Chapter 18, Problem 91P

(a)

To determine

The center temperature of the cylinder.

Explanation of Solution

Calculation:

The Biot number for the plane wall is,

  $\begin{array}{c}Bi=\frac{hL}{k}\\ =\frac{\left(40\text{W}/{\text{m}}^2\cdot °\text{C}\right)\left(0.075\text{ m}\right)}{110\text{W}/{\text{m}}^2}\\ =0.02727\end{array}$

Hence the constants ${\lambda }_1\text{and }{A}_1$ are $\text{0}\text{.1620 and 1}\text{.0045}$ respectively.

The Biot number for the long cylinder is,

  $\begin{array}{c}Bi=\frac{h{r}_o}{k}\\ =\frac{\left(40\text{W}/{\text{m}}^2\cdot °\text{C}\right)\left(0.04\text{ m}\right)}{110\text{W}/{\text{m}}^2}\\ =0.01455\end{array}$

Hence the constants ${\lambda }_1\text{and }{A}_1$ are $\text{0}\text{.1677 and 1}\text{.0036}$ respectively.

The center temperature is determined using the product solution method as follows:

  $\begin{array}{l}\theta {(}_0=\left[\theta {(0,t)}_{\text{wall }}\right]\left[\theta {(0,t)}_{\text{cyl }}\right]\\ \frac{T(0,0,t)-T_{\infty }}{T_i-T_{\infty }}=\left(A_1{e}^{-{\lambda }_1^2\tau }\right)\left(A_1{e}^{-{\lambda }_1^2\tau }\right)\\ \frac{T(0,0,t)-20}{150-20}=\left[(1.0045)e^{-{(0.1620)}^2\tau }\right]\left[(1.0036)e^{-{(0.1677)}^2\tau }\right]\end{array}$

The Fourier number for the plane wall is,

  $\begin{array}{c}\tau =\frac{\alpha t}{L^2}\\ =\frac{\left(3.39\times {10}^{-5}{\text{m}}^2/\text{s}\right)\left(15\times 60\text{ s}\right)}{{\left(0.075\text{m}\right)}^2}\\ =5.434>0.2\end{array}$

The Fourier number for the cylinder is,

  $\begin{array}{c}\tau =\frac{\alpha t}{r_o{}^2}\\ =\frac{\left(3.39\times {10}^{-5}{\text{m}}^2/\text{s}\right)\left(15\times 60\text{ s}\right)}{{\left(0.04\text{m}\right)}^2}\\ =19.07>0.2\end{array}$

The center temperature is,

  $\begin{array}{l}\frac{T(0,0,15)-20}{150-20}=\left[(1.0045)e^{-{(0.1620)}^2\left(5.424\right)}\right]\left[(1.0036)e^{-{(0.1677)}^2\left(19.07\right)}\right]\\ T(0,0,15)=86.5°\text{C}\end{array}$

Thus, the center temperature of the cylinder is $86.5°\text{C}$.

(b)

To determine

The center temperature of the top surface of the cylinder.

Explanation of Solution

Calculation:

The center temperature of the top surface of the cylinder is determined using the product solution method as follows:

  $\begin{array}{l}\theta {(}_L=\left[\theta {(L,t)}_{\text{wall }}\right]\left[\theta {(0,t)}_{\text{cyl }}\right]\\ \frac{T(L,0,t)-T_{\infty }}{T_i-T_{\infty }}=\left[\left(A_1{e}^{-{\lambda }_1^2\tau }\right)\cos \left(\frac{{\lambda }_{1}L}{L}\right)\right]\left(A_1{e}^{-{\lambda }_1^2\tau }\right)\\ \frac{T(L,0,t)-20}{150-20}=\left[(1.0045)e^{-{(0.1620)}^2\tau }\cos \left(0.162\right)\right]\left[(1.0036)e^{-{(0.1677)}^2\tau }\right]\\ \frac{T(L,0,15)-20}{150-20}=\left[(1.0045)e^{-{(0.1620)}^2\left(5.424\right)}\cos \left(0.162\right)\right]\left[(1.0036)e^{-{(0.1677)}^2\left(19.07\right)}\right]\\ T(L,0,15)=85.6°\text{C}\end{array}$

Thus, the center temperature at the top surface of the cylinder is $85.6°\text{C}$.

(c)

To determine

The total heat transfer from the cylinder after 15 min from the start of cooling.

Explanation of Solution

Calculation:

The maximum heat that can be transferred from the cylinder is,

  $\begin{array}{c}{Q}_{\max }=m{c}_p\left(T_i-T_{\infty }\right)\\ =\rho \text{V}{c}_p\left(T_i-T_{\infty }\right)\\ =\left(8530\text{kg}/{\text{m}}^3\right)\left[\frac{\pi }{4}{\left(0.08\text{ m}\right)}^2\left(0.15\text{ m}\right)\right]\left(0.389\text{kJ}/\text{kg}\cdot °\text{C}\right)\left(150-20\right)°\text{C}\\ =325.2\text{ kJ}\end{array}$

The dimensionless heat transfer ratio for the plane wall is,

  $\begin{array}{c}{\left(\frac{Q}{Q_{\max }}\right)}_{wall}=1-{\theta }_{o,wall}\frac{\sin \left({\lambda }_1\right)}{{\lambda }_1}\\ =1-\left[(1.0045)e^{-{(0.1620)}^2\left(5.424\right)}\right]\frac{\sin (0.1620)}{0.1620}\\ =0.1326\end{array}$

The dimensionless heat transfer ratio for the cylinder is,

  $\begin{array}{c}{\left(\frac{Q}{Q_{\max }}\right)}_{cyl}=1-2{\theta }_{o,cyl}\frac{J_1\left({\lambda }_1\right)}{{\lambda }_1}\\ =1-2\left[(1.0036)e^{-{(0.1677)}^2\left(19.07\right)}\right]\frac{0.08348}{0.1677}\\ =0.4156\end{array}$

The heat transfer rate for the short cylinder is,

  $\begin{array}{c}{\left(\frac{Q}{Q_{\max }}\right)}_{\begin{array}{c}short\\ cylinder\end{array}}={\left(\frac{Q}{Q_{\max }}\right)}_{\begin{array}{l}plane\\ wall\end{array}}+{\left(\frac{Q}{Q_{\max }}\right)}_{\begin{array}{c}long\\ cylinder\end{array}}\left[1-{\left(\frac{Q}{Q_{\max }}\right)}_{\begin{array}{l}plane\\ wall\end{array}}\right]\\ =\text{0}\text{.1326}+\text{(0}\text{.4156)(1}-\text{0}\text{.1326)}\\ =\text{0}\text{.4931}\end{array}$

Calculate the total heat transfer from the short cylinder.

  $\begin{array}{c}Q=\text{0}\text{.4931}{Q}_{\max }\\ =\text{0}\text{.4931}\left(325.2\text{ kJ}\right)\\ =160\text{ kJ}\end{array}$

Thus, the total heat transfer from the cylinder after 15 min from the start of cooling is $160\text{ kJ}$.