Question

Two wall of thickness in 1 : 3 and thermal conductivity in ratio 3 : 2 form a composite wall of building. if the free surface of the wall he at temperature 30°c and 20°c respectively. what is the temperature of inference.

Answer 1

the average temperature is 25deg cel

Answer 2

1 st method : If the temperature diffrence across first wall be ΔT∘C, then that across that second wall will (10−ΔT)∘C (note)

In steady state, the rate of heat flow across the two walls will be the same.

So, considering an area A normal to the flow of heat,

dQdt=K1A(ΔT)l1=K2A(10−ΔT)l2

where K1=K,K2=3K and l1=3l,l2=2l

KAΔT3l=3KA(10−ΔT)2l⇒2ΔT=9(10−ΔT)

ΔT=8.18∘C

So, the temperature of the interface will be

30−ΔT=30−8.18=21.822C

2nd method : Using the equation, temperature of the interface

T=(K1T1/l1+K2T2/l2)(K1/l1+K2/l2)

We have T=(K(30)3l+3K(20)2l)/(K3l+3K2l)

=(10+30)(12+32)=21.82∘C

3rd method : R1=l1K1A=3lKA and R2=l2K2A=2l3KA

Effective thermal resistance R=R1+R2=113lKA

Now, thermal current i=ΔTR=30−2011l3KA=30KA11l

Also, thermal current through the first slab will be

i=(30−T)R1

i.e., 30KA11l=(30−T)KA3l⇒90=330−11T

⇒T=21.82∘C