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

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

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

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

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

\dot{m} = \rho \dot{V} = (1.20  kg/m^3)(9.6 \times 10^{-3}  m^3/s) = 0.0115  kg/s

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

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

We draw the control volume around the fan such that both the inlet and the outlet are at atmospheric pressure (P_1 = P_2 = P_{atm}), 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 (V_1 ≅ 0). Noting that z_1 = z_2 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

\dot{m} \left(\frac{P_1}{\rho } + \alpha _1\frac{V^2_1}{2} + gz_1 \right) + \dot{W}_{pump} = \dot{m} \left(\frac{P_2}{\rho } +\alpha _2 \frac{V^2_2}{2} + gz_2 \right) + \dot{W}_{turbine} + \dot{E}_{mech, loss}               (5.76)

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

Solving for \dot{W}_{fan}  –  \dot{E}_{mech  loss,  fan} = \dot{W}_{fan, u}  and substituting,

\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} \right) = 0.152  W

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

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

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 z_3 = z_4 again and V_3 = V_4 since the fan is a narrow cross section, and the energy
equation reduces to

\dot{m} \frac{P_3}{\rho } +\dot{W}_{fan} = \dot{m}\frac{P_4}{\rho } + \dot{E}_{mech  loss,  fan} → \dot{W}_{fan,  u} = \dot{m} \frac{P_4  –  P_3}{\rho }

Solving for P_4  −  P_3 and substituting,

P_4  –  P_3 = \frac{\rho  \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} \right) = 15.8  Pa

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 \dot{W}_{electric}  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).