The temperature of the two outer surfaces of a composite slab, consisting of two materials having coefficients of thermal conductivity $K$ and $2 K$ and thickness $x$ and $4 x$ , respectively, are $T_{2}$ and $T_{1}$ $(T_{2} > T_{1})$ . The rate of heat transfer through the slab, in a steady state is $(\frac{A (T_{2} - T_{1}) K}{x}) f .$ Find the value of 100f.

Question

Smart Achievers · Accepted Answer

(33)

The thermal resistance is given by

$\frac{x}{K A} + \frac{4 x}{2 K A} = \frac{x}{K A} + \frac{2 x}{K A} = \frac{3 x}{K A}$

$∴$ $\frac{d Q}{d t} = \frac{Δ T}{\frac{3 x}{K A}} = \frac{(T_{2} - T_{1}) K A}{3 x} = \frac{1}{3} {\frac{A (T_{2} - T_{1}) K}{x}}$

$∴$ $f = \frac{1}{3}$

Solution

Ans.
(33)

The thermal resistance is given by

$\frac{x}{K A} + \frac{4 x}{2 K A} = \frac{x}{K A} + \frac{2 x}{K A} = \frac{3 x}{K A}$

$∴$ $\frac{d Q}{d t} = \frac{Δ T}{\frac{3 x}{K A}} = \frac{(T_{2} - T_{1}) K A}{3 x} = \frac{1}{3} {\frac{A (T_{2} - T_{1}) K}{x}}$

$∴$ $f = \frac{1}{3}$

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Solution

Ans. (33) The thermal resistance is given by xKA+4x2KA=xKA+2xKA=3xKA ∴ dQdt=ΔT3xKA=(T2-T1)KA3x=13{A(T2-T1)Kx} ∴ f=13

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Ans.
(33)

The thermal resistance is given by

$\frac{x}{K A} + \frac{4 x}{2 K A} = \frac{x}{K A} + \frac{2 x}{K A} = \frac{3 x}{K A}$

$∴$ $\frac{d Q}{d t} = \frac{Δ T}{\frac{3 x}{K A}} = \frac{(T_{2} - T_{1}) K A}{3 x} = \frac{1}{3} {\frac{A (T_{2} - T_{1}) K}{x}}$

$∴$ $f = \frac{1}{3}$