---
title: "Double U-tube manometer, oil and brine pipes"
subject: "Fluid Mechanics"
chapter: 2
key: "semester_3/fluid_mechanics:double-u-tube-oil-brine-pressure"
source: https://www.bctengineer.com/q/semester_3/fluid_mechanics/double-u-tube-oil-brine-pressure
site: BCT Engineering
---

# Double U-tube manometer, oil and brine pipes

Fluid Mechanics · Chapter 2: Pressure and Head

- Asked in: 2079 Baisakh, 5+5 marks
- Video: [Thermodynamics - Test 1 Problem 1 - Multifluid manometer](https://www.youtube.com/watch?v=P6Hqdv12iFc) — Engineering Deciphered
- Page: https://www.bctengineer.com/q/semester_3/fluid_mechanics/double-u-tube-oil-brine-pressure
- Whole chapter: https://www.bctengineer.com/md/semester_3/fluid_mechanics/chapter-2

## Question

The two pipes are connected by a double U-tube manometer as shown in figure where the brine pipe is connected to a tank filled with different fluids. Oil and brine are flowing in parallel horizontal pipes. The pressure at the centre of oil pipe is 200 kPa. Calculate pressures at point 2 & 3. [Figure: oil pipe (S.G. = 0.85) on the left connected through a mercury (S.G. = 13.6) U-tube and air (S.G. = 0.001) segments to the brine pipe (S.G. [?]) at point 2; the brine pipe is connected by a second U-tube to a closed tank with air (S.G. = 0.001) at the top (point 3) and layers of water and mercury (S.G. = 13.6) below, each layer 30 cm; marked heights include 20 cm, 50 cm, 60 cm, 50 cm, 10 cm, 30 cm, 10 cm, 30 cm]

## Short answer

Stepwise manometer (add γh downward, subtract upward) using 20 cm oil, 60 cm Hg, 50 cm air, 10 cm brine gives p₂ ≈ 120.5 kPa. From p₂ through 50 cm brine, 10+30 cm Hg, 30 cm water yields p₃ ≈ 69.6 kPa (gauge). Assumed brine SG = 1.1.

## Answer

The figure does not give every number legibly, so the reading used is stated below. Pressures are in gauge kPa, $g=9.81\ \text{m/s}^2$, and moving **down** a fluid adds $\gamma h$, moving **up** subtracts $\gamma h$.

**Specific weights:** oil $\gamma_o=0.85\times9.81=8.34$; mercury $\gamma_{Hg}=13.6\times9.81=133.4$; brine (assumed S.G. 1.1) $\gamma_b=10.79$; air $\gamma_a=0.00981$; water $\gamma_w=9.81$ kN/m³.

**Assumed reading (heights from the figure):**

- pipe 1 (oil) $\to$ down 20 cm in oil $\to$ up 60 cm in mercury $\to$ down 50 cm in air $\to$ up 10 cm in brine to pipe 2 (net rise = 0, since both pipe centres are at the same level)
- pipe 2 $\to$ down 50 cm in brine $\to$ up 10 cm in mercury $\to$ up 30 cm in mercury and 30 cm in water inside the tank, to the air at point 3

```text
 (1) oil pipe          (2) brine pipe
   |                      |
   oil 20cm  Hg 60cm  air 50cm  brine 10cm
```

#### Pressure at point 2
$$
\begin{aligned}
p_2&=p_1+\gamma_o(0.20)-\gamma_{Hg}(0.60)+\gamma_a(0.50)-\gamma_b(0.10)\\
&=200+1.668-80.05+0.005-1.079\\
&=120.54\ \text{kPa}
\end{aligned}
$$

#### Pressure at point 3
$$
\begin{aligned}
p_3&=p_2+\gamma_b(0.50)-\gamma_{Hg}(0.10)-\gamma_{Hg}(0.30)-\gamma_w(0.30)\\
&=120.54+5.395-13.34-40.02-2.943\\
&=69.63\ \text{kPa}
\end{aligned}
$$

**Answer: $p_2\approx120.5$ kPa and $p_3\approx69.6$ kPa (gauge).** For any other reading of the figure, use the same step-by-step rule with the correct heights and brine S.G.

## Check yourself

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

1. In the step-by-step manometer rule, what happens moving down in a fluid?
   A. Pressure decreases by gamma h
   B. Pressure is unchanged
   C. Pressure increases by gamma h
   D. Pressure doubles
   Answer: C. Moving down a fluid adds gamma h; moving up subtracts gamma h. (Key point: Sign rule)

2. Why does p2 fall far below p1 (120.5 vs 200 kPa)?
   A. Oil is heavier than brine
   B. The 60 cm rise through mercury subtracts a large term, about 80 kPa
   C. Air adds a large pressure
   D. The pipes are at different levels
   Answer: B. Mercury has gamma = 133.4 kN/m^3, so rising 0.6 m subtracts about 80 kPa. (Key point: Mercury leg effect)

3. Why is the air segment contribution (0.005 kPa) almost negligible?
   A. Air column height is zero
   B. Air has a very small specific weight
   C. Air pressure is always zero
   D. It cancels with oil
   Answer: B. Air has S.G. 0.001, so gamma h over 0.5 m is tiny. (Key point: Negligible air term)

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