Notes: Hydrodynamics Level 1: The Continuity Equation In this first set of notes, hydrostatics, we discussed fluids at rest. In this section, we look at

fluids in motion. This sounds more complicated, but most students find this section a lot

easier and more fun.

A Quick Note on Ideal Fluids A quick note: most of the problems we will deal with in this section ignore a little something

called viscosity. As we discussed briefly in the previous section, fluids are actually quite sticky-

they like to stick to the material they are in flowing past, and the molecules in the fluid like to

stick to one another. This is a very physics-y way of describing something much more complex

and chemical-y. Suffice it to say, fluids with high viscosity like honey or syrup are slow to flow or

pour. Water and brake fluid have low viscosity. For the sake of the physics problems you will be

doing in this unit, the fluids have virtually no viscosity- an artificial idea like say something has

“no friction.” Not real but hey, this is physics, we can do what we want.

The fluids we are dealing with are also non-compressible. You can’t squeeze the molecules

closer together.

When we ignore viscosity and compressibility, we deal with ideal fluids- in other words, fluids at

condition that have low viscosity and are incompressible. Little like a frictionless table (not real)

but close enough to work.

Flow Rate If we are going to discuss fluids in motion, we better have a way of measuring how they much

they move. Imagine you are sitting on a pipe and you want to measure how much water is

passing you every second. A logical measurement would be the volume of water (in m3) that

passes you per second (s). This is flow rate (Q), and it is measured (of course) in m3/s. Every

now and then, you will see flow rate in something heinous like liters/minute, and then you will

have to convert (ugh).

𝑄 =𝑉

𝑡

That’s easy enough, if we happen to have volume and time. But if you really were sitting on top of

a pipe, watching water rush by below you, it would be pretty hard to count the volume of water.

What if we could calculate this using the speed of the rushing water, though? Then we could just

drop a leaf in, what how fast it moved, and find out how much water moved past us every

second? That might be convenient. So let’s try to relate flow rate to velocity.

Imagine the pipe is cylindrical. The volume is therefore:

𝑉 = 𝐴𝑑

Apply this to the flow rate equation:

𝑄 =𝐴𝑑

𝑡

Remember that 𝑣 =𝑑

𝑡 so:

𝑄 = 𝐴𝑣

The velocity here is the flow speed, v, or how fast the fluid is rushing by any given point. This is

different than flow rate, Q,which tells us how much fluid is racing by every moment.

Practice Problem Water in a circular pipe flows by at a rate of 4 m/s. The pipe has a radius of 0.4 m. What is the

flow rate of this pipe? How long would it for 1000 m3 of water to flow past?

Answer Q=2.01 m3/s

t= 497 seconds, or 8.28 minutes

Practice Problem: Flow Rate and Flow Speed A circular pipe with a diameter of 4 cm is half filled with flowing water. A toy boat moving on the

water is seen to move 0.5 meters in 2 seconds. An empty bucket is placed at the end of the pipe

to collect the water. How long would it take for the bucket to have 0.3 m3 in it?

Answer t= 1910 s or 31.8 minutes

The Continuity Equation Fluid flowing through perfectly uniform pipes is kind of boring. Imagine instead that we know

had a pipe that changed diameter. This is a lot more like real life, where rivers run into boulders,

where IVs send fluids into veins, where breezes move through hallways. We already showed

that there is a relationship between velocity, area and flow rate.

Here is an example of a non-uniform pipe:

One key idea here: whatever mass flows into the pipe through the Hole 1 has to come out

through Hole 2. Otherwise, we would have a magic pipe that makes matter, and physics would

be broken and everyone would cry. This is the Law of Conservation of Matter, which states that

matter cannot be created or destroyed.

This also means that the volume of water flowing into Hole 1 has to be the same as the volume of

water flowing out every second at Hole 2. If this wasn’t true, it would mean that matter was either

being created in the tube (again, broken physics, tears, death) or that there was a black hole

eating matter up in the tube (which is much more exciting but just as bad). Neither is happening

in our normal tube, so the flow rate coming into the tube is the same as the flow rate of fluid

coming out. That’s an ugly sentence, so let’s say it with math:

𝑄1 = 𝑄2

Remember that 𝑄 = 𝐴𝑣 so we can say:

𝐴1𝑣1 = 𝐴2𝑣2

This is the continuity equation. Try out the problem below to get some practice using it.

Practice Problem: Plug n’ Chug Continuity

Solution

Practice Problem: Straight Forward Continuity Equation

Solution

Practice Problem: Continuity and Changing Variables

Try solving the problem above using proportional reasoning. It is a lot

easier.

Solution

Video Animation: Watch this short video, which goes more in depth regarding the

conservation of mass in the continuity equation. It is short and boring, but almost

exactly the sort of thing you will see in college science classes.

http://www.youtube.com/watch?v=BmrhXf-rrY0

Supplemental Lecture Video: Looking for more help with these? You can watch Ms.

Twu solve a few flow rate and continuity problems here (the first few videos). This is

not required, but if you are struggling or want to make sure you have it down, this

might be a good idea. You could also save these and try them out for a test review.

https://sites.google.com/site/twuphysicslessons/home/fluids/fluids-page-4

Even without the math, you had a sense of the continuity

equation long before you entered this classroom. Children

figure this out pretty quickly once they are given a garden

hose. If you decrease the diameter of the nozzle but covering

part of it with your finger, you can increase the flow rate,

making the water rush out and spray everywhere. Similarly, if

you’ve ever entered a tunnel (like in breezeway) on a windy

day, you might suddenly find the wind whirling around you

quickly. This is because the wind entered a space with a

smaller area, so the velocity increased.

Level 2: Bernoulli’s Principle In the last section, we dealt with fluids moving through pipes that changed diameter, and that

was fun. What if we also had the pipes move up and down- for example in a house. What if we

wanted to have a fluid flow through something like this:

Notice we aren’t just dealing with a pipe whose diameter is changing, but one whose diameter

and height is changing. I’m going to tell you right now, if you have flowing fluid and it is

changing height in any way, you are going to need to use the Bernoulli equation.

The continuity equation applies no matter what the height of the pipe is doing. But Bernoulli’s can

help us discuss other things- such as the pressure in the pipe at different point, and what the

energy within the pipe is like.

The continuity equation explored how the Law of Conservation of Mass effects flowing fluids.

In this section, we are going to look at Bernoulli’s Principle, which focuses on how the Law of

Conservation of Energy effects flowing fluids.

Blast From the Past: The Law of Conservation of Energy You probably remember The Law of Conservation of Energy from last year (that or you just

made a snarky joke about how you don’t remember it all). It’s not a difficult concept. The Law of

Conservation of Energy states that the energy in a system cannot change. You probably saw

it written this way:

𝐾𝑖 + 𝑃𝑖 = 𝐾𝑓 + 𝑃𝑓

or, in more detail:

𝑚𝑣12

2+𝑚𝑔ℎ1 =

𝑚𝑣22

2+𝑚𝑔ℎ2

Bernoulli’s Equation is an extremely useful application of the law of conservation of energy to

fluids. Here it is:

𝑃1 +𝜌𝑣1

2

2+ 𝜌𝑔𝑦1 = 𝑃2 +

𝜌𝑣22

2+ 𝜌𝑔𝑦2

When you use this equation, it is important to keep in mind that one of the symbols is 𝜌, which

stands for density, and the other is P, which stands for pressure. Occasionally, students mix these

up, fail and end up miserable and unemployed, forsaken by all who knew them. Don’t let this

happen to you!

You might notice that there is a lot of similarity between this equation and the Law of

Conservation equation you were familiar with last year. Yup, that’s because they stand for

roughly the same things. The 𝜌𝑣2

2 stands for the kinetic energy of the fluid. The 𝜌𝑔ℎ stands for

the potential energy of the equation. The P accounts for differences in pressures at different

points.

As scary as it looks, the Bernoulli equation is actually quite easy to use. Many of these problems

are simple plug n’ chugs, and if you see it on the AP test, you will probably feel grateful.

Okay, all that said, let’s look at a practice problem.

Video Practice Problem: Ms. Twu is going to show you how to solve a very

typical Bernoulli problem. Note that some of the variables (which she will show

you) are zero. This frequently happens in Bernoulli problems, which is why

students love them. Also, she uses 10 instead of 9.8 for g. Some physics teachers

prefer to do this. I like 9.8, but on the AP test, they accept either 9.8 or 10.

https://www.youtube.com/watch?v=vOGY31CMnn4#t=143

Practice: AP MC Bernoulli

1. The pressure at A is 9.5 atm and the water velocity is 10 m/s. What is the water velocity at point C?

(a) 2.5 m/s (b) 5 m/s (c) 10 m/s (d) 20 m/s (e) 40 m/s

Solution Answer: C Use flow continuity. A1v1 = A2v2 Since the area is the same at both locations the

speed would also have to be the same.

2. The pressure at C is

(a) 0 N/m2 (b) 100,000 N/m2 (c) 150,000 N/m2 (d) 800,000 N/m2 (e) 1,100,000 N/m2

Solution Answer: D

Practice Problem: AP Bernoulli

1. A cylindrical tank containing water of density 1000 kg/m3

as shown in the cross section above. A hole of

radius 0.0010 m in the bottom of the tank is opened. Water then is filled to a height of 0.70 m and placed

on a stand flows through the hole and through an opening in the stand and is collected in a tray 0.30 m

below the hole. At the same time, water is added to the tank at an appropriate rate so that the water level in

the tank remains constant.

(a) Calculate the speed at which the water flows out from the hole.

(b) Calculate the volume rate at which water flows out from the hole.

(c) Calculate the volume of water collected in the tray in t = 2.0 minutes.

(d) Calculate the time it takes for a given droplet of water to fall 0.25 m from the hole.

Solution a) To make life easier, let’s say the top of the tank is h=0.7 m, the bottom is 0 m. We can also assume the

fluid at the top is at rest.

𝑃1 +𝜌𝑣1

2

2+ 𝜌𝑔𝑦1 = 𝑃2 +

𝜌𝑣22

2+ 𝜌𝑔𝑦2

𝑃1 +𝜌𝑣1

2

2+ 𝜌𝑔𝑦1 = 𝑃2 +

𝜌𝑣22

2+ 𝜌𝑔𝑦2

𝜌𝑔𝑦1 =𝜌𝑣2

2

2

Plug n’chug

v=3.7 m/s

b) Volume flow rate Q = Av = π(0.001)2(3.7) = 1.16x10–5 m3/s

c) Q = V / t 1.16x10–5 = V / (2min * 60 s/min) V = 0.0014 m3

d) Free fall. d = vit + ½ gt2 – 0.25 = (– 3.7 t) + ½(– 9.8) t2

Alternatively, first determine v solve quadratic t = 0.062 s f at the 0.25 m location then use

vf = vi + at to solve for t.

Let’s leave the numbers behind for a second and talk more conceptually about Bernoulli’s

equation. When Bernoulli first purposed his bid idea, he said it this way:

Bernoulli’s Principle

For an ideal fluid, an increase in the speed of the fluid will result in a decrease pressure or

a decrease in the fluid’s potential energy.

Let me say that in simpler words: if you increase the speed at a certain point, the pressure will

decrease. If you decrease the pressure, the pressure will increase.

Or, even simpler:

Velocity goes up, pressure goes down. And vice versa.

Video Demonstration: Ms. Twu is going to show you some great demos to help

reinforce this idea.

Practice Problem: Bernoulli’s Principle A fluid is forced through a pipe of changing cross section as shown. In which section would the pressure of the

fluid be a minimum

A) I B) II C) III D) IV E) all section have the same pressure.

Solution Answer: B

This is based on two principles. 1 – Bernoulli’s principle says that when speed increases

pressure drops. Second, continuity says more area means less speed based on A1v1 = A2v2

So the smallest area would have the largest speed and therefore most pressure drop.

Connecting Back to Mechanics AP Physics problems love to cover more than one topic. If they can come up with some way to

combine fluids and electricity (and they have) they will put it on the test. The most common

physics mash-up, however, it to combine Bernoulli with mechanics problems, particularly free

fall problems, horizontal projectile problems, and projectile at an angle problems. If you don’t

remember these at all, you are going to need to relearn them. Don’t worry-it’s like riding a bike.

Once you learn them, it is pretty easy to get them back. If you need a refresher, you can relearn

the material by reading through Lesson 2 on this site. You will see some of these in your

practice.

Supplemental: Going into medicine? Wonder how this fluid stuff relates? Here is a

medical lecture that discusses the use of Bernoulli equations (and the Continuity

Equation) are used to monitor the human body.

Level 3: Torricelli One problem that we often see in Bernoulli problems are tanks with holes poked in the side of

them. They look a lot like this:

You can (and have) figured out how fast the water is rushing out the side

using Bernoulli’s. But there is actually a short cut- a short cut everyone

loves. Not all teachers teach their students this short cut because it can

make them lazy, but I’m all about doing things the easy way whenever

possible. I will let you do this, but there is a small catch, which I will

explain in a moment. First, let me show you where this short cut comes

from.

Derivation of Torricelli’s Theorem We have water flowing out of a tank, so we will definitely need to start

with Bernoulli.

𝑃1 +𝜌𝑣1

2

2+ 𝜌𝑔𝑦1 = 𝑃2 +

𝜌𝑣22

2+ 𝜌𝑔𝑦2

The water on the top of the tank is pretty much still, so v1=0.

𝑃1 + 𝜌𝑔𝑦1 = 𝑃2 +𝜌𝑣2

2

2+ 𝜌𝑔𝑦2

Take a look at the pressures. At the top of the tank, there is air, so the pressure should be 1 atm.

What about the pressure of the fluid coming out the side of the tank? It is also exposed to the

atmosphere, so it is also 1 atm. Since they are the same, they subtract one another out:

𝜌𝑔𝑦1 =𝜌𝑣2

2

2+ 𝜌𝑔𝑦2

Whew! Getting exciting. You might think we’re done… but take a look at the ρ. That’s right, we

get to say:

𝑔𝑦1 =𝑣22

2+ 𝑔𝑦2

Do a little arranging and we get:

𝑣2 = √2𝑔(𝑦1 − 𝑦2)

even easier:

Torricelli’s Theorem 𝑣2 = √2𝑔ℎ

This is called Torricelli’s Theorem. If you had trouble following all that, what it basically says is

that the velocity of the water leaving the tank is equal to the square root of gravity times the

height of the water above the hole.

You will not be given this on the test, since some teachers frown on teaching it and it isn’t in the

official AP curriculum. But- you are allowed to use it. You will need to have all the important

equations memorized anyway, so you might as well start with this one.

Here is the catch: if you want to use it on the AP test (or my tests) you need to state the

following phrase before you use it: “Torricelli, by way of Bernoulli”. If you write this down,

I will let you use this easy equation and these problems will suddenly be incredibly simple. If

you don’t write this phrase, you will not get any credit for using Torricelli’s. Why am I being this

harsh? Because the scoring guidelines on the AP test literally state that if you don’t state this

phrase (or something similar) the grader should give you no credit for your work.

One other thing- these problems are often combined with horizontal projectile problems or

projectile at an angle problems. If you have trouble remembering these types of problems, you

might want to review your notes from last year or talk to your group members. If you are still

confused, you can look at the following pages to help you out.

Ms. Twu Teaches Projectiles (Supplemental): On this page, there are a ton of

videos where Ms. Twu will teach you projectiles, both horizontal and at an angle.

Great review for everyone. Designed for AP students, so it goes fast, but she’s a

great teacher.

Classroom Physics: Horizontal Projectiles (Supplemental): Great, straight

forward site that review problem solving with horizontal projectiles.

Classroom Physics: Projectiles at an Angle (Supplemental): More from this site.

This page focuses on projectiles at an angle.

Practice Problem: Hole Problem

Solution

AP Practice Problem: Hole in the Side of a Container

2007B4. The large container shown in the cross section is filled with a liquid of density 1.1 x 103 kg/m3. A small

hole of area 2.5 x 10 -6 m2 allows a stream of liquid to flow through the hole and into a beaker placed to the

right of the container. At the is opened in the side of the container a distance h below the liquid surface,

which same time, liquid is also added to the container at an appropriate rate so that h remains constant.

The amount of liquid collected in the beaker in 2.0 minutes is 7.2 x 10–4 m3

(a) Calculate the volume rate of flow of liquid from the hole in m3/ s.

(b) Calculate the speed of the liquid as it exits from the hole.

(c) Calculate the height h of liquid needed above the hole to cause the speed you determined in part (b).

(d) Suppose that there is now less liquid in the container so that the height h is reduced to h/2. In relation

to the

collection beaker, where will the liquid hit the tabletop?

____ Left of the beaker ____ In the beaker ____ Right of the beaker Justify your answer.

Solution

Level 4: Applications of Bernoulli’s

Entrainment This is a weird and interesting use of Bernoulli’s Principle. Image we were to blow air through a

tube. According to the continuity equation, the flow rate of material entering the tube would have

to be equal to the flow rate of fluid exiting the tube. If we were to make the tube so it had a

skinny middle like this:

…we can see that the pressure is lower in the middle area. Why? The continuity equation tells us

that the fluid has to flow faster in areas where the area decreases. Because the fluid is moving

faster in the middle, Bernoulli’s Principle tells us that the pressure decreases here.

How is this useful? Ever used a perfume bottle or a mister? Here’s the idea behind how they

work. (Note: take this with a grain of salt. I noticed that some teachers teach this, and others say it

is a misconception. I’m still sorting out what the truth is. This has appeared on the AP test though,

so…good enough).

Here’s the idea: You squeeze the bulb of the atomizer, and it forces air down a tube. The tube

narrows, creating an area of low pressure. Because the pressure is lower here (and higher at the

surface of the perfume at the bottom) the fluid is forced upward, where it is then sprayed

outward.

Flying Here’s another fun one. Imagine you had a tube of air. You push air into the tube, and the

continuity equation tells you that air needs to leave out the other end of the tube. Great. Now,

let’s stick this strange shape into the tube.

As the picture shows, the air that goes along the path needs to go slightly farther than the air that

just gets to travel along the bottom. They both need to meet up at about the same time at the end

to support the continuity equation. That means the air on the top needs to travel much faster.

Faster moving air means a decrease in pressure. If the pressure on the bottom of the wing is

higher than the pressure on the top, we get an increase in pressure and the wing goes up. We

have flight. Ta-da. It only took us a few thousand years to figure that out. And you learned it in

one paragraph. Science.

Video Animation: How planes fly. Super short, but it does a good job.

Video Discussion: That said, this explanation above is a bit of an oversimplification.

Here is a more accurate (if more confusing) description. Just so you know there is

more to this story.

Career Option (Supplemental): Not sure what you want to do in life yet? Think this

stuff is interesting? Want to make money doing it? This video does a great job showing

you a career path you may want to consider. This video is so useful I almost made it

required.

Tornado Safety Short safety note here. You may have heard that in a tornado, the safest place you can go is an

underpass. This is completely wrong. It is actually one of the most dangerous things you can do.

It’s true that the underpass will stop some of the debris from above falling on you, and

underpasses are actually pretty strong and unlikely to collapse. Think about what the problem

is….

An underpass creates a constricted area. The air flowing through it must therefore speed up. You

can experience massive wind speeds in an underpass. Instead, if you can, enter any sturdy

building you can find. Do not worry about opening doors and window. It is true that the high

speed of the wind outside creates low pressure. That means the pressure in your house is

higher- which could (technically) make your house explode in an attempt to equalize the

pressure differences. However, running around opening windows wastes precious time and

could, inadvertently create pathways of enormous airspeed through your house (due to the

continuity equation). You may end up with massive quantities of high speed debris moving

through your house. If you can’t get to a building, get away from any trees and debris. Park

somewhere safe, buckle up and cover your body with as much padding as possible (to offer

debris a means of reducing its inertia without going through your body)