If you're seeing this message, it means we're having trouble loading external resources on our website.

If you're behind a web filter, please make sure that the domains *.kastatic.org and *.kasandbox.org are unblocked.

Main content

Bernoulli's equation derivation part 1

Bernoulli's equation is an equation from fluid mechanics that describes the relationship between pressure, velocity, and height in an ideal, incompressible fluid. Learn how to derive Bernoulli’s equation by looking at the example of the flow of fluid through a pipe, using the law of conservation of energy to explain how various factors (such as pressure, area, velocity, and height) influence the system. Created by Sal Khan.

Want to join the conversation?

  • piceratops ultimate style avatar for user Hemanth
    Isn't the work in putted into the fluid manifested in the form of kinetic energy? Why do we need to have two separate terms indicating KE and W? I can understand that KE1 + PE1 = KE2 + PE2 , by law of conservation of energy.... So, why do we need a work term? Please help.
    (68 votes)
    Default Khan Academy avatar avatar for user
    • spunky sam blue style avatar for user Aleks
      Bernoulli's expression (ρ+ρgh+1/2ρv^2) states that for an incompressible fluid not experiencing friction forces (low viscosity), the sum of the static pressure and dynamic pressure will be constant within a closed container.

      The term 1/2ρv^2 looks similar to Kinetic energy and is actually known as "dynamic pressure", which is the pressure associated with the movement of the fluid.

      The term ρgh looks a lot like the expression for gravitational potential of an object. It is the pressure associated with the mass of a fluid sitting above some position of depth.

      The term P is derived from what Sal refers to as Work. Specifically, it is derived from "energy density".
      We know that Pressure = Force/Area (N/m^2); if we manipulate the formula by multiply force and area by meters: N/m^2 x m/m; we get N·m/m^3; which is actually J/m^3 (energy density). A system at a higher pressure cotain a greater density of energy than those systems at lower pressure.

      Recall that adding P + ρgh gives us the static pressure (looks just like the absolute pressure formula).

      Ultimately, Bernoulli's principle says more energy dedicated towards fluid movement (higher 1/2ρv^2 value) means less energy dedicated towards fluid pressure (lower P + ρgh values). The opposite can be true too, less movement means more static pressure.

      Sorry for the long post, hoped it helped.
      (71 votes)
  • starky sapling style avatar for user Katie
    At , Sal divides Force by area than multiplies by area. Why is he multiplying by area? Aren't we just looking for pressure which is simply Force/Area?
    (9 votes)
    Default Khan Academy avatar avatar for user
  • leaf green style avatar for user tahseen1995
    Maybe this is a really stupid question but wouldn't the Kinetic Energy of the liquid at the start of the system be a result of/induced by the work inputted? So adding up the work inputted along with the KE of the liquid being inputted would give too large a total result for total energy at start of system wouldn't it?

    Or have I lost the plot lol..
    (14 votes)
    Default Khan Academy avatar avatar for user
    • female robot grace style avatar for user Vruta Gupte
      The kinetic energy will be induced by a force (and not the work done). Since for a mass to have kinetic energy, it needs to have a velocity, but if the initial velocity is zero, we would need to accelerate the mass--and so we require a force. Here the forces are mg on the two sections of water (which gives us the potential energy for the two sections of water we are considering here) and we also need to apply a force at the bottom of the pipe to make sure the water reaches the top (like a pump).

      Sal has just written Work done by all forces = change in kinetic energy in a different way.
      (3 votes)
  • mr pink red style avatar for user Christopher Ball
    how is it possible for liquid to go up?
    (6 votes)
    Default Khan Academy avatar avatar for user
  • mr pants teal style avatar for user isaac
    this question may not be related here but here it goes. viscosity is inversely proportional to temperature i.e if temp increases viscosity decreases and if temp decreases viscosity increases. but lava is a very hot liquid. and also it has a very high viscosity. but it violates the above law of temperature viscosity relationship . how is this explained??
    (5 votes)
    Default Khan Academy avatar avatar for user
  • marcimus pink style avatar for user Raymond Greenwood
    If the pipe narrows in the middle, wouldn't this equation be obsolete? I am thinking about when I kink a hose, the pressure is not the same at the delivery end as the production end, so should there not be a caveat that this equation only works if the pipe does not narrow in the middle?
    (5 votes)
    Default Khan Academy avatar avatar for user
    • blobby green style avatar for user Harsha Bacham R
      Your question makes alot of sense. So let us consider a hose with water flowing through it. Say somewhere around the middle of the hose, you kink it. Have you noticed that at the kink, from the source side, the presence of a very hard bulge, tight would be a better word. This bulge is solely due to a large increase in pressure. This pressure develops because of all that water trying to fit into a tiny opening. On the other side of the kink, the water would come out at an increased velocity, but much reduced volume. This is in accordance with the equation.
      Then one would have to keep in mind that, a hose is an enclosed pipe of small diameter rarely ever placed in a linear fashion, this water exiting the kink, doesn't really have much space to 'be fast'(couldn't think of a better word.), its speed would be reduced due to collisions with hose walls (I hope you can visualize this.) And the water to reach the output end, would require more water to raise it to the end of the hose(like filling a glass of water, you need more water to reach its brim). And so we find a reduced pressure instead of an increased pressure. This is not due to inadequacy of the given equation but due to the 'bendy' nature and length of a hose.
      (7 votes)
  • leafers sapling style avatar for user DEEPAK TRIPATHI
    why here we add extra term as work in conservation of energy?
    (7 votes)
    Default Khan Academy avatar avatar for user
  • blobby green style avatar for user Ananya Bharadwaj
    Is it possible to re-record this topic again by Sal, the quality is so bad...and the Hindi version doesn't suit...please...
    (6 votes)
    Default Khan Academy avatar avatar for user
  • primosaur seedling style avatar for user Abhilash
    Just to confirm this..These said equations are applicable only for a STEADY flowing fluids right..?
    (3 votes)
    Default Khan Academy avatar avatar for user
  • male robot donald style avatar for user sreekar
    when we are considering a portion fluid that is entering the pipe it receives force from both the direction why aren't we considering the work done by force on right side of the fluid portion that's
    entering into the pipe
    (3 votes)
    Default Khan Academy avatar avatar for user
    • blobby green style avatar for user Andreas
      My understanding is that positive work is done on the fluid and negative work is done by the fluid. The fact that velocities and pressures on the ends are given, saves us from the headache of asking who is doing the work on whom. In school, the teacher told us that all work is done by the fluid to the fluid. Well, this is hard to conciliate with other physics lessons but this explanation "works" for some until they come up with their own :). Anyway, the work is done by pressure for an arbitrary small distance and the fluid propagates this action further through it. At the end we subtract the work done by the last arbitrary small volume of fluid on the "rest" of the fluid considered. Thus the work resembles a state parameter of the fluid, which is obviously an incorrect explanation for the correct math :) .
      (1 vote)

Video transcript

Let's say we have a pipe again-- this is the opening-- and we have fluid going through it. The fluid is going with a velocity of v1, the pressure entering the pipe is P1, and then the area of this opening of the pipe is A1. It could even go up, and the other end is actually even smaller. The fluid-- the liquid-- is exiting the pipe with velocity v2, the pressure that it exerts as it goes out. If there was a membrane on the outside, how much pressure would it exert on it as it pushes it out on the adjacent water is P2, and the area of the smaller opening-- it doesn't have to be smaller-- is A2. Let's say that this opening is at a height, on average, of h1, and the water exiting this opening is on average at a height of h2. We won't worry too much about the differential between the top of the pipe and the bottom of the pipe-- we'll assume that these h's are much bigger relative to the size of the pipe. With that set up-- and remember, there's fluid going through this thing-- let's go back to what keeps showing up, which is the law of conservation of energy, which is in any closed system, the amount of energy that you put into something is equal to the amount of energy that you get out. So energy in is equal to energy out. What's the energy that you put into a system, or that the system starts off with at this end? It's the work that you input plus the potential energy at that point of the system, plus the kinetic energy at that point of the system. Then we know from the conservation of energy that that has to equal the output work plus the output potential energy plus the output kinetic energy. A lot of times in the past, we've just said that the potential energy input plus the kinetic energy input is equal to the potential energy output plus the kinetic energy output, but the initial energy in the system can also be done by work. So we just added work to this equation that says that the energy in is equal to the energy out. With that information, let's see if we can do anything interesting with this pipe that I've drawn. So what's the work that's being put into this system? Work is force times distance, so let's just focus on this. It's the force in times the distance in, and so over a period of time, t, what has been done? We learned in the last video that over a period of time, t, the fluid here might have moved this far. What is this distance? This distance is the input velocity times whatever amount of time we're dealing with, so T-- so that's the distance. What's the force? The force is just pressure times area, and we can figure that out by just dividing force by, area and then multiply by area, so we get input force divided by area input, times area input. It's divided and multiplied by the same number-- that's pressure, that's area. It's equal to the input distance over that amount of time, and that's velocity times time, so the work input is equal to the input pressure times the input area times input velocity times time. What is this area times velocity times time, times this distance? That's the volume of fluid that flowed in over that amount of time. So that equals the volume of fluid over that period of time, so we could call that volume in, or volume i-- that's the input volume. We know that density is just mass per volume, or that volume times density is equal to mass, or we know that volume is equal to mass divided by density. The work that I'm putting into the system-- I know I'm doing a lot of crazy things, but it'll make sense so far-- is equal to the input pressure times the amount of volume of fluid that moved over that period of time. That volume of fluid is equal to the mass of the fluid that went in at that period of time, and we'll call that the input mass, divided by the density. Hopefully, that makes a little bit of sense. As we know, the input volume is going to be equal to the output volume, so the input mass-- because the density doesn't change-- is equal to the output mass, so we don't have to write an input and output for the mass. The mass is going to be constant; in any given amount of time, the mass that enters the system will be equivalent to the mass that exits the system. There we go: we have an expression, an interesting expression, for the work being put into the system. What is the potential energy of the system on the left-hand side? The potential energy of the system is going to be equal to that same mass of fluid that I talked about times gravity times this input height-- the initial height-- times h1. The initial kinetic energy of the fluid equals the mass of the fluid-- this mass right here, of that same cylinder volume that I keep pointing to-- times the velocity of the fluid squared. We remember this from kinetic energy divided by 2. So what's the total energy at this point in the system over this period of time? How much energy has gone into the system? It's going to be the work done, which is the input pressure-- I'm running out of space, so let me erase all of this. I'll probably have to run out of time, too, but that's OK-- it's better than being confused. Back to what we were doing. So, the total energy going into the system is the work being done into the system, and I rewrote it in this format, which is the input pressure-- we'll call that P1-- times the mass divided by the density of the liquid, whatever it is. This is work in plus-- and what's the potential energy? I wrote it right here-- that's just mgh, where m is the mass of this volume of fluid, h is its average height, and you could almost think of how high the center of mass above the surface of the planet. Since we have a g here, we assume we're on Earth, so this is h1, because the height actually changes, so this is potential energy input plus the kinetic energy mv1 squared over 2. That is the kinetic energy input. We know that this has to equal the energy coming out of the system. This is going to be equal to the same thing on the output side. This is going to be equal to the work out, so that'll be the output pressure times the mass divided by the density plus the output potential energy, which will just be mg h2, plus the outbound kinetic energy, which will be mv2 squared divided by 2. I just realized I'm out of time. I will continue this in the next video. See you soon.