Chapter 17, Problem 131RQ

(a)

To determine

The surface temperature on the two sides of the circuit board.

Explanation of Solution

Given:

Thermal conductivity $\left(k\right)$ is $\text{12 }\text{W}/\text{m}\cdot \text{K}$.

Heat transfer coefficient $\left(h\right)$ is $\text{45 }\text{W}/{\text{m}}^2\cdot \text{K}$.

Thickness of the circuit board $\left(t\right)$ is 0.2 cm.

Height of the circuit board $\left(H\right)$ is 10 cm.

Length of the circuit board $\left(L\right)$ is 15 cm.

Calculation:

Determine the thermal resistance of the board.

  $\begin{array}{c}{R}_{\text{board }}=\frac{L}{kA}\\ =\frac{0.002\text{ m}}{\left(12\text{W}/\text{m}\cdot °\text{C}\right)\left(0.1\text{ m}\right)\left(0.15\text{ m}\right)}\\ =0.011°\text{C}/\text{W}\end{array}$

Determine the thermal resistance of the convection.

  $\begin{array}{c}{R}_{\text{conv}}=\frac{1}{hA}\\ =\frac{1}{\left(45\text{W}/\text{m}\cdot °\text{C}\right)\left(0.1\text{ m}\right)\left(0.15\text{ m}\right)}\\ =1.481°\text{C}/\text{W}\end{array}$

Determine the total thermal resistance.

  $\begin{array}{c}{R}_{\text{total }}=R_{\text{board }}+R_{\text{conv }}\\ =0.011°\text{C}/\text{W}+1.481°\text{C}/\text{W}\\ =1.492°\text{C}/\text{W}\end{array}$

Determine the surface temperature on one side.

  $\begin{array}{c}\dot{Q}=\frac{T_1-T_{\infty }}{R_{\text{total }}}\\ T_1=T_{\infty }+\dot{Q}{R}_{\text{total }}\\ =37°\text{C}+(15\text{ W})\left(1.492°\text{C}/\text{W}\right)\\ =59.4°\text{C}\end{array}$

Thus, the surface temperature on one side of the circuit board is $\underset{\_}{59.4°\text{C}}$.

Determine the surface temperature on the other side.

  $\begin{array}{c}\dot{Q}=\frac{T_1-T_2}{R_{\text{board }}}\\ T_2=T_1-\dot{Q}{R}_{\text{bourd }}\\ =59.4°\text{C}-(15\text{ W})\left(0.011°\text{C}/\text{W}\right)\\ =59.2°\text{C}\end{array}$

Thus, the surface temperature on the other side of the circuit board is $\underset{\_}{59.2°\text{C}}$.

(b)

To determine

The new temperature on the two sides of the circuit board.

Explanation of Solution

Given:

Thermal conductivity of the aluminum plate $\left(k\right)$ is $\text{237 }\text{W}/\text{m}\cdot \text{K}$.

Thickness of the aluminum plate $\left(t\right)$ is 0.1 cm.

Height of the aluminum plate $\left(H\right)$ is 10 cm.

Length of the aluminum plate $\left(L\right)$ is 15 cm.

Thermal conductivity of epoxy adhesive $\left(k\right)$ is $\text{1}\text{.8 }\text{W}/\text{m}\cdot \text{K}$.

Calculation:

Determine the value of m.

  $\begin{array}{c}m=\sqrt{\frac{hp}{k{A}_c}}\\ \cong \sqrt{\frac{h(2w)}{k(tv)}}\\ =\sqrt{\frac{2h}{kt}}\\ =\sqrt{\frac{2\left(45\text{W}/{\text{m}}^2\cdot °\text{C}\right)}{\left(237\text{W}/\text{m}\cdot °\text{C}\right)\left(0.002\text{ m}\right)}}\end{array}$

  $m=13.78{\text{m}}^{-1}$

Determine the efficiency of fin.

  $\begin{array}{c}{\eta }_{\text{fin }}=\frac{\tanh mL}{mL}\\ =\frac{\tanh \left(13.78{\text{ m}}^{-1}\times 0.02\text{ m}\right)}{13.78{\text{ m}}^{-1}\times 0.02\text{ m}}\\ =0.975\end{array}$

Determine the finned surface area.

  $\begin{array}{c}{A}_{\text{finned }}=\left(20\right)2w\left(L+\frac{t}{2}\right)\\ =20\times 2\left(0.15\text{ m}\right)\left(0.02\text{ m}+\frac{0.002\text{ m}}{2}\right)\\ =0.126{\text{ m}}^2\end{array}$

Determine the unfinned surface area.

  $\begin{array}{c}{A}_{\text{unfinned }}=\left(0.1\text{ m}\right)\left(0.15\text{ m}\right)-20\left(0.002\text{ m}\right)\left(0.15\text{ m}\right)\\ =0.0090{\text{ m}}^2\end{array}$

Determine the expression of finned heat transfer.

  $\begin{array}{c}{\dot{Q}}_{\text{finned }}={\eta }_{\text{fin},\text{ max}}\\ ={\eta }_{\text{fin }}h{A}_{\text{fin }}\left(T_{\text{base }}-T_{\infty }\right)\end{array}$

Determine the expression of unfinned heat transfer.

  ${\dot{Q}}_{\text{unfinned }}=h{A}_{\text{unfinned }}\left(T_{\text{base }}-T_{\infty }\right)$

Determine the expression of total heat transfer.

  ${\dot{Q}}_{\text{total }}={\dot{Q}}_{\text{unfinned }}+{\dot{Q}}_{\text{finned }}=h\left(T_{\text{base }}-T_{\infty }\right)\left({\eta }_{\text{fin }}{A}_{\text{fin }}+A_{\text{unfinned }}\right)$

Determine the base temperature of the finned surface.

  $\begin{array}{c}{T}_{\text{base}}=T_{\infty }+\frac{{\dot{Q}}_{\text{total}}}{h\left({\eta }_{\text{fin}}{A}_{\text{fin}}+A_{\text{unfinmed}}\right)}\\ =37°\text{C}+\frac{15\text{ W}}{\left(45\text{W}/{\text{m}}^2\cdot °\text{C}\right)\left[\left(0.975\right)\left(0.126{\text{ m}}^2\right)+\left(0.0090{\text{ m}}^2\right)\right]}\\ =39.5°\text{C}\end{array}$

Determine the thermal resistance of aluminum surface.

  $\begin{array}{c}{R}_{\text{aluminum }}=\frac{L}{kA}\\ =\frac{0.001\text{ m}}{\left(237\text{W}/\text{m}\cdot °\text{C}\right)\left(0.1\text{ m}\right)\left(0.15\text{ m}\right)}\\ =0.00028°\text{C}/\text{W}\end{array}$

Determine the thermal resistance of epoxy surface.

  $\begin{array}{c}{R}_{\text{epoxy }}=\frac{L}{kA}\\ =\frac{0.0003\text{ m}}{\left(1.8\text{W}/\text{m}\cdot °\text{C}\right)\left(0.1\text{ m}\right)\left(0.15\text{ m}\right)}\\ =0.01111°\text{C}/\text{W}\end{array}$

Determine the new temperature on one side of the circuit board.

  $\begin{array}{l}\dot{Q}=\frac{T_1-T_{\text{base}}}{R_{\text{aluminum }}+R_{\text{epoxy }}+R_{\text{board }}}\\ 15\text{ W}=\frac{\left(T_1-39.5°\text{C}\right)}{0.00028°\text{C}/\text{W}+0.01111°\text{C}/\text{W}+0.011°\text{C}/\text{W}}\\ T_1=39.8°\text{C}\end{array}$

Thus, the new temperature on one side of the circuit board is $\underset{\_}{39.8°\text{C}}$.

Determine the new temperature on the other side of the circuit board.

  $\begin{array}{l}\dot{Q}=\frac{T_1-T_2}{R_{\text{board }}}\\ T_2=T_1-\dot{Q}{R}_{\text{board }}\\ T_2=39.8°\text{C}-\left(15\text{ W}\right)\left(0.011°\text{C}/\text{W}\right)\\ T_2=39.6°\text{C}\end{array}$

Thus, the new temperature on the other side of the circuit board is $\underset{\_}{39.6°\text{C}}$.