---
title: "Pressure diagram in rotating closed tank"
subject: "Fluid Mechanics"
chapter: 2
key: "semester_3/fluid_mechanics:closed-rotating-tank-pressure-diagram"
source: https://www.bctengineer.com/q/semester_3/fluid_mechanics/closed-rotating-tank-pressure-diagram
site: BCT Engineering
---

# Pressure diagram in rotating closed tank

Fluid Mechanics · Chapter 2: Pressure and Head

- Asked in: 2074 Asoj, 6 marks
- Video: [A closed cylinderical tank completely filled with water,(speed of rotation=w), draw pressure diagram](https://www.youtube.com/watch?v=9ju_GvyxwHA) — Civil Engg Network
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_3/fluid_mechanics/closed-rotating-tank-pressure-diagram
- Whole chapter: https://www.bctengineer.com/md/semester_3/fluid_mechanics/chapter-2

## Question

A closed cylindrical tank of 1 m diameter and 2 m high is completely filled with water. If it is being rotated about its vertical axis with uniform speed of 100 rpm, draw pressure intensity diagram along surface AB and AC with values. [Figure: cylinder with A and C at the top corners (A on the left wall, C on the right), B and D at the bottom, height 2 m, diameter 1 m]

## Short answer

Pressure on top AC is parabolic, from 0 at centre to 13.71 kPa at corners; on vertical wall AB it increases linearly from 13.71 kPa at A to 33.33 kPa at B.

## Answer

**Assumption:** the tank is completely full and closed, and the pressure at the centre of the top cover (on the axis) remains atmospheric (0 gauge). Then the parabolic pressure variation starts from zero at the axis on the top.

**Data:** $R=0.5$ m, height 2 m, $N=100$ rpm so $\omega=\dfrac{2\pi\times100}{60}=10.472$ rad/s, $\rho=1000$ kg/m³.

Pressure at a point at radius $r$ and depth $z$ below the top:

$$
p=\tfrac{1}{2}\rho\omega^{2}r^{2}+\rho g z
$$

#### Along AC (top cover, $z=0$)
$$
p=\tfrac{1}{2}(1000)(10.472)^{2}r^{2}=54\,831\,r^{2}\ \text{Pa}
$$

| Position | $r$ (m) | $p$ (kPa) |
|---|---|---|
| Centre | 0 | 0 |
| $r=0.25$ m | 0.25 | 3.43 |
| A or C | 0.5 | 13.71 |

The diagram is a parabola, zero at the centre and 13.71 kPa at A and at C (symmetrical).

#### Along AB (vertical wall, $r=0.5$ m)
$$
p=13.71+9.81\,z\ \text{kPa}
$$

| Depth $z$ (m) | 0 (A) | 1 | 2 (B) |
|---|---|---|---|
| $p$ (kPa) | 13.71 | 23.52 | 33.33 |

The diagram is a straight line (trapezium) from 13.71 kPa at A to 33.33 kPa at B.

```text
  A |->13.71 kPa          C
    |  \     parabola    /
    |   \_______________/  (0 at centre)
    |->23.52 (1 m)
    |->33.33 kPa at B
```

**Answer: at A and C, 13.71 kPa; at B and D, 33.33 kPa; AC is parabolic (0 at the centre) and AB linear.**

## Check yourself

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

1. What condition fixes the pressure variation on the top cover AC?
   A. Pressure at the centre of the top cover (on the axis) is zero gauge
   B. Pressure at A and C is zero gauge
   C. Pressure at the bottom centre is zero gauge
   D. Pressure is uniform over the top cover
   Answer: A. The parabola p = 1/2 ρ ω^2 r^2 starts from zero at the axis on the top. (Key point: Starting assumption)

2. For 100 rpm, ω = 10.472 rad/s. What is the pressure at A and C on the top cover (r = 0.5 m)?
   A. 33.33 kPa
   B. 13.71 kPa
   C. 3.43 kPa
   D. 27.4 kPa
   Answer: B. p = 1/2 (1000)(10.472)^2 (0.5)^2 = 13.71 kPa. (Key point: Speed conversion and top-cover pressure)

3. What is the shape of the pressure diagram along AC and along AB?
   A. AC is a straight line; AB is a parabola
   B. Both are parabolas
   C. AC is a parabola; AB is a straight line
   D. Both are straight lines
   Answer: C. Along AC (z = 0) p varies with r^2; along the wall (r constant) p = 13.71 + 9.81 z varies linearly with depth. (Key point: Shape of the diagrams)

4. What is the pressure at B, 2 m below A on the wall?
   A. 33.33 kPa
   B. 19.62 kPa
   C. 13.71 kPa
   D. 47.04 kPa
   Answer: A. p = 13.71 (rotation) + 9.81 x 2 (depth) = 33.33 kPa. (Key point: Pressure at the bottom of the wall)

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