A centrifugal pump operating at $N = 1450\text{ rpm}$ delivers a discharge of $Q = 0.04\text{ m}^3/\text{s}$ against a total head of $H = 16.0\text{ m}$. The specific speed ($N_s$) of the pump in SI metric units ($\text{rpm}, \text{m}^3/\text{s}, \text{m}$) is ________ (round off to one decimal place).
Answer Key:
36.2
Step-by-Step Solution & Explanation
The specific speed is defined as:
$N_s = \frac{N \sqrt{Q}}{H^{3/4}}$
Given $N = 1450\text{ rpm}$, $Q = 0.04\text{ m}^3/\text{s}$, and $H = 16.0\text{ m}$ ($16^{0.75} = (16^{0.25})^3 = 2^3 = 8$):
$N_s = \frac{1450 \times \sqrt{0.04}}{8} = \frac{1450 \times 0.2000}{8} = 36.2$