The figure shows a system of two concentric s | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) Consider a concentric spherical shell of radius $r$ and thickness $dr$ as shown in the figure.In a steady state,the radial rate of flow of heat $H

Vedclass · Accepted Answer

$\frac{{{r_1}\,{r_2}}}{{({r_2} - {r_1})}}$

Vedclass · Answer

(A) Consider a concentric spherical shell of radius $r$ and thickness $dr$ as shown in the figure.In a steady state,the radial rate of flow of heat $H$ through this shell is given by Fourier's law of heat conduction:$H = \frac{{dQ}}{{dt}} = - KA\frac{{dT}}{{dr}}$Since the area of the spherical shell is $A = 4\pi {r^2}$,we have:$H = - K(4\pi {r^2})\frac{{dT}}{{dr}}$Rearranging the terms to integrate:$\frac{{dr}}{{{r^2}}} = - \frac{{4\pi K}}{H}dT$Integrating from $r_1$ to $r_2$ and $T_1$ to $T_2$:$\int_{{r_1}}^{{r_2}} \frac{{dr}}{{{r^2}}} = - \frac{{4\pi K}}{H} \int_{{T_1}}^{{T_2}} dT$$\left[ - \frac{1}{r} \right]_{{r_1}}^{{r_2}} = - \frac{{4\pi K}}{H} (T_2 - T_1)$$\left( \frac{1}{{{r_1}}} - \frac{1}{{{r_2}}} \right) = \frac{{4\pi K}}{H} (T_1 - T_2)$$\frac{{{r_2} - {r_1}}}{{{r_1}{r_2}}} = \frac{{4\pi K}}{H} (T_1 - T_2)$Solving for $H$:$H = \frac{{4\pi K{r_1}{r_2}({T_1} - {T_2})}}{{{r_2} - {r_1}}}$Thus,the rate of heat flow is proportional to $\frac{{{r_1}{r_2}}}{{{r_2} - {r_1}}}$.

The figure shows a system of two concentric spheres of radii $r_1$ and $r_2$ kept at temperatures $T_1$ and $T_2$,respectively. The radial rate of flow of heat in a substance between the two concentric spheres is proportional to

Similar Questions

The two ends of a rod of length $L$ and a uniform cross-sectional area $A$ are kept at two temperatures $T_{1}$ and $T_{2}$ $(T_{1} > T_{2})$. The rate of heat transfer,$\frac{dQ}{dt}$ through the rod in a steady state is given by:

The two ends of a rod of length $L$ and a uniform cross-sectional area $A$ are kept at two temperatures $T_1$ and $T_2$ $(T_1 > T_2)$. The rate of heat transfer,$\frac{dQ}{dt}$,through the rod in a steady state is given by:

$A$ cylindrical rod has temperatures $\theta_1$ and $\theta_2$ at its ends. The rate of heat flow is $Q \ J s^{-1}$. All the linear dimensions of the rod are doubled while keeping the temperatures constant. What is the new rate of flow of heat?

Vedclass Test Series

Exam Paper Generator

Online Exam Module