Speed of Efflux: Torricelli’s Law

The word efflux means fluid outflow. Torricelli discovered that the speed of efflux from an open tank is given by a formula identical to that of a freely falling body. 

Fig. 1 Torricelli’s law. The speed of efflux, v$_1$ , from the side of the container is given by the application of Bernoulli’s equation. If the container is open at the top to the atmosphere then $v=\sqrt{2gh}$

Consider a tank containing a liquid of density ρ with a small hole in its side at a height y$_1$  from the bottom (see Fig. 1). The air above the liquid, whose surface is at height y$_2$ , is at pressure P. From the equation of continuity, we have

$v_1A_1=v_2A_2$

$v_2=\frac{A_1}{A_2}v_1$

If the cross-sectional area of the tank A$_2$  is much larger than that of the hole (A$_2$  >> A$_1$), then we may take the fluid to be approximately at rest at the top, i.e., v$_2$  = 0. Now, applying the Bernoulli equation at points 1 and 2 and noting that at the hole P$_1$ = Pa, the atmospheric pressure, we have from Eq. 

$P_1+\frac{1}{2}\rho v_1^2+\rho gh_1=P_2+\frac{1}{2}\rho v_2^2 +\rho gh_2$

we have

$P_a+\frac{1}{2} \rho v_1^2+ \rho gy_1=P + \rho gy_2$

Taking y$_2$  – y$_1$  = h we have

$v_1=\sqrt{2gh+\frac{2(P-P_a)}{\rho}}$     (1)

When P >> Pa and 2gh may be ignored, the speed of efflux is determined by the container pressure. Such a situation occurs in rocket propulsion. On the other hand, if the tank is open to the atmosphere, then P = Pa and

$v_1=\sqrt{2gh}$            (2)

This is also the speed of a freely falling body. Equation (2) represents Torricelli’s law.