Chapter 20, Problem 20.82CP

A spherical shell has inner radius 3.00 cm and outer radius 7.00 cm. It is made of material with thermal conductivity k = 0.800 W/m · °C. The interior is maintained at temperature 5°C and the exterior at 40°C. After an interval of time, the shell reaches a steady state with the temperature at each point within it remaining constant in time. (a) Explain why the rate of energy transfer P must be the same through each spherical surface, of radius r, within the shell and must satisfy

$\frac{dT}{dr}=\frac{P}{4\pi k{r}^2}$

(b) Next, prove that

${\displaystyle {\int }_5^{\infty }dT}=\frac{P}{4\pi k}{\displaystyle {\int }_{0.03}^{0.07}{r}^{-2}dr}$

where T is in degrees Celsius and r is in meters. (c) Find the rate of energy transfer through the shell. (d) Prove that

${\displaystyle {\int }_5^{T}dT}=1.84{\displaystyle {\int }_{0.03}^r{r}^{-2}dr}$

where T is in degrees Celsius and r is in meters. (e) Find the temperature within the shell as a function of radius. (f) Find the temperature at r = 5.00 cm, halfway through the shell.

To determine

The reason for the same rate of energy transfer through spherical surface.

Answer to Problem 20.82CP

The rate of energy transfer through the spherical surface is same because the temperature gradient is constant.

Explanation of Solution

Let the rate of energy transfer is $P$, radius of the spherical surface is $r$, the elemental temperature is $dT$ and elemental distance is $dr$.

The law of thermal conduction is,

$P=kA\frac{dT}{dr}$

Here,

$P$ is the power or rate of energy transfer.

$k$ is the coefficient of thermal conductivity.

$A$ is the surface area of surface.

$T$ is the temperature in Kelvin.

$r$ is the radial distance.

The expression for the surface area of the sphere ism,

$A=4\pi r^2$

Substitute $4\pi r^2$ for a in above equation,

$P=k4\pi r^2\frac{dT}{dr}$

$\frac{dT}{dr}=\frac{P}{4\pi k{r}^2}$ (1)

Since the value of the coefficient of thermal conductivity is constant and the radius of the spherical surface is also constant. The thermal gradient becomes constant.

The rate of energy transfer is directly proportional to the thermal gradient. Since the thermal gradient is constant so the rate of energy transfer through the spherical surface is same.

Conclusion:

Therefore, the rate of energy transfer through the spherical surface is same because the temperature gradient is constant

To determine

To show: The given relation, ${\displaystyle \underset{5}{\overset{\infty }{\int }}dT}=\frac{P}{4\pi k}{\displaystyle \underset{0.03}{\overset{0.07}{\int }}{r}^{-2}dr}$.

Answer to Problem 20.82CP

The given relation, ${\displaystyle \underset{5}{\overset{\infty }{\int }}dT}=\frac{P}{4\pi k}{\displaystyle \underset{0.03}{\overset{0.07}{\int }}{r}^{-2}dr}$ is valid.

Explanation of Solution

Let the temperature is $T$ in degrees Celsius and the radius of the spherical surface is $r$.

Rearrange the equation (1) to prove the relation,

$dT=\frac{P}{4\pi r^{2}k}dr$

Integrate at both sides for temperature from $5°\text{C}$ to $\infty °\text{C}$ and for radial distances from $0.03\text{m}$ to $0.07\text{m}$.

$\begin{array}{l}{\displaystyle \underset{5}{\overset{\infty }{\int }}dT}=\frac{P}{4\pi k}{\displaystyle \underset{0.03}{\overset{0.07}{\int }}\frac{dr}{r^2}}\\ {\displaystyle \underset{5}{\overset{\infty }{\int }}dT}=\frac{P}{4\pi k}{\displaystyle \underset{0.03}{\overset{0.07}{\int }}{r}^{-2}dr}\end{array}$

Conclusion:

Therefore, the equation, ${\displaystyle \underset{5}{\overset{\infty }{\int }}dT}=\frac{P}{4\pi k}{\displaystyle \underset{0.03}{\overset{0.07}{\int }}{r}^{-2}dr}$ is valid.

To determine

The rate of energy transfer through the shell.

Answer to Problem 20.82CP

The rate of energy transfer through the shell is $18.47\text{W}$.

Explanation of Solution

From the equation (1),

$\begin{array}{l}\frac{dT}{dr}=\frac{P}{4\pi k{r}^2}\\ dT=\frac{P}{4\pi r^{2}k}dr\end{array}$

Integrate at both sides for temperature from $5°\text{C}$ to $40°\text{C}$ and for radial distances from $0.03\text{m}$ to $0.07\text{m}$.

$\begin{array}{c}{\displaystyle \underset{5}{\overset{40}{\int }}dT}=\frac{P}{4\pi k}{\displaystyle \underset{0.03}{\overset{0.07}{\int }}\frac{dr}{r^2}}\\ {\left[T\right]}_5^{40}=\frac{P}{4\pi \times 0.8}{\left[-\frac{1}{r}\right]}_{0.03}^{0.07}\\ P=18.47\text{W}\end{array}$

Conclusion:

Therefore, the rate of energy transfer through the shell is $18.47\text{W}$.

To determine

To show: The given equation, ${\displaystyle \underset{5}{\overset{T}{\int }}dT}=\frac{P}{4\pi k}{\displaystyle \underset{0.03}{\overset{r}{\int }}{r}^{-2}dr}$.

Answer to Problem 20.82CP

The equation, ${\displaystyle \underset{5}{\overset{T}{\int }}dT}=1.84{\displaystyle \underset{0.03}{\overset{r}{\int }}{r}^{-2}dr}$ is valid.

Explanation of Solution

Let the temperature is $T$ in degrees Celsius and the radius of the spherical surface is $r$.

Rearrange the equation (1) to prove the relation,

$dT=\frac{P}{4\pi r^{2}k}dr$

Integrate at both sides for temperature from $5°\text{C}$ to $T°\text{C}$ and for radial distances from $r\text{m}$ to $0.07\text{m}$,

${\displaystyle \underset{5}{\overset{T}{\int }}dT}=\frac{P}{4\pi k}{\displaystyle \underset{0.03}{\overset{r}{\int }}\frac{dr}{r^2}}$

Substitute $40°\text{C}$ for $T$ and $0.07\text{m}$ for r in above equation to calculate the value of power,

$\begin{array}{l}{\displaystyle \underset{5}{\overset{40}{\int }}dT}=\frac{P}{4\pi k}{\displaystyle \underset{0.03}{\overset{0.07}{\int }}{r}^{-2}dr}\\ {\left[T\right]}_5^{40}=\frac{P}{4\pi k}{\left[-\frac{1}{r}\right]}_{0.03}^{0.07}\\ \frac{P}{4\pi k}=1.8375\end{array}$

$\frac{P}{4\pi k}=1.84$ (3)

Put the value of the $\frac{P}{4\pi k}$ in equation (2),

${\displaystyle \underset{5}{\overset{T}{\int }}dT}=1.84{\displaystyle \underset{0.03}{\overset{r}{\int }}{r}^{-2}dr}$ (4)

Conclusion:

Therefore, the equation, ${\displaystyle \underset{5}{\overset{T}{\int }}dT}=1.84{\displaystyle \underset{0.03}{\overset{r}{\int }}{r}^{-2}dr}$ is valid.

To determine

The temperature within the cell as a function of radius.

Answer to Problem 20.82CP

The temperature within the cell as a function of radius is $T=-56.33+\frac{1.84}{r}$

Explanation of Solution

From the equation (1),

${\displaystyle \underset{5}{\overset{T}{\int }}dT}=1.84{\displaystyle \underset{0.03}{\overset{r}{\int }}{r}^{-2}dr}$

Integrate the above equation,

$\begin{array}{l}\left[T-5\right]=1.84\left[-\frac{1}{0.03}+\frac{1}{r}\right]\\ T=-56.33+\frac{1.84}{r}\end{array}$ (5)

Conclusion:

Therefore, the temperature within the cell as a function of radius is $T=-56.33+\frac{1.84}{r}$

To determine

The temperature at radius. $5.00\text{cm}$.

Answer to Problem 20.82CP

The temperature at radius. $5.00\text{cm}$. is $-19.53°\text{C}$

Explanation of Solution

From the equation (5),

$T=-56.33+\frac{1.84}{r}$

Substitute $0.05\text{cm}$ for $r$ in above equation,

$\begin{array}{l}T=-56.33+\frac{1.84}{5\text{cm}}\\ T=-56.33+\frac{1.84}{5\text{cm}}\times \frac{100\text{cm}}{1\text{m}}\\ T=-19.53°\text{C}\end{array}$

Conclusion:

Therefore, the temperature in spherical shell at radius. $5.00\text{cm}$ is $-19.53°\text{C}$.