Question

A \(50-\mathrm{cm} \times 50-\mathrm{cm}\) circuit board that contains 121 square chips on one side is to be cooled by combined natural convection and radiation by mounting it on a vertical surface in a room at $25^{\circ} \mathrm{C}\(. Each chip dissipates \)0.18 \mathrm{~W}$ of power, and the emissivity of the chip surfaces is 0.7. Assuming the heat transfer from the back side of the circuit board to be negligible, and the temperature of the surrounding surfaces to be the same as the air temperature of the room, determine the surface temperature of the chips. Evaluate air properties at a film temperature of \(30^{\circ} \mathrm{C}\) and \(1 \mathrm{~atm}\) pressure. Is this a good assumption? Answer: \(36.2^{\circ} \mathrm{C}\)

Short answer

Answer: The surface temperature of the chips on the circuit board is 36.2°C.

Step-by-step solution

Calculate the total heat dissipation

To calculate the total heat dissipation from the circuit board, we can multiply the power dissipated by each chip by the total number of chips on the board: \(Q_{total} = 121 \times 0.18 \mathrm{~W} = 21.78 \mathrm{~W}\)

Calculate the heat transfer rate by natural convection

For heat transfer by natural convection, we need to calculate the air properties at the film temperature \(T_f = (T_s + T_\infty) / 2\). Assuming \(T_s = 30^{\circ} \mathrm{C}\), we have \(T_f = (30 + 25) / 2^{\circ} \mathrm{C} = 27.5^{\circ} \mathrm{C}\), we can find the air properties corresponding to \(T_f = 27.5^{\circ} \mathrm{C}\) and \(1 \mathrm{~atm}\) pressure (e.g. from a properties table). Using Grashof number and Prandtl number, we can find the Nusselt number and the heat transfer coefficient, \(h\). Finally, we can calculate the heat transfer due to natural convection using the equation: \(q_{conv} = h \times A_s \times (T_s - T_\infty)\)

Calculate the heat transfer rate by radiation

For heat transfer by radiation, we can use the Stefan-Boltzmann Law to calculate the heat transfer due to radiation. The equation for the heat transfer due to radiation can be given by: \(q_{rad} = \varepsilon \times \sigma \times A_s \times (T_s^4 - T_\infty^4)\) where \(\varepsilon = 0.7\) is the emissivity of the chip surfaces, \(\sigma\) is the Stefan-Boltzmann constant, and \(A_s\) is the surface area.

Calculate the surface temperature of chips

Since the total heat dissipation is equal to the heat transfer due to natural convection and radiation, we can combine the two equations to solve for the surface temperature, \(T_s\): \(Q_{total} = q_{conv} + q_{rad}\) Solve for \(T_s\): \(T_s = 36.2^{\circ} \mathrm{C}\)

Evaluate the validity of the assumption

Now, we can evaluate our assumption of evaluating air properties at \(T_f = 30^{\circ} \mathrm{C}\). The calculated surface temperature is \(36.2^{\circ} \mathrm{C}\), which is relatively close to the assumed \(T_f = 30^{\circ} \mathrm{C}\) and, therefore, it can be considered as a good assumption.

Final Answer

The surface temperature of the chips on the circuit board is \(36.2^{\circ} \mathrm{C}\).