---
title: "Pressure and velocity measurement devices"
subject: "Fluid Mechanics"
chapter: 7
key: "semester_4/fluid_mechanics_mechanical:static-pressure-and-pitot-tube"
source: https://www.bctengineer.com/q/semester_4/fluid_mechanics_mechanical/static-pressure-and-pitot-tube
site: BCT Engineering
---

# Pressure and velocity measurement devices

Fluid Mechanics · Chapter 7: Flow Measurement

- Asked in: year not given, 6 marks
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_4/fluid_mechanics_mechanical/static-pressure-and-pitot-tube
- Whole chapter: https://www.bctengineer.com/md/semester_4/fluid_mechanics_mechanical/chapter-7

## Question

Describe the methods of measuring static pressure in a flowing fluid. With a neat sketch, explain the working of a Pitot tube and a Pitot-static tube, and derive the expression for velocity of flow.

## Short answer

Static pressure is measured via wall taps, piezometer tubes, or static probes. A Pitot tube senses stagnation pressure; Bernoulli gives V = sqrt(2gh). A Pitot-static tube combines stagnation and static ports, yielding V = Cv sqrt(2(p0-p)/ρ).

## Answer

#### Measurement of static pressure
- **Wall tap / piezometer:** a small hole drilled flush with and normal to the wall, connected to a manometer or gauge. It reads static pressure because the streamlines at the wall run parallel to the wall, so a small flush hole does not disturb the flow. The hole must be small and burr-free.
- **Piezometer tube:** a vertical open tube connected to the wall tap; the liquid height gives the gauge pressure head. It is suitable only for moderate positive pressures in liquids.
- **Static probe:** a slender probe aligned with the flow, with small holes on its side, used inside the stream.
- **Manometers and Bourdon gauges:** read the pressure transmitted from the tap.

#### Pitot tube
A Pitot tube is a tube bent at right angles, with its open end facing the flow. The fluid at the tip is brought to rest (**stagnation point**), where its kinetic energy is converted into pressure.

Applying Bernoulli's equation between a point 1 upstream, where the velocity is $V$ and the pressure is $p$, and the stagnation point 0, where the velocity is zero:

$$
\frac{p}{\rho g} + \frac{V^2}{2g} = \frac{p_0}{\rho g}
$$

So the rise of liquid in the tube above the static level equals the velocity head:

$$
h = \frac{V^2}{2g} \quad \Rightarrow \quad V = \sqrt{2gh}
$$

In practice a coefficient $C_v$ (0.98 to 1.0) is included: $V = C_v\sqrt{2gh}$.

#### Pitot-static tube
It combines the Pitot tube (stagnation pressure) and a concentric outer tube with side holes (static pressure) in one probe, and connects both to a differential manometer.

```text
                  stagnation hole
                        |
   ---------------------+-----> flow
   ===========[inner tube]=====> p0  to manometer
   ==[outer tube, side holes]==> p   to manometer
                   o  o  static holes
```

$$
p_0 - p = \tfrac12\rho V^2 \quad \Rightarrow \quad V = C_v\sqrt{\frac{2(p_0 - p)}{\rho}}
$$

For a differential manometer with reading $h_m$ of a fluid of density $\rho_m$, $p_0 - p = (\rho_m - \rho)g h_m$. Because the instrument gives the velocity at a point, the mean velocity in a pipe is found by traversing the cross-section.

## Check yourself

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

1. Why does a wall tap read static pressure correctly?
   A. It faces the flow and stops it
   B. The small burr-free hole is flush and normal to the wall, so streamlines are parallel and undisturbed
   C. It averages the pressure over the section
   D. It measures velocity head
   Answer: B. Parallel streamlines at the wall mean the hole senses only static pressure, provided it is small and burr-free. (Key point: Wall tap)

2. What does the Pitot tube open end facing the flow measure, and what velocity formula follows?
   A. Static pressure; V = 2gh
   B. Stagnation pressure; V = sqrt(2gh) where h is the rise above static level
   C. Dynamic viscosity; V = sqrt(gh)
   D. Stagnation pressure; V = gh
   Answer: B. Fluid is brought to rest at the tip; Bernoulli gives h = V^2/2g, so V = sqrt(2gh) (times C_v in practice). (Key point: Pitot tube principle)

3. For a Pitot-static tube with differential manometer, which expression is correct?
   A. p0 - p = rho V^2 and V is mean velocity
   B. p0 - p = (1/2) rho V^2, giving point velocity V = C_v sqrt(2(p0 - p)/rho)
   C. p0 + p = (1/2) rho V^2
   D. p0 - p = rho g h with no velocity link
   Answer: B. The difference between stagnation and static pressure equals the dynamic pressure; it gives a point velocity, so the pipe must be traversed for the mean. (Key point: Pitot-static tube)

## Related in Fluid Mechanics

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- [Pressure variation in a static fluid](https://www.bctengineer.com/question.md?key=semester_4%2Ffluid_mechanics_mechanical%3Apressure-variation-static-fluid) — Fluid Mechanics, chapter 2
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## Related in other subjects

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