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Accelerated vertically

Fluid masses subjected to vertical acceleration

Contents

Overview
Page Comments

Overview

Consider a tank open at top, containing a liquid and moving vertically upwards with a uniform acceleration. Since the tank is subjected to an acceleration in the vertical direction only, therefore the liquid surface will remain horizontal.

23547/fluid_subjected_to_vertical_acceleration_1.png

Now consider a small column of the liquid of height h and area dA in the tank as sown in fig-1.

Let, $\inline p$ = Pressure due to vertical acceleration

We know that the forces acting on this column are :

1. Weight of the liquid column $\inline [W = w(h.dA)]$ acting vertically downwards,

2. Acceleration force, $\inline F = ma = \frac{w}{g}(h.dA)a$

3. Pressure $\inline (P = p.dA)$ exerted by the liquid particles on the column.

Now resolving the forces vertically,

$P = W + F$

$\Rightarrow p.dA = w(h.dA) + \frac{w}{g}(h.dA)a = wh.dA(1 + \frac {a}{g})$

$\therefore p = wh(1 + \frac {a}{g})$

Example:

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Example - Fluid masses subjected to vertical acceleration

Problem

An open rectangular tank 4m long and 2.5m wide contains an oil of specific gravity 0.85 up to a depth of 1.5m. Determine the total pressure on the bottom of the tank, when the tank is moving with an acceleration of of g/2 m/s² (i) vertically upwards (ii) vertically downwards.

Workings

Given,

$\inline l$ = 4 m
$\inline b$ = 2.5 m
$\inline d$ = 1.5 m
$\inline a$ = g/2 m/s²
Specific gravity of liquid = 0.85

(i) Total pressure on the bottom of the tank, when it is vertically upwards

Specific weight of oil, $\inline w = 0.85\times 9.81 = 8.34\;KN/m^3$

Intensity of pressure at the bottom of the tank,

$p_1 = wh(1 + \frac{a}{g})$

$\Rightarrow p_1 = 8.34\times 1.5(1 + \frac{g}{2g})$

$\therefore p_1 = 18.765\;KN/m^2$

Total pressure on the bottom of the tank,

$P_1 = p_1.A = 18.765\times (4\times 2.5) = 187.65\;KN$

(i) Total pressure on the bottom of the tank, when it is vertically downwards

Intensity of pressure at the bottom of the tank,

$p_2 = wh(1 + \frac{a}{g})$

$\Rightarrow p_2 = 8.34\times 1.5(1 + \frac{g}{2g})$

$\therefore p_2 = 6.255\;KN/m^2$

Total pressure on the bottom of the tank,

$P_2 = p_2.A = 6.255\times (4\times 2.5) = 62.55\;KN$