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 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

Answer: The flow rate of the fan should be 0.0178 m³/s, and the diameter of the fan casing should be 0.267 m.

Step-by-step solution

Calculate the specific heat capacity of air

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.

Calculate the mass flow rate of air

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 \(m\): \[m = \frac{P}{c_p \times (T_{exit} - T_{inlet})} \] \[m = \frac{80 \mathrm{~J/s}}{1005 \mathrm{~J/kg K} \times (60 - 50)}\] \[m = 0.016 \mathrm{~kg/s}\]

Calculate the volumetric flow rate

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, \(Q\): \[Q = \frac{m}{\rho}\] \[Q = \frac{0.016 \mathrm{~kg/s}}{0.899 \mathrm{~kg/m^3}}\] \[Q = 0.0178 \mathrm{~m^3/s}\]

Determine the diameter of the fan casing

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 \(0.267\ \mathrm{m}\).

Key concepts

Specific Heat Capacity

The specific heat capacity is a property that describes how much heat energy is required to raise the temperature of a substance by a certain temperature interval. In our case, it's about determining how much energy is needed to increase the temperature of air by one degree Celsius per unit mass. For dry air, the specific heat at constant pressure (\(c_p\)) is approximately 1005 Joules per kilogram per Kelvin (J/kg.K). This value is crucial for cooling systems as it helps us understand how effective the fan must be to absorb and dissipate the heat generated by the computer components.

To illustrate, if we look back at the exercise, we know that the computer emits 80 Watts of power, which all need to be dissipated to keep the interior temperature within safe limits. Understanding the specific heat capacity of air allows us to calculate the mass flow rate required to prevent overheating.

Mass Flow Rate of Air

The mass flow rate of air is a measure indicating how much mass of air passes through a given point per unit time. It's typically measured in kilograms per second (kg/s). In cooling systems, this rate determines the fan's ability to maintain the desired temperatures by replacing the heated air inside the computer case with cooler air from the environment.

In the example provided, we want to ensure the air exiting the computer does not exceed 60 degrees Celsius. By using the formula \(m \times c_p \times (T_{exit} - T_{inlet}) = P\), we calculate that the required mass flow rate to dissipate the 80 W of power without exceeding the maximum exit temperature is 0.016 kg/s. This figure plays a vital part in the design and selection of the appropriate cooling fan.

Volumetric Flow Rate

Volumetric flow rate, represented by the letter \(Q\), defines the volume of air flowing through the fan per unit of time, usually expressed in cubic meters per second (m³/s). It is determined by the mass flow rate of the air and its density (\(\rho\)); the formula is \(Q = \frac{m}{\rho}\).

In the exercise, we've calculated the mass flow rate and air density at the given conditions (altitude and pressure using the ideal gas law). With these values, we determine the volumetric flow rate needed for the cooling fan to sustain the computer's operation without overheating. The calculated volumetric flow rate, 0.0178 m³/s, will guide us in determining the fan size and speed required to achieve the desired cooling effect.

Fan Casing Diameter

The fan casing diameter is an important aspect of fan design, as it dictates the size of the fan and, consequently, how much space it occupies within the computer case. It also influences the fan's capability to deliver the required volumetric flow rate while adhering to noise constraints.

In this scenario, the maximum air velocity is restricted to keep the noise level down. Thus, using the formula \(2 = \frac{Q}{\frac{\pi}{4} \times \varnothing ^2}\), we find a diameter that will allow us to meet both the volumetric flow rate and velocity requirements. The calculation reveals that the fan casing diameter should be 0.267 meters. By correctly sizing the fan casing, we ensure that the fan will not be excessively loud while providing adequate cooling to the computer.