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### Course: UP Class 11 Physics > Unit 5

Lesson 2: Streamline flow# Volume flow rate and equation of continuity

A fluid’s motion is affected by its speed, density, and viscosity, and weight, as wells as drag and lift. Use an example of a pipe with different sized openings on either end to observe and quantify laminar flow of liquids. Learn about the concept of flux, and how it is used to calculate the power of a system with moving fluid.
Created by Sal Khan.

## Want to join the conversation?

- can you please tell me what does a venturi meter mean?(16 votes)
- A venturi meter is a way to measure flow in a pipe. The venturi is a section of pipe where the diameter is gradually reduced to some smaller area then gradually increased to the initial diameter

To measure the flow in the pipe you measure the pressure at the inlet and then again at the narrowest section of the venturi.

Assuming no friction. Conservation of energy tells you that the pressure in the reduced area will be lower because the velocity is increased (speeding a fluid up lowers it pressure, some what counter intuitive because we think of pressure in terms of force not potential energy)

Flow rate (Q) = velocity * Area

Q1 = Q2 v1 * A1 = v2 * A2

Potential Energy + Kinetic Energy remains constant. P1 + (1/2density *v1^2) = P2 + (1/density *v2^2)

you can solve these 2 equations for Q (refer to http://en.wikipedia.org/wiki/Venturi_effect)

(43 votes)

- You appear to be saying that volume in = volume out only applies if laminar flow exists. Does this mean that fluid is lost or gained in the case of turbulent flow?(13 votes)
- Regardless of whether a flow is laminar or turbulent, if it is incompressible then volume in will always be equal to volume out. Since we're dealing with an incompressible liquid, this is always the case.

The example about emptying a bottle is a little confusing, because in that case the incompressible water is being replaced by air, which is very much compressible.(8 votes)

- Could Khan Academy consider adding a unit quiz just like all other courses in Physics II? It would be really helpful to take some quizzes after watching the material. It helps to enhance the concepts and help students see how they are doing in the course.(17 votes)
- Could you please tell me what kind of Flux Sal meant?because if he meant Momentum flux, the rate of transfer of momentum across a unit area, or Volumetric flux, the rate of volume flow across a unit area,why its far from his explaination about flux which he said its volume over time and its unit is m^3/s(10 votes)
- Hi there, I believe Sal has mixed up his terminology here.

He says that V/t is flux, but actually it is Volume Flow Rate, Q (units m^3/s).

Volumetric Flux is actually this flow rate per unit area, i.e. V/t.A, symbol q (units m^3/s.m^2).(14 votes)

- At2:07, we should consider the time taken as dt 'coz the velocity changes as soon as the cross sectional area changes but for dt time we can assume that to be constant.(7 votes)
- He is simplifying the problem. He is considering the inlet velocity to be constant over the time of interest.(6 votes)

- Minute4:43. I think it is actually in-viscid flow not laminar flow. As with laminar flow you can still have viscosity effects,...?(5 votes)
- Laminar flow is when a fluid flows in parallel layers with no disruption between the layers. Unless the wals of the container are frictionless the fluid next to the wall of the container that it is flowing through causes it to flow slower than the middle of the container causing a difference in velocity of the fluid. With low velocity and/or viscosity you can have laminar flow, above a critical velocity which is inversely dependent on the viscosity the flow will become turbulent. This ratio of a fluids velocity vs viscosity is related to the Reynolds Number for the fluid.(4 votes)

- how do we know that water is leaving from the entire output area of the cylinder as shown in the diagram? there aren't any forces pushing the water up so shouldn't the depth of the water in respect to the bottom of the cylinder remain constant throughout the cylinder?(5 votes)
- I think we are meant to imagine we are looking down at the pipe bird's eye view rather than from the side.(2 votes)

- The pipe is tilting downward but according to the equation Ai*Vi=Ao*Vo shouldn't the liquid decelerate instead. How is that possible?(1 vote)
- He is not considering gravity. You interpret this problem as happening in a space station. Or that the pipe section shown is not vertical, but horizontal, so we see the pipe from above.(8 votes)

- Could you tell me how to deal with the question if the tube splits into two paths with smaller radii? Thanks.(3 votes)
- Volume flow rate would still remain constant. The sum of the flow rates in the two tubes is still equivalent to the flow rate in the original tube.(3 votes)

- The problem is the tube is slanted.So the Area and Volucity will keep changings( increase).Do we have to use integral to figure out those thing?(3 votes)

## Video transcript

Everything we've done so far has
been stationary fluids, or static fluids, and we've been
dealing with static pressure. We were trying to figure out
what happens when everything's in a steady state. Now let's work on what
happens when the fluid is actually moving. Let's imagine a pipe. Let's say one end of the pipe
has a larger area than the other end, or at least
a different area. So this is one end of the pipe,
and this is the other end of a pipe. It's filled with some fluid,
some liquid, actually, in our example, so there's just a bunch
of liquid in this fluid. Let's say this area
at the entrance is called the area in. That's the area of the opening
into the pipe, and let's call this area out. It's the area of the opening
coming out of the pipe. Let's think about what
happens if this liquid is actually moving. Let's say it's moving into the
pipe with the velocity V in. Let's think about how much
volume moves into the pipe after T seconds. After T seconds, if you
think about it, you'd have this much area. If you think about what was
right here, it will then be moved to the right
by how much? We could just go back to our
basic kinematic formula: distance is equal to
rate times time. The distance something travels
equals velocity times time, so after T seconds, whatever fluid
was here, it would have an area of about that much. Whatever fluid was there
would have traveled how much to the right? It would have traveled-- let's
assume that the pipe doesn't change too much in diameter or
in radius from here to here. It would have traveled velocity
times time, so V in times time. It could be meters or whatever
our length units are. After T seconds, essentially
this much water has traveled into the pipe. You could imagine a cylinder
of water here. Once again, I know I made it
look like it's getting wider the whole time, but let's assume
that its width doesn't change that much over the T
seconds or whatever units of time we're looking at. What is the volume of this
cylinder of water? The volume-in over the T seconds
is equal to the area, or the left-hand side
of the cylinder. Let me draw the cylinder in a
more vibrant color so you can figure out the volume. So it equals this side, the left
side of the cylinder, the input area times the length
of the cylinder. That's the velocity of the
fluid times the time that we're measuring, times the input
velocity times time. That's the amount of volume
that came in. If that volume came into the
pipe-- once again, we learned several videos ago that the
definition of a liquid is a fluid that's incompressible. It's not like no fluid could
come out of the pipe and all of the fluid just
gets squeezed. The same volume of fluid would
have to come out of the pipe, so that must equal
the volume out. Whatever comes into the pipe has
to equal the volume coming out of the pipe. One assumption we're assuming in
this fraction of time that we're dealing with is also that
there's no friction in this liquid or in this fluid,
that it actually is not turbulent and it's
not viscous. A viscous fluid is really just
something that has a lot of friction with itself and that
it won't just naturally move without any resistance. When something is not viscous
and has no resistance with itself and moves really without
any turbulence, that's called laminar flow. That's just a good word to
know about and it's the opposite of viscous flow. Different things have different
viscosities, and we'll probably do
more on that. Like syrup or peanut
butter has a very, very high viscosity. Even glass actually
is a fluid with a very, very high viscosity. I think there's some kinds of
compounds and magnetic fields that you could create that have
perfect laminar flow, but this is kind of a perfect
situation. In these circumstances, the
volume in, because the fluid can't be compressed, it's
incompressible, has to equal the volume out. What's the volume out over
that period of time? Similarly, we could draw this
bigger cylinder-- that's the area out-- and after
T seconds, how much water has come out? Whatever water was here at the
beginning of our time period will have come out and we can
imagine the cylinder here. What is the width
of the cylinder? What's going to be the velocity
that the liquid is coming out on the
right-hand side? Capital V is volume, and
lowercase v is for velocity, so it's going to be the output
velocity-- that's a lowercase v-- times the same time. So what is the volume that has
come out in our time T? It's just going to be this area
times this width, so the output volume over that same
period of time is equal to the output area of this pipe
times the output velocity times time. Once again, I know I keep saying
this, but this is kind of the big ah-hah moment, is
in that amount of time, the volume in this cylinder
has to equal the volume in this cylinder. Maybe it's not as wide, or
something like that, but their volumes are the same. You can't get more water here
all of a sudden than what's going in, and likewise, you
can't put more water into the left side than what's coming out
of the right side, because it's incompressible These two volumes equal each
other, so we know the area of the opening onto to the left
hand of the pipe times the input velocity times the
duration of time we're talking about is equal to the output
area times the output velocity times the duration of time
we're talking about. It's the same time on both sides
of this equation, so we could say that the input area
times the input velocity is equal to the output area times
the output velocity. This is actually called in fluid
motion the equation of continuity, and it leads to
some interesting things. We'll do some problems
with it in a second. One thing that I want to
introduce at this point as well is what is the
volume per second? Because this is also something
we're going to deal with in a second, probably in the next
video, because I'm about to run out of time. We said that in T seconds we
have this amount of volume coming in and it's the same
coming in as coming out. So what is the volume
per second? It's this big capital Vi
per amount of time, and we call that flux. We'll learn a lot about flux,
especially when we start doing vector calculus, but flux is
just how much of something crosses a surface in
an amount of time. It's how much a volume
crosses a surface in an amount of time. So in this case, the
surface is the left-hand side of cylinder. And we're saying how much
crosses in amount of time? We figured out it's that input
volume, which crosses in every T seconds, and this
is called flux. You've probably heard of the
flux capacitor in Back To The Future, and maybe we can think
about what they were trying to hint at. Let's see if we can use flux
and these ideas to come up with some other interesting
equations. We know that the volume per
time is equal to flux. This is a big V. V is equal to flux, and actually
the variable people generally use for flux is R. Of course, it's in meters
cubed per second. That's its unit. We also know that the input area
times input velocity-- that's a lowercase v-- is equal
to the output area times output velocity, and this is
called the equation of continuity. It holds true whenever
we have laminar flow. Actually, I'm about to
run out of time. In the next video, I'm actually
going to use some of this information to figure out
how much power is there in a system where we have fluid
going through a pipe. See you soon.