---
title: "Gauge pressure in cleaning fluid tank"
subject: "Fluid Mechanics"
chapter: 2
key: "semester_3/fluid_mechanics:cleaning-fluid-tank-gauge-pressure"
source: https://www.bctengineer.com/q/semester_3/fluid_mechanics/cleaning-fluid-tank-gauge-pressure
site: BCT Engineering
---

# Gauge pressure in cleaning fluid tank

Fluid Mechanics · Chapter 2: Pressure and Head

- Asked in: 2078 Kartik, 5 marks
- Video: [Problem No 1 on Simple U-Tube Monometer - Pressure and Pressure Measurement - Fluid Mechanics](https://www.youtube.com/watch?v=PndIPTVYSRU) — Ekeeda
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_3/fluid_mechanics/cleaning-fluid-tank-gauge-pressure
- Whole chapter: https://www.bctengineer.com/md/semester_3/fluid_mechanics/chapter-2

## Question

A cylindrical tank contains water at a height of 55 mm as shown in figure below. Inside a smaller open cylinder tank containing cleaning fluid (S = 0.8) at a height h. If $P_B$ = 13.4 kPa and $P_C$ = 13.42 kPa gauge, what are gauge pressure $P_A$ and height h of cleaning fluid? Assume that the cleaning fluid kerosene is prevented from moving to the top of the tank. [Figure: closed tank with air at the top (gauge $P_A$), water 55 mm deep, a smaller open cylinder of kerosene (height h) inside; gauges $P_B$ and $P_C$ at the bottom]

## Short answer

pA = pB – γw×0.055 m = 12.86 kPa gauge. Then h = (pC – pA)/(0.8×γw) = 0.0713 m = 71.3 mm.

## Answer

**Reading of the figure:** gauge B is at the bottom of the main tank (under 55 mm of water). Gauge C is at the bottom of the small cylinder (under a depth $h$ of cleaning fluid, $S=0.8$). The air above both liquids is at gauge pressure $p_A$.

**Gauge pressure $p_A$.** From B:

$$
p_B=p_A+\gamma_w(0.055)\ \Rightarrow\ p_A=13.4-9.81(0.055)=12.86\ \text{kPa}
$$

**Height $h$.** From C:

$$
p_C=p_A+\gamma_f h,\qquad \gamma_f=0.8\times9.81=7.848\ \text{kN/m}^3
$$

$$
h=\frac{p_C-p_A}{\gamma_f}=\frac{13.42-12.860}{7.848}=0.0713\ \text{m}
$$

**Answer: $p_A\approx12.86$ kPa (gauge) and $h\approx71.3$ mm.**

## Check yourself

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

1. How is the air pressure p_A found from gauge B under 55 mm water?
   A. p_A = p_B + gamma_w x 0.055
   B. p_A = p_B x 0.055
   C. p_A = p_B - gamma_w x 0.055
   D. p_A = gamma_w x 0.055
   Answer: C. p_B = p_A + gamma_w(0.055), so p_A = 13.4 - 0.54 = 12.86 kPa. (Key point: Gauge A from B)

2. How is h obtained?
   A. h = (p_C + p_A)/gamma_f
   B. h = p_C/gamma_f
   C. h = (p_C - p_A)/gamma_f
   D. h = (p_C - p_B)/gamma_w
   Answer: C. p_C = p_A + gamma_f h gives h = (13.42 - 12.86)/7.848 = 71.3 mm. (Key point: Height h)

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