Question Video: Determining the Exit Speed of Water Flowing through a Pipe Physics • 9th Grade

Water flows smoothly through a pipe. The water enters the pipe at a point where its cross-sectional area is 2.12 × 10⁻⁴ m² and leaves the pipe at a point where its cross-sectional area is 1.41 × 10⁻⁵ m². What is the ratio of the water’s speed where it enters the pipe to its speed where it exits the pipe?

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Video Transcript

Water flows smoothly through a pipe. The water enters the pipe at a point where its cross-sectional area is 2.12 times 10 to the negative four meters squared and leaves the pipe at a point where its cross-sectional area is 1.41 times 10 to the negative five meters squared. What is the ratio of the water’s speed where it enters the pipe to its speed where it exits the pipe?

Let’s say that this is the pipe that our water travels through. It enters the pipe here with a speed we’ll call 𝑣 one and exits here with a speed we’ll call 𝑣 two. We’ve drawn the entrance area to the pipe as larger than the exit area because we’re told that the cross-sectional area of the pipe at its entrance point is indeed larger than its cross-sectional area at its exit. We’ll call these areas 𝐴 one and 𝐴 two, respectively. What we want to do is solve for the ratio of the water’s speed where it enters the pipe to its speed where it exits the pipe. In other words, we want to solve for 𝑣 one divided by 𝑣 two.

We can begin to do this by using what’s called the continuity equation for fluids. This equation says that if we consider two cross sections through which a fluid, such as water, flows smoothly, then the density of the fluid at the first cross section multiplied by that area multiplied by the speed of the fluid equals the density of the fluid at the second cross section multiplied by that area times the speed of the fluid there. When we apply this equation to our situation, note that the density on either side is just the density of water, which is constant because water is an incompressible fluid. Therefore, we can divide both sides of the equation by that constant density and cancel it out entirely.

We’re now close to the answer we want to solve for. Recall that we want to calculate what 𝑣 one divided by 𝑣 two is. If we divide both sides of this equation by the product 𝐴 one and 𝑣 two, then on the left-hand side, 𝐴 one cancels out, and on the right-hand side, 𝑣 two cancels. What we have then is 𝑣 one divided by 𝑣 two being equal to 𝐴 two divided by 𝐴 one. Our last step is to substitute in the known values for our areas and then calculate that fraction. The cross-sectional area of our pipe’s exit is 1.41 times 10 to the negative five square meters. And the cross-sectional area of its entrance is 2.12 times 10 to the negative four square meters. This is equal to the decimal value 0.0665. That’s the ratio of the water’s entrance speed to its exit speed.

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