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Steady Heat Conduction(Chapter 3)Zan Wu [email protected] Room: 5113
Objectives
Steady-state heat conduction• Without internal heat generation
- Derive temperature profile for a plane wall
- Derive temperature profile for a circular layer
- Interpret thermal resistance and apply this concept to calculate heat transfer rate
• With internal heat generation
- Derive temperature profile for a plane wall with Q’
Heat conduction equation (isotropic material)
If constant
Thermal diffusivity
+ + 1 18
1 19
Simple plane wall
BC: x = 0, t = t1; x = b, t = t2
General solution: t = c1x + c2
0
0
0
constant0 (1-19)
Simple plane wall
Heat flow, Fourier’s law
Alternate formulation
potential = resistance · current
)33(xb
tttt 121
(3-4)
·
Composite wall
“Serial circuit” 1 4 ( thermal resistence)t t Q
QA
bA
bA
btt3
3
2
2
1
141
)63(
Ab
Ab
Ab
ttQ
3
3
2
2
1
141
Convective thermal resistance
A
Newton’s law of cooling
1A ·
Convective thermal resistance:
Composite wall, convective BC
tf2
tf1
b1
1 2 3
b2 b3
1
2
)73(
A1
Ab
A1
ttQ
2
3
li i
i
1
2ff1
Hot liquid
Cold liquid
Circular tube or layer (Shell)
(3 10)
1 2 lni o
o i
t tQL r r
Composite circular walls
i of f
32
1 1 1 2 2 3 0
1 1 1 1ln ln2 2 2 2i
t tQ rr
r L L r L r r L
Contact resistanceTemperature drop due to thermal contact resistance
Fouling
The accumulation and formation of unwanted materials on the surfaces of processing equipmentOne of the major unsolved problems in heat transfer
Plane wall with internal heat generation
b
x
btf tf
Q'
Uniform heat generation per unit volume
Governing equation
)181(
Qzt
z
yt
yxt
xtc
2
2 0d t Qdx
21 2
'2Qt x c x c
General solution
Boundary conditions
0x 0dtdx
bx wall fdt t tdx
At the plane of symmetry
Adiabatic or insulated BC
Solution
2 2f( )
2Q Q bt b x t
bQbQtt 2
fmax 2
Recap: steady-state heat conduction
• Start with the heat conduction equation, simply it with proper assumptions
• Then get a general solution, combining with BCs to obtain a specific solution for temperature distribution
• Use the Fourier’s Law to obtain the heat transfer rate based on the temperature distribution
Heat Transfer from Fins, Extended Surfaces
Objectives
• Derive governing equations and formulate boundary conditions for rectangular and triangular fins
• Calculate fin efficiency and fin effectiveness
• Understand optimal fin criteria for rectangular and triangular fins
• Apply fin efficiency in heat transfer rate calculations
Example fins
(a) Individually finned tubes(b) flat (continuous) fins on an array of tubes
Example fins
Example microfins
Microfin copper tubeCarbon nanotube microfins
on a chip surface
Fins on Stegosaurus
Absorb radiation from the sun or cool the blood?
Rectangular fin
Rectangular fin
Boundary conditions:
long and thin fin, heat transfer at the fin tip is negligible
)313(0AC
dxd
2
2
)tt( fb
2bZ
Z2ACm 2
)tt(dxdt:Lx fLx
0dxdt
Lx
f111 tttt:0x
x d x
L
t 1
Q 1
.
t f
bZ
Rectangular fin
General solution:
(co sh2
sin h )2
m x m x
m x m x
e em x
e em x
1 2
3 4c o s h s in h
m x m xC e C e
C m x C m x
Hyperbolic functions
For x = L = 2
1 1
c o s h ( ) ( 3 3 8 )c o s h
f
f
t t m L xt t m L
2
1
1c o s h m L
heat transfer from the fin ?Q
1 10
s in h ( )co shx
d m LQ A A mdx m L
CmA
Rectangular fin
· tanh 2 · tanh 3 40
Rectangular fin
= 25 W/m2K, b = 2 cm, L = 10 cmEq. (3-38)
Fin performance evaluation
1: Fin effectiveness
2: Fin efficiency
1
1
from the finfrom the base area w ithout the fin
fromthefinfromasimilarfinbutwithλ ∞
Optimal fin: maximum heat transfer at a given fin weight
M = b L Z = Z A1
A1 = b L, Z, are given.
Find maximum for A1 = bL, constant.
(3-40)
C 2Z , A = bZ (3-35)
(3-52)
mLtanhACQ 11
b2
ACm 2
1Q
b
Ab
2tanhZb2Q 111
LZb
Optimal rectangular fin
Condition
1 0 gives optimum
after some algebra one obtains
21.419 (3 55)/ 2
dQdb
Lb b
Straight triangular fin
= t tf
Heat balance
Solution:
K0 as x 0 B = 0 because is finite for x = 0
x = L = 1
)623(0bL2
x1
dxd
x1
dxd
2
2
bL2
)x2(BKx2AI 00
L2AI 01
LxbZA
L
d x
x
b t 1
t f
1Q
Bessel differential equation
I0 and K0 are the modified Bessel functions of order zero
Triangular fin
)L2(IA
0
1
)653(
)L2(I)x2(I
0
0
1
LxdxdtAQ
1
Lx
0
011 dx
)x2(dI)L2(I
1bZQ
)663()L2(I)L2(Ib2ZQ
0
111
Table 3.2 for numerical values of I0 and I1
)673(b
2309.12/b
L
Optimal triangular fin: maximum heat transfer at given weight
Summary of formulae for rectangular and triangular fins
optimal fin optimal fin
)383(mLcosh
)xL(mcosh
ft1tftt
1
b22m
)403(mLtanh1Zb21Q
mLtanhb
2
mLmLtanh
)553(b
2419.12/b
L
)653()L2(0I)x2(0I
1
bL2
)L2(0I)L2(1I
1Zb21Q
)L2(0I)L2(1I
b2
L)L2(0I/)L2(1I
)673(b
2309.12/b
L
Formulas for fin performance
Some simple calculations give:
Rectangular fin
Triangular fin
mLtanhb
2
)L2(I)L2(I
b2
0
1
L)L2(I/)L2(I 01
mLmLtanh
Fin effectiveness
Fin efficiency
37
Circular or annular fins
Heat conducting area
A = 2r b
Convective perimeter
C = 2 2r = 4r
r 1 r 2
b
Fin efficiency for circular fins
How to use the fin efficiency in engineering calculations
s
flänsarareaoflänsad
QQQ
QQQ
finareaunfinned
( )
( )
fins b f
b b f fins
A t t
Q A t t Q
a
( ) (3 7 1)b f b f in sQ t t A A a
