---
title: "Pressure difference between containers A and B"
subject: "Hydro Power"
chapter: 1
key: "semester_6/hydro_power:manometer-inclined-containers"
source: https://www.bctengineer.com/q/semester_6/hydro_power/manometer-inclined-containers
site: BCT Engineering
---

# Pressure difference between containers A and B

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

- Asked in: 2071 Magh, 7 marks
- Page: https://www.bctengineer.com/q/semester_6/hydro_power/manometer-inclined-containers
- Whole chapter: https://www.bctengineer.com/md/semester_6/hydro_power/chapter-1

## Question

Find the pressure difference between the container A and B as shown in figure. [Figure: container A holds liquid of sp. gr. 0.8 and container B holds water; they are connected by a U-tube with mercury at the bottom; the limb from A is inclined at 45° and the limb to B at 60°; lengths marked are 130 mm and 230 mm on the A side, 230 mm on the B side, and vertical heights 550 mm (left limb) and 300 mm (right limb).]

## Short answer

Only vertical heights contribute; convert inclined lengths using sinθ. Balancing pressures at the lower mercury surface gives pA − pB = ρg[0.30 + 13.6(0.1073) − 0.8(0.55)] ≈ 12.94 kN/m², A higher.

## Answer

In an inclined limb only the **vertical** height of a column produces pressure: $h = L\sin\theta$.

**Assumed reading of the figure:** liquid of $S = 0.8$ stands 550 mm vertically above the mercury surface in A's limb, and water stands 300 mm vertically above the mercury surface in B's limb. The mercury surface lies 130 mm along the 45° limb on the A side and 230 mm along the 60° limb on the B side, measured from the common bottom of the U-tube.

#### Vertical heights of the mercury surfaces
$$
\begin{aligned}
h_A &= 0.130\sin 45^\circ = 0.0919\ \text{m} \\
h_B &= 0.230\sin 60^\circ = 0.1992\ \text{m} \\
\Delta h &= 0.1992 - 0.0919 = 0.1073\ \text{m}
\end{aligned}
$$
So mercury on the B side stands 0.1073 m higher.

#### Pressure balance at the level of the lower (A-side) mercury surface
$$
p_A + \rho_{0.8}g(0.55) = p_B + \rho_w g(0.30) + \rho_{Hg}g(0.1073)
$$
$$
\begin{aligned}
p_A - p_B &= 9.81\,[0.30 + 13.6(0.1073) - 0.8(0.55)] \\
&= 2.943 + 14.310 - 4.316 \\
&= 12.94\ \text{kN/m}^2
\end{aligned}
$$

**Answer:** $p_A - p_B \approx 12.94$ kN/m² (container A at the higher pressure), for the reading above. Use the same $L\sin\theta$ method if the figure places the lengths differently.

## Check yourself

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

1. In an inclined limb, the pressure produced by a column of length L at angle theta to the horizontal comes from:
   A. The full length L
   B. The vertical height L sin(theta)
   C. L cos(theta) always
   D. The cross-sectional area
   Answer: B. Only vertical height produces hydrostatic pressure, so h = L sin(theta). (Key point: Inclined limbs)

2. The vertical heights of the mercury surfaces are h_A = 0.130 sin 45 and h_B = 0.230 sin 60. The difference is about:
   A. 0.0919 m
   B. 0.1992 m
   C. 0.1073 m
   D. 0.360 m
   Answer: C. 0.1992 - 0.0919 = 0.1073 m, with mercury higher on the B side. (Key point: Vertical heights)

3. Taking the level of the lower mercury surface (A side), the balance is:
   A. p_A + rho_0.8 g(0.55) = p_B + rho_w g(0.30) + rho_Hg g(0.1073)
   B. p_A = p_B
   C. p_A + rho_w g(0.30) = p_B + rho_0.8 g(0.55)
   D. p_A + rho_Hg g(0.1073) = p_B
   Answer: A. On B's side the extra mercury column of 0.1073 m lies between the two mercury surfaces, plus water. (Key point: Balance equation)

4. The pressure difference is about:
   A. 1.29 kN/m^2
   B. 12.94 kN/m^2
   C. 129.4 kN/m^2
   D. 7.26 kN/m^2
   Answer: B. 9.81[0.30 + 13.6(0.1073) - 0.8(0.55)] = 2.943 + 14.310 - 4.316 = 12.94 kN/m^2. (Key point: Result)

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