Question
The temperature of the two outer surfaces of a composite slab, consisting of two materials having coefficients of thermal conductivity K and 2K and thickness x and 4x, respectively are T2 and T1(T2>T1). The rate of heat transfer through the slab, in a steady state is (A(T2−T1)Kx)f , with f which equal to
Answer: Option D
:
D
Equation of thermal conductivity of the given combination Keq=l1+l2l1K1+l2K2=x+4xxK+4x2K=53K . Hence rate of
flow of heat through the given combination is Qt=Keq.A(T2−T1))(x+4x)=53KA(T2−T1)5x=13KA(T2−T1)x
On comparing it with given equation we get f=13
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:
D
Equation of thermal conductivity of the given combination Keq=l1+l2l1K1+l2K2=x+4xxK+4x2K=53K . Hence rate of
flow of heat through the given combination is Qt=Keq.A(T2−T1))(x+4x)=53KA(T2−T1)5x=13KA(T2−T1)x
On comparing it with given equation we get f=13
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