A 0.3-cm-thick, 12-cm-high, and 18-cm-long circuit board houses 80 closely spaced logic chips on one side, each dissipating 0.06 W. The board is impregnated with copper fillings and has an effective thermal conductivity of [latex] 16 W / m \cdot{ }^{\circ} C [/latex]. All the heat generated in the chips is conducted across the circuit board and is dissipated from the back side of the board to the ambient air. Determine the temperature difference between the two sides of the circuit board.

A 0.3-cm-thick, 12-cm-high, and 18-cm-long circ...

A 0.3-cm-thick, 12-cm-high, and 18-cm-long circuit board houses 80 closely spaced logic chips on one side, each dissipating 0.06 W. The board is impregnated with copper fillings and has an effective thermal conductivity of $16 W / m \cdot{ }^{\circ} C$ . All the heat generated in the chips is conducted across the circuit board and is dissipated from the back side of the board to the ambient air. Determine the temperature difference between the two sides of the circuit board.

Question Data is a breakdown of the data given in the question above.

Thickness of circuit board: 0.3 cm
Height of circuit board: 12 cm
Length of circuit board: 18 cm
Number of logic chips: 80
Power dissipated by each logic chip: 0.06 W
Thermal conductivity of the board: 16 W/(m·°C)

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Step 1:

We are given that there are 80 closely spaced logic chips on one side of the circuit board, and each chip dissipates 0.06 W of heat. We need to determine the temperature difference between the two sides of the circuit board.

Step 2:

We can start by calculating the total rate of heat dissipated by the chips. This can be done by multiplying the power dissipated by each chip (0.06 W) by the number of chips (80). This gives us a total heat dissipation of 4.8 W.

Step 3:

Next, we need to calculate the area of the circuit board. We are given the dimensions of the board as 0.12 m by 0.18 m. The area can be calculated by multiplying these two dimensions, giving us 0.0216 m^2.

Step 4:

Using the formula for heat conduction, which states that the rate of heat transfer (Q) is equal to the thermal conductivity (k) times the area (A) times the temperature difference (ΔT) divided by the thickness (L) of the material, we can rearrange the formula to solve for ΔT. This gives us ΔT = (Q L) / (k A).

Step 5:

Plugging in the values we have, we can calculate the temperature difference. The rate of heat transfer (Q) is 4.8 W, the thickness of the board (L) is given as 0.003 m, and the thermal conductivity (k) is given as 16 W / m * °C. The area (A) is 0.0216 m^2.

Step 6:

Solving the equation, we find that the temperature difference (ΔT) is equal to (4.8 W × 0.003 m) / ((16 W / m °C) * 0.0216 m^2), which simplifies to 0.042 °C.

Step 7:

Therefore, the temperature difference between the front and back surfaces of the circuit board is 0.042 °C.

Note: It is important to note that the circuit board is nearly isothermal, meaning that the temperature difference across the board is very small.

Final Answer

A circuit board houses 80 closely spaced logic chips on one side, each dissipating 0.06 W. All the heat generated in the chips is conducted across the circuit board. The temperature difference between the two sides of the circuit board is to be determined.

Assumptions 1 Steady operating conditions exist. 2 Thermal properties of the board are constant. 3 All the heat generated in the chips is conducted across the circuit board.

Properties The effective thermal conductivity of the board is given to be $k = 16 W / m \cdot{ }^{\circ} C$ .

Analysis The total rate of heat dissipated by the chips is

\dot{Q} = 80 \times(0.06  W ) = 4.8  W

Then the temperature difference between the front and back surfaces of the board is

A=(0.12  m )(0.18  m ) = 0.0216  m ^{2}

\dot{Q}=k A \frac{\Delta T}{L} \longrightarrow \Delta T=\frac{\dot{Q} L}{k A}=\frac{(4.8  W )(0.003  m )}{\left(16  W / m \cdot{ }^{\circ} C \right)\left(0.0216  m ^{2}\right)}=0.042^{\circ} C

Discussion Note that the circuit board is nearly isothermal.

Question 1.92: A 0.3-cm-thick, 12-cm-high, and 18-cm-long circuit board hou...

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