# Lesson Video: Flow Rate and Continuity

In this video we learn fluid flow rate and continuity equations for incompressible liquids and compressible gases.

13:25

### Video Transcript

In this video, we’re going to learn about flow rate and continuity. We’ll learn what these terms mean. We’ll see how they’re different for gases and liquids. And we’ll learn how to apply them practically. To start out, imagine that, in celebration of your best friend’s birthday, you’ve decided to make an indoor skating rink in the basement of your parents’ house. Your plan is to fill the unfinished basement six inches deep with water from the garden hose. You then hope to freeze this quantity of water using a collection of air-conditioning units you’ve borrowed from your friends. You’d like to know how long you can leave the hose running in order to fill the basement with six inches of water. In order to understand that, we’ll want to know something about flow rate and continuity.

But before talking about these two terms, it’s helpful to get a bit of background on the topic. When we talk about fluids, we’re speaking of something that has the capacity to flow. And this could include liquids as well as gases. There’s an important difference between these two. Gases are compressible, while liquids are effectively incompressible. This means that the volume of space a quantity of gas occupies can change, while the volume of space a liquid occupies stays the same. The terms flow rate and continuity apply to fluids in general, that is, both liquids and gases. But this difference between gases and liquids is significant enough that we can formulate separate flow rate and continuity expressions for these two states of matter.

Imagine that we have two identical pipes. In the first pipe, a liquid flows. And in the second pipe, a gas flows through. A basic question we’d be interested in understanding is how much of these fluids flow through the pipe. That is, what is their flow rate? When it comes to an incompressible liquid, we know that flow rate will be determined by the cross-sectional area of the pipe as well as the rate at which the liquid flows. If we call that cross-sectional area 𝐴 and the rate of liquid flow 𝑣, then we can say that 𝐴 times 𝑣 is equal to the volume flow rate. That is, the amount of liquid that moves through the pipe over a certain time interval.

On the other hand, if we’re working with a gas that can be compressed, then if the gas has a density we can call 𝜚 and we’re referring to a volume of that gas we can call capital 𝑉. Then the density 𝜚 multiplied by the volume 𝑉 is equal to the mass flow rate of the gas. Here, we’ve put the mass flow rate under the heading of working with gases and the volume flow rate under the heading of working with liquids. But it’s helpful to see that we can use either type of flow rate for either type of fluid.

It’s often just more helpful when talking of liquids to speak of a volume flow rate in terms of cross-sectional area and fluid speed 𝑣. While when working with gases, keeping the density 𝜚 forefront as well as the volume the gas takes up 𝑣 is very helpful. So when it comes to flow rate, there are two different ways we can express flow rate, one in terms of volume and one in terms of mass. And which one we choose just depends on the particulars of our situation.

Now imagine that our two identical pipes change, so that though they’re still identical, now they open up as the pipe moves along to a wider diameter farther downstream. Let’s look at the pipe with liquid in it for a moment and consider how this liquid will flow. You may have seen something like this before. Have you ever seen a river that gets wide at some sections and then narrower at others? Well, what does the water look like in those wide versus the narrow sections? Typically, we see that, in narrower sections, the water moves more quickly. And in wider sections, it slows down. If we call the cross-sectional area of the pipe at the narrow section 𝐴 one and the cross-sectional area at the wider part 𝐴 two, there’s actually a relationship between 𝐴 one 𝑣 one and 𝐴 two 𝑣 two.

Here’s one way to think about it. Since a liquid is incompressible, if we put a certain amount of liquid in at the narrow end of the pipe, then that same amount must come out the wider end of the pipe. After all, there’s no space for that liquid to go anywhere in between the intake and the output. Mathematically, that means that if we take the cross-sectional area of the pipe at any point along its length and multiply that by the fluid’s speed at that same point, that equals the cross-sectional area of the pipe times the speed of the fluid at any other point. This is called the continuity equation. It says that the fluid flow must be continuous in this mathematical way. One thing that makes this equation easier to recall is that it agrees with our intuition. Like we saw earlier, the wider a pipe gets, the slower the flow through it. And the narrower it gets, the faster the fluid will flow.

Now that’s the continuity equation for liquids. But what about for gases? How does it change when our fluid is compressible? Just like with liquids, the continuity equation for gases involves cross-sectional areas and fluid speeds. But for gases, we’ll include the density of the gas, which itself can vary across the pipe length. Taking the product of gas density, cross-sectional pipe area, and gas speed, we find that multiplying these three terms together gives us a value that’s constant across the length of the pipe. This also is a continuity equation. And, actually, we could have included densities in our expression for liquids. We could have written 𝜚 one, our initial density of our liquid, times 𝐴 one times 𝑣 one equals 𝐴 two 𝑣 two times 𝜚 two. But the property of liquids being incompressible means that 𝜚 one will always equal 𝜚 two. So we may as well cancel that out from either side of our expression. But for gases, where density truly can vary, we include it. Let’s get a bit of practice working with flow rate and continuity through a few examples.

Angioplasty is a technique in which arteries partially blocked with plaque are dilated to increase blood flow. By what factor must the radius of an artery be increased to increase blood flow by a factor of 7.0?

In this situation, if we imagine that the artery initially has a radius 𝑟 sub one and after the operation has a radius 𝑟 sub two, then we want to figure out what is 𝑟 two divided by 𝑟 one. We can recall that, for a liquid, the flow rate of that liquid equals the cross-sectional area of the tube or pipe it passes through multiplied by the speed of the liquid. In this scenario, we can assume that the speed of the blood flow in the original artery is the same as the blood flow speed in the extended artery. This means we can write that 𝐴 two, the cross-sectional area of the dilated artery, times 𝑣 is equal to 7.0 times 𝐴 one times 𝑣. The fluid flow speed 𝑣 cancels out. And if we assume that the artery has a circular cross section, we can write that 𝜋 times 𝑟 two squared is equal to 7.0 times 𝜋 times 𝑟 one squared. We see the factors of 𝜋 cancel out.

If we divide both sides by 𝑟 one squared and take the square root of both sides, we see that 𝑟 two over 𝑟 one equals the square root of 7.0, or to two significant figures 2.6. That’s the factor by which the artery radius must be dilated in order to increase the flow rate by a factor of 7.0.

Now let’s look at an example involving flow rate and continuity.

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.

Now let’s summarize what we’ve learned about flow rate and continuity. We’ve seen that the concepts of flow rate and continuity apply to fluids that are flowing through pipes. We’ve also seen that when it comes to flow rate, speaking in terms of mass flow rate, that is, density times volume, is often helpful for gases. While volume flow rate, cross-sectional area times liquid’s speed, is often helpful for liquids. And finally, we learned a generalized continuity equation. Which says that if we consider the density, cross-sectional area, and fluid’s speed at a particular point in a pipe, then their product is equal to the fluid density, cross-sectional area, and speed at any other point. This relationship helps us understand fluid flow through pipes of varying diameters.