>> Okay, we talked about vary area flows and compressible flow before, previous lecture. Today, we're going to look at what happens in a converging nozzle. Converging nozzle has a reservoir where the pressure and a temperature are P-knot and T-knot. The area gets smaller to an exit area, and the flow exhausts to a back pressure region where the pressure is P-back. You can plot that on a graph of the X direction versus the pressure ratio, P over P-knot. If there is no flow if P-back is P-knot, there will be no flow. So this is no flow. As we start to reduce the back pressure, we get flow. And the curves go down something like this. And I'll just draw a few of these to see how all these guys go. One more. And then one right here. When we get down to here at the magic number, P over P-knot, is .528 for any gas with a K of 1.4. Then that's going to reach Mach-1. So at that point, we reached Mach number at the exit equal 1. If we reduce the back pressure still more, reduce that pressure still more, nothing happens in the nozzle. The nozzle stays the same. What happens is there is expansion waves outside the nozzle where the fluid, let's say air, where the air expands to P-back and expansion waves. So that particular picture, let's just make this a little bit clearer here, once you reach Mach number at the exit equal 1, that's called choke flow. So if you reach Mach number at the exit equal 1, that's called choke flow. Choke flow means no more mass flow will be going through the nozzle no matter how much more you reduce the back pressure. So you can draw another graph, which is useful. It's in our textbook. In this graph, we're plotting on the X-axis, P-back over P-knot. We're plotting on this axis M-dot. This is M-dot max. It comes over here flat, and then it drops down like this. This is 1. And this is P-star over P-knot, and this is 0. Any conditions from our past lecture, any condition where the Mach number is 1 are the start conditions. So P start is the static pressure where the Mach number is 1. If we reduce the back pressure beyond that point, let's start off like this. Let's say P-back equal P-knot. P-back equal P-knot is 1. Do we get any flow, M-dot? No, zero flow. Start reducing the back pressure. Do we get some flow? Yeah, we get some flow. We get more and more flow when we reach the Mach number of 1 at the exit, okay? Beyond that point, no more mass can be passed by the nozzle if the back pressure is reduced further. It's called choke flow. It's like the lines of communication have been cut. The nozzle doesn't know the back pressure has gone down. Why? Because the speed of sound has been reached at the throat. So the signal can't be taken back to the reservoir. You reduce the back pressure. The reservoir didn't know what happened. So it keeps putting out the same mass flow rate no matter how much more you reduce the back pressure. Okay, all right, this is all isentropic. Forget this. We're looking at what's inside the nozzle. What's inside the nozzle is all isentropic. In our appendix, we have Table B1, which is called Isentropic Flow. Okay, so this is the right able. This is our Mach number. This is a table we use for the converging nozzle. Or if you'd rather, you can use the equations. There are equations for all of these things in the textbook, PO or P-knot, T over T-knot, row over row-knot, A over star. It's written in terms of the Mach number. So you have equations you can use. Or if you'd rather, you can use the table itself. The equations, I'll put them up here, equations 934. These guys right here are equations 934. This guy right here is equation 935. Only if K is 1.4. K, 1.4. Air, nitrogen, oxygen, carbon dioxide, nitrogen oxide. So, yeah, there's lots of gases that have K 1.4, not just air. But it's very common. So the table would fork for any of those gases. What if K is not 1.4? Don't use table B1. Use equations 928. And 944. Those have K in them. So a K is not 1.4. You've got to use those equations. You can't use the table. Okay, so now we have our picture here. And let's go ahead and this is your roadmap. You want to find out which line you're on. This tells you something about the mass flow rate in the nozzle, okay? So we're going to use this now to solve a problem. All right, here's my first problem. I'm going to have P-knot. I'm going to do it in English because English is British gravitational is harder to do than SI. SI is pretty straightforward. The British gravitational is a little bit more difficult in units. So I'll go through this because it's more difficult. All right, P-knot, 100 PSIA. T-knot, given 600 degrees R. The fluid is air, K, 1.4. The back pressure is given, 59.1, PSIA, all pressures, all temperatures in absolute. The exit area is .01 square foot. Okay, okay, find M at the exit and M-dot. Find the Mach number at the exit and the mass flow rate, M-dot. All right, well, we know, one thing we know, the back pressure. We know P-knot too. So, okay, I know P-back, I know P-knot. I'm going it take P-back over P-knot, 59.1 divided by 100, 591. Greater than 0.528. Here's my graph. At the exit, here is .528. My problem, it's .591. I'm right here. There's my roadmap. There I am. Now, I didn't get down to 528, where the Mach number at the exit is equal to 1. I didn't get down there. I'm on that curve. Conclusion, it's all subsonic. I never reach Mach 1 at the throat. All right, so let's go to .591, the Mach number. Here's the pressure ratio, 591, and the Mach number is 0.90 from Table B1, from Table B1. It's not choke flow. Okay, I'm going to calculate--I've got part of the problem now. Find the Mach number at the exit. Find M-dot. So next, find M-dot. Row, exit, A, exit, V, exit. Okay, row AV. I'm going to get V exit first. V exit, Mach number at the exit, times the speed of sound at the exit. Okay, I can get the temperature at the exit from here, T over--TX over T-knot. I know what T-knot is. So let's make sure I get that number right. There it is. Okay, so I'll put that number up there, .9, temperature ratio, .98606. Okay, 8606. And so we go ahead and get our temperature. Table B1 at Mach number exit equal .9, T at the exit over T-knot is equal to 8606. So our temperature at the exit plane comes out to be 517 degrees R. Got it. I put that 517 degrees R in there for our--I'll write it out, 0.90. The square root of K, 1.4. Put in our value of 1716 for R times T exit 517 square root. So the velocity at the exit, 1,000 feet per second. Velocity at the exit, 1,000 feet per second. I'm going to get the density at the exit, P over RT. So our P exit over RT exit. Pressure at the exit, I found that. It's, here, P-back, 59.1. We don't want inches anywhere. So 144, to get it in square feet, R1716, T exit 517. Density at the exit, 00960. Slugs per cubic foot. So we got that, and we got velocity. Solve for M-dot. M-dot equal row. Our area at the exit, 0.01. Our velocity at the exit, 1,000. M-dot then is 0.0960 slugs per second. Where are we on this roadmap? Okay, if K is 1.4, this is .528, where you would reach Mach number 1 at the exit. Our Mach number, our ratio was .591. So where are we? We're right here. I'm sorry. We're over here. [ Inaudible ] 591 right there. I'll squeeze it in there. There's .591. Our mass flow rate, .096. Is that the maximum mass flow rate? No. Why not? Because the flow is not choked. That's why not. We didn't reach Mach 1 at the exit plane. Okay, so that's a non-choke case. Now, we're going to do another one. So for this example, my back pressure is going to be 30. And, let's see, I know not that, get that out of there, get that out of there. We're going to redo this guy. No flow. .528. Okay, let's clean him up. Okay, all right, now back to here. P, okay, some location, call it A1. P1 equals 60 PSIA. T1 equals 40 degrees Fahrenheit. Mach number at 1 equals 0.52. Area at 1 equals 0.013 square feet. Find Mach number at the exit. Find the area of the exit. Find the mass flow rate through the nozzle. Okay, let's put that over here. All right, so-- It's air. I don't know. I don't know P-knot. I don't know T-knot. Okay, right away, we have trouble. We say, okay, the last way it worked, you were given P-knot and T-knot. This problem, you're not given P-knot and T-knot, but you're given one important thing. You're given that Mach number somewhere. Okay, .52 there, I know that, okay. So Table B1 at a Mach number, .52. Got it. 08, round to 832. Temperature, 949. All right, I know the pressure and temperature there. So T1 over T-knot, .949, which gives me T-knot. Since I know T1, be careful, 460 plus 40, 500R. T-knot, then come up to be 527 degrees R. P1 over P-knot, 832. So P-knot has to be--I know what P1 is, 60, 72.1, PSIA. Then P-back over P-knot is equal to back pressure was given, 30 PSIA, 30 over P-knot, 72.1 equals 0.416. Uh-oh. That's less than .528. Uh-oh. That's choke flow. Mach number at the exit has to be 1. Okay, so now I've answered the first question. What's the Mach number at the exit? It's 1. Why? Choke flow. Why? Because P-back over P-knot was less than the magic number, .528. Okay, here's a roadmap. All right, so now I know. It actually does this. It's below there, so it does that, okay? Okay, so from here to here, subsonic. Table B1. Okay, got it. Let's go over here. The pressure ratio now, .416. .416. There. Oh, okay. I've got M-dot max. I'll use that for the third part of this. But you'll go ahead then and, yeah, let's get A exit next. >> Professor? >> Mhmm? >> So does that mean for the [inaudible] nozzle, you can never go past or any higher than Mach 1? >> Exactly. A converging nozzle will always be subsonic. The best you can get would be sonic speed, Mach number 1, at the throat, and only at the throat. Right. >> And so to go higher than Mach 1, you'd have to be converging, diverging? >> That's right. You've got to put the diverging section on the converging nozzle. Right, uh-huh, uh-huh, that's right. Okay, so now we're going to get A exit. Okay, so, again, the Mach number is .52, 52. Table B1, Mach number .52, Table B1. Area ratio, 1.303. A exit over A star equal 1.303. But A exit, yeah, over A star. A, I was saying A1, A1 over A star. Pardon me. A1 over A star. But A star is equal to A exit since the Mach number at the exit is 1. Anywhere the Mach number is 1, it's automatically the standard quantities. Because we've reached a Mach number of 1 at the exit, that's A star. It's also A exit. Okay, so now solve for A exit. A exit is equal to A1 divided by 1.303. And I know A1 because I was given A1, .013 squared feet. So A exit comes out to be 0, .010 square feet. Okay, now my last part is find M-dot. Okay, one way to find M-dot, M-dot is equal to row AV. Where? At 1R. It's equal to row AV at where? V exit. Is it the same? Yes, steady flow problem. The mass flow rate, of course, is equal anywhere along the nozzle itself. So the value is the same no matter where, at 1 or the exit. >> Thank you, Professor. >> Mhmm. Take care. Okay, so here's what I got out of this. I used, I think I used this guy up here. Okay, so I used this one. So V1, just like we did before, is equal to Mach number 1 times the speed of sound at 1, the Mach number at 1 times the square root of KRT1. And that's equal to the Mach number 0.52. Square root K is 1.4. R is 1716. T1 is degree R. Oh, I'm sorry, that was in degrees R, but let's get our temperature. T1, 500, yeah, 500. Square root. So the velocity at location 1, 570 feet per second. I need row 1. Okay, row 1. P1 over R, T1. P1. Okay, my pressure, P1, given to me, 60 times 144 divided by R, 1716 divided by T1, 500, is equal to 0.01007 slugs per cubic foot. Got it. Put it in the equation. M-dot equal row 1, A1, V1. And when you do that, you're going to get 0.0745 slugs per second. Okay, if you don't want to do it that way, there's also an equation in the textbook which is only use this if it's choke flow. Use only if choked flow. M-dot max is equal to 0.6847 P-knot A-star A-star divided by RT-knot square root. Okay, it is choke flow. We found out choke flow, okay. I can use it. 0.6847, P-knot was given to me over there. P-knot found P-knot. 72.1 times 144. Divided by--oh, times A star, A star, 0.010 divided by R, 1716, times T-knot, 527. Take the square root, square root, M-dot max by this method, 0.0745 slugs per second. Yeah, it doesn't matter. They give the same answer. Row AV, this equation always works, choke flow or not choke flow. These equations always work, choke flow or not choke flow. The only time you use this equation is if the flow is choked. In this problem, choke flow. I can use it. When I do, I get the same answer I would have gotten by taking row AV at 1. Okay, now, the last thing. We're going to look at the units, because the units can really cause people--it does. It causes people tremendous problems, the actual units that we have here. So I'm going to go through some unit stuff now. All right, let me see here now. Okay, let's do the velocity. For the velocity, V equals M times C equals M times the square root of KRT. All right, so I do this for a purpose. I didn't put units in there because I'm doing that now. I put numbers in, but not units. You better check these units, because they better come out to be feet per second or you've got trouble. Okay, the Mach number is dimensionless. K is dimensionless. R, this 1716, it's in this, foot pounds per slug degree R. Foot pounds per slug degree R. So I put that guy in here. That value is not in the back of our book because he only puts SI values in the appendices. You've got to go to another textbook to find that R value with the British gravitational. But that's the units anyway. Temperature, degrees R. Square root. Degrees R cancel out. You say, okay, so the velocity is 570, the square root of foot pounds her slug. A guy says, huh? I thought velocity is in miles per hour or feet per second. How can the velocity be in foot pounds per slug? Well, you've got to keep going. Don't stop there. Remember, in British gravitational, what is a pound? A pound, F equal MA, a pound is a slug [inaudible] at the rate of 1 foot per second squared. So slug foot per second squared divided by slug square root slug slug feet feet feet squared second squared square root. Yeah, it comes out to be feet per second. Don't worry. You want to, you want to do it in SI. Okay, Mach number, dimensionless. K, dimensionless. The value of R, joules per kilogram K. Temperature, degree K. Square root. Cancels, cancels. What is a joule? A newton meter. For a kilogram, square root. No, not kilogram, kilogram. What's a newton in SI? Okay, F equal MA. One newton equal one kilogram, accelerate at the rate of one meter per second squared. So replace the newton with a kilogram meter per second squared times meter divided by kilograms, square root, cancel, cancel, meter square per second square square root meters per second. Checks out in SI. So you always want to be able to do that like here. This guy is in PSI. This converts it to pounds per square foot. This is degrees K, 1716, foot pounds per slug degree R. It's going to come out to be slugs per cubic foot. Here's what the textbook has for the R value. The textbook has, in the text, in the appendix, he has R equal 287 meters square per second squared degree Kelvin in the back of the book in the appendix. That's what he has. I picked up two or three other textbooks. They always give R a net. In thermal, that's very common. R is what? Energy per mass degree change in temperature. Here it is here. What's foot pounds? Energy per mass change in temperature. That's how R is normally specified. Well, our textbook does it this way, okay? Does it this way. Meters squared and so on and so forth. Are they the same thing? Because this is also 287. Get a thermal textbook. It will say R is equal to 287 joules per kilogram. You need K for air. Our textbook says R is 287 square meters per square second per degree K. Well, how come? You've got to be able to go from one to the other. So start off with this guy here. R equals 287 joules per kilogram K. Okay, 287. What's a joule? A newton meter. Kilogram, degree K. Okay, 287. What's a newton? A newton is a kilogram meter per second squared times meter divided by kilograms degree K kilogram kilogram. Meter square, 287 meter square, per second square degrees K. Oh, yeah, they're the same thing. One of those guys is the same as one of those guys. You have to be able to do stuff like that to develop confidence in the unit systems. Because in the world of engineering, we engineers operate internationally, and we must know both unit systems. English engineering, British gravitational, and, of course, SI, and you've got to be able to convert stuff like this to find the velocity, to find the density, so you feel more confident about what you're doing. That gives you confidence. Because otherwise, you think, wow, this stuff is magic. No, it's not magic. It's just a bunch of definitions. Follow the rules, and you'll get the right thing. Follow the rules. Follow the rules. You'll get feet per second. Follow the rules. You'll get over here meters per second. But you've got to be able to convert these guys, go from newtons to kilograms or slugs to foot pounds to get what you want to get. That's why we work the problems in English engineering, because, of course, they're much harder to work in the British gravitational than they are over here in the SI system. The SI system, don't we all wish we lived in the world of SI only? No, we don't. We live in a dual language world, us engineers. British gravitational or English engineering, and then the SI system. Okay, good stopping point for today. We'll continue on then with converging, diverging nozzles our next class meeting.
