Chapter 18, Problem 101RQ

a)

To determine

The average heat transfer coefficient at the surface of the turkey.

Explanation of Solution

Given:

Mass of the stuffed turkey $\left(m\right)$ is $14\text{lbm}$.

Initial temperature $\left(T_i\right)$ is $40°\text{F}$.

Temperature maintained at the oven $\left(T_{\infty }\right)$ is $325°\text{F}$.

Temperature at the meat $\left(T\left(x,t\right)\right)$ is $185°\text{F}$.

Time taken to roast the turkey $\left(t\right)$ is $5\text{h}$.

Calculation:

Write the given properties of the turkey.

  $\begin{array}{l}\rho =75\text{lbm}/{\text{ft}}^3\\ c_p=0.98\text{Btu}/\text{lbm}\cdot °\text{F}\\ k=0.26\text{Btu}/\text{h}\cdot \text{ft}\cdot °\text{F}\\ \alpha =0.0035{\text{ft}}^2/\text{s}\end{array}$

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

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

  $\begin{array}{c}{r}_0={\left(\frac{3m}{4\pi \rho }\right)}^{\frac{1}{3}}\\ ={\left(\frac{3\left(14\text{lbm}\right)}{4\pi \left(75\text{lbm}/{\text{ft}}^3\right)}\right)}^{\frac{1}{3}}\\ =0.3545\text{ft}\end{array}$

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

  $\begin{array}{c}\tau =\frac{\alpha t}{r_0^2}\\ =\frac{\left(0.0035{\text{ft}}^2/\text{s}\right)\left(5\text{h}\right)}{{\left(0.3545\text{ft}\right)}^2}\\ =\frac{\left(0.0035{\text{ft}}^2/\text{s}\right)\left(5\times 3600\text{s}\right)}{{\left(0.3545\text{ft}\right)}^2}\\ =0.1392\end{array}$

The Fourier number is nearly close to 0.2$\left(0.1392\approx 0.2\right)$. Therefore, the one-term approximate solution can be applicable.

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

  $\begin{array}{l}\theta {\left(x,t\right)}_{\text{sph}}=A_1{e}^{-{\lambda }_1^2\tau }\frac{\sin \left(\frac{{\lambda }_{1}r}{r_0}\right)}{\frac{{\lambda }_{1}r}{r_0}}\\ \frac{T\left(x,t\right)-T_{\infty }}{T_i-T_{\infty }}=A_1{e}^{-{\lambda }_1^2\tau }\frac{\sin \left(\frac{{\lambda }_{1}r}{r_0}\right)}{\frac{{\lambda }_{1}r}{r_0}}\\ \frac{185°\text{F}-325°\text{F}}{40°\text{F}-325°\text{F}}=A_1{e}^{-{\lambda }_1^2\left(0.1392\right)}\frac{\sin \left(0.333{\lambda }_1\right)}{0.333{\lambda }_1}\\ 0.491=A_1{e}^{-{\lambda }_1^2\left(0.1392\right)}\frac{\sin \left(0.333{\lambda }_1\right)}{0.333{\lambda }_1}\end{array}$        (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, $\text{Bi}=20$. It corresponds to the constants values given as follows:

  $\begin{array}{l}{\lambda }_1=2.9857\\ A_1=1.9781\end{array}$

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

  $\text{Bi}=\frac{h{r}_0}{k}$

  $\begin{array}{c}h=\frac{k\left(\text{Bi}\right)}{r_0}\\ =\frac{\left(0.26\text{Btu}/\text{h}\cdot \text{ft}\cdot °\text{F}\right)\left(20\right)}{\left(0.3545\text{ft}\right)}\\ =14.7\text{Btu}/\text{h}\cdot {\text{ft}}^2\cdot °\text{F}\end{array}$

Thus, the average heat transfer coefficient at the surface of the turkey is $14.7\text{Btu}/\text{h}\cdot {\text{ft}}^2\cdot °\text{F}$.

b)

To determine

The temperature of the skin of the turkey.

Explanation of Solution

Calculation:

Calculate the temperature of the skin of the turkey $\left(T\left(r_0,t\right)\right)$.

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

  $\begin{array}{c}\frac{T\left(r_0,t\right)-325°\text{F}}{40°\text{F}-325°\text{F}}=\left(1.9781\right)e^{-{\left(2.9857\right)}^2\left(0.1392\right)}\left[\frac{\sin \left(2.9857\right)}{\left(2.9857\right)}\right]\\ =0.02953\end{array}$

  $T\left(r_0,t\right)=317°\text{F}$

Thus, the temperature of the skin of the turkey is $317°\text{F}$.

c)

To determine

The amount of heat transferred to the turkey in the oven.

Explanation of Solution

Calculation:

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

  $\begin{array}{c}{Q}_{\max }=m{c}_p\left(T_{\infty }-T_i\right)\\ =\left(14\text{lbm}\right)\left(0.98\text{Btu}/\text{lbm}\cdot °\text{F}\right)\left(325°\text{F}-40°\text{F}\right)\\ =3910\text{Btu}\end{array}$

Calculate the amount of heat transferred to the turkey in the oven $\left(Q\right)$.

  $\begin{array}{c}{\left(\frac{Q}{Q_{\max }}\right)}_{\text{cyl}}=1-3\left({\theta }_{\text{0,sph}}\right)\left(\frac{\sin {\lambda }_1-{\lambda }_1\cos {\lambda }_1}{{\lambda }_1^3}\right)\\ =1-3\left(0.495\right)\left(\frac{\sin \left(2.9857\right)-\left(2.9857\right)\cos \left(2.9857\right)}{{\left(2.9857\right)}^3}\right)\\ =0.828\end{array}$

  $\begin{array}{c}Q=\left(0.828\right)Q_{\max }\\ =\left(0.828\right)\left(3910\text{Btu}\right)\\ =3240\text{Btu}\end{array}$

Thus, the amount of heat transferred to the rib is $3240\text{Btu}$.

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 $\text{3}\text{hr}$.