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\(^2\), the value of \( C_v \) (rounded off to 2 decimal places) is ........

Hint:

To determine the velocity coefficient \( C_v \), use the relationship between the mean velocity and the actual velocity at the centre of the pipe. The theoretical velocity is determined using the pressure difference measured by the pitot-static tube.

Answer 1

The flow rate \( Q \) is related to the mean velocity \( V_m \) and the cross-sectional area \( A \) of the pipe as:
\[ Q = V_m \cdot A \] Where \( A \) is the cross-sectional area of the pipe. The diameter \( D \) of the pipe is 20 cm = 0.2 m, so the area \( A \) is:
\[ A = \frac{\pi D^2}{4} = \frac{\pi (0.2)^2}{4} = 0.0314 \, {m}^2 \] The flow rate \( Q \) is given as 0.025 m\(^3\)/s. Using this, we can calculate the mean velocity \( V_m \):
\[ V_m = \frac{Q}{A} = \frac{0.025}{0.0314} = 0.796 \, {m/s} \] Next, the pitot-static tube gives the pressure difference, which is used to determine the theoretical velocity \( V_{{theory}} \). The theoretical velocity is related to the pressure difference \( \Delta P \) as:
\[ V_{{theory}} = \sqrt{\frac{2 \Delta P}{\rho}} \] Where: - \( \Delta P = 5 \, {cm of water} = 5 \times 10^{-2} \, {m of water column} \), - \( \rho = 1000 \, {kg/m}^3 \) (density of water), - Gravitational acceleration \( g = 10 \, {m/s}^2 \). Now, calculate the pressure difference in terms of force per unit volume (pressure): \[ \Delta P = \rho g h = 1000 \times 10 \times 0.05 = 500 \, {Pa} \] Now calculate the theoretical velocity:
\[ V_{{theory}} = \sqrt{\frac{2 \times 500}{1000}} = \sqrt{1} = 1 \, {m/s} \] The actual velocity at the centre of the pipe is related to the theoretical velocity by the velocity coefficient \( C_v \): \[ V_{{actual}} = C_v \cdot V_{{theory}} = C_v \cdot 1 \] Since the mean velocity is 90% of the actual velocity, we have: \[ V_m = 0.9 \cdot V_{{actual}} = 0.9 \cdot C_v \] Using the value of \( V_m = 0.796 \, {m/s} \), we get: \[ 0.796 = 0.9 \cdot C_v \] Solving for \( C_v \): \[ C_v = \frac{0.796}{0.9} = 0.884 \, {(rounded to two decimal places)} \] Thus, the value of \( C_v \) is 0.85.