---
title: "Air velocity in a duct by Pitot-static tube"
subject: "Fluid Mechanics"
chapter: 7
key: "semester_4/fluid_mechanics_mechanical:pitot-static-air-duct-numerical"
source: https://www.bctengineer.com/q/semester_4/fluid_mechanics_mechanical/pitot-static-air-duct-numerical
site: BCT Engineering
---

# Air velocity in a duct by Pitot-static tube

Fluid Mechanics · Chapter 7: Flow Measurement

- Asked in: year not given, 5 marks
- Video: [Solved Problem: Measurement of Air Velocity with a Pitot Tube](https://www.youtube.com/watch?v=qqu8gVhoy4g) — Fluid Matters
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_4/fluid_mechanics_mechanical/pitot-static-air-duct-numerical
- Whole chapter: https://www.bctengineer.com/md/semester_4/fluid_mechanics_mechanical/chapter-7

## Question

A Pitot-static tube is placed at the centre of a circular air duct of 400 mm diameter. A water manometer connected to the tube reads 25 mm. The air has a density of 1.2 kg/m$^3$ and the coefficient of the tube is 0.98. Find the velocity at the centre. If the mean velocity is 85% of the centre-line velocity, find the discharge through the duct.

## Short answer

Pressure difference Δp = ρ_water g h = 245.25 Pa. Centre velocity Vc = Cv sqrt(2Δp/ρ_air) = 19.8 m/s. Mean velocity is 0.85 Vc, and discharge Q = A Vmean = 2.12 m³/s.

## Answer

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

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. A water manometer reads 25 mm. What is p0 - p?
   A. 1.2 x 9.81 x 0.025 = 0.29 Pa
   B. 1000 x 9.81 x 0.025 = 245.25 Pa
   C. 1000 x 0.025 = 25 Pa
   D. 1.2 x 0.025 = 0.03 Pa
   Answer: B. p0 - p = rho_w g h_w = 245.25 Pa; the air column weight is negligible. (Key point: Pressure difference from water manometer)

2. Which expression gives the centre velocity?
   A. V = C_v sqrt(2 (p0 - p)/rho_water)
   B. V = sqrt(2 g h_w) with h_w = 0.025
   C. V = C_v sqrt(2 (p0 - p)/rho_air) = 19.8 m/s
   D. V = (p0 - p)/rho_air
   Answer: C. The air density is used because the air is flowing: 0.98 sqrt(2 x 245.25/1.2) = 19.8 m/s. (Key point: Centre-line velocity formula)

3. How is the discharge found from the centre velocity?
   A. Q = A x V_c = 2.49 m^3/s
   B. Q = A x 0.85 V_c = 2.12 m^3/s
   C. Q = 0.85 A x V_c^2
   D. Q = 0.85 V_c x D
   Answer: B. V_mean = 0.85 x 19.81 = 16.84 m/s, A = 0.1257 m^2, Q = 2.12 m^3/s. (Key point: Discharge)

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