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Current time:0:00Total duration:7:46

Finding flow rate from Bernoulli's equation

Video transcript

let's say I have a horizontal pipe that and at this at the left end of the pipe let's say at the left end of the pipe the area the cross-sectional area so I'll call it area one let's say that that is I don't know it's equal to two meters yeah let's view simple numbers two meters squared let's say it tapers off tapers off so that the area the cross-sectional area at this end of the pipe so area 2 is equal to I don't know half a meter half a square meter and let's say that you know just the typical you know we have some velocity at this point and the pipe which is v1 and then the velocity exiting the pipe is v2 and let's say that the external pressure at this point essentially being applied rightwards into the pipe let's say that pressure one is I don't know let's say it's it's 10 let me write in a bold color let's say it's 10,000 Pascal's now let's say that the pressure at this end so the pressure to and so that's the external pressure at that point in the pipe let's say that that is equal to I don't know let's say it's equal to 6,000 Pascal's right and so given this information let's say we have water in this pipe there's just generally so I could draw water and we're assuming that's laminar flow so there's no friction within the pipe and there's no turbulence and all of that so using that what I want to do is I want to figure out I want to figure out how much what is the flow or the flux of the water in this pipe so how much volume goes either into the pipe per second or out of the pipe per second and we know that those are the going to be the same numbers because of the the content the equation of continuity so we know that the flow which is our which is volume per amount of time that is the same thing as the input velocity times the input area right so the input area is 2 so it's 2 V 1 and that also equals the output area times the output velocity so it equals 1/2 V 2 and then we could we could rewrite this that V 1 is equal to 1/2 R and that V 2 is equal to 2r so this immediately tells us that V 2 is is coming out it up at a faster rate right just based on the the size of the openings and then we also know because V 2 is coming out at faster rate but we also know because we have much higher pressure this ended at the sense 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 spread out of this end and it's coming in at this end so let's use Bernoulli's equation to figure out what the flow through this pipe is so Bernoulli's equation let's just write it down P 1 plus Rho Rho G H 1 plus 1/2 Rho V 1 squared is equal to P 2 plus Rho G H 2 plus 1/2 Rho V 2 squared well this pipe is level right the height at either end is the same so these H 1 is going to be equal to H 2 so 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 P 1 and what's P 1 P 1 is 10,000 Pascal's so let's start substituting numbers 10,000 Pascal's plus 1/2 Rho times v1 squared well what's v1 well that's R over 2 we figure that out here so the times are over two squared is equal to p2 that's 6,000 Pascal's plus one half Rho times V 2 squared we figure out V 2 is V 2 is 2 R 2 R squared let's just do some simplification so let's subtract 6,000 from both sides and we're left with 4,000 plus let's see get R squared over 4 and then a 1/2 so we'd have Rho R squared over this is a over 4 over 1/2 so over 8 is equal to we ready to subtract the 6,000 from both sides so we'll have 1/2 times R squared times 4 so this is 2 Rho R squared see we could multiply both sides of this equation by 8 just to get rid of this in the denominator so we'd get 32,000 plus Rho R squared is equal to 16 Rho R squared and subtract Rho R squared from both sides this equation we get 32,000 is equal to well this is 1 Rho R squared this is 16 Rho R squared so this is 15 Rho R squared and then what's Rho what's the density of water roll the density of water is is 1,000 kilograms per meter cube so this is 1,000 so let's divide both sides by 15 times Rho so let me draw let me switch colors so we get R squared is equal to 32 thousand divided by 15 Rho Rho is a thousand so R squared is equal to 32,000 over 15,000 which is the same thing as 32 over 15 so R is equal to the square root of 32 fifteen it's going to be meters cubed per second let's see let me get my calculator and D calculator did I actually remove my calculator I think I did get my calculator going so I get 32 divided by 15 is equal to two point one and then take the square root of that one point four six so the answer is R is equal to one point four six meters cubed per second so that is how much what is that's the volume of water that is either entering the system in any given second or exiting the system in any given second and we can figure out the velocities too right what's the velocity exiting the system what's two times that so it's like two point eight meters per second exiting the system and going in it is it is half that so it's like point eight meters per second but anyway hopefully that that gives you a actually 0.7 meter per turn that gives you a bit of an intuition more intuition on fluids and that that's all I'm going to do for today I'll see in the next video and we're going to do some stuff on thermodynamics