Question : Choosing a Fan to Cool a Computer The desktop computer shown...

Choosing a Fan to Cool a Computer

The desktop computer shown in Figure 15–53 is to be cooled by a fan. The electronics of the computer consume 75 W of power under full-load conditions. The computer is to operate in environments at temperatures up to 40^{\circ}C and at elevations up to 2000 m, where the atmospheric pressure is 79.50 kPa. The exit temperature of air is not to exceed 70^{\circ}C to meet reliability requirements. Also, the average velocity of air is not to exceed 75 m/min at the exit of the computer case, where the fan is installed, to keep the noise level down. Determine the flow rate of the fan that needs to be installed and the diameter of the casing of the fan.

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SOLUTION  A desktop computer is to be cooled by a fan safely in hot environments and high elevations. The airflow rate of the fan and the diameter of the casing are to be determined.

Assumptions  1 Steady operation under worst conditions is considered. 2 Air is an ideal gas.

Properties  The specific heat of air at the average temperature of \left ( 40 + 70 \right )/2 = 55^{\circ}C is 1007  J/kg \cdot ^{\circ}C (Table A–15).

Analysis  We need to determine the flow rate of air for the worst-case scenario. Therefore, we assume the inlet temperature of air to be 40^{\circ}C and the atmospheric pressure to be 79.50 kPa and disregard any heat transfer from the outer surfaces of the computer case. Note that any direct heat loss from the computer case will provide a safety margin in the design. Noting that all the heat dissipated by the electronic components is absorbed by air, the required mass flow rate of air to absorb heat at a rate of 75 W can be determined from

                        \dot{Q} = \dot{m}  C_{p} \left ( T_{out}  –  T_{in} \right )

 

Solving for \dot{m} and substituting the given values, we obtain

\dot{m} = \frac{\dot{Q}}{C_{p} \left ( T_{out}  –  T_{in} \right )} = \frac{75  J/s}{\left ( 1007  J/kg \cdot  ^{\circ}C \right )\left ( 70    40 \right )^{\circ}C}

 

                             = 0.00249  kg/s = 0.149  kg/min

 

In the worst case, the exhaust fan will handle air at 70°C. Then the density of air entering the fan and the volume flow rate become

\rho = \frac{P}{RT} = \frac{79.50  kPa}{\left ( 0.287  kPa \cdot m^{3}/kg \cdot K \right )\left ( 70 + 275 \right )  K} = 0.8076  kg/m^{3}

 

  \dot{V} = \frac{\dot{m}}{\rho } = \frac{0.149  kg/min}{0.8076  kg/m^{3}} = 0.184  m^{3}/min

 

Therefore, the fan must be able to provide a flow rate of 0.184  m^{3}/min or 6.5 cfm (cubic feet per minute). Note that if the fan were installed at the inlet instead of the exit, then we would need to determine the flow rate using the density of air at the inlet temperature of 40°C, and we would need to add the power consumed by the motor of the fan to the heat load of 75 W. The result may be a slightly smaller or larger fan, depending on which effect dominates. For an average velocity of 75 m/min, the diameter of the duct in which the fan is installed can be determined from

                     \dot{V} = A_{c}  ^{\circ}V = \frac{1}{4} \pi D^{2} \ ^{\circ}V

 

Solving for D and substituting the known values, we obtain

D = \sqrt{\frac{4 \dot{V}}{\pi  ^{\circ}V}} = \sqrt{\frac{\left (4 \times 0.184  m^{3}/min \right )}{\pi \left ( 75  m/min \right )}} = 0.056  m = 5.6  cm

 

Therefore, a fan with a casing diameter of 5.6 cm and a flow rate of 0.184  m^{3}/min will meet the design requirements.