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Question: Two walls of thicknesses d<sub>1</sub> and d<sub>2</sub> and thermal conductivities k<sub>1</sub> an...

Two walls of thicknesses d1 and d2 and thermal conductivities k1 and k­2 are in contact. In the steady state, if the temperatures at the outer surfaces are T1T _ { 1 } and T2T _ { 2 } , the temperature at the common wall is

A

k1T1d2+k2T2d1k1d2+k2d1\frac { k _ { 1 } T _ { 1 } d _ { 2 } + k _ { 2 } T _ { 2 } d _ { 1 } } { k _ { 1 } d _ { 2 } + k _ { 2 } d _ { 1 } }

B

k1T1+k2d2d1+d2\frac { k _ { 1 } T _ { 1 } + k _ { 2 } d _ { 2 } } { d _ { 1 } + d _ { 2 } }

C

(k1d1+k2d2T1+T2)T1T2\left( \frac { k _ { 1 } d _ { 1 } + k _ { 2 } d _ { 2 } } { T _ { 1 } + T _ { 2 } } \right) T _ { 1 } T _ { 2 }

D

k1d1T1+k2d2T2k1d1+k2d2\frac { k _ { 1 } d _ { 1 } T _ { 1 } + k _ { 2 } d _ { 2 } T _ { 2 } } { k _ { 1 } d _ { 1 } + k _ { 2 } d _ { 2 } }

Answer

k1T1d2+k2T2d1k1d2+k2d1\frac { k _ { 1 } T _ { 1 } d _ { 2 } + k _ { 2 } T _ { 2 } d _ { 1 } } { k _ { 1 } d _ { 2 } + k _ { 2 } d _ { 1 } }

Explanation

Solution

In series both walls have same rate of heat flow. Therefore

dQdt=K1A(T1−θ)d1=K2A(θ−T2)d2\frac { d Q } { d t } = \frac { K _ { 1 } A \left( T _ { 1 } - \theta \right) } { d _ { 1 } } = \frac { K _ { 2 } A \left( \theta - T _ { 2 } \right) } { d _ { 2 } }

⇒K1d2(T1−θ)=K2d1(θ−T2)\Rightarrow K _ { 1 } d _ { 2 } \left( T _ { 1 } - \theta \right) = K _ { 2 } d _ { 1 } \left( \theta - T _ { 2 } \right)

⇒θ=K1d2T1+K2d1T2K1d2+K2d1\Rightarrow \theta = \frac { K _ { 1 } d _ { 2 } T _ { 1 } + K _ { 2 } d _ { 1 } T _ { 2 } } { K _ { 1 } d _ { 2 } + K _ { 2 } d _ { 1 } }