The figure shows the cross-section of a tank full of water under pressure. The length of the tank is 2 m. An empty cylinder lies along the length of the tank on one of its corners. Find the horizontal and vertical components of the force acting on the curved surface ABC of the cylinder.

Leave a Comment / Hydrostatic Forces on Surfaces / By
Cylinder in Pressurized Tank Problem

Problem Statement

The figure shows the cross-section of a tank full of water under pressure. The length of the tank is 2 m. An empty cylinder lies along the length of the tank on one of its corners. Find the horizontal and vertical components of the force acting on the curved surface ABC of the cylinder.

Given Data

  • Radius of Cylinder, \( R = 1 \, \text{m} \)
  • Length of Tank, \( L = 2 \, \text{m} \)
  • Pressure, \( p = 0.2 \, \text{kgf/cm}^2 \)
  • Density of Water, \( \rho = 1000 \, \text{kg/m}^3 \)
  • Acceleration due to Gravity, \( g = 9.81 \, \text{m/s}^2 \)

Diagram of Cylinder in Pressurized Tank

Diagram of the cylinder in the pressurized tank

Solution

1. Calculate Equivalent Pressure Head

First, we convert the gauge pressure to an equivalent height of water (pressure head) which represents an imaginary free surface.

$$ p = 0.2 \, \frac{\text{kgf}}{\text{cm}^2} \times 9.81 \, \frac{\text{N}}{\text{kgf}} = 1.962 \, \frac{\text{N}}{\text{cm}^2} $$ $$ = 1.962 \times 10^4 \, \frac{\text{N}}{\text{m}^2} $$
$$ \text{Pressure Head, } h_{pressure} = \frac{p}{\rho g} $$ $$ = \frac{1.962 \times 10^4}{1000 \times 9.81} $$ $$ = 2 \, \text{m} $$

This means the imaginary free surface is 2 m above the top of the tank.

2. Horizontal Force Component (\(F_x\))

The horizontal force is the pressure on the vertically projected area of the curved surface ABC.

$$ A_{proj} = \text{Projected Height} \times L $$ $$ = 1.5 \, \text{m} \times 2 \, \text{m} = 3.0 \, \text{m}^2 $$

The depth of the centroid of this area from the imaginary free surface is:

$$ \bar{h} = h_{pressure} + \frac{1.5}{2} $$ $$ = 2 \, \text{m} + 0.75 \, \text{m} = 2.75 \, \text{m} $$

The horizontal force is:

$$ F_x = \rho g A_{proj} \bar{h} $$ $$ = 1000 \times 9.81 \times 3.0 \times 2.75 $$ $$ = 80932.5 \, \text{N} $$

3. Vertical Force Component (\(F_y\))

The vertical force is the weight of the imaginary volume of water vertically above the curved surface ABC.

$$ F_y = \text{Weight(CODFBC)} - \text{Weight(AEFB)} $$

Weight of water in volume CODFBC:

$$ V_1 = \left( \text{Area(Quadrant COB)} + \text{Area(Rect. ODFB)} \right) \times L $$ $$ = \left( \frac{\pi R^2}{4} + (BO \times OD) \right) \times 2 $$ $$ = \left( \frac{\pi (1)^2}{4} + (1 \times (2+0.5)) \right) \times 2 $$ $$ \approx (0.7854 + 2.5) \times 2 \approx 6.57 \, \text{m}^3 $$ $$ W_1 = \rho g V_1 = 1000 \times 9.81 \times 6.57 \approx 64458.5 \, \text{N} $$

Weight of water in volume AEFB:

$$ \text{Area(AEFB)} = \text{Area(Rect. AEFG)} + \text{Area(Rect. AGBH)} - \text{Area(Segment AHB)} $$ $$ = (2 \times 0.134) + (0.134 \times 0.5) - 0.0453 $$ $$ = 0.268 + 0.067 - 0.0453 = 0.2897 \, \text{m}^2 $$ $$ V_2 = \text{Area(AEFB)} \times L = 0.2897 \times 2 = 0.5794 \, \text{m}^3 $$ $$ W_2 = \rho g V_2 = 1000 \times 9.81 \times 0.5794 \approx 5684 \, \text{N} $$

Net Vertical Force:

$$ F_y = W_1 - W_2 $$ $$ = 64458.5 - 5684 $$ $$ = 58774.5 \, \text{N} $$
Final Result:

The horizontal component of the force is \( F_x = 80932.5 \, \text{N} \) or \( 80.93 \, \text{kN} \).

The vertical component of the force is \( F_y = 58774.5 \, \text{N} \) or \( 58.77 \, \text{kN} \).

Explanation of Concepts

Pressure Head: In a closed or pressurized tank, the pressure above the liquid surface can be converted into an equivalent height of the same liquid. This creates an "imaginary" or "piezometric" free surface. All depth calculations for hydrostatic forces (\(\bar{h}\)) must be measured from this imaginary surface, not from the actual liquid level inside the tank.

Scroll to Top