Steady flow—Bernoulli’s theorem
Now we want to return to the equation of motion, Eq. (40.8), but limit ourselves to situations in which the flow is “steady.” By steady flow we mean that at any one place in the fluid the velocity never changes. The fluid at any point is always replaced by new fluid moving in exactly the same way. The velocity picture always looks the same—$\FLPv$ is a static vector field. In the same way that we drew “field lines” in magnetostatics, we can now draw lines which are always tangent to the fluid velocity as shown in Fig. 40-5. These lines are called streamlines.  For steady flow, they are evidently the actual paths of fluid particles. (In unsteady flow the streamline pattern changes in time, and the streamline pattern at any instant does not represent the path of a fluid particle.)  A steady flow does not mean that nothing is happening—atoms in the fluid are moving and changing their velocities. It only means that $\ddpl{\FLPv}{t}=\FLPzero$. Then if we take the dot product of $\FLPv$ into the equation of motion, the term $\FLPv\cdot(\FLPOmega\times\FLPv)$ drops out, and we are left with \begin{equation} \label{Eq:II:40:12} \FLPv\cdot\FLPgrad{\biggl\{ \frac{p}{\rho}+\phi+\frac{1}{2}\,v^2 \biggr\}}=0. \end{equation} This equation says that for a small displacement in the direction of the fluid velocity the quantity inside the brackets doesn’t change. Now in steady flow all displacements are along streamlines, so Eq. (40.12) tells us that for all the points along a streamline, we can write \begin{equation} \label{Eq:II:40:13} \frac{p}{\rho}+\frac{1}{2}\,v^2+\phi=\text{const}\:(\text{streamline}). \end{equation} This is Bernoulli’s theorem. The constant may in general be different for different streamlines; all we know is that the left-hand side of Eq. (40.13) is the same all along a given streamline. Incidentally, we may notice that for steady irrotational motion for which $\FLPOmega=\FLPzero$, the equation of motion (40.8) gives us the relation \begin{equation*} \FLPgrad{\biggl\{ \frac{p}{\rho}+\frac{1}{2}\,v^2+\phi \biggr\}}=0, \end{equation*} so that \begin{equation} \label{Eq:II:40:14} \frac{p}{\rho}+\frac{1}{2}\,v^2+\phi=\text{const}\:(\text{everywhere}). \end{equation} It’s just like Eq. (40.13) except that now the constant has the same value throughout the fluid.  The theorem of Bernoulli is in fact nothing more than a statement of the conservation of energy. A conservation theorem such as this gives us a lot of information about a flow without our actually having to solve the detailed equations. Bernoulli’s theorem is so important and so simple that we would like to show you how it can be derived in a way that is different from the formal calculations we have just used. Imagine a bundle of adjacent streamlines which form a stream tube as sketched in Fig. 40-6. Since the walls of the tube consist of streamlines, no fluid flows out through the wall. Let’s call the area at one end of the stream tube $A_1$, the fluid velocity there $v_1$, the density of the fluid $\rho_1$, and the potential energy $\phi_1$. At the other end of the tube, we have the corresponding quantities $A_2$, $v_2$, $\rho_2$, and $\phi_2$. Now after a short interval of time $\Delta t$, the fluid at $A_1$ has moved a distance $v_1\,\Delta t$, and the fluid at $A_2$ has moved a distance $v_2\,\Delta t$ [Fig. 40-6(b)]. The conservation of mass requires that the mass which enters through $A_1$ must be equal to the mass which leaves through $A_2$. These masses at these two ends must be the same: \begin{equation*} \Delta M=\rho_1A_1v_1\,\Delta t=\rho_2A_2v_2\,\Delta t. \end{equation*} So we have the equality \begin{equation} \label{Eq:II:40:15} \rho_1A_1v_1=\rho_2A_2v_2. \end{equation} This equation tells us that the velocity varies inversely with the area of the stream tube if $\rho$ is constant.  Now we calculate the work done by the fluid pressure. The work done on the fluid entering at $A_1$ is $p_1A_1v_1\,\Delta t$, and the work given up at $A_2$ is $p_2A_2v_2\,\Delta t$. The net work on the fluid between $A_1$ and $A_2$ is, therefore, \begin{equation*} p_1A_1v_1\,\Delta t-p_2A_2v_2\,\Delta t, \end{equation*} which must equal the increase in the energy of a mass $\Delta M$ of fluid in going from $A_1$ to $A_2$. In other words, \begin{equation} \label{Eq:II:40:16} p_1A_1v_1\,\Delta t-p_2A_2v_2\,\Delta t=\Delta M(E_2-E_1), \end{equation} where $E_1$ is the energy per unit mass of fluid at $A_1$, and $E_2$ is the energy per unit mass at $A_2$. The energy per unit mass of the fluid can be written as \begin{equation*} E=\tfrac{1}{2}v^2+\phi+U, \end{equation*} where $\tfrac{1}{2}v^2$ is the kinetic energy per unit mass, $\phi$ is the potential energy per unit mass, and $U$ is an additional term which represents the internal energy per unit mass of fluid. The internal energy might correspond, for example, to the thermal energy in a compressible fluid, or to chemical energy. All these quantities can vary from point to point. Using this form for the energies in (40.16), we have  \begin{equation*} \frac{p_1A_1v_1\,\Delta t}{\Delta M}- \frac{p_2A_2v_2\,\Delta t}{\Delta M}= \frac{1}{2}\,v_2^2+\phi_2+U_2- \frac{1}{2}\,v_1^2-\phi_1-U_1. \end{equation*} 
 \begin{align*} \frac{p_1A_1v_1\,\Delta t}{\Delta M}&- \frac{p_2A_2v_2\,\Delta t}{\Delta M}\,=\\[1ex] \frac{1}{2}\,v_2^2+\phi_2+U_2&- \frac{1}{2}\,v_1^2-\phi_1-U_1. \end{align*}  But we have seen that $\Delta M=\rho Av\,\Delta t$, so we get  \begin{equation} \label{Eq:II:40:17} \frac{p_1}{\rho_1}+\frac{1}{2}\,v_1^2+\phi_1+U_1= \frac{p_2}{\rho_2}+\frac{1}{2}\,v_2^2+\phi_2+U_2, \end{equation} 
 \begin{equation} \begin{aligned} &\frac{p_1}{\rho_1}+\frac{1}{2}\,v_1^2+\phi_1+U_1\,=\\[1ex] &\frac{p_2}{\rho_2}+\frac{1}{2}\,v_2^2+\phi_2+U_2, \end{aligned} \label{Eq:II:40:17} \end{equation}  which is the Bernoulli result with an additional term for the internal energy. If the fluid is incompressible, the internal energy term is the same on both sides, and we get again that Eq. (40.14) holds along any streamline.  We consider now some simple examples in which the Bernoulli integral gives us a description of the flow. Suppose we have water flowing out of a hole near the bottom of a tank, as drawn in Fig. 40-7. We take a situation in which the flow speed $v_{\text{out}}$ at the hole is much larger than the flow speed near the top of the tank; in other words, we imagine that the diameter of the tank is so large that we can neglect the drop in the liquid level. (We could make a more accurate calculation if we wished.) At the top of the tank the pressure is $p_0$, the atmospheric pressure, and the pressure at the sides of the jet is also $p_0$. Now we write our Bernoulli equation for a streamline, such as the one shown in the figure. At the top of the tank, we take $v$ equal to zero and we also take the gravity potential $\phi$ to be zero. At the hole $v=v_{\text{out}}$ and $\phi=-gh$, so that \begin{equation*} p_0=p_0+\tfrac{1}{2}\rho v_{\text{out}}^2-\rho gh, \end{equation*} or \begin{equation} \label{Eq:II:40:18} v_{\text{out}}=\sqrt{2gh}. \end{equation} This velocity is just what we would get for something which falls the distance $h$.  It is not too surprising, since the water at the exit gains kinetic energy at the expense of the potential energy of the water at the top. Do not get the idea, however, that you can figure out the rate that the fluid flows out of the tank by multiplying this velocity by the area of the hole. The fluid velocities as the jet leaves the hole are not all parallel to each other but have components inward toward the center of the stream—the jet is converging. After the jet has gone a little way, the contraction stops and the velocities do become parallel. So the total flow is the velocity times the area at that point. In fact, if we have a discharge opening which is just a round hole with a sharp edge, the jet contracts to $62$ percent of the area of the hole.  The reduced effective area of the discharge varies for different shapes of discharge tubes, and experimental contractions are available as tables of efflux coefficients.  If the discharge tube is re-entrant, as shown in Fig. 40-8, it is possible to prove in a most beautiful way that the efflux coefficient is exactly $50$ percent. We will give just a hint of how the proof goes. We have used the conservation of energy to get the velocity, Eq. (40.18), but there is also momentum conservation to consider. Since there is an outflow of momentum in the discharge jet, there must be a force applied over the cross section of the discharge tube. Where does the force come from? The force must come from the pressure on the walls. As long as the efflux hole is small and away from the walls, the fluid velocity near the walls of the tank will be very small. Therefore, the pressure on every face is almost exactly the same as the static pressure in a fluid at rest—from Eq. (40.14). Then the static pressure at any point on the side of the tank must be matched by an equal pressure at the point on the opposite wall, except at the points on the wall opposite the charge tube. If we calculate the momentum poured out through the jet by this pressure, we can show that the efflux coefficient is $1/2$. We cannot use this method for a discharge hole like that shown in Fig. 40-7, however, because the velocity increase along the wall right near the discharge area gives a pressure fall which we are not able to calculate.  Let’s look at another example—a horizontal pipe with changing cross section, as shown in Fig. 40-9, with water flowing in one end and out the other. The conservation of energy, namely Bernoulli’s formula, says that the pressure is lower in the constricted area where the velocity is higher. We can easily demonstrate this effect by measuring the pressure at different cross sections with small vertical columns of water attached to the flow tube through holes small enough so that they do not disturb the flow. The pressure is then measured by the height of water in these vertical columns. The pressure is found to be less at the constriction than it is on either side. If the area beyond the constriction comes back to the same value it had before the constriction, the pressure rises again. Bernoulli’s formula would predict that the pressure downstream of the constriction should be the same as it was upstream, but actually it is noticeably less. The reason that our prediction is wrong is that we have neglected the frictional, viscous forces which cause a pressure drop along the tube. Despite this pressure drop the pressure is definitely lower at the constriction (because of the increased speed) than it is on either side of it—as predicted by Bernoulli. The speed $v_2$ must certainly exceed $v_1$ to get the same amount of water through the narrower tube. So the water accelerates in going from the wide to the narrow part. The force that gives this acceleration comes from the drop in pressure.  We can check our results with another simple demonstration. Suppose we have on a tank a discharge tube which throws a jet of water upward as shown in Fig. 40-10. If the efflux velocity were exactly $\sqrt{2gh}$, the discharge water should rise to a level even with the surface of the water in the tank. Experimentally, it falls somewhat short. Our prediction is roughly right, but again viscous friction which has not been included in our energy conservation formula has resulted in a loss of energy.  Have you ever held two pieces of paper close together and tried to blow them apart? Try it! They come together. The reason, of course, is that the air has a higher speed going through the constricted space between the sheets than it does when it gets outside. The pressure between the sheets is lower than atmospheric pressure, so they come together rather than separating. 