## 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**

**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**

**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