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Fluid masses subjected to vertical acceleration

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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, $p$ = Pressure due to vertical acceleration

We know that the forces acting on this column are :

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

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

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

Now resolving the forces vertically,

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

Blank

Example 1 [metric]

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/s\^2 (i) vertically upwards (ii) vertically downwards.

Workings

Given,

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

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

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

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

Accelerated vertically

Example 1 [metric]

Example 2