---
title: "Use of manometer and inverted manometer"
subject: "Hydro Power"
chapter: 1
key: "semester_6/hydro_power:inverted-manometer"
source: https://www.bctengineer.com/q/semester_6/hydro_power/inverted-manometer
site: BCT Engineering
---

# Use of manometer and inverted manometer

Hydro Power · Chapter 1: Introduction to Basic Fluid Mechanics and Hydraulics

- Asked in: 2070 Bhadra, 2+4 marks
- Video: [Pressure Measurement Manometers](https://www.youtube.com/watch?v=oXXGIFSmZRE) — TutorialsPoint
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_6/hydro_power/inverted-manometer
- Whole chapter: https://www.bctengineer.com/md/semester_6/hydro_power/chapter-1

## Question

What is the use of manometer? Show the working principle of inverted manometer with neat sketch.

## Short answer

An inverted U-tube manometer measures small pressure differences using a light fluid at the top. Pressure difference is pA − pB = ρ1 g h1 − ρ2 g h2 − ρ_s g h. With air on top the light-liquid term is negligible, giving precise readings for low differentials.

## Answer

#### Use of a manometer
A **manometer** measures pressure at a point (simple manometer) or the difference of pressure between two points (differential manometer) by balancing it against a column of liquid. Used for pipes, venturimeters, orifice meters and laboratory flow measurement; it is simple, cheap and accurate.

#### Inverted U-tube differential manometer
It is an inverted U-tube whose top part contains a **light** fluid (air or a light oil) lighter than the liquid in the pipes. It is used to measure **small pressure differences**, where a heavy liquid like mercury would give too small a reading.

```text
        .---------------.
        | light liquid  |  (air or oil, rho_s)
  X ----+------.        |
        |      |  h     |
        |      '--------+---- right interface
   h_1  | liquid        |
        | rho_1         | h_2   liquid rho_2
        o A             |
                        o B
```
Let X–X be the left (higher) interface, A be $h_1$ below it, the right interface be $h$ lower than X–X, and B be $h_2$ below the right interface.

#### Working principle
Pressure at level X–X in the left limb:
$$
p_A - \rho_1 g h_1
$$
Pressure at level X–X in the right limb (go up $h_2$ of pipe liquid from B, then up $h$ of light liquid):
$$
p_B - \rho_2 g h_2 - \rho_s g h
$$
Equating:
$$
p_A - p_B = \rho_1 g h_1 - \rho_2 g h_2 - \rho_s g h
$$
If air is used at the top, $\rho_s g h \approx 0$.

**Example:** water in both pipes, oil $S = 0.8$ at top, $h_1 = 0.30$ m, $h_2 = 0.20$ m, $h = 0.12$ m: $p_A - p_B = 9.81(0.30 - 0.20 - 0.8\times0.12) = 0.039$ kN/m², a very small difference that an inverted manometer can read well.

## Check yourself

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

1. A differential manometer measures:
   A. Pressure at a point only
   B. The difference of pressure between two points by balancing it against a liquid column
   C. Velocity directly
   D. Discharge directly
   Answer: B. Simple manometers measure pressure at a point; differential ones measure the difference between two points. (Key point: Use of manometer)

2. The top fluid of an inverted U-tube manometer must be:
   A. Heavier than the pipe liquid
   B. Lighter than the pipe liquid (air or light oil)
   C. Mercury
   D. Same as the pipe liquid
   Answer: B. A light fluid is used so small pressure differences give a measurable reading. (Key point: Inverted manometer fluid)

3. The inverted manometer equation is:
   A. p_A - p_B = rho_1 g h_1 - rho_2 g h_2 - rho_s g h
   B. p_A - p_B = rho_1 g h_1 + rho_2 g h_2 + rho_s g h
   C. p_A - p_B = rho_s g h
   D. p_A - p_B = rho_2 g h_2 - rho_1 g h_1
   Answer: A. Equating pressures at X-X gives p_A - rho_1 g h_1 = p_B - rho_2 g h_2 - rho_s g h. (Key point: Equation)

4. If air is the top fluid, the term rho_s g h is:
   A. Large and must be kept
   B. Approximately zero and neglected
   C. Negative
   D. Equal to p_B
   Answer: B. Air density is very small, so rho_s g h is approximately 0. (Key point: Air at top)

## Related in Hydro Power

- [Working principle of centrifugal and reciprocating pumps](https://www.bctengineer.com/question.md?key=semester_6%2Fhydro_power%3Acentrifugal-reciprocating-working) — Hydro Power, chapter 7
- [Manometer between oil and water pipes](https://www.bctengineer.com/question.md?key=semester_6%2Fhydro_power%3Amanometer-oil-water-pipes) — Hydro Power, chapter 1
- [Pressure difference using compound manometer](https://www.bctengineer.com/question.md?key=semester_6%2Fhydro_power%3Amanometer-water-oil-mercury) — Hydro Power, chapter 1

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