Test: Liquids in Relative Equilibrium - Civil Engineering (CE) MCQ

In this case the surface of constant pressure is a downward sloping plane surface. The free surface of the liquid is a surface of cohstant pressure, because the pressure at the free surface is equal to atmospheric pressure which is constant. As such under the effect of the horizontal acceleration the free surface of the liquid will become a downward sloping inclined plane, with the liquid rising at the back end and the liquid falling at the front end.

The shape of the free surface is paraboloid.

Pressure at bottom of tank
= p (g - a) h = 1000 x (9.81 - 2) x 1.5
= 11.7 kN/m²

height of paraboloid = Z = 1 m = r²w²/2g
VoI. of water = V_AB + V_BC
V_AB = Volume of paraboloid
= 1/2 volume of circumscribing cylinder
V_BC = Cross- sectional area x 1
1/2 x 1 x 1 = 0.5 m³
Total volume of water = 1.5 m³

After half the water has spilled out the vertex of the paraboloid will touch the bottom of cylinder thus pressure will be zero.

Variation in pressure in horizontal direction in a rotating cylindrical vessel containing fluid is given by,
∂P/∂r = pw²r

⇒ tan 45º = a/g ⇒  a = g

In the case of maximum acceleration, the free surface will be such that one end will have 2.5m depth and the other will have zero depth. Equating volume contained by free surface to earlier volume;
l x 2.5/2 x 3 = 3 x 10 x 1
I = 20/2.5 = 8m
tanθ = 2.5/8 = a/g
a = 0.31g