Question Video: Using Continuity Equations to Calculate Flow Rate and Channel Dimensions

Video Transcript

Water emerges vertically downward from a faucet that has a 1.600-centimeter diameter, moving at a speed of 0.400 meters per second. Because of the construction of the faucet, there is no variation in speed across the stream. What is the flow rate of water from the faucet? What is the diameter of the water stream at a point 0.200 meters vertically below the faucet? Neglect any effects due to surface tension.

In this two-part exercise, we want to solve first for the flow rate of the water coming from the faucet. And then we wanna solve for the diameter of the water stream a certain distance below the faucet. We’ll call this flow rate 𝑓 sub 𝑟. And the diameter we’ll name 𝑑. Let’s start by drawing a diagram of the situation. Here we have water coming out of a faucet, falling vertically downward, where the output diameter of the faucet we’ve called 𝑑 one. It’s 1.600 centimeters. The water comes out with an initial speed we’ve called 𝑣 one, 0.400 meters per second. And the first thing we want to solve for is the flow rate of the water as it leaves the faucet.

We can recall that the volume flow rate of a liquid is equal to the cross-sectional area of the pipe or tube it moves through multiplied by its speed. We recall further that the cross-sectional area of a circle is equal to 𝜋 divided by four times the diameter of that circle squared. So we can write that 𝑓 sub 𝑟 is equal to 𝜋 divided by four times 𝑑 one squared times 𝑣 one. When we plug in for these two values, we keep 𝑑 one in units of centimeters and convert 𝑣 one to units of centimeters per second. Entering these values on our calculator, we find that 𝑓 sub 𝑟, to three significant figures, is 80.4 cubic centimeters per second. That’s the flow rate of the water from the faucet.

Next, we want to solve for the diameter of the water stream 𝑑, a distance of 0.200 meters below the faucet mouth. At first, we might think that 𝑑 is equal to 𝑑 one, that the diameter of the stream doesn’t change. But we realize that the water, once it leaves the faucet, is falling under the acceleration of gravity and will speed up. The continuity equation tells us that the cross-sectional area along some point of a fluid’s flow times its speed is equal to the cross-sectional area at another point times its speed there. We can write then that 𝜋 over four times 𝑑 one squared times 𝑣 one is equal to 𝜋 over four times 𝑑 squared times the speed of the falling water at that point, which we can call 𝑣 two. We see the factor of 𝜋 over four cancels from this expression. And if we rearrange it to solve for 𝑑, we find that 𝑑 is equal to 𝑑 one, the given diameter of the faucet, multiplied by the square root of 𝑣 one divided by 𝑣 two, the speed of the falling water at 𝑑.

Since we’re given 𝑑 one and 𝑣 one, the question now becomes how do we solve for 𝑣 two. Since the water as it falls is under a constant acceleration, that is, the acceleration due to gravity, the kinematic equations apply for describing its motion. In particular, we can make use of the kinematic equation that says final speed squared equals initial speed squared plus two times acceleration times displacement. In our case, we can write that 𝑣 two squared equals 𝑣 one squared plus two times 𝑔 times ℎ. Where 𝑔, the acceleration due to gravity, we’ll treat as exactly 9.8 meters per second squared. Now that we have an expression for 𝑣 two, we can substitute this in for 𝑣 two in our equation for 𝑑. And being given 𝑑 one and 𝑣 one, as well as ℎ, and knowing that 𝑔 is a constant, we’re ready to plug in and solve for 𝑑. When we do and enter this expression on our calculator, we find that 𝑑 is 0.712 centimeters. That’s the diameter of the stream of water after it’s fallen from the faucet a distance of 0.200 meters.