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## MCAT

### Course: MCAT > Unit 8

Lesson 7: Fluids in motion- Fluids in motion questions
- Volume flow rate and equation of continuity
- Bernoulli's equation derivation part 1
- Bernoulli's equation derivation part 2
- Finding fluid speed exiting hole
- More on finding fluid speed from hole
- Finding flow rate from Bernoulli's equation
- Viscosity and Poiseuille flow
- Turbulence at high velocities and Reynold's number
- Surface Tension and Adhesion
- Venturi effect and Pitot tubes
- Two circulations in the body
- Arteries vs. veins - what's the difference?
- Resistance in a tube
- Putting it all together: Pressure, flow, and resistance

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# 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?

- 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)
- 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.(70 votes)

- At4:00, 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)
- Because when F is divided by Area it also needs to be multiplied by it as it is being introduced by us and hence to maintain the equality in the equation

.(7 votes)

- 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)- 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)

- how is it possible for liquid to go up?(6 votes)
- By definition it can not become more dense so it has to go wherever is available when the force on one end becomes greater than the resistance on the other end(7 votes)

- 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)
- That relation has to do with the change in viscosity of a single substance at different temperatures. As lava increases in temperature it decreases in viscosity so the law is not violated.(8 votes)

- 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)
- 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)

- why here we add extra term as work in conservation of energy?(7 votes)
- 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)
- Just to confirm this..These said equations are applicable only for a
**STEADY**flowing fluids right..?(3 votes)- Correct, when you have to deal with turbulent flow it becomes much more complex.(2 votes)

- 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)- 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.