FLUIDS CYU 18

Fluid mechanics

Rotating Liquid Draining Through a Central Orifice

The final free surface must meet the bottom at the edge of the orifice. This boundary condition fixes the vertical shift of the paraboloid.

Vout = πR2h − [πω2/(4g)](R2 − r2)2
Problem

A cylindrical vessel of radius R contains an ideal liquid of density ρ, initially filled to height h. The vessel rotates steadily about its vertical axis with angular velocity ω; the liquid neither overflows nor leaves any part of the bottom dry. A central circular orifice of radius r is then opened in the bottom. Find the volume of liquid that can flow out.

1

Shape of the rotating free surface

Let x be the radial distance from the rotation axis and z(x) the free-surface height above the bottom. In steady rotation,

∂p/∂x = ρω2x,   ∂p/∂z = −ρg.

Pressure is constant along the free surface. Hence dp = 0, giving

0 = ρω2x dx − ρg dz  ⟹  dz/dx = ω2x/g.

Therefore, every steady free surface has the form

z(x) = C + ω2x2/(2g).

where C is determined by the amount of liquid and the relevant boundary condition.

Initial rotating liquid Cross-section before the orifice is opened. The parabolic free surface remains above the entire bottom. ω R
Initial state: the complete bottom remains covered.
Final rotating liquid Cross-section after draining. The free surface meets the bottom at both edges of the central orifice. ω x = r z(r)=0
Final state: the surface meets the bottom at the orifice edge.
2

Initial volume

Rotation changes the shape of the surface but not the amount of liquid. Hence

Vi = πR2h.

For completeness, volume conservation gives the initial surface:

zi(x) = h − ω2R2/(4g) + ω2x2/(2g).

The condition that the bottom is initially wet everywhere is zi(0) > 0, or h > ω2R2/(4g).

3

Final boundary condition

Key condition

Liquid keeps draining while it covers any part of the opening. Flow ceases when the free surface reaches the bottom at the rim of the orifice. Therefore, zf(r) = 0.

Using zf(x) = Cf + ω2x2/(2g),

Cf = −ω2r2/(2g).

Thus, for r ≤ x ≤ R, the final liquid depth is

zf(x) = [ω2/(2g)](x2 − r2).
4

Volume remaining

The remaining liquid occupies the annulus r ≤ x ≤ R. Using cylindrical shells,

Vf = ∫rR 2πx zf(x) dx
= (πω2/g) ∫rR x(x2 − r2) dx
= (πω2/g) [x4/4 − r2x2/2]rR
= [πω2/(4g)](R2 − r2)2.
5

Volume discharged

The volume that flows out is Vi − Vf. Therefore,

Exact finite-radius result
Vout = πR2h − [πω2/(4g)](R2 − r2)2
Small-orifice limit

For r ≪ R, neglecting all terms involving r/R,

Vout ≃ πR2h − πω2R4/(4g).
Why density cancels

Both pressure gradients are proportional to ρ, so the free-surface shape is independent of liquid density.

Boundary check

The final profile satisfies zf(r) = 0. A profile with z(r) > 0 still leaves liquid over the opening.

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Assumes steady rigid-body rotation at constant ω, an ideal incompressible liquid, and a final equilibrium in which the central orifice is just uncovered.
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