---
title: "Differential U-tube manometer between two pipes"
subject: "Fluid Mechanics"
chapter: 2
key: "semester_4/fluid_mechanics_mechanical:differential-manometer-pipes"
source: https://www.bctengineer.com/q/semester_4/fluid_mechanics_mechanical/differential-manometer-pipes
site: BCT Engineering
---

# Differential U-tube manometer between two pipes

Fluid Mechanics · Chapter 2: Fluid Statics

- Asked in: year not given, 5 marks
- Video: [Fluid Mechanics - Two Pipes are Connected by a Manometer](https://www.youtube.com/watch?v=uxWogQ2KB9Q) — R12 Engineering
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_4/fluid_mechanics_mechanical/differential-manometer-pipes
- Whole chapter: https://www.bctengineer.com/md/semester_4/fluid_mechanics_mechanical/chapter-2

## Question

Two pipes A and B carry water. The centre of pipe B is 0.5 m above the centre of pipe A. They are connected by an U-tube differential manometer (mercury in the bottom of the U-tube) containing mercury (specific gravity 13.6), the rest of the connecting tubes being filled with water. The mercury level in the limb connected to B is 0.3 m higher than in the limb connected to A. Find the pressure difference $p_A - p_B$, and the gauge pressure in pipe A if the gauge pressure in pipe B is 50 kPa.

## Short answer

Balancing pressures at the mercury datum level gives p_A – p_B = ρ_w g z + (ρ_m – ρ_w) g h. Substituting yields 42.0 kPa. With p_B=50 kPa gauge, p_A = 92.0 kPa gauge.

## Answer

#### Setup
Let the mercury surface in the A limb be at depth $a$ below the centre of A. This is the datum plane. The mercury in the B limb stands $h = 0.3\ \text{m}$ higher, and B is $z = 0.5\ \text{m}$ above A.

```text
                   o B
   o A             |
   |               | water
   | water      ===+=== Hg, B limb
   |               |   ^ h
 ==+== Hg, A limb  |   v
   |  datum -------|---
   +---- mercury --+
```

#### Pressure balance at the datum
Pressure at the datum, from the A side:

$$
p_A + \rho_w g a
$$

From the B side, the water column above the mercury is $(a + z - h)$ and the mercury column is $h$:

$$
p_B + \rho_w g (a + z - h) + \rho_m g h
$$

Equating and simplifying:

$$
\begin{aligned}
p_A - p_B &= \rho_w g (z - h) + \rho_m g h \\
&= \rho_w g z + (\rho_m - \rho_w) g h
\end{aligned}
$$

#### Substitution
$$
\begin{aligned}
p_A - p_B &= 1000 \times 9.81 \times 0.5 + (13600 - 1000) \times 9.81 \times 0.3 \\
&= 4905 + 37081.8 \\
&= 41\,986.8\ \text{Pa} \approx 41.99\ \text{kPa}
\end{aligned}
$$

#### Pressure in A
$$
p_A = 50 + 41.99 = 91.99\ \text{kPa (gauge)}
$$

**Answer: $p_A - p_B = 41.99\ \text{kPa}$; $p_A = 91.99\ \text{kPa}$ (gauge)**

## Check yourself

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

1. How is the manometer equation set up?
   A. Equate pressures at the same level in the same fluid (mercury surface in the lower limb) computed from both sides
   B. Add the two pipe pressures
   C. Equate the heights of the two mercury surfaces
   D. Ignore the water columns above the mercury
   Answer: A. The pressure at the datum plane is computed down the A side and down the B side and equated. (Key point: Datum method)

2. Which expression gives p_A - p_B (B is z above A, mercury difference h)?
   A. rho_w g z + (rho_m - rho_w) g h
   B. rho_w g z - (rho_m - rho_w) g h
   C. rho_m g z + rho_w g h
   D. (rho_m - rho_w) g z
   Answer: A. Equating the datum pressures gives p_A - p_B = rho_w g (z - h) + rho_m g h = rho_w g z + (rho_m - rho_w) g h. (Key point: Manometer result formula)

3. With z = 0.5 m, h = 0.3 m and p_B = 50 kPa gauge, the gauge pressure in A is about:
   A. 8.0 kPa
   B. 42.0 kPa
   C. 92.0 kPa
   D. 141.9 kPa
   Answer: C. p_A - p_B = 4.905 + 37.08 = 41.99 kPa, so p_A = 50 + 41.99 = 91.99 kPa. (Key point: Numerical result)

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