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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
• 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.
• 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.
• 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.
• Is pressure always supposed to be in Pascals and does it matter if it's in torr or atm?
• 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.
• Should v1=0.25v2 (0.5/2)
• 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 at to its said that h1=h2
• 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.
• WHAT IS R? I cant seem to figure out what R is?
• 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)
• in . does that mean that, v1 times A1 = v2 times A2 (Law of Continuity) is the same as R1 = R2 ? (Incoming Flux = Outgoing Flux)
• 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)
• 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
• 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)
• at , 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?