Affinity
Consuming 10 W of power, a fan with a 1.5 m diameter impeller rotates at 500 rpm to cool a warehouse in the summer. If the same fan were to increase its rotational speed to 3,000 rpm, how much power is the fan now using? Assume the density of air is 101 kPa.
Expand Hint
$$$\left ( \frac{\dot{W}}{\rho N^3D^5} \right )_2= \left ( \frac{\dot{W}}{\rho N^3D^5} \right )_1$$$
where
$$\dot{W}$$
is the power,
$$\rho$$
is the fluid density,
$$N$$
is the rotational speed, and
$$D$$
is the impeller diameter.
Hint 2
Since both diameter and density are constant, the equation becomes:
$$$\left ( \frac{\dot{W}}{N^3} \right )_2= \left ( \frac{\dot{W}}{N^3} \right )_1$$$
Using the scaling/affinity laws:
$$$\left ( \frac{\dot{W}}{\rho N^3D^5} \right )_2= \left ( \frac{\dot{W}}{\rho N^3D^5} \right )_1$$$
where
$$\dot{W}$$
is the power,
$$\rho$$
is the fluid density,
$$N$$
is the rotational speed, and
$$D$$
is the impeller diameter.
Since both diameter and density are constant, the equation becomes:
$$$\left ( \frac{\dot{W}}{N^3} \right )_2= \left ( \frac{\dot{W}}{N^3} \right )_1$$$
$$$\left ( \frac{\dot{W_1}}{N_{1}^{3}} \right )= \left ( \frac{\dot{W_2}}{N_{2}^{3}} \right )$$$
$$$\left ( \frac{10W}{(500rpm)^{3}} \right )= \left ( \frac{\dot{W_2}}{(3,000rpm)^{3}} \right )$$$
$$$\dot{W_2}=\frac{10W}{(500rpm)^{3}}\cdot (3,000rpm)^{3}=2,160\:W$$$
2,160 W
Time Analysis
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