Chapter 18, Problem 32P

To determine

The time taken for the temperature of each geometry to rise to $25°\text{C}$.

Explanation of Solution

Given:

Diameter of the sphere $\left(D\right)$ is $\text{5}\text{cm}$.

Side of the cube $\left(L\right)$ is $\text{5}\text{cm}$.

Volume of the rectangular prism $\left(V\right)$ is $\text{4}\text{cm}\times \text{5}\text{cm}\times \text{6}\text{cm}$.

Ambient air temperature $\left(T_{\infty }\right)$ is $33°\text{C}$.

Initially temperature of the geometry $\left(T_i\right)$ is $0°\text{C}$.

Temperature of the geometry $\left(T\left(t\right)\right)$ is $25°\text{C}$.

Convection heat transfer coefficient $\left(h\right)$ is $12\text{W}/{\text{m}}^{\text{2}}\cdot \text{K}$.

Calculation:

Write the given properties of the silver.

  $\begin{array}{l}\rho =10,500\text{kg}/{\text{m}}^3\\ c_p=0.235\text{kJ}/\text{kg}\cdot \text{K}\\ k=429\text{W}/\text{m}\cdot \text{K}\end{array}$

Sphere:

Calculate the characteristics length of the sphere $\left(L_c\right)$.

  $\begin{array}{c}{L}_c=\frac{V}{A_s}\\ =\frac{\left(\frac{\pi D^3}{6}\right)}{\pi D^2}\\ =\frac{D}{6}\end{array}$

  $\begin{array}{c}=\frac{0.05\text{m}}{6}\\ =0.008333\text{m}\end{array}$

Calculate the Biot number $\left(\text{Bi}\right)$.

  $\begin{array}{c}\text{Bi}=\frac{h{L}_c}{k}\\ =\frac{\left(12\text{W}/{\text{m}}^{\text{2}}\cdot \text{K}\right)\left(0.008333\text{m}\right)}{429\text{W}/\text{m}\cdot \text{K}}\\ =0.00023\end{array}$

The Biot number is less than 0.1$\left(0.00023<0.1\right)$. Therefore the lumped system analysis is applicable.

Calculate the value of exponent $\left(b\right)$.

  $\begin{array}{c}b=\frac{h{A}_s}{\rho c_{p}V}\\ =\frac{h}{\rho c_p{L}_c}\\ =\frac{\left(12\text{W}/{\text{m}}^{\text{2}}\cdot \text{K}\right)}{\left(10,500\text{kg}/{\text{m}}^3\right)\left(0.235\text{kJ}/\text{kg}\cdot \text{K}\right)\left(0.008333\text{m}\right)}\\ =0.0005836{\text{s}}^{-1}\end{array}$

Calculate the time taken for the temperature of the sphere to rise to $25°\text{C}$$\left(t\right)$.

  $\begin{array}{l}\frac{T\left(t\right)-T_{\infty }}{T_i-T_{\infty }}=e^{-bt}\\ \frac{25°\text{C}-33°\text{C}}{0°\text{C}-33°\text{C}}=e^{-\left(0.0005836{\text{s}}^{-1}\right)t}\end{array}$

  $\begin{array}{c}t=2428\text{s}\\ =\text{40}\text{.5}\text{min}\end{array}$

Thus, the time taken for the temperature of the sphere to rise to $25°\text{C}$ is $\text{40}\text{.5}\text{min}$.

Cube:

Calculate the characteristics length of the cube $\left(L_c\right)$.

  $\begin{array}{c}{L}_c=\frac{V}{A_s}\\ =\frac{L^3}{6L^2}\\ =\frac{L}{6}\end{array}$

  $\begin{array}{c}=\frac{0.05\text{m}}{6}\\ =0.008333\text{m}\end{array}$

Calculate the Biot number $\left(\text{Bi}\right)$.

  $\begin{array}{c}\text{Bi}=\frac{h{L}_c}{k}\\ =\frac{\left(12\text{W}/{\text{m}}^{\text{2}}\cdot \text{K}\right)\left(0.008333\text{m}\right)}{429\text{W}/\text{m}\cdot \text{K}}\\ =0.00023\end{array}$

The Biot number is less than 0.1$\left(0.00023<0.1\right)$. Therefore the lumped system analysis is applicable.

Calculate the value of exponent $\left(b\right)$.

  $\begin{array}{c}b=\frac{h{A}_s}{\rho c_{p}V}\\ =\frac{h}{\rho c_p{L}_c}\\ =\frac{\left(12\text{W}/{\text{m}}^{\text{2}}\cdot \text{K}\right)}{\left(10,500\text{kg}/{\text{m}}^3\right)\left(0.235\text{kJ}/\text{kg}\cdot \text{K}\right)\left(0.008333\text{m}\right)}\\ =0.0005836{\text{s}}^{-1}\end{array}$

Calculate the time taken for the temperature of the cube to rise to $25°\text{C}$$\left(t\right)$.

  $\begin{array}{l}\frac{T\left(t\right)-T_{\infty }}{T_i-T_{\infty }}=e^{-bt}\\ \frac{25°\text{C}-33°\text{C}}{0°\text{C}-33°\text{C}}=e^{-\left(0.0005836{\text{s}}^{-1}\right)t}\end{array}$

  $\begin{array}{c}t=2428\text{s}\\ =\text{40}\text{.5}\text{min}\end{array}$

Thus, the time taken for the temperature of the cube to rise to $25°\text{C}$ is $\text{40}\text{.5}\text{min}$.

Rectangular prism:

Calculate the characteristics length of the rectangular prism $\left(L_c\right)$.

  $\begin{array}{c}{L}_c=\frac{V}{A_s}\\ =\frac{\left(\text{0}\text{.04}\text{m}\times 0.0\text{5}\text{m}\times 0.0\text{6}\text{m}\right)}{2\left(\text{0}\text{.04}\text{m}\times 0.0\text{5}\text{m}\right)+2\left(\text{0}\text{.04}\text{m}\times 0.0\text{6}\text{m}\right)+2\left(\text{0}\text{.05}\text{m}\times 0.0\text{6}\text{m}\right)}\\ =0.008108\text{m}\end{array}$

Calculate the Biot number $\left(\text{Bi}\right)$.

  $\begin{array}{c}\text{Bi}=\frac{h{L}_c}{k}\\ =\frac{\left(12\text{W}/{\text{m}}^{\text{2}}\cdot \text{K}\right)\left(0.008108\text{m}\right)}{429\text{W}/\text{m}\cdot \text{K}}\\ =0.00023\end{array}$

The Biot number is less than 0.1$\left(0.00023<0.1\right)$. Therefore the lumped system analysis is applicable.

Calculate the value of exponent $\left(b\right)$.

  $\begin{array}{c}b=\frac{h{A}_s}{\rho c_{p}V}\\ =\frac{h}{\rho c_p{L}_c}\\ =\frac{\left(12\text{W}/{\text{m}}^{\text{2}}\cdot \text{K}\right)}{\left(10,500\text{kg}/{\text{m}}^3\right)\left(0.235\text{kJ}/\text{kg}\cdot \text{K}\right)\left(0.008108\text{m}\right)}\\ =0.0005998{\text{s}}^{-1}\end{array}$

Calculate the time taken for the temperature of the rectangular prism to rise to $25°\text{C}$$\left(t\right)$.

  $\begin{array}{l}\frac{T\left(t\right)-T_{\infty }}{T_i-T_{\infty }}=e^{-bt}\\ \frac{25°\text{C}-33°\text{C}}{0°\text{C}-33°\text{C}}=e^{-\left(0.0005998{\text{s}}^{-1}\right)t}\end{array}$

  $\begin{array}{c}t=2363\text{s}\\ =\text{39}\text{.4}\text{min}\end{array}$

Thus, the time taken for the temperature of the rectangular prism to rise to $25°\text{C}$ is $\text{39}\text{.4}\text{min}$.