welcome to lesson 15c compressible flow and converging diverging ducts in this lesson we'll discuss why for flow from a pressurized tank supersonic flow can occur only if there is a throat in the duct I'll derive the area ratio versus Mach number relationship I'll discuss what happens in a converging diverging duct when we vary the back pressure and I'll do some example problems recall the conservation Mass equation from a previous lesson what happens when the Mach number is exactly equal to one well this term becomes zero so d a over a equals 0 or D A equals zero so the answer is d a must equal zero when the flow is Sonic in other words when Mach number is one so area a must be either a maximum or a minimum when Mach number is one I sketch these two possibilities here where the flow is from left to right and let's examine whether it's possible to have Mach number one when the area is a maximum and when the area is a minimum in both case will consider the left side of the flow if the flow is subsonic in this diverging portion of the duct we know that Mach number goes down as area goes up it's a subsonic diffuser so Mach number cannot approach one since it's already subsonic and it's decreasing if the flow is Supersonic in this diverging duct we have the opposite behavior and Mach number goes up as area goes up it's a supersonic nozzle so again Mach number cannot approach one so this case is impossible now let's do the same analysis for the second case if the flow is subsonic in a converging duct again in this region The Mach number goes up as area goes up this is a subsonic nozzle so Mach number does indeed approach one if the flow is Supersonic in this converging duct Mach number goes down as area goes up so Mach number can indeed approach one this one is possible we conclude that for steady one-dimensional isentropic adiabatic duct flow from an upstream pressurized tank Sonic or Star conditions critical conditions can occur only at a throat a minimum area in the duct as in this sketch you cannot have Mach number one in a flow like this now let's discuss what's known as a converging diverging duct or a CD duct or a CD nozzle suppose there's some large Upstream pressurized tank with properties P naught T naught Etc the flow accelerates through this converging diverging duct to the left of the minimum area Mach number is less than one the flow in this region is subsonic right at the minimum area which we call the throat the flow is Sonic or critical and redenoted by an asterisk and this throat area is a star I need to make a caveat here we're assuming that the back pressure PB is low enough to generate this flow where PB is the pressure outside the duct the back pressure at the exit plane this area is AE later we'll be able to calculate pressure p e temperature te Etc once we go Downstream of the throat Mach number is greater than one flow is supersonic we note that both V and Mach number increase continuously through the CD nozzle all the way from zero accelerating through the subsonic portion Sonic at the throat and then continuously accelerating through the supersonic portion I'll make another statement supersonic flow can occur only Downstream of the throat you can't have supersonic flow here only here again that's assuming that the back pressure is low enough to generate this flow now I want to talk about the area ratio versus The Mach number relationship consider steady adiabatic isentropic 1D flow in a ductive variable area from a previous lesson we had these two equations for mass flow rate through a converging duct this is the general case where the flow can be choked or not this is the simplified case when the flow is choked and mass flow is maximum if you think about the converging part of this and ignore the diverging part those equations must still apply and once we reach Sonic conditions or choked flow nothing that happens down here influences anything that happens Upstream so both of these equations must hold in our converging diverging duct since both equations hold in our CD nozzle case we equate them in other words set this right hand side equal to this right hand side and solve for a over a star the area ratio as a function of Mach number and K the ratio of specific Heats after a little bit of algebra we solve for a over a star and we get this area ratio versus Mach number relationship this is a Workhorse equation for duct flows this is best Illustrated with an example problem air flows from a pressurized tank through a CD nozzle we assume that the back pressure is low enough that we have subsonic flow Upstream of the throat Sonic flow at the throat and supersonic flow Downstream of the throat as we've been discussing I also give the throat area we're asked to calculate the area at the location where Mach number is this value or this value I picked one subsonic and one supersonic case here's our list of assumptions and approximations here's the area ratio versus Mach number relationship we just derived we simply plug in our values of Mach number to four digits I get a over a star is 2.000 so a is a over a star times a star and the throat area in this case is a star so the area is 2.000 times a star I get 0.272 square meters and that's my answer for the subsonic case I do the same thing with the supersonic case and I get the same value of a over a star full disclosure I picked these Mach numbers such that this would happen again we use our ratio times a star to get the area and I get the exact same answer for the supersonic case since a over a star is the same for both of these Mach numbers notice that both of these Mach numbers give the same area ratio it turns out that for any a over a star ratio greater than one there are two roots for Mach number a subsonic root and a supersonic root graphically if we plot a over a star versus Mach number the Curve will be u-shaped with the dip at Mach number one since a star is the minimum area or the throat area a over a star cannot be less than one everything to the left of Mach number one is subsonic and everything to the right is Supersonic in our example problem when a over a star was 2 we get two Roots 0.3059 the subsonic root and 2.197 the supersonic root what if we try to go the other way in other words suppose we know this area ratio show and we want to find these two Roots here's our equation again this equation is explicit if we're given Mach number and we calculate a over a star this is what we just did in the example problem but this equation is implicit if given a over a star and we need to calculate Mach number you can see that Mach number occurs twice in this equation and this equation is ugly enough that you cannot solve for Mach number explicitly you're welcome to try so we must find a way to solve for Mach number implicitly how you can do it graphically you can generate this plot and zero in on these two values you can use trial and error you can use excel's what if analysis that can get you pretty close but I found out that it's not always real accurate if you have a fancy calculator or a software package that can solve implicit equations you can use that you can use Newton's method which is an iteration scheme that uses the derivative of the equation you can use the false position method which I'll abbreviate fpm this is the one I recommend for this equation false position method is a smart trial and error iteration technique that uses linear interpolation to converge more rapidly than simple trial and error keep in mind that with any of these methods there are two Roots one subsonic and one supersonic so you need to know where you are in this flow if you're Upstream of the throat you pick the subsonic route and if you're Downstream of the throat you pick the supersonic root if you're not familiar with the false position method I have a short YouTube video called false position method in Excel the URL is given here it's a three and a half minute video where I explain the false position method and I show how to set it up in Excel once you master this false position method you'll find yourself using it for all kinds of implicit equations I find myself using it a lot and once I set up my false position method in Excel I can copy and paste all this changing only the equation so far it has never failed me let's demonstrate this false position method with an example this example is actually the same as the previous example except in Reverse I give the same throat area and this time I give the cross-sectional area at a certain location and I want to calculate the Mach number there's actually two locations where where this cross-sectional area is this value since it's a converging diverging nozzle we have the same assumptions and approximations and we apply the same equation but this time in Reverse in the first case we were given Mach number and we just plugged it in explicitly to get a over a star now we know a over a star and I picked the answer that we got from the previous problem and now we need to solve this equation implicitly for Mach number first I calculate a over a star and I get 2.00 which is the same value we got in the previous example I do this for illustrative purposes we set this ratio as the goal in the false position method I used Excel and I got these two answers Mach number is 0.3059 for the subsonic root and Mach number equal 2.197 for the supersonic root here's a screenshot from my Excel spreadsheet and now I'll do a live demo of this here's my Excel spreadsheet where I use the false position method these highlighted cells are ones that I enter I entered the ratio specific Heats a star and a i define the goal as a over a star now I need to find the Mach number that gives me this area ratio for the subsonic case I pick two guesses here 0.3 and 0.7 that are subsonic this a over a star cell is just our equation for a over a star and I fill that down here once I have my two guesses the third one is an interpolation here's the equation for the interpolation using these two values in these two values I describe this in more detail in my short video about the false position method once I have that set up I can fill that one down as well at any cell it interpolates from the latest two guesses you can see that it quickly converges to our desired goal of two I copied and pasted for the supersonic case the only difference is that I guess two supersonic values and I get the supersonic root this root and this root agree with the values we calculated the other way around I want to show you now that the initial guesses are not very critical let's change this to 0.4 I still get the right answer even though my answer does not lie in between these two values I can change this to 0.1 and I still get the same value if I guess something real close like 0.3 and 0.35 it converges much more rapidly this also works with the supersonic case suppose I pick 2.5 and 3 it converges if I pick 2.5 and 5 it still converges if I pick 2 and 2.2 it converges very rapidly since we're very close to the converged value finally I want to discuss what happens when we vary the back pressure we'll do a thought experiment we slowly lower back pressure PB and discuss qualitatively what happens in this CD duct here's the setup converging diverging nozzle we plot pressure as a function of X where we align the plot vertically with the nozzle this red line is the throat and the black line is the exit area we also plot Mach number similarly aligned we label cases a through G and I'll make some comments a is the case with no flow nothing's happening at all P at the exit plane P A is equal to PB is equal to P naught this is the trivial case case b is subsonic PB is equal to P little B Which is less than P naught as you can see here so we get a flow but this flow is subsonic everywhere the pressure decreases and then increases through the throat similarly Mach number Rises and then decreases this flow is subsonic everywhere Case C is Sonic at the throat but just barely so PC equal PB equal just low enough to choke the flow pressure goes down through the converging portion hits P star at the throat but PC is up here and the flow remains subsonic all the way to the exit plane Mach number reaches one at the throat but then decays again in the diverging part the flow is subsonic everywhere except right at the throat let's jump to condition F and then I'll go back and fill in d and e f is the case where the flow is supersonic everywhere through the nozzle we call this ideal expansion where PF equal PB and this back pressure is so low that the flow can go supersonic all the way through the nozzle pressure decreases continuously and Mach number increases continuously this flow is supersonic through the diverging part of the duct what about case d the pressure here is in between PF and PC and we get a normal shock somewhere in the diverging part of the duct a normal shock has a sudden rise in pressure and a sudden decrease in Mach number so D has a shock in the diverging portion it turns out that the location of this shock will continuously move Downstream as you continuously decrease back pressure there's a kind of special case we call E where the shock is right at the exit plane again we see a pressure rise and a Mach number drop at condition e finally condition G is the case where PB is very low in fact PG is less than PF so PG is somewhere down here in this case you get shocks and complicated flow Downstream of the exit and you have a supersonic jet we'll talk more about shocks in the next lesson I want to make one final comment namely for any case with back pressure less than PC the flow is choked nothing we do Downstream of this throat affects anything Upstream of the throat so once we reach condition C the mass flow rate has reached its maximum regardless of whether there are shocks back here or not thank you for watching this video please subscribe to my YouTube channel for more 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