Efflux and Dynamic Lift

Efflux: how fast does liquid leave a hole?

A hole below a tank’s free surface produces a jet. The word efflux means outflow. Choose point s at the free surface and point o at the jet just outside the hole. Let their heights above one datum be y_s and y_o, with h = y_s − y_o positive. Let the gas above the liquid have pressure P_s; the jet is exposed to atmospheric pressure P_atm. For a small hole in a large tank, the free surface falls slowly compared with the jet speed.

Continuity gives A_s v_s = A_o v_o. If A_s is much larger than A_o, then v_s = (A_o/A_s)v_o is small. This is a reasoned approximation, not a statement that the water surface literally never moves. The water level changes slowly enough to treat the outflow as approximately steady over a short interval. Apply Bernoulli along a streamline from the surface region to the outlet, neglecting viscous losses.

For a large tank we neglect v_s² and take the positive square root for outward flow. For an open tank P_s = P_atm, so the two pressures cancel. Liquid density also cancels in this ideal open-tank result: both gravitational energy and inertia are proportional to the liquid mass. A denser liquid does not automatically leave the same-height ideal tank faster.

The open-tank speed equals the speed acquired by a body falling freely from rest through height h. Both results follow from gravitational potential energy converting into kinetic energy. This does not mean the jet falls vertically inside the tank. If the tank pressure excess dominates ρgh, v_o is approximately the square root of 2(P_s − P_atm)/ρ. This pressure-driven limit helps explain powerful expelled jets; a real rocket also requires momentum and propulsion analysis.

Why a jet can push its container backward

Liquid leaving a hole carries momentum outward. The container and remaining fluid experience a reaction in the opposite direction. This backward thrust follows from momentum conservation or Newton’s third law. The pressure-energy calculation determines a possible jet speed; the momentum flow explains the recoil. An unsupported vessel can therefore move opposite to the jet even when the hole is horizontal.

Partly cover a tap with your fingers and small openings can produce narrow fast jets. For a given volume flow, continuity makes a smaller total outlet area correspond to a higher speed. A real tap’s pressure supply and losses also affect the resulting flow rate, so it is not safe to assume Q remains exactly fixed for every degree of obstruction. Distinguish a continuity relationship from a guarantee that the pump supplies unlimited speed.

Dynamic lift on balls and wings

Dynamic lift is a force perpendicular to the relative flow that arises when a body moves through a fluid or a fluid flows past it. Work in the body’s rest frame so the relative air motion is clear. For an ideal nonspinning sphere with a symmetric flow pattern, corresponding upper and lower speeds are equal. There is no upper–lower pressure difference to provide lift in that symmetric model. This statement does not deny real drag or possible asymmetry in other circumstances.

A spinning ball drags nearby air because of the interaction at its surface. Spin and relative translation can make the surrounding flow asymmetric. For the illustrated clockwise ball with air moving to the right, the upper surface moves with the nearby flow and the lower surface against it. In the simplified picture the speed is greater above and smaller below. The resulting pressure difference gives upward lift. Reversing the spin reverses that asymmetry. This spin-dependent transverse effect is called the Magnus effect.

The lift changes a spinning ball’s acceleration as well as its weight-driven motion. Its trajectory therefore need not be the simple parabola predicted when gravity is the only force. Real sport-ball motion also involves drag, seams, surface roughness and flow separation. Bernoulli relates speed and pressure in an appropriate outer-flow approximation; it does not alone calculate the entire viscous flow around a spinning ball.

A wing’s shape and orientation establish a flow that can be faster over its upper surface than below. At approximately the same height, Bernoulli associates the faster upper flow with lower pressure. Integrating the pressure distribution produces lift. A simple estimate assumes a uniform average pressure difference ΔP over wing area A and uses L ≈ ΔP A. For level flight, lift approximately balances weight. The wing also turns air downward; momentum and pressure descriptions are compatible.

There is no rule requiring air parcels separated at the wing’s front to meet again at its rear. Avoid an “equal travel time” explanation. Bernoulli does not determine the two speeds from the two path lengths alone. In the following estimates the speeds, or their approximate mean, are given. Air is treated as approximately incompressible; this is a model limitation, especially at very high aircraft speeds.

Blowing across the top of a light paper strip can reduce the pressure there, so the higher pressure below lifts the strip. This is a qualitative application with a real jet and surrounding air entrainment; the comparison is useful only when the flow geometry is specified. Blowing underneath is not the same setup and does not test the intended upper-surface pressure reduction.

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