Application of Steady-State Heat Transfer

Steady-state heat transfer

• Temperature in a system remains constant with time.

• Temperature varies with location.

Conductive heat transfer in a rectangular slab

dx

dTkAq

x

T

TkdTdx

x

x A

qx

11

T

TdTk

x

xdx

A

qX

11

x

TkAq

x

T1 > T2

T1

T2

ExampleFor the stainless steel plate 1 cm thick is

maintained at 110C, while the other face is at 90 C. Calculate temperature at 0.5 cm from the

110C-temperature face.

Given :

heat flux = 34,000 W/m2

thermal conductivity of stainless steel = 17 W/m C

Conductive Heat Transfer through a Tubular Pipe

• Consider a long hollow cylinder

l

r

rodr

rl)r(A 2

Tiri

To

T

Conductive Heat Transfer through a Tubular Pipe

• Consider a long hollow cylinder

Example

A 2 cm thick steel pipe (k= 43 W/mC) with 6 cm inside diameter

is being used to convey steam from a boiler to process

equipment for a distance of 40 m. The inside pipe surface

temperature is 115C, and the outside pipe surface temperature

is 90C. Under steady state conditions, calculate total heat

loss to the surrounding.

Heat conduction in multilayered systems

Composite rectangular wall (in series)

x3

k3

T3

x2

k2

T2

x1

k1

T1

q

q

R3

R2

R1

= composite thermal resistance

q

q

R3

R2

R1

x1 k1

T1x2 k2

T2x3 k3

T3

q = A T k / x = A T / (x/k)

T1 + T2 + T3 = T

R = Resistance = x/k = 1/C 1/RT = 1/R1+1/ R2+1/ R3

= (1/(x1 / k 1))+ (1/(x2 / k 2))+ (1/(x3 / k 3))

Composite rectangular wall (in parallel)

and it is resistance which is additive when several conducting layers lie between the hot and cool regions, because A and Q are the same for all layers. In a multilayer partition, the total conductance is related to the conductance of its layers by:

So, when dealing with a multilayer partition, the following formula is usually used:

ExampleA cold storage wall (3m X 6m) is constructed of a 15 cm

thick concrete (k = 1.37 W/mC). Insulation must be provided to maintain a heat transfer rate through the wall at or below 500 W. If k of insulation is 0.04 W/mC. The outside surface temperature of the wall is 38C and the

inside wall temperature is 5C.

Example

How many joules of thermal energy

flow through the wall per second? -------------------------------------------

Heat is like a fluid:  whatever flows through the insulation must also flow

through the wood. 

        

            k (insulation) = 0.20 J/(s-m-C)k (wood)      = 0.80 J/(s-m-C)

Across insulation: Hins = (0 .2 0 )(4 0 )(2

-   5 T)/0.076      -2631 6 105= .

          3. T Across wood: 0 80 40Hwood = ( . )( )(

- T 4 )/0 .0 1 9       - = 1 6 8 4 .2 T 6 736.8

  Heat is like afluid: whateverffff f fffffff fff ffffffffff f fff ffff ffff fff

ough thewood:

      Hwood = Hins - 1684.2 T 6736.8 = 2631.

-     6 105.3 T                 1789.5 T = 9368.4

                                     5 235T = . C

                       H=Hwood=Hins             

1684 2 5 23H= . ( . -     5 67368 2080) . = J/s

- 2631 6 105H= . .      3 (5.235) = 2080 J/s

1 32 54

B

A

D

C E

G

F

RB

RE

RA

RD

RC

RG

RF

Series and parallel one-dimensional heat transfer through a composite wall and

electrical analog

Composite cylindrical tube(in series)

r1

r2

r3

ExampleA stainless steel pipe (k= 17

W/mC) is being used to convey heated oil. The inside surface

temperature is 130C. The pipe is 2 cm thick with an inside diameter

of 8 cm. The pipe is insulated with 0.04 m thick insulation (k=

0.035 W/mC). The outer insulation temperature is 25C. Calculate the temperature of interface between steel and

isulation. Assume steady-state conditions.

THERMAL CONDUCTIVITY CHANGE WITH TEMPERATURE

k = k0(1+T)

dx

dTkAqx

)(21TT

X

Akq m

x

221

TT km is thermal conductivity at T =

Heat transfer through a slab

))(2

)(( 21

2212

0 TTTTx

Akq

THERMAL CONDUCTIVITY CHANGE WITH TEMPERATURE

Heat transfer through a cylindrical tube

dr

kAdTqr

dr

dTrLTkqr )2))(1(( 0

dTTLkr

drqr )1(2 0

)))((1(/ln

200

0 TTTTrr

Lkq ii

ior

Problem1. Find the heat transfer per unit area through the composite

wall. Assume one-dimensional heat flow.

Given:

kA = 150 W/mC

kB = 30 W/mC

kC = 50 W/mC

kD = 70 W/mC

AB = AD

T = 370C

CA

D

Bq

AA = AC = 0.1 m2

T = 66C2.5 cm

7.5 cm

5.0 cm

Problem

2. One side of a copper block 5 cm thick is maintained at 260C. The other side is covered with a layer of fiber glass 2.5 cm thick. The outside of the fiber glass is

maintained at 38C, and the total heat flow through the copper-fiber-glass combination is 44 kW. What is the

area of the slab?

3. A wall is constructed of 2.0 cm of copper, 3.0 mm of asbestos, and 6.0 cm of fiber glass. Calculate the heat

flow per unit area for an overall temperature difference of 500C.

Problem

4. A certain material has a thickness of 30 cm and a thermal conductivity of 0.04 W/mC. At a particular

instant in time the temperature distribution with x, the distance from the left face, is T = 150x2 - 30x, where x is in meters. Calculate the heat flow rates at x = 0 and x =

30 cm. Is the solid heating up or cooling down?5. A certain material 2.5 cm thick, with a cross-sectional area of 0.1 m2, has one side maintained at 35C and the other at 95C. The temperature at the center plane of the material is 62C, and the heat flow through the material is 1 kW. Obtain an expression for the thermal conductivity

of the material as a function of temperature.