Fan Selection for Air Cooling of a Computer A fan is to be selected to cool a computer case whose dimensions are 12 cm × 40 cm × 40 cm (Fig. 5–60). Half of the volume in the case is expected to be filled with components and the other half to be air space. A 5-cm-diameter hole is available at the

Question

Fan Selection for Air Cooling of a Computer

A fan is to be selected to cool a computer case whose dimensions are 12 cm × 40 cm × 40 cm (Fig. 5–60). Half of the volume in the case is expected to be filled with components and the other half to be air space. A 5-cm-diameter hole is available at the back of the case for the installation of the fan that is to replace the air in the void spaces of the case once every second. Small lowpower fan–motor combined units are available in the market and their efficiency is estimated to be 30 percent. Determine (a) the wattage of the fan–motor unit to be purchased and (b) the pressure difference across the fan. Take the air density to be 1.20 kg/m³.

Accepted Answer

A fan is to cool a computer case by completely replacing the air inside once every second. The power of the fan and the pressure difference across it are to be determined.
Assumptions 1 The flow is steady and incompressible. 2 Losses other than those due to the inefficiency of the fan–motor unit are negligible. 3 The flow at the outlet is fairly uniform except near the center (due to the wake of the fan motor), and the kinetic energy correction factor at the outlet is 1.10.
Properties The density of air is given to be 1.20 kg/m³.
Analysis (a) Noting that half of the volume of the case is occupied by the components, the air volume in the computer case is

[latex]ΔV_{air}[/latex] = (Void fraction)(Total case volume)
[latex]= 0.5(12 cm × 40 cm × 40 cm) = 9600 cm^3[/latex]

Therefore, the volume and mass flow rates of air through the case are

[latex] \dot{V} = \frac{\Delta V_{air}}{\Delta t} = \frac{9600 cm^3}{1 s} = 9600 cm^3/s = 9.6 imes 10^{-3} m^3/s[/latex]

[latex] \dot{m} = ho \dot{V} = (1.20 kg/m^3)(9.6 imes 10^{-3} m^3/s) = 0.0115 kg/s[/latex]

The cross-sectional area of the opening in the case and the average air velocity through the outlet are

[latex]A = \frac{\pi D^2}{4} = \frac{\pi (0.05 m)^2}{4} = 1.96 imes 10^{-3} m^2[/latex]
[latex]V = \frac{\dot{V} }{A} = \frac{9.6 imes 10^{-3} m^3/s}{1.96 imes 10^{-3 }m^2} = 4.90 m/s[/latex]

We draw the control volume around the fan such that both the inlet and the outlet are at atmospheric pressure ([latex]P_1 = P_2 = P_{atm}[/latex]), as shown in Fig. 5–60, where the inlet section 1 is large and far from the fan so that the flow velocity at the inlet section is negligible ([latex]V_1 ≅ 0[/latex]). Noting that [latex]z_1 = z_2[/latex] and frictional losses in the flow are disregarded, the mechanical losses consist of fan losses only and the energy equation (Eq. 5–76) simplifies to

[latex] \dot{m} \left(\frac{P_1}{ ho } + \alpha _1\frac{V^2_1}{2} + gz_1 ight) + \dot{W}_{pump} = \dot{m} \left(\frac{P_2}{ ho } +\alpha _2 \frac{V^2_2}{2} + gz_2 ight) + \dot{W}_{turbine} + \dot{E}_{mech, loss}[/latex] (5.76)

[latex] \dot{m} \left(\cancel{\frac{P_1}{ ho } } + \alpha _1 \overset{0}{\frac{\cancel{V^2_1}}{2} } +\cancel{gz_2} ight) +\dot{W}_{fan} = \dot{m}\left(\cancel{\frac{P_2}{ ho } } +\alpha _2 \frac{V^2_2}{2} + \cancel{gz_2} ight) +\overset{0}{\cancel{W_{turbine}}} + \dot{E}_{mech loss, fan} [/latex]

Solving for [latex] \dot{W}_{fan} - \dot{E}_{mech loss, fan} = \dot{W}_{fan, u} [/latex] and substituting,

[latex]\dot{W}_{fan, u} = \dot{m}\alpha _2 \frac{V^2_2}{2} = (0.0115 kg/s)(1.10) \frac{(4.90 m/s)^2}{2}\left(\frac{1 N}{1 kg.m/s^2} ight) = 0.152 W[/latex]

Then the required electric power input to the fan is determined to be

[latex] \dot{W}_{elect} = \frac{\dot{W}_{fan, u} }{\eta _{fan−motor}}= \frac{0.152 W}{0.3} = 0.506 W[/latex]

Therefore, a fan–motor rated at about a half watt is adequate for this job (Fig. 5–61). (b) To determine the pressure difference across the fan unit, we take points 3 and 4 to be on the two sides of the fan on a horizontal line. This time [latex]z_3 = z_4[/latex] again and [latex]V_3 = V_4[/latex] since the fan is a narrow cross section, and the energy
equation reduces to

[latex] \dot{m} \frac{P_3}{ ho } +\dot{W}_{fan} = \dot{m}\frac{P_4}{ ho } + \dot{E}_{mech loss, fan} → \dot{W}_{fan, u} = \dot{m} \frac{P_4 - P_3}{ ho }[/latex]

Solving for [latex]P_4 − P_3[/latex] and substituting,

[latex] P_4 - P_3 = \frac{ ho \dot{W}_{fan, u} }{\dot{m} } = \frac{(1.2 kg/m^3)(0.152 W)}{0.0115 kg/s}\left(\frac{1 Pa.m^3}{1 Ws} ight) = 15.8 Pa[/latex]

Therefore, the pressure rise across the fan is 15.8 Pa.
Discussion The efficiency of the fan–motor unit is given to be 30 percent, which means 30 percent of the electric power [latex] \dot{W}_{electric}[/latex] consumed by the unit is converted to useful mechanical energy while the rest (70 percent) is “lost” and converted to thermal energy. Also, a more powerful fan is required in an actual system to overcome frictional losses inside the computer case. Note that if we had ignored the kinetic energy correction factor at the outlet, the required electrical power and pressure rise would have been 10 percent lower in this case (0.460 W and 14.4 Pa, respectively).

Question 5.14: Fan Selection for Air Cooling of a Computer A fan is to be s...

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