Bernoulli’s equation, a statement about the energy of flowing fluids, is equivalent to our earlier discovery that mechanical energy is conserved for a system that has no dissipative forces. But nearly all real-world systems do have dissipative forces that generate thermal energy. Flowing fluids dissipate energy due to their viscosity.

Viscosity $\eta$, a fluid’s resistance to flow, was introduced in 《 SECTION 6.5 when we looked at the drag force on an object moving through a stationary fluid. Now we want to analyze a moving fluid in a stationary tube. We’ll restrict our analysis to tubes that have circular cross sections. TABLE 14.3 gives values of $\eta$ for some common fluids. Note that viscosity decreases very rapidly with an increase in temperature.

Viscosity has a profound effect on how a fluid moves through a tube. FIGURE 14.33a shows that in an ideal fluid, which has no viscosity, all the fluid particles move with the same speed v, the speed that appears in Bernoulli’s equation. For a viscous fluid, seen in Figure 14.33b, the fluid moves fastest in the center of the tube. The speed decreases away from the center until it reaches zero on the walls of the tube; that is, the fluid in contact with the walls of the tube does not move at all. Whether it is water moving through pipes or blood through arteries, the fact that the fluid at the outer edges “lingers” and barely moves allows deposits to build up on the inside walls of a tube.

FIGURE 14.33 Viscosity alters the speed profile of a fluid flowing through a tube.

We can’t characterize the flow of a viscous fluid by a single speed v, but we can define an average flow speed $v_{\text{avg}}$. The volume flow rate $Q=\text{Δ}V/\text{Δ}t$ is still a well-defined quantity, so, corresponding to $Q=vA$ for an ideal fluid, we define the average flow speed of a viscous fluid through a tube with cross-section area $A$ by

FIGURE 14.34 shows fluid flowing smoothly (i.e., laminar flow) with average speed $v_{\text{avg}}$ through a horizontal tube that has constant radius r. There’s no change of height $(y_2=y_1)$ and no change of speed along a streamline between points 1 and 2 $(v_2=v_1)$. If this is an ideal fluid, then we learn from Bernoulli’s equation that $\text{Δ}p=p_1-p_2=0$. That is, there’s no net force on this segment of fluid; it simply “coasts” through the tube at constant speed with no change in pressure. It’s equivalent to a puck gliding at constant speed across a frictionless surface with no applied force.

FIGURE 14.34 Laminar flow of a viscous fluid requires a pressure difference.

If there’s friction, however, something has to apply a steady force—equal and opposite the friction force—to keep the puck moving at a steady speed. Likewise, something has to apply a steady force—equal and opposite the viscous drag—to push a viscous fluid through a tube at a steady speed. That “something” is the pressure difference $\text{Δ}p$ between the ends of the tube. A pressure difference is needed to keep a viscous fluid flowing, whereas no pressure difference is needed for an ideal fluid. The net force on the segment of fluid in the tube of Figure 14.34 is $F_{\text{net}}=A\text{Δ}p$.

We can use the energy principle to determine how large the pressure difference must be. Our starting point for finding Bernoulli’s equation was $W_{\text{ext}}=\text{Δ}K+\text{Δ}U$, which is valid for a system that has no dissipative forces. Dissipative forces transform mechanical energy into thermal energy, which leads to $W_{\text{ext}}=\text{Δ}K+\text{Δ}U+\text{Δ}{E}_{\text{th}}$. In 《 SECTION 9.5, we found that $\text{Δ}{E}_{\text{th}}=f_{\text{k}}\text{Δ}x$ when friction is the dissipative force. This is easily generalized: If $F_{\text{drag}}$ is the “friction” due to a fluid’s viscosity, then pushing the fluid a distance $\text{Δ}x$ against the viscous drag generates thermal energy $\text{Δ}{E}_{\text{th}}=F_{\text{drag}}\text{Δ}x$.

We will limit our analysis to the horizontal tube of Figure 14.34, in which case $\text{Δ}{U}_{\text{G}}=0$ and also $\text{Δ}K=0$ because $v_{\text{avg}}$ doesn’t change in a tube of constant diameter. We’ve already calculated the net work done by a pressure difference: Equation 14.23 found $W_{\text{ext}}=\text{Δ}pV=\text{Δ}pA\text{Δ}x$ for the work done to push the fluid through $\text{Δ}x$. Thus the energy principle for a viscous fluid flowing through this horizontal tube is

The work done by the pressure difference does not increase the fluid’s kinetic energy or potential energy, only the thermal energy of the fluid and the tube.

In Section 6.5 we found that the drag on a sphere of radius $r$ moving through a fluid with viscosity $\eta$ is $F_{\text{drag}}=6\pi \eta vr$. We see that drag depends linearly on the viscosity, the speed, and the distance over which viscous forces act. Thus it would seem likely that the viscous drag on a segment of fluid flowing through a tube of length $L$ is $F_{\text{drag}}=c\eta v_{\text{avg}}L$, where $c$ is an unknown constant. Delving deeper into the fundamentals of fluid mechanics, it can be shown that $c=8\pi$ and thus $F_{\text{drag}}=8\pi \eta v_{\text{avg}}L$. Then, using this in Equation 14.35, we find that the pressure difference needed to push a viscous fluid through a tube with average speed $v_{\text{avg}}$ is

where $A=\pi r^2$ is used as the cross-section area. A larger pressure difference is needed for a more viscous fluid, a longer tube, or a narrower tube.

Solving for $v_{\text{avg}}$, we find that a pressure difference $\text{Δ}p$ across a tube of length $L$ causes a fluid to flow with average speed

Not surprisingly, a more viscous fluid flows more slowly.

The more practical quantity, because it’s easily measured, is the volume flow rate Q. Because $Q=v_{\text{avg}}A=\pi r^2{v}_{\text{avg}}$, the flow rate is

This is called Poiseuille’s equation for viscous flow, named for a 19th-century French scientist who studied fluids and was especially interested in blood flow.

The quantity $\text{Δ}p/L$ is called the pressure gradient. It is the pressure change per unit length or, equivalently, the slope of a pressure-versus-distance graph. A large pressure change over a small distance is a large pressure gradient. We can interpret Poiseuille’s equation as saying that a pressure gradient drives a fluid flow. This is our first introduction to the idea that gradients drive flows, but it won’t be our last.

One surprising consequence of Poiseuille’s equation is the very strong dependence of the flow on the tube’s radius: The flow rate is proportional to the fourth power of r. Fairly small changes in the radius have a large effect on the flow rate.

EXAMPLE 14.13 Blood flow through a clogged artery

Atherosclerosis is a condition in which an artery is narrowed by the buildup of fatty plaques on its inner wall (as shown in the photo). The process restricts blood flow to organs, including the heart, and can lead to serious health issues. Suppose the diameter of an artery is reduced by 25%, a not untypical value. By what percent is the blood flow reduced if the pressure difference stays constant? By what factor would the pressure difference established by the heart have to increase to keep the blood flow rate constant?

MODEL Model the blood flow through the artery as viscous flow through a tube of constant diameter.

SOLVE Suppose the flow rate is $Q_1$ through an unclogged artery of radius $r_1$ and $Q_2$ through a clogged artery with a smaller radius $r_2=0.75r_1$. From Poiseuille’s equation, Equation 14.38, we find

$$\frac{Q_2}{Q_1}=\frac{r_2^4}{r_1^4}={(0.75)}^4=0.32$$

A 25% reduction in diameter leads to a huge 68% reduction in flow if the pressure difference is unchanged. Conversely, for a constant flow rate, the pressure difference $\text{Δ}p$ is inversely proportional to $r^4$. Thus

$$\frac{\text{Δ}{p}_2}{\text{Δ}{p}_1}=\frac{(1/r_2{)}^4}{(1/r_1{)}^4}={\left(\frac{1}{0.75}\right)}^4=3.2$$

Blood pressure would have to increase by a factor of 3.2 to maintain a constant flow rate.

REVIEW The fourth-power dependence on $r$ has profound consequences for the flow of a viscous fluid. Physiologically, the heart is not able to substantially increase the pressure, so atherosclerosis causes a significant reduction of blood flow. It is the leading cause of death in the United States.

EXAMPLE 14.14 Measuring viscosity

A liquid with density $1200{\text{kg/m}}^3$ flows through the horizontal tube shown in FIGURE 14.35. The flow rate is measured to be $7.5\text{L}\text{/}\text{min}$. What is the liquid’s viscosity?

FIGURE 14.35 Fluid flowing through a constant-diameter tube.

FIGURE 14.35 Full Alternative Text

MODEL Treat the flow as the flow of a viscous fluid.

VISUALIZE The 8.1 cm difference in liquid heights in the vertical tubes shows that there’s a pressure difference and thus a pressure gradient between points $L=75\text{cm}=0.75\text{m}$ apart. The tube’s radius is $r=0.50\text{cm}=5.0\times {10}^{-3}\text{m}$.

SOLVE Poiseuille’s equation can be solved for the viscosity once all the other terms are known. We don’t need to know the absolute pressures, just the pressure difference. The liquid in the vertical tubes is stationary, so we can use the hydrostatic pressure to find

$$\text{Δ}p=\rho g\text{Δ}h=(1200{\text{kg/m}}^3)(9.80{\text{m/s}}^2)(0.081\text{m)}=953\text{Pa}$$

Thus the pressure gradient is $\text{Δ}p/L=1270\text{Pa}\text{/}\text{m}$. The volume flow rate is a measured quantity that must be converted to SI units:

$$Q=7.5\frac{\text{L}}{\text{min}}\times \frac{1{\text{m}}^3}{1000\text{L}}\times \frac{1\text{min}}{60\text{s}}=1.25\times {10}^{-4}{\text{m}}^3/\text{s}$$

With this information, we find

$$\eta =\frac{\pi r^4\text{Δ}p}{8LQ}=\frac{\pi (5.0\times {10}^{-3}{\text{m)}}^4(953\text{Pa)}}{8(0.75\text{m})(1.25\times {10}^{-4}{\text{m}}^3/\text{s)}}=2.5\times {10}^{-3}\text{Pa}\text{s}$$

REVIEW The result is reasonable because it is similar to the viscosities given in Table 14.3.