---
title: "Pressure on circular plate (2.2 m)"
subject: "Hydro Power"
chapter: 1
key: "semester_6/hydro_power:circular-plate-2-2m"
source: https://www.bctengineer.com/q/semester_6/hydro_power/circular-plate-2-2m
site: BCT Engineering
---

# Pressure on circular plate (2.2 m)

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

- Asked in: 2082 Chaitra, 6 marks
- Video: [Problem on Vertical Circular Plate Immersed in Water - Fluid Mechanics 1](https://www.youtube.com/watch?v=vJcFaNG9QPA) — Ekeeda
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_6/hydro_power/circular-plate-2-2m
- Whole chapter: https://www.bctengineer.com/md/semester_6/hydro_power/chapter-1

## Question

A circular plate of diameter 2.2 m is immersed in a liquid of specific gravity 1.2. The highest and lowest points of the plate are at depths of 2.5 m and 4.5 m below the free surface, respectively. Calculate the total pressure force on one face of the plate and the location of its center of pressure.

## Short answer

Inclined circular plate: sinθ = 0.9091, centroid depth 3.5 m. Total pressure F = ρg A h̅ = 156.62 kN. Centre of pressure depth h* = h̅ + (I_G sin²θ)/(A h̅) = 3.571 m.

## Answer

The plate is not vertical because its depth range (4.5 − 2.5 = 2 m) is less than its diameter, so it is inclined. Use the formulas for an inclined plane surface.

**Given:** $d = 2.2$ m, $S = 1.2$ so $\rho = 1200$ kg/m³, depths of top and bottom points $= 2.5$ m and $4.5$ m.

```text
  free surface =====================
         :  2.5 m
         :     * top point
         :    / \
         :   /   \   plate, d = 2.2 m
         :  /  G  \   inclined at theta
         :   \   /
         :    \ /
         :     * bottom point (4.5 m)
```

#### Inclination and centroid depth
$$
\begin{aligned}
\sin\theta &= \frac{4.5 - 2.5}{2.2} = 0.9091 \\
\bar h &= \frac{2.5 + 4.5}{2} = 3.5\ \text{m}
\end{aligned}
$$

#### Area and second moment of area
$$
\begin{aligned}
A &= \frac{\pi}{4}(2.2)^2 = 3.8013\ \text{m}^2 \\
I_G &= \frac{\pi d^4}{64} = \frac{\pi (2.2)^4}{64} = 1.1499\ \text{m}^4
\end{aligned}
$$

#### Total pressure
$$
\begin{aligned}
F &= \rho g A \bar h = 1200 \times 9.81 \times 3.8013 \times 3.5 \\
&= 156\,622\ \text{N} = 156.62\ \text{kN}
\end{aligned}
$$

#### Centre of pressure
$$
\begin{aligned}
h^* &= \bar h + \frac{I_G \sin^2\theta}{A \bar h} \\
&= 3.5 + \frac{1.1499 \times 0.8264}{3.8013 \times 3.5} \\
&= 3.5 + 0.0714 = 3.571\ \text{m}
\end{aligned}
$$

**Answer:** total pressure $F \approx 156.62$ kN; centre of pressure at a depth of $3.571$ m below the free surface (on the plate's vertical diameter, below the centroid).

## Check yourself

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

1. The plate is inclined (not vertical) because:
   A. Its depth range (4.5 - 2.5 = 2 m) is less than its diameter 2.2 m
   B. It is circular
   C. The liquid is dense
   D. Its centroid depth is 3.5 m
   Answer: A. A vertical plate would span its full 2.2 m diameter in depth; here sin(theta) = 2/2.2 = 0.9091. (Key point: Plate inclination)

2. Total pressure uses centroid depth h-bar and area A of:
   A. h-bar = 3.5 m; A = pi/4 (2.2)^2 = 3.80 m^2
   B. h-bar = 2.5 m; A = 3.80 m^2
   C. h-bar = 4.5 m; A = 7.6 m^2
   D. h-bar = 3.5 m; A = pi (2.2)^2
   Answer: A. Centroid depth is the mean of 2.5 and 4.5, 3.5 m, giving F = 1200 x 9.81 x 3.8013 x 3.5 = 156.6 kN. (Key point: Total pressure)

3. For an inclined plane the centre of pressure depth is:
   A. h-bar + I_G/(A h-bar)
   B. h-bar + I_G sin^2(theta)/(A h-bar)
   C. h-bar - I_G sin(theta)/(A h-bar)
   D. h-bar sin(theta)
   Answer: B. For an inclined surface, h* = h-bar + I_G sin^2(theta)/(A h-bar). (Key point: Centre of pressure formula)

4. The centre of pressure is at depth about:
   A. 3.571 m
   B. 3.5 m
   C. 4.57 m
   D. 3.07 m
   Answer: A. 3.5 + (1.1499 x 0.8264)/(3.8013 x 3.5) = 3.5 + 0.0714 = 3.571 m, below the centroid. (Key point: Result)

## Related in Hydro Power

- [Pressure on circular plate (2.5 m)](https://www.bctengineer.com/question.md?key=semester_6%2Fhydro_power%3Acircular-plate-2-5m) — Hydro Power, chapter 1, asked 2×
- [Pressure on circular plate (2.6 m)](https://www.bctengineer.com/question.md?key=semester_6%2Fhydro_power%3Acircular-plate-2-6m) — Hydro Power, chapter 1, asked 2×
- [Pressure on isosceles triangular plate](https://www.bctengineer.com/question.md?key=semester_6%2Fhydro_power%3Atriangular-plate-pressure) — Hydro Power, chapter 1
- [Pressure on inclined rectangular plate](https://www.bctengineer.com/question.md?key=semester_6%2Fhydro_power%3Ainclined-rectangular-plate) — Hydro Power, chapter 1
- [Centre of pressure below centroid proof](https://www.bctengineer.com/question.md?key=semester_6%2Fhydro_power%3Acentre-pressure-below-centroid) — Hydro Power, chapter 1
- [Pressure on vertical rectangular plate 2 m × 3 m](https://www.bctengineer.com/question.md?key=semester_6%2Fhydro_power%3Arectangular-plate-2x3) — Hydro Power, chapter 1

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