AP Physics 1 & 2 Continuity Equation for Fluids

Hi everyone and welcome back to Educator.com.0000

Today we are going to continue our study of fluids as we talk about the continuity equation.0003

Our objectives are going to be to apply the continuity equation to fluids and motion.0008

To explain the continuity equation in terms of conservation of mass flow rate.0012

Conservation of mass for fluid flow. When fluids move through a full pipe, the volume of fluid entering the pipe must be equal to the volume of fluid leaving the pipe.0017

The Law of Conservation of Mass for Fluids.0027

This holds true even if the diameter of the pipe changes.0030

In short, what we call the volume flow rate remains constant throughout the pipe.0034

And we will look through a couple of applications of that here.0055

Volume flow rate. The volume of fluid moving through the pipe can be quantified in terms of volume flow rate.0059

The volume flow rate is the area of the pipe times the velocity of the fluid, and it must be constant throughout the pipe.0065

So over here on the left-hand side, if we are looking at a pipe with a changing diameter, we have Area1, where the fluid has Velocity1.0072

Over here on the right-hand side, we have Area2 and Velocity2.0079

A1V1, the volume flow rate on the left-hand side, must be equal to A2V2, the volume flow rate on the right-hand side.0083

What that means practically is that you must have a higher velocity or a faster flow over here and a slower flow over here.0091

Let us look at some examples and applications.0104

Water runs through a water main of cross-sectional area 0.4 square meters with a velocity of 6 meters per second.0107

Calculate the velocity of the water in the pipe when the pipe tapers down to a cross-secitonal area of 0.3 meters squared.0114

Well, continuity equation for fluid says A1V1 must equal A2V2.0122

Therefore, Velocity2 at the skinnier section of the pipe must be equal to A1 over A2 times V1.0128

Or 0.4 divided by 0.3 square meters times that 6 meters per second or 8 meters per second.0129

It gets a little narrower, it gets a little faster.0150

Let us take a look at the garden hose example.0156

A lot of folks have probably done this before.0158

As you are watering the garden or playing with the hose, you want the water to come out a little bit faster so you cover up the end of the nozzle with your thumb a little bit.0160

You decrease that cross-sectional area so that the water has to come out faster to maintain that volume flow rate.0167

In this problem, the water enters a typical garden hose of diameter 1.6 centimeters with the velocity of 3 meters per second.0174

Calculate the exit velocity of water from the garden hose when a nozzle of diameter half a centimeter is attached to the end.0181

First let us figure out what the cross-sectional areas are.0188

When it is entering the pipe, A1 is πr1², or π times. If our diameter is 1.6 centimeters, our radius must be 0.8 centimeters.0192

So that is 0.008 square meters, or an area of about 2.01 times 10^-4 square meters.0204

Area 2 at the nozzle is πr2² or π times. Well its diameter is 0.5 centimeters so its radius is half of that, 0.25 centimeters, 0.0025 meters squared.0215

Which is 1.96 times 10^-5 square meters.0231

Now we can apply our continuity equation for fluids.0239

A1V1 equals A2V2. This implies then that V2 equals A1 over A2 times V1.0245

Or 2.01 times 10^-4 square meters over 1.96 times 10^-5 square meters.0259

All times V1 which was 3 meters per second, for a total of about 30.8 meters per second.0270

It comes out a lot faster when you decrease that area.0282

Let us take a look at an oil pipe line problem.0287

Oil flows through a pipe of radius (r) with speed (v).0291

Some distance down the pipe line, the pipe narrows to half its original radius.0294

What is the speed of the oil in the narrow region of the pipe?0299

Well, A1 we will call πR². A2 is going to be πR/2², which is going to be πR²/4 or π/4R².0303

Now as we apply the continuity equation for fluids, A1V1 = A2V2, which implies then that V2 = A1/A2(V1).0325

A1 = πR², A2 = π/4R² times V1 which we will just call V.0341

We are going to have some simplifications, R², R², π, π.0356

We have 1/, 1/4 times V, which is going to be equal to 4V.0361

That is 4 times faster.0370

One last problem here.0375

So we look at the roots of the continuity equation, which statement below best describes the continuity equation for fluids?0377

Energy is conserved in a closed system? Mass is conserved in a closed system? Linear momentum is conserved in a closed system?0384

Angular momentum is conserved in a closed system? Or charge is conserved in a closed system?0391

Well, we are really talking about a mass conservation here.0398

The volume flow rate is basically saying, the continuity equation is saying that the mass that goes in must come out.0404

Therefore, mass is conserved in a closed system.0409

Hopefully this gets you a good start on the continuity equation for fluids.0414

Thanks for watching and make it a great day.0418