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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. 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.