>> Maybe instead of coupling here, maybe there's.a flange, a bolted flange with bolts going.through there. Same thing. What force must.the bolts have, the bolts in place? That's.a big question. What-- if this is a big steam.pipe on a power plant, nuclear power plant,.what is the force in the bolts to hold this.thing in place? [Inaudible] very important,.very important. Once you find the force divided.by the bolt area that gives the bolt stress.and now you pick a bolt out with the right.stress, factor of safety. All these things.are coupled together, you know, when you're.building things. OK. So angular, we'll come.back to that now, 536, 539, we'll do 550 next..But I want to finish up the last part of Chapter.6 first. OK. Rewind and we had a talk about.in Chapter 6, we divide the continuity equation,.we divide the differential form of a continuity.equation. OK. There's-- The only other equation.in Chapter 6 which goes through derivation.is going to lead us to-- it's really linear.momentum, linear momentum equation in differential.form. We start off with a differential system.of mass delta M. So, a differential system.of mass delta M. Now, here's the flow field.and we take a little differential system of.mass delta M and we watch it as it moves through.the flow field. Now, that was fruitful, applied.Newton's law to it. The summation of the force,.it was differential forces. Differential forces.on that little element of mass. Here is a.time rate of change, look at the big D, OK?.The big D, OK, material derivative of velocity.times delta M. What's this momentum? Momentum,.mass times velocity. The forces on the object,.the time rate of change of momentum that the.object has. OK. It can be considered to be.[inaudible] because if you replace-- pull.mass outside the material group, pull the.mass outside, why can you do that? Well, don't.forget, a system has a constant mass. So mass.is a constant, pull the mass outside the derivative..Then we have delta M times capital DvDt. Now.we've defined capital DvDt previously as acceleration..It looks like Newton's Law, F equal ma. That's.why I set up here Newton's Law. F equal ma..Where did it come from? A change in momentum.caused by the forces acting on the different.mass delta M. These forces on the left hand.side can be, what we call body forces, that.means it's distributed throughout the body.uniformly. Typically, the weight, the weight.is a body force. And these forces on the surface.of that little differential mass delta M..Surface forces can be pressured, and what's.pressure of normal stress. And if there's.friction, what are those guys called? Shearing.stresses. So the forces on the surface can.be either normal stress like pressure or forces.like shearing stresses that act on the sides.of this little object right here. So, those.guys there can be put together in the equation.and we know what the capital DMVT is we had.that DVD-- capital DvDt, we've have that previously..And so if we do that, so this gives rho gx.partial with respect to x, sigma xx, shear.stress partial tau yx with respect to y, shear.stress tau zx with respect to z plus rho..We had this before. This guy here is just.DvDt. That was in chapter, I think 4. What's.on the left hand side? The forces. What kind.of forces? And [inaudible] that gravity and.mass, rho is mass, that's gravity. That's.the weight. That's not the weight. This guy.right here, sigma is a compressive stress,.that's the pressure. This guy right here,.a shearing stress, on the top and bottom..This guy right here, a shearing stress, on.the front and back sides, all those subscripts.mean something. You know that if you took.MA 218, strict materials, you know what those.subscripts mean. The first one means the shearing.stress acts on a surface perpendicular to.the y direction and in acts in a direction.in the plot in the x direction. So one subscript.tells you what surface it acts on, the other.one tells you what direction it acts in. That's.the double subscript. The normal stress sigma,.x in a plane perpendicular to the x direction,.that's the x direction, my hand is the plane..Second subscript, x in the x direction. Yeah,.there it is. First subscript, x in a plane.normal to the x direction, my hand. Second.subscript, it acts in a direction, x direction,.that way. That tells us how the stresses act.on a surface. So and then, I'm back to remind.them all up, there is one for y and there's.one for z. Three equations. One, the x direction..One, the y direction. One the z direction..They are equation 6.15 in textbook. OK. These.are called general differential-- -- equations.of motion. There's three. One x direction,.one y, one z. OK. Let's see. Anyone saw that.right now? I don't think so. OK. Now, we're.going to make some assumptions. Number one.assumption. If the flow is inviscid, don't.forget the word inviscid means frictionless--.-- .or nonviscous. They are all the same thing.in our nomenclature. If there's no friction,.all these shear stress guys here go, everything.with the shear stress goes. So we're left.within-- -- where you replace the sigma xx.which is a normal stress and what's pressure?.Pressure is a compressive normal stress. What's.compressive mean? It's a negative sign on.sigma. A compressive stress is a negative.sign, so the stress on the x phase is compressive,.the minus means that's equal to pressure..So in a fluid, like that, that's what we have..If it's inviscid now, just the inviscid, OK..There is three of these guys, of course, there's.one x direction, there's one y direction,.one z direction. See equations, these guys.are 6.51. OK. So these are called Euler's.equations of motion. These are nonlinear PDEs..And they're not easy to solve. They're, you.know, it takes a little bit of math. There's.some simple geometries and some simple approaches..So, yeah. By the way, when you solve this,.guys, what you're doing is you're solving.for, look over here what you've got, pressure.may not be known. Little u, little v, little.w. So for little u, little v, little w and.pressure. How many equations of the Euler's.equations of motion? One in the x, one in.the y, one in the z, that's three. How many.unknowns? Four, I need one more. Guess what.we did in Chapter 6, first-- first part of.the chapter, continuity, you got it. Continuity.and the three Euler's, four equations, four.unknowns, it could be solved. It could be.solved. It might be typical, but it could.be solved. OK. Oh, by the way, as a function.of-- Look at the variables, time x, y, z..So you can solve for the pressure as a function.of time x, y, z. You can solve for the x [inaudible].of the velocity as a function of time x, y.and z and so on and so forth. OK. You can.also reduce this down. They can be-- or let's.just say, it can be reduced to the Bernoulli.equation. It can be reduced to the Bernoulli.equation. It's very, very similar to Bernoulli.equation in the one dimension. In Bernoulli's,.is there a z term? Yeah, g tells you how g.is a function of how up or down you are or.you could-- There's a pressure term, that's.in Bernoulli's. If it's only a function of.x, we're the y and z. If a steady state, where.the t function. This guy here, or is this.guy here, partial u squared with respect to.x. OK, 1/2. OK. So that's 2u, 2u divided by.2 DmDx, those guys are the same. Does that.look like Bernoulli's? Of course, it does..V squared divided what? Guess what's divided.by 2. Yeah, you can reduce it down to Bernoulli's..Some people that teach the course derive Bernoulli's.this way. We derived Bernoulli's back in week.three just to kind of get a flavor for energy.in the fluid mechanics. But you can derive.Bernoulli's-- and back in Chapter 3, we derived.Bernoulli's along a streamline and the whole.fluid element along the streamline. Remember,.we took a little fluid element along a streamline.and we derived Bernoulli's equation from that..Some people don't cover that in Chapter 3.and they do the same thing in Chapter 6. They.stop here and now they derive Bernoulli's.and textbook [inaudible] again. He says, you.want to see us derive Bernoulli's, read that.thing in the textbook. And they derive Bernoulli's.from this and they end up with the same Bernoulli.we have back in Chapter 3. OK. Now, one more.step. Now, this is frictionless. Now consider.friction. OK. It can be shown-- -- that--.Those are all in equation 6.125. They're sometimes.called the constitutive equations. They relate.the stress terms to the velocity field and.the pressure and the viscosity. And the textbook.says, if you want to see how-- where they.came from, here's three references. Go find.and read them. In other words, it's pretty.tough stuff. It's not something you want to.put in undergraduate textbook. OK. So you.have to say, OK, I believe it. There they.are. That's what they are. So we take those.guys there and where do you see these guys.up here, for sigma xx? You put this one. Tau.xy, you put this one. Tau yx, this one, and.so on and so forth. You solve stuff in. That.is complicated. OK. When you're done, you.need equations like this. Et cetera and yes,.the y direction and z direction. Three of.them, x direction, y direction, z direction..See equations 6.127. These guys are even worse..These guys now are non-linear second order.partial differential equation, second order,.partial differential equations. Oh yeah, that's.for the advanced group, OK? Yeah, those guys,.because they include viscosity. Euler's do.not include viscosity. The ones that work.all the time are the general differential.equations of motion. If you neglect viscosity,.you get these guys. If you include viscosity,.you get these guys. These guys are called.the Navier-Stokes equations. So now we .have three equations on the board, the general.equations of motion of a fluid in a fluid.field. The equations which are inviscid which.you're-- end up what we call Euler's equation.of motions. And if you include viscosity,.you end up with what's called the Navier-Stokes.equations. Oh yeah. Oh yeah. You know now,.if you finished, typically, civil engineers.and [inaudible] take two quarters of fluids,.OK. But if this is going to be your area of.interest and you want to get a master's degree,.let's say, here at Cal Poly, you would first.of all, in a master's program, take Advanced.Fluids, EGR 535, Advanced Fluids, four unit.class, one quarter. Then if you want to go.further, you would take potential flow four.unit class in the graduate program. Potential.flow neglects viscosity, OK. And then if you.want to look at these guys, you take a course.called boundary layer theory, which talks.about the boundary layers on objects, between.an object in a fluid like an air flow or a.turbine blade. You take a course which focuses.on these guys. So, there's at least three.graduate courses worth 4 units each, 12 units.of graduate work, OK, in Advanced Fluids..And what did we-- done this thing here, probably.about 12 to 15 minutes. We just, you know,.when we scratch like that. Nothing, nothing..It's advanced stuff, it's heavy stuff. OK..We got our own problems, OK. So anyway, just.so you know. That's what you get into if you're.interested in fluid mechanics, thermo sciences,.you get equations like that. They're called.field equations. They apply at every point.in a field. They're not a controlled volume.approach, there is no controlled surfaces..You solve for the velocities and the pressure.as a function of time x, y and z analytically.if you can, which you many times can. Numerically,.if you can, many times you can. OK. That's.where numerical methods comes in, courses.like that. OK. So, I just want to give a flavor.of that so you know what those words mean..If somebody throws out the word Navier-Stokes,.you say, who's that guy? No, no, it's not.a guy's name, OK? They're advanced and the.key thing is to know when they apply. Friction,.considering friction, Euler's, frictionless,.these guys, that's why they call it general..You start here and you derive these two guys.from that one. And the textbook takes about.eight pages with that stuff, eight pages..We don't have time in a quarter course. OK..Now let's go back, so if we did that, let's.go back to our problems we were working and.you've got four problems for homework in Chapter.6, all right? That homework will now be due.with the Chapter 5 homework on Monday. So,.add the Chapter 6 homework. So the Chapter.5 homework that was due anyway on Monday..OK. We're going through the problems I assigned.for homework in Chapter 5 talking about the.energy equation. We discussed two already,.test homework of that, it's over there if.you missed it. We talked about 36 and 39,.so we'll quickly look at 50 if I think a test--.something is important, then we just look.at it real quick, 6.50. Down the-- probably.not, but I'll do additional. Oh yeah, we will--.6, that's quite difficult. OK. I'll read it..And now to attach to a virtual plane, discharges.water to the atmosphere, all these were major.case. Discharge is we know Q, we know the.pressure at the plant, so discharge and we.know the Q. We know the pressure at the plant..We know Q. The pressure here, we know, zero..He said atmosphere. Those were all his. Determine.the vertical component of the anchoring force.required to hold the nozzle in place. The.nozzle has a weight of 200 Newtons and the.volume of water in the nozzle is 0.012, is.the anchoring force of horizontal. OK. So.again, we always show our points through all.your control line, OK, control surface. Call.this 1, call this 2. OK now but, you know,.we got to be careful now because there is.a weight acting down of the nozzle and there's.a weight acting down of the water. OK. We.got to include those two guys in there. And.now you might say, well, I'm going to assume.that the force here is a positive to make.it easy in my equation. Here is my Fy. So,.and my x and y, like this, left hand side.of momentum, plus Fy minus weight of nozzle.minus weight of the water plus the pressure.force up here in the y direction. Zero, zero,.equal to change in momentum, I'm not going.to move to it again but, you know, rho B times.B dot A at one, rho B times B dot A at two..You know the Q, you know, big times A because.it's given. If he gives you the areas A1 and.A2, done deal. Plug them in, crank it up,.you got it. And as I told you, when you're.studying for exam, don't stop there. What.if I ask you to find a force in the x direction?.You know? [Inaudible] you know, that's a y.direction, but in the x direction, what's.the pressure on the top there? Zero. You know,.so let's take [inaudible]. If you got momentum.leading but what's the velocity, you know,.the velocity leaving there is at-- also needed.is rho q times v. That's the easy one there..All right. Let's go to 5.61, 5.61. Oh yeah,.OK, 5.61 is like this. Stuff comes in, it's.going to go on two locations now. One location.is it goes up and back, turns around your.back and the other location down here. Like.this, out, out, in. Control volume, control.surface inside. Pressure we know, velocity.we know, velocity we know, OK. We call this.one, we know V1. Call this two, we're given.V2. Call this three, we don't know V3. We.weren't given V3. We know P1 and here's a.key thing. Water discharges into the atmosphere,.thank you very much. P3 is zero, P2 is zero..He told me it discharged atmosphere and I.know P1. I know P1, I know V1, I know V2..Continuity gives V3. Momentum gives support.in the x direction. OK. Just so you know how.to solve the V3, continuity. V times A, V1A1.equal V2A2 plus V3A3, the only unknown is.V3, solve for it. OK. Let's go on--.>> Professor?.>> Yeah. Oh. Let's see [inaudible] 6.51, OK..OK, let's do this next one, 65. And then here's.extra plate and let's go down here, here's.weight here and [inaudible]. OK. A horizontal.circular jet of air, air, OK, strikes a stationary.flat place indicating jet velocity is 40 meters.per second, the jet diameter is 30. If the.air velocity magnitude remains constant over.the-- as the air flows over the plate surface.in the direction shown, determine the magnitude.of F, the anchoring force required to hold.in place stationary and then do part B and.then part C. That's air to air now, OK. A.horizontal circular jet of air. Horizontal.tells you, don't worry about the delta Z's,.don't worry about Z1 and Z2 and Z3. These.nicely labeled, 0.1, 0.2 and 0.3. So this.thing is here's my horizontal jet of air and.it hits this thing, see here it is right here..Here's a horizontal jet of air. The jet of.air hits this, he's saying some goes up, some.goes down. So that's how it looks, OK. It's.something like the one we've worked before.where the jet of water hit the vertical plate,.it's on the ground, the jet of water hits.here and it spreads out in all direction uniformly,.except now the plate is air now, the plate.is like this, it hits here and goes up that.way and down there. He tells you B1, B2 and.B3 I guess. But the key of that problem to.make life easier, to find your coordinate.system this way. And then momentum in the.y direction use the force F. So it changed.your coordinate system to make life easy for.yourself..[ Inaudible Remark ].OK. That's the key. That's the key. It's part.C I think, maybe, yeah, maybe. In the x direction,.OK, you're going to find out it's zero. He.told you that there is no foreseen x direction.which means it's frictionless flow. It's frictionless.flow, part C is frictionless flow, OK? OK..I think that was the last one. Yeah, that.was the last one, momentum. Let's do the energy.ones [inaudible] time. All right, 102, 5.102..OK. OK. This big diameter going down to a.small diameter, yeah. And we've got the flow.going down, the problem says it flows downward..Vertical plate now, so you're like the [inaudible].Z's. Mercury manometer, OK, get a manometer.up here down to here. We have-- This is--.He says oil, that's oil, not water, so oil..And this guy goes down there and comes up.H. So down here comes up to here. This distance.here is H. And that's gamma of mercury. Yeah..And it's 0.6 meters from here to here. I'm.going to call this location 1 and I'll call.this location 2. OK. And I'm going to write.energy 1 to 2. I think I'm going to write.P1 over gamma plus V1 squared over 2g plus.Z1. P2 over gamma plus V2 squared over 2g.plus Z2. Let's see what's applying here now..Determine the volumetric flow rate for our.frictionless flow, OK. So I put the loss is.zero in here. Because it said frictionless.flow, put the last term in the x equation.zero. There is no pump, HP is zero. There.is no turbine, HT is zero. There is no losses,.HL, HL, losses. In what, feet or meters? What's.this in? Feet or meters. What's that in? Feet.or meters. What's that in? Feet or meters..Everything in there is in feet or meters..OK. The reason why I assigned this because.I want to start-- I know you got test Monday.but the final exam is two weeks later. OK,.[inaudible] I'm sure. Yeah, you got to go.back and review manometers again. Come back.in week two, might as well do it now. That.way-- Let's study for, for the final. So,.let's just read in there and see, we're supposed.to find Q. Of course, Q1 and Q2, so don't.even call it Q1. I know they're both the same..Why? Steady flow, incompressible. OK. Can.I find P1 minus P2? Of course, I can, manometer,.Chapter 2. Can I find V1 in terms of V2? Of.course, V1A1 equal V2A2. Got it? Can I find.Z1 minus Z2? Of course, 0.6. What are you.going to solve for? I don't know, V1 or V2,.take your choice. Once you get V1, where do.you most plot by? A1. To get what? Q, end.of problem. But he's not going to be super.kind to you. He's not going to tell you that.you're supposed to call this distance here.x. It's not in the sketch, but he hopes you.realize it, because if you-- I think, he made.a mistake, he didn't put it [inaudible]--.I think he meant to-- I think he meant to.put-- to put this thing up to there. Oh no,.he didn't. No, not way up there, no. Well,.you know, you don't have to do that. If you.just realize it. If you go down in water,.that distance x on the left hand side, going.down is what? Plus gamma x. And go up on this.side, going up and use what? Minus sign, gamma.times x, plus gamma times x of oil minus gamma.times x of oil cancels up, it's gone, x disappears..But it's nice to know why it disappears. Well,.you know, that's-- The point there was you.shouldn't put an x there when you do the problem.so that I know that you know what you're doing..That's the point. OK. Now let's take the next.one. 102, we're going to go to 105 I guess..105, yeah 105. All right 105, we have a siphon.in a tank. And I guess this is water, we'll.see. Yeah, water. OK. We know the diameter,.given. We know all the distances from here..Let's put it down [inaudible] shown down here..So here to here, to here. We know the distance.from here to here. Those are on one. Check,.check, check. We know it's water, gamma water..Got it? And the friction loss is 0.8 V squared.divided by 2. OK. We're supposed to find the.flow rate. OK, which means find the velocity?.What velocity? The velocity in the pipe. OK,.so we start out and say, you know what, I'm.going to call this point 1 and I'm going to.call that point 2. OK, got it. Let's go ahead.and write the energy equation down. Here it.is. Here we go. There is no pump. There is.no turbine. Zero, zero. Pick out the right.equation. I put three on the board for you..I said sometimes it's easier to use one equation.than it is the other equation. Which equation.am I going to pick now? Here, I'll tell you.which one. What did he give me? He gave me.the losses in terms of what? B squared or.2g? No. That'd be the third equation. B squared--.it's called the head equation. No. He gave.me the loss in terms of V squared over 2..I try and find in those three equations a.V squared over 2 sitting all by itself. Oh.there, just right there. The second down,.the middle one. So I choose the middle one.here. Why? Because then I'm getting difference.of losses and I better have what's the units.here? What's the units here? What's the units.here? What's the units here? When you can.derive it. This guy better have the same units.that those guys have, OK? Oh, I did again..Let's say you put him right here. Now we can..Just so you know, there are three velocities,.V1, V2, and V. They're all different generally..V1 has a velocity F point 1, on the control.surface. If it's-- if you draw them. V2, the.velocity F point 2, and V, the velocity in.the height. Yeah, got [inaudible]..>> Yeah, so on the midterm problem that you.put us in online for the energy problem, you.gave us the losses had a like-- as a coefficient..So, what-- when you give us the coefficient,.what is it usually in terms of? And P squared.over 2 and--.>> Coefficients like, let us say P sub 2,.that's the coefficient 2. Do you like mean.coefficient, a subscript? You mean something.in front of something..>> Yeah, like--.>> Give, give me an example here. What was.it? It was like it-- It gave the-- the lot--.the coefficient of losses due to the typing.in the 2 the 0.4..>> OK. I don't recall that. I'll look at--.Yeah. Give it some more and look at afterwards..>> Yeah..>> Online something..>> I'll look at as the--.>> OK..>> Yeah. Normally, it's given like that. Often.with that with [inaudible]. OK, now we're.supposed to find V, which mean that V. How.about this guy? He's gone. This guy, he's.gone. This guy, he's gone. Z1 minus Z2, I'm.right here, Z2, Z1. This distance minus that.distance. Got it. Got it. Got it. What's V2?.The velocity leaving the pipe. What's V? The.velocity in the pipe. Are they the same? Of.course they have to be. So the equation for.V. Once you get V, multiply it by the area.of the pipe. Pi E squared divided by 4. End.of problem. You've got Q. But make sure you.know what those V's are. Is it V1, is it V2,.and is it V, OK. On Monday, oh hey, I'm telling.you, things like that [inaudible]. What if.I asked you for the pressure here? And the.losses in the pipe up to that point A, the.losses in the pipe up to point A were 0.3.V squared over 3. OK, I'm not going to solve..I'm just telling you how I think. What if.I asked you for the pressure point B? Oh I'm.going to stop here. You get the point. Don't.stop. Just didn't work the problem you're.assigned. Use your mind to think ahead. What.can the instructor ask me by tweaking the.problem a little bit? Yeah, I'm interested..Up here, maybe the water's going to vaporize.and stop deciding. That's what I'm worried.about, the vapor pressure of water up there..And when-- how about the entrance here. Is.the pressure-- Does gamma times that distance?.Oh, no. No. No. Because if there's a velocity.there, what's that velocity? V. OK. So, I.can go from 1 to B or I can go from 1 to A.or from B to A. So, we'll be thinking about.things like that too. All right. Let's go.on, that was 105. Now let's go to 114..>> Professor [inaudible]. So, the question.was asking what is the last coefficient for.[inaudible] is K equals 0.4..>> Say it again now, the last one?.>> The last coefficient we're [inaudible].is K equals 0.4. On this problem in the board.or--.>> No, no, [inaudible]..>> Oh, K equal-- what again now?.>> K equals 0.4..>> OK. That's all it says?.>> The last coefficient for [inaudible] is.K equal 0.4. Determine--.>> OK?.>> -- our output on the turbine..>> All right. Typically, that's the coefficient.in front of the V squared or 2g [inaudible]..>> OK..>> The coefficient in front of the V. I would.be more specific next time I tell you that.because you couldn't tell from that. You couldn't.tell from that. I would put it back, yes,.to make sure there's no air interpretation..>> [Inaudible] this will be 0.8?.>> Yeah, I would put 0.4 V squared 2g probably..Because normally it's specified in feet, feet.loss. Yeah. That's a good point. That's for.defining it for us. OK, 114. All right. This.must have a [inaudible]. This one would happen.real quick, 114. Oh, the fire truck, OK. See,.the pressure is there, the diameter is there..So, we have this fire truck with the pump.here and the water comes in and then it goes.out here. Here is the fire hose. And then.the water goes up here, fire hose and-- The.pumper truck show and delivers-- Q is given..Sixty feet is given. So, Q-- Q. We have 60.feet. Let's see what we know about the problem..The pressure. We know the pressure and we.know the diameter up of the hydrant is 10..The head losses in the pipe are negligibly.small. Determine the power that the pump provides..So, find W dot comma. OK, let's write down.the energy equation. Leave hp, that's the.pump head. And write the equation in terms.of head. At least I'll do it that way. I prefer.that. If you want to use the one that has.the power in it directly, you can do that.too. I know they look like this though. So,.it is equal to P2 gamma P2 squared over 2g.plus Z2. Let's see, find-- Neglect the losses.and go ahead and find out the power. OK. P1..Yes, I know what V1, OK, Z,1 I'll call it.zero, Z2 is 60, V2 I don't know, I don't know.that guy there. Determine there. Yeah, just.like the P2 is atmosphere, zero. This point.2. Say you know what point 2 is. And I'll.see what it does there. Yeah, we know Q here..So, we know Q so we get V1 equal Q over A..I know that. Q is good. Solve for hp. Then.W dot power of the pump equal gamma Q hp divided.by-- I want this. Let's do it like that. If.you want it in horsepower divided by 550 so.that's in [inaudible] computational. Otherwise.it comes out to be in kilowatt-- kilowatts..OK..>> Is it OK if we leave it as pounds per second.or which is you prefer [inaudible]?.>> I will say the problem, find the power.and horsepower, what kind of power [inaudible].to be specific..>> OK. All right..>> OK, 5.116. OK. Again, you know, I can tweak.this problem so you-- of losses in the hose.or want to take these [inaudible] too and.stuff like that and all the-- It doesn't change.the problem much, you know. I can say, if.the pump efficiency is 80%, how much power.must be provided to the pump because I'll.tell you, this power I get out of here is.the power that the pump blades provide to.the water. But what's the power required to.rotate the pump blades in a centrifugal pump?.OK. You got to divide that guy by the efficiency..The equations are in your notes, the efficiency.equations, they're in your notes. So, I can.put efficiency in there. I can put losses.in there. Those are things I can do to the.same problem assigned for homework. OK, 5.116..Now we got a reservoir with a pump pumping.to another reservoir on top of this reservoir.that's open to the atmosphere. And we take.water running here and it goes through a pump,.then it goes up into a second reservoir which.has its capped on pump. Has air up there at.2 atmospheres. So we know the pressure up.there. So, here is the water and it's capped.on top. This is air, and this is water. OK..So, we know the flow rise that way. We're.specifically given those [inaudible] what.we know. OK. We know the flow rate, the atmosphere--.We know gpm. We know the Q, so we know Q..It has 3 horsepower, so. OK, losses. P1, zero,.I'm going to call it point 1 here. I'm going.to call it point 2 up here. P1 zero, V1 zero,.Z1 zero..>> [Inaudible] the pump..>> Oh yeah, yeah. Over the pump here [inaudible].two good things. P2 is not [inaudible] 2 atmospheres,.OK, got it. V2, yeah, that's zero. Z2 got.it. Hp I know what it is. It's right here..OK. What's the power, somebody, in kilowatts?.>> Three horsepower..>> OK, three horsepower. Now you got to play.the 550 game. I can just write 50. I know.this value, I've got it. I know gamma of water,.got it. He gave me Q. He gave me Q, got it..Solve for hp, for hp, there. OK, not hp. I.know, check mark. Now, I solve for the losses..OK. Now, if I get the loss is positive, it's.OK. If I get a negative, it's impossible..So, that's what you do. You solve for the.losses. If you get a positive sign, yeah,.it can happen. If you get a negative sign,.it can happen. Losses can never be negative..Losses are always positive. They take energy.up. Losses don't put energy into a fluid..They take energy out of the fluid. OK. I'm.not going to do 117 and give you everything.you need. Let's see. Yeah, OK, good. Let's.read 120. Water flows by gravity from one.side of the lake to another, from one lake.to another lake at a steady rate 80. Good,.we know Q. What is the loss of available energy.associated with this flow? And he gives us--.He gives us delta Z. OK. Flow down, losses.equal 50. And this is in feet, so losses are.50 feet of water. P1 zero, P2 zero. There's.no pump. There's no turbine. P1 zero, P2 zero,.V1 zero, V2 zero, Z1 50, Z2 zero. There's.only one thing left in the equation, Z2 equal.losses. OK, easy, you got it, 50. Now comes.in the other side of the coin. Keep reading..Now, if we want to reverse this flow and pump.water up in the lower part of the higher reservoir,.estimate the pumping power required. So, now,.I put a pump in here and now I change the.flow direction. Now, you better change the.subscripts, if you don't, you're going to.hit really bad. This has to be 1 because that's.where the flow comes from. This has to be.2. That's where the flow ends up. P1 zero,.Z1 zero, V1 zero. Two, P2 zero, V2 zero, Z2.50, Z2 50. Here's the equation. Hp equals.zero, zero, zero equal zero, zero, plus 50.plus 50. Fifty plus 50. OK. Power, you want.power, hp times gamma Q hp, that's the power..OK. So, what [inaudible] my gravity. That's.how these-- How many stations in the [inaudible].work sometimes. They let the water flow all.the way down to a generator, in a lower elevation.during the day to generate electricity. Then.when it's cheap to run electricity, they used.electricity to pump the water back up late.at night, and [inaudible] pump storage. And.a lot of-- a lot of utilities use that because.electricity at night is cheap because nobody.has the lights on in that. So, they pump the.water back up to the upper way in the [inaudible].at night. They let it run down the day to.power the lights in Fresno and blah blah blah.and so on and so forth. So it happens, and.it happens all the time. That's why it's an.important [inaudible]. OK. Last one, we're.going to finish on time, that's great. All.right, 5.131. OK. Let's look at this guy now..All right, 130-- Boy, that's [inaudible]..I got to take my book and sketch [inaudible]..It's a little bit different than we've done.in the past. It looks like this. All right,.so we have water. Water, we think it's water..Yes, it is. Water flows steadily down. Down,.down. There is a manometer. So, practice again.from Chapter 2. OK, like this, like this,.and this and this, like this. And we've got.the fluid level here and here and we know.this distance from here to here, checked..We know this is mercury, gamma Hg. We know.this is water, gamma water. The diameter is.the same. The difference in the first part.is it says go back to Chapter 2 and find P1.minus P2. There's 1, there's 2. Part A, find.P1 minus P2, Chapter 2. I'm not going to do.this, we've done enough. OK, part B, the log.between sections 1 and 2. OK, it is over there.again. It's over there. Is there a pump? No..Is there a turbine? No. Are there losses?.Yes, leave it in. Do I know P1 minus P2? Of.course I do, I found that from part A. .Conservation of mass. Q1 equal Q2. V1A1 equal.V2A2 and it's the same diameter inclusion.V2 equal V1, OK. Cancel, cancel. Difference.in elevation, it tells us it's 5, 5 feet..It tells us 30 degrees. Yeah, I got it. The.vertical, the vertical part of that, OK. Five.times the sign of 30, I dropped the delta.Z. Matter of fact, I'm going to call Z2 zero,.my choice, I call it zero. So solve for Z1,.5 sign 30. That's it, solve for losses, part.B. OK, done. Got that. We got that. Now let's.do part C. OK, here's my control volume. Find.the friction velocity in the pipe. OK..>> Is it part C asked for RX?.>> Yeah..>> What did it ask for?.>> For the extra--.>> The force in [inaudible]..>> OK. That's right. Thank you. Yeah. OK..OK, it comes in, [inaudible] dot product dot..I'm writing it this way. OK. In the x direction,.that's what I want. I want the force to the.friction in the x direction called the axial.direction. OK, top product out, air vector.out, plus vector in minus dot product, area.vector out, velocity vector out positive,.velocity here V1, velocity there V2, OK. This.sign negative, that sign positive. Let's see.here. OK, we got that guy and then we've got.the V2 rho 2 A2V2. Force be the x direction,.of course the friction force. Which way you.want to assume friction acts? Well, you know,.most folks believe friction acts against the.flow. Of course it does, so I'm plugging it.to minus F. My intuition tells me that's the.way it goes. It acts against the flow. It.tends to slow the flow down. Pressure force.here. Don't try and get. I told you, don't.try and do too much on one picture. Three.forces, there they are. Don't try to [inaudible].your mind. You put too much stuff on one diagram,.you're asking for trouble, I guarantee you..P1, the force of pressure force-- Your pressure,.P1. P1A1, positive, P2 minus P2A2. The weight,.the actual weight times the sign of 30 acting.in the axial direction. The weight times the.weight of water. The weight of what? Don't.include the pipe, look at my dash lines. Did.my dash lines include the pipe wall? No, they.only included the water. They didn't include.the pipe wall. There is an identical problem.of this. It's a long problem. It's the longest.problem in Chapter 5. It starts on page 216,.it goes through page 217, and ends up on page.218. And on page 218, they draw a nozzle with.just the water shown. So the example to follow.is on page 218, it's example 511. That's what.you need to do to follow to do what I did.to the problem right here. OK, this is an.interesting problem. It's a great review problem..You know, chapter-- We would go to Chapter.2 for part A, we would go to Chapter 5 on.energy for part B and we go into Chapter 5.for momentum on part C. Oh yeah, it gets you.ready for-- for the exam, yeah, it will. OK,.let's not be blank so see you then..
