Chapter 18, Problem 61P

To determine

The center temperature of the apples, the surface temperature of the apples and the amount of heat transfer from the apple.

Explanation of Solution

Given:

Diameter of the apples $\left(D\right)$ is $9\text{cm}$.

Initial temperature of the apple $\left(T_i\right)$ is $20°\text{C}$.

Temperature of the freezer $\left(T_{\infty }\right)$ is $-15°\text{C}$.

Time taken to cool the apples $\left(t\right)$ is $1\text{h}$.

Convection heat transfer coefficient $\left(h\right)$ is $8\text{W}/{\text{m}}^{\text{2}}\cdot \text{K}$.

Calculation:

Write the given properties of the apples.

  $\begin{array}{l}\rho =840\text{kg}/{\text{m}}^3\\ c_p=3.81\text{J}/\text{kg}\cdot \text{K}\\ k=0.418\text{W}/\text{m}\cdot \text{K}\\ \alpha =1.3\times {10}^{-7}{\text{m}}^2/\text{s}\end{array}$

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

  $\begin{array}{c}\text{Bi}=\frac{h{r}_0}{k}\\ =\frac{h\left(\frac{D}{2}\right)}{k}\\ =\frac{\left(8\text{W}/{\text{m}}^{\text{2}}\cdot \text{K}\right)\left(0.045\text{m}\right)}{0.418\text{W}/\text{m}\cdot \text{K}}\\ =0.861\end{array}$

Refer Table 18-2, “Coefficients used in the one-term approximate solution of transient one-dimensional heat conduction in plane walls, cylinders, and spheres”, obtain the constants ${\lambda }_1$ and $A_1$ corresponding to the Biot number of $0.861$.

  $\begin{array}{l}{\lambda }_1=1.476\\ A_1=1.2390\end{array}$

Calculate the Fourier number $\left(\tau \right)$.

  $\begin{array}{c}\tau =\frac{\alpha t}{L^2}\\ =\frac{\left(1.3\times {10}^{-7}{\text{m}}^2/\text{s}\right)\left(1\text{h}\right)}{{\left(0.045\text{m}\right)}^2}\\ =\frac{\left(1.3\times {10}^{-7}{\text{m}}^2/\text{s}\right)\left(1\times 3600\text{s}\right)}{{\left(0.045\text{m}\right)}^2}\\ =0.231\end{array}$

The Fourier number is greater than 0.2$\left(0.231>0.2\right)$. Therefore, the one-term approximate solution is applicable.

Calculate the temperature at the center of the apples $\left(T_0\right)$.

  $\begin{array}{l}{\theta }_{\text{0,sph}}=A_1{e}^{-{\lambda }_1^2\tau }\\ \frac{T_0-T_{\infty }}{T_i-T_{\infty }}=A_1{e}^{-{\lambda }_1^2\tau }\end{array}$

  $\begin{array}{c}\frac{T_0-\left(-15°\text{C}\right)}{20°\text{C}-\left(-15°\text{C}\right)}=\left(1.2390\right)e^{-{\left(1.476\right)}^2\left(0.231\right)}\\ =0.749\end{array}$

  $T_0=11.2°\text{C}$

Thus, the center temperature of the apples is $11.2°\text{C}$.

Calculate the temperature at the surface of the apples $\left(T\left(r_0,t\right)\right)$.

  $\begin{array}{l}\theta {\left(r_0,t\right)}_{\text{sph}}=A_1{e}^{-{\lambda }_1^2\tau }\left[\frac{\sin \left(\frac{{\lambda }_1{r}_0}{r_0}\right)}{\left(\frac{{\lambda }_1{r}_0}{r_0}\right)}\right]\\ \frac{T\left(r_0,t\right)-T_{\infty }}{T_i-T_{\infty }}=A_1{e}^{-{\lambda }_1^2\tau }\left[\frac{\sin \left(\frac{{\lambda }_1{r}_0}{r_0}\right)}{\left(\frac{{\lambda }_1{r}_0}{r_0}\right)}\right]\end{array}$

  $\begin{array}{c}\frac{T\left(r_0,t\right)-\left(-15°\text{C}\right)}{20°\text{C}-\left(-15°\text{C}\right)}=\left(1.2390\right)e^{-{\left(1.476\right)}^2\left(0.231\right)}\left[\frac{\sin \left(1.476\right)}{\left(1.476\right)}\right]\\ =0.505\end{array}$

  $T\left(r_0,t\right)=2.7°\text{C}$

Thus, the center temperature of the apples is $2.7°\text{C}$.

Calculate the mass of the apples $\left(m\right)$.

  $\begin{array}{c}m=\rho V\\ =\rho \left(\frac{4}{3}\pi r_0^3\right)\\ =\left(840\text{kg}/{\text{m}}^3\right)\left(\frac{4}{3}\pi {\left(0.045\text{m}\right)}^3\right)\\ =0.3206\text{kg}\end{array}$

Calculate the maximum amount of heat transferred to the apples $\left(Q_{\max }\right)$.

  $\begin{array}{c}{Q}_{\max }=m{c}_p\left(T_i-T_{\infty }\right)\\ =\left(0.3206\text{kg}\right)\left(3.81\text{J}/\text{kg}\cdot \text{K}\right)\left(20°\text{C}-\left(-15°\text{C}\right)\right)\\ =42.8\text{kJ}\end{array}$

Calculate the amount of heat transfer from the apple $\left(Q\right)$.

  $\begin{array}{c}{\left(\frac{Q}{Q_{\max }}\right)}_{\text{cyl}}=1-3\left({\theta }_{\text{0,sph}}\right)\left(\frac{\sin {\lambda }_1-{\lambda }_1\cos {\lambda }_1}{{\lambda }_1^3}\right)\\ =1-3\left(0.749\right)\left(\frac{\sin \left(1.476\right)-\left(1.476\right)\cos \left(1.476\right)}{{\left(1.476\right)}^3}\right)\\ =0.402\end{array}$

  $\begin{array}{c}Q=\left(0.402\right)Q_{\max }\\ =\left(0.402\right)\left(42.8\text{kJ}\right)\\ =17.2\text{kJ}\end{array}$

Thus, the amount of heat transfer from the apple is $17.2\text{kJ}$.