Scaling Law

Consuming 20 W of power, a pump with a 0.75 m diameter impeller rotates at 100 rpm to produce a 2 m head. If the same pump were to increase its impeller diameter to 1 m and its rotational speed to 500 rpm, how much head (m) can the new system produce?

Expand Hint
$$\left ( \frac{H}{N^2D^2} \right )_2= \left ( \frac{H}{N^2D^2} \right )_1$$$where $$H$$ is the head, $$N$$ is the rotational speed, and $$D$$ is the impeller diameter. Hint 2 A pump’s head is the max height a pump can vertically push a fluid against gravity. A pump’s head is the max height a pump can vertically push a fluid against gravity. $$\left ( \frac{H}{N^2D^2} \right )_2= \left ( \frac{H}{N^2D^2} \right )_1$$$
where $$H$$ is the head, $$N$$ is the rotational speed, and $$D$$ is the impeller diameter.
$$\left ( \frac{2m}{(100rpm)^2(0.75m)^2} \right )_2= \left ( \frac{H}{(500rpm)^2(1m)^2} \right )_1$$$$$\frac{2m}{(100)^2(0.75)^2} = \frac{H}{(500)^2(1)}$$$
$$\frac{2m}{5,625} = \frac{H}{250,000}$$$$$H=\frac{2m}{5,625}\times 250,000 = 88.9\:m$$$
88.9 m
Similar Problems from FE Sub Section: Performance of Components

Similar Problems from FE Section: Fluid Flow Machinery