditions are: at r = R, T = Tw and at r = 0, dT/dr = 0.
From equation 2.59 at r = 0
(2.61) 0C1 =
From equation 2.60 at r = R
k4
qRTC 2w2
⋅
′′′⋅+= (2.62)
The solution is:
−
⋅
′′′=−
2
22
wR
r1k4
qRTT (2.61)
Example 2.1
Walls of a cold-storage room are constructed of 10 cm thick brick on the outside and 1 cm thick plywood on the inside. Sandwiched between the brick and the plywood is glass fiber insulation that is 7 cm thick. The inside surface temperature is to be maintained at -5°C and the outside surface temperature is 32°C. In order to estimate the cooling capacity of a refrigeration system, the heat flow through the wall per square meter must be determined.
1 cm 7 cm 10 cm
glass fiber wool
plywood brick
-5°C 32°C
Example 2.2
A wall as shown in the figure is a typical wall construction for a convention house. Determine the heat flow of 1 section if the wall is 2 m long.
8 cm 10 cm
air space
25°C
plaster board (1.5 cm thick)
wall stud (8 cm × 5 cm)
40 cm
brick
Example 2.3 A steel pipe with 5 cm OD and 4 cm ID carries hot steam at temperature of 120°C. The pipe is insulated with 3 cm thick of glass wool. The convective heat transfer coefficient between the steam and the pipe is 300 W/m2.K and 60 W/m2.K for the insulation and the air. Determine the overall heat transfer coefficient for this system and the heat transfer coefficient for this system and the heat loss for 1 meter if the air temp is 30°C.
R1 = 2 cm R2 = 2.5 cm R3 = 5.5 cm
R1 R2
R3
T = 30°C T∝-1 = 120°C
h∝-1 = 300 /m2.K
∝-2h∝-2 = 60 W/m2.K
Example 2.4 There is a 1 kW heating element with diameter of 10 mm and 1 m long. The heater is used in an electric kettle. Assuming that the heater outer surface temperature is 10°C above water temperature (because the convective resistance between the surface and the water is very low due to an extremely high convective heat transfer coefficient for boiling). Determine the maximum of the heater.
10 cm silica
stainless-steel tube (thickness of 0.5 mm)
heating wire
CHAPTER III
HEAT TRANSFER FROM EXTENDED SURFACE
One example of an extended surface is a spoon placed in a cup of hot coffee. The handle
extended beyond the hot coffee. Heat is conducted along the spoon handle, causing the
handle to become warmer than the surrounding air. The heat conducted to the handle is
then transfer to the air by convection.
Figure 3.1 A spoon and a cup of coffee.
The purpose of adding an extended surface is to help dissipate heat. Fins are
usually added to a heat transfer device to increase the rate of heat removal. This is because
of the increase of the heat transfer area.
3.1 General equation for extended surface
dqc x-section area = A
dz
z
qz qz+dz
Figure 3.1 An extended surface of aebitrary shape and cross section
Apply the first law of thermodynamics to the system:
Rate of energy convection out from the control
Rate of energy conducted out off the control volume
Rate of energy conducted into the control volume
dzzzc dqdqdq +−= (3.1)
+−= dz
dzdq
dqdqdq zzzc (3.2)
dzdz
dqdq z
c = (3.3)
From Fourier’s law for heat conduction:
dz
dTAkq x ⋅⋅−= (3.4)
Where A is the cross-section and varies along the length, then:
⋅⋅−=
dz
dTAdz
dkdz
dq z (3.5)
For convection:
)TT(Adhdq scc ∞−⋅= (3.6)
By substituting equations 3.5and 3.6 into equation 3.3 then:
dzdz
dTAdz
dk)TT(Adh sc
⋅⋅−=−⋅ ∞ (3.7)
Then divide by dzk ⋅ :
)TT(zd
Ad
k
h
Zd
TdAzd
Td
zd
Ad sc
2
2
∞−=+ (3.8)
0)TT(zd
Ad
A
1
k
h
zd
Td
zd
Ad
A
1
Zd
Td sc
2
2=−−+ ∞ (3.9)
If we define
(3.10) ∞−=θ TT
Then
zd
d
Zd
Td θ= (3.11)
and
2
2
2
2
zd
d
Zd
Td θ= (3.12)
Substitute equations 3.10 to 3.12 into 3.9 then:
0zd
Ad
A
1
k
h
zd
d
zd
Ad
A
1
zd
d sc
2
2=θ−
θ+
θ (3.13)
Equation 3.13 is the general equation for extended surface.
3.2 Uniform fin
Assumptions used in the analysis
• Fin is uniform or constant cross-sectional area, thus 0zd
Ad= .
• There is no heat convection out at the fin tip. The fin tip is insulated or the area is
very small compared with the total surface area.
w
z
Tw
L
dz
dδ
Figure 3.2 A uniform fin.
Based on these assumptions equation 3.13 is simplified to:
0zd
Ad
A
1
k
h
zd
d sc
2
2=θ−
θ (3.14)
If we define pin perimeter as:
zd
AdP s= (3.15)
Therefore, equation 3.14 is:
0Ak
Ph
zd
d c
2
2=θ
⋅
⋅−
θ (3.16)
Then we define
Ak
Phm c
⋅
⋅= (3.17)
Equation 3.16 is then become
0mzd
d 2
2
2=θ⋅−
θ (3.18)
By solving equation 3.18, we have:
)zm(sinhC)Zm(coshC 21 ⋅+⋅=θ (3.19)
or
(3.20) zm4
zm3 eCeC −+=θ
As we assume that the fin’s tip is insulated thus:
at root
z = 0, T = Tw θw = Tw - T∞
at tip
z = L, 0zdTd
= 0zd
d=
θ
from equation 3.19, at root cosh (0) = 1 and sinh (0) = 0, thus
w1C θ= (3.21)
then
)zm(sinhC)zm(cosh 2w ⋅+⋅θ=θ (3.22)
by differentiating
)zm(coshmC)zm(sinhmzd
d2w ⋅⋅+⋅⋅θ=
θ (3.23)
from equation 3.23, at root
)Lm(coshmC)Lm(sinh0 2w ⋅⋅+⋅θ= (3.24)
thus
)Lm(cosh
)Lm(sinhC w
2⋅
⋅θ−= (3.25)
the solution is then:
)Lm(cosh
)zm(sinh)Lm(sinh)zm(cosh)Lm(coshTTTT
ww
z
⋅⋅⋅−⋅⋅
=θθ
=−−
∞
∞ (3.26)
[ ])Lm(cosh
)L/z1(Lmcosh
⋅
−⋅= (3.27)
Equation 3.27 can be used to determine temperature (Tz) at distance z from the root. In
order to fine the heat rejected out from the fin, Fourier’s law of heat conduction may be
applied. We also know that, the heat conducted through the root is equal to the heat
rejected out by convection, thus:
0z
zzd
TdAkq
=
⋅⋅−= (3.28)
0zzd
dAk
=
θ⋅⋅−= (3.29)
[
] 0z
w
)zm(cosh)Lm(sinhm
.....)zm(sinh)Lm(coshm)Lm(cosh
Ak
=⋅⋅
−⋅⋅⋅×⋅
θ⋅⋅−=
(3.30)
as sinh 0 = 0 and cosh 0 = 1 then
)Lm(tanhmAkq wz ⋅θ⋅⋅⋅= (3.31)
3.3 Fin efficiency and effectiveness
The fin efficiency is defined as:
etemperaturwallatisfinentireifdtransferrebewouldthatheat
attachedfinwithwallfromdtransferreheatactualfin =η (3.32)
thus
)TT)(LP(h
)Lm(tanh)TT(PhAk
wc
wcfin
∞
∞
−⋅⋅
⋅−⋅⋅⋅⋅=η (3.33)
Lm
)Lm(tanhfin
⋅
⋅=η (3.34)
(a) (b)
Figure 3.3 a.) Dimensionless graph of heat flow as a function of length for a uniform fin.
b) Efficiency of a fin.
It can be seen that the heat rejected through the fin cannot be substantially increased past
. Practically a fin length of over L3Lm =× m/3= will not improve the performance
( m is over designed). 3L >×
The fin effectiveness is defined as:
finaddingbeforewallfromfluxheat
finaddingafterwallfromfluxheatfin =ε (3.35)
thus
)Lm(tanhAh
Pk
c
fin ⋅×⋅
⋅=ε (3.36)
It can be seen that, fin will increase heat transfer rate when 1fin >ε and the effectiveness
increase when k is high and ch is low. Thus, install fin may not increase the heat transfer
rate if the value of ch is large and the material k is low.
Figure 3.4 Effectiveness of a fin.
Figure 3.5 Efficiency of a fin.
Figure 3.6 Efficiency of a circular fin.
Example 3.1 Calculate heat transfer rate through a pot handle as shown in the figure. Also find temperature distribution along it. Assuming that the handle material is stainless steel 304, or brass.
∅ 3 cm
15 cm
100°C
Example 3.2 Hot steam flows through a tube whose outer diameter is 3 cm and whose wall are maintained at 120°C. Circular aluminum fins of outer diameter of 6 cm and constant thickness of 2 mm are attached to the tube. The space between the fin is 3 mm and thus there are 200 fins/meter. The surrounding air temperature is 25°C and the convective heat transfer coefficient is 60 W/m2.K. Determine the increase in heat transfer rate per meter as a result of adding these fins.
CHAPTER IV
STEADY-STATE CONDUCTION IN MULTIPLE DIMENSIONS
Figure 4.1 Two-dimensions heat transfer.
• Heat flow line and isothermal line are perpendicular to each other.
• A heat flow line is parallel to an insulated surface or a line of symmetry
• An isotherm will intersect an insulated surface or a line of symmetry
Figure 4.2 A chimney quarter cross-section.
Figure 4.3 A heat flow lane from figure 4.2
For each heat flow lane:
)TT(y
Lxkq aba1L −∆
⋅∆⋅= (4.1)
)TT(y
Lxkacab −
∆
⋅∆⋅= (4.2)
Where L is the dept into the paper. Then we have:
)TT(yn
Lxkq 21L −∆⋅
⋅∆⋅= (4.3)
Where n is number of block. If we assume yx ∆≈∆ then:
)TT(n
Lkq 21L −⋅
= (4.4)
If m is a total number of the heat flow lane, then the total heat flow is:
Lqmq ⋅= (4.5)
)TT(n
Lmk 21 −⋅
= (4.6)
(4.7) )TT(Sk 21 −⋅⋅=
where n
LmS ⋅= is defined as the conduction shape factor.
Table 5.1 Shape factors for a number of conduction heat-transfer system.
Example 4.1 A heat-treating furnace has outside dimensions of 15 mm ×150 mm×200 mm. The walls are 6 mm thick and made of fireclay brick. For an inside wall temperature of 550°C and an outside wall temperature of 30°C, determine the heat lost through the walls, by using shape factor method.
Example 4.2 It is proposed to cool a certain volume of air by piping it underground. The cooled air would then supplement the air-condition system of a dwelling and reduce costs. Determine the conduction shape factor for the underground portion of the configuration if the pipe is 4 nominal , schedule 40.
CHAPTER V
UNSTEADY-STATE HEAT CONDUCTION
In this chapter we consider transient-conduction problems in which there is no internal heat
generation. Temperature will therefore vary with location within the system and with time.
Temperature and heat transfer variation of the system are dependent on its internal
resistance and surface resistance.
If we have a slab with initial temperature of Ti and it is left in fluid stream at T∝.
Heat is transferred by convection at the surface. As the surface temperature decreases, heat
is transfer from the center of the slab to the surface, then to the fluid. Now, if the system
itself is copper or the volume is small, the temperature response within the slab is
considerably different from that if it is glass or the volume is large. The response has to do
with what is called the internal resistance of the material. Further, if the convection
coefficient is very high, then the surface temperature almost becomes identical to the fluid
temperature quickly. Alternatively, for a low convection coefficient a large temperature
difference exists between the surface and the fluid. The value of the convection coefficient
controls what is known as the surface resistance to heat transfer.
Thus, the temperature variation within the system is dependent on the internal and
surface resistances. The larger internal resistance or the smaller surface resistance, the
larger temperature variation within the system, and vice versa. A Biot number is defined
as:
bodythewithinconduction
bodytheinwithsurfacetheatconvectionBi = (5.1)
T)L/k(
ThBic ∆⋅
∆⋅= (5.2)
Figure 5.1 Relationship between the Biot number and the temperature profile.
5.1 System with negligible internal resistance
For this case Bi 1.0≤ and the temperature profile within the body is quite uniform.
T = T(t)
fluid
h∝, T∞
Figure 5.2 An object be cooled in a fluid.
The rate of change in internal energy of the body is equal to the rate of heat taken away
from the surface by convection:
)TT(Ahdt
dTc tsc ∞−⋅⋅−=⋅⋅∀⋅ρ (5.3)
At the initial state t = 0, T = Ti, and we also define:
∞
∞
−
−=θ
TT
TT
i
t (5.4)
and then:
dt
d)TT(dt
dTi
θ−= ∞ (5.5)
substitue equation 5.5 in to 5.3 then:
)TT(Ahdt
d)TT(c tsci ∞α −⋅⋅−=θ
−⋅⋅∀⋅ρ (5.6)
θ=θ
⋅⋅
⋅∀⋅ρ−
dt
d
Ah
c
sc
(5.7)
If we integrate equation 5.7 from T = Ti to T∞ and θ = 1 to 0, then we have:
∫⋅∀⋅ρ
⋅=
θ
θ−
0
1
t
0
sc dtc
Ahd∫ (5.8)
finally we have:
⋅
⋅∀⋅ρ
⋅−=
−
−
∞
∞ tc
Ahexp
TT
TT sc
i
t (5.9)
[ FoBiexpTT
TT
i
t ⋅−=−
−
∞
∞ ] (5.10)
Where Bi is Biot number:
s
c
Ak
hBi
⋅
∀⋅= (5.11)
and Fo is Fourier number:
2
2sAt
Fo∀
⋅⋅α= (5.12)
For the heat transfer rate at t, it can be obtained from:
dt
dTcq ⋅⋅∀⋅ρ= (5.13)
and dt
dT can be obtained by integrating equation 5.9, thus:
[ ]FoBiexp)TT(Ahq isct ⋅−−⋅⋅= ∞ (5.14)
the total cumulative heat transfer for period of t second is:
)TT(CQ ti −⋅⋅∀⋅ρ= (5.15)
5.3 System with finite internal and external resistances
For this case, the problem is to complicate to solve therectically, it must be solved based on
graphical solutions. The solution provided here are:
Figure 5.3 to 5.5 for a simi-infinite plate,
Figure 5.6 to 5.8 for an infinite cylinder, and
Figure 5.9 to 5.11 for a sphere.
Figure 5.3 Dimensionless temperature history at the center of semi-infinite plate.
Figure 5.4 Dimensionless temperature distribution in a semi-infinite plate.
Figure 5.5 Change in internal-energy ratio as a function of dimesionless time for a semi-inite plate.
Figure 5.6 Dimesionless temperature history at the centerline of an inifinite cylinde.
Figure 5.7 Dimesionless temperature distribution in an infinite cylinder.
Figure 5.8 Change in internal-energy ratio as a function of dimensionless time for an infinite cylinder.
Figure 5.9 Dimensionless temperature history at the center of a sphere.
Figure 5.10 Dimensionless temperature distribution in a sphere.
Figure 5.11 Change in internal-energy ratio as a function of dimensionless time for a sphere.
5.4 System witth negligible surface resistance
In this category of problems we will deal with system where the film resistance is
negligible. We therefore simply assign a temperature at the surface of the object that in
essence will equal the surrounding temperature (T∞). Similar to the previous, this kind of
problem is solved graphically.
Figure 5.12 Central dimensionless temperature variation witth dimensionless time for an infinite plate, an infinite square rod, an infinite cylinder, a cube, a finite cylinder, and a sphere.
Figure 5.13 Cumulative (Q) and instantaneous (q) heat transfer rates in various solids for tthhe case of negligible surface resistance.
Example 5.1 A 5 cm, 60 cm long aluminum cylinder initially at 50°C is submerged in an ice-water bath at 2°C. The unit surface conductance between the metal and thebath is 550 W/m2.K. Determine the temperature of the aluminum after 1 minute. Also calculate the cumulative heat transfer for 1 minute.
Example 5.2 Orange are usually refrigerated as a preservative measure. However, some people prefer to eat oranges that are a little cooler than room temperature but not as cold as the refrigerator makes them. Determine the time it takes for an orange removed from a refrigerator to reach 20°C.
Refrigerated temperature = 4°C Ambient room temperature = 23°C Surface heat transfer coefficient = 6 W/m2.K Thermal conductivity of the orange 0.431 W/m.K Density of orange = 998 kg/m3 Specific heat of orange = 2000 J/kg.K Orange diameter = 105 mm
Example 5.3
A concrete wall is 15 cm thick. The outside wall surface is heated to a temperature of 200°C by fire. If the initial temperature of the wall is 25°C, how long will it take for the inside wall surface to reach 100°C ? What would the time required become if the wall were made of common brick instead ?