Chapter 18, Problem 88P

To determine

The center temperature after 10, 20 and 60 min for each geometry.

Explanation of Solution

Calculation:

For cubic block:

The Biot number corresponding to $h=40\text{W}/{\text{m}}^2\cdot °\text{C}$ is,

  $\begin{array}{c}Bi=\frac{hL}{k}\\ =\frac{\left(40\text{W}/{\text{m}}^2\cdot °\text{C}\right)\left(0.025\text{m}\right)}{2.5\text{W}/{\text{m}}^2}\\ =0.400\end{array}$

Hence the constants ${\lambda }_1\text{and }{A}_1$ are $0.5932\text{ and 1}\text{.0580}$ respectively.

The Biot number corresponding to $h=80\text{W}/{\text{m}}^2\cdot °\text{C}$ is,

  $\begin{array}{c}Bi=\frac{hL}{k}\\ =\frac{\left(80\text{W}/{\text{m}}^2\cdot °\text{C}\right)\left(0.025\text{m}\right)}{2.5\text{W}/{\text{m}}^2}\\ =0.800\end{array}$

Hence the constants ${\lambda }_1\text{and }{A}_1$ are $0.7910\text{ and 1}\text{.1016}$ respectively.

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

  $\begin{array}{l}\theta {(}_0={\left[\theta {(0,t)}_{\text{wall }}\right]}^2\left[\theta {(0,t)}_{\text{wall }}\right]\\ \frac{T(0,0,0,t)-T_{\infty }}{T_i-T_{\infty }}={\left(A_1{e}^{-{\lambda }_1^2\tau }\right)}^2\left(A_1{e}^{-{\lambda }_1^2\tau }\right)\\ \frac{T(0,0,0,t)-500}{20-500}={\left[(1.0580)e^{-{(0.5932)}^2\tau }\right]}^2\left[(1.1016)e^{-{(0.7910)}^2\tau }\right]\end{array}$

After 10 min:

The Fourier number is,

  $\begin{array}{c}\tau =\frac{\alpha t}{L^2}\\ =\frac{\left(1.15\times {10}^{-6}{\text{m}}^2/\text{s}\right)\left(10\times 60\text{ s}\right)}{{\left(0.025\text{ m}\right)}^2}\\ =1.104>0.2\end{array}$

The center temperature is,

  $\begin{array}{l}\frac{T(0,0,0,10)-500}{20-500}={\left[(1.0580)e^{-{(0.5932)}^2\left(1.104\right)}\right]}^2\left[(1.1016)e^{-{(0.7910)}^2\left(1.104\right)}\right]\\ T(0,0,0,10)=364°\text{C}\end{array}$

After 20 min:

The Fourier number is,

  $\begin{array}{c}\tau =\frac{\alpha t}{L^2}\\ =\frac{\left(1.15\times {10}^{-6}{\text{m}}^2/\text{s}\right)\left(20\times 60\text{ s}\right)}{{\left(0.025\text{ m}\right)}^2}\\ =2.208>0.2\end{array}$

The center temperature is,

  $\begin{array}{l}\frac{T(0,0,0,20)-500}{20-500}={\left[(1.0580)e^{-{(0.5932)}^2\left(2.208\right)}\right]}^2\left[(1.1016)e^{-{(0.7910)}^2\left(2.208\right)}\right]\\ T(0,0,0,20)=469°\text{C}\end{array}$

After 60 min:

The Fourier number is,

  $\begin{array}{c}\tau =\frac{\alpha t}{L^2}\\ =\frac{\left(1.15\times {10}^{-6}{\text{m}}^2/\text{s}\right)\left(60\times 60\text{ s}\right)}{{\left(0.025\text{ m}\right)}^2}\\ =6.624>0.2\end{array}$

The center temperature is,

  $\begin{array}{l}\frac{T(0,0,0,60)-500}{20-500}={\left[(1.0580)e^{-{(0.5932)}^2\left(6.624\right)}\right]}^2\left[(1.1016)e^{-{(0.7910)}^2\left(6.624\right)}\right]\\ T(0,0,0,60)=500°\text{C}\end{array}$

Thus, the center temperature after 10, 20 and 60 min for cubic block is $364°\text{C}$, $\text{469}°\text{C}$ and $500°\text{C}$ respectively.

For cylinder:

The Biot number is,

  $\begin{array}{c}Bi=\frac{h{r}_o}{k}\\ =\frac{\left(40\text{W}/{\text{m}}^2\cdot °\text{C}\right)\left(0.025\text{m}\right)}{2.5\text{W}/{\text{m}}^2}\\ =0.400\end{array}$

Hence the constants ${\lambda }_1\text{and }{A}_1$ are $0.8516\text{ and 1}\text{.0931}$ respectively.

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

  $\begin{array}{l}\theta {(}_0=\left[\theta {(0,t)}_{\text{wall }}\right]\left[\theta {(0,t)}_{\text{cyl }}\right]\\ \frac{T(0,0,t)-T_{\infty }}{T_i-T_{\infty }}={\left(A_1{e}^{-{\lambda }_1^2\tau }\right)}_{\text{wall}}\left(A_1{e}^{-{\lambda }_1^2\tau }\right)\\ \frac{T(0,0,t)-500}{20-500}=\left[(1.1016)e^{-{(0.7910)}^2\tau }\right]\left[(1.0931)e^{-{(0.8516)}^2\tau }\right]\end{array}$

After 10 min:

The Fourier number is,

  $\begin{array}{c}\tau =\frac{\alpha t}{L^2}\\ =\frac{\left(1.15\times {10}^{-6}{\text{m}}^2/\text{s}\right)\left(10\times 60\text{ s}\right)}{{\left(0.025\text{ m}\right)}^2}\\ =1.104>0.2\end{array}$

The center temperature is,

  $\begin{array}{l}\frac{T(0,0,t)-500}{20-500}=\left[(1.1016)e^{-{(0.7910)}^2\left(1.104\right)}\right]\left[(1.0931)e^{-{(0.8516)}^2\left(1.104\right)}\right]\\ T(0,0,10)=370°\text{C}\end{array}$

After 20 min:

The Fourier number is,

  $\begin{array}{c}\tau =\frac{\alpha t}{L^2}\\ =\frac{\left(1.15\times {10}^{-6}{\text{m}}^2/\text{s}\right)\left(20\times 60\text{ s}\right)}{{\left(0.025\text{ m}\right)}^2}\\ =2.208>0.2\end{array}$

The center temperature is,

  $\begin{array}{l}\frac{T(0,0,20)-500}{20-500}=\left[(1.1016)e^{-{(0.7910)}^2\left(2.208\right)}\right]\left[(1.0931)e^{-{(0.8516)}^2\left(2.208\right)}\right]\\ T(0,0,20)=471°\text{C}\end{array}$

After 60 min:

The Fourier number is,

  $\begin{array}{c}\tau =\frac{\alpha t}{L^2}\\ =\frac{\left(1.15\times {10}^{-6}{\text{m}}^2/\text{s}\right)\left(60\times 60\text{ s}\right)}{{\left(0.025\text{ m}\right)}^2}\\ =6.624>0.2\end{array}$

The center temperature is,

  $\begin{array}{l}\frac{T(0,0,60)-500}{20-500}=\left[(1.1016)e^{-{(0.7910)}^2\left(6.624\right)}\right]\left[(1.0931)e^{-{(0.8516)}^2\left(6.624\right)}\right]\\ T(0,0,60)=500°\text{C}\end{array}$

Thus, the center temperature after 10, 20 and 60 min for cylinder is $370°\text{C}$, $\text{471}°\text{C}$ and $500°\text{C}$ respectively.