AP EAMCET Engineering 26 Apr 2017 Shift 1 Paper

The figure below represents the wall P and Q.It is given that,
kQ=

kp

2

kp=2kQ
And
T1−T2=24°C
In steady state, the rate of flow heat through layer P and Q issame,

kpa(T1−T3)

x

=

kQa(T3−T2)

x

Substitute the values in the above equation.

2kQa(T1−T3)

x

=

kQa(T3−T2)

x

2(T1−T3)=(T3−T2)
2T1−3T3+T2=0
From equation (I),
2T1−3T3+T1−24=0
3T1−3T0=24
T1−T0=8°C