The equation of continuity is one of two important relationships for ideal fluids. The other is a statement of energy conservation. The general statement of the energy principle that you learned in Chapter 10 is

where $W_{\text{ext}}$ is the work done by any external forces.

Let’s see how this applies to the fluid flowing through the tube of FIGURE 14.28. This is the situation we considered in Figure 14.26, but now the tube changes height. The more darkly shaded volume of fluid is the system to which we will apply Equation 14.20. Work is done on this system by pressure forces from the surrounding fluid in the tube. The unseen fluid to the left of our system exerts force $\vec{F}{}_1=(p_1{A}_1\text{,}\text{to the right})$, where $p_1$ is the fluid pressure at this point in the tube and $A_1$ is the cross-section area. During a small time interval $\text{Δ}t$, this force pushes the fluid through displacement $\text{Δ}\vec{r}{}_1=(\text{Δ}{x}_1\text{,}\text{to the right})$ and does work

The $A_1$ and $\text{Δ}{x}_1$ enter the equation from different terms, but they conveniently combine to give $\text{Δ}V=A_1\text{Δ}{x}_1$, the volume “vacated” by the shaded fluid as it’s pushed forward.

FIGURE 14.28 Energy analysis of fluid flow through a tube.

The situation is much the same on the right edge of the system, where the surrounding fluid exerts a pressure force $\vec{F}{}_2=(p_2{A}_2\text{,}\text{to the left})$. However, force $\vec{F}{}_2$ is opposite the displacement $\text{Δ}\vec{r}{}_2$, which introduces a minus sign into the dot product for the work, giving

Because the fluid is incompressible, the volume $\text{Δ}V=A_2\text{Δ}{x}_2$ “gained” on the right side is exactly the same as that lost on the left. Altogether, the work done on the system by the surrounding fluid is

The work depends on the pressure difference $p_1-p_2$.

Now let’s see what happens to the system’s potential and kinetic energy. Most of the system does not change during time interval $\text{Δ}t\text{;}$ it’s fluid at the same height moving at the same speed. All we need to consider are the volumes $\text{Δ}V$ at the two ends. On the right, the system gains kinetic and gravitational potential energy as the fluid moves into volume $\text{Δ}V$. Simultaneously, the system loses kinetic and gravitational potential energy on the left as fluid vacates volume $\text{Δ}V$.

The mass of fluid in volume $\text{Δ}V$ is $m=\rho \text{Δ}V$, where $\rho$ is the fluid density. Thus the net change in the system’s gravitational potential energy during $\text{Δ}t$ is

Similarly, the system’s change in kinetic energy is

Combining Equations 14.23, 14.24, and 14.25 gives us the energy equation for the fluid in the flow tube:

The volume $\text{Δ}V$ cancels out of all the terms. Regrouping terms, we have

Equation 14.27 is called Bernoulli’s equation. It is named for the 18th-century Swiss scientist Daniel Bernoulli, who made some of the earliest studies of fluid dynamics.

Bernoulli’s equation is really nothing more than a statement about work and energy. It is sometimes useful to express Bernoulli’s equation in the alternative form

This version of Bernoulli’s equation tells us that the quantity $p+\frac{1}{2}\rho v^2+\rho \text{g}y$ remains constant along a streamline.

EXAMPLE 14.11 An irrigation system

Water flows through the pipes shown in FIGURE 14.29. The water’s speed through the lower pipe is $5.0\text{m}\text{/}\text{s}$ and a pressure gauge reads 75 kPa. What is the reading of the pressure gauge on the upper pipe?

FIGURE 14.29 The water pipes of an irrigation system.

FIGURE 14.29 Full Alternative Text

MODEL Treat the water as an ideal fluid obeying Bernoulli’s equation. Consider a streamline connecting point 1 in the lower pipe with point 2 in the upper pipe.

SOLVE Bernoulli’s equation, Equation 14.27, relates the pressure, fluid speed, and heights at points 1 and 2. It is easily solved for the pressure $p_2$ at point 2:

$$\begin{array}{ll}{p}_2\hfill & =p_1+\frac{1}{2}\rho v_1^2-\frac{1}{2}\rho v_2^2+\rho g{y}_1-\rho g{y}_2\hfill \\ \hfill & =p_1+\frac{1}{2}\rho \left(v_1^2-v_2^2\right)+\rho g\left(y_1-y_2\right)\hfill \end{array}$$

All quantities on the right are known except $v_2$, and that is where the equation of continuity will be useful. The cross-section areas and water speeds at points 1 and 2 are related by

$$v_1{A}_1=v_2{A}_2$$

from which we find

$$v_2=\frac{A_1}{A_2}{v}_1=\frac{r_1{}^2}{r_2{}^2}{v}_1=\frac{{(0.030\text{ m)}}^2}{{(0.020\text{ m)}}^2}(5.0\text{ m}\text{/}\text{s})=11.25\text{ m}\text{/}\text{s}$$

The pressure at point 1 is $p_1=75\text{kPa}+1\text{atm}=176\text{,}300\text{P}\text{a}$. We can now use the above expression for $p_2$ to calculate $p_2=105\text{,}900\text{Pa}$. This is the absolute pressure; the pressure gauge on the upper pipe will read

$$p_2=105\text{,}900\text{Pa}-1\text{atm}=4.6\text{kPa}$$

REVIEW Reducing the pipe size decreases the pressure because it makes $v_2>v_1$. Gaining elevation also reduces the pressure.

EXAMPLE 14.12 Hydroelectric power

Small hydroelectric plants in the mountains sometimes bring the water from a reservoir down to the power plant through enclosed tubes. In one such plant, the 100-cm-diameter intake tube in the base of the dam is 50 m below the reservoir surface. The water drops 200 m through the tube before flowing into the turbine through a 50-cm-diameter nozzle.

1. What is the water speed into the turbine?

2. By how much does the inlet pressure differ from the hydrostatic pressure at that depth?

MODEL Treat the water as an ideal fluid obeying Bernoulli’s equation. Consider a streamline that begins at the surface of the reservoir and ends at the exit of the nozzle. The pressure at the surface is $p_1=p_{\text{atmos}}$ and $v_1\approx 0\text{m}\text{/}\text{s}$. The water discharges into air, so $p_3=p_{\text{atmos}}$ at the exit.

VISUALIZE FIGURE 14.30 is a pictorial representation of the situation.

FIGURE 14.30 Pictorial representation of the water flow to a hydroelectric plant.

FIGURE 14.30 Full Alternative Text

SOLVE a. Bernoulli’s equation, with $v_1=0\text{m}\text{/}\text{s}$ and $y_3=0\text{m,}$ is

$$p_{\text{atmos}}+\rho g{y}_1=p_{\text{atmos}}+\frac{1}{2}\rho v_3{}^2$$

The power plant is in the mountains, where $p_{\text{atmos}}<1\text{atm,}$ but $p_{\text{atmos}}$ occurs on both sides of Bernoulli’s equation and cancels. Solving for $v_3$ gives

$$v_3=\sqrt{2g{y}_1}=\sqrt{2(9.80{\text{m/s}}^2)(250\text{m})}=70\text{m}\text{/}\text{s}$$

b. You might expect the pressure $p_2$ at the intake to be the hydrostatic pressure $p_{\text{atmos}}+\rho gd$ at depth d. But the water is flowing into the intake tube, so it’s not in hydrostatic equilibrium. We can find the intake speed $v_2$ from the equation of continuity:

$$v_2=\frac{A_3}{A_2}{v}_3=\frac{r_3{}^2}{r_2{}^2}\sqrt{2g{y}_1}$$

The intake is along the streamline between points 1 and 3, so we can apply Bernoulli’s equation to points 1 and 2:

$$p_{\text{atmos}}+\rho g{y}_1=p_2+\frac{1}{2}\rho v_2{}^2+\rho g{y}_2$$

Solving this equation for $p_2$, and noting that $y_1-y_2=d$, we find

$$\begin{array}{lll}\hfill p_2& =\hfill & p_{\text{atmos}}+\rho g\left(y_1-y_2\right)-\frac{1}{2}\rho v_2{}^2\hfill \\ \hfill & =\hfill & p_{\text{atmos}}+\rho gd-\frac{1}{2}\rho {\left({\displaystyle \frac{r_3}{r_2}}\right)}^4\left(2g{y}_1\right)\hfill \\ \hfill & =\hfill & p_{\text{hydrostatic}}-\rho g{y}_1{\left({\displaystyle \frac{r_3}{r_2}}\right)}^4\hfill \end{array}$$

The intake pressure is less than hydrostatic pressure by the amount

$$\rho g{y}_1{\left(\frac{r_3}{r_2}\right)}^4=153\text{,}000\text{ Pa}=1.5\text{ atm}$$

REVIEW The water’s exit speed from the nozzle is the same as if it fell 250 m from the surface of the reservoir. This isn’t surprising because we’ve assumed a nonviscous (i.e., frictionless) liquid. “Real” water would have less speed but still flow very fast.