## Centrifugal Fan

Consider you are conducting an experiment, and need to run a test specimen on a centrifugal fan. After turning on the 10 kW machine with the 0.25 m diameter fan running at 2,000 rpm for 5 minutes, what is the fan's efficiency if the measured pressure rise was 1 kPa and the fluid (air at 300 K, 101 kPa) flow rate was 6 m^3/s?

Hint
Pump Power Equation:
$$\dot{W}=\frac{Q\gamma h}{\eta }=\frac{Q\rho gh}{\eta}$$$where $$Q$$ is the volumetric flow, $$h$$ is the head for the fluid to be lifted, $$\eta$$ is the total efficiency, and $$\dot W$$ is the power. Hint 2 Since $$\Delta P=\gamma h$$ , our equation for a typical backward curved fan is: $$\dot{W}=\frac{\Delta PQ}{\eta }$$$
where $$\Delta P$$ is the pressure rise.
Pump Power Equation:
$$\dot{W}=\frac{Q\gamma h}{\eta }=\frac{Q\rho gh}{\eta}$$$where $$Q$$ is the volumetric flow, $$h$$ is the head for the fluid to be lifted, $$\eta$$ is the total efficiency, and $$\dot W$$ is the power. Since $$\Delta P=\gamma h$$ , our equation for a typical backward curved fan is: $$\dot{W}=\frac{\Delta PQ}{\eta }$$$
where $$\Delta P$$ is the pressure rise. Solving for $$\eta$$ :
$$\eta =\frac{\Delta PQ}{\dot{W}}=\frac{(1kPa)(6m^3/s)}{10kW}$$$Since efficiency is unitless, remember to convert the pressure and power units to base units: $$1kPa=\frac{kN}{m^2}$$$
$$1kW=\frac{kN\cdot m}{s}$$$Thus, $$\eta =\frac{(1kN)(6m^3)\cdot s}{10kN\cdot m\cdot m^2\cdot s}=0.6$$$
0.6