Lesson Video: The Continuity Equation for Fluids Physics • 9th Grade

In this video, we will learn how to calculate the rate of transfer of smoothly flowing fluids through channels with varying cross sections.

14:05

Video Transcript

In this video, we will be discussing what’s known as the continuity equation for fluids. The equation in question is often used to tell us how certain fluids can flow from one container to another where the volumes of the two containers are different.

So to understand the continuity equation, let’s first imagine the following scenario. Let’s imagine that we’ve got a cylindrical container here. And within this cylinder, we’ve got a gas that’s flowing from left to right. Now, right now, we happen to be considering a gas. But remember, the continuity equation that we’ll be studying applies to fluids, not just gases. And remember that fluids flow. Therefore, they could be liquids or gases.

But anyway, for now, we’ll just consider a gas in our cylindrical container. Now, an important assumption that we’ll make here is that the motion of all the gas molecules is mainly toward the right. The gas is flowing smoothly from left to right. In other words, we will ignore any motion of molecules that is not toward the right, since this will be minimal and not affect our fluid flow. Let’s also imagine that for every second of time that passes, one kilogram of gas moves from left to right within our container.

Let’s imagine, for example, that we were to consider this position along the length of our container. We can measure the amount of gas passing this cross section per unit time. And in this case, we happen to find that one kilogram of gas is flowing per second past this cross section. This is an important quantity to remember for later because what we’ve actually found is the mass flow rate of our gas, which sounds quite complicated. But it’s just a fancy way of saying the amount of mass of gas flowing per second from left to right.

And we can actually quantify the mass flow rate of our gas as the mass of the gas that’s flowing from left to right divided by the time taken for that particular amount of mass to flow. And in this particular situation, the way that we did that was to measure the amount of gas passing this point in our container. And in this particular scenario, we imagine that we found one kilogram of gas flowing per second.

Now, this whole idea about mass flow rate is quite important because now let’s imagine that our container actually does this. Let’s imagine that as the gas flows from left to right, the container itself narrows. In other words, the gas is shoved into a region with a smaller cross sectional area, where specifically the cross sectional area is the area of the circle that we’ve drawn in here. As we can see, earlier on in the container, the cross sectional area was much larger. We can say that this area is 𝐴 subscript one. And now, in the smaller region of the container, we can see that the cross sectional area is different. So we will call this cross sectional area 𝐴 subscript two.

Now, here’s the interesting thing. Our gas is still flowing from left to right in our container despite the fact that it suddenly narrows. So how does our gas respond to this narrowing of the container? Well, because we’re imagining a scenario where the gas molecules are simply left to their own devices traveling from left to right and we don’t exert any external forces on these gas molecules. This means that the mass flow rate in the second part of the container must be the same as the mass flow rate in the first part.

The reason for this is that in the first part of the container, we’ve got a certain amount of mass flowing per second from left to right. And that mass cannot suddenly disappear at the point where the size of the container changes. That amount of mass must still flow from left to right in the narrower part of the container. Therefore, if we’ve got one kilogram of gas flowing from left to right per second in the wider part of the container, then we’ve also got one kilogram of gas flowing from left to right per second in the narrower part of the container.

Now, if we give our mass flow rate a symbol, let’s say we give it the symbol 𝑅, then we can say that the mass flow rate in region one, we will call this 𝑅 subscript one, must be equal to the mass flow rate in the second region 𝑅 subscript two. At which point, we’ve arrived at a nice little consequence of the fact that a certain amount of gas flowing from left to right in the first part of the container cannot change as the shape of the container changes. And therefore, the mass flow rates in both regions must be the same because particles of gas cannot randomly appear or disappear as they move across the boundary between the first and second regions.

Now, instead of considering specific value for the mass flow rate of our gas, so instead of saying one kilogram of gas flows from left to right per second. Let’s say that in the first region, a mass 𝑚 one of gas moves from left to right in a time interval 𝑡. Now, interestingly, we can imagine that at the beginning of our time interval 𝑡, if we think about gas molecules that started here, at the end of the time interval 𝑡, those gas molecules would end up further along the cylinder. In other words, they’d end up somewhere like here. And hence, we could say that over our time interval 𝑡, the molecules have moved this distance here. Let’s call this distance 𝑑 subscript one.

At which point, we can think about the fact that if our gas molecules are moving this distance here, 𝑑 one over a time interval of 𝑡, then they must be moving with a speed 𝑣 one of 𝑑 one divided by 𝑡. Because, remember, the speed of an object is equal to the distance moved divided by the time taken to move that distance. Now, it’s worth noting that 𝑣 one is the speed at which the bulk of the gas moves from left to right because individual molecules may have slightly slower or slightly faster speeds as well. But, on average, they’ll be moving with a speed of 𝑣 one.

And another interesting thing to note is that the gas which initially started out here at the beginning of our time interval and finished up here has actually occupied this amount of volume. This is a cylindrical volume naturally because our container is a cylinder with a cross sectional area of 𝐴 one, same as this cross sectional area here, and a length of 𝑑 one. At which point, we can recall that the volume of a cylinder given by what we will call capital 𝑉 is equal to 𝐴, the cross sectional area of the cylinder, multiplied by 𝑑, the length of the cylinder.

And hence, we can say that the volume occupied by the gas that we happen to be considering, specifically the gas with mass 𝑚 one moving from left to right a distance 𝑑 one over a time interval 𝑡, is given by capital 𝑉 subscript one. Is equal to 𝐴 one, that’s the cross sectional area of the cylinder, multiplied by 𝑑 one, that’s the distance the gas travels from left to right.

And by the way, we need to be careful in differentiating between lowercase 𝑣, which is the speed of the gas as it moves from left to right, and uppercase 𝑉, which is the volume that the gas occupies. Now, at this point, we might be wondering, what’s the point of writing down all of these equations? Well, remember that we’re trying to link back our gas’s behavior to its mass flow rate. And we’re considering the mass of gas that moves from left to right over a time interval 𝑡.

We know that the particular amount of gas that we’re considering has a mass of 𝑚 one and occupies a volume of 𝑉 one, which means that at this point, we can recall a quantity known as density, signified by the letter 𝜌. The density of an object is equal to its mass divided by the volume that it occupies. And so we can quantify the mass of our gas, in this case, 𝑚 one, in terms of the density of the gas and the volume that it occupies. Because we can also say that in the first region of the cylinder, the gas has a density 𝜌 subscript one.

So taking this equation and rearranging it by multiplying both sides of the equation by the volume 𝑉, we find that the mass of the gas, 𝑚 one, is equal to the density of the gas, 𝜌 one, multiplied by the volume it occupies, 𝑉 one. So we’re getting closer to quantifying our mass flow rate in the first region of the container in terms of certain properties of the gas and the container. Specifically the density of the gas and the volume of the container occupied by that gas. Which we will now express in terms of the cross sectional area of the cylinder and the distance that the gas moves. In other words, we can say that the volume 𝑉 one is equal to 𝐴 one 𝑑 one.

And now, finally, we can substitute into our mass flow rate equation that the mass flow rate on the left-hand side of the cylinder, which is 𝑅 one, and we know this to be equal to 𝑚 one divided by 𝑡, is actually equal to 𝜌 one 𝐴 one 𝑑 one divided by 𝑡. So we’ve expressed the mass flow rate in the first part of the container in terms of the density of the gas, the cross sectional area of the cylinder, and the distance that the gas moves in a time interval 𝑡.

But then, at this point, we can also look at this part of our fraction here, 𝑑 one divided by 𝑡. And recall that 𝑑 one divided by 𝑡 is actually equal to lowercase 𝑣 one, the speed at which the gas moves from left to right. And so we find that the mass flow rate in the first part of the container is equal to the density of the gas multiplied by the cross sectional area of the cylinder multiplied by the speed at which the gas is moving from left to right. All of which are known quantities on the right-hand side now.

And here’s where things get interesting. Remember we arrived at a conclusion early on that the mass flow rate in the second part of our container must be the same as the mass flow rate in the first part of the container. Therefore, if we also say that the mass flow rate in the second part of the container, 𝑅 two, is equal to the density of the gas in the second part of the container. Multiplied by the cross sectional area of the second part of the container multiplied by the speed with which the gas moves from left to right in the second part of the container. And then we use the equation 𝑅 one is equal to 𝑅 two and substitute in the right-hand sides of both equations. Then we arrive at this equation here. This is known as the continuity equation for fluids.

This is a very useful and important equation when it comes to the analysis of a smoothly flowing fluid. And although we don’t need to know off by heart how we derive this equation. It’s important to have seen how it’s derived so that we understand the deeper meaning behind these symbols. Now, interestingly, this equation applies fairly generally, and it can tell us how our fluid must behave as it flows. In this particular diagram, we’re considering a gas flowing from left to right through a container that has a smaller cross sectional area in the second region compared to the first.

So one thing that we can say is that the cross sectional area of the first region of the container, 𝐴 one, is larger than the cross sectional area of the second region, 𝐴 two. But then, if 𝜌 one 𝐴 one 𝑣 one has to be equal to 𝜌 two 𝐴 two 𝑣 two and 𝐴 one is bigger than 𝐴 two. Then this must be compensated for by an increase in the value of 𝜌 two, the density of the gas, or an increase in value of 𝑣 two, the speed at which the gas moves, or both. In other words, either we find that 𝜌 one is less than 𝜌 two or 𝑣 one is less than 𝑣 two or some element of both of these above conditions.

And if we think about it, this makes intuitive sense as well. We’ve got a larger container on the left-hand side. And a gas is moving at a certain speed from left to right through this container. And as it enters a narrower container, either the gas must flow faster so that the same amount of mass is transferred from left to right in the same amount of time. Or all the particles must be squished in to a smaller volume, so the density increases, or some element of both. Now, in this particular situation because we’re considering a gas, it’s likely that some element of both will happen. In the narrower part of the container, the density and the speed with which the gas moves will increase.

However, there is one particular type of material that behaves in a way that would only result in one of these changes occurring. One example of such a material is, in fact, water. Water is what’s known as an incompressible fluid. And essentially, what this means is that it cannot be compressed. If we look on a molecular level so that we’re looking at individual water molecules here, then, as it turns out, water molecules cannot be forced together into a smaller volume.

In other words, let’s say we’ve got 100 molecules of water here, and they occupy a specific volume. We cannot exert a force on this volume here to make these water molecules occupy a smaller volume. We cannot shove these water molecules closer together. That’s why water is incompressible. And a consequence of this is that a certain number of molecules will always occupy a certain volume. And that volume cannot be compressed so that those molecules occupy a smaller volume. And because that certain number of molecules has a certain mass, essentially, what we’re saying is that for a particular mass of water molecules, the volume occupied by that mass must be constant. Or, in other words, the mass per unit volume is constant for water, and therefore the density of water is constant.

What this means is that as water flows from left to right, from the wider part of the cylinder to the narrower part of the cylinder, the density cannot change. So what we’ve got in this particular scenario, as we move from left to right, is that the cross sectional area of the first part of the cylinder is greater than the cross sectional area of the second part of the cylinder. But because we’ve got water flowing through this container, the density on both sides of the container must be the same. It’s an incompressible fluid. And hence, the only way to ensure that 𝜌 one 𝐴 one 𝑣 one is equal to 𝜌 two 𝐴 two 𝑣 two is to result in a corresponding increase in 𝑣 two, the speed with which the water moves in this part of the container.

Since the water cannot be compressed, the same mass of water must occupy the same volume. But the cross sectional area is decreasing. So the length along the cylinder occupied by the water must increase. The water must flow faster to the right. In other words, what we find is that 𝑣 one, the speed with which the water moved in the first part of the container, is less than 𝑣 two, the speed with which it moves in the second part of the container. And so that is what happens when we apply the continuity equation to an incompressible fluid, such as water. At which point, we’ve seen the derivation of the continuity equation and seen how it can be applied to a couple of different fluids.

So let’s now summarize what we’ve talked about in this lesson. Firstly, we saw that the continuity equation relates the mass flow rates of a smoothly flowing fluid at different points along its flow path. Specifically, we saw that for a fluid flow being compared at two different positions along its flow, positions one and two, the continuity equation tells us that the density of the fluid multiplied by the cross sectional area of the container it’s flowing through multiplied by the speed at which the fluid flows in position one must be equal to the density of the fluid multiplied by the cross sectional area multiplied by the speed at which the fluid flows in position two.

And we saw it used in a context where positions one and two were actually regions one and two. And region two featured a change in the cross sectional area of the container compared to region one. Which brings us on to our final point, which is that the equation is commonly used to describe the effects of a change in the cross sectional area of the container that the fluid is flowing through. But the equation is not limited to this purpose. The continuity equation can be used to describe changes in the properties of a fluid flow when any of these quantities change. And that is an overview of the continuity equation for fluids.