Question

The temperature of the two outer surface of a composite slab consisting of two materials having coefficient of thermal conductivity k and 2k and thickness x and 4x respectively. T2 and t1 (t2 > t1). The rate of heat transfer through the slab in steady state is y, then y is equal to

Answer 1

Answer:

Let the temperature of common interface be TK rate of heat flow H=

t

Q

=

l

KAΔT

∴H

1

=(

t

Q

)

1

=

4x

2KA(T−T

1

)

and H

2

=(

t

Q

)

2

=

x

KA(T

2

−T)

In steady-state, the rate of heat flow should be the same in the whole system i.e.,

H

1

=H

2

or

4x

2KA(T−T

1

=

x

KA(T

2

−T)

2

T−T

1

=T

2

−T or T−T

1

=2T

2

−2T

T=

3

2T

2

+T

1

........(i)

Hence heat flow from composite slab is

H=[

x

KA(T

2

−T)

]=

x

KA

(T

2

−

3

2T

2

+T

1

) [from eqn.(i)]

=

3x

KA

(T

2

−T

1

) ..........(ii)

According to question,H=[

x

A(T

2

−T

1

)K

]f ...........(iii)

by comparing eqs.(ii) and (iii) we get f=

3

1

Answer