Question 8.11: A wind tunnel has a wooden (ε = 0.0001 m) rectangular sectio...
A wind tunnel has a wooden (ε = 0.0001 m) rectangular section 40 cm by 1 m by 50 m long. The average velocity is 45 m/s for air at sea-level standard conditions. Find the power required if the fan has 65 percent efficiency. For air, ρ 1.2 kg / m ^{3}, \mu=1.81 \times 10^{-5} kg/m-s.
The wetted perimeter of the duct is
P_{w}=\left[\frac{40}{100}+\frac{40}{100}+1+1\right]=2.8 m
and the flow area is A_{w}=\frac{40}{100} \times 1=0.4 m ^{2}.
Hence, hydraulic diameter D_{h}=\frac{4 \times 0.4}{2.8}=0.5714 m
\operatorname{Re}=\frac{\rho D_{h} V_{ av }}{\mu}=\frac{(1.20)(0.5714)(45)}{\left(1.81 \times 10^{-5}\right)}=1.7 \times 10^{6}
Now, \varepsilon / D_{h}=\frac{0.0001}{0.5714}=0.000175
From Moody′s chart for \operatorname{Re}=1.7 \times 10^{6} \text { and } \varepsilon / D_{h}=0.000175
f = 0.0140
h_{f}=f\left(\frac{L}{D_{h}}\right) \frac{V^{2}}{2 g}=(0.0140)\left(\frac{50}{0.5714}\right)\left[\frac{(45)^{2}}{2 \times 9.8}\right]
= 126.5 m
and pressure drop, \Delta p=\rho g h_{f}=(1.20)(9.8)(126.5)=1489 pa
Q=A V=\left(\frac{40}{100}\right)(1)(45)=18.0 m^{3} / s
Pumping power, P=\frac{\rho g Q h_{f}}{\eta}=\frac{(1.20)(9.8)(18.0)(126.5)}{0.65}
= 41200 W