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

Wind turbine cooling

Heating of wind turbine blades