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Unformatted text preview: Lecture 1 John McRaven Physics. It is AWESOME 7B Topics Fluid Transport Circuits Linear Transport Exponential Change Vectors Linear Momentum Angular Momentum Newton’s Laws Motion Simple Harmonic Motion 7B Topics Vectors http://sdsu
physics.us/sdsu_physics/sdphys_images/ vectors_large.jpg Syllabus It is online http://smartsite.ucdavis.edu Buy clickers. I will start using them next class. I hope they help prevent sleeping The Goal of my Lectures I expect you to read the course notes. There are many relevant details that I do not have time to cover in Lecture I have two goals for my lectures Illustrate the concepts behind the models we use Teach you how to use the models, with an emphasis on the types of things you will see in DL and on quizzes I have taught the 7B DL about 20 times, so I know what you do in there, and quizzes will be written with DLs in mind. Practicing and understanding concepts in DL will be essential to your success in this class. Steady
State Energy Density Model We will attempt to understand multiple phenomenon with one basic model We will start by understanding ﬂuid ﬂow (today’s lecture) in depth, then seeing how the same concepts apply to (In future lectures): Flow of Electricity Flow of Heat Diﬀusion Fluid Flow: What I am going to teach you Energy Density Equation ΔP + ΔPE /V + ΔKE /V = E Pump /V − E Th /V
€ Each of the terms on the left is an energy system, so we now have 3 energy systems Each of the terms on the right is a way of adding or subtracting energy from the system Matter Conservation I = I
1
2
€ I = Area x Velocity All the ﬂuid that goes in has to go out Fluid Flow: What I am going to teach you Example Pipe: Height goes up, so PE goes up Area goes down, so velocity goes up, so KE goes up To understand this, you must understand matter conservation Since no energy is entering or exiting the system through a pump or resistance, pressure must go down for energy to be conserved KE and PE went up, so P went down. To understand this, you must understand energy conservation with multiple energy systems I do not expect you to understand this right now, but it is what I hope you understand by the end of the lecture f i ﬂow Why we care? Circulatory System House Pipes Later lectures will draw analogies to ﬂuid ﬂow with other things (circuits, heat ﬂow, diﬀusion) Steady
State Energy Density Model: The Most Important Concepts in Fluid Flow Bernoulli’s Equation Energy Density Conservation Matter Conservation Pressure A Typical Fluid Circuit: Important ProperJes The Pipe is entirely ﬁlled with ﬂuid Intuition is diﬃcult to use in this section, because most of the time when you SEE water ﬂowing, it is not in a fully ﬁlled pipe The ﬂuid is incompressible This is important because it means that when the pipe gets smaller, the ﬂuid does not get denser Fluid has a constant density ρ (“rho”) Matter Conservation All the water going in has to come out, so I1 = I2 I1 ALWAYS EQUALS I2 (unless the pipe splits) We usually consider the ﬂow of liquids, but we sometimes consider the ﬂow of gasses as well MaKer ConservaJon Current ﬂowing through a pipe is labeled I = volume/sec I1: Volume of water ﬂowing in per second I2: Volume of water ﬂowing out per second Must be the same or you are gaining/
losing water I = A * v A = area v = velocity of ﬂuid Units: m2 * m/s = m3 / s I1=I2 A1v1 = A2v2 v1 I1 If the area goes down, the velocity goes up, and vice versa v2 A2 A1 If A1>A2 then v1<v2 I2 SpliMng the Pipe If one pipe splits into two I1=I2+I3 Slightly diﬀerent, but still saying the same thing: “All the ﬂuid that goes in has to go out” We won’t see this much in ﬂuid circuits, but it will be a very important concept in electrical circuits I2 I1 I3 Highway Traﬃc Analogy 6 lanes converge into 3 lanes Assume bumper to bumper traﬃc (but the traﬃc is still moving) This is the analogy of assuming the pipe is entirely ﬁlled If there were only one car on the highway, the car could do whatever it wanted, and we couldn’t say anything interesting about it Assume the width of one car is about the width of one lane, and the width of the car cannot get any smaller. Also, the length of each car is the same and does not change I1 This is the analogy to saying the ﬂuid is incompressible. Traﬃc would not build up if we could just make our cars smaller and ﬁt 6 cars into 3 lanes side by side I2 Highway Traﬃc Analogy “Current” in this case is cars per second entering and leaving this part of the highway You could express current as I= (width of the highway in lanes) * (speed of the cars) / (Length of a car) So, if cars are leaving the highway (3 lanes) at 20 m/s, and each car is 2m long, then the current is: I = W * v / L = 3*20 / 2 = 30 cars per second (leaving the highway) Don’t write this equation down. I just made it up and it is only applicable to this example Since the cars are packed as tight as they can be (ﬂuid is incompressible and ﬁlls the pipe), cars must enter the highway at the same rate they leave the highway (conservation of matter) So the speed at which cars enter the highway (6 lanes) is: v = I * L / W = 30 * 2 / 6 = 10 meters per second So, the cars go twice as fast through the part of the highway that is half as wide Don’t write down any equations from this slide, they are made up for the sake of an example. Be careful of your intuiJon in ﬂuid ﬂow In the example we just saw, cars traveled faster on the narrower section of the highway This would tend to go against your intuition, which will tell you that cars travel faster on wider highways Cars travel faster on wider highways in reality because they reduce the chance of high traﬃc conditions occurring. The high traﬃc condition was an important assumption in our analogy because it mirrors the ﬂuid ﬂow conditions If you have ever been in high traﬃc while lanes are merging, you should know that you DO in fact travel faster AFTER the lanes have completed merging and you are in the more narrow part of the highway The point here is that it will be diﬃcult to intuitively understand the answer to many ﬂuid ﬂow problems. Use the Energy Conservation and Matter Conservation models to solve problems rather than trying to guess at the solution Benoulli’s Energy
Density EquaJon ΔP + ΔPE /V + ΔKE /V = E Pump /V − E Th /V
3 Energy Systems € Pressure, Potential Energy, Kinetic Energy V stands for volume and will be explained later EPump adds energy to the system ETh removes energy from the system Energy ConservaJon with 2 energy systems: A simple example Start with a simple example involving only 2 energy systems A car rolls down a hill ΔPE + ΔKE = 0
+ Since the potential energy decreased, the Kinetic Energy must increase and the car will be going faster This is ONLY true because we had ONLY two energy systems Don’t write down any equations from this slide, they are made up for the sake of an example. Energy ConservaJon with 3 energy systems € € Add a 3rd energy system, Egas Look at a car driving downhill Δ
E Gas + ΔPE + ΔKE = 0
++ Car was going faster at the bottom of the hill because it gained energy from PE and Egas Look at a car driving uphill Δ
E Gas + ΔPE + ΔKE = 0
+ ? We can’t say if KE went up or down unless we know the relative value of the changes in PE and Egas Depending on how much gas you used, your ﬁnal speed could be faster, slower, or the same speed compared to your initial speed Don’t write down any equations from this slide, they are made up for the sake of an example. Driving the point home When you do ﬂuid ﬂow problems, you have multiple energy systems You must look at what information you have on each one before you solve the problem We will look at how to determine this in a few slides Using Bernoulli’s EquaJon: IniJal and Final Points Initial and Final points are in position not in time The car example we just did looked at the car at some initial time vs the car at some ﬁnal time The highway traﬃc example looked at the highway at some initial position vs some ﬁnal position In a ﬂuid circuit, we compare two points at the same time but diﬀerent positions Bernoulli’s equation considers ﬂuids in a steady state, so they do not change over time The two points must be in the same ﬂuid system They are considered to be in the same ﬂuid system if:
Fluid can ﬂow directly from one point to the other within the same ﬂuid Or, the ﬂuid is stationary and the two points are in the same ﬂuid with nothing between them They are in the same pipe Energy Density Why do I keep saying “Energy Density” instead of just “Energy”? We can’t talk about what one water atom is doing Did enough of that in 7A We look at what the average atom is doing So, at each “point”, we are considering some really small volume We consider some mass of ﬂuid (m) in some small volume (V) All energies are now Energy per Volume, “Energy Density” Replace m by m/V = ρ, the density of the ﬂuid All terms are J/m3 Bernoulli’s Equation becomes: 1
ΔP + ρgΔh + ρΔv 2 = E Pump − IR
2 A Typical Circuit: PE ΔPE V = ρgΔh You can look directly at your circuit diagram to see if the height changes Only the diﬀerence between the initial and ﬁnal heights matter Any height variation in between the initial and ﬁnal point does not matter In this example, height clearly increases, so PE clearly increase f Δh i ﬂow A Typical Circuit: KE ΔKE 1
2
= ρΔv
V
2 Δv 2 = v 2 − v i2
f
You cannot see velocity changes directly, but you can € tell how the velocity changes because the cross sectional area of the pipe changes i I1=I2 A1v1 = A2v2 If the area goes down, the velocity goes up, and vice versa Only diﬀerences in the initial and ﬁnal area matter, not what happens in between So, in this example, KE goes up because velocity went up, which happened because area went down This is the ONLY reason v1 and v2 will ever be diﬀerent Pumps/Resistance Add/Subtract Energy, but they do NOT make v1 and v2 diﬀerent Seriously f ﬂow Pressure P = F / A This does NOT imply that if the area of a pipe gets bigger, the pressure gets smaller, because you do not know how the force changes This deﬁnition is not used much in ﬂuid circuits Standard units are Pa (Pascals) 1atm = 105 Pa Absolute Pressure vs ΔP 105 Pa is the absolute pressure in air. The term in Bernoulli’s Equation (ΔP) requires two pressures to get a diﬀerence in pressure. A Typical Circuit: Pressure Pressure cannot be determined directly from a ﬂuid diagram
Many problems ask you to solve for pressure. You must then determine it from Bernoulli’s Equation Sometimes you are given information about pressure If you see “open to atmosphere,” this means at a certain point the ﬂuid is in contact with air outside the pipe, and so its pressure is 1atm = 105 Pa In this example, Pressure goes down because both PE and KE went up 1
ΔP + ρgΔh + ρΔv 2 = E Pump − IR
2
+ + 0 0 f i € ﬂow Standing Pipes 1
ΔP + ρgΔh + ρΔv 2 = E Pump − IR
2 +
0 = 0 0 € P1 = 1atm = 105 Pa ΔP=
ρgΔh =
ρg(y2
y1) (P2
P1)=ρgd P2=P1+ρgd Pressure is higher at the bottom Pressure at an Interface At the interface between two ﬂuid systems, the pressure will be the same on each side of the interface PA=PB<PC=PD A Typical Circuit: Pump The Pump Adds Energy to the System Epump is always positive if you are going in the direction of ﬂow Epump is zero if the pump is not between your two points With A as the initial point, and B or C as the ﬁnal point, Epump is positive With B as the initial point, and C as the ﬁnal point, Epump is zero I A B C A Typical Circuit: Resistance
“Dissipation” Refers to the loss of energy in the system to thermal energy “Resistance” is a property of the pipe (this is R in the “
IR” term) The I in “
IR” is the current. I = I1=I2 Resistance takes energy out of the system A problem will typically say something about dissipation or resistance A statement like “Assume no dissipation” means R=0 There is no value of Resistance at a point, there is only resistance between two points It would be incorrect to say something like “Point A has a higher resistance than point B” It would be correct to say there is some amount of resistance “between points A and B”
IR will always be negative if you are going in the direction of ﬂow, and will be zero if there is no resistance Even if there is resistance, there will be no dissipation if the ﬂuid is not moving (I=0) Resistance per unit length is usually constant if the pipe is the same, so in this example: RAC=2 RAB=2 RBC A B C Comparing two pipes Bernoulli’s Equation can be used between 1 and 2, or 3 and 4, but NOT between, for example, 1 and 3, since they are separate ﬂuid systems So, I1=I2 and I3=I4 but I1≠I3 If ΔP12=ΔP34 then I1 = 2 I3, and v1=v2=2v3=2v4 So resistance would slow water down, but it would slow it coming in as well as going out I1 I3 1 A=2, R= 4 2 3 A=2, R= 8 4 I2 I4 The general method for solving a ﬂuid problem – Part 1 Pick two points Picking the correct two points is often the key to solving diﬃcult problems Does height change? If yes, then PE changes Is the ﬂuid moving? If not, then there is no KE or dissipation. Does area change? If yes, and the ﬂuid is moving, then KE changes The general method for solving a ﬂuid problem – Part 2 Is there resistance/dissipation? If yes, and the ﬂuid is moving, add a –IR term. Is there a pump? If yes, add an Epump term. Were you given information about pressure at each point or the change in pressure between two points? If not, or if you were only given information for one point, you can probably now solve for pressure If yes, use it Write down all the terms you know, and solve for the ones you don’t, ie: 1
ΔP + ρgΔh + ρΔv 2 = E Pump − IR
2
0 = + 0 Solve for ΔP + € Further Examples KE goes up, so P goes down P2 < P1 v1 I1 v2 I2 A2 A1 If A1>A2 then v1<v2 1
ΔP + ρgΔh + ρΔv 2 = E Pump − IR
2 0 + = 0 0 Solve for ΔP
€ Further Examples Add a Pump Can’t solve for ΔP unless we know the relative size of ΔKE and EPump v1 I1 v2 I2 A2 A1 If A1>A2 then v1<v2 1
ΔP + ρgΔh + ρΔv 2 = E Pump − IR
2 0 + = + 0 Solve for ΔP ? € Further Examples Add Dissipation KE goes up, but energy is being removed from the system through dissipation, so P must decrease more than KE increases v1 I1 v2 I2 A2 A1 If A1>A2 then v1<v2 1
ΔP + ρgΔh + ρΔv 2 = E Pump − IR
2 0 + = 0
Solve for ΔP
€ Summary When describing ﬂuid ﬂow, we use Bernoulli’s equation This is an Energy Conservation Equation Matter Conservation Determines if the ﬂuid changes speed Many of the concepts will be used to understand other topics in future lectures ...
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 Spring '10
 McRaven
 Momentum

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