Bernoulli's equation

What is Bernoulli's principle?

Bernoulli's principle is a seemingly counterintuitive statement about how the speed of a fluid relates to the pressure of the fluid. Many people feel like Bernoulli's principle shouldn't be correct, but this might be due to a misunderstanding about what Bernoulli's principle actually says. Bernoulli's principle states the following,

Bernoulli's principle: Within a horizontal flow of fluid, points of higher fluid speed will have less pressure than points of slower fluid speed. 

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So within a horizontal water pipe that changes diameter, regions where the water is moving fast will be under less pressure than regions where the water is moving slow. This sounds counterintuitive to many people since people associate high speeds with high pressures. But, we'll show in the next section that this is really just another way of saying that water will speed up if there's more pressure behind it than in front of it. In the section below we'll derive Bernoulli's principle, show more precisely what it says, and hopefully make it seem a little less mysterious.

How can you derive Bernoulli's principle?

Incompressible fluids have to speed up when they reach a narrow constricted section in order to maintain a constant volume flow rate. This is why a narrow nozzle on a hose causes water to speed up. But something might be bothering you about this phenomenon. If the water is speeding up at a constriction, it's also gaining kinetic energy. Where is this extra kinetic energy coming from? The nozzle? The pipe? 

[The energy fairy?]

The only way to give something kinetic energy is to do work on it. This is expressed by the work energy principle.

$W_{external}=\Delta K=\dfrac{1}{2}mv_f^2-\dfrac{1}{2}mv_i^2$W, start subscript, e, x, t, e, r, n, a, l, end subscript, equals, delta, K, equals, start fraction, 1, divided by, 2, end fraction, m, v, start subscript, f, end subscript, squared, minus, start fraction, 1, divided by, 2, end fraction, m, v, start subscript, i, end subscript, squared

So if a portion of fluid is speeding up, something external to that portion of fluid must be doing work it. What force is causing work to be done on the fluid? Well, in most real world systems there are lots of dissipative forces that could be doing negative work, but we're going to assume for the sake of simplicity that these viscous forces are negligible and we have a nice continuous and perfectly laminar (streamline) flow. Laminar (streamline) flow means that the fluid flows in parallel layers without crossing paths. In laminar streamline flow there is no swirling or vortices in the fluid. 

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$^{[1]}$start superscript, open bracket, 1, close bracket, end superscript

$^{[1]}$start superscript, open bracket, 1, close bracket, end superscript

OK, so we'll assume we have no loss in energy due to dissipative forces. In that case, what non-dissipative forces could be doing work on our fluid that cause it to speed up? The pressure from the surrounding fluid will be causing a force that can do work and speed up a portion of fluid.

Consider the diagram below which shows water flowing along streamlines from left to right. As the outlined volume of water enters the constricted region it speeds up. The force from pressure $P_1$P, start subscript, 1, end subscript on the left side of the shaded water pushes to the right and does positive work since it pushes in the same direction as the motion of the shaded fluid. The force from pressure $P_2$P, start subscript, 2, end subscript on the right side of the shaded fluid pushes to the left and does negative work since it pushes in the opposite direction as the motion of the shaded fluid.

We know that the water must speed up (due to the continuity equation) and therefore have a net positive amount of work done on it. So the work done by the force from pressure on the left side must be larger than the amount of negative work done by the force from pressure on the right side. This means that the pressure on the wider/slower side $P_1$P, start subscript, 1, end subscript has to be larger than the pressure on the narrow/faster side $P_2$P, start subscript, 2, end subscript. 

[Wait, does that really follow?]

$W=Fd$W, equals, F, d$F=PA$F, equals, P, A

$W=Fd=PAd$W, equals, F, d, equals, P, A, d

$P_1$P, start subscript, 1, end subscript$A$A$d$d$P_2$P, start subscript, 2, end subscript$P_1$P, start subscript, 1, end subscript$P_1$P, start subscript, 1, end subscript$P_2$P, start subscript, 2, end subscript

$P_1$P, start subscript, 1, end subscript$P_2$P, start subscript, 2, end subscript

This inverse relationship between the pressure and speed at a point in a fluid is called Bernoulli's principle.

Bernoulli's principle: At points along a horizontal streamline, higher pressure regions have lower fluid speed and lower pressure regions have higher fluid speed.

It might be conceptually simplest to think of Bernoulli's principle as the fact that a fluid flowing from a high pressure region to a low pressure region will accelerate due to the net force along the direction of motion.

The idea that regions where the fluid is moving fast will have lower pressure can seem strange. Surely, a fast moving fluid that strikes you must apply more pressure to your body than a slow moving fluid, right? Yes, that is right. But we're talking about two different pressures now. The pressure that Bernoulli's principle is referring to is the internal fluid pressure that would be exerted in all directions during the flow, including on the sides of the pipe. This is different from the pressure a fluid will exert on you if you get in the way of it and stop its motion. 

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Note that Bernoulli's principle does not say that a fast moving fluid can't have significantly high pressures. It just says that the pressure in a slower region of that same flowing system must have even larger pressure than the faster moving region.

What is Bernoulli's equation?

Bernoulli's equation is essentially a more general and mathematical form of Bernoulli's principle that also takes into account changes in gravitational potential energy. We'll derive this equation in the next section, but before we do, let's take a look at Bernoulli's equation and get a feel for what it says and how one would go about using it.

Bernoulli's equation relates the pressure, speed, and height of any two points (1 and 2) in a steady streamline flowing fluid of density $\rho$rho. Bernoulli's equation is usually written as follows,

$\Large P_1+\dfrac{1}{2}\rho v^2_1+\rho gh_1=P_2+\dfrac{1}{2}\rho v^2_2+\rho gh_2$P, start subscript, 1, end subscript, plus, start fraction, 1, divided by, 2, end fraction, rho, v, start subscript, 1, end subscript, squared, plus, rho, g, h, start subscript, 1, end subscript, equals, P, start subscript, 2, end subscript, plus, start fraction, 1, divided by, 2, end fraction, rho, v, start subscript, 2, end subscript, squared, plus, rho, g, h, start subscript, 2, end subscript

The variables $P_1$P, start subscript, 1, end subscript, $v_1$v, start subscript, 1, end subscript, $h_1$h, start subscript, 1, end subscript refer to the pressure, speed, and height of the fluid at point 1, whereas the variables $P_2$P, start subscript, 2, end subscript, $v_2$v, start subscript, 2, end subscript, and $h_2$h, start subscript, 2, end subscript refer to the pressure, speed, and height of the fluid at point 2 as seen in the diagram below. The diagram below shows one particular choice of two points (1 and 2) in the fluid, but Bernoulli's equation will hold for any two points in the fluid.

When using Bernoulli's equation, how do you know where to choose your points? Choosing one of the points at the location where you want to find an unknown variable is a must. Otherwise how will you ever solve for that variable? You will typically choose the second point at a location where you have been given some information, or where the fluid is open to the atmosphere, since the absolute pressure there is known to be atmospheric pressure $P_{atm}=1.01\times 10^5Pa$P, start subscript, a, t, m, end subscript, equals, 1, point, 01, times, 10, start superscript, 5, end superscript, P, a.

Note that the $h$h refers to the height of the fluid above an arbitrary level that you can choose in any way that is convenient. Typically it is often easiest to just choose the lower of the two points (1 or 2) as the height where $h=0$h, equals, 0. The $P$P refers to the pressure at that point. You can choose to use gauge pressure or absolute pressure, but whichever kind of pressure you choose (gauge or absolute) must also be used on the other side of the equation. You can't insert the gauge pressure at point 1, and the absolute pressure at point 2. Similarly, if you insert the gauge pressure at point 1 and solve for the pressure at point 2, the value you obtain will be the gauge pressure at point 2 (not the absolute pressure).

The terms $\dfrac{1}{2}\rho v^2$start fraction, 1, divided by, 2, end fraction, rho, v, squared and $\rho gh$rho, g, h in Bernoulli's equation look just like kinetic energy $\dfrac{1}{2}m v^2$start fraction, 1, divided by, 2, end fraction, m, v, squared and potential energy $mgh$m, g, h, only with the mass $m$m replaced with density $\rho$rho. So it may not come as much of a surprise that Bernoulli's equation is the result of applying conservation of energy to a flowing fluid. We'll derive Bernoulli's equation using conservation of energy in the next section.

How do you derive Bernoulli's equation?

Consider the following diagram where water flows from left to right in a pipe that changes both area and height. As before, water will speed up and gain kinetic energy $K$K at constrictions in the pipe, since the volume flow rate must be maintained for an incompressible fluid even if those constricted sections move upward. But now since the constriction also causes the fluid to move upward, the water will be gaining gravitational potential energy $U_g$U, start subscript, g, end subscript as well as kinetic energy $K$K. We will derive Bernoulli's equation by setting the energy gained by the fluid equal to the external work done on the fluid. 

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Let's assume the energy system we're considering is composed of the volumes of water 1 and 2 as well as all the fluid in between those volumes. If we assume the fluid flow is streamline, non-viscous, and there are no dissipative forces affecting the flow of the fluid, then any extra energy $\Delta ({K+U})_{system}$delta, left parenthesis, K, plus, U, right parenthesis, start subscript, s, y, s, t, e, m, end subscript added to the system will be caused by the external work $(W_{external})$left parenthesis, W, start subscript, e, x, t, e, r, n, a, l, end subscript, right parenthesis done on the fluid from pressure forces surrounding it. 

$(K+U_g)$left parenthesis, K, plus, U, start subscript, g, end subscript, right parenthesis

We can express this mathematically as,

$W_{external} = \Delta ({K+U})_{system}$W, start subscript, e, x, t, e, r, n, a, l, end subscript, equals, delta, left parenthesis, K, plus, U, right parenthesis, start subscript, s, y, s, t, e, m, end subscript

First we'll try to find the external work done $W_{external}$W, start subscript, e, x, t, e, r, n, a, l, end subscript on the water. None of the water between points 1 and 2 can do external work since that water is all part of our energy system. The only pressures that can directly do external work on our system are $P_1$P, start subscript, 1, end subscript and $P_2$P, start subscript, 2, end subscript as shown in the diagram. The water at $P_1$P, start subscript, 1, end subscript to the left of volume 1 will do positive work since the force points in the same direction as the motion of the fluid. The water at $P_2$P, start subscript, 2, end subscript to the right of volume 2 will do negative work on our system since it pushes in the opposite direction as the motion of the fluid.

For simplicity's sake we'll consider the case where the force from water pressure to the left of volume 1 pushes volume 1 through its entire width $d_1$d, start subscript, 1, end subscript. Assuming the fluid is incompressible, this must displace an equal volume of water everywhere in the system, causing volume 2 to be displaced through its length a distance $d_2$d, start subscript, 2, end subscript.

Work can be found with $W=Fd$W, equals, F, d. We can plug in the formula for the force from pressure $F=PA$F, equals, P, A into the formula from work to get $W=PAd$W, equals, P, A, d. So, the positive work done on our system by the water near point 1 will be $W_1=P_1A_1d_1$W, start subscript, 1, end subscript, equals, P, start subscript, 1, end subscript, A, start subscript, 1, end subscript, d, start subscript, 1, end subscript and the work done by the water near point 2 will be $W_2=-P_2A_2d_2$W, start subscript, 2, end subscript, equals, minus, P, start subscript, 2, end subscript, A, start subscript, 2, end subscript, d, start subscript, 2, end subscript. 

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