# Czym jest równanie Bernoulliego?

This equation will give you the powers to analyze a fluid flowing up and down through all kinds of different tubes.

## 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.
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 only way to give something kinetic energy is to do work on it. This is expressed by the work energy principle.
$W_{external}=\Delta K=\dfrac{1}{2}mv_f^2-\dfrac{1}{2}mv_i^2$
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.
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 $P_1$ 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 $P_2$ 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 $P_1$ has to be larger than the pressure on the narrow/faster side $P_2$.
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.
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 $\rho$. Bernoulli's equation is usually written as follows,
$\Large P_1+\dfrac{1}{2}\rho v^2_1+\rho gh_1=P_2+\dfrac{1}{2}\rho v^2_2+\rho gh_2$
The variables $P_1$, $v_1$, $h_1$ refer to the pressure, speed, and height of the fluid at point 1, whereas the variables $P_2$, $v_2$, and $h_2$ 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 $P_{atm}=1.01\times 10^5Pa$.
Note that the $h$ 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 $h=0$. The $P$ 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 $\dfrac{1}{2}\rho v^2$ and $\rho gh$ in Bernoulli's equation look just like kinetic energy $\dfrac{1}{2}m v^2$ and potential energy $mgh$, only with the mass $m$ replaced with density $\rho$. 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 $K$ 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 $U_g$ as well as kinetic energy $K$. We will derive Bernoulli's equation by setting the energy gained by the fluid equal to the external work done on the fluid.
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 $\Delta ({K+U})_{system}$ added to the system will be caused by the external work $(W_{external})$ done on the fluid from pressure forces surrounding it.
We can express this mathematically as,
$W_{external} = \Delta ({K+U})_{system}$
First we'll try to find the external work done $W_{external}$ 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 $P_1$ and $P_2$ as shown in the diagram. The water at $P_1$ 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 $P_2$ 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 $d_1$. 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 $d_2$.
Work can be found with $W=Fd$. We can plug in the formula for the force from pressure $F=PA$ into the formula from work to get $W=PAd$. So, the positive work done on our system by the water near point 1 will be $W_1=P_1A_1d_1$ and the work done by the water near point 2 will be $W_2=-P_2A_2d_2$.
Plugging these expressions for work into the left side of our work energy formula $W_{net} = \Delta ({K+U})_{system}$ we get,
$P_1A_1d_1-P_2A_2d_2=\Delta ({K+U})_{system}$
But the terms $A_1d_1$ and $A_2d_2$ have to be equal since they represent the volumes of the fluid displaced near point 1 and point 2. If we assume the fluid is incompressible, an equal volume of fluid must be displaced everywhere in the fluid, including near the top. So, $V_1=A_1d_1=A_2d_2=V_2$. We can just write the volume term simply as $V$ since the volumes are equal. This simplifies the left side of the work energy formula to,
$P_1V-P_2V= \Delta ({K+U})_{system}$
That takes care of the left hand side. Now we have to address the right hand side of this equation. This is a crucial and subtle part of the derivation. Remember that our system includes not only the shaded portions of water near point 1 and 2, but also all the water in between those two points. How will we ever be able to account for all the change in kinetic energy and gravitational potential energy of all parts of that large and winding system?
Well, we have to make one more assumption to finish the derivation. We're going to assume that the flow of the fluid is steady. By "steady flow" we mean that the speed of the fluid passing by a particular point in the pipe doesn't change. In other words, if you stood and stared at any one particular section of the transparent pipe, you would see new water moving past you every moment, but if there's steady flow, then all the water would have the same speed when it moves past that particular point.
So how does the idea of steady flow help us figure out the change in energy of the big winding system of fluid? Consider the diagram below. Our energy system consists of the greyed out fluid (volume 1, volume 2, and all fluid in between). In the first image, the system has some amount of total energy $(K+U)_{initial}$. In the second image the entire system had work done on it, gains energy, shifts to the right, and now has some different total energy $(K+U)_{final}$. But notice that the energy of the fluid between the dashed lines will be the same as it was before the work was done assuming a steady flow. Water changed position and speed in the region between the dashed lines, but it did so in such a way that it will be moving with the exact same speed (e.g. $v_a$ and $v_b$) and have the same height as the previous water had in that location. The only thing that's different about our system is that volume 2 now extends into a section of the pipe it wasn't in previously, and now nothing in our system is occupying the old position behind volume 1.
Overall this means that the total change in the energy of the system can be found by simply considering the energies of the end points. Namely, we can take the kinetic and potential energy $(K_2+U_2)$ that now exists in volume 2 after the work was done and subtract the kinetic and potential energy $(K_1+U_1)$ that no longer exists behind volume 1 after the work was done. In other words, $\Delta (K+U)_{system}=(K_2+U_2)-(K_1+U_1)$.
Plugging this into the right hand side of the work energy formula $P_1V-P_2V= \Delta ({K+U})_{system}$ we get,
$P_1V-P_2V= (K_2+U_2)-(K_1+U_1)$
Now we'll substitute in the formulas for kinetic energy $K=\dfrac{1}{2}mv^2$ and gravitational potential energy $U_g=mgh$ to get,
$P_1V-P_2V= (\dfrac{1}{2}m_2v^2_2+m_2gh_2)-(\dfrac{1}{2}m_1v^2_1+m_1gh_1)$
In this equation $P_1$ and $P_2$ represent the pressures of the fluid in volumes 1 and 2 respectively. The variables $v_1$ and $v_2$ represent the speeds of the fluid in volumes 1 and 2 respectively. And $h_1$ and $h_2$ represent the height of the fluid in volumes 1 and 2 respectively.
But since we are assuming the fluid is incompressible, the displaced masses of volumes 1 and 2 must be the same $m_1=m_2=m$. So getting rid of the subscript on the $m$'s we get,
$P_1V-P_2V= (\dfrac{1}{2}mv^2_2+mgh_2)-(\dfrac{1}{2}mv^2_1+mgh_1)$
We can divide both sides by $V$ and drop the parenthesis to get,
$P_1-P_2= \dfrac{\dfrac{1}{2}mv_2^2}{V}+\dfrac{mgh_2}{V}-\dfrac{\dfrac{1}{2}m v^2_1}{V}-\dfrac{mgh_1}{V}$
We can simplify this equation by noting that the mass of the displaced fluid divided by volume of the displaced fluid is the density of the fluid $\rho=\dfrac{m}{V}$. Replacing $\dfrac{m}{V}$ with $\rho$ we get,
$P_1-P_2= \dfrac{1}{2}\rho v_2^2+\rho gh_2-\dfrac{1}{2}\rho v^2_1-\rho gh_1$
Now, we're just going to rearrange the formula using algebra to put all the terms that refer to the same point in space on the same side of the equation to get,
$\Large P_1+\dfrac{1}{2}\rho v^2_1+\rho gh_1=P_2+\dfrac{1}{2}\rho v^2_2+\rho gh_2$
And there it is, finally. This is Bernoulli's equation! It says that if you add up the pressure $P$ plus the kinetic energy density $\dfrac{1}{2}\rho v^2$ plus the gravitational potential energy density $\rho gh$ at any 2 points in a streamline, they will be equal.
Bernoulli's equation can be viewed as a conservation of energy law for a flowing fluid. We saw that Bernoulli's equation was the result of using the fact that any extra kinetic or potential energy gained by a system of fluid is caused by external work done on the system by another non-viscous fluid. You should keep in mind that we had to make many assumptions along the way for this derivation to work. We had to assume streamline flow and no dissipative forces, since otherwise there would have been thermal energy generated. We had to assume steady flow, since otherwise our trick of canceling the energies of the middle section would not have worked. We had to assume incompressibility, since otherwise the volumes and masses would not necessarily be equal.
Since the quantity $P+\dfrac{1}{2}\rho v^2+\rho gh$ is the same at every point in a streamline, another way to write Bernoulli's equation is,
$P+\dfrac{1}{2}\rho v^2+\rho gh=\text{constant}$
This constant will be different for different fluid systems, but for a given steady state streamline non-dissipative flowing fluid, the value of $P+\dfrac{1}{2}\rho v^2+\rho gh$ will be the same at any point along the flowing fluid.

## How is Bernoulli's principle a result of Bernoulli's equation?

We should note here that Bernoulli's principle is contained within Bernoulli's equation. If we start with,
$P_1+\dfrac{1}{2}\rho v^2_1+\rho gh_1=P_2+\dfrac{1}{2}\rho v^2_2+\rho gh_2$
and assume that there is no change in the height of the fluid, the $\rho gh$ terms cancel if we subtract them from both sides.
$P_1+\dfrac{1}{2}\rho v^2_1=P_2+\dfrac{1}{2}\rho v^2_2$
Or we could write it as,
$P+\dfrac{1}{2}\rho v^2=\text{constant}$
This formula highlights Bernoulli's principle since if the speed $v$ of a fluid is larger in a given region of streamline flow, the pressure $P$ must be smaller in that region (which is Bernoulli's principle). An increase in speed $v$ must be accompanied by a simultaneous decrease in the pressure $P$ in order for the sum to always add up to the same constant number.

## What do solved examples involving Bernoulli's equation look like?

### Example 1: Root Beer blueprints

You own a restaurant that is investigating new ways to deliver beverages to customers. One proposal is for a tube that will deliver root beer of density $1,090\dfrac{kg}{m^3}$ throughout the restaurant. A section of the tube is shown below. The blueprints say that the speed and gauge pressure of the root beer at point 1 are $3.00\text{ m/s}$ and $12,300\text{ Pa}$ respectively. The root beer at point 2 is $1.20\text{ m}$ higher than the fluid at point 1 and is traveling at a speed of $0.750\text{ m/s}$. You can't make out the number on the blueprints for the pressure of the root beer at point 2.
Use Bernoulli's equation to figure out the gauge pressure of the root beer at point 2.
$P_1+\dfrac{1}{2}\rho v^2_1+\rho gh_1=P_2+\dfrac{1}{2}\rho v^2_2+\rho gh_2 \quad\text{(first, start with Bernoulli's equation)}$
$P_2=P_1+\dfrac{1}{2}\rho v^2_1+\rho gh_1-\dfrac{1}{2}\rho v^2_2-\rho gh_2 \quad\text{(algebraically solve Bernoulli's equation for }P_2)$
At this point we need to choose an $h=0$ reference line. We'll choose the height at point 1 to be $h=0$. This makes the value $h_1=0$ and $h_2=1.2\text{ m}$. Plugging in these values for the height we get,
$P_2=P_1+\dfrac{1}{2}\rho v^2_1+\rho g(0\text{ m})-\dfrac{1}{2}\rho v^2_2-\rho g(1.2{ m}) \quad\text{(we plug in the values of h_1 and h_2)}$
We can get rid of the term with the zero in it and plug in numerical values for the other variables to get,
$P_2=12,300\text{ Pa}+\dfrac{1}{2}(1,090\dfrac{kg}{m^3}) (3.00\text{ m/s})^2-\dfrac{1}{2} (1,090\dfrac{kg}{m^3})(0.750\text{ m/s})^2-(1,090\dfrac{kg}{m^3})g(1.20\text{ m})$
$P_2= 4,080 \text{ Pa}\quad{\text{(calculate and celebrate)}}$
Note: We know this is the gauge pressure at point 2, rather than the absolute pressure, since we plugged in the gauge pressure for point 1. If we wanted the absolute pressure we could add atmospheric pressure $(1.01\times 10^5 \text{ Pa})$ to our answer.

### Example 2: Water fountain engineering

A large hotel has asked you build a water fountain that is fed by a $15\text{ cm}$ diameter cylindrical pipe that carries water horizontally $8.00\text{ m}$ below the ground. The pipe turns upwards and eventually fires water out of the $5.00\text{ cm}$ diameter end of the cylindrical pipe, which is located $1.75\text{ m}$ above the ground, with a speed of $32.0 \text{ m/s}$. Water has a density of $1,000 \dfrac{kg}{m^3}$.
What gauge pressure is required in the large underwater horizontal pipe for this fountain?
These Bernoulli's equation problems are complicated so we should draw a diagram of the situation and pick two points of interest. (this diagram is not to scale)
We'll pick the point near the bottom of the pipe as point 1, since that's where we want to determine the pressure, and we'll pick the top of the pipe where the water emerges as point 2 since we have been given information about the speed of the water at that point.
$P_1+\dfrac{1}{2}\rho v^2_1+\rho gh_1=P_2+\dfrac{1}{2}\rho v^2_2+\rho gh_2 \quad\text{(first, start with Bernoulli's equation)}$
$P_1=P_2+\dfrac{1}{2}\rho v^2_2+\rho gh_2-\dfrac{1}{2}\rho v^2_1-\rho gh_1\quad\text{(algebraically solve for the pressure P_1)}$
We don't know the speed of the water at point 1. We'll need to figure out the speed $v_1$ first before we can use Bernoulli's equation to solve for unknown pressure at point 1.
We can do this by using the equation of continuity $A_1v_1=A_2v_2$ since water is incompressible. We know the cross sectional area of a cylindrical pipe can be found with $A=\pi r^2$ so plugging areas into the equation of continuity we get,
$(\pi r^2_1)v_1=(\pi r^2_2)v_2$
When we solve this for the speed $v_1$ the $\pi$'s cancel and we're left with,
$v_1=(\dfrac{r^2_2}{r^2_1})v_2$
Plugging in the radii of the pipes we can solve for the speed at point 1 to get,
$v_1=\dfrac{(2.50\text{ cm})^2}{(7.50\text{ cm})^2}(32.0\text{ m/s})=3.56\text{ m/s} \quad$
Now that we have the speed at point 1, we can plug this into our rearranged Bernoulli's equation to get,
$P_1=P_2+\dfrac{1}{2}\rho (32\text{ m/s})^2+\rho gh_2-\dfrac{1}{2}\rho (3.56\text{ m/s})^2-\rho gh_1\quad\text{(we plugged in the speeds)}$
We can choose the $h=0$ reference line at point 1, which makes $h_1=0\text{ m}$ and $h_2=8.00\text{ m}+1.75\text{ m}=9.75\text{ m}$.
Plugging these into our rearranged Bernoulli equation makes the $\rho gh_1$ term go away (since it's zero) and we get,
$P_1=P_2+\dfrac{1}{2}\rho (32\text{ m/s})^2+\rho g(9.75\text{ m})-\dfrac{1}{2}\rho (3.56\text{ m/s})^2 \quad\text{(we plugged in the h values)}$
All we have to do now is figure out the pressure $P_2$ at point 2. We are going to argue that the pressure at point 2 must be atmospheric pressure since the water emerged out into the atmosphere. This is an assumption that needs to be made in many Bernoulli equation problems. Whenever a point is open to the atmosphere, that point should be at atmospheric pressure. We can either use absolute pressures in Bernoulli's equation and say that $P_2=1.01\times 10^5 Pa$, or we can use gauge pressures and say that $P_2=0$ (since gauge pressure measures pressure above atmospheric pressure). Anytime we can include zeros it makes our life easier so we'll use gauge pressure and use $P_2=0$. This makes our rearranged Bernoulli equation look like,
$P_1=\dfrac{1}{2}\rho (32\text{ m/s})^2+\rho g(9.75\text{ m})-\dfrac{1}{2}\rho (3.56\text{ m/s})^2 \quad\text{(we plugged in P_2=0)}$
Now we can plug in the density of water $\rho=1,000 \dfrac{kg}{m^3}$ and the magnitude of the acceleration due to gravity $g=+9.8\dfrac{m}{s^2}$ to get,
$P_1=\dfrac{1}{2}(1,000 \dfrac{kg}{m^3}) (32\text{ m/s})^2+(1,000 \dfrac{kg}{m^3})(+9.8\dfrac{m}{s^2})(9.75\text{ m})-\dfrac{1}{2}(1,000 \dfrac{kg}{m^3}) (3.56\text{ m/s})^2$
$P_1=6.01\times 10^5 Pa \quad\text{(calculate and celebrate)}$
Note: What we found was the gauge pressure since we plugged in $P_2=0$. If we would have plugged in $P_2=1.01\times 10^5 \text{ Pa}$ we would have solved for the absolute pressure at point 1.