Question

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; this is 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.

Short answer

The flow rate required for the fan is 0.00796 kg/s, and the diameter of the fan casing is approximately 94.9 mm.

Step-by-step solution

Find the specific heat of air at constant pressure

First, we should find the specific heat of air at constant pressure \((c_p)\). It can vary slightly depending on the temperature. However, we can approximate it by taking an average value for the given temperature range. \(c_p\) for air is around \(1005 \mathrm{~J/(kg\cdot K)}\).

Calculate the cooling capacity required

The cooling capacity required is the amount of heat that the fan needs to dissipate from the computer system. As per the problem, the electronic components are consuming 80 W of power under full-load conditions. This implies that the cooling capacity required is also 80 W.

Determine the temperature difference

The problem states that the maximum exit temperature for air is \(60^{\circ}\mathrm{C}\) and the computer needs to operate in environments with temperatures up to \(50^{\circ}\mathrm{C}\). The temperature difference \((\Delta T)\) is the difference between the maximum exit temperature and the environmental temperature. Hence, \(\Delta T = 60^{\circ}\mathrm{C} - 50^{\circ}\mathrm{C} = 10^{\circ}\mathrm{C}\).

Calculate the mass flow rate of the air

Given the cooling capacity required and the temperature difference, we can find the mass flow rate of the air \((\dot{m})\) required for cooling. We can use the following equation to calculate the mass flow rate: $$ \dot{m} = \frac{Q}{c_p\Delta T} $$ where \(Q = 80\mathrm{~W}\), \(c_p = 1005\mathrm{~J/(kg\cdot K)}\), and \(\Delta T = 10\mathrm{~K}\). Plugging in the values, we get: $$ \dot{m} = \frac{80}{1005 \cdot 10} = 0.00796\mathrm{~kg/s} $$

Calculate the exit area of air

Given the maximum average velocity of air at the exit of the computer case \((120\mathrm{~m/min})\), we can calculate the exit area of the air \((A)\). Converting the maximum average velocity to m/s \((V_{\mathrm{max}})\), we get: $$ V_{\mathrm{max}} = \frac{120\mathrm{~m/min}}{60\mathrm{~s/min}} = 2\mathrm{~m/s} $$ We can now find the exit area using the equation: $$ A = \frac{\dot{m}}{\rho V_{\mathrm{max}}} $$ where \(\rho\) is the air density. We can find the air density using the ideal gas law: $$ \rho = \frac{P}{RT} $$ where \(P = 70.12\mathrm{~kPa} = 70120\mathrm{~Pa}\), \(R = 287\mathrm{~J/(kg\cdot K)}\) (specific gas constant for air), and \(T = 50^{\circ}\mathrm{C} + 273.15 = 323.15\mathrm{~K}\). Then we can calculate the air density as: $$ \rho = \frac{70120}{287 \cdot 323.15} = 0.897\mathrm{~kg/m^3} $$ Using the air density and maximum velocity, we can now find the exit area: $$ A = \frac{0.00796}{0.897 \cdot 2} = 4.438 \times 10^{-3}\mathrm{~m^2} $$

Calculate the diameter of the fan

Now that we have the exit area, we can calculate the diameter of the fan casing required. Since the fan casing is expected to be circular, we can use the following formula for the area of a circle: $$ A = \pi r^2 $$ where \(r\) is the radius of the fan casing. Rearranging and solving for the diameter \((D = 2r)\), we get: $$ D = 2\sqrt{\frac{A}{\pi}} = 2\sqrt{\frac{4.438 \times 10^{-3}}{\pi}} = 0.0949\mathrm{~m} = 94.9\mathrm{~mm} $$ Thus, the flow rate of the fan that needs to be installed is \(0.00796\mathrm{~kg/s}\), and the diameter of the fan casing is approximately \(94.9\mathrm{~mm}\).