## 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**

**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**

**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

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