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

Water flows smoothly through a pipe that changes thickness along its length. The water enters the pipe at a point where its cross-sectional area is 1.24 × 10⁻⁵ m² and leaves the pipe at a point where its cross-sectional area is 6.35 × 10⁻⁶ m². The water increases in speed by 2.50 m/s between entering and exiting the pipe. At what speed does the water exit the pipe?

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### Video Transcript

Water flows smoothly through a pipe that changes thickness along its length. The water enters the pipe at a point where its cross-sectional area is 1.24 times 10 to the negative fifth meters squared and leaves the pipe at a point where its cross-sectional area is 6.35 times 10 to the negative six meters squared. The water increases in speed by 2.50 meters per second between entering and exiting the pipe. At what speed does the water exit the pipe?

Let’s say that this is our pipe, where this end is the entrance and this end is the exit. We can call the cross-sectional area of the entrance 𝐴 one, and we’re told that value in our problem statement. Likewise, if we call the exit area of our pipe 𝐴 two, we’re given that value as well. We can see that 𝐴 one is larger than 𝐴 two. And this is why our pipe’s diameter decreases as we move from the entrance to the exit. Water enters this pipe at a speed we’ll call 𝑣 one. It exits the pipe at a speed 𝑣 two, and it’s this exit speed that we want to solve for.

To help us do that, let’s recall the continuity equation for fluids of constant density. Whenever an incompressible fluid such as water flows through a volume that it fills, then the cross-sectional area of that volume at one point multiplied by the speed of the flowing fluid equals the cross-sectional area of that volume at another point multiplied by the fluid speed at that point. Here, our two points of interest are the entrance and the exit of our pipe. In terms of the variables in this equation, we know 𝐴 one and 𝐴 two. Those are the cross-sectional areas of the pipe at our points of interest.

We don’t know either 𝑣 one or 𝑣 two, the speed we want to solve for. But we do know this, that the water as it passes through the pipe increases in speed by 2.50 meters per second. That means that if we take 𝑣 one, the speed of the water as it enters the pipe, and add this speed of 2.50 meters per second to it, we’ll get the speed of the water as it exits the pipe, 𝑣 two. Let’s now clear some space to work and use this equation here to help us begin solving for 𝑣 two using the continuity equation.

Using this equation, one thing we could do is replace 𝑣 two with 𝑣 one plus 2.50 meters per second. If we did that though, notice that it gives us equation where the unknown is now 𝑣 one, the entrance velocity of our water. Once we solve for 𝑣 one, we could then use that information to solve for 𝑣 two. But a bit of a more direct route is to replace 𝑣 one in this equation with an expression in terms of 𝑣 two, the variable we want to solve for. We can do that by subtracting 2.50 meters per second from both sides of this equation. When we do so, on the left side, negative 2.50 meters per second plus 2.50 meters per second adds up to zero. What remains is an expression for 𝑣 one in terms of 𝑣 two. This means we can replace 𝑣 one in our continuity equation with 𝑣 two minus 2.50 meters per second.

Now, the only unknown in this entire equation is the variable we want to solve for. To begin doing that, let’s multiply 𝐴 one through both terms it multiplies in these parentheses. That gives us this result. And we can then add 𝐴 one times 2.50 meters per second to both sides of the equation. On the left-hand side, subtracting and adding the same quantity causes them to sum to zero. And as a next step, if we subtract 𝐴 two times 𝑣 two from both sides, then that quantity by being subtracted and added on the right-hand side disappears. That is, the sum of these two terms is zero.

The right-hand side of our expression simplifies to 2.50 meters per second times 𝐴 one. And then on the left, notice that both of these terms have a common factor, 𝑣 two. If we factor out 𝑣 two, we get 𝑣 two times the quantity 𝐴 one minus 𝐴 two. Lastly, to isolate 𝑣 two, let’s divide both sides by 𝐴 one minus 𝐴 two. On the left side, that difference cancels in numerator and denominator. We have finally that 𝑣 two, the exit velocity of water from this pipe, equals 2.50 meters per second times 𝐴 one divided by 𝐴 one minus 𝐴 two. Since we know the values of 𝐴 one and 𝐴 two, we can substitute them into this expression. And we then have this result.

Notice that since in the denominator we’re subtracting one term from another, the similar units of those two terms, meters squared, will be the units of the denominator overall. However, our numerator includes the units of meters squared. So when we calculate this fraction, those units will cancel out entirely. We’ll be left with final units simply of meters per second, units of speed. Entering this expression on our calculator, to three significant figures, 𝑣 two is 5.12 meters per second. That’s the speed of water as it exits the pipe, which is 2.50 meters per second faster than the water is traveling when it enters.