1
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Dr Mazlan - SME 4463
HEAT TRANSFERHEAT TRANSFERSME 4463SME 4463
LECTURER: PM DR MAZLAN ABDUL WAHIDhttp://www.fkm.utm.my/~mazlan
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Dr Mazlan - SME 4463
CChapter hapter 33
One Dimensional One Dimensional SteadySteady--State State
Heat ConductionHeat ConductionPM Dr Mazlan Abdul Wahid
Faculty of Mechanical EngineeringUniversiti Teknologi Malaysiawww.fkm.utm.my/~mazlan
2
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Dr Mazlan - SME 4463
One-Dimensional Steady-State Conduction
• Conduction problems may involve multiple directions and time-
dependent conditions
• Inherently complex – Difficult to determine temperature distributions
• One-dimensional steady-state models can represent accurately
numerous engineering systems
• In this chapter we will:
� Learn how to obtain temperature profiles for common geometries
with and without heat generation.
� Introduce the concept of thermal resistance and thermal circuits
� Introduce to the analysis of one dimensional conduction analysis
on extended surfaces
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Dr Mazlan - SME 4463
Assumptions
- 1 dimensional heat transfer
- Isothermal surfaces
- Steady state
The Plane Wall
3
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The Plane Wall
Consider a simple case of one-dimensional conduction in a plane wall, separating two fluids of different temperature, without energy generation
• Temperature is a function of x
• Heat is transferred in the x-direction
Must consider– Convection from hot fluid to wall
– Conduction through wall
– Convection from wall to cold fluid
� Begin by determining temperature distribution within the wall
qx
1,∞T
1,sT
2,sT
2,∞T
x
x=0 x=L
11, ,hT∞
22, ,hT∞
Hot fluid
Cold fluid
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TRANSFER
CCCCCCCCHHHHHHHH
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PPPPPPPP
TTTTTTTT
EEEEEEEE
RRRRRRRR
33333333
Dr Mazlan - SME 4463
Temperature Distribution – Plane Wall
• Heat diffusion equation in the x-direction for steady-state conditions, with no energy generation:
0=
dx
dTk
dx
d
• Boundary Conditions:2,1, )(,)0( ss TLTTT ==
• Temperature profile, assuming constant k:
1,1,2, )()( sss TL
xTTxT +−=
� Temperature varies linearly with x
� qx is constant
(3.1)
t
Tcq
z
Tk
yy
Tk
yx
Tk
x p ∂∂=+
∂∂
∂∂+
∂∂
∂∂+
∂∂
∂∂ •
ρ
4
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Dr Mazlan - SME 4463
Thermal Resistance
Based on the previous solution, the conduction heat transfer rate can be calculated:
( ) ( )kAL
TTTT
L
kA
dx
dTkAq ss
ssx /2,1,
2,1,
−=−=−=
� Recall electric circuit theory - Ohm’s law for elect rical resistance:
Similarly for heat convection, Newton’s law of cooling applies:
Resistance
e DifferencPotentialcurrent Electric =
hA
TTTThAq S
Sx /1
)()( ∞
∞−=−=
And for radiation heat transfer:
Ah
TTTTAhq
r
surssursrrad /1
)()(
−=−=
HEAT
TRANSFER
CCCCCCCCHHHHHHHH
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33333333
Dr Mazlan - SME 4463
Thermal Resistance
� The temperature difference is the “potential” or driving force for the heat flow and the combinations of thermal conductivity, convection coefficient, thickness and area of material act as a resistance to this flow:
• We can use this electrical analogy to represent heat transfer problems using the concept of a thermal circuit (equivalent to an electrical circuit).
AhR
hAR
kA
LR
rradtconvtcondt
1,
1, ,,, ===
∑∆==
R
Tq overall
Resistance
Force Driving Overall
5
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Thermal Resistance for Plane Wall
In terms of overall temperature difference:qx
1,∞T
1,sT
2,sT
2,∞T
xx=0 x=L
11, ,hT∞
22, ,hT∞
Hot fluid
Cold fluid
AhkA
L
AhR
R
TTq
tot
totx
21
2,1,
11 ++=
−= ∞∞
Ah
TT
kAL
TT
Ah
TTq ssss
x2
2,2,2,1,
1
1,1,
/1//1∞∞ −
=−
=−
=
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Composite Walls
Express the following geometry in terms of a an equivalent thermal circuit.
6
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Dr Mazlan - SME 4463
Composite Walls
What is the heat transfer rate for this system?
AlternativelyUAq
TRR
TUAq
ttot
x
1=∆==
∆=
∑
where U is the overall heat transfer coefficient and ∆T the overall temperature difference.
)]/1()/()/()/()/1[(
11
41 hkLkLkLhARU
CCBBAAtot ++++==
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Example – single layer wall
7
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TRANSFER
CCCCCCCCHHHHHHHH
AAAAAAAA
PPPPPPPP
TTTTTTTT
EEEEEEEE
RRRRRRRR
33333333
Dr Mazlan - SME 4463
Example – multi layer wall
HEAT
TRANSFER
CCCCCCCCHHHHHHHH
AAAAAAAA
PPPPPPPP
TTTTTTTT
EEEEEEEE
RRRRRRRR
33333333
Dr Mazlan - SME 4463
8
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TRANSFER
CCCCCCCCHHHHHHHH
AAAAAAAA
PPPPPPPP
TTTTTTTT
EEEEEEEE
RRRRRRRR
33333333
Dr Mazlan - SME 4463
Example – single layer window
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TRANSFER
CCCCCCCCHHHHHHHH
AAAAAAAA
PPPPPPPP
TTTTTTTT
EEEEEEEE
RRRRRRRR
33333333
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Example – two layer window
9
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TRANSFER
CCCCCCCCHHHHHHHH
AAAAAAAA
PPPPPPPP
TTTTTTTT
EEEEEEEE
RRRRRRRR
33333333
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Thermal Resistance concept - Convection & Radiation
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(a) Surfaces normal to the x-direction are isothermal
(b) Surfaces parallel to x-direction are adiabatic
� For resistances in series: Rtot=R1+R2+…+Rn
� For resistances in parallel:
1/Rtot=1/R1+1/R2+…+1/Rn
Composite Walls – with parallel resistances
10
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TRANSFER
CCCCCCCCHHHHHHHH
AAAAAAAA
PPPPPPPP
TTTTTTTT
EEEEEEEE
RRRRRRRR
33333333
Dr Mazlan - SME 4463
Composite Walls – with parallel resistances
HEAT
TRANSFER
CCCCCCCCHHHHHHHH
AAAAAAAA
PPPPPPPP
TTTTTTTT
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RRRRRRRR
33333333
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� For resistances in series: Rtot=R1+R2+…+Rn
� For resistances in parallel:1/Rtot=1/R1+1/R2+…+1/Rn
Composite Walls – with parallel resistances
11
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CCCCCCCCHHHHHHHH
AAAAAAAA
PPPPPPPP
TTTTTTTT
EEEEEEEE
RRRRRRRR
33333333
Dr Mazlan - SME 4463
Parallel heat conduction
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AttentionThe result obtained will be somewhat approximate, since the surfaces of thethird layer will probably not be isothermal, and heat transfer between the firsttwo layers is likely to occur.
12
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Heat loss through a composite wall
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CCCCCCCCHHHHHHHH
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PPPPPPPP
TTTTTTTT
EEEEEEEE
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Contact Resistance
13
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Contact Resistance
The temperature drop across the interface between materials may be appreciable, due to surface roughness effects, leading to air pockets. We can define thermal contact resistance:
cct hR /1"
, =
",intint
" / ctcx RTThq ∆=∆=
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14
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TRANSFER
CCCCCCCCHHHHHHHH
AAAAAAAA
PPPPPPPP
TTTTTTTT
EEEEEEEE
RRRRRRRR
33333333
Dr Mazlan - SME 4463
HEAT
TRANSFER
CCCCCCCCHHHHHHHH
AAAAAAAA
PPPPPPPP
TTTTTTTT
EEEEEEEE
RRRRRRRR
33333333
Dr Mazlan - SME 4463
Contact resistance of transistors
15
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TRANSFER
CCCCCCCCHHHHHHHH
AAAAAAAA
PPPPPPPP
TTTTTTTT
EEEEEEEE
RRRRRRRR
33333333
Dr Mazlan - SME 4463
HEAT
TRANSFER
CCCCCCCCHHHHHHHH
AAAAAAAA
PPPPPPPP
TTTTTTTT
EEEEEEEE
RRRRRRRR
33333333
Dr Mazlan - SME 4463
Composite Walls – with contact resistances
16
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TRANSFER
CCCCCCCCHHHHHHHH
AAAAAAAA
PPPPPPPP
TTTTTTTT
EEEEEEEE
RRRRRRRR
33333333
Dr Mazlan - SME 4463
Problem: Thermal Barrier Coating
Problem 3.30: Assessment of thermal barrier coating (TBC) for protectionof turbine blades. Determine maximum blade temperaturewith and without TBC.
Schematic:
ASSUMPTIONS: (1) One-dimensional, steady-state conduction in a composite plane wall, (2) Constant properties, (3) Negligible radiation.
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Problem: Thermal Barrier Coating (cont..)
( )-3 -4 -4 -4 -3 2 -3 2tot 10 3.85 10 10 2 10 2 10 m K W 3.69×10 m K W ,wR = =′′ + × + + × + × ⋅ ⋅
( ) ( )3 4 2 5 2400 K 2 10 2 10 m K W 3.52 10 W m 1174 K− −= + × + × ⋅ × =( ) ( )(w) Ins,o ,i i wT = T + 1 h + L k q∞ ′′
5 23 2
tot
1300K3.52 10 W m
3.69 10 m K W
,o ,iw
,w
T - Tq = =
R
∞ ∞−
′′ = ×′′ × ⋅
With a heat flux of
the inner and outer surface temperatures of the Inconel are
( )(w)s,i ,i w iT = T + q h∞ ′′5 2
400 K 1104 K2
3.52 10 W m
500W m K / W= + =
× ⋅
ANALYSIS: For a unit area, the total thermal resistance with the TBC is
( ) ( )1 1tot Zr In
- -,w o t,c iR = h + L k + R + L k + h′′ ′′
<
17
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CCCCCCCCHHHHHHHH
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Problem: Thermal Barrier Coating (cont..)
( )1 1 3 2tot wo In 3 20 10 m K W
- -, o iR = h + L k + h .
−′′ = × ⋅
The inner and outer surface temperatures of the Inconel are then
( )(wo) 1212Ks,i ,i wo iT T q h∞ ′′= + =
( ) ( )[ ](wo) In1 1293 Ks ,o ,i i woT T h L k q∞ ′′= + + =
Use of the TBC facilitates operation of the Inconel below Tmax = 1250 K.
COMMENTS: Since the durability of the TBC decreases with increasing temperature, which increases with increasing thickness, limits to its thickness are associated with reliability considerations.
( )wo tot,wo,o ,iq T T R∞ ∞′′ ′′= − = 4.06×105 W/m2
Without the TBC,
<
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Radial Systems-Cylindrical Coordinates
Consider a hollow cylinder, whose inner and outer surfaces are exposed to fluids at different temperatures
Temperature distribution
18
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• Heat diffusion equation (eq. 2.5) in the r-direction for steady-state conditions, with no energy generation:
t
Tcq
z
Tk
z
Tk
rr
Tkr
rr p ∂∂=+
∂∂
∂∂+
∂∂
∂∂+
∂∂
∂∂ •
ρφφ2
11
01 =
dr
dTkr
dr
d
r
• Boundary Conditions: 2,21,1 )(,)( ss TrTTrT ==
• Temperature profile, assuming constant k:
2,221
2,1, ln)/ln(
)()( s
ss Tr
r
rr
TTrT +
−= � Logarithmic temperature
distribution
• Fourier’s law: constdr
dTrLk
dr
dTkAqr =−=−= )2( π
01 =
dr
dTr
dr
d
r
for constant k
Temperature Distribution - Thermal resistance for cylinder
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dr
dTrLk
dr
dTkAqr −=−= )2( π
2,221
2,1, ln)/ln(
)()( s
ss Tr
r
rr
TTrT +
−=
)/ln(
)(2
12
21
rr
TTkLq r
−= π
kLrr
TTq r
π2/)/ln( 12
21 −=
kLrrR cyl π2/)/ln( 12=therefore
Temperature Distribution - thermal resistance for cylinder
3.33
19
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TRANSFER
CCCCCCCCHHHHHHHH
AAAAAAAA
PPPPPPPP
TTTTTTTT
EEEEEEEE
RRRRRRRR
33333333
Dr Mazlan - SME 4463
Temperature Distribution - thermal resistance for cylinder
HEAT
TRANSFER
CCCCCCCCHHHHHHHH
AAAAAAAA
PPPPPPPP
TTTTTTTT
EEEEEEEE
RRRRRRRR
33333333
Dr Mazlan - SME 4463
Thermal resistance for cylinder
Based on the previous solution, the conduction hear transfer rate can be calculated:
( ) ( ) ( )condt
ssssssx R
TT
Lkrr
TT
rr
TTLkq
,
2,1,
12
2,1,
12
2,1,
)2/()/ln()/ln(
2 −=
−=
−=
ππ
� In terms of equivalent thermal circuit:
)2(
1
2
)/ln(
)2(
1
22
12
11
2,1,
LrhkL
rr
LrhR
R
TTq
tot
totx
πππ++=
−= ∞∞
• Fourier’s law: constdr
dTrLk
dr
dTkAqr =π−=−= )2(
20
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TRANSFER
CCCCCCCCHHHHHHHH
AAAAAAAA
PPPPPPPP
TTTTTTTT
EEEEEEEE
RRRRRRRR
33333333
Dr Mazlan - SME 4463
Composite Walls
Express the following geometry in terms of a an equivalent thermal circuit.
HEAT
TRANSFER
CCCCCCCCHHHHHHHH
AAAAAAAA
PPPPPPPP
TTTTTTTT
EEEEEEEE
RRRRRRRR
33333333
Dr Mazlan - SME 4463
Composite Walls
What is the heat transfer rate?
where U is the overall heat transfer coefficient. If A=A1=2πr1L:
44
1
3
41
2
31
1
21
1
1lnlnln
11
hr
r
r
r
k
r
r
r
k
r
r
r
k
r
h
U
CBA
++++=
alternatively we can use A2=2πr2L, A3=2πr3L etc. In all cases:
∑====
tRAUAUAUAU
144332211
21
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CCCCCCCCHHHHHHHH
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TTTTTTTT
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RRRRRRRR
33333333
Dr Mazlan - SME 4463
Spherical Coordinates
Fourier’s law:
dr
dTrk
dr
dTkAqsph
)4( 2π−=
−=
• Starting from Fourier’s law, acknowledging that q r is constant, independent of r, and assuming that k is constant, derive the equation describing the conduction heat transfer ra te. What is the thermal resistance?
Need temperature profile, then differentiate
t
Tcq
Tk
r
Tk
rr
Tkr
rr p ∂∂=+
∂∂
∂∂+
∂∂
∂∂+
∂∂
∂∂ •
ρθ
θθθφφθ
sinsin
1
sin
11222
22
Assume:1) Steady state2) Constant properties3) 1 dimensional coordinate
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Dr Mazlan - SME 4463
Spherical Coordinates
01 22
=
∂∂
∂∂
r
Tkr
rr
2,21,1 )(,)( ss TrTTrT ==
2,221
2,1, ln)/ln(
)()( s
ss Tr
r
rr
TTrT +
−=
)11
(4
1
21 rrkR sph −=
π
22
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CCCCCCCCHHHHHHHH
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33333333
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dTkrdrqsph ∫∫ −=2/4/ π
)/1()/1(
)(4
21
21
rr
TsTskq sph −
−= π
)11
(4
1
21 rrkR sph −=
π
Temperature Distribution - thermal resistance for sphere
3.41
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TRANSFER
CCCCCCCCHHHHHHHH
AAAAAAAA
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TTTTTTTT
EEEEEEEE
RRRRRRRR
33333333
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Conduction with Thermal Energy Generation
Thermal energy may be generated or consumed due to conversion from some other energy form.
� If thermal energy is generated in the material at the expense of some other energy form, we have a source: is +ve
– Deceleration and absorption of neutrons in a nuclear reactor
– Exothermic reactions
– Conversion of electrical to thermal energy:
V
RI
V
Eq eg
2
==&
where I is the current, Re the electrical resistance, V the volume of the medium
� If thermal energy is consumed we have a sink: is -ve– Endothermic reactions
q
q
24
.
.
.
23
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33333333
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The Plane Wall
Consider one-dimensional, steady-state conduction in a plane wall ofconstant k, with uniform generation, and asymmetric surface conditions:
• Heat diffusion equation (eq. 2.21) :
02
2
=+k
q
dx
Td
• Boundary Conditions:
2,1, )(,)( ss TLTTLT ==−
• General Solution:
212
2CxCx
k
qT ++−=
25
.
.
HEAT
TRANSFER
CCCCCCCCHHHHHHHH
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TTTTTTTT
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33333333
Dr Mazlan - SME 4463
Temperature Profile
22
)(1
2)( 2,1,1,2,
2
22ssss TT
L
xTT
L
x
k
qLxT
++
−+
−=
� Profile is parabolic.
� Heat flux not independent of x
What happens when:
(3.46)
?0 increases, ,0 <= qqq
26
.
. . .
24
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33333333
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Symmetrical Distribution
• When both surfaces are maintained at a common temperature, Ts,1= Ts,2 = Ts
sTL
x
k
qLxT +
−=
2
22
12
)(
What is the location of the maximum temperature?
2
max
max)(
=−
−∴L
x
TT
TxT
s
27
.(3.48)
(3.49)
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33333333
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Symmetrical Distribution
� Note that at the plane of symmetry:
0q" 00
0
=⇒=
=
=x
xdx
dT
� Equivalent to adiabatic surface
28
25
HEAT
TRANSFER
CCCCCCCCHHHHHHHH
AAAAAAAA
PPPPPPPP
TTTTTTTT
EEEEEEEE
RRRRRRRR
33333333
Dr Mazlan - SME 4463
Calculation of surface temperature T s
In equations (3.48) and (3.49) the surface temperature, Ts is needed.
� Boundary condition at the wall:
)( ∞=
−=− TThdx
dTk s
Lx
Substituting (dT/dx)x=L from equation(3.47):
h
qLTTs += ∞
29
.
HEAT
TRANSFER
CCCCCCCCHHHHHHHH
AAAAAAAA
PPPPPPPP
TTTTTTTT
EEEEEEEE
RRRRRRRR
33333333
Dr Mazlan - SME 4463
Example
A plane wall of thickness 0.1 m and thermal conductivity 25 W/m.K having uniform volumetric heat generation of 0.3 MW/m3 is insulated on one side, while the other side is exposed to a fluid at 92°C. The convection heat transfer coefficient between the wall and the fluid is 500 W/m2.K. Determine the maximum temperature in the wall.
31
26
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TRANSFER
CCCCCCCCHHHHHHHH
AAAAAAAA
PPPPPPPP
TTTTTTTT
EEEEEEEE
RRRRRRRR
33333333
Dr Mazlan - SME 4463
Radial Systems
Cylindrical (Tube) Wall Spherical Wall (Shell)
Solid Cylinder (Circular Rod) Solid Sphere
32
HEAT
TRANSFER
CCCCCCCCHHHHHHHH
AAAAAAAA
PPPPPPPP
TTTTTTTT
EEEEEEEE
RRRRRRRR
33333333
Dr Mazlan - SME 4463
Radial Systems• Heat diffusion equation in
the r-direction for steady-state conditions:
01 =+
k
q
dr
dTr
dr
d
r
• Boundary Conditions: sor
TrTdr
dT ===
)( ,00
• Temperature profile:
so
o Tr
r
k
qrrT +
−=
2
22
14
)(
L
hT ,∞
))(2()( 2
∞−= TTLrhLrq soo ππ h
qrTT o
s 2+= ∞
• Calculation of surface temperature:
and
• General Solution:
33
212 ln
4CrCr
k
qT ++−=
.
.
.
Cylinder
..
27
HEAT
TRANSFER
CCCCCCCCHHHHHHHH
AAAAAAAA
PPPPPPPP
TTTTTTTT
EEEEEEEE
RRRRRRRR
33333333
Dr Mazlan - SME 4463
Example
A cylindrical shell of inner and outer radii ri and ro, respectively, is filled with a heat-generating material that provides a uniform volumetric generation rate. The inner surface is insulated, while the outer surface of the shell is exposed to a fluid with a convection coefficient h.
a) Obtain an expression for the steady-state temperature distribution T(r) in the shell.
b) Determine an expression for the heat rate q’ (ro) at the outer radius of the shell in terms of the heat generation rate and the shell dimensions
34
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TRANSFER
CCCCCCCCHHHHHHHH
AAAAAAAA
PPPPPPPP
TTTTTTTT
EEEEEEEE
RRRRRRRR
33333333
Dr Mazlan - SME 4463
28
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TRANSFER
CCCCCCCCHHHHHHHH
AAAAAAAA
PPPPPPPP
TTTTTTTT
EEEEEEEE
RRRRRRRR
33333333
Dr Mazlan - SME 4463
CRITICAL THICKNESS OF INSULATION
For smaller diameter tubes it is sometimes possible to increase heat loss by adding insulation on the outer surface
HEAT
TRANSFER
CCCCCCCCHHHHHHHH
AAAAAAAA
PPPPPPPP
TTTTTTTT
EEEEEEEE
RRRRRRRR
33333333
Dr Mazlan - SME 4463
29
HEAT
TRANSFER
CCCCCCCCHHHHHHHH
AAAAAAAA
PPPPPPPP
TTTTTTTT
EEEEEEEE
RRRRRRRR
33333333
Dr Mazlan - SME 4463
HEAT
TRANSFER
CCCCCCCCHHHHHHHH
AAAAAAAA
PPPPPPPP
TTTTTTTT
EEEEEEEE
RRRRRRRR
33333333
Dr Mazlan - SME 4463
30
HEAT
TRANSFER
CCCCCCCCHHHHHHHH
AAAAAAAA
PPPPPPPP
TTTTTTTT
EEEEEEEE
RRRRRRRR
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Dr Mazlan - SME 4463
Critical radius of insulation
HEAT
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Optimum thickness of insulation
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HEAT
TRANSFER
CCCCCCCCHHHHHHHH
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Dr Mazlan - SME 4463
Heat Transfer from Extended SurfacesHeat Transfer from Extended Surfaces-- FinsFins
There are two ways to increase the rate of heat transfer: to increase the convection heat transfer coefficient h or to increase the surface area A .
Fins enhance heat transfer from a surface by exposing a larger surface area toconvection and radiation.
qconv =
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Dr Mazlan - SME 4463
Extended surfacesExtended surfaces-- FinsFins
An extended surface (also know as a combined conduction-convection system or a fin) is a solid within which heat transfer by conduction is assumed to be one dimensional, while heat is also transferred by convection (and/or radiation) from the surface in a direction transverse to that of conduction
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CCCCCCCCHHHHHHHH
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HEAT
TRANSFER
CCCCCCCCHHHHHHHH
AAAAAAAA
PPPPPPPP
TTTTTTTT
EEEEEEEE
RRRRRRRR
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HEAT
TRANSFER
CCCCCCCCHHHHHHHH
AAAAAAAA
PPPPPPPP
TTTTTTTT
EEEEEEEE
RRRRRRRR
33333333
Dr Mazlan - SME 4463
HEAT
TRANSFER
CCCCCCCCHHHHHHHH
AAAAAAAA
PPPPPPPP
TTTTTTTT
EEEEEEEE
RRRRRRRR
33333333
Dr Mazlan - SME 4463
Extended surfacesExtended surfaces-- FinsFins
Extended surfaces may exist in many situations but are commonly used as fins to enhance heat transfer by increasing the surface area available for convection (and/or radiation). They are particularly beneficial when h is small, as for a gas and natural convection.
� Solutions for various fin geometries can be found in the literature (see for example in textbook).
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Dr Mazlan - SME 4463
A general conduction analysis for an extended surfaces
Applying the conservation of energy
Using,
Then, the heat equation becomes:
General form of the energy equation for an extended surface
Eq. (3.66)
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Dr Mazlan - SME 4463
Heat transfer from extended surface
The Fin Equation� Assuming 1-D case, steady state conduction
in an extended surface, constant k, uniform cross sectional area, negligible generation and radiation.
� Cross section area, Ac is constant and fin surface area, As = Px, this mean dAc/dx = 0 and dAs/dx = P
� General equation becomes:
Eq. (3.67)
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Heat transfer from extended surface
� To simplify the equation, we define an excess temperature ( the reduced temperature) as:
� The previous equation becomes:
where,
* m also known as fin parameter
Eq. (3.68)
Eq. (3.70)
P is the fin perimeter
HEAT
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Dr Mazlan - SME 4463
Extended Surface: Fin equation
0)(2
2
=−− ∞TTkA
hp
dx
Td
c
022
2
=− θθm
dx
d
2
ckA
hpm =
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)( 21mxmx eCeCx −+=θ
Two unknowns, C1 and C2, therefore
require 2 BC to solve
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CCCCCCCCHHHHHHHH
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PPPPPPPP
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Dr Mazlan - SME 4463
Heat transfer from extended surface
By referring to Table 3.4 : at different case of heat transfer analysis
• Temperature distribution, θ/θb
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D
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CCCCCCCCHHHHHHHH
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EEEEEEEE
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B
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A
39
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CCCCCCCCHHHHHHHH
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TTTTTTTT
EEEEEEEE
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HEAT
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CCCCCCCCHHHHHHHH
AAAAAAAA
PPPPPPPP
TTTTTTTT
EEEEEEEE
RRRRRRRR
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Dr Mazlan - SME 4463
C
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HEAT
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CCCCCCCCHHHHHHHH
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PPPPPPPP
TTTTTTTT
EEEEEEEE
RRRRRRRR
33333333
Dr Mazlan - SME 4463
HEAT
TRANSFER
CCCCCCCCHHHHHHHH
AAAAAAAA
PPPPPPPP
TTTTTTTT
EEEEEEEE
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Dr Mazlan - SME 4463
Proper length of a fin
41
Problem: Turbine Blade CoolingProblem: Turbine Blade Cooling
Problem 3.126: Assessment of cooling scheme for gas turbine blade.Determination of whether blade temperatures are lessthan the maximum allowable value (1050°C) for prescribed operating conditions and evaluation of bladecooling rate.
Assumptions: (1) One-dimensional, steady-state conduction in blade, (2) Constant k, (3)Adiabatic blade tip, (4) Negligible radiation.
Analysis: Conditions in the blade are determined by Case B of Table 3.4.
(a) With the maximum temperature existing at x = L, Eq. 3.80 yields
Schematic:
( ) 1
coshb
T L - T
T - T mL∞
∞=
( ) ( )1/22 4 2250 W/m K 0.11m/20W/m K 6 10 m1/ 2cm hP/kA −= = ⋅ × ⋅ × × = 47.87 m-1
mL = 47.87 m-1 × 0.05 m = 2.39
From Table B.1 (or by calculation), Hence, cosh 5.51.mL =
and, subject to the assumption of an adiabatic tip, the operating conditions are acceptable.
Eq. 3.81 and Table B.1 yield
Hence,
Comments: Radiation losses from the blade surface contribute to reducing the blade temperatures, but what is the effect of assuming an adiabatic tip condition? Calculatethe tip temperature allowing for convection from the gas.
( ) o o o1200 C (300 1200) C/5.51 1037 CT L = + − =
(b) With ( ) ( ) ( )1/22 4 2 o1/2 250W/m K 0.11m 20W/m K 6 10 m 900 C 517Wc bM hPkA θ −= = × × ⋅ × × − = −⋅ ,
( )tanh 517W 0.983 508Wfq M mL= = − = −
508Wb fq q= − =
Problem: Turbine Blade Cooling (cont.)Problem: Turbine Blade Cooling (cont.)
<
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Fin effectiveness
Fin Performance
HEAT
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CCCCCCCCHHHHHHHH
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Fin efficiency
43
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CCCCCCCCHHHHHHHH
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PPPPPPPP
TTTTTTTT
EEEEEEEE
RRRRRRRR
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HEAT
TRANSFER
CCCCCCCCHHHHHHHH
AAAAAAAA
PPPPPPPP
TTTTTTTT
EEEEEEEE
RRRRRRRR
33333333
Dr Mazlan - SME 4463
44
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CCCCCCCCHHHHHHHH
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Dr Mazlan - SME 4463
Performance ParametersPerformance Parameters Fin Performance• Fin Efficiency:
, max
f ff
f f b
q q
q hAη
θ≡ = (3.91)
How is the efficiency affected by the thermal conductivity of the fin?Expressions for are provided in Table 3.5 for common geometries.fη
( )1/ 2222 / 2fA w L t = +
( )/ 2pA t L=( )( )
1
0
21
2f
I mL
mL I mLη =
• Fin Effectiveness:
Consider a triangular fin:
,
ff
c b b
q
hAε
θ≡
• Fin Resistance:
with , and /f ch k A Pε ↑ ↓ ↑ ↓(3.86)
,
1bt f
f f f
Rq hA
θη
≡ = (3.97)
where 0 1fη≤ ≤
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CCCCCCCCHHHHHHHH
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PPPPPPPP
TTTTTTTT
EEEEEEEE
RRRRRRRR
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Dr Mazlan - SME 4463
HEAT
TRANSFER
CCCCCCCCHHHHHHHH
AAAAAAAA
PPPPPPPP
TTTTTTTT
EEEEEEEE
RRRRRRRR
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Dr Mazlan - SME 4463
Efficiency of fins (I)
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CCCCCCCCHHHHHHHH
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Dr Mazlan - SME 4463
Efficiency of fins (II)
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CCCCCCCCHHHHHHHH
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TRANSFER
CCCCCCCCHHHHHHHH
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TTTTTTTT
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(cont.)
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CCCCCCCCHHHHHHHH
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49
ArraysArrays Fin Arrays
• Representative arrays of(a) rectangular and(b) annular fins.
– Total surface area:
t f bA NA A= + (3.104)
Number of fins Area of exposed base (prime surface)– Total heat rate:
,
bt f f b b b o t b
t o
q N hA hA hAR
θη θ θ η θ= + ≡ =(3.105)
– Overall surface efficiency and resistance:
( )1 1fo f
t
NA
Aη η= − −
(3.107)
Arrays (Cont.)Arrays (Cont.)
• Equivalent Thermal Circuit:
• Effect of Surface Contact Resistance:
( )( ),
bt t bo c
t o c
q hAR
θη θ= =
( )1
1 1f fo c
t
NA
A C
ηη
= − −
(3.110a)
( )1 , ,1 /f f t c c bC hA R Aη ′′= + (3.110b)
( )( )
,
1t o c
to c
RhAη
= (3.109)
,1b
t ot o t
Rq hA
θη
= = (3.108)
50
Problem 3.144: Determination of maximum allowable power for a 20 mm 20 mm electronic chip whose temperature is not to exceedwhen the chip is attached to an air-cooled heat sink with N = 11 fins of prescribed dimensions.
cq85 C,cT = o
Schematic:
Assumptions: (1) Steady-state, (2) One-dimensional heat transfer, (3) Isothermal chip, (4)Negligible heat transfer from top surface of chip, (5) Negligible temperature rise for air flow,(6) Uniform convection coefficient associated with air flow through channels and over outersurface of heat sink, (7) Negligible radiation, (8) Adiabatic fin tips.
Problem: Chip Heat SinkProblem: Chip Heat Sink
×
Analysis: (a) From the thermal circuit,
tot
c cc
t,c t,b t,o
T -T T -Tq = =
R R + R + R∞ ∞
( )22 6 2m K/W, / W 2 10 / 0.02m 0.005 K/Wt,c t cR R −′′= = × =⋅
( )2/ Wt,b bR L k= ( )2W/m K0.003m / 180 0.02m 0.042 K/W⋅= =
From Eqs. (3.108), (3.107), and (3.104)
( )1, 1 1 ,
ft,o o f t f b
o t t
NAR A NA A
h A Aη η
η= = − − = +
Af = 2WLf = 2 × 0.02m × 0.015m = 6 × 10-4
m2
Ab = W2 – N(tW) = (0.02m)
2 – 11(0.182 × 10
-3 m × 0.02m) = 3.6 × 10
-4 m
2
With mLf = (2h/kt)1/2
Lf = (200 W/m2⋅K/180 W/m⋅K × 0.182 × 10
-3m)
1/2 (0.015m) =
1.17, tanh mLf = 0.824 and Eq. (3.94) yields tanh 0.824
0.7041.17
ff
f
mL
mLη = = =
At = 6.96 × 10-3
m2
ηo = 0.719,
Rt,o = 2.00 K/W, and
( )( )
85 20 °C31.8 W
0.005 0.042 2.00 K/Wcq−
= =+ +
Problem: Chip Heat Sink (cont.)Problem: Chip Heat Sink (cont.)
<
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Comments: The heat sink significantly increases the allowable heat dissipation. If it were not used and heat was simply transferred by convection from the surface of the chip with
from Part (a) would be replaced by 2tot100 W/m K, 2.05 K/W= ⋅ =h R
2conv 1 / 25 K/W, yielding 2.60 W.cR hW q= = =
Problem: Chip Heat Sink (cont.)Problem: Chip Heat Sink (cont.)
HEAT
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CCCCCCCCHHHHHHHH
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Dr Mazlan - SME 4463 102