Home Repair. You need to extend a 2.50-inch-diameter pipe, but you have only a 1.00-inch-diameter pipe on hand. You make a fitting to connect these pipes end to end. If the water is flowing at 6.00 cm/s in the wide pipe, how fast will it be flowing through the narrow one?

Video transcript

Hey everyone in this problem, we're told that reducers are devices that connect pipes of different cross sections. A pipeline has a reducer that connects a 3" diameter pipe to a 2" diameter pipe. If water in the large pipe flows at 10 cm/s, how quickly does water flow in the narrow pipe? Alright, so we're given information about the diameter of these pipes. Okay, we're giving information about one of the speeds are one of the flows and we want to find information about the speed of the other pipe. Okay, so let's recall our continuity equation for in compressible fluid. K. We're dealing with water so we can approximate or assume that that's an in compressible fluids are continuity equation tells us that a one V one is equal to a two B two. Okay, so we have the cross sectional area of the first pipe or the first section of that pipe? Times the speed of the water flowing through. It is gonna be equal to the cross sectional area of the second part of the pipe, times the speed of the water flowing through that part of the pipe. Okay, so we want to know how quickly does water flow in the narrow pipe. So we're going to say that pipe to is going to be the narrow pipe part, one is going to be that large pipe. So what we want to do is find this quantity V two. Alright. So let's start from left to right. A one. Do we know a one? Well, a one is going to be the cross sectional area of the pipe. We know the diameter of the pipe. The cross sectional area is going to be a circle so we know how to find that. So a one is going to be equal to pi times R one squared OK. The radius of that first pipe. So we have pie. Now the diameter of the pipe is 3". The radius is going to be half of that. Okay, so the radius is going to be 1.5", this is all gonna be squared, This is going to give us a cross sectional area of that pipe of 2.25 Hi, inches squared. Alright now we have V one. Okay, the speed through that first Pipe, we're told that that's 10 cm/s. We have 10 cm per second and then the other quantity we need to find is a two way to just like a one. That cross sectional area of our pipe is going to be a circle. So the area is going to be given by pi r squared and in this case it's our to the radius of that second pipe. So we have pie. Now the diameter of this pipe is two inches Again the radius is going to be half of the diameter, so the radius is just going to be one inch all squared and so the area is going to be pi inches squirt. Alright now you'll notice that our areas are an inches squared and our speed is in centimeters per second. Okay now in most cases or most problems you would need to convert these so that they are all in like units. Okay we're gonna I'm gonna show you why in this case it's okay to leave them as they are. Okay, so we have a one. V one is equal to a two V two and we're trying to solve for V two. Okay, so we isolate V two. We have that V two is equal to a one over a two times V one. Okay now you'll see that we're going to divide area one and area too, so as long as a one and a two have the same dimension as each other. Okay, they both are in inches squared. Then their units are going to divide out and we're just going to be left with the units of V one. So because a one and a two have the same units, We're okay to leave it even though V1 doesn't have the same unit. Alright, so let's substitute in the information we have and solve for V two. So V two, it's going to be equal to a one that first Area 2.25 pi inches squared divided by a two which is pi inches squared all times V one which is going to be 10 centimeters per second. Okay, now we see that the pie will divide out and again the unit of inches squared will divide out. And we're gonna be left with just 2. Times 10 cm/s. Oops, not per two per second. There we go. Which is gonna give us a speed of 22.5 cm/s in the second pipe or in that narrow pipe. Okay. Alright. So if we go back up to our answer choices, We can see the rounding to the nearest centimeter per second. The water is going to flow through that narrow pipe at cm/s, which is answer choice. C. Thanks everyone for watching. I hope this video helped see you in the next one.