(A) In this system,the two materials are arranged in parallel with respect to the direction of heat flow.For parallel arrangement,the equivalent therm

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(A) In this system,the two materials are arranged in parallel with respect to the direction of heat flow.For parallel arrangement,the equivalent thermal conductivity $K_{eq}$ is given by $K_{eq} = \frac{K_1 A_1 + K_2 A_2}{A_1 + A_2}$.The area of the inner cylinder is $A_1 = \pi R^2$.The area of the outer cylindrical shell is $A_2 = \pi (2R)^2 - \pi R^2 = 4\pi R^2 - \pi R^2 = 3\pi R^2$.The total area is $A = A_1 + A_2 = \pi R^2 + 3\pi R^2 = 4\pi R^2$.Substituting these values into the formula:$K_{eq} = \frac{K_1(\pi R^2) + K_2(3\pi R^2)}{4\pi R^2}$.$K_{eq} = \frac{K_1 + 3K_2}{4}$.

Class 11 Physics 10-2.Heat Transfer Thermal Resistance and it's Combination

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$A$ cylinder of radius $R$ is made of a material with thermal conductivity $K_1$. It is surrounded by a cylindrical shell of a material with thermal conductivity $K_2$,having an inner radius $R$ and an outer radius $2R$. The two ends of this composite system are maintained at two different temperatures. There is no heat loss from the surface of the cylinder in the steady state. Find the equivalent thermal conductivity of the system.

A
$\frac{K_1 + 3K_2}{4}$
B
$K_1 + K_2$
C
$\frac{K_1 + 8K_2}{9}$
D
$\frac{8K_1 + K_2}{9}$

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10-2.Heat Transfer

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Thermal Resistance and it's Combination

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MediumAIPMT 2003

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The temperatures of the two outer surfaces of a composite slab are $T_2$ and $T_1$ $(T_2 > T_1)$. The thermal conductivities of the materials are $K$ and $2K$,and their thicknesses are $x$ and $4x$,respectively. In the steady state,the rate of heat flow through the slab is given by $f \left( \frac{A(T_2 - T_1)K}{x} \right)$. Find the value of $f$.

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Three rods of the same material,same length,and same cross-sectional area are connected as shown in the figure. Find the temperature of their junction in $^oC$.

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