---
title: "Pressure difference using compound manometer"
subject: "Hydro Power"
chapter: 1
key: "semester_6/hydro_power:manometer-water-oil-mercury"
source: https://www.bctengineer.com/q/semester_6/hydro_power/manometer-water-oil-mercury
site: BCT Engineering
---

# Pressure difference using compound manometer

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

- Asked in: 2075 Bhadra, 5 marks
- Video: [Differential Manometer Numerical | PU 2015 Qn2b | Fluid Mechanics | BE Civil |3rd Sem| Prashant YT |](https://www.youtube.com/watch?v=_-b5tFbv-D4) — Prashant YT
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_6/hydro_power/manometer-water-oil-mercury
- Whole chapter: https://www.bctengineer.com/md/semester_6/hydro_power/chapter-1

## Question

Calculate the pressure difference between the points A and B. Specific gravity of oil is 0.8. [Figure: pipes A and B both contain water and are connected by two mercury U-tubes joined at the top by an inverted U-tube containing oil; marked heights are 90 cm (centre of A above the bottom datum), 60 cm and 40 cm (mercury columns on the left), 50 cm (mercury column on the right) and 80 cm (centre of B above the right mercury level).]

## Short answer

Go from A to B through each column: add ρgh going down, subtract going up. The pressure difference is pA − pB ≈ 41.4 kN/m², with A at higher pressure.

## Answer

Go from A to B through each column: add $\rho g h$ going down, subtract going up.

**Assumed reading of the figure (heights above the bottom datum):** centre of A = 0.90 m. In the left mercury U-tube, the mercury surface is at 0.40 m in A's limb and 0.60 m in the other limb. Oil fills the inverted U-tube, down to the right U-tube's mercury surface, taken at 0.40 m. The right U-tube's mercury in B's limb is at 0.50 m. B is 0.80 m above that, i.e. at 1.30 m.

**Given:** $S_{oil} = 0.8$, $S_{Hg} = 13.6$, water $S = 1$, $\rho_w g = 9.81$ kN/m³.

| Step | Column | Height (m) | Pressure change (kN/m²) |
|---|---|---|---|
| A down to Hg (left) | water | 0.50 | $+9.81(0.50) = +4.905$ |
| Up in Hg to 0.60 | mercury | 0.20 | $-13.6(9.81)(0.20) = -26.683$ |
| Down through oil to 0.40 | oil | 0.20 | $+0.8(9.81)(0.20) = +1.570$ |
| Up in Hg to 0.50 | mercury | 0.10 | $-13.6(9.81)(0.10) = -13.342$ |
| Up in water to B | water | 0.80 | $-9.81(0.80) = -7.848$ |

#### Manometer equation
$$
p_A + 4.905 - 26.683 + 1.570 - 13.342 - 7.848 = p_B
$$
$$
p_A - p_B = 41.398\ \text{kN/m}^2
$$

**Answer:** $p_A - p_B \approx 41.4$ kN/m² (A at higher pressure), for the heights read as above. The method is the same for any other placement of the marked heights.

## Check yourself

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

1. In the A-to-B traverse, a column traversed upward contributes:
   A. +rho g h
   B. -rho g h
   C. 0
   D. +2 rho g h
   Answer: B. Pressure decreases going up and increases going down. (Key point: Method)

2. Why is the oil column term positive (+1.570)?
   A. The path goes down through the oil
   B. Oil is heavier than mercury
   C. The path goes up through the oil
   D. Oil has no weight
   Answer: A. The path descends 0.20 m through oil in the inverted U, adding 0.8 x 9.81 x 0.20 = 1.570. (Key point: Oil section)

3. The two mercury rises contribute:
   A. -26.683 and -13.342 kN/m^2
   B. +26.683 and +13.342
   C. -2.68 and -1.33
   D. -26.683 and +13.342
   Answer: A. Going up 0.20 m and 0.10 m of mercury gives -13.6(9.81)(0.20) and -13.6(9.81)(0.10). (Key point: Mercury terms)

4. The pressure difference p_A - p_B is about:
   A. 41.4 kN/m^2
   B. 12.4 kN/m^2
   C. 80.6 kN/m^2
   D. 4.14 kN/m^2
   Answer: A. 4.905 - 26.683 + 1.570 - 13.342 - 7.848 = -41.398, so p_A - p_B = 41.4 kN/m^2. (Key point: Result)

## Related in Hydro Power

- [U-tube manometer with oil, mercury and water](https://www.bctengineer.com/question.md?key=semester_6%2Fhydro_power%3Au-tube-oil-mercury-water) — Hydro Power, chapter 1
- [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
- [Differential manometer between CCl4 and oil pipes](https://www.bctengineer.com/question.md?key=semester_6%2Fhydro_power%3Amanometer-ccl4-oil-pipes) — Hydro Power, chapter 1
- [Differential manometer pressure at point B](https://www.bctengineer.com/question.md?key=semester_6%2Fhydro_power%3Amanometer-air-water-oil) — Hydro Power, chapter 1
- [Pressure difference between containers A and B](https://www.bctengineer.com/question.md?key=semester_6%2Fhydro_power%3Amanometer-inclined-containers) — Hydro Power, chapter 1
- [Use of manometer and inverted manometer](https://www.bctengineer.com/question.md?key=semester_6%2Fhydro_power%3Ainverted-manometer) — Hydro Power, chapter 1

## Related in other subjects

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