Lecture 1 - Lecture 1 John McRaven Physics. It is...

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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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