Chapter 18, Problem 64P

a)

To determine

The heat transfer coefficient at the surface of the rib.

Explanation of Solution

Given:

Mass of the rib $\left(m\right)$ is $3.2\text{kg}$.

Initial temperature $\left(T_i\right)$ is $4.5°\text{C}$.

Temperature maintained at the oven $\left(T_{\infty }\right)$ is $163°\text{C}$.

Temperature at the center of the meat $\left(T_0\right)$ is $60°\text{C}$.

Time taken to roast the rib $\left(t\right)$ is $2\text{h 45}\text{min}$.

Calculation:

Write the given properties of the rib.

  $\begin{array}{l}\rho =1200\text{kg}/{\text{m}}^3\\ c_p=4.1\text{J}/\text{kg}\cdot \text{K}\\ k=0.45\text{W}/\text{m}\cdot \text{K}\\ \alpha =0.914\times {10}^{-7}{\text{m}}^2/\text{s}\end{array}$

Calculate the radius of the roast $\left(r_0\right)$.

  $\begin{array}{l}m=\rho V\\ V=\frac{m}{\rho }\\ \frac{4}{3}\pi r_0^3=\frac{m}{\rho }\end{array}$

  $\begin{array}{c}{r}_0={\left(\frac{3m}{4\pi \rho }\right)}^{\frac{1}{3}}\\ ={\left(\frac{3\left(3.2\text{kg}\right)}{4\pi \left(1200\text{kg}/{\text{m}}^3\right)}\right)}^{\frac{1}{3}}\\ =0.08603\text{m}\end{array}$

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

  $\begin{array}{c}\tau =\frac{\alpha t}{r_0^2}\\ =\frac{\left(0.91\times {10}^{-7}{\text{m}}^2/\text{s}\right)\left(2\text{h 45}\text{min}\right)}{{\left(0.08603\text{m}\right)}^2}\\ =\frac{\left(0.91\times {10}^{-7}{\text{m}}^2/\text{s}\right)\left(2\times 3600\text{s}+\text{45}\times 60\text{s}\right)}{{\left(0.08603\text{m}\right)}^2}\\ =0.1217\end{array}$

The Fourier number is nearly close to 0.2$\left(0.1217\approx 0.2\right)$. Therefore, the one-term approximate solution can be applicable.

Calculate the dimensionless temperature of the roast $\left({\theta }_{0,\text{sph}}\right)$.

  $\begin{array}{l}{\theta }_{0,\text{sph}}=A_1{e}^{-{\lambda }_1^2\tau }\\ \frac{T_0-T_{\infty }}{T_i-T_{\infty }}=A_1{e}^{-{\lambda }_1^2\tau }\\ \frac{60°\text{C}-163°\text{C}}{4.5°\text{C}-163°\text{C}}=A_1{e}^{-{\lambda }_1^2\left(0.1217\right)}\end{array}$        (I)

Solve Equation (I) by trial and error method using Table 18-2, Coefficients used in the one-term approximate solution of transient one-dimensional heat conduction in plane walls, cylinders, and spheres”.

Equation (I) is satisfied when the Biot number, $\text{Bi}=30$. It corresponds to the constants values given as follows:

  $\begin{array}{l}{\lambda }_1=3.0372\\ A_1=1.9898\end{array}$

Calculate the heat transfer coefficient at the surface of the rib $\left(h\right)$.

  $\text{Bi}=\frac{h{r}_0}{k}$

  $\begin{array}{c}h=\frac{k\left(\text{Bi}\right)}{r_0}\\ =\frac{\left(0.45\text{W}/\text{m}\cdot \text{K}\right)\left(30\right)}{\left(0.08603\text{m}\right)}\\ =156.9\text{W}/{\text{m}}^{\text{2}}\cdot \text{K}\end{array}$

Thus, the heat transfer coefficient at the surface of the rib is $156.9\text{W}/{\text{m}}^{\text{2}}\cdot \text{K}$.

b)

To determine

The temperature at the surface of the rib.

Explanation of Solution

Calculation:

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

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

  $\begin{array}{c}\frac{T\left(r_0,t\right)-163°\text{C}}{4.5°\text{C}-163°\text{C}}=\left(1.9898\right)e^{-{\left(3.0372\right)}^2\left(0.1217\right)}\left[\frac{\sin \left(3.0372\right)}{\left(3.0372\right)}\right]\\ =0.0222\end{array}$

  $T\left(r_0,t\right)=159.5°\text{C}$

Thus, the temperature at the surface of the rib is $159.5°\text{C}$.

c)

To determine

The amount of heat transferred to the rib.

Explanation of Solution

Calculation:

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

  $\begin{array}{c}{Q}_{\max }=m{c}_p\left(T_i-T_{\infty }\right)\\ =\left(3.2\text{kg}\right)\left(4.1\text{J}/\text{kg}\cdot \text{K}\right)\left(163°\text{C}-4.5°\text{C}\right)\\ =2080\text{kJ}\end{array}$

Calculate the amount of heat transferred to the rib $\left(Q\right)$.

  $\begin{array}{c}{\left(\frac{Q}{Q_{\max }}\right)}_{\text{cyl}}=1-3\left({\theta }_{\text{0,sph}}\right)\left(\frac{\sin {\lambda }_1-{\lambda }_1\cos {\lambda }_1}{{\lambda }_1^3}\right)\\ =1-3\left(0.65\right)\left(\frac{\sin \left(3.0372\right)-\left(3.0372\right)\cos \left(3.0372\right)}{{\left(3.0372\right)}^3}\right)\\ =0.783\end{array}$

  $\begin{array}{c}Q=\left(0.783\right)Q_{\max }\\ =\left(0.783\right)\left(2080\text{kJ}\right)\\ =1629\text{kJ}\end{array}$

Thus, the amount of heat transferred to the rib is $1629\text{kJ}$.

d)

To determine

The time taken to cook the medium-done rib.

Explanation of Solution

Calculation:

It is given that the innermost temperature of the rib is $71°\text{C}$.

  $T_0=71°\text{C}$

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

  $\begin{array}{l}{\theta }_{0,\text{sph}}=A_1{e}^{-{\lambda }_1^2\tau }\\ \frac{T_0-T_{\infty }}{T_i-T_{\infty }}=A_1{e}^{-{\lambda }_1^2\tau }\\ \frac{71°\text{C}-163°\text{C}}{4.5°\text{C}-163°\text{C}}=\left(1.9898\right)e^{-{\left(3.0372\right)}^2\tau }\\ \tau =0.1336\end{array}$

Calculate the time taken to cook the medium-done rib $\left(t\right)$.

  $\begin{array}{c}t=\frac{\tau r_0^2}{\alpha }\\ =\frac{\left(0.1336\right){\left(0.08603\text{m}\right)}^2}{0.91\times {10}^{-7}{\text{m}}^2/\text{s}}\\ =10,866\text{s}\end{array}$

  $\begin{array}{c}=181\text{min}\\ \simeq \text{3}\text{hr}\end{array}$

Thus, the time taken to cook the medium-done rib is $\text{3}\text{hr}$.

The calculated cooking time $\left(\text{3}\text{hr}\right)$ is nearer to the given time of $\text{3}\text{hr}20\text{min}$. The difference between the two values is because of the Fourier number being less than 0.2 and therefore, the error in the one-term approximation method.