Air velocity in a duct by Pitot-static tube

Given data

$h_w = 0.025\ \text{m}$ of water, $\rho_a = 1.2\ \text{kg/m}^3$, $\rho_w = 1000\ \text{kg/m}^3$, $C_v = 0.98$, $D = 0.4\ \text{m}$.

Pressure difference (velocity pressure)

The manometer reads the difference between stagnation and static pressures:

$$p_0 - p = \rho_w g h_w = 1000 \times 9.81 \times 0.025 = 245.25\ \text{Pa}$$

(The weight of the air column is negligible compared with water.)

Centre-line velocity

$$\begin{aligned} V_c &= C_v\sqrt{\frac{2(p_0 - p)}{\rho_a}} \\ &= 0.98\sqrt{\frac{2 \times 245.25}{1.2}} \\ &= 0.98\sqrt{408.75} = 0.98 \times 20.22 \\ &= 19.81\ \text{m/s} \end{aligned}$$

Mean velocity and discharge

$$\begin{aligned} V_{mean} &= 0.85 \times 19.81 = 16.84\ \text{m/s} \\ A &= \frac{\pi}{4}(0.4)^2 = 0.1257\ \text{m}^2 \\ Q &= A V_{mean} = 0.1257 \times 16.84 = 2.12\ \text{m}^3/\text{s} \end{aligned}$$

Answer: $V_c = 19.8\ \text{m/s}$; $Q = 2.12\ \text{m}^3/\text{s}$