Chapter 17, Problem 113P

(a)

To determine

The temperature on the two sides of the circuit board.

Explanation of Solution

Given:

Thickness of the board is 0.3 cm.

The height of the board is 12 cm.

The length of the board is 18 cm.

Number of logic chips is 80.

Heat dissipated by each chip is $0.04\text{}\text{W}$.

The thermal conductivity of the board is $30\text{W}/\text{m}\cdot \text{K}$.

Temperature of the medium is $40°\text{C}$.

The heat transfer coefficient is $40\text{W}/{\text{m}}^2\cdot \text{K}$.

Thickness of the aluminum plate is 0.2 cm.

The height of the aluminum plate is 12 cm.

The length of the aluminum plate is 18 cm.

The thermal conductivity of the aluminum plate is $237\text{W}/\text{m}\cdot \text{K}$.

Number of aluminum fins is 864.

Length of the fins is 2 cm.

Diameter of the fin is 0.25 cm.

Thickness of the epoxy adhesive is 0.02 cm.

Thermal conductivity of the epoxy is $1.8\text{W}/\text{m}\cdot \text{K}$.

Calculation:

The total heat transfer rate by the chips is,

  $\begin{array}{c}\dot{Q}=80\left(0.04\text{}\text{W}\right)\\ =3.2\text{ W}\end{array}$

The total thermal resistance is,

  $\begin{array}{c}{R}_{total}=R_{board}+R_{conv}\\ =\frac{L}{kA}+\frac{1}{hA}\\ =\frac{0.003\text{ m}}{\left(30\text{W}/\text{m}\cdot \text{K}\right)\left(0.12\text{ m}\times \text{0}\text{.18 m}\right)}+\frac{1}{\left(40\text{W}/{\text{m}}^2\cdot \text{K}\right)\left(0.12\text{ m}\times \text{0}\text{.18 m}\right)}\\ =1.1620°\text{C}/\text{W}\end{array}$

Calculate the temperature on the two sides of the board.

  $\begin{array}{l}\dot{Q}=\frac{T_1-T_{\infty 2}}{R_{total}}\\ 3.2\text{W}=\frac{T_1-40°\text{C}}{1.1620°\text{C}/\text{W}}\\ T_1=43.72°\text{C}\end{array}$

Similarly,

  $\begin{array}{l}\dot{Q}=\frac{T_1-T_2}{R_{board}}\\ 3.2\text{W}=\frac{43.72°\text{C}-T_2}{\frac{0.003\text{ m}}{\left(30\text{W}/\text{m}\cdot \text{K}\right)\left(0.12\text{ m}\times \text{0}\text{.18 m}\right)}}\\ T_2=43.71°\text{C}\end{array}$

Thus, the temperature on the two sides of the circuit board is $43.72°\text{C}$ and $43.71°\text{C}$.

(b)

To determine

The temperature on the two sides of the circuit board after the aluminum plate is attached.

Explanation of Solution

Calculation:

Calculate the value of m.

  $\begin{array}{c}m=\sqrt{\frac{hp}{k{A}_c}}\\ =\sqrt{\frac{4h}{kD}}\\ =\sqrt{\frac{4\left(40\text{W}/{\text{m}}^2\cdot \text{K}\right)}{\left(237\text{W}/\text{m}\cdot \text{K}\right)\left(0.0025\text{ m}\right)}}\\ =16.43{\text{ m}}^{-1}\end{array}$

The efficiency of the fin is,

  $\begin{array}{c}{\eta }_{fin}=\frac{\tanh mL}{mL}\\ =\frac{\tanh \left(16.43\times 0.02\text{ m}\right)}{\left(16.43\times 0.02\text{ m}\right)}\\ =0.9655\end{array}$

Calculate the total area with fins,

  $\begin{array}{c}{A}_{total}=A_{finned}+A_{unfinned}\\ ={\eta }_{fin}n\pi DL+\left(0.12\text{ m}\times \text{0}\text{.18 m}\right)-864\frac{\pi D^2}{4}\\ =\left(0.9655\right)\left(864\right)\pi \left(0.0025\text{ m}\right)\left(0.02\text{ m}\right)+0.0216{\text{ m}}^2-864\frac{\pi {\left(0.0025\text{ m}\right)}^2}{4}\\ =0.1484{\text{ m}}^2\end{array}$

The total thermal resistance is,

  $\begin{array}{c}{R}_{total}=R_{board}+R_{conv}+R_{epoxy}+R_{aluminum}\\ ={\left(\frac{L}{kA}\right)}_{board}+\frac{1}{h{A}_{total}}+{\left(\frac{L}{kA}\right)}_{epoxy}+{\left(\frac{L}{kA}\right)}_{aluminum}\\ =\frac{0.003\text{ m}}{\left(30\text{W}/\text{m}\cdot \text{K}\right)\left(0.12\text{ m}\times \text{0}\text{.18 m}\right)}+\frac{1}{\left(40\text{W}/{\text{m}}^2\cdot \text{K}\right)\left(0.1484{\text{ m}}^2\right)}\\ +\frac{0.0002\text{ m}}{\left(1.8\text{W}/\text{m}\cdot \text{K}\right)\left(0.12\text{ m}\times \text{0}\text{.18 m}\right)}+\frac{0.002\text{ m}}{\left(237\text{W}/\text{m}\cdot \text{K}\right)\left(0.12\text{ m}\times \text{0}\text{.18 m}\right)}\\ =0.1787°\text{C}/\text{W}\end{array}$

Calculate the temperature on the two sides of the board.

  $\begin{array}{l}\dot{Q}=\frac{T_1-T_{\infty 2}}{R_{total}}\\ 3.2\text{W}=\frac{T_1-40°\text{C}}{0.1787°\text{C}/\text{W}}\\ T_1=40.57°\text{C}\end{array}$

Similarly,

  $\begin{array}{l}\dot{Q}=\frac{T_1-T_2}{R_{board}}\\ 3.2\text{W}=\frac{40.57°\text{C}-T_2}{\frac{0.003\text{ m}}{\left(30\text{W}/\text{m}\cdot \text{K}\right)\left(0.12\text{ m}\times \text{0}\text{.18 m}\right)}}\\ T_2=40.56°\text{C}\end{array}$

Thus, the temperature on the two side of the circuit board after the aluminum plate is attached is $40.57°\text{C}$ and $40.56°\text{C}$.