1

Chapter 4 Transient heat conduction

4-1 introduction

In this chapter we will study

- Lumped system method

- Use Lumped system method to solve some transient heat conduction

problems

T

time

2

4-1 Lumped system method • The Boit number

A plate has thermal conductivity k, thickness L, and temperatureT1 at the left-hand side. The right-hand side temperature is T2 and it is under convection boundary condition with heat transfer coefficient h and ambient temperature T∞.

Under steady state,

If hL/k is much smaller than unity, T1 –T2 will be much small than T2 –T∞

The temperature inside the plate is approximately uniform or constant. The Boit number, Bi is defined as

Theoretically, if Bi is equal to 0, T1 = T2.

T1

T2

T∞

1 2

1 22( )

T TQ kA hA T T

L ∞−

= = − ⇒&1 2 2( )

hLT T T T

k ∞− = −

hLBi

k=

L

kA

T1 T2 T∞

1

hA

3

4-1 Lumped system method

• If Boit number is less than 0.1, . It is reasonable to assume the temperature in the plane wall is approximately constant. For

an arbitrary shaped solid with no unique thickness the characteristic length L is defined as

For a sphere of diameter d,

• Note V is the volume of the object and A is the area which is responsible to heat transfer. If part of the surface area is insulated, that part cannot be counted as heat transfer area.

No insulation part Two ends are insulated

. . ..

VolumeL

heat transfer surface area=

1 2 2T T T T∞− −pp

3

2

1166

dVL d

A d

π

π= = =

4

4-2 Temperature variation of an object without internal heat generation

• A device, initially, has a temperature Ti in an oven. It is suddenly taken out from the oven and put it in a fluid medium with temperature T∞ and convection heat transfer coefficient h. The device has a mass m and specific heat c. During a time interval dt, the amount of heat loss by convection is equal to the change of internal energy of the device.

T is the instant temperature of the solid at time t and A is the heat transfer area.

• Rearranging the above equation

• Integrating over the time

interval from 0 to t

is called time constant.

( ) ( )hA T T dt mcdT mcd T T∞ ∞− =− =− −

( )d T T hAdt

T T mc∞

∞

−=−

−

hA tt

mc

i

T Te e

T Tτ

− −∞

∞

−= =

−

hA

mc=τ

time

Ti

T∞

dT

5

4-3 Temperature variation of an object with internal heat generation

• An electronic device has mass m, specific heat c, surface area exposed to the fluid medium A. It is initially in equilibrium with the ambient temperature T∞. Suddenly, the device is energized and initiates a constant heat generation. The energy balance relationship is

• Rearranging

• The initial and steady state conditions

• The general solution consists of two parts;

- the complementary solution and the particular solution. it is obtained from

b is the integration constant

( ) ( )in in

dT dTQ mc hA T T mc Q hA T T

dt dt∞ ∞= + − ⇒ = − −& &

( )( ) inQd T Tmc Q dT T

hA dt hA dt hA

θτ θ∞∞

−= − − ⇒ + =

&&

0, 0

, s

t

t T T

θθ ∞

≤ =→ ∞ = −

0 ln ln lnt

c c cc c c

c

d d dt t tb be

dt bτθ θ θτ θ θ θ

θ τ τ τ−

+ = ⇒ =− ⇒ =− + ⇒ =− ⇒ =

T

time

T∞

Ts

6

4-3 Temperature variation of an object with internal heat generation

- The particular solution

• The general solution is

when t = 0, T = T∞ and Θ =0

• When time is very long, the rate of heat generation is equal to the heat dissipation

Ts is the steady state operation temperature

inp

Q

hAθ =

&

( )in sQ hA T T∞= −&

tinQbehA

τθ−

= +&

(1 )t

inQ ehA

τθ−

= −&

inQbhA

=−&

Ts

T∞

Time t

7

Example 4-1 thermocouple

• Given conditions : A thermocouple, d = 1mm, k = 35W/mK, ρ = 8500kg/m3, c = 320J/kgK, h = 210W/m2K, • Find : time when the thermocouple to read 99% of the (T∞ - Ti)

• Solution - The Boit number

the Lumped system method is applicable

hLBi

k=

0.99(T∞ - Ti)

T∞

Ti

Time t

0.01(T∞ - Ti)

3

2

1210 0.000166 0.00016 0.1

6 35

dV d xL m Bi

A d

π

π= = = = ⇒ = pp

8

• The time constant

• The time to reach 1% error

• Conclusion: this means that when you use this thermocouple to measure temperature under these conditions, you still get 1% error after 10 sec of waiting. For smaller error, you need to wait longer.

2.160.01 10sect t

i

T Te e t

T Tτ

− −∞

∞

−= ⇒ = ⇒ =

−

3

2

1 18500 (0.001) 320

6 6 2.16210

d c x xmcs

hA h d

ρ πτ

π= = = =

9

Example 4-2 Temperature variation of a semiconductor device

Example 2 An electronic device generates 4W of heat when it is energizedhas dimensions 10mm x 10mm x 1mm. The device is mounted on a lowthermal conductivity epoxy glass printed circuit board. The heat transfer rate form the device to the board can be neglected. Direct air cooling has to be applied to maintain the operation temperature of the device to below to the manufacturer’s recommended value. The convection heat transfer coefficient is 400 W/m2K. The thermo physical properties of the device are thermal conductivity =20W/mK, specific heat= 4kJ/kgK, and density= 2000kg/m3. The initial temperatures of the device and the ambient fluid temperature are 30oC.Determine(a) The time constant of the device(b) The steady state operation temperature ©The temperature of the device after 5 sec when the power is on.

• Assumptions - Constant thermo physical properties - Radiation effect is neglected• Solutions

10

Example 4-2 Temperature variation of a semiconductor device

- Calculate the Boit number

Lumped system method is applicable (a) The time constant

(b) The steady state operation temperature

© The temperature after 5 sec operation

After the power is switched off, the temperature will reduce gradually.

2000(0.01 0.01 0.001)4000 0.814.3sec

400(0.01 0.01 4 0.01 0.001) 0.056

mc Vc x x

hA hA x x x

ρτ = = = = =+

430 101.4

0.056o

s i

QT T C

hA= + = + =

&

5

14.35

4(1 ) 30 (1 ) 51.4

0.056

to

i

QT T e e C

hAτ

− −= + − = + − =

&

Ts

T∞

Time t

11

Review

Lumped system method

• Boit number

• Temperature variation of an object with internal heat generation

• Temperature variation of an object without internal heat generation

hLBi

k= . . ..

VolumeL

heat transfer surface area=

( )in

dTQ mc hA T T

dt ∞= + −& (1 )t

inQT T ehA

τ−

∞− = −&

( )dT

hA T T mcdt∞− =−

''

hA tt

mc

i

T Te e

T Tτ

− −∞

∞

−= =

−