Discussion
A pitot static tube placed in the center of a 200 mm pipe line, has one orifice pointing upstream and the other perpendicular to it. If the pressure difference between the two orifices is 40 mm of water when the discharge through the pipe is 1365 liters/min, then the coefficient of the Pitot tube is:
(Take the mean velocity in the pipe to be 0.83 of the central velocity)
Answer: A
The coefficient of the Pitot tube is C = 0.984.
First find the mean velocity from the discharge, then the centre velocity:
\[ \begin{aligned} Q &= 1365/60000 = 0.02275\ \text{m}^3/\text{s} \\ A &= \pi (0.1)^2 = 0.03142\ \text{m}^2 \\ V_m &= 0.02275/0.03142 = 0.724\ \text{m/s} \\ V_c &= 0.724/0.83 = 0.872\ \text{m/s} \end{aligned} \]
The Pitot tube gives \(V_c = C\sqrt{2gh}\) with \(h = 0.04\) m of water:
\[ \begin{aligned} \sqrt{2 \times 9.81 \times 0.04} &= 0.886 \\ C &= 0.872/0.886 = 0.984 \end{aligned} \]
Using the mean velocity instead of the centre velocity gives about 0.82, which is not offered; the other options match no step.
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