>> One last time, Chapter 5. OK. We’re coveringthree different basic laws of fluid mechanics in Chapter 5. Using the Reynold’s transporttheorem, taking the basic laws written for systems and converting those laws using engineeringbased on the control volume and control surface approach. All right, we talked about continuity.We talked about momentum. So today we start energy. We’ll kind of derive the correct equationsto use with energy. And then maybe later today, a brochure on Monday will include exampleson energy. The last two homeworks that you have go through the end of Chapter 5 whichis the energy problems. OK. Obviously, a very important equation, we got a little bit ofthat back in Chapter 3 in Bernoulli’s [inaudible] was something about energy in fluids. Butthis is the next step in that. We start off again– Let’s just get a show of hands. Iknow some of you had, some haven’t. How many of you in here have had ME301? Show of hands.Let me just do a quick count. Keep up. So I have 2, 4, 6, 7, 8, 9, 10, 11, 12, 13, 14,15. All right, roughly 17 out 34 people.That’s about a half, OK. I understand that. You reallyget a good dose of that in ME301. You get a really good derivation. So, for some people,it’s going to be totally new in this class and for some people it will be a review onsome of that. We’re not going to go through as much in detail of this derivation as wedo in thermo that we see half the class have been through this thing all the time. So,we’re going to do maybe what you call a more casual derivation of this, not mathematicallyrigorous, OK? Now, we start out with the energy equation written in system terms. This iswhat you get in physics in ME301, OK? This is the equation written in terms of a systemwhere the rate of change of the energy in the system is equal to– we’re going to shiftgears temporary for about two class periods.This q, q dot now stands for heat transfer.No longer by the [inaudible] flow rate. So, it’s q [inaudible] flow rate. So you got tokind of shift gears. W is work. So q is heat transfer, w is work. The dot above them meanswith respect to time. W is work. Work is Newton meters divided by time, seconds. Newton metersper second, what’s a Newton meter, a joule? What’s a joule per second, a watt? OK. It’swatts. If it’s a British computation of the work, pound foot of work, pound feet of workdivided by time, seconds. Foot pounds per second. That’s the power. You convert it intohorsepower divided by 550, divided by 550.You did in horsepower if you’re in Britishcomputation. If you’re in SI which we said joules per second, what’s that? Watts. OK.So, those are power terms. W all by itself stands for work. W dot is the rate at whichwork is done and that’s power. OK, that– the left hand side looks something– thatshould be done. It looks something like the left hand side of our previous Reynolds transporttheorem.OK. So, here they are right here. So, we’re going to use Reynolds transporttheorem now. So, this right hand side here equals this right hand side here. Since theleft hand sides are the same, system– e stands for energy, e stands for energy. What’s thatenergy? Joules, OK, foot pounds, OK. So now we equate the left hand– the right hand sideof the top equation with the right hand side of the bottom equation. OK, we do that andwe end up with this equation. OK. So we have left hand side. So again, the system equationsays the time rate of change of energy of a system is equal to how much heat transferoccurs, and net in, net in means in minus out. The net heat transfer in plus the network out. OK. Oops, that should be in. Sorry about that. We’re going to define that aspositive, some which call it negative, we’ll call it positive. OK. Now, remember what wedid when we had– when we had continuity.This was written originally, Reynolds transporttheorem with a capital v, capital v was here, capital v was here, capital v was here, andwe said let v equal mass. Little v was big v divided by mass. So when we did continuity,that became 1. Mass divided by mass with 1. When we had momentum that was mass times velocity,big v. Divide that by mass. That was just the velocity. When we have energy, capitale is the energy. If we take and divide e by m, we get little e. So that’s energy per mass.So little e is the energy per mass, but we know it is. OK. And what’s rho dv? That’smass. That’s the mass. Energy per mass times the mass give me what energy with respectto time gives me what? The rate of change of energy in the system with respect to timedown here.And over here, it’s control volume. OK. Yeah. Let’s go ahead and look at whatkind– In the third one, they talk about work for about a week and describe different kindsof work. I’ll just give you a little flavor to what kind of work we’re talking about here.One thing, work could be a rotating shaft. How do you drive a pump with a motor? Howare they connected by a shaft? What’s the shaft do? It rotates. It powers the pump topump for instant water. So one way work can cross the boundaries or the control service,is by a rotating shaft crossing the boundaries.A turbine or a pump, those are examples ofthat. There’s other kinds of work– pressure work. It requires work to push mass into thecontrol volume through the control service. It requires a work. The pressure work pushesthe fluid let’s say into the pump that’s going to pump. There might be friction work. Wehave friction. Let’s say on a conveyer belt with the fluid, there’s friction between thebelt and the fluid. So friction work, those are all examples of what this term here mightbe. OK. Now, what kind of energy can there be? We’ll consider several. We’ll start offwith what’s called internal energy. We’ll add p over rho, energy and the pressure plusv squared over 2 and gz. Kinetic energy, potential energy, energy due to pressure, and this guyu, little u is called internal energy.Pretty much it’s everything you can’t see. If youheat a cup of coffee up by 30 degrees Fahrenheit, it doesn’t start to go red, [inaudible], OK?But do you think it has more energy? Oh yeah. Why? Because the molecules are moving fasterin there. Oh yeah. It’s invisible, it’s microscopic. I can’t see it. Now, if I throw a baseball,oh, you better believe, I’ll see this guy right here, kinetic energy.I drop a brickfrom the building, you better believe, I’ll see this guy here, potential energy, but Idon’t see the internal energy. That’s why it’s called that, it happens internally. OK.So, that’s what this guy in this equation is. We’re going to sum that in now where wesee little e. So, using this gives the following ddt. And now we sub in little e. And now,we do– it says up here. That’s e and we have rho and we have our v.n, dA. And then thathas to be equal to– way out here– Q net in plus W dot net in.>> Professor? >> Yeah. >> Is there a g net under the v squared overit? >> Let’s see, under the– Let’s check. Let’scheck and see. What’s u into v squared? What’s u into v? What’s u into d? What’s u into z?No, they’re the same. That’s why you do it. I’ll just check it that way. OK. Now, in Chapter5 we pretty much tell– yeah, I’m sorry. >> Is there another g? >> Yeah. Oh, there’s another G here? Oh yeah,I got to two Gs.Thank you very much. Bernoulli came back in my mind. Chapter 5 says in herbook pretty much the first introductory through its course, we normally are going to considerthis guy to be 0. So normally we consider steady state except sometimes the continuitywe consider increase in mass by water going in and coming out. But generally in Chapter5, we look at steady state. So we’re going to assume make matters simple. I’ll starthere. So, assume steady state. We’re also going to assume incompressible. We’re alsogoing to assume inviscid are frictionless. So we’re not going to worry about this frictionalwork right now. You’re going to worry about the work done by the pressure forces and thenthe work done by rotating shafts. OK, so we have all that in here. And then, so doingthat, we get– now, this is guy is going to go out, OK? And we’re also going to assumeone more, or right now we’re going to assume no network in.We’ll come back later to putthat in there, but right now assume it’s going to 0. So we end up with this, m dot– now,the m dot comes from the rho av. This is m dot right here. Rho av, m dot, OK. So we pullthat outside m dot. Rearrange the terms. Rearrange it one more time. We just separate the termswith some on the left hand side equation, some on the right hand side of the equation.So I’m going to write here. We have q net in, q dot net in divided by m dot. So, that’show we got that. >> Professor. >> Hmm? >> Quick question. >> Yeah. >> If I could use the [inaudible]? >> I’m sorry, which one? This one? >> Over there. >> That one? Oh, let’s see that.Is therea gz in here? I’ve got my Us, my Ps, my Vs. I got my [inaudible]. The elevation can’tchange, thank you so much. So, we got plus gz out minus z in and that whole rho is equalto q dot net. OK. Thanks. I think you’ll work again here. OK, we’re getting there, we’regetting there. Now, we say– you know what, that thing looks a little bit like the endof Chapter 3, something called Bernoulli equation. It looks like it almost. There’re some differences.I’ll tell you where it is here. The p2 over rho plus v2 squared divided by 2 plus gz2equal p1 divided by rho plus v1 squared divided by 2 plus gz1. Now, in Bernoulli’s, that wasit, stop, that’s Bernoulli right there. When we derive the energy equation from the systemthrough the Reynolds transport there, we get this equation.This term suddenly appears,OK? Well, this is then similar to Bernoulli, OK? So, I’m going to put that over here, Ithink. And remember, Bernoulli was pretty far steady, incompressible, and inviscid.So, neglecting viscosity, neglecting friction due to viscosity. They can call the inviscidfrictionless. Of course, we derived this thing already for incompressible because we saidrho is a constant, rho is a constant.So we’ve already said incompressible. We’ve alreadysaid steady, that term went to 0. Oh, what’s left over here? Oh, the third one, frictionless.You know what, that term there must be somehow tied into friction. And if we have frictionin something, like a turbine or a pump, what happens? Friction eats up the available energy.That’s why pump or turbine is not 100% efficient. The friction, what’s it do? Well, it heatsthe pump case in, it heats the water inside the pump, is that useful energy? No, I don’twant that.The bearings on your– wheel of your car, the engine drives that, heat upwith friction, what does that do? That takes power away from your engine to drive the car.Do you want that? No. So what would we call this thing here again? We call that the lossof energy. That’s the loss of energy due to friction. OK. So this guy here is called aloss of energy. So comparing to Bernoulli is this u out minus u in minus q net in isequal to 0, OK, compared to Bernoulli. If there is friction– it’s the difference now.If there is friction, that is not 0. U out minus u in minus q net in is greater than0. And that’s tied into the second law of thermo. When we derive this thing we saidthis is called the first law of thermo or the energy equation. We get to this pointhere, we say, “You know what, this term always has to be greater than 0.” Loss is due tofriction always use up energy, use up available energy.The loss always has to be a positiveterm. This thing here always has to be greater than 0. A loss to the w, friction here inyour car gas doesn’t make the car go faster. Never, never, never, never. Why not? Well,I’ll tell you, right here. Second law of thermo. That’s why we know greater than 0. OK. So,we can go ahead and say that this is the energy that leaves, this is the energy that comesin, and this is the energy that’s lost due to frictional effects. So we rewrite thatguy over there. I’ll put down again at the bottom here. So we have p out over rho plusv out squared over 2 plus gz out. Now, everything with 1 are in. This is in. We’ll change in1 and 2 in just a minute.We get tired of running in and out of these things. OK, andthen we have losses. So, this thing right here is called the losses. Right there. That’sthe losses. And then once we do that, we say, “You know what, let’s bring back in the workterm now.” OK, so we bring back in the work term. I’ll just give you a [inaudible] here.This pressure came in because of the work done by pressure forces on the boundary ofthe other system. So where did that work come from? We said there’s three kinds of work,shaft work, friction work, and pressure work. Where is the pressure work in here? Rightthere and right there. Where’s the shaft work? Right there. Where’s the friction work? Therewould be no frictional work on the boundaries, on the boundaries. Friction inside, yeah,on the boundaries, no. >> Should it be plus losses or minus losses? >> This is minus losses. And I’ll tell youwhy I have a plus there. I’d rather write this thing the other way around. You knowhow it goes.What comes in this [inaudible] in a second. So what comes in– Now, I’llbe back in a minute. But anyway, when I write it down, that thing is always a plus sign.I’ll show you why in a minute. OK. So let’s go back and include shaft work. So now wehave a p out, rho, v out squared over 2, gz out and now we have our in terms. And that’sequal to w shaft net in minus the losses. That’s an equal sign there. Let’s get ridof all of these. This guy up here is equal, this guy is plus right here. There we go.Well, that’s what we end up with. Sometimes we call it mechanical energy equation. Sometimeswe call the extended Bernoulli equation Because it includes now losses and work that’s whyit’s called extended Bernoulli’s. And now, once we’ve got that basic equation, now westart to look at different forms of it. OK. So, let’s take a look at the first form.Bythe way, just so you know, where you see a little w, there’s a big w, there’s a big q,and there’s a little q. They’re a different of course. They’re different. You see whatyou do to get little q? You take q dot divided by m dot. To get little w, make sure you,you know, use your lower case w and capital w. Little w is equal to w dot net in dividedby m dot. You divide it by m dot to get the little, lower case w and the lower case q,OK? Now we divide that equation there by g. This is this. Divide that guy by G. Hs. Wecall that w shaft net in. And this divided by g. Hl is losses divided by g. Well, yougot to watch the units. You got to really watch the units, OK? Everything in this equationhere, that guy is in meters. That means every term in there better be in meters. That meanshs and hl are in meters. You finish that. This guy up here is power, Newton meters persecond or joules per second.That’s power. Newton meters per second divided by kilogramsper second. Newton meters per kilogram. What’s Newton meters? What’s that thing there overthere? Work per mass, work per mass. What’s this guy here? Little w, OK, little w here,same units. Newton meters per kilogram divided by g meters per second squared. OK. Take thatlittle w, little q, they’re the same thing. Little w and little q, same thing, OK, dividedby g, meters per second squared, what do I get? Cancel, cancel, Newton seconds squaredper kilogram, the same. What is f? That’s one of our engineers, OK? That’s why we canthink of big box, OK? You say, “What is a Newton?” Let me go back to basics.What isa Newton? A Newton is a kilogram meter per second squared. Multiplied by what? Secondsquared. Divided by what? Kilogram. And what do you get? Guess what you get, of course,but nobody’s going to tell you that. You’re expected to know that. That’s why we drillyou with units all the time. Units, units, units. I’m telling you right now, this isthe easy one. What is a kilogram? A thousand grams. What’s a kilometer? A thousands meters.What’s a centimeter? About 10 millimeters. What’s a liter? A thousand cubic centimeters.Everything in SI structure of 1 times 10 to some power, everything.Easy to memorize?Of course it is. I thought about it before. And how about [inaudible]? Barrel, 55 gallons.How many quarts in a galloon? How many pints in a quart? You go on and on. How many feetin a mile? How many square feet in acre? Oh my gosh, nothing’s structured to 1. Nothingstructured to 1. What do you want to work in? Of course, you want to work in SI. Badnews, not in this country.Some places, oh yeah, some places. It’s getting there, butreally, really slowly. A gallon of milk and a liter of soda, right? We’re getting therebut it’s slow. So anyway, what I’m telling you is when you’re doing homework or exam,you better pay attention to units because they will kill you. They will kill you. Andyou better know how to do stuff if I say “What in the world is a Newton second squared perkilogram?” “I must have made mistake somewhere.” Oh no, you didn’t.It’s perfectly right. Yeah,it’s important. Don’t play it. That’s the easy one. That’s the easy one. You know, tryto go to g sub z in English Engineering, use pounds, mass and pounds force. Oh wow. Nowlet’s [inaudible]. What do I divide by 32.2? What do I multiply by 32.2? You better knowthat or it can get embarrassing, I guarantee you that. It can get downright embarrassing.OK, back to here again. Here we are. Everything is meters now. Isn’t that nice? So what’sBernoulli? And that’s what a nice form of Bernoulli’s. OK. So, we end up with– let’sput down the w dot. If we got pump– now, this guy up here, mean shaft, OK? That’s thisguy right here, OK? Let’s do a pump first or a pump. Because pretty much in our class,in these energy equations, we probably will have a pump or turbine or at least the possibilityof that in a problem.So, for a pump, w dot pump is equal to gamma qhp. For our turbine,we have w dot t– t for turbine, p for pump– gamma qht. OK, try it again. I’ll do the SI.Gamma. Anytime you see gamma is Newtons. If you see rho, it’s kilograms. Newtons per cubicmeter, cubic meters per second. Ht is– all these hs are in meters.Oh, guess what we’vegot here. Newton meters per second. Guess what that is, joule per second. Guess whatthat is, a watt. One Newton meter equal 1 joule, 1 joule per second equal 1 watt. Oh,1, 1, 1 everywhere, a wonderful system. Now, [inaudible] English Engineering, probablybetter convert it to horsepower because this will [inaudible] horsepower. Most pumps arespecified in that, your power, maybe horsepower. OK. Anyway, these guys now are all going tobe in watts.OK. I’ll erase that right now. OK. And then, the efficiency of the pump,we’ll talk about pump efficiency. Pump efficiency, the power is the gamma qhp divided by thepower into the pump. It always takes more power in and comes out. This is the powerthat comes out, but– the power into the pump, yeah, right. So, if you’re driving a pump,the pump starts to get warm.The water starts to get warm. There are losses in their friction.So, turbine different. Well, the turbine– we have w dot turbine. That’s the out dividedby the gamma qh turbine. That’s what you put in. This is what comes out. It’s always lessthan 1. This is what comes out. That’s what you put in. It’s always less than 1. The efficiencyis always less than 1. OK. And now, I’m going to write the three– there’s different equationsyou can write into the energy equation. So, remember Bernoulli’s, we have three Bernoulli’sback in Chapter 3. I put on the board, three forms of the same equation, the Bernoulliequation. Here are three forms of the energy equation. OK. The first form. This is on theboard awhile ago. We’re going to use one now. Yeah, this is what I want to do. We’re nowgoing to use subscript 1 and 2, not in and out.Here’s the second form. And finallythe third form. OK. So the first equation, these terms should be Newton meters per second.That’s joules per second, that’s watts. Everything is in watts. Second one, we divide it by mass.We divide it by the m dot. Everything now is in Newton meters per kilogram which meanswork per kilogram of fluid flowing. Or if you try to think about it, divide both numeratorand denominator by time. Newton meters per second divided by kilograms per second. Thisthen becomes power per mass flow rate or you use this one where now everything is in metersor feet. This is called the head loss in feet, the turbine, the pump. If there’s no pump,hp goes to 0. If there’s no turbine, ht goes to 0. If there’s no losses, hl goes to 0.Yeah. >> Will those m dots always be the same value? >> In Chapter 5, what we’re doing, yeah. Itcan get– It doesn’t have to be, but normally what comes in goes out.Take a pump. Of course,what goes in goes out. A turbine, what goes in goes out, of course. Sometimes, no. Youcan have mixing chambers. Two things go in, they mix together, one thing goes out. Uhoh,changing now. How do you change it? What do we do? We went around the control surface,didn’t we? The control surface, there might be other terms in there, yeah. OK. So nowthe bottom line is [inaudible], the bottom line is you’re probably going to choose oneof these three to work most of the problems on energy in Chapter 5, OK? And what– Andwe’ll work different forms, we’ll probably mostly use the last one. Just like in Bernoulli,we mostly use the last one. It was kind of convenient and simple. Everything is in metersor feet. That makes life a little bit easier. When you can do it, then normally you do itthat way. OK. So, I’m going to do one problem today and then we’re going to do– next classmeeting, we’re going to do– that day going through a lot of those problems in energy,and maybe even start in Chapter 6.And by the end of next Monday’s lecture, that willbe the end of what you’re responsible for, for the next exam which is a week from nextMonday. I give you a week to study after the exam after we finish and say stop here. OK,that’s our plan. Anyway, let’s try to see your homework. You’ve got 102 and 105. I’mgoing work 103. You have 102 and 105 homework. I’m going to work 103. All right, there’sa picture in the textbook on this one. We don’t need this anymore. All right, here’san inclined pipe over here and here.There’s a manometer here. It’s a piece of metric tube.And it goes up. There’s a piece of metric tube here. It goes out, it goes up. It’s aclear tube. This is a review of Chapter 2. We use symmetric tubes to measure static pressure.If we can’t, sometimes you have to do it if you can.We’ve got several of those of withthe fluids hydraulics with, several of them fluids hydraulics with. OK. The left handone– let me see I got this thing right. And by the way, there’s again, ton of work examplesin your textbooks. So, take advantage of the textbook and look at those examples especiallywhen you’re studying for the midterm because it really help, it quite varies. OK. So goingfrom that left hand point to the right hand point, we have1.5 meters. So from here tohere is 1.5 meters.Now, let’s see where our manometer levels are. What’s the [inaudible]the fluid. It doesn’t say what the fluid is on that page. So then we got– there it is–1 meter and 3 meters. So, the 3 meters is up high. We got 1, also 3 meters, 1 meterand 3 meter. One meter here, 3 meters here, atmosphere. This is 1 meter. This is 3 meters.OK. Incompressible liquid. We don’t know what it is, water boil, whatever.Flow is alongthe pipe. Determine– Oh, let’s see. I don’t know if you need– 0.75 meter diameter. PartB, [inaudible] part B. It’s OK. Determine the direction of flow. So my question is,is it going down the pipe or is it going up the pipe? That’s what we have to find, thedirection of flow. OK. Let’s see which one [inaudible], but I’d say, normally with thisguy, so I’ll start with him.That’s fine. Now, you have to assume– can’t assume point1 somewhere. So, my assumption is the flow goes from point 1 to 2. And here’s point 1and point 2. So, I’m going to see if that’s right, am I right? OK. So, p1 over gamma plusv1 squared over 2g plus z1. No pump, hp is 0. No turbine, ht is 0. P2 over gamma plusv2 squared over 2g plus z2 plus losses. There were losses, yeah. A matter of fact he says,what’s the head loss over a 6meter pipe? So the distance from 1 to 2 is 6 feet.That’s6 meters apart. Yeah. So there’s losses, impulsive losses. Now, you know, we’re really doingthis for our control line. We derive this thing for our control line. So, if you want–if you want, you could draw the dash lines around your control volume and all the controlsurface. But it’s not nearly as critical here as it is on momentum. There, you got to drawthe control volume and control surface. Here, it’s like a modified Bernoulli equation. Soas we write it from point 1 to point 2, that’s why it’s called modified Bernoulli. But ifyou want to be official and do it by control volume or a control surface which you probablywould for a pump returning, then it’s still [inaudible] there’s a point 1 and point 2.OK. P1 over gamma minus p2 over gamma plus v1 squared over 2g minus v2 squared over 2gplus z1 minus z2 equal losses. First of all, v1 equal v2. OK, so why? Let’s see, momentumequation. No, of course not. Momentum is for forces. Continuity? Oh, of course it is. Steadystate? Yeah. Incompressible? Yeah. Continuity? M dot 1 equal m dot 2.What’s m dot? Rho 1q1equal rho 2q2, but the rhos are equal. Cancel out, q1 equal q2. What’s q? A1v1 equal a2v2.A1 though is equal to a2. Conclusion, v1 equal v2, got it. Z1 minus z2. Z1 is down here minusz2. OK. So, that’s 1.5. Let’s [inaudible]. OK. Let’s see– Oops. There, it’s right there.So, my z1 minus z2 minus 1.5. OK. P1 over gamma, p1 here over gamma. Now remember, howwe get p1 from Chapter 2. We did this in Chapter 2. OK. P1 is equal to gamma of the fluid timesthe height from where you are to the free surface.Three– OK, so that guy is 3– sotake 3 gamma divided by gamma and you get 3 minus p2. P2 is gamma times 1, 1 times gamma.One times gamma divided by gamma, 1 minus 1. Three minus 2.5. OK. The losses have tobe positive. It can never be negative. If you would go the other way, go and try it.Start up here and go down. Call this point up here 1. Call this point down here 2. Usethe same equation and you find out that this number is negative. So impossible, lossescan’t be negative. Friction never does create more energy. Friction dissipates energy. That’swhy [inaudible] thermo, it has to be held on. Yeah. >> Was p1 given to you or was that derived? >> No. It was from the picture. >> OK. >> P1 is gamma of the fluid in the pipe timesthe distance from the pipe centerline to the top of the fluid out there. How far is that?One. OK.Now, just so we know over here that’s why I like to write this equation this way.The one on the left side and two on the right, because this is the energy that comes in withthe fluid, the energy that comes in with the fluid. This is the energy added to the fluidfrom the pump. The pump adds energy to fluid. This is the energy that the fluid has. Thisis where the– this is where the energy goes. It goes out, the location too, it can drivea turbine to create power, and it has to overcome losses. So, this is the available energy firstthree terms. It’s used up, it’s used up. Some of it at least, some of it maybe of turbineand the rest of it losses by friction. So it’s easy to think that way. The losses areunavailable energy. You run the car it gets hot.It powers the car. You know, can younot use that heat from the engine getting warm and the fluid getting warm to drive yourcar? >> No. >> No. Gosh, no. It’s lost. It’s lost. Don’ttry and say, “Well, yeah. No, I could put that back in [inaudible] desk.” No, we can’t.It’s lost to what? The atmosphere and the air around the engine in the hood of the car,it’s lost. You can’t recover it. So that’s why this is called the available energy. Thisis the loss of energy. L stands for loss. OK. So, we’ll come back and we’ll work problemsnext time on that. But, I want to go over a couple other problems that I’ve had in officehour and also after class. We didn’t get it just by the windows over there. All right,let’s do problem 5. OK. Water flows through– out through a thin closely space [inaudible]shown with a speed around the entire circumference of the outlet. Determine the mass flow ratethrough the end of the pipe.Look like this. Fluid comes in here and it comes out here.I’ll look down and up above now. Looking down and up above. We have the picture given toyou in the textbook showing this. And the angle here that goes out is 60 degrees. That’sthe velocity. It gives that 10 feet per second. Go back to continuity equation in Chapter5 like [inaudible] board. OK. Rho av, rho av is m dot. So, if I want to know how muchmass flow rate comes out of this thing, I do that. Here’s my picture. My control volumeis inside here. Where is my area vector? Take the dot product. V times A times 60. Is–Continuity says steady state m dot in equal m dot out. You’re supposed to solve for mdot in. That’s what you do, OK? Oh by the way, remember now, the area vector alwayspoints outward from the control surface, the control volume, it does. What is the area?Well, I’ll make one.There’s the area, the fluid comes out. What is this area? Circumferencetimes w. So that’s the area, 2 pi r times w dimension, OK? So, just so you know howto properly do it efficiently the correct way, use the product. OK. See that one. Oh,and by the way, the second problem just so you know that’s this container with water coming in and there’s water in here and there’s gasolineup here. Gasoline, the water comes in, it pushes the gasoline out. I think they giveyou water in. They want to know how much– yeah, 100 liter per second, something likethat coming in. They want to know how much gasoline goes out. Well, you know, if I pusha galloon of water in here, guess what’s going to happen, a galloon of gasoline goes outhere. That’s the equation you need. That’s the only equation you need. M dot gasolineequal m dot water, and the water they gave you q. Rho times q, OK? Rho of water timesq, OK. And that’s the change of the weight of gasoline in the tank.That’s the weightof change of the gasoline– this is gasoline by the way, fuel water. It’s kind of– youknow, fuel water, fuel gasoline, OK. OK. So that’s the kind of an easier way out of thatproblem just to realize that if you push a galloon of water in the bottom, a galloonof gasoline is coming out the top. All right, now, let’s take the last few minutes and youget the problem for, well, for homework. And I had two or three people coming on officehour want to know. >> Professor? >> Yes. >> So, my question on my pipe flow is thatpipe conservation for the continuity for the steady state, m dot gasoline and m dot waterthat since the different densities you’ll have a larger– >> Yeah. Yeah, yeah, yeah. Yeah, yeah, yeah.This is a Q– pardon me. Thanks. Q water. Now, once you get q water– once you get qgasoline [inaudible].OK. Now, I’ve got q gas. Now, I put q gas down here multiply itby rho gas, now I get that gas now. >> OK. All right. >> Thanks. OK, work 12, it’s a thinking problem[inaudible]. I did one in class, go simple. I said, “I’ll do a simple one in class andlet the hard one– you try it.” All right, I’m going to do an intermediate one now. Hardin class but simple in my homework, very similar though. Here we go. One more time. Just–Again, this is problem 4.12 modified. All right, smoke plume lasts for five hours. Smokecomes out of a smokestack for five hours. I’m going to put several perhaps on the board.So, [inaudible] phase. All right, there are three different segments. First, for the time,t for three hours. The wind speed is in vector terms 10i minus 5j.Plot where the plume is, show the streakline of the plume at three hours.OK. The origin.And by the way, the plume in three hours is emitted at 0, 0. Here’s what’s emitted at,I don’t know, minus something and plus something. That’s OK. All right, here’s 0, 0. 5, 10,15, 20, 25, 30, OK, that’s fine. 10, 20, 30. Let’s see here, 5, 10, 15, 20, 5, 10, 15,20. All right, the first bit of smoke that comes out of a smokestack when it comes out,it has that velocity. It goes to the right and it goes down. OK, got it. After threehours, where is that first smoke that came out, the first smoke emitted after three hours?Velocity in the x direction, 10 miles per hour. This one, 5 miles per hour.After threehours, 3 times 10, 30 miles, OK? Yeah, yeah, 30 miles. OK. [Inaudible], that’s OK. No,that’s way too big. Sorry about that. I’ll be out of the window if I do that. Make this20, 40, 60, 10, 30, 50. Oh we got– OK, that’s fine. All right, after three hours, 10 times3, 30, 30 miles it went. Which way? East, 30 miles here. But also has southeast recordsof plume. Plume is moving southeast. It went down 5 miles per hour times 3. 15, 15 down.30 to the right, 15 down. There is the first bit of smoke that left a chimney at time equals0 after three hours. Where is the smoke that left the chimney at three hours? Right here,connect the dots. There’s the streakline of the plume after three hours. OK. I’ll erasethis. For time three to five hours. Now, the wind speed changes. Now, it’s a 15 miles anhour i, that’s x direction, plus 10 miles an hour in the j direction. Show the screen–show the streakline at t equal five hours.All right, 5, 10, 15, 20. OK. Let’s see. Ijust want to go up to maybe 8, 10, 20, 30, 40, 50, 60, 70, there goes 80. First of all,where did this line of smoke go? Well, at time equal three hours, it was here. Thenthe wind change for how long? From three hours to five hours. What’s the difference in time?Two hours. So, the smoke particle right here. In two hours, two hours times 15 miles anhour, it went 30 miles to east, 30 miles to east. He’s sitting on 30 right now. He wentto 30. He’s out there now at 60. How about the y direction? He went north 10 miles anhour times how many hours? Two hours, two hours because 3 to 5, two hours. Twenty, whichway? North, 20 north. Where did he start? Fifteen. Where is he now? Up here. Where isthat? That’s 5. And so right up here.Just so you know where it is. OK. So over here.And by the way, where did this guy here go? He started at 0. In two hours he went 30 up–30 to the right, 20 up, 30 to the right, 20 up, 30, 20. He’s right here. This guy wentup here. This guy went up over here. These are molecules of smoke. So here he is. Thatline if I had [inaudible] the same scope with my scope is correct. It moved up here. Thisline moved up here. But wait, there’s still smoke coming out after three hours. It goesfor five hours, OK? We’ve got five hours smoke [inaudible]. OK. Well, when the smoke cameout of stack at three hours and one second, three hours and one second, which way it go?Fifteen miles an hour times how long? For two hours. Thirty– two hours times 10? Twenty.It went to 30 and 20.Thirty and– Oh, he’s sitting right there now, right next to hisbuddy right there. That smoke left after three hours and one second. How about the one wholeft after three hours and 30 seconds? He’s here. Three hours? Three hours, 30. Threeand a half hours, there. Twenty and half hours, there. Five. The last guy comes out, lastguy comes out, he’s right here. OK, over here. OK.Now, it’s the same problem, right? So,60 up to 5, here. OK. He goes at 20 and 30. So here’s 30 and there’s 20. He’s up here.And this one up is right here. There’s the whole smoke in five hours, five hours. Smokewent out that way for a while, went southeast, then the wind shifted, it went up that way.And the smoke leaving here in 3 to 5, went straight out that way, northeast. North–that’s north, that’s east. OK, after 10. For 5 to 10 hours, v is equal to 5i, the windis dying down. It’s 5 miles an hour east now, 5 miles an hour east. Show streakline at 10hours. OK, here we go again. Five, 10, 15, 20. Five, 10, 15, 20. And over here. Let’ssee, 10, 20, 30, 40, 50, 60, 70, 80. OK. Take the one at 0. Just make that easy on ourselves.Take the one at 0. That’s the last smoke came out. This was the last smoke particle thatcame out of the chimney. That wind, five hours. For the next five hours, what he do? He driftedeast at 5 miles an hour times how long? Five hours.Five times 5, 25 miles. He went toeast 25 miles. He’s out here now at 85. So, there he is right there. That’s the point.That was right there. He moved down to here, OK? And then let’s see. OK. Up there. Let’ssee, the scale of light, I guess I’m out of proportion here. Yeah. I’m out of proportionhere. OK, because– oh, I’m sorry 25. I’m going to go 25, excuse me.He moved 25 andgone from 60. He moved 25. He’s out here. Let’s take this guy up here. Now, he’s at60. He’s going to go east 5 miles an hour times five hour, 25, 60, 70, 85. Eighty five.Did he go anywhere in the y direction? No, he didn’t. Where was he? Here at 5. Whereis he here? Five. In the end, [inaudible] down first. Left hand side, right hand side.Where did this guy go up here? He was at 20 and 30, 20 and 30. He moved due east 5 milesan hour times five hours, 25 miles from where he was.He was at 30 plus 25, 55. There heis. Once you got those three guys pinned down, now it’s simple. Connect the dots. There itis, the streakline after 10 hours. Oh, big long explanation, but people want to knowhow to do these things, so I thought I spent some time on it. That’s how you do them, OK?Yeah. >> So three separate wind velocity, you haveto make separate graph. >> You don’t have to, but life is simplerif you do.>> OK. >> You go get– I mean ME214 from staticsand try represent one sketch. I guarantee you will suffer sometimes by doing that. It’salways better to draw diagrams. That’s why I did this way. You try do [inaudible] likelet’s say this solution [inaudible], oh no you guys– confusion, mass confusion. Tryingto put these three guys on this over here, my gosh, you’re asking for confusion. Yousimplify something if you spread out and do a graph for each particular time interval.Yeah, at the time interval. Uhhuh. >> So you put in the initial just 5 into theequation and you get– >> I plug in the– nope, nope, nope. I pluggedin the differential time. How many hours between these two? >> Five. >> Times what? We got 25 miles.Every pointgoes 25 miles to the right. Every point on this line and this line go 25 miles to theright. I made it a little bit easier than the problem you did with the assignment problem.But, if you can understand this and do it this way, you’ll get that problem fine, Iguarantee you. OK. Good stopping point. So we’ll see you on Monday. Homework is due onMonday.
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