Practice: A cylindrical pipe with inner diameter of 4 cm is used to fill up a 10,000 L tank with a 700 kg/m^{3} oil. If it takes one hour to fill up the tank, calculate the speed, in m/s, with which the oil travels inside of the pipe.

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Density | 34 mins | 0 completed | Learn |

Intro to Pressure | 76 mins | 0 completed | Learn |

Pascal's Law & Hydraulic Lift | 28 mins | 0 completed | Learn |

Pressure Gauge: Barometer | 14 mins | 0 completed | Learn |

Pressure Gauge: Manometer | 15 mins | 0 completed | Learn |

Pressure Gauge: U-shaped Tube | 23 mins | 0 completed | Learn |

Buoyancy & Buoyant Force | 64 mins | 0 completed | Learn |

Ideal vs Real Fluids | 5 mins | 0 completed | Learn |

Fluid Flow & Continuity Equation | 22 mins | 0 completed | Learn |

Additional Practice |
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Bernoulli's Equation |

Viscous Flow & Poiseuille's Law |

Concept #1: Flow & Continuity Equation

**Transcript**

Hey guys. So, in this video we're going to start talking about flow continuity, which is one of the most important topics in fluid flow, let's check it out. Alright, so first I want to go over the two main terms that deal with how quickly a fluid flows, and you need to know the difference between the two, it could be tricky because they sound similar. So, you have fluid speed and volume flow rates, and it's kind of tricky because speed and rate sort of tend to mean the same thing. Now, fluid speed is just basically distance over time, it's meters per second, how many meters, how many meters, well, let's say, a water molecule travel in a pipe over the course of one second. So, it's just Delta X over Delta t or distance over time. So, let's say, we have two pipes and in this pipe water, a little water molecule goes from here to here, in 1 second and another, and the other pipe or water molecule goes from here to here in 1 second, which one has a greater fluid speed, hopefully you got that v2 would be greater than v1 because this is a greater distance in the same amount of time, okay? So, it just has to do with how quickly is move, water or whatever fluid is moving. So, if you have this one moving like this and the other one moving like this getting ahead of it then the bottom one is faster, volume flow rate is a little bit different, it's not distance, it is volume hence the name volume, right? So, you can see in the unit's meters per second and then this is meters, cubic meters per second, this is not the distance but the amount of volume over time. Now, before we get confused. Notice that I wrote this as a lowercase v, which is speed, and this is uppercase V, which is volume, okay? Don't get those mixed up, so the idea is, let's say you have two pipes and one is thicker than the other. So, in this pipe of the molecule, water goes from here to here in one second, and then in this pipe you're going to get the same thing, molecule water goes from here to here in one second as well, so the distances and the times are the same. So, v1 equals v2, however, if you have a ton of water going through this pipe even though they're moving at the same fluid speed there's a lot more water that is traveling from here to here, there's a lot more volume of water here. So, because this has the greater volume, we're going to say that this here, has the greater volume flow rates, which is given by the letter Q. So, I'm going to say that Q2 is greater than Q1, that should make sense, it's a bigger pipe so it carries more volume of water per second, if the speed is the same, okay? I want to do one more thing with this equation here, I'm going to rewrite Delta V over Delta t over here and I want to remind you that volume is a three-dimensional measure and in a pipe like this volume would be the cross-sectional area here, times the distance, okay? So, let's say, a little distance here, this volume here would be the cross-sectional area A times this little distance, let's call this distance Delta X, so the volume there is a Delta X. So, I can rewrite Delta V as A, cross-sectional area, which is constant in this pipe, times Delta X, and notice that I end up with Delta X, over Delta t, which I can rewrite as A, V, this is just speed, okay? So, I'm going to put this over here and I need you to know that fluid speed is Delta X over Delta t and volume flow rate is Q equals Delta V over Delta t but can also be written as A, little v, okay? This is little v, this is speed, let's do a quick example to highlight the difference between these two.

So, a long horizontal pipe has a 2 square meter cross-sectional area. So, let's draw a pipe and that means that this area here is 2, and it says, it takes, it takes water 5 seconds to traverse, to travel an 80 meter segment of the pipe. So, let's say that from here to here, water moves, this is a distance or Delta X of 80 meters and it takes time T or Delta t of 5 seconds and we want to know what is the fluid speed, fluid speed is just little v, Delta X over Delta t, this is 80 divided by 5, which is 16 meters per second, if you want to know the volume flow rate, that's Q, which is volume over time. Now, I don't have the volume, I could calculate the volume but I don't have to because I also know that I can rewrite this as area times speed, area times speed, I have the area is 2 square meters and the speed is 16 meters per second. Notice, what happens here, meter square with meters become cubic meters per second you multiply the two numbers you end up with 32. So, 32 cubic meters travel every second. So, that's how those two things work, and the reason why this distinction between v and Q is important, is because it's going to play into the next thing we're going to talk about, which is flow continuity and flow continuity is a pretty important principle and it's the idea that because an ideal fluid is incompressible its volume flow rates, volume flow rates, which takes letter, remember, Q, it is never going to change. So, I need you to remember that Q never changes, okay? So, for example, let's say that you are looking through a pipe like this that changes area, right? So, you have area 1 and you have area 2 which is obviously smaller than area 1, and let's say that there's a certain amount of volume that goes through this cross-sectional area, right? So, if you sort of pick a point in the pipe a certain amount of volume goes through that point? Well, if you pick a point on a second point, at this thinner part of the pipe here, the same amount of volume has to go through that second point every second as well, so the rate has to be the same, but now there's a much smaller area so, how do you get the same amount of water per second to go through this piece that this piece is much skinnier, the only way to do that is if the water is up moving much faster, and that's the idea of continuity, that the flow is continuous and it never changes. So, if the area becomes smaller then the speed has to increase to make up for that, okay? So, let's write this out real quick, and by the way, you may have done this, if you've ever played with a with a water faucet, right? So, there's a little bit of other some speed with which the water is coming out but then if you cover this, maybe like with your thumb, I'm going to draw an ugly thumb here, if you cover this with your thumb, that looks terrible, let's put a little hair there, what's going to happen is that the water is going to now come out faster because you closed it, okay? So, Q never changes. Remember, Q can be written as the volume, change in volume over time, it could also be written as area times speed, okay? It can also be written as area times speed, so the idea, and this by the way, this is called the continuity equation, continuity equation, and the fact that A, v is constant means that, if the area changes then the velocity or the speed has to change in the opposite direction. So, greater area would mean slower speed. So, if area changes speed changes, and this segment here that I circled can be sort of extracted out of the equation and written like this, area 1, speed 2 equals area 2 speed 2, and that's because A, v never changes. So, if one goes up the other one goes down, the product of those two is constant. Last point I want to make before we make it do an example here is that pipes are usually going to be cylindrical, so the area with this area here, this is cylindrical, is going to be pi, R squared, it's the area of a circle.

Let's do a quick example. So, a garden house with radius 2 and at the end there's a nozzle of radius 1. So, it's something like this, radius 2, and then at the end there's a nozzle with radius 1. So, here I'm going to call this R1 is 2 centimeters and R2 is 1 centimeter, water flows into it with two meters per second. So, speed 1 is 2 meters per second and we want to know what is the speed over here. So, what is speed 2, so this is part A so, how do we solve this? Well, whenever we have a pipe and the area is changing, we're always going to be able to use this equation to figure out one of the speeds or one of the areas, okay? So, we're going to go straight through this, we're not going to use the sort of bigger equation that has all the variables, we're just going to use this one. So, we're going to write that A1, v1 equals A2 v2 I'm looking for v2. So, I just have to move A2 to the other side. So, v2 is going to be v1 times A1 over A2, okay? V1 we have, it's 2 meters per second, A1 remember, this is a cylindrical pipe, the way I know that is because it says it's got a radius, so the area is going to be pi, R1 squared divided by pi, R2 squared. Notice that the pi's cancel, that's cool. And then I have 2 meters per second, the radia are 2 and 1. So, 2 centimeters and 1 centimeter. Now, you might be thinking, shouldn't I replace or convert from centimeters to meters? You don't have to because they're just going to cancel each other out. So, for this part you don't have to convert anything, okay? So, this is going to be 2 square which is 4, 4 times 2, 8. So, 8 meters per second is the speed with which water will come out here. So, let me write that here, 8 meters per second. And then for part B we want to know in how many minutes does it fill up a 350 liter bathtub? So this is giving us the volume, 350 liters, and it's asking for the time but it's asking for the time in minutes. So, what's gonna happen is you're going to find time in seconds and then you're going to convert it to minutes at the very end, let me out of the way, and we are going to keep going here. So, what do we do? Well, now we are going to go back to this big equation, this big equation sort of has all the variables and you can see how it has a Delta t in it, right? So, I'm going to write the whole thing just real quick, Delta V over Delta t, I like to think of this equation, I like to sort of remember this whole thing together okay. Notice that what I have here is the definition of Q, the two definitions of Q and the fact that A1, v1 equals A2, v2, all into one equation. Now, there are three equal signs here, there are four parts to this. So, what we do is we just pick two of them, I want time so I'm going to pick this and this and I'm going to say the Delta V over Delta t equals A1, v1, I could have said equals A2, v2, just the same. Alright, so let's start plugging some stuff, so the volume is going to be, the volume is going to be 350 liters divided by time, which is what I want, area is going to be pi, R1 squared. Now, I actually have to convert because I'm going to be calculating stuff, there's nothing to cancel centimeters with. So, I have to convert this into meters and the R1, R1 is 2 centimeters or 0.02 meters and that's going to be times the first velocity, which is 2 meters per second, okay? And, if you put this whole thing in a calculator, if you put this whole thing in the calculator, you get that this is 0.0025 cubic meters per second. Notice that I have meter square and then meter. So, that's 3 m's. So, that's cubic meters, okay? That's what that's going to be but I'm looking for time. So, I'm going to move time up here and then this guy is going to come to the bottom of the 350, okay? So, please do this very carefully, it's going to come out like this, Delta t equals 350 liters divided by 0.0025 meter cubed divided by seconds, getting tightened there but we'll make it work. Now, notice that you have liters and, which is the volume in meter cube which is also volume but they can't directly cancel because one meter is not equal to one, I'm sorry, 1 liter does not equal 1 cubic meter, so, hopefully you remember that one cubic meter, put this a little closer, one cubic meter is actually 1,000 liters, okay? So, what I can do is I can write over here that one cubic meter is 1000 liters and in doing this what happens, I can cancel this and this, okay? And, if I multiply this whole thing here, I get that this is 140 seconds, okay? But, we want this in minutes, last step is to divide by 60 seconds because that's what one minute is and long story short, we end up with 2.33 minutes. So, this water pipe would take 2.3 minutes to fill up this bathtub and that's it for this one, let's keep going.

Example #1: Continuity / Proportional Reasoning

**Transcript**

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 big V and point B has double 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 diameter of B will be twice that to D and we want to know what is the volume, what is the 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 air the D 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 A times speed is a constant. So, if the area increases which it does here the speed has decreased so that the products A V stays the same, okay?

So, one way to think about, this is, 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, 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 over 4 or V over 2 and what you can do is you can just write, you can write A1, V1 equals A2 V2, right? And we're solving for or I guess I could say a, a and a B, right? And we're writing, we're solving for VB. So, VB is the first area times the first speed divided by the second area. Now, the area of a cylindrical pipe is pi, r squared. So, I can write pi, r square divided by pi, r squared times the first velocity which is V, okay? Now, I don't have the radius, I have the, 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 going to call this r and this is going to be 2r, okay? So, if the diameter doubles the radius doubles and all these questions whenever you have diameter, pretty much in all of physics whenever you see diameters are supposed to change that into radius, okay? So, one is double the other so the pi's 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 r square, this 2 here becomes a 4, 4r squared so the rs squares cancel and you're left with V over 4, okay? 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 over 4, cool? These are pretty popular, hopefully this makes sense, let's keep going.

Practice: A cylindrical pipe with inner diameter of 4 cm is used to fill up a 10,000 L tank with a 700 kg/m^{3} oil. If it takes one hour to fill up the tank, calculate the speed, in m/s, with which the oil travels inside of the pipe.

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Concept #1: Flow & Continuity Equation

Example #1: Continuity / Proportional Reasoning

Practice #1: Speed to Fill Up Tank

A hose is connected to a faucet and used to fill a 7.0-L container in a time of 45 s. Determine the velocity of the water in the hose in the first part if it has a radius of 1 cm.

A cylindrical blood vessel is partially blocked by the buildup of plaque. At one point, the plaque decreases the diameter of the vessel by 65.0%. The blood approaching the blocked portion has speed v0 Just as the blood enters the blocked portion of the vessel, what will be its speed in terms of v0 ?

An incompressible fluid flows steadily through a pipe that has a change in diameter. The fluid speed at a location where the pipe diameter is 8.0 cm is 1.28 m/s. What is the fluid speed at a location where the diameter has narrowed to 4.0 cm?(A) 0.32 m/s(B) 0.64 m/s(C) 5.08 m/s(D) 2.56 m/s(E) 1.28 m/s

Compared to the speed of the water in the 0.5-cm pipe, the speed in the 1-cm pipe isa. one-quarter the speed in the 0.5-cm pipe.b. one-half the speed in the 0.5-cm pipe.c. the same as the speed in the 0.5-cm pipe.d. double the speed in the 0.5-cm pipe.e. quadruple the speed in the 0.5-cm pipe.

Blood flows through an artery that is partially blocked. As the blood moves from the wider region into the narrow region,the blood speed
a. increases.
b. decreases.
c. stays the same.
d. drops to zero.
e. alternately increases and then decreases.

An incompressible fluid is flowing through a horizontal tube which, at some point has a constriction such that the area of the tube becomes much smaller.
How do the fluid pressure and speed of flow compare at point B in the constricted region to their values at point A in the normal part of the tube?
A. The pressure and speed of flow are both much greater at point A than at point B.
B. The pressure and speed of flow are both much greater at point B than at point A.
C. At point B, the speed of flow is less but the pressure is greater than at point A.
D. Because the fluid is incompressible, the pressure and speed of flow must be constant throughout the tube.
E. At point A, the speed of flow is less but the pressure is greater than at point B.

A hose is connected to a faucet and used to fill a 7.0-L container in a time of 45 s. Determine the volume flow rate in m3/s. (Express your answer to two significant figures.)

You’re holding a hose at waist height and spraying water horizontally with it. The hose nozzle has a diameter of 1.80 cm, and the water splashes on the ground a distance of 0.950 m horizontally from the nozzle. Suppose you now constrict the nozzle to a diameter of 0.750 cm; how far horizontally from the nozzle will the water travel before hitting the ground? (Ignore air resistance.)

a) Calculate the mass flow rate (in grams per second) of blood (ρ = 1.0 g/cm 3) in an aorta with a cross-sectional area of 2.0 cm2 if the flow speed is 40 cm/s.b) Assume that the aorta branches to form a large number of capillaries with a combined cross-sectional area of 3.0 x 103 cm2. What is the flow speed in the capillaries?

The approximate diameter of the aorta is 0.50 cm; that of a capillary is 10 μm. The approximate average blood flow speed is 1.0 m/s in the aorta and 1.0 cm/s in the capillaries. If all the blood in the aorta eventually flows through the capillaries, estimate the number of capillaries in the circulatory system.

Water flows from the pipe shown in the figure with a speed of 3.0m/s.What is the height h of the standing column of water?

River Pascal with a volume flow rate of 5.0 x 105 L/s joins with River Archimedes, which carries 10.0 x 105 L/s, to form the Bernoulli River. The Bernoulli River is 150m wide and 10 m deep. What is the speed of the water in the Bernoulli River?

A nozzle with a radius of .33 cm is attached to a garden hose with a radius of .85 cm. the flow rate through hose and nozzle is .55 L/s.a. Calculate the maximum height to which water could be squirted with the hose if it emerges from the nozzle in m.b. Calculate the maximum height (in cm) to which water could be squirted with the hose if it emerges with the nozzle removed, assuming the same flow rate.

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