Welcome to Lesson 15b, Compressible Flow and Choking in Converging Ducts. In this lesson we'll discuss compressible flow in a converging duct, how to calculate critical conditions, and mass flow rate. We'll define choking and we'll do an example problem.
Let's consider compressible flow from a pressurized tank into a converging duct as sketched. The stagnation conditions are in the tank and we label the exit area. the Mach number at the exit and the pressure at the exit. In the air outside of this duct, the back pressure is Pb. Now let's think about this flow.
It starts at zero speed and accelerates into the converging duct. The flow here has to be subsonic and as discussed in a previous lesson, for a subsonic converging duct, speed goes up, Mach number goes up, pressure goes down and temperature goes down. So while the flow is subsonic, the Mach number is approaching 1. Now suppose that here the Mach number becomes greater than 1. In other words, supersonic. Well for a supersonic converging duct, the behavior is opposite.
So V and mA go down, pressure and temperature go up. So as soon as the Mach number is greater than 1, it has to start going back down. Well this doesn't make any sense.
Therefore we conclude that the flow cannot go supersonic in this duct. In other words, the flow must remain subsonic in the converging duct, and the Mach number approaches 1, but cannot exceed 1. So we also conclude that the only place where mA can equal 1 is at the exit plane. Here's where the back pressure comes in.
If back pressure Pb is not low enough, the flow is subsonic, and Mach number is less than 1 at the exit plane. If Pb is low enough, the flow is still subsonic, except Mach number equal 1 at the exit plane. For this second case, when Mach number equal 1 at the exit plane, we call this critical or sonic conditions. And we denote this by an asterisk. So t equal t star, p equal p star, rho equal rho star, etc.
But how low of a pb is low enough to make the flow sonic at the exit plane? Well let's go back to our equations for steady adiabatic isentropic flow. Suppose Mach number is 1 at the exit plane. So let's set mA equal 1 in these equations and set p equal p star, t equal t star, and rho equal rho star. So these equations become t naught over t star is 1 plus k minus 1 over 2 times 1 squared, which is just 1. And similarly p0 over p star is 1 plus k minus 1 over 2 to the k over k minus 1 exponent.
And a similar equation for rho0 over rho star. Well in all the textbooks people like to calculate the reciprocal of these equations. So we write this as t star over t0. And we simply take this quantity to the negative 1. We do the same with pressure and density.
For air, k is 1.40. So T star over T naught is 0.833333. And to four digits P star over P naught is 0.5283. And rho star over rho naught is 0.6339.
When the Mach number equals 1 at the exit plane, we say that the flow is choked. Which means that further decrease in Pb, the back pressure, has no effect on the flow. Choked flow is an important concept in compressible flow.
So let's go back to our question. How can we tell if the flow is choked? In other words, how low does Pb need to be?
I'll give three cases here. If Pb is greater than P star, the flow is not choked, Mach number at the exit is less than one, and Pe equal Pb. Any subsonic jet exiting into the air has the same pressure as the surrounding back pressure of the air. If Pb equals P star, the flow is choked, just barely.
The Mach number at the exit is equal to 1, and Pe is still equal to Pb, but is also equal to P star, the critical or sonic condition. If Pb is less than P star, the flow is choked, Ma at the exit is 1, and Pe equals P star. But P star is greater than Pb as we said here.
Once the flow is choked, you can lower Pb all the way to zero, in other words a perfect vacuum, and nothing in the nozzle will change. So for any condition in which Pb is less than P star, the Mach number at the exit plane is 1, the pressure at the exit plane is P star, Te equals T star, and rho E equals rho star, no matter how far you lower the back pressure. The flow in here remains constant and choked.
Why? Because once the flow is sonic here, the flow upstream cannot tell that anything is changing downstream. Because of this so-called sonic barrier here.
Pressure changes downstream cannot affect the flow upstream. Let's do an example problem. Air flows from this large pressurized tank through a converging nozzle. We give P0, T0 and the back pressure and also the exit area.
We want to know if the flow at the exit is subsonic, sonic or supersonic. If we keep lowering the back pressure, at what value of Pb will the nozzle exit become sonic? And we want to calculate Mach number and temperature at the exit plane.
And for part c I mean at the given conditions, not when the flow becomes sonic. Well let's list our assumptions and approximations. 1d flow, ideal gas, steady, nearly isentropic, adiabatic.
We also assume that the tank is so large that these conditions do not change. as flow is escaping. Or imagine that we have a compressor that keeps compressing the air so that T0 and P0 remain the same. Part A, we know that the flow cannot be supersonic as we discussed above.
So that one is definitely not the case. But is it subsonic or sonic? To find out, we compare Pb to P star. For air, P star over P0 is 0.5283. So P star is 0.5283.
times our given value of P0. So P star is 83.47 kPa. But our back pressure is larger than this. So we have the case of Pb being greater than P star.
So this flow is subsonic everywhere and not choked. For part B, as we discussed, if Pb were less than or equal to P star, the flow would choke. So the conditions will be sonic or critical at the exit plane.
So the answer to part B At what value of Pb does the nozzle exit become sonic? At Pb equals P star, which is 83.47 kPa. For part C we are to calculate Mach number and temperature at the exit plane for the given conditions. I start by using our stagnation equations for P0 over P and T0 over T. We apply these at the exit plane, where P equals Pb equals Pe, since the flow is subsonic at the exit plane.
So I let p equal pe and ma equal mae at the exit plane. I can solve this equation now for mae, the Mach number at the exit plane. This involves a little bit of algebra.
I take both sides to the k minus 1 over k power to get rid of this exponent on the right hand side. Then I solve for mae squared and take the square root to get mae. This is my answer in variables. We plug in the numbers, 2 over k minus 1. p0 was given and pe equal pb which was also given and then k minus 1 over k minus 1 square root. I get.82286 and I report my final answer to three digits 0.823.
To calculate temperature at the exit plane te I write te equal t0 times te over t0 or t0 times t0 over te to the minus 1 since we have an equation for t0 over t. So let t equal te and ma equal mae and I get this equation which is our answer in variables for te. Plugging in the numbers, t naught was given 1 plus k minus 1 over 2 times mae squared and I'll use all five digits I have there.
Then the whole thing to the minus 1 and I get 457.98k. My final answer to three digits is te equal 458k. Notice that TE is less than T0.
The temperature has dropped by 62 K or 62 degrees C. That's a significant drop in temperature. That's why, for example, if you push the little valve in on your bicycle tire and let the air out, the air will feel very cold to your finger.
Since we can approximate the tire as our pressurized tank and that little protrusion with the valve as a converging duct. Now let's consider the mass flow rate through a converging duct. Again assuming that this flow comes from a large tank that is pressurized. While m dot is rho VA, anywhere in the nozzle we can apply that at any x location.
We typically take the exit plane since we usually know the conditions there. Well let's work on this equation from the ideal gas law. P equal rho RT, so rho equal P over RT. So we write that in place of rho. Mach number is v over c.
So v equals m a times c. And c is the square root of k r t. So we replace v by this expression. And A just comes along for the ride. We can combine the R and the T terms under one square root.
So m dot is PAMA, square root of K over R, times temperature to the minus one half. We can massage this equation by using our ratios. Instead of P we use P over P naught times P naught. And instead of T we use T over T naught times T naught. And that raised to the minus one half power.
Now we can use our stagnation equations for these two ratios. This is our equation for p over p naught. And this is our equation for t over t naught. And we finish off the equation. We can combine these two similar terms since the part in parentheses here and here is the same.
We just have different exponents. You should remember this identity from math class. So finally we have, rearranging, m dot equal p naught a m a square root of k over r. 1 plus k minus 1 over 2 ma squared.
And I put the t knot under the square root sign. And the combination of these two exponents, after a little bit of algebra, is negative k plus 1 over 2 k minus 1. If the flow is choked, ma equal 1. And we use a equal ae equal a star at the exit plane. Where we define a star as the critical area where the Mach number is 1, since the flow is choked.
Note that this equation is a general equation whether the flow is choked or not. When the flow is choked, m dot is m dot max. And we can simplify this equation.
I type these equations up. Here's the general case, which is the equation I derived here. And when we set mA equal 1 and A equal A star, it simplifies to this equation. Finally, let's do an example, namely calculate the mass flow rate for the above example problem.
I repeat the equation here. So we plug in the given values and let k equal 1.40 for air. m dot equal p knot. The area at the exit. The Mach number at the exit plane.
I should mention that we're using the general equation since this flow is not choked. k over r and t knot. And we must be very careful with our units.
I'll put in two unity conversion ratios under the square root sign. A kilojoule is a kilonewton meter. And 1000 kilogram meters is a kilonewton second squared. Now we have this part of the equation left.
1 plus k minus 1 over 2. Our Mach number at the exit plane squared. And the exponent where k is 1.40. I need one more unity conversion factor. And we check our units. kPa cancels.
Meters cancel. Kilojoules cancel. k cancels. These two kilonewtons cancel this one.
This meter squared cancels this one. And we're left with a kilogram and another kilogram square root which is kilogram over second squared square root which is second. So we get 3.5365 kilograms per second. This is an exercise in unity conversion factors.
But if you do everything correctly, the units all work out and we get kilograms per second as our final answer. Finally, to three digits, we get m dot equal 3.54 kilograms per second. I'll leave it as an exercise to calculate m dot max. You should get 3.64 kilograms per second. That would be using this equation and being similarly careful with your units.
So this mass flow rate is our actual flow rate and this m dot max would occur if Pb were less than P star. In other words for choked flow. Thank you for watching this video. Please subscribe to my YouTube channel for more videos. Thank