WEBVTT physics/ap-physics-1-2/fullerton
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Hi everyone and welcome back to Educator.com.
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Today we are going to continue our study of fluids as we talk about the continuity equation.
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Our objectives are going to be to apply the continuity equation to fluids and motion.
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To explain the continuity equation in terms of conservation of mass flow rate.
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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.
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The Law of Conservation of Mass for Fluids.
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This holds true even if the diameter of the pipe changes.
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In short, what we call the volume flow rate remains constant throughout the pipe.
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And we will look through a couple of applications of that here.
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Volume flow rate. The volume of fluid moving through the pipe can be quantified in terms of volume flow rate.
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The volume flow rate is the area of the pipe times the velocity of the fluid, and it must be constant throughout the pipe.
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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.
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Over here on the right-hand side, we have Area2 and Velocity2.
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A1V1, the volume flow rate on the left-hand side, must be equal to A2V2, the volume flow rate on the right-hand side.
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What that means practically is that you must have a higher velocity or a faster flow over here and a slower flow over here.
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Let us look at some examples and applications.
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Water runs through a water main of cross-sectional area 0.4 square meters with a velocity of 6 meters per second.
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Calculate the velocity of the water in the pipe when the pipe tapers down to a cross-secitonal area of 0.3 meters squared.
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Well, continuity equation for fluid says A1V1 must equal A2V2.
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Therefore, Velocity2 at the skinnier section of the pipe must be equal to A1 over A2 times V1.
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Or 0.4 divided by 0.3 square meters times that 6 meters per second or 8 meters per second.
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It gets a little narrower, it gets a little faster.
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Let us take a look at the garden hose example.
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A lot of folks have probably done this before.
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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.
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You decrease that cross-sectional area so that the water has to come out faster to maintain that volume flow rate.
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In this problem, the water enters a typical garden hose of diameter 1.6 centimeters with the velocity of 3 meters per second.
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Calculate the exit velocity of water from the garden hose when a nozzle of diameter half a centimeter is attached to the end.
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First let us figure out what the cross-sectional areas are.
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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.
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So that is 0.008 square meters, or an area of about 2.01 times 10^-4 square meters.
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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.
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Which is 1.96 times 10^-5 square meters.
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Now we can apply our continuity equation for fluids.
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A1V1 equals A2V2. This implies then that V2 equals A1 over A2 times V1.
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Or 2.01 times 10^-4 square meters over 1.96 times 10^-5 square meters.
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All times V1 which was 3 meters per second, for a total of about 30.8 meters per second.
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It comes out a lot faster when you decrease that area.
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Let us take a look at an oil pipe line problem.
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Oil flows through a pipe of radius (r) with speed (v).
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Some distance down the pipe line, the pipe narrows to half its original radius.
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What is the speed of the oil in the narrow region of the pipe?
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Well, A1 we will call πR². A2 is going to be πR/2², which is going to be πR²/4 or π/4R².
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Now as we apply the continuity equation for fluids, A1V1 = A2V2, which implies then that V2 = A1/A2(V1).
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A1 = πR², A2 = π/4R² times V1 which we will just call V.
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We are going to have some simplifications, R², R², π, π.
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We have 1/, 1/4 times V, which is going to be equal to 4V.
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That is 4 times faster.
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One last problem here.
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So we look at the roots of the continuity equation, which statement below best describes the continuity equation for fluids?
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Energy is conserved in a closed system? Mass is conserved in a closed system? Linear momentum is conserved in a closed system?
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Angular momentum is conserved in a closed system? Or charge is conserved in a closed system?
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Well, we are really talking about a mass conservation here.
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The volume flow rate is basically saying, the continuity equation is saying that the mass that goes in must come out.
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Therefore, mass is conserved in a closed system.
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Hopefully this gets you a good start on the continuity equation for fluids.
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Thanks for watching and make it a great day.