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Course: 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
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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. Understand how pressure differentials and pipe dimensions influence flow rate and velocity. Created by Sal Khan.
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I'm a little bit lost with understanding the whole Pressure in/Pressure out thing. In the fluids 1 video we are told that Pressure in = Pressure out. When the velocity increases, is it the pressure IN which drops? ie. Is this saying that the force/A behind the molecules reduces because the moving particles carry their own weight through momentum/intertia? If P + 1/2pv^2 = k along a straight line, then as v increases, P must decease. Is this refering to the the input pressure or the pressure out against the spurting flow? I think some consolidation on what which direction the pressures are going in and what is happening on a molecular level would help.
Thanks
Max(22 votes)
Unfortunately I couldn't clearly understand your doubt. However I will explain to you what happens and hopefully it'll be useful.
First, you have to understand that Sal is using and ideal fluid. This fluid, besides being imaginary, has the properties of having zero viscosity and being incompressible. Therefore, if you apply a certain force (pressure) on one edge it will be distributed along the fluid without ANY lost of energie and in every direction. So, if a certain amount of fluid (volume) passes through area 1 in a certain amount of time (flux - R) the same volume will cross section 2. In a molecular level, you can imagine (just imagine it because such perfect fluid doesn't exist) some fluid molecules align in a stream line, when you push one (by applying a force) all of them will move. Just like a Newton's pendulum. And if you want to understand this in terms os momentum you just have to consider all the collisions between the molecules as perfectly elastic (watch some videos about this topic).
The velocity in the section 2 in the video is bigger because the same amount of molecules that passed through section 1 now have to pass section 2 in the exact same amount of time.
If this wasn't helpful please fell free to specify your questions again.(8 votes)
- I didn't quite get the ending of the video. Sal gets R=1.46 m^3/s (I understood that). Then he said to find the velocity of the fluid coming out, multiply R by 2. But if you do that don't you get 2.92 m^5/s? or 2.92 m^3/s and not 2.92 m/s since R has the units in m^3/s. I am confuse on the unit part only. He ends up saying that the velocity is 2.8 "m/s." So do you completely ignore the units and use the numbers only and then plug the right units back in? I'm confuse.(6 votes)
For the exiting opening you have an area of 1/2 m^2. R is 1.46 (m^3)/s. For the velocity of the fluid exiting the tube you divide R by the area of the opening which gives you 1.46/(1/2) which = 2.92. For the unite you have (m^3/s) / (m^2) giving you m/s.(6 votes)
Is pressure always supposed to be in Pascals and does it matter if it's in torr or atm?(5 votes)
Pascals, Torr, and atm are conversions of each other, so leaving your answers in these values should be O.K. However, this would be like leaving your answers in pounds or Fahrenheit. This would be inconvenient to work with when solving questions because you would just have to convert back to the metric units. If you are doing physics problems, leave them in Pascals. Chemistry classes tend to like atm, and weather classes tend to like Torr.(8 votes)
i did not understand how the heights cancel out, isn't the right side at a higher height relative to the height from the left. I dont understand the concept behind that please help at time approximately at3:43to3:52its said that h1=h2(3 votes)
The midpoint of the pipe on either side is level in this example. Therefore, h1 = h2 and the terms cancel out when you subtract them from each side.
Alternatively, you can give them the value of 0m and the term ρgh on either side equals 0.(3 votes)
WHAT IS R? I cant seem to figure out what R is?(2 votes)
R(flow rate) = A(area) * v(velocity) of a fluid
which means an A area of fluid travel with v velocity per second, you can visualize it a circle (or whatever 2-dimensional shape) runs through a tunnel (with the same area of the shape). and water (or whatever fluid) follows its path for a time. then R is the filled "volume" that water filled in that tunnel
of course the shape can be shifted on journey as the case in video
in short, R(flow rate) is a volume filled with a running fluid for a time(1 vote)
- in2:18. does that mean that, v1 times A1 = v2 times A2 (Law of Continuity) is the same as R1 = R2 ? (Incoming Flux = Outgoing Flux)(3 votes)
- you are right
cause R=v*A
by the way, R is (volumetric) flow rate. and flux is a bit different from that (you can see it as a unit flow rate or a slice of it)(1 vote)
- Should v1=0.25v2 (0.5/2)(3 votes)
- I have a doubt in biology, related to pressure, in my book it says that "Atrial natriuretic factor cause dilation of blood vessels and thereby decrease in blood pressure ". I don't get this
if vessel crossection increase then pressure should rise.. I think so(3 votes)- A1*v1 = A2*v2 (by continuity equation)
say
6*5 = 3*10
1. this means A1 grows to 2A1 = A2
2. this gives v1 shirnks 1/2v1 = v2
3. (blood) pressure = F/area = m*a/area = m*v / area*second
1) this area is the whole area meeting the blood inside the vessel
2) which is different from the areas above (that is the dissected 2-d circle)
3) when dilation happens, the area of 2-d circle is growing. while the whole area of 1) stays still
if A(not area) increases,
velocity decreases,
acceleration decreases,
force decreases,
and thus pressure decreases (by the equation 3)(1 vote)
- at6:30, how did Sal get the units m3/s if dividing Pa/rho which is kgm/s2/m2 over kg/m3? I keep getting sqrt m2/s2?(3 votes)
is flow and flux the same?(2 votes)
No, the integral of (volumetric) flux over a given area is the (volumetric) flow rate. In simpler words, flux is the flow rate in unit area.(1 vote)
3:43Video 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.