Question

Water flows through a pipe of diameter 20 cm at a flow rate of 0.025 m$^3$/s. A pitot-static tube is placed at the centre of the pipe and indicates the pressure difference of 5 cm of water column. Theoretical velocity measured through pitot-static tube when multiplied with velocity coefficient $C_V$ gives the actual velocity of the flow. If the mean velocity in the pipe is 90% of the actual velocity at the centre of the pipe and the gravitational acceleration is 10 m/s$^2$, find $C_V$ (rounded off to 2 decimal places).

Hint:

For Pitot tube calibration: actual velocity is obtained by multiplying theoretical velocity with $C_V$. If mean velocity is related to actual centreline velocity, use continuity to link flow rate and then adjust.

Answer 1

Correct Answer: 0.85

Step 1: Pipe flow mean velocity.
\[ Q = 0.025 \, \text{m}^3/s, D = 0.2 \, \text{m}. \] Cross-sectional area: \[ A = \frac{\pi D^2}{4} = \frac{\pi (0.2)^2}{4} = 0.031416 \, \text{m}^2. \] Mean velocity: \[ V_{mean} = \frac{Q}{A} = \frac{0.025}{0.031416} = 0.796 \,\text{m/s}. \]

Step 2: Relation between mean and actual velocity.
\[ V_{mean} = 0.9 V_{actual}. \] So, \[ V_{actual} = \frac{0.796}{0.9} = 0.884 \,\text{m/s}. \]

Step 3: Theoretical velocity from pitot tube.
Differential head: $h = 0.05 \, \text{m}$, $g = 10$. \[ V_{theo} = \sqrt{2 g h} = \sqrt{2 \times 10 \times 0.05} = \sqrt{1} = 1 \,\text{m/s}. \]

Step 4: Velocity coefficient.
\[ C_V = \frac{V_{actual}}{V_{theo}} = \frac{0.884}{1} = 0.884. \] Rounded: \[ C_V = 0.88. \] Final Answer:
\[ \boxed{0.88} \]