Main content

## Physics library

### Unit 9: Lesson 3

Fluid Dynamics- Volume flow rate and equation of continuity
- What is volume flow rate?
- 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
- What is Bernoulli's equation?
- Viscosity and Poiseuille flow
- Turbulence at high velocities and Reynold's number
- Venturi effect and Pitot tubes
- Surface Tension and Adhesion

© 2022 Khan AcademyTerms of usePrivacy PolicyCookie Notice

# Finding flow rate from Bernoulli's equation

Sal solves a Bernoulli's equation example problem where fluid is moving through a pipe of varying diameter. Created by Sal Khan.

## Video transcript

Let's say I have a horizontal
pipe that at the left end of the pipe, the cross-sectional
area, area 1, which is equal to 2 meters squared. Let's say it tapers off so that
the cross-sectional area at this end of the pipe,
area 2, is equal to half a square meter. We have some velocity at this
point in the pipe, which is v1, and the velocity exiting
the pipe is v2. The external pressure at this
point is essentially being applied rightwards
into the pipe. Let's say that pressure
1 is 10,000 pascals. The pressure at this end, the
pressure 2-- that's the external pressure at that point
in the pipe-- that is equal to 6,000 pascals. Given this information,
let's say we have water in this pipe. We're assuming that it's laminar
flow, so there's no friction within the pipe, and
there's no turbulence. Using that, what I want to do
is, I want to figure out what is the flow or the flux of the
water in this pipe-- how much volume goes either into the pipe
per second, or out of the pipe per second? We know that those are the going
to be the same numbers, because of the equation
of continuity. We know that the flow, which
is R, which is volume per amount of time, is the same
thing as the input velocity times the input area. The input area is 2, so it's
2v1, and that also equals the output area times output
velocity, so it equals 1/2 v2. We could rewrite this, that v1
is equal to 1/2 R, and that v2 is equal to 2R. This immediately tells us that
v2 is coming out at a faster rate, and this is based on
the size of the openings. We know, because V2 is coming
out at a faster rate, but we also know because we have much
higher pressure at this end than at this end, that the water
is flowing to the right. The pressure differential, the
pressure gradient, is going to the right, so the water
is going to spurt out of this end. And it's coming in this end. Let's use Bernoulli's equation
to figure out what the flow through this pipe is. Let's just write it down: P1
plus rho gh1 plus 1/2 rho v1 squared is equal to
P2 plus rho gh2 plus 1/2 rho v2 squared. This pipe is level, and the
height at either end is the same, so h1 is going
to be equal to h2. These two terms are going to be
equal, so we can cross them out-- we can subtract that value
from both sides, and we're just left with P1. What's P1? P1 is 10,000 pascals plus 1/2
rho times v1 squared. What's v1? That's R over 2-- we figured
that out up here. v2 times R over 2 squared is
equal to P2, and that's 6,000 pascals plus 1/2 rho
times v2 squared. We figured out what v2 is--
v2 is 2R squared. Let's just do some
simplification, and so let's subtract 6,000 from both sides,
and we're left with 4,000 plus rho R squared over
8 is equal to 1/2 times R squared times 4. So this is 2 rho R squared. We could multiply both sides of
this equation by 8, just to get rid of this in the
denominator, so we would get 32,000 plus rho R squared is
equal to 16 rho R squared. Subtract rho R squared from both
sides of this question, and we get 32,000 is equal
to 15 rho R squared. Then what's rho? What's the density of water? The density of water is 1,000
kilograms per meter cubed, so this is 1,000. Let's divide both sides
by 15 times rho. We get R squared is equal to
32,000 divided by 15 rho-- rho is 1,000, so R squared is equal
to 32,000 over 15,000, which is the same thing
is 32 over 15. R is equal to the square root
of 32 over 15, and that's going to be meters
cubed per second. I get 32 divided by 15 is equal
to 2.1, and the square root of that 1.46. So the answer is R is equal to
1.46 meters cubed per second. That is the volume of water that
is either entering the system in any given second, or
exiting the system in any given second. We can figure out the
velocities, too-- what's the velocity exiting the system? What's two times that? It's 2.8 meters per second
exiting the system, and going in it is half that, so it's
0.8 meters per second. Hopefully, that gives you--
actually, 0.7 meters per second-- a bit more intuition on
fluids, and that's all I'm going to do for today. I'll see you in the next video,
and we're going to do some stuff on thermodynamics.