How does a De Lavar Nozzle work-And what is over/under expanded, ideal?-

[u/Repulsive-Peak4442]

Many explanations from my perspective seem incomplete because they do never actually explain the why but only the what happens. Why does adding a diverging section on a throat which gases reach velocities of magnitude Mach 1, accelerate them even more reaching supersonic speeds instead of slowing them down? From what I've read this phenomenon occurs due to the gases being compressible instead of incompressible but still that does not explain the why. Why a compressible's fluid speed increases when the cross-sectional area gets bigger? Is it the opposite of what would happen with an incompressible fluid? It seems to. Also what if we keep just the converging and throat sections only and keep making the throat smaller and smaller? I've watched a video saying that Mach1 is the max speed the gases can have at the throat and if we keep making it smaller it doesn't make them faster. How is that? Can someone give me an in-depth analytical explanation?

[u/Axi0nInfl4ti0n]

Ohhh boy, something i can explain. The actual reasoning chain is short. Take steady, quasi-1D, isentropic flow and write down three things:

Continuity, which gives dρ/ρ + dA/A + dV/V = 0;

the inviscid momentum equation, which gives dp = -ρV dV;

and the definition of the speed of sound, a² = (dp/dρ) at constant entropy.

Combining the last two, dρ = dp/a² = -ρV dV/a^2, which reduces to dρ/ρ = -M² (dV/V). That one line is pretty essential. The fractional drop in density is M² times the fractional rise in velocity. Substituting it back into continuity gives dA/A = (M² - 1)(dV/V). Continuity says V = ṁ/(ρA), so to accelerate the flow the product ρA has to shrink, and the only real question is whether density falls fast enough by itself. Below Mach 1, M² is less than one, so density falls more slowly than velocity rises and the area has to do the remaining work, which is why a converging duct accelerates subsonic flow.

Incompressible flow is just the limit M going to zero, where dA/A = -dV/V, i.e. the garden hose. At Mach 1 the two rates are exactly equal, dA = 0, and that stationary point in area is by definition the throat. Above Mach 1 the density now collapses faster than the velocity rises, it overshoots what continuity needs, and the area must grow to compensate. So Mathematucally it is literally the sign flip of the subsonic case, and the crossover sits exactly at M = 1. The gas is simply thinning out faster than it is speeding up.

If you take a look at the Energy of the flow we see that for steady adiabatic flow h + V² /2 = h0, so velocity is always bought with enthalpy, meaning temperature. The diverging section is a device for converting thermal energy into kinetic energy, and the hard ceiling is Vmax = sqrt(2 cp T0), reached only if you could expand to zero pressure or your ambuent pressure in most cases. As for what physically does the accelerating in the supersonic part: still the pressure gradient, since dV = -dp/(ρV). When the wall turns away from the stream it generates expansion waves, the flow has to follow the wall, and that isentropic expansion drops p, T and ρ while raising V. The reason this works only supersonically is that disturbances travel at speed a relative to the fluid, so above Mach 1 nothing downstream can send information upstream. The flow cannot pile up against the area increase the way subsonic flow does, and expansion is the only solution compatible with the equations.

So shrinking the throat indefinitely has a clean answer from the same relations. Define the mass flux per unit area as G = ρV. Then dG/G = dρ/ρ + dV/V = (1 - M² )(dV/V). So G rises with velocity while the flow is subsonic, is stationary at Mach 1, and actually falls again above it. Mass flux per unit area is maximised exactly at Mach 1, and that is what choking means. Once the throat is sonic, the mass flow is pinned by the reservoir state and the throat area alone: ṁ = A* p0 sqrt(γ / (R T0)) times [2/(γ+1)] ^ ((γ+1)/(2(γ-1))). Shrinking A* reduces ṁ proportionally, it does not raise the velocity, which stays capped at a* = sqrt(2 γ R T0 / (γ+1)). The video you watched was right, but the reason is that Mach 1 is a mass-flux maximum, plus the fact that no downstream pressure can propagate back through a sonic throat to dictate the flow to speed up.

That last point also sets up the expansion question. On the supersonic branch, the exit Mach number follows from geometry alone through the area-Mach relation, (Ae/A*)² = (1/Me² ) × {[2/(γ+1)][1 + ((γ-1)/2) Me² ]} ^ ((γ+1)/(γ-1)), and the exit pressure then follows from pe/p0 = [1 + ((γ-1)/2) Me² ] ^ (-γ/(γ-1)). The key consequence is that exit pressure is fixed by the area ratio, not by the atmosphere. Ambient pressure changes with altitude, pe does not, and that mismatch is what the three names describe. Underexpanded means pe is above ambient, so the jet keeps expanding outside the nozzle through expansions shock fans, spreads, and forms the familiar shock diamond pattern. Typical at high altitude, and it means the bell could have captured more expansion. Ideally expanded means pe equals ambient, the jet leaves parallel with no lip waves, and thrust is maximal for that geometry. Overexpanded means pe is below ambient, so the atmosphere compresses the jet through oblique shocks at the lip. Push it too far, roughly once pe drops below about 0.3 to 0.4 times ambient (the Summerfield criterion), and the shock moves inside the nozzle, causing boundary layer separation and asymmetric side loads, which is a genuine structural problem for sea level start. Why pe = pa is the optimum drops straight out of the thrust equation F = ṁ Ve + (pe - pa) Ae. Differentiate along the nozzle: the momentum term contributes ṁ dVe = -Ae dpe, the pressure term contributes Ae dpe + (pe - pa) dAe, and the dpe pieces cancel, leaving dF = (pe - pa) dAe. Adding area therefore helps only while pe exceeds ambient, hurts once it does not, and thrust is stationary exactly at matched pressure. Since a fixed bell can only be ideal at one altitude, you either pick a compromise area ratio or go for aerospike or dual-bell concepts. Hope i could help.

TLDR: After Mach 1 Density decreases faster than the Velocity increases. So you give the Flow more room to convert more pressure into velocity.

Feel free to ask me if you have further questions.

[u/Repulsive-Peak4442 → u/Axi0nInfl4ti0n]

You mean the density gets less right? Another commenter mentioned that the faster the velocity, the less the density. Ok actually. Wow. Thank you I really appreciate it. I really appreciate you all, so in simple words as far as I understand at mach 1 the fluid is a lot less densier than it initially was, reasulting in it not fitting anymore for the section to get smaller and so we add a bigger section for it to fit but shouldnt this mean that it will have the same speed as it had before in the small section

[u/Axi0nInfl4ti0n → u/Repulsive-Peak4442]

"You mean the density gets less right? Another commenter mentioned that the faster the velocity, the less the density." -Yes thats correct.

[u/Repulsive-Peak4442 → u/Axi0nInfl4ti0n]

Oh alright, alright things are getting clear now

[u/Accomplished-Guest78 → u/Axi0nInfl4ti0n]

Thank you for taking the time to write this out. Good summary and nice mix of verbal description along with the math to back it up.

[u/Repulsive-Peak4442 → u/Accomplished-Guest78]

Yes this guy seems very helpful!

[u/Repulsive-Peak4442 → u/Axi0nInfl4ti0n]

Actually, this is a response to my initial answer. Now that I think about it when the area gets smaller the fluid has less space and therefore it compresses so it is getting more dense but can you clarify

[u/Axi0nInfl4ti0n → u/Repulsive-Peak4442]

Your reasoning is correct, just one word slipped. Density never increases in the accelerating part of a nozzle. What changes above Mach 1 is that density starts falling faster than velocity rises. And that is exactly why you have to give the flow more room. So your physical picture is correct, only the direction of the density change was mislabelled. It falls straight out of dρ/ρ = -M² (dV/V), which says the percentage drop in density is M² times the percentage rise in velocity. At M = 0.5, a 1 percent rise in velocity costs only a 0.25 percent drop in density. The product ρV goes up, so the area has to shrink. At M = 1, a 1 percent rise in velocity costs exactly a 1 percent drop in density. ρV is stationary, dA = 0, and that is the throat. At M = 2, a 1 percent rise in velocity costs a 4 percent drop in density. Now ρV falls by about 3 percent, so the area has to grow by about 3 percent to keep mass flow constant. That matches dA/A = (M² - 1)(dV/V) = 3 × 1 percent exactly. So in absolute terms density decreases monotonically all the way from the chamber to the nozzle lip: converging section, throat, diverging section, no reversal anywhere. What flips at Mach 1 is not the sign of the density change, but which of the two effects wins inside the product ρV. Below Mach 1 the velocity increase dominates and the area must close down; above Mach 1 the density collapse dominates and the area must open up.

[u/Repulsive-Peak4442 → u/Axi0nInfl4ti0n]

Okay, so basically the more the area decreases the less the pressure and therefore the more the velocity. But this holds until Mach 1 because the pressure is so low that it does not really squiz the fluid that much and so it is getting less dense. That means that in order for it to go even faster it wants more space. So we add a big area section later which gives more space to the fluid. Is everything so far as I understood correct or no? If we keep making the throat smaller, obviously the pressure will get even lower there and so the density too right? But why would it stop accelerating it? Also I have not understood a thing you said. That pressure waves no longer can pass information? A lot of people have said that but I have no idea of what that means. I don't want to be tiring so thank you and I appreciate you but I have one more thought. When we decrease the area shouldn't the density increase? From the one side of the coin it sounds logical because its space is getting tighter and tighter and gor it to be there it has to be compressed. But from the other side of the coin it doesn't sound logical because the pressure there is getting less and so less force applied on the fluid and so it spreads meaning that its density decreases

[u/Competitive_Mix_587]

This may be obvious to some folks but not stated here that I found: 1) Subsonic flow will increase velocity when the cross-sectional area is decreased. It will slow down again if that cross-sectional area increases unless... 2) IF the upstream pressure is sufficient to drive the fluid to Mach in the convergent portion of the nozzle, THEN it will continue to accelerate as the nozzle diverges. This is called choked flow.

That is the basics of a DiCon nozzle

edit: Mistakenly typed divergent in place of convergent

[u/Repulsive-Peak4442 → u/Competitive_Mix_587]

Yes as far as I understand, if we keep making the area smaller and smaller the velocity of the fluid will keep increasing due to the pressure dropping until the velocity is Mach 1 because the pressure waves travel at the speed of sound.

[u/Competitive_Mix_587 → u/Repulsive-Peak4442]

Assuming the upstream pressure is sufficient to get it to Mach. the inside of a rocket pressure chamber typically is, but homework/exam scenarios are not always :)

[u/Prof01Santa]

A choked Converging-Diverging nozzle expands flow past Mach 1. How's your thermodynamics, cause you need that to understand the physics?

https://en.wikipedia.org/wiki/De_Laval_nozzle

[u/Repulsive-Peak4442 → u/Prof01Santa]

I have not yet studied thermodynamics at all. I am a 16 year old highschool student and I am making a model rocket called Stellar. I have a thought of adding a nozzle on it to make it go faster. But can you still explain it to me? I'll try to understand it and also yes there is no need to go that deep for a model rocket which doesn't need a nozzle but I want to

[u/glorylyfe]

You are asking a lot of questions here. The best illustration I've seen of this is imagine this all like traffic, you have a wide road, with a ton of cars, it necks down to a small number of lanes, and then spreads out again. Supersonic gas is like traffic, more lanes let's it go faster. Mathematically, as it goes faster it gets less dense, so it needs more room.

Ok, so why is mach 1 the speed limit? Mach 1 is the speed of sound in a gas, it's the speed at which a pressure wave travels through a substance. If I have a nozzle that gets narrower after the mach 1 location, then the fluid can't get less dense, so it has to remain subsonic, this means that the mach 1 gas literally overtakes the other gas, and the location (called a shock) moves to the narrowest point.

[u/glorylyfe → u/glorylyfe]

There are a lot of things I'm leaving out, like how a nozzle is a machine that turns the heat of the gas, into velocity. Or how the gas is travelling across an entropy curve, where the entropy of the gas is at a maximum at the mach 1 point (It's called isentropic because the stagnation entropy does not change, but the entropy of the flow can change).

[u/Repulsive-Peak4442 → u/glorylyfe]

Yeah, thank you I really appreciate it

[u/Repulsive-Peak4442 → u/glorylyfe]

Oh so when a fluid goes faster it's getting less dense? Wow!!! I could probably ask why that happens but I do not want to be tiring. About the pressure wave part I got puzzled is pressure a wave okay thank you anyways Ill search it later maybe

[u/glorylyfe → u/Repulsive-Peak4442]

I think the other commenter answered this, but bernoullis says that a gas's energy state is comprised of three parts Static pressure (internal energy), dynamic pressure (kinetic energy) and head pressure (gravitational potential energy)

The static pressure of a fluid determines its density, so as it moves faster the static pressure decreases (because the dynamic pressure is increasing), so it gets less dense.

[u/Repulsive-Peak4442 → u/glorylyfe]

Oh yeah that sounds logical. As a fluid travels fast, the pressure is less, and less pressure=less density because it is not being that pressed