The Traditional Bernoulli Eqn Derivation of Rocket Exhaust Velocity

[BrandonInFlorida]

If you've seen it, they chose one point in the combustion chamber and the other in the exhaust nozzle. I think they're assuming that we have a gas both places. They say that the pressure in the nozzle is atmospheric pressure, or it you're in outer space, zero. That makes perfect sense. That's not my problem. My problem is that they assume that the gas density is the same in the nozzle as in the tank. They only use that one density variable.

This doesn't seem remotely obvious to me. I would think that the gas density might be less in the nozzle. Why should I think it's the same as in the tank? And if you tell me to pick a point that's just differentially out of the combustion chamber, wouldn't that undermine the idea that we have atmospheric (or zero) pressure there?

 

[Astronuc]

They say

Who they?

The pressure in the nozzle is certainly greater than the environment into which the propellant is exhausted. At some boundary of the 'plume' outside the nozzle exit, the pressure must equal the ambient environment.

 

[BrandonInFlorida]

Who they?

The pressure in the nozzle is certainly greater than the environment into which the propellant is exhausted. At some boundary of the 'plume' outside the nozzle exit, the pressure must equal the ambient environment.

It's certainly greater than the environment, but in every derivation of this type, they assume that it's identical to the pressure in the tank and that's by no means obvious.

 

[Astronuc]

but in every derivation of this type, they assume that it's identical to the pressure in the tank and that's by no means obvious.

Please provide an example or link to such a statement.

Is one referring to liquid propellants, e.g., liquid oxygen (LOX) and liquid hydrogen? Or LOX and RP-1?

In the Saturn V F-1 main engine, the chamber pressure is 70 bars (1,015 psi; 7 MPa), or ~69 atm, which is much greater than 1 atm at sea level.
https://en.wikipedia.org/wiki/Rocketdyne_F-1

The chamber pressure on the Space Shuttle Main Engine, Rocketdyne's RS-25, is even greater at 2,994 psi (20.64 MPa), or 203.6 atm. https://en.wikipedia.org/wiki/RS-25

If a tank sits vertically, and is vented to the atmosphere, then the pressure at the top of the liquid oxidant and fuel is at atmospheric pressure. At the bottom of the take, it is greater due to the weight of the liquid above wherever a measurement is taken. If a rocket accelerates, the pressure in the liquid increases as gravity and the acceleration of the rocket (tank) act on the liquid.

 

[BrandonInFlorida]

Let me make a more modest claim - in the book I'm reading, "Physics" by Resnick and Halliday (I don't have it in front of me this moment), they use the Bernoulli equation with the two points chosen being (1) a random place in the section before the nozzle, which I guess is the combustion chamber, and(2) somewhere in the nozzle. They use only one variable (rho) for density in both places. It seems to me that the density in the nozzle might instead be less.

 

[Astronuc]

Let me make a more modest claim - in the book I'm reading, "Physics" by Resnick and Halliday (I don't have it in front of me this moment), they use the Bernoulli equation with the two points chosen being (1) a random place in the section before the nozzle, which I guess is the combustion chamber, and(2) somewhere in the nozzle. They use only one variable (rho) for density in both places. It seems to me that the density in the nozzle might instead be less.

Does it look like the discussion in the following?
https://www.grc.nasa.gov/www/k-12/airplane/bern.html
https://www.grc.nasa.gov/www/k-12/rocket/nozzle.html

See also - https://www.grc.nasa.gov/www/k-12/airplane/mass.html

Liquids are mostly incompressible, while gases are compressible. A compressed gas will have a greater density than an uncompressed gas (conservation of mass).

See also this discussion - https://sites.astro.caltech.edu/~srk/ay125/Nozzle.pdf

 

[BrandonInFlorida]

Actually, it doesn't look like my book's derivation. I'm hindered somewhat by not being able to write equations here with Tex or Lex or whatever, but as well as I can type it, its:

p1 + 1/2(rho)(v1)^2 = p2 + 1/2(rho)(v2)^2

Note that they use only a single variable, rho, for the density, on both sides, which is equivalent to saying that the density in the nozzle is equal to the density in the combustion chamber, which is by no means clear to me.

 

[anorlunda]

I'm hindered somewhat by not being able to write equations here with Tex or Lex or whatever

That's not true. You can use LaTex. There is a LaTex guide button at the lower left corner of the post edit window.

 

[Astronuc]

Actually, it doesn't look like my book's derivation. I'm hindered somewhat by not being able to write equations here with Tex or Lex or whatever, but as well as I can type it, its:

p1 + 1/2(rho)(v1)^2 = p2 + 1/2(rho)(v2)^2

Note that they use only a single variable, rho, for the density, on both sides, which is equivalent to saying that the density in the nozzle is equal to the density in the combustion chamber, which is by no means clear to me.

That is the form in https://www.grc.nasa.gov/www/k-12/airplane/bern.html, where they have a statement:

Assuming that the flow is incompressible, the density is a constant. Multiplying the energy equation by the constant density . . . . (equation).

The article then states, "This is the simplest form of Bernoulli's equation and the one most often quoted in textbooks. If we make different assumptions in the derivation, we can derive other forms of the equation." In other words, if ρ is a function of position, the equation can be modified.

Earlier in the article, one finds the statement:

In the 1700s, Daniel Bernoulli investigated the forces present in a moving fluid. This slide shows one of many forms of Bernoulli's equation. The equation appears in many physics, fluid mechanics, and airplane textbooks. The equation states that the static pressure ps in the flow plus the dynamic pressure, one half of the density r times the velocity V squared, is equal to a constant throughout the flow. We call this constant the total pressure pt of the flow.

So the total pressure at any point in the control volume is assumed constant (from outlet of turbopump to nozzle exit. As the dynamic pressure increases, the static pressure decreases, assuming the total pressure is constant.

Consider a solid rocket. At the closed end the velocity of the propellant gases is zero, so at that location all the pressure is static. At the exhaust location, much of the pressure is dynamic, where velocity is the greatest.

 

[BrandonInFlorida]

A lot of the things you're telling me are new information for me and I appreciate your ongoing effort to help. First of all, though, my question involves only density, not pressure. I don't mean to seem dense (no pun intended), but my intuitive feeling is that in a flowing (or non-flowing) incompressible fluid, the density must be constant, but in the nozzle, the fluid is sort of sprayed into the air (yes, I know burning gases are expanding too) and when a flowing liquid gets sprayed out, the density would likely become less.

 

[Astronuc]

First of all, though, my question involves only density, not pressure.

Equation of state (ideal gas): P/ρ = RT, where P is pressure, ρ = density, R = gas constant for specific gas, T = Temperature (abs.) , or ρ = P/(RT), so local density and pressure are interrelated.
https://www.grc.nasa.gov/www/k-12/airplane/eqstat.html

Note the equations for Conservation of momentum and isentropic flow.
https://www.grc.nasa.gov/www/BGH/nozzled.html
https://www.grc.nasa.gov/www/BGH/isentrop.html

Non-isentropic flow
http://mae-nas.eng.usu.edu/MAE_5420_Web/section5/section5.2.pdf

 

[Astronuc]

The chamber pressure on the Space Shuttle Main Engine, Rocketdyne's RS-25, is even greater at 2,994 psi (20.64 MPa), or 203.6 atm. https://en.wikipedia.org/wiki/RS-25

The same article indicates an operating temperature up to 3300°C.

Note: At the critical point there is no change of state when pressure is increased or if heat is added. At the critical point water and steam can't be distinguished and there is no point referring to water or steam.

The critical point of water is achieved at

• Water vapor pressure of 217.75 atm = 220.64 bar = 22.064 MPa = 3200.1 psi
• Temperature of 647.096 K = 373.946 °C = 705.103 °F
• Critical point density: 0.322 g/cm³ = 0.6248 slug/ft³ = 20.102 lbm/ft³

Ref: https://www.engineeringtoolbox.com/critical-point-water-steam-d_834.html

Although the chamber pressure of the SSME (2994 psi (20.64 MPa)) is less than the critical pressure, 3200.1 psi (22.064 MPa), the maximum temperature (3300°C) is considerably greater than the critical temperature ~374°C. Supercritical water is compressible.

https://water.lsbu.ac.uk/water/supercritical_water.html

It should be noted that the HPFTP, a three-stage centrifugal pump driven by a two-stage hot-gas turbine, boosts the pressure of the liquid hydrogen from 1.9 to 45 MPa (276 to 6,515 psia). The HPOTP second-stage pre-burner pump to boost the liquid oxygen's pressure from 30 to 51 MPa (4,300 psia to 7,400 psia). Both systems are well above the combustion chamber pressure of 20.64 MPa (2,994 psi), so they have to equilibrate at some point.

 

[BrandonInFlorida]

This seems pointless because your answers never address my one and only question. If a fluid which has been enclosed is sprayed out into the air, one cannot claim that the density won't be reduced. Please don't link me to the great library of Alexandria or something. My question is very specific.

 

[256bits]

If you've seen it, they chose one point in the combustion chamber and the other in the exhaust nozzle. I think they're assuming that we have a gas both places. They say that the pressure in the nozzle is atmospheric pressure, or it you're in outer space, zero. That makes perfect sense. That's not my problem. My problem is that they assume that the gas density is the same in the nozzle as in the tank. They only use that one density variable.

This doesn't seem remotely obvious to me. I would think that the gas density might be less in the nozzle. Why should I think it's the same as in the tank? And if you tell me to pick a point that's just differentially out of the combustion chamber, wouldn't that undermine the idea that we have atmospheric (or zero) pressure there?

Do you have choked flow in the nozzle?
Or sub-sonic?
If subsonic, the bernouilli equation would be more a first equation calculation, and then refine from there.
If sonic flow at the throat, and further down the divergent region bernouilli is not quite applicable due to the choked flow.

BYW, the nozzle, if it is a Laval nozzle, convergent-divergent includes both sections, and not just the outer expanding part.

Not really an expert on this, but just wanted to point out a few things.

 

[BrandonInFlorida]

You sound like an expert. What do you do?

On Oct. 1, 2020, I retired from my job as a software engineer, which I'd done for several decades. Before that I was a physicist or engineer at various jobs. I have a BS and MS in physics.

My retirement project is to learn the entirety of physics, and I work on it seven days a week (usually). Obviously, the goal is unattainable, but I'm doing as much as I can.

 

[anorlunda]

My retirement project is to learn the entirety of physics, and I work on it seven days a week (usually).

Are you familiar with Leonard Susskind's video lectures on YouTube?
It was an evening course specifically designed for senior technical scientists and engineers who wanted to understand modern physics. Not for grad students or practicing physicists working toward a degree or a job.

Susskind said, "It's called the theoretical minimum because the students are seniors who can't be certain how much time they have left to learn."

https://theoreticalminimum.com/courses

He also has videos on particle physics, string theory, M theory, and super symmetry.

In total it is roughly 90 two hour lectures.

 

[BrandonInFlorida]

I'll take a look. My review isn't quite up to modern physics yet. I wish I could find the Feynman lectures (books) on physics. I glanced through them about 50 years ago, but I don't remember anything. I'm not going to shell out the money a copy costs today with no memory of how good they were.

 

[Frabjous]

Caltech put them online
https://www.feynmanlectures.caltech.edu/

 

[BrandonInFlorida]

Caltech put them online
https://www.feynmanlectures.caltech.edu/

How wonderful! Thanks. I wonder about the copyright.

 

[Astronuc]

Thanks. I wonder about the copyright.

I believe Caltech has the copyright.

 

[berkeman]

How wonderful! Thanks. I wonder about the copyright.

Click the copyright link at the bottom of the Caltech web page. It looks like they are fine for "personal review" use, but forbid any other redistribution type of use. Thank you Caltech.

https://www.feynmanlectures.caltech.edu/III_copyright.html

 

[Benjies]

I'd really venture to say that applying Bernoulli's principle to modern rockets is actually the crux of the entire issue and confusion. As some others above me have stated, this is a principle that can be applied to incompressible flows. So, two main issues trying to get Bernoulli's principle to "make sense" with a rocket:

1. Flow is compressible. There is a normal shock at the throat, and supersonic flow downstream. The moment flow encounters a normal shock the density is absolutely changing.
2. Combustion is imperfect and non-frozen. Frozen flow is an assumption that is made for rocket modeling where it is assumed that there is no further chemical reactions occurring downstream of the combustion chamber. This is in fact not realistic, so if our chemical makeup is changing across our flow, then our density is changing.

SO- let's wind it back to the title of this thread. Deriving the exhaust velocity using Bernoulli's principle is insufficient due to the two points discussed above. I would echo what 256bits said above me and ask whether or not station 1 in the system you are describing is actually a point after the throat normal shock in the nozzle, rather than the combustion chamber. If we assume frozen flow and change station 1 to this location, then Bernoulli's principle can be fielded in a theoretically perfect system.

 

[cjl]

1. Flow is compressible. There is a normal shock at the throat, and supersonic flow downstream. The moment flow encounters a normal shock the density is absolutely changing.

I certainly hope there's not a normal shock at the throat - that would be terrible for the rocket motor efficiency. You also don't need a shock to have a density change - merely a flow acceleration from stagnation(ish) up to any speed above about mach 0.3 will give a non-negligible density change. In this case, the flow is accelerating to several times the speed of sound (and is at mach 1 at the throat), so you definitely cannot assume anything close to a constant density here (not to mention the fact that the flow isn't entirely frozen, as you stated, and that there's actual energy release happening as the combustion continues to occur through the flow).

 

[Benjies]

I certainly hope there's not a normal shock at the throat - that would be terrible for the rocket motor efficiency. You also don't need a shock to have a density change - merely a flow acceleration from stagnation(ish) up to any speed above about mach 0.3 will give a non-negligible density change. In this case, the flow is accelerating to several times the speed of sound (and is at mach 1 at the throat), so you definitely cannot assume anything close to a constant density here (not to mention the fact that the flow isn't entirely frozen, as you stated, and that there's actual energy release happening as the combustion continues to occur through the flow).

Good catch. No normal shock, just sonic condition at the throat. Sorry, that is a pretty gross miss-remembrance of the function of a nozzle.

 

[FluidDroog]

Hi Brandon,

Let me give it a shot.

Bernoulli's equation is a conservation of energy statement for fluids. The write up in reference [1] is quite good and brief. Take a look at the list of assumptions needed to achieve such a beautifully simple (and powerful) equation.

Many textbooks are teaching you to start with the simplest equation, since most of the time it is accurate 'enough' for most problems. The important matter for the Bernoulli equation is to understand the conservation of energy statement for a fluid flow stream.

If any of those assumptions is not true, then the equation becomes more complex to account for that. For example, the Wikipedia describes the Bernoulli equation for compressible fluids [2]. As Benjies has stated, Bernoulli's principle with the incompressible assumption is not valid for nozzles since the Mach # is large.

Astronuc has sent the relevant equations for Isentropic flow through nozzles, and as you can see it will get complicated. If you want to go learn more about this subject, I enjoyed Genick Bar–Meir's textbook [4]. Otherwise the Wikipedia might suffice [3].

References:
[1] What is Bernoulli's Equation? https://www.thermal-engineering.org... law?msclkid=09d77c14ae4911ecb02ad3cdcb6575a3
[2] Wikipedia Bernoulli's Principle https://en.wikipedia.org/wiki/Berno...eca3fba2c417b0eaae#Compressible_flow_equation
[3] Compressible Flow https://en.wikipedia.org/wiki/Compr...se).?msclkid=c8acffbcae4e11ec920a1b6581e637e9
[4] Fundamentals of Compressible Fluid Mechanics http://www.bigbook.or.kr/bbs/data/f....pdf?msclkid=e313660fae5011ecb55be72def8570d2