Question

Consider a hollow-core printed circuit board \(9 \mathrm{cm}\) high and \(18 \mathrm{cm}\) long, dissipating a total of \(15 \mathrm{W}\). The width of the air gap in the middle of the \(\mathrm{PCB}\) is \(0.25 \mathrm{cm}\). If the cooling air enters the 12 -cm-wide core at \(25^{\circ} \mathrm{C}\) and 1 atm at a rate of \(0.8 \mathrm{L} / \mathrm{s}\), determine the average temperature at which the air leaves the hollow core.

Short answer

Answer: The average temperature of the air leaving the hollow core of the PCB is approximately 298.16 K (25.16°C).

Step-by-step solution

Determine the mass flow rate of the air

To determine the mass flow rate (m_dot) of the cooling air, we will first convert the volume flow rate (given as 0.8 L/s) to a mass flow rate, using the Ideal Gas Law: \(PV = mRT\) Where: P: pressure (1 atm, which is equivalent to 101325 Pa) V: volume flow rate (0.8 L/s, which is equivalent to 0.0008 m³/s) m: mass flow rate (to be determined) R: specific gas constant for air (287 J/kg-K) T: temperature (25°C, which is equivalent to 298 K) First, we rearrange the Ideal Gas Law to solve for m: \(m = \frac{PV}{RT}\) Next, we substitute the given values into the equation: \(m = \frac{101325 \times 0.0008}{287\times 298}\) Now, we can calculate the mass flow rate: \(m \approx 0.0942 \, \text{kg/s}\)

Identify the energy conservation equation

As energy conservation dictates that the energy entering a system equals the energy leaving the system, we can write this balance equation: \(m\dot c_p \Delta T_{air} = Q_{dissipated}\) Where: \(c_p\): specific heat capacity of air at constant pressure (approximately 1005 J/kg-K) \(\Delta T_{air} = T_{out} - T_{in}\): The temperature difference between the incoming and outgoing air \(Q_{dissipated}\): The power dissipation of the PCB (15 W) We are interested in finding \(T_{out}\), which is the average temperature of the air leaving the hollow core.

Solve for the ΔT of air

We can use the energy conservation equation to solve for the temperature difference in the air: \(\Delta T_{air} = \frac{Q_{dissipated}}{m\dot c_p}\) Substituting given values: \(\Delta T_{air} = \frac{15}{0.0942 \times 1005}\) Calculating the temperature difference: \(\Delta T_{air} \approx 0.158 \, \text{K}\)

Calculate the temperature of the air leaving the hollow core

Now that we have the temperature difference, we can use this value to calculate the temperature of the air leaving the hollow core: \(T_{out} = T_{in} + \Delta T_{air}\) Substituting given values: \(T_{out} = 298 + 0.158\) Calculating the final temperature: \(T_{out} \approx 298.16 \, \text{K}\) The cooling air leaves the hollow core of the PCB at an average temperature of approximately 298.16 K (25.16°C).