The discharge \( Q = 0.02 \, \text{m}^3/\text{s} \).
The diameter of the pipe \( D = 0.15 \, \text{m} \).
The cross-sectional area \( A = \frac{\pi D^2}{4} \).
\[ A = \frac{\pi (0.15)^2}{4} = 0.01767 \, \text{m}^2 \]
The velocity \( V \) is given by:
\[ V =
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