---
title: "Inverted U-tube and inclined manometer use"
subject: "Fluid Mechanics"
chapter: 2
key: "semester_3/fluid_mechanics:inverted-u-tube-single-column-condition"
source: https://www.bctengineer.com/q/semester_3/fluid_mechanics/inverted-u-tube-single-column-condition
site: BCT Engineering
---

# Inverted U-tube and inclined manometer use

Fluid Mechanics · Chapter 2: Pressure and Head

- Asked in: 2078 Kartik, 3 marks
- Video: [Fluid Mechanics - Inverted U-Tube Manometer Contains Oil and Water](https://www.youtube.com/watch?v=XIZib7uzdU8) — R12 Engineering
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_3/fluid_mechanics/inverted-u-tube-single-column-condition
- Whole chapter: https://www.bctengineer.com/md/semester_3/fluid_mechanics/chapter-2

## Question

Under what condition inverted U-tube manometer and single column inclined manometer are used to measure pressure.

## Short answer

Inverted U-tube measures small pressure differences in low-density liquid pipelines using a lighter fluid to amplify height difference. Single-column inclined manometer measures very small gauge pressures (e.g. gas draught) by magnifying the reading via the inclined tube’s 1/sinθ factor.

## Answer

#### Inverted U-tube manometer
Used to measure the **small pressure difference between two points in a pipeline carrying a low-density liquid such as water**. The space above the two liquid columns is filled with a lighter fluid (air or oil), so a small pressure difference gives a large, easily read difference in column heights:

$$
p_A-p_B=(\gamma_w-\gamma_m)\,h
$$

where $\gamma_m$ is the specific weight of the light fluid (A and B at the same level). It is chosen when the fluids are liquids of low density and the pressure difference is very small.

#### Single-column inclined manometer
Used to measure **very small gauge pressures, mostly of gases (low draught pressures)**, where ordinary manometer readings would be too small to read accurately. One limb is a large reservoir and the other a narrow tube inclined at angle $\theta$ to the horizontal. The liquid moves a length $L$ along the incline for a vertical rise $L\sin\theta$, so the scale is magnified by $1/\sin\theta$. The reservoir level falls by $L\,a/A$, so

$$
p=\gamma\,L\left(\sin\theta+\frac{a}{A}\right)\approx\gamma L\sin\theta
$$

where $a/A$ is the ratio of the tube area to the reservoir area. It is therefore preferred where high sensitivity is needed.

## Check yourself

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

1. When is an inverted U-tube manometer used?
   A. Measuring high gas pressures
   B. Measuring small pressure differences in a light liquid like water
   C. Measuring atmospheric pressure
   D. Measuring very high liquid pressures
   Answer: B. A lighter fluid above the columns magnifies small pressure differences in liquid lines. (Key point: Inverted U-tube use)

2. What is the pressure difference in an inverted U-tube?
   A. (gamma_w + gamma_m) h
   B. gamma_w h
   C. gamma_m h
   D. (gamma_w - gamma_m) h
   Answer: D. p_A - p_B = (gamma_w - gamma_m) h, so a small difference gives a large h when gamma_m is small. (Key point: Inverted U-tube formula)

3. Why does the single-column inclined manometer suit very small gas pressures?
   A. Length along the incline L = vertical rise / sin(theta), magnifying the scale
   B. It uses mercury
   C. The tube is wider
   D. It reads absolute pressure
   Answer: A. A small vertical rise L sin(theta) appears as a longer movement L, so the scale is magnified by 1/sin(theta). (Key point: Inclined manometer magnification)

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