Chapter 9, Problem 9.1P

The one-dimensional plane wall of Figure 3.1 is of thickness $\text{L = 75 mm}$ and thermal conductivity $k\text{ = 5 W/m}\cdot \text{K}$ . The fluid temperatures are $T_{\infty ,1}=200°C\text{and}{T}_{\infty ,2}=100°C$ , respectively. Using the minimum and maximum typical values of the convection heat transfer coefficients listed in Table 1.1, determine the minimum and maximum steady-state heat fluxes through the wall for (i) free convection in gases, (ii) free convection in liquids, (iii) forced convection in gases, (iv) forced convection in liquids, and (V) convection with phase change.

To determine

The minimum and maximum heat fluxes through the wall for free convection in gases.

Answer to Problem 9.1P

The minimum heat fluxes through the wall for free convection in gasses is $98.52\text{W}/{\text{m}}^2$ and maximum is $1052.63\text{W}/{\text{m}}^2$ .

Explanation of Solution

Given:

The thickness of the length is $75\text{ mm}$ .

The thermal conductivity is $5\text{ W}/\text{m}\cdot \text{K}$ .

The fluid temperature is $200°\text{C and 100}°\text{C}$ .

Minimum heat transfer coefficient is $2\text{W}/{\text{m}}^2\cdot \text{K}$ .

Maximum heat transfer coefficient is $25\text{W}/{\text{m}}^2\cdot \text{K}$ .

Concept used:

Write the expression for heat flux.

  $q^{″}=\frac{T_{\infty ,1}-T_{\infty ,2}}{\frac{1}{h_1}+\frac{L}{k}+\frac{1}{h_2}}$ ...... (1)

Here, $q^{″}$ is heat transfer coefficient, $h_1\text{ and }{h}_2$ heat transfer coefficient, $T_{\infty ,1}\text{ and }{T}_{\infty ,2}$ are temperatures, L is the length, and k is thermal conductivity.

Calculation:

Substitute $200°\text{C}$ for $T_{\infty ,1}$ , $75\text{ mm}$ for L, $h_1$for $h_2$ and $100°\text{C}$ for $T_{\infty ,2}$ in equation (1)

  $\begin{array}{c}{q}^{″}=\frac{200-100}{\frac{1}{h_1}+\frac{\left(75\text{ mm}\left(\frac{1\text{ m}}{1000\text{ mm}}\right)\right)}{5}+\frac{1}{h_1}}\\ q^{″}=\frac{100h_1}{2+0.015h_1}\end{array}$

Therefore, the expression for heat flux is shown below.

  $q^{″}=\frac{100h_1}{2+0.015h_1}$ ....... (2)

The minimum heat flux is obtained below.

Substitute $2\text{W}/{\text{m}}^2\cdot \text{K}$ for $h_1$ , in equation (2)

  $\begin{array}{c}{q}^{″}=\frac{100\left(2\right)}{2+0.015\left(2\right)}\\ =98.52\text{W}/{\text{m}}^2\end{array}$

The maximum heat flux is obtained below.

Substitute $25\text{W}/{\text{m}}^2\cdot \text{K}$ for $h_1$ , in equation (2)

  $\begin{array}{c}{q}^{″}=\frac{100\left(25\right)}{2+0.015\left(25\right)}\\ =1052.63\text{W}/{\text{m}}^2\end{array}$

Conclusion:

Thus, the minimum heat fluxes through the wall for free convection in gases is $98.52\text{W}/{\text{m}}^2$ and maximum is $1052.63\text{W}/{\text{m}}^2$ .

To determine

The minimum and maximum heat fluxes through the wall for free convection in liquids.

Answer to Problem 9.1P

The minimum heat fluxes through the wall for free convection in liquids is $1818.2\text{W}/{\text{m}}^2$ and maximum is $5882.35\text{W}/{\text{m}}^2$ .

Explanation of Solution

Given:

Minimum heat transfer coefficient is $50\text{W}/{\text{m}}^2\cdot \text{K}$ .

Maximum heat transfer coefficient is $1000\text{W}/{\text{m}}^2\cdot \text{K}$ .

Calculation:

Substitute $50\text{W}/{\text{m}}^2\cdot \text{K}$ for $h_1$ , in equation (2)

  $\begin{array}{c}{q}^{″}=\frac{100\left(50\right)}{2+0.015\left(50\right)}\\ =1818.2\text{W}/{\text{m}}^2\end{array}$

The maximum heat flux is obtained below.

Substitute $1000\text{W}/{\text{m}}^2\cdot \text{K}$ for $h_1$ , in equation (2)

  $\begin{array}{c}{q}^{″}=\frac{100\left(1000\right)}{2+0.015\left(1000\right)}\\ =5882.35\text{W}/{\text{m}}^2\end{array}$

Conclusion:

Thus, the minimum heat fluxes through the wall for free convection in liquids is $1818.2\text{W}/{\text{m}}^2$ and maximum is $5882.35\text{W}/{\text{m}}^2$ .

To determine

The minimum and maximum heat fluxes through the wall for forced convection in gases.

Answer to Problem 9.1P

The minimum heat fluxes through the wall for forced convection in gases is $1052.63\text{W}/{\text{m}}^2$ and maximum is $4347.63\text{W}/{\text{m}}^2$ .

Explanation of Solution

Given:

Minimum heat transfer coefficient is $25\text{W}/{\text{m}}^2\cdot \text{K}$ .

Maximum heat transfer coefficient is $250\text{W}/{\text{m}}^2\cdot \text{K}$ .

Calculation:

Substitute $25\text{W}/{\text{m}}^2\cdot \text{K}$ for $h_1$ , in equation (2)

  $\begin{array}{c}{q}^{″}=\frac{100\left(25\right)}{2+0.015\left(25\right)}\\ =1052.63\text{W}/{\text{m}}^2\end{array}$

The maximum heat flux is obtained below.

Substitute $250\text{W}/{\text{m}}^2\cdot \text{K}$ for $h_1$ , in equation (2)

  $\begin{array}{c}{q}^{″}=\frac{100\left(250\right)}{2+0.015\left(250\right)}\\ =4347.63\text{W}/{\text{m}}^2\end{array}$

Conclusion:

Thus, the minimum heat fluxes through the wall for forced convection in gases is $1052.63\text{W}/{\text{m}}^2$ and maximum is $4347.63\text{W}/{\text{m}}^2$ .

To determine

The minimum and maximum heat fluxes through the wall for forced convection in liquids.

Answer to Problem 9.1P

The minimum heat fluxes through the wall for forced convection in liquids is $2857.14\text{W}/{\text{m}}^2$ and maximum is $6622.52\text{W}/{\text{m}}^2$ .

Explanation of Solution

Given:

Minimum heat transfer coefficient is $100\text{W}/{\text{m}}^2\cdot \text{K}$ .

Maximum heat transfer coefficient is $2\times {10}^4\text{W}/{\text{m}}^2\cdot \text{K}$ .

Calculation:

Substitute $100\text{W}/{\text{m}}^2\cdot \text{K}$ for $h_1$ , in equation (2)

  $\begin{array}{c}{q}^{″}=\frac{100\left(100\right)}{2+0.015\left(100\right)}\\ =2857.14\text{W}/{\text{m}}^2\end{array}$

The maximum heat flux is obtained below.

Substitute $2\times {10}^4\text{W}/{\text{m}}^2\cdot \text{K}$ for $h_1$ , in equation (2)

  $\begin{array}{c}{q}^{″}=\frac{100\left(2\times {10}^4\right)}{2+0.015\left(2\times {10}^4\right)}\\ =6622.52\text{W}/{\text{m}}^2\end{array}$

Conclusion:

Thus, the minimum heat fluxes through the wall for forced convection in liquids is $2857.14\text{W}/{\text{m}}^2$ and maximum is $6622.52\text{W}/{\text{m}}^2$ .

To determine

The minimum and maximum heat fluxes through the wall for convection with phase change.

Answer to Problem 9.1P

The minimum heat fluxes through the wall for convection with phase change is $6329.11\text{W}/{\text{m}}^2$ and maximum is $6657.8\text{W}/{\text{m}}^2$ .

Explanation of Solution

Given:

Minimum heat transfer coefficient is $2500\text{W}/{\text{m}}^2\cdot \text{K}$ .

Maximum heat transfer coefficient is $1\times {10}^5\text{W}/{\text{m}}^2\cdot \text{K}$ .

Calculation:

Substitute $2500\text{W}/{\text{m}}^2\cdot \text{K}$ for $h_1$ , in equation (2)

  $\begin{array}{c}{q}^{″}=\frac{100\left(2500\right)}{2+0.015\left(2500\right)}\\ =6329.11\text{W}/{\text{m}}^2\end{array}$

The maximum heat flux is obtained below.

Substitute $1\times {10}^5\text{W}/{\text{m}}^2\cdot \text{K}$ for $h_1$ , in equation (2)

  $\begin{array}{c}{q}^{″}=\frac{100\left(1\times {10}^5\right)}{2+0.015\left(1\times {10}^5\right)}\\ =6657.8\text{W}/{\text{m}}^2\end{array}$

Conclusion:

Thus, the minimum heat fluxes through the wall for convection with phase change is $6329.11\text{W}/{\text{m}}^2$ and maximum is $6657.8\text{W}/{\text{m}}^2$ .