At one point in a pipeline the water's speed is 3.00 m/s and the gauge pressure is 5.00×10^4 Pa. Find the gauge pressure at a second point in the line, 11.0 m lower than the first, if the pipe diameter at the second point is twice that at the first.

Video transcript

Hey everyone. So this problem is dealing with fluid flow. Let's see what they're asking us for a city Water company pipes water from a raised area using pipes of different diameters. So that's our first clue at a point near the source water flows at a speed of five m per second and the pressure measured is 4.6 times 10 to the fourth pascal's. We know that there's a second point, that's eight m below the first point, and the company has a pipe that is 1.5 times the diameter of the first pipe. We are asked to determine the pressure at this second point. So this is a pretty straightforward Bernoulli equation problem. So the first thing that we need to do is recall the Bernoulli equation so we can figure out what we're missing and therefore what we need to solve for. So the Bernoulli equation is P one plus one half row v one squared plus ro G h one equals P two plus one half row V two squared plus ro G h squared. Alright, so let's work through all of the givens that we have from this problem. We know that the speed at the first point Is five m/s. We know the pressure at that first point is 4. times 10 to the four pascal's. We know that the height of the second point is eight m below the first point. So the way we're going to write that is H1 equals zero m and H two equals negative eight m. We also know that the second diameter D two is 1.5 T one And we are asked to solve for P two. So we also need to recall the continuity equation which is a one. V one equals a two V two. And that's important because what we're going to do is use this diameter that we know from the problem to solve for the second velocity term and then plug that back in to the to the Bernoulli equation. Okay, so let's recall that the area is given as pi R squared where R is just the diameter over to. So we can use this to rewrite this second equation. This continuity equation as pie. The 1/2 squared times V one equals pi D 2/2 Squared Times B two. Now we know D two is 1.5 times d one. So we're gonna use that here. So 1.5 D 1/2 squared B two. Okay, so from there the D1's cancel and we can solve for V two high cancels. And so plug that into our calculators. V two equals 2. m per second. So now we're gonna go back to the Bernoulli equation and solve use That V 2 to solve for P two. So let's rewrite the Bernoulli equation down here. P one plus one, half row V one squared plus ro G H one equals P two plus one half row V two squared plus Ro G H two. So H. One we know is zero. So that term goes to zero. And from there we can just plug and chug 4.6 times 10 to the fourth pascal's plus one half. Um That's right. We need to recall that the density of water is 1000 kg per meter cubed. So I'm gonna use that here so one half times kg per meter cubed. That Constant Times 2.22. Oops. Sorry B one was given to us in the problem as five m per second. That term squared equals p. To remember. That's the term we're solving for plus one half times bro. 1000 kg per meter cubed times V two squared. That's what we just saw for 2.2 m per second plus Row G. H. two. So that's again 1000 kg per meter cubed Times 9.8 m/s squared. And recall that H. two from our problem is negative eight m. So we're gonna plug that into our calculator and find that we have 5.85 times 10 to the 4th Pascal's equals p two minus 7. times 10 to the fourth pascal's. And that gives us ap two of 1.35 times 10 to the fifth pass scouts. And so we're going to go back up to our choices. Look at our answers. The closest one is D. 1.34 times 10 to the fifth pascal's That is the answer for this problem. That's all We'll see you in the next video.