Question 14.9: Torricelli’s Law An enclosed tank containing a liquid of den...
Torricelli’s Law
An enclosed tank containing a liquid of density ρ has a hole in its side at a distance y_1 from the tank’s bottom (Fig. 14.20). The hole is open to the atmosphere, and its diameter is much smaller than the diameter of the tank. The air above the liquid is maintained at a pressure P. Determine the speed of the liquid as it leaves the hole when the liquid’s level is a distance h above the hole.
Conceptualize Imagine that the tank is a fire extinguisher. When the hole is opened, liquid leaves the hole with a certain speed. If the pressure P at the top of the liquid is increased, the liquid leaves with a higher speed. If the pressure P falls too low, the liquid leaves with a low speed and the extinguisher must be replaced. Because A_2>>A_1, the liquid is approximately at rest at the top of the tank, where the pressure is P, so v_2=0. At the hole, the liquid is open to the external atmosphere, so P_1 is equal to atmospheric pressure P_0.
Categorize Looking at Figure 14.20, we know the pressure at two points and the velocity at point 2. We wish to find the velocity at point 1. Therefore, we can categorize this example as one in which we can apply Bernoulli’s equation.
Analyze
Apply Bernoulli’s equation between points 1 and 2:
P_0+\frac{1}{2} \rho v_1{}^2+\rho g y_1=P+\rho g y_2Solve for v_1, noting that y_2-y_1=h :
v_1=\sqrt{\frac{2\left(P-P_0\right)}{\rho}+2 g h}Finalize When P is much greater than P_0 (so that the term 2gh can be neglected), the exit speed of the water is mainly a function of P. If the tank is open at the top to the atmosphere, then P=P_0 and v_1=\sqrt{2 g h}. In other words, for an open tank, the speed of the liquid leaving a hole a distance h below the surface is equal to that acquired by an object falling freely through a vertical distance h. This phenomenon is known as Torricellis law.