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Unsteady Heat Conduction(Chapter 4)Zan Wu zan.wu@energy.lth.se
Recap of the previous lecture
To derive the governing equation for rectangular fin To calculate temperature distribution, heat transfer rate, fin
effectiveness, fin efficiency- the meanings of C, A, m and beta
Fin optimization at a given fin mass- rectangular fin: (mL)optimum =1.419- triangular fin: (mL)optimum =1.309
The total heat transfer rate from fin arrays, Eq. (3.71)
Three clarifications
• Fin?
• Eq. (15.1b) thermal conductivity of plate material unimportant?
• Ex. 122
+ + e.g., fouling ignored
Outline
Lumped capacitance method: object with very high thermal conductivity
- Derive the function of temperature vs. time- Bi, Fo- Validity of the lumped capacitance method
Unsteady heat conduction in- infinite plane walls- long cylinders- spheres- 2D and 3D cases
Unsteady heat conduction in semi-infinite bodies
Unsteady heat conduction with very high thermal conductivity
First law of thermodynamics
Solution
0
expA
Vc AeVc
Set a characteristic length Lc
cVLA
20
exp exp exp Bi Focc
LA aVc L
= t tf
Dimensionless numbers: Biot number and Fourier number
conductive resistance within the solidBi1 convective resistance across the fluid boundary layer
cc L ALA
2Foc
aL
If Bi < 0.1, the body temperature can be assumed to be uniform
Outline
Lumped capacitance method: object with very high thermal conductivity
- Derive the function of temperature vs. time- Bi, Fo- Validity of the lumped capacitance method
Unsteady heat conduction in- infinite plane walls- long cylinders- spheres- 2D and 3D cases
Unsteady heat conduction in semi-infinite bodies
Infinite plate with moderate thermal conductivity
2
2
xa
= t tf
1D; Edge effects negligible
Infinite plate with moderate thermal conductivity
= t tf
Initial condition (I.C.) : = 0 ; (x, 0) = 0(x) = t0(x) - tfBoundary conditions:
2
2
xa
x
ttxt:Lx
0x
0xt:0x
f
(x
L
x
tf
Infinite plate, limited thermal conductivity
aeCF2
)(
Assumption: (x, ) = F() G(x)
xsinCxcosC)x(G 21
)184(sincos),( 2 xBxAex a
x, are independent variables
Cont’d
20
, , (4 34)a xfunctionL L L
Graphical representation of solution
0 0.5 1 1.5 20
0.2
0.4
0.6
0.8
1
m = 0.00.100.25
0.50
0.75
1.00
1.502.00
3.004.006.0010.020.0
x ' = 0.4
0
2/ La
Fig. 4.4
Long circular cylinder
r
z
ro
Figs. long circular cylinder
0 0.5 1 1.5 20
0.2
0.4
0.6
0.8
1
m = 0.0 0.250.50
0.751.00
1.50
2.00
3.00
4.00
6.00
10.020.050.0
r ' = 0.0
2o/ra
0
f2
0 0 f o o o
function ( , , )t t a rt t r r r
Fig. 4.6
Figs. long circular cylinder
0 0.5 1 1.5 20
0 .2
0 .4
0 .6
0 .8
1
m = 0 .25 0 .500 .751 .00
1 .50
2 .00
4 .00
3 .00
6 .00
10.0
20.050.0
r ' = 1 .0
2o/ ra
0
f2
0 0 f o o o
f u n c t i o n ( , , )t t a rt t r r r
Two- and three-dimensional solutions
Product of two one-dimensional solutions
Three-dimensional case
2
2
2
2
2
2
zyxa
0 0 0x y
0 0 0 0x y z
A
Bx
ytf
Semi-infinite bodies
Initial condition = 0 : t(x, 0) = t0BC
The solution is given by
error function
2
2
xtat
st),0(t:00x
x finitet
0
(4.51)2
s
s
t t xerft t a
de2a2
xerfa2/x
0
2
x
Heat flux: heat transfer rate/surface area
a
ttAQ s
x
)( 00
20( )1 e x p
4st tQ d Q x
A A d aa
(4.53)
(4.54)
Total heat amount
= QQ d
Recap• Solid with very high thermal conductivity or Bi < 0.1
- First law of thermodynamics
-
• Infinite plate, cylinder, sphere- Figs. 4.4, 4.6, 4.7
- extended to 2D, 3D cases
• Semi-infinite bodies- Eqs. (4.51), (4.53), (4.54)
- Gaussian error function, Table 4.1
0 exp Bi Fo
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