Question

Figure (20) shows the composite wall of a freezing plant which consists of a 10 a layer of corkboard on the inside and 14 cm layer of concrete on the outside, with an area of 1 m'. The temperature on the inside corkboard surface is -20 °C, and the temperature on the outside concrete surface is 24 °C. The thermal conductivity for corkboard is 0.04 Wm.K and for concrete is 0.8 Wm K Figure Q2(b) Page 1 of 2 Calculate the temperature of the interface bergart: 0 Material: and the rate of heat los: for both the co.kboard and concrete

Figure Q2(b) shows the composite wall of a freezing plant which consists of a 10 cm layer of corkboard on the inside and 14 cm layer of concrete on the outside, with an area of 1 m2. The temperature on the inside corkboard surface is -20 °C, and the temperature on the outside concrete surface is 24 °C. The thermal conductivity for corkboard is 0.04 W/m.K and for concrete is 0.8 W/m.K.

Calculate the temperature of the interface between the two materials.(3marks)(ii)Find the rate of heat loss for both the corkboard and concrete.

Answer 1

The composite wall is shown.

K2 A=im? corkboard .24°C * -2007 Concrete *10*14 cm ki = 0.04 W/mk K2=0.8 W/mk

For cork-board,

Thermal conductivity K1=0.04 W/mK

Area A=1m²

length dx=10 cm=0.1 m

Temperature on inside T1=-20^oC=253K

For concrete,

Thermal conductivity K2=0.8 W/mK

Area A=1m²

length dx=14 cm=0.14 m

Temperature on inside T1=24^oC=297 K

Let the temperature at the interface be T K

(i) The rate of heat flow is given as Q/t=-KAdT/d

where dT/dx=temperature gradient

Equating the heat loss and heat gained for corkboard and concrete,

-0.04*1(T-253)/0.1=-0.8*1(297-T)/0.14

=0.07T-17.71=297-T

Thus, T=294.12K

This is the temperature at the interface

(ii) The rate of heat loss is same for both the materials=0.04*1(294.12-253)/0.1=16.45 W/s