Impeller Diameter
Consider a pump with a 5 m diameter impeller moves water through a pipe by spinning at 1,000 rpm. If the same pump were to decrease its speed by 200 rpm, what new diameter impeller size (m) should the pump be modified with to maintain the same mass flowrate?
Expand Hint
$$$\left ( \frac{\dot{m}}{\rho ND^3} \right )_2= \left ( \frac{\dot{m}}{\rho ND^3} \right )_1$$$
where
$$\dot m$$
is the mass flowrate,
$$\rho$$
is the fluid density,
$$N$$
is the rotational speed, and
$$D$$
is the impeller diameter.
Hint 2
Since fluid density and mass flowrate are constant, the equation becomes:
$$$\left ( \frac{1}{ND^3} \right )_2= \left ( \frac{1}{ND^3} \right )_1$$$
Using the scaling/affinity laws:
$$$\left ( \frac{\dot{m}}{\rho ND^3} \right )_2= \left ( \frac{\dot{m}}{\rho ND^3} \right )_1$$$
where
$$\dot m$$
is the mass flowrate,
$$\rho$$
is the fluid density,
$$N$$
is the rotational speed, and
$$D$$
is the impeller diameter.
Since fluid density and mass flowrate are constant, the equation becomes:
$$$\left ( \frac{1}{ND^3} \right )_2= \left ( \frac{1}{ND^3} \right )_1$$$
$$$\left ( \frac{1}{N_2D_2^3} \right )= \left ( \frac{1}{N_1D_1^3} \right )$$$
$$$\frac{1}{(1,000rpm-200rpm)D_2^3} = \frac{1}{(1,000rpm)(5m)^3} $$$
$$$\frac{1}{(800)D_2^3} = \frac{1}{(1,000)125m^3} $$$
$$$D_2^3=\frac{125,000m^3}{800}=156.25\:m^3$$$
$$$D_2=(156.25m^3)^{1/3}=5.4\:m$$$
5.4 m
Time Analysis
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