Question

State the law of efflux. Derive an expression for the speed of efflux for a tank discharging through an opening at a depth h below the liquid surface. Hence or otherwise show that the speed of efflux for an open tank is √{2gh}

Answer 1

Law of efflux (Torricelli’s theorem) : The speed of efflux for an open tank through an orifice at a depth h below the liquid surface is equal to the speed acquired by a body falling freely through a vertical distance h.

Consider a tank with cross-sectional area A holding a static liquid of density ρ. The tank discharges through an opening (of crosssectional area A ) in the side wall at a depth h below the surface of the liquid. The flow speed at which the liquid leaves the tank is called the speed of efflux.

The pressure at point 2 it is the atmospheric pressure p₀ Let the pressure of the air above the liquid at point 1 be ρ. We assume that the tank is large in cross section compared to the opening (A₁ >> A₂) so that the upper surface of the liquid will drop very slowly. That is, we may regard the liquid surface to be approximately at rest v₁ ≈ 0). Bernoulli’s equation, in usual notation, states p₁ + 1/2 ρv₁² + ρgy₁ = p₂ + 1/2 pv₂²+ ρgy₂ Substituting p₁ = p, p₂ = p₀ , v₁ ≈ 0 and (y₁ – y₂) = h,

v₂²= 2 \(\frac{p-p_0}p\) + 2gh

If the tank is open to the atmosphere, then p = p₀

v₂ = √{2gh}

which is the law of efflux.

[Note : For an open tank, the speed of the liquid, v₂ , leaving a hole a distance h below the surface is equal to that acquired by an object falling freely through a vertical distance h.]