A desktop computer is to be cooled by a fan. The electronic components of the computer consume \(80 \mathrm{~W}\) of power under full-load conditions. The computer is to operate in environments at temperatures up to \(50^{\circ} \mathrm{C}\) and at elevations up to \(3000 \mathrm{~m}\) where the atmospheric pressure is \(70.12 \mathrm{kPa}\). The exit temperature of air is not to exceed \(60^{\circ} \mathrm{C}\) to meet the reliability requirements. Also, the average velocity of air is not to exceed \(120 \mathrm{~m} / \mathrm{min}\) at the exit of the computer case, where the fan is installed to keep the noise level down. Specify the flow rate of the fan that needs to be installed and the diameter of the casing of the fan.

Using the information given, we can determine the specific heat at constant pressure for air, assuming it is an ideal gas. For dry air, the specific heat at constant pressure \(c_p\) is approximately 1005 J/kg K.

We need to find the mass flow rate of air, which can be found using the given power dissipation and temperature limits. Let the mass flow rate be \(m\), then at steady state, we get: \[m \times c_p \times (T_{exit} - T_{inlet}) = P\] Given : Power dissipation, \(P = 80 \mathrm{~W} = 80 \mathrm{~J/s}\), Inlet temperature, \(T_{inlet}= 50 ^{\circ} \mathrm{C}\), Exit temperature \(T_{exit} = 60 ^{\circ} \mathrm{C}\), and Specific heat of air, \(c_p = 1005 \mathrm{~J/kg K}\). Plug in the values in the equation and solve for \(m\): \[m = \frac{P}{c_p \times (T_{exit} - T_{inlet})} \] \[m = \frac{80 \mathrm{~J/s}}{1005 \mathrm{~J/kg K} \times (60 - 50)}\] \[m = 0.016 \mathrm{~kg/s}\]

Now we need to find the volumetric flow rate of the air. Let the volumetric flow rate be \(Q\). We know that: \[Q = \frac{m}{\rho}\] Where, \(\rho\) is the air density. We can now calculate the air density at the given condition (altitude 3000 m, pressure: 70.12 kPa) using the ideal gas law for air: \[\rho = \frac{P_{atm}}{R \times T}\] Given: Atmospheric pressure, \(P_{atm} = 70.12 \mathrm{kPa} = 70.12 \times 10^3 \mathrm{N/m^2}\), Gas constant for air, \(R = 287 \mathrm{J/kg K}\), and Temperature at height, \(T = T_{inlet} + 273.15 = 50 + 273.15 = 323.15 \mathrm{K}\). Plug in the values in the equation and solve for \(\rho\): \[\rho = \frac{70.12 \times 10^3 \mathrm{N/m^2}}{ 287 \mathrm{J/kg K} \times 323.15 \mathrm{K}}\] \[\rho = 0.899 \mathrm{~kg/m^3}\] Now, calculate the volumetric flow rate, \(Q\): \[Q = \frac{m}{\rho}\] \[Q = \frac{0.016 \mathrm{~kg/s}}{0.899 \mathrm{~kg/m^3}}\] \[Q = 0.0178 \mathrm{~m^3/s}\]

To find the diameter of the casing, we need to use the maximum average velocity allowed at the exit of the fan. The area of exit is given by: \[A = \frac{\pi \times \varnothing^2}{4}\] Where \(\varnothing\) is the diameter of the casing. The maximum exit velocity \(V_{exit} = 120 \mathrm{~m/min} = 2 \mathrm{~m/s}\). We substitute the values in the equation and solve for the diameter: \[2 = \frac{0.0178}{\frac{\pi}{4} \times \varnothing ^2}\] \[\varnothing^2 = \frac{4 \times 0.0178}{2 \times \pi}\] \[\varnothing = \sqrt{\frac{4 \times 0.0178}{2 \times \pi}}\] \[\varnothing = 0.267 \mathrm{~m}\] So, the fan casing diameter is 0.267 m. In conclusion, the flow rate of the fan needs to be \(0.0178\ \mathrm{m^3/s}\), and the diameter of the fan casing must be \(0.267\ \mathrm{m}\).