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Question

The 60-W fan of a central heating system is to circulate air through the ducts. The analysis of the flow shows that the fan needs to raise the pressure of air by 50 Pa to maintain flow. The fan is located in a horizontal flow section whose diameter is 30 cm at both the inlet and the outlet. Determine the highest possible average flow velocity in the duct.

Solution

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Givens:\textbf{Givens:}

W˙=60  W\dot{W}=60\; W

Δp=50  Pa\Delta p=50\; Pa

D=30  cm=0.3  m        A=πD24=0.0707  m2D=30\; cm=0.3\; m\;\; \to \;\; A=\frac{\pi D^2}{4}=0.0707\; m^2

Solution:\textbf{Solution:}

Power (W˙)=m˙Δpρ=ρAVavg×Δpρ=AVavgΔp\text{Power }(\dot{W})=\dot{m} \frac{\Delta p}{\rho}=\rho A V_{avg}\times \frac{\Delta p}{\rho}=A V_{avg} \Delta p

So,

Vavg,max=Wmax˙AΔp=600.0707×50=17  m/sV_{avg,max}=\frac{\dot{W_{max}}}{A \Delta p}=\frac{60}{0.0707\times 50}=\mathbf{17}\; m/s

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