Fluid Flow & Continuity Equation - Online Tutor, Practice Problems & Exam Prep

Fluid Speed & Volume Flow Rate

Folks, in this video, we're going to talk about 2 important terms when it comes to fluids in motion. We're going to talk about fluid speed and volume flow rates. Both of them are really important; they sound kind of similar. I'm going to go ahead and break down the difference between the 2 and we'll do a quick example, alright? So these two main terms deal with basically how quickly a fluid flows. So, fluid speed is pretty straightforward because we've dealt with speed before. Remember, speed is always just meters per second. It's a distance over time. So this would be literally like how fast a water molecule is traveling through a pipe. So the velocity is given as ΔxΔt. Basically, if you could track a little water molecule as it flows through this pipe like this and if one of them was traveling faster, then basically this would have Δx1, this would have Δx2 and if both of these water molecules had the same Δt, let's say it was just one second, then what this means here is that the velocity of the second molecule because Δx2>Δx1 would be greater than the velocity of the first one. It's pretty straightforward. Basically, it's just how fast something is moving through an actual pipe or something like that. Alright. Now the other one is going to be volume flow rate, which is a little different. We haven't heard that one before, but it's tricky because sometimes rate and speed are sort of generally used to mean the same thing. But speed is going to be meters per second, but rate is basically how fast something is changing. The thing that's changing here is going to be the volume. So the equation for this is actually going to be or the units for this is actually going to be meters cubed per second. Notice how this is meters per second, right? It's distance over time, but this is actually units of volume per seconds. So the unit or the letter that we use for the volume flow rate is actually going to be Q and what this is ΔVΔt. Alright. So this would be like if you had one water molecule that was traveling through a skinnier pipe like this, and then you had another water molecule that was traveling through a much larger thicker pipe like this, even though their speeds are the same. So they're so in this case, the Δx1, Δx2 are the same. It's just the same Δx, so therefore, v1=v2. There's something that's significantly different about this thicker pipe here, which is basically the volume of water that's passing through it in a certain amount of time is much larger. If you take a cross sectional area of this pipe like this, then basically all the water that's going through this pipe is everything that I have shaded in this region over here. Right? I'm going to go ahead and label this or highlight this in yellow. Alright. So all of this water here is passing through. In the same exact amount of time in the skinnier pipe, you have a much smaller volume of water because the cross sectional area is smaller. Notice how the amount of water that flows through this one second is only just this piece over here. So basically, that's what's different about these 2. Even though their speeds are the same, their fluid speeds, their volume flow rates are different. ΔV2>ΔV1, and that's basically just because you have a bigger cross sectional area. Alright? So one thing I want to do with this equation here is sort of rewrite this. Remember that ΔV or remember that volume can generally be written as area times height. Right? So in other words, this is the Δx and to this cross sectional area over here of this cylindrical pipe is going to be A. These two things combined, the area times the Δx is what your volume is equal to. So this volume here is equal to A×Δx. So you can rewrite this equation as A×ΔxΔT. Now notice something weird happens because we have an area times ΔX over ΔT. Remember, that's just the definition of velocity. So in another way, you can rewrite this equation is that it's actually equal to A×v. So I want to point out here that this capital V, this ΔV for volume is an uppercase V. So this is uppercase V, whereas this one is a lowercase v. So this has to do with the velocity. Alright. So don't get those confused. But basically, what this means is that you can actually rewrite this ΔVΔT as the area times the velocity. Alright? So it's kind of a sort of a little confusing there because you have a big V and a little v, but you can essentially rewrite the volume flow rate in terms of the fluid speed. Alright? So these are going to be the equations that you use, basically throughout your fluid flow problems. Alright? So let's go ahead and just take a look at an example real quick. It's pretty straightforward. So we have a long horizontal pipe, it's got a 2 meter section a cross sectional area. So in other words, we've got this area over here. This A is equal to 2. Alright, so we've got, it takes 5 seconds to travel an 80 meter segment of the pipe. So basically, I'm going to say is there's a water molecule here. It's going to travel some distance and I have that this distance here is going to be Δx=80. Alright, and so what happens here, I want to calculate 2 things. I want to calculate the fluid speed and then the volume flow rate. Let's take a look at the first one here. We're going to calculate the fluid speed here. Remember that fluid speed V is just ΔxΔt, so straight up distance over time. It doesn't take into account the amount of water it's flowing. It's just how long does it take one molecule to get from one place to the other. And it's pretty straightforward. We have that Δt is just equal to let's see. It was 80 meters in 5 seconds. So in other words, this is just going to be 80 divided by 5 and what you're going to get is 16 meters per second. Alright, so that's just straight up what the speed of this liquid running through is. Now, let's take a look at the second part here. We're going to calculate the volume flow rate. Remember that that's going to be Q and Q is equal to 2 things. We have ΔVΔT, or you can rewrite this in terms of area times the velocity, terms of basically area times this fluid speed over here, which we just found is just 16 meters per second. So in other words, we can rewrite this and say we have the cross sectional area. The area is just 2 over here, and that's just going to represent all of the volume over here. That's ΔV and this is just going to be 2 times 16. Notice how let's see. Let's write out the units. We're going to have 2 meters squared we're going to have 16 meters per second, and what you end up with here is you end up with 32 meters cubed per second. Alright? So even though the fluid speed is only 16 meters per second, the amount of volume that passes through this pipe in that time is 32 meters cubed per second. Alright, so that's the distinction. Hopefully that makes sense the difference between speed and volume flow rates.

Flow Continuity

Hey everyone. So now that we've talked about the difference between fluid speed and volume flow rates, in this video we're going to discuss an important consequence of that, which is called fluid or flow continuity. This happens whenever you have some liquid that's traveling in a region in which the area is going to change, like a pipe that gets skinnier or fatter or something like that. Alright? We're going to discuss a really important equation, and then I'll show you an example. Alright? So continuity basically says that because ideal fluids are incompressible because their densities can't change, they can't get more or less dense, and they can't get squeezed, then the volume flow rates of a liquid have to remain the same. So volume flow rates, which remember is given by the letter Q, is never going to change. So we can write this as an equation. We can say that remember Q, which is given as ΔV over ΔT . We saw that in the last couple of videos. We can also rewrite this in terms of area times speed. All these equations are sort of equal to each other. That has to remain constant. Alright, so this important relationship here that I've highlighted is oftentimes called the continuity equation. So let me go ahead and show you in this diagram what's going on here. If you have some kind of a pipe or you have some water that's flowing through this pipe over here and it's got a cross-sectional area A1, let's say the fluid speed is traveling over here. Then basically, what we know is that area times speed is going to give you a volume, and I'll highlight this in yellow over here. So you have, somewhat of volume that's flowing through this pipe. Let's just go ahead and call this 10 liters, just to give it some numbers. Right? What continuity says is that if you have 10 liters of water that's flowing through this pipe every second, then later on if the pipe changes its area, its cross-sectional area, you still have to have 10 liters that are traveling through the pipe. You can't have 5 liters because then where did all the extra water go. And you also can't have 20 liters because then all of a sudden you've just created a bunch of liquid out of nowhere. So basically, all of this stuff that's flowing through this pipe has to remain flowing at the exact same rates. The volume flow rate always has to remain the same. Okay? Now what we can see here is that in this pipe, the cross-sectional area has gone down. This cross-sectional area over here, A2, is much less. So what this equation tells us, this continuity equation is if the area changes like the area goes down, then the only way that you can have this Q that remains the same is if the speed increases. So in other words, the speed in this skinnier section of the pipe has to go up like this. Alright. So also this works vice versa. If the area goes up, the speed has to go down. So basically what continuity says is that if the area changes then the speed also has to change. Alright. And here's probably the most important equation that you need to know. The A 1 × V 1 = A 2 × V 2 . Okay. This is basically the equation that you're going to use when you talk about continuity. Alright. The last thing I want to mention here is that pipes are usually cylindrical, which means that their cross-sectional areas are going to be or π r 2 . Alright? Now if you've ever sort of messed with like a garden hose or something like this, you've probably sort of like stuck your thumb at the end of the garden hose. Right? So you've got some liquid that's coming out, You stick your thumb over here. This is going to be your thumb like this, and basically what happens is the water comes shooting out much much faster when you do that, and that's basically what's going on here. Right? So you have some cross-sectional area. Let me go ahead and do this in red. So you've got an '..

Continuity / Proportional Reasoning

Hey guys. So in this example, I want to quickly show how to deal with proportional reasoning or proportional change questions involving continuity. Let's check it out. Alright. So here we have water flow in a horizontal cylindrical pipe, something like this. And it says water has a speed of v at point A. So let's say that this is point A, and at this point speed at A will be v. And point B has doubled the diameter. So somewhere over here, this thing grows to have double the diameter. Right? So point B is somewhere over here. B and the diameter, let's say that the diameter of A will be d and the diameter of B will be twice that, 2d. And we want to know what is the volume, what is the velocity, the speed of water at this point. Okay. And first I want you to think about this, in conceptual terms. Do you think the water will be faster or slower here? And hopefully you pick that the water will be slower. Remember, if water is going into a tighter pipe part or a tighter segment of the pipe, it's going to go faster. So if it's going to a wider section, it's going to go slower. And that's because water, or fluid flow rates q, which equals area times speed, is a constant. So if, if the area increases, which it does here, the speed has to decrease so that the product AV stays the same. Okay. So one way to think about this is if this is a 2 and this is a 10, right? And this grows to a 4, this is 20. This has to decrease to a 5 so that this is still 20. Cool. So it should be slower, which means it can't it's not going to be the same. It's not going to be faster. So it's now down to whether it's v÷4 or v÷2. And what you can do is you can just write, you can write A1V1=A2V2. Right? And we're solving for, or I guess I could say AAandAB. Right? And we're writing, we're solving for VB. So VB is the first area, time, times the first speed divided by the second area. Now the area of a cylindrical pipe is πr2. So I can write πr2 divided by πr2 times the first velocity, which is v. Okay? Now I don't have the radii. I have the, the diameter. But diameter is half the radius. And if the diameter is doubling, that means that the radius doubles as well. So I can simplify this whole thing by saying, I'm just gonna call this r, and this is going to be 2r. K? So if the diameter doubles, the radius doubles. And in all these questions, whenever you have diameter, pretty much in all of physics, whenever you see diameter, you're supposed to change that into radius. K? So one is double the other. So the pies will cancel. And I can say that a is r. And then this guy here is 2r times v. So look what happens. I have, I have r2. This 2 here becomes a 4. 4r2. So the r2 cancel, and you're left with v÷4. K? So if the radius becomes twice as big, then the speed will become 4 times smaller. And that's because the area, depends on the square of the radius. So if the radius becomes twice as big, then the area becomes 4 times greater, which means that the speed has to go down by a factor of 4. So the answer will be v÷4. Cool? These are pretty popular. Hopefully, this made sense. Let's keep going.