---
title: "Differential manometer pressure at point B"
subject: "Hydro Power"
chapter: 1
key: "semester_6/hydro_power:manometer-air-water-oil"
source: https://www.bctengineer.com/q/semester_6/hydro_power/manometer-air-water-oil
site: BCT Engineering
---

# Differential manometer pressure at point B

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

- Asked in: 2082 Kartik, 5 marks
- Video: [Differential manometer numerical | Pressure difference calculation | 2015 PU | Full concept](https://www.youtube.com/watch?v=Ls3aPFLLa_0) — Sudip Dhakal
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_6/hydro_power/manometer-air-water-oil
- Whole chapter: https://www.bctengineer.com/md/semester_6/hydro_power/chapter-1

## Question

A differential manometer is connected to points A and B, as shown in the figure below. Given that the air pressure at A is 60 kN/m², determine the pressure at point B. [Figure: pipe A contains air above water (S₁ = 1) and pipe B contains oil (S₂ = 0.9); they are connected by a U-tube with mercury (S = 13.6) in its bottom; the vertical distances marked on the figure are 200 mm, 550 mm and 100 mm, measured relative to the datum X–X at the mercury surface.]

## Short answer

Pressure at X–X from left: 60 + 9.81×0.2 = 61.962 kN/m². Right: p_B + 0.9×9.81×0.45 + 13.6×9.81×0.1. Equating: p_B = 61.962 − 17.315 = 44.65 kN/m².

## Answer

Write the pressure balance at the datum X–X (mercury surface in the left limb), moving through each fluid column.

**Assumed reading of the figure:** the air–water surface on the A side is 200 mm above X–X. On the B side, mercury stands 100 mm above X–X, and B is 550 mm above X–X, so the oil column is $550 - 100 = 450$ mm.

**Given:** $p_A = 60$ kN/m² (air; its weight is neglected), water $S_1 = 1$, oil $S_2 = 0.9$, mercury $S = 13.6$, $\rho_w g = 9.81$ kN/m³.

```text
   A (air)               B (oil)
   |~~~~~| 200 mm water    |  450 mm oil
   |     |                 |------ Hg top (+100)
 X-|-----|------------ X   | 100 mm Hg
   |__ mercury ____________|
```

#### Pressure at X–X, left limb
$$
p_L = p_A + \rho_w g(0.20) = 60 + 9.81(0.20) = 60 + 1.962 = 61.962\ \text{kN/m}^2
$$

#### Pressure at X–X, right limb
$$
p_R = p_B + 0.9(9.81)(0.45) + 13.6(9.81)(0.10) = p_B + 3.973 + 13.342
$$

#### Equate
$$
\begin{aligned}
61.962 &= p_B + 17.315 \\
p_B &= 44.647\ \text{kN/m}^2
\end{aligned}
$$

**Answer:** $p_B \approx 44.65$ kN/m² (gauge), i.e. about 15.35 kN/m² less than at A. (If the figure's heights are placed differently, use the same column-by-column balance with those heights.)

## Check yourself

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

1. At X-X in the left limb, pressure is p_A plus:
   A. The water column 0.20 m: rho_w g (0.20) = 1.962 kN/m^2
   B. The air column weight, which is large
   C. The oil column
   D. Nothing
   Answer: A. Air weight is neglected; the 200 mm water column adds 9.81 x 0.2 = 1.962 kN/m^2, giving 61.962. (Key point: Left-limb pressure)

2. On the right limb, the oil column length used is 550 - 100 = 450 mm because:
   A. Oil lies only above the mercury top, which is 100 mm above X-X
   B. The oil is lighter
   C. Mercury is 550 mm deep
   D. The datum is at B
   Answer: A. B is 550 mm above X-X and mercury top is 100 mm above X-X, leaving a 450 mm oil column, plus 100 mm mercury. (Key point: Right-limb columns)

3. The right-limb pressure at X-X is:
   A. p_B + 0.9(9.81)(0.45) + 13.6(9.81)(0.10)
   B. p_B - 0.9(9.81)(0.45) - 13.6(9.81)(0.10)
   C. p_B + 13.6(9.81)(0.45)
   D. p_B + 0.9(9.81)(0.55)
   Answer: A. Going down from B adds oil head (3.973) and mercury head (13.342), totalling 17.315. (Key point: Equation)

4. Equating gives p_B of about:
   A. 44.65 kN/m^2
   B. 61.96 kN/m^2
   C. 17.3 kN/m^2
   D. 79.3 kN/m^2
   Answer: A. p_B = 61.962 - 17.315 = 44.647 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
- [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
- [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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