>> 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.
