The rate of heat flow through the cross-secti | vedclass

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(A) For a conical rod of length $L$ with radii $r_1$ and $r_2$ at the ends,the radius at a distance $x$ from the end with radius $r_1$ is given by $r(

Vedclass · Accepted Answer

$\frac{K\pi r_1 r_2 (T_2 - T_1)}{L}$

Vedclass · Answer

(A) For a conical rod of length $L$ with radii $r_1$ and $r_2$ at the ends,the radius at a distance $x$ from the end with radius $r_1$ is given by $r(x) = r_1 + \frac{r_2 - r_1}{L}x$.The thermal resistance $dR$ of a small element of thickness $dx$ is $dR = \frac{dx}{K A(x)} = \frac{dx}{K \pi (r(x))^2}$.The total thermal resistance $R$ is the integral of $dR$ from $0$ to $L$:$R = \int_{0}^{L} \frac{dx}{K \pi (r_1 + \frac{r_2 - r_1}{L}x)^2}$.Solving this integral gives $R = \frac{L}{K \pi r_1 r_2}$.The rate of heat flow is $\frac{dQ}{dt} = \frac{T_2 - T_1}{R} = \frac{K \pi r_1 r_2 (T_2 - T_1)}{L}$.

The rate of heat flow through the cross-section of the rod shown in the figure is ($T_2 > T_1$ and the thermal conductivity of the material of the rod is $K$).

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