Question Video: Calculating the Time Taken for Water to Pass through Pipes of Different Radii | Nagwa Question Video: Calculating the Time Taken for Water to Pass through Pipes of Different Radii | Nagwa

Question Video: Calculating the Time Taken for Water to Pass through Pipes of Different Radii Physics • Second Year of Secondary School

Water with a speed 𝑣₁ = 3.15 m/s flows smoothly through a cylindrical pipe of radius π‘Ÿβ‚ = 1.25 m and then flows smoothly through a second, connected cylindrical pipe of radius π‘Ÿβ‚‚ = 0.951 m, as shown in the diagram. The first pipe has a length 𝐿₁ = 1.44 m and the second pipe has a length 𝐿₂ = 1.21 m. What is the ratio of the time the water takes to pass through the first pipe to the time it takes to pass through the second pipe?

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Video Transcript

Water with a speed 𝑣 one equals 3.15 meters per second flows smoothly through a cylindrical pipe of radius π‘Ÿ one equals 1.25 meters and then flows smoothly through a second connected cylindrical pipe of radius π‘Ÿ two equals 0.951 meters, as shown in the diagram. The first pipe has a length 𝐿 one equals 1.44 meters, and the second pipe has a length 𝐿 two equals 1.21 meters. What is the ratio of the time the water takes to pass through the first pipe to the time it takes to pass through the second pipe?

Here in our diagram, we see the first pipe with length 𝐿 one and the second pipe with length 𝐿 two. These pipes have different radii, π‘Ÿ one and π‘Ÿ two. And we’re told that the speed of water as it enters the first pipe with a speed called 𝑣 one is 3.15 meters per second. Knowing this, along with the values of the pipe lengths, 𝐿 one and 𝐿 two, we want to solve for the ratio of the time it takes water to pass through this first pipe to the time it takes water to pass through this second pipe. Let’s clear some space on screen, and we can start by assigning variables to these two time values.

Let’s say that the time it takes water to go from one end of pipe one to the other is 𝑑 one, and the time it takes water to travel through pipe two is 𝑑 two. What we’re trying to solve for is the ratio 𝑑 one to 𝑑 two. Whenever an incompressible fluid, like the water that we have here, flows through a container with rigid walls, when that fluid flows smoothly, it follows what is called the continuity equation. This equation is a statement about how a volume of fluid is transmitted through a container. It says that the rate at which a volume of fluid passes through one cross section of a container equals the rate at which that same volume passes through another cross section of the container. In our case, those cross sections that we’re thinking about are the cross sections of our two pipes with radii π‘Ÿ one and π‘Ÿ two.

The continuity equation says that the volume of water that passes through this cross section of our pipe over some amount of time, say, one second, is equal to the volume of water that passes through this section of our pipe with a different radius over that same amount of time. At first, it may not be clear that the continuity equation will help us though because we want to solve for a ratio of times. But the continuity equation does include speeds. 𝑣 one and 𝑣 two are the entrance and exit speeds of fluid through this pipe system, respectively. And we can recall that, in general, the speed 𝑣 of an object equals the distance traveled by that object divided by the time taken to travel that distance.

For our water as it travels through, say, the first pipe, the distance it travels is 𝐿 one, and the time it takes to travel that distance is 𝑑 one. Likewise, 𝑣 two, the exit speed of water through the system of pipes, is equal to 𝐿 two divided by 𝑑 two. All this means we can write the continuity equation this way. Here, 𝑣 one is replaced by 𝐿 one over 𝑑 one, and 𝑣 two is replaced by 𝐿 two over 𝑑 two. As we’ve seen, this is in agreement with the general formula for speeds. Since we’re trying to solve for 𝑑 one divided by 𝑑 two, let’s rearrange this equation so 𝑑 one over 𝑑 two is on one side by itself.

We can begin doing that by multiplying both sides by 𝑑 one, canceling that factor on the left. Next, we divide both sides of the equation by 𝐴 two times 𝐿 two. Doing that causes 𝐴 two and 𝐿 two both to cancel out of the right-hand side. And when all the cancelation dust settles, we have just what we wanted, 𝑑 one divided by 𝑑 two by itself on one side of our equation.

To solve for this ratio, we’ll need to know the values of these variables. 𝐿 one and 𝐿 two are given to us in the problem statement. And though we don’t know 𝐴 one and 𝐴 two directly, we do know the radii π‘Ÿ one and π‘Ÿ two. And we also know that the cross sections of these pipes are circular. In terms of its radius π‘Ÿ, the area of a circle equals πœ‹ times π‘Ÿ squared. We can write then that 𝐴 one is equal to πœ‹ times π‘Ÿ one squared and 𝐴 two equals πœ‹ times π‘Ÿ two squared.

Notice that in this fraction, the factor of πœ‹ cancels out. Substituting in for π‘Ÿ one, 𝐿 one, π‘Ÿ two, and 𝐿 two gives us this fraction. In both the numerator and denominator, we have units of meters cubed, so they all cancel out. And our final answer will be unitless. Rounding this fraction to three significant figures, we get 2.06. This is the ratio of the time it takes water to travel through the first pipe to the time it takes water to travel through the second pipe. And note that our answer tells us that it takes a little bit more than twice as long for water to travel through the first pipe as it does for it to travel through the second pipe.

We see then that the water must increase in speed as it travels through the second pipe. And this is because that second pipe has a smaller radius than the first one.

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