# Learning Objectives

### Learning Objectives

By the end of this section, you will be able to do the following:

- Calculate using Torricelli's theorem
- Calculate power in fluid flow

The information presented in this section supports the following AP® learning objectives and science practices:

**5.B.10.2**The student is able to use Bernoulli's equation and/or the relationship between force and pressure to make calculations related to a moving fluid.**(S.P. 2.2)****5.B.10.3**The student is able to use Bernoulli's equation and the continuity equation to make calculations related to a moving fluid.**(S.P. 2.2)**

# Torricelli's Theorem

### Torricelli's Theorem

Figure 12.8 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

Both ${P}_{1}$ and ${P}_{2}$ equal atmospheric pressure. ^{}${P}_{1}$ is atmospheric pressure because it is the pressure at the top of the reservoir. ${P}_{2}$ must be atmospheric pressure, since the emerging water is surrounded by the atmosphere and cannot have a pressure different from atmospheric pressure. Subtract out of the equation, leaving

Solving this equation for ${v}_{2}^{2}$, noting that the density $\rho $ cancels because the fluid is incompressible, yields

We let $h={h}_{1}-{h}_{2}\mathrm{;}$ the equation then becomes

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.

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 12.9

### Example 12.5 Calculating Pressure: A Fire Hose Nozzle

Fire hoses used in major structure fires have inside diameters of 6.40 cm. Suppose such a hose carries a flow of 40 L/s starting at a gauge pressure of $1\text{.}\text{62}\times {\text{10}}^{6}\phantom{\rule{0.25em}{0ex}}{\text{N/m}}^{2\mathrm{.}}$ The hose goes 10 m up a ladder to a nozzle having an inside diameter of 3 cm. Assuming negligible resistance, what is the pressure in the nozzle?

**Strategy**

Here we must use Bernoulli's equation to solve for the pressure, since depth is not constant.

**Solution**

Bernoulli's equation states

where the subscripts 1 and 2 refer to the initial conditions at ground level and the final conditions inside the nozzle, respectively. We must first find the speeds ${v}_{1}$ and ${v}_{2}$. Since $Q={A}_{1}{v}_{1\mathrm{,}}$ we get

Similarly, we find

This rather large speed is helpful in reaching the fire. Now, taking ${h}_{1}$ to be zero, we solve Bernoulli's equation for ${P}_{2.}$

Substituting known values yields

**Discussion**

This value is a gauge pressure, since the initial pressure was given as a gauge pressure. Thus, the nozzle pressure equals atmospheric pressure, as it must because the water exits into the atmosphere without changes in its conditions.

### Making Connections: Squirt Toy

A horizontally oriented squirt toy contains a one-cm-diameter barrel for the water. A 2.2-N force on the plunger forces water down the barrel and into a 1.5-mm-diameter opening at the end of the squirt gun. In addition to the force pushing on the plunger, pressure from the atmosphere is also present at both ends of the gun, pushing the plunger in and also pushing the water back in to the narrow opening at the other end. Assuming that the water is moving very slowly in the barrel, with what speed does it emerge from the toy?

**Solution**

First, find the cross-sectional areas for each part of the toy. The wider part is

Next, we will find the area of the narrower part of the toy.

The pressure pushing on the barrel is equal to the sum of the pressure from the atmosphere $\mathrm{(1.0}\text{\hspace{0.17em}}\text{atm}=101,300\text{\hspace{0.17em}}{\text{N/m}}^{2}\mathrm{)}$ and the pressure created by the 2.2-N force.

The pressure pushing on the smaller end of the toy is simply the pressure from the atmosphere.

Since the gun is oriented horizontally ${\mathrm{(}}_{h}={h}_{2}\mathrm{),}$ we can ignore the potential energy term in Bernoulli's equation, so the equation becomes

The problem states that the water is moving very slowly in the barrel. That means we can make the approximation that ${v}_{1}\approx \mathrm{0,}$ which we will justify mathematically.

How accurate is our assumption that the water velocity in the barrel is approximately zero? Check using the continuity equation.

How does the kinetic energy per unit volume term for water in the barrel fit into Bernoulli's equation?

As you can see, the kinetic energy per unit volume term for water in the barrel is very small (14) compared to the other terms, which are all at least 1000 times larger. Another way to look at this is to consider the ratio of the two terms that represent kinetic energy per unit volume.

Remember that from the continuity equation

Thus, the ratio of the kinetic energy per unit volume terms depends on the fourth power of the ratio of the diameters.

In this case, the diameter of the barrel (*d*_{2}) is 6.7 times larger than the diameter of the opening at the end of the toy (*d*_{1}), which makes the kinetic energy per unit volume term for water in the barrel ${(6.7)}^{4}\approx \mathrm{2,000}$ times smaller. We can usually neglect such small terms in addition or subtraction without a significant loss of accuracy.

# Power in Fluid Flow

### Power in Fluid Flow

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.

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

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

### Making Connections: Power

Power is defined as the rate of energy transferred, or *$E/\mathrm{t.}$* Fluid flow involves several types of power. Each type of power is identified with a specific type of energy being expended or changed in form.

### Example 12.6 Calculating Power in a Moving Fluid

Suppose the fire hose in the previous example is fed by a pump that receives water through a hose with a 6.40-cm diameter coming from a hydrant with a pressure of $0\text{.}\text{700}\phantom{\rule{0.25em}{0ex}}\times \phantom{\rule{0.25em}{0ex}}{\text{10}}^{6}\phantom{\rule{0.25em}{0ex}}{\text{N/m}}^{2\mathrm{.}}$ What power does the pump supply to the water?

**Strategy**

Here we must consider energy forms as well as how they relate to fluid flow. Since the input and output hoses have the same diameters and are at the same height, the pump does not change the speed of the water nor its height, and so the water's kinetic energy and gravitational potential energy are unchanged. That means the pump only supplies power to increase water pressure by $0\text{.}\text{92}\phantom{\rule{0.25em}{0ex}}\times \phantom{\rule{0.25em}{0ex}}{\text{10}}^{6}\phantom{\rule{0.25em}{0ex}}{\text{N/m}}^{2}$ (from $0.700\phantom{\rule{0.25em}{0ex}}\times \phantom{\rule{0.25em}{0ex}}{\text{10}}^{6}\phantom{\rule{0.25em}{0ex}}{\text{N/m}}^{2}$ to $1.62\phantom{\rule{0.25em}{0ex}}\times \phantom{\rule{0.25em}{0ex}}{\text{10}}^{6}\phantom{\rule{0.25em}{0ex}}{\text{N/m}}^{2}$).

**Solution**

As discussed above, the power associated with pressure is

**Discussion**

Such a substantial amount of power requires a large pump, such as is found on some fire trucks. This kilowatt value converts to about 50 hp. The pump in this example increases only the water's pressure. If a pump—such as the heart—directly increases velocity and height as well as pressure, we would have to calculate all three terms to find the power it supplies.