Torricelli’s Theorem

Figure 1. (a) Water gushes from the base of the Studen Kladenetz dam in Bulgaria. (credit: Kiril Kapustin; http://www.ImagesFromBulgaria.com) (b) In the absence of significant resistance, water flows from the reservoir with the same speed it would have if it fell the distance h without friction. This is an example of Torricelli’s theorem.

Figure 1 shows water gushing from a large tube through a dam. What is its speed as it emerges? Interestingly, if resistance is negligible, the speed is just what it would be if the water fell a distance h from the surface of the reservoir; the water’s speed is independent of the size of the opening. Let us check this out. Bernoulli’s equation must be used since the depth is not constant. We consider water flowing from the surface (point 1) to the tube’s outlet (point 2). Bernoulli’s equation as stated in previously is

[latex]{P}_{1}+\frac{1}{2}{{\rho v}_{1}}^{2}+\rho gh_{1}={P}_{2}+\frac{1}{2}{{\rho v}_{2}}^{2}+\rho gh_{2}\\[/latex].

Both P₁ and P₂ equal atmospheric pressure (P₁ is atmospheric pressure because it is the pressure at the top of the reservoir. P₂ must be atmospheric pressure, since the emerging water is surrounded by the atmosphere and cannot have a pressure different from atmospheric pressure.) and subtract out of the equation, leaving

[latex]\frac{1}{2}{{\rho v}_{1}}^{2}+\rho gh_{1}=\frac{1}{2}{{\rho v}_{2}}^{2}+\rho gh_{2}\\[/latex].

Solving this equation for [latex]{{v}_{2}}^{2}[/latex], noting that the density ρ cancels (because the fluid is incompressible), yields

[latex]{{v}_{2}}^{2}={{v}_{1}}^{2}+2g\left({h}_{1}-{h}_{2}\right)\\[/latex].

We let h = h₁ − h₂; the equation then becomes

[latex]{{v}_{2}}^{2}={{v}_{1}}^{2}+2gh\\[/latex]

where h is the height dropped by the water. This is simply a kinematic equation for any object falling a distance h with negligible resistance. In fluids, this last equation is called Torricelli’s theorem. Note that the result is independent of the velocity’s direction, just as we found when applying conservation of energy to falling objects.

Figure 2. Pressure in the nozzle of this fire hose is less than at ground level for two reasons: the water has to go uphill to get to the nozzle, and speed increases in the nozzle. In spite of its lowered pressure, the water can exert a large force on anything it strikes, by virtue of its kinetic energy. Pressure in the water stream becomes equal to atmospheric pressure once it emerges into the air.

All preceding applications of Bernoulli’s equation involved simplifying conditions, such as constant height or constant pressure. The next example is a more general application of Bernoulli’s equation in which pressure, velocity, and height all change. (See Figure 2.)

Power is the rate at which work is done or energy in any form is used or supplied. To see the relationship of power to fluid flow, consider Bernoulli’s equation:

[latex]P+\frac{1}{2}{\rho v}^{2}+\rho{gh}=\text{ constant}\\[/latex]

All three terms have units of energy per unit volume, as discussed in the previous section. Now, considering units, if we multiply energy per unit volume by flow rate (volume per unit time), we get units of power. That is, (E/V)(V/t)=E/t. This means that if we multiply Bernoulli’s equation by flow rate Q, we get power. In equation form, this is

[latex]\left(P+\frac{1}{2}{\rho v}^{2}+\rho gh\right)Q=\text{ power}\\[/latex].

Each term has a clear physical meaning. For example, PQ is the power supplied to a fluid, perhaps by a pump, to give it its pressure P. Similarly, [latex]\frac{1}{2}{\rho v}^{2}Q\\[/latex]  is the power supplied to a fluid to give it its kinetic energy. And pghQ is the power going to gravitational potential energy.