I Rocket Fuel Ejection: Intuitive & Math Explained

[Filip Larsen]

Rockets combust propellent and expel exhaust gases from within the rocket's frame of reference not the instantaneous CM frame

As jbriggs444 already mentioned it doesn't really matter which frame you choose, as long as you then stay in that frame for both the pre- and post-ejection momentum and energy equations. But if you for the single-chunk ejection choose to go with the rocket frame (i.e. the frame in which the rocket is at rest after the ejection) that means you will have to include the pre-ejection speed full rocket, as its speed will not be zero in that frame.

And correcting for relative speed is very simple even in the CM frame: $m_r v_r = m_f (v_e - v_r)$. There, that's all you need for the speed constrained model.

 

[gary350]

Liquid fuel rocket engines have a fuel pump they pump about 30 tons of fuel & oxygen through the engine in about 65 seconds. At 4000 mph when engine stops about 65 miles up the rocket will coast on up another 70 miles. There is a very good British YouTube video on the detailed workings of the V2 engines and how it works.

 

[bob012345]

Things happen in all reference frames, not just in some.

Sure, I didn't contradict that at all. I could compute things from the surface of the Moon but it would complicate the problem.

The invariants of the problem such as the total increment to kinetic energy as a result of burning a prescribed quantity of fuel and ejecting a particular sized packet of exhaust are invariant. They do not depend on the choice of reference frame. The change in payload velocity as a result of a particular burn is also an invariant.

The choice of reference frame within which to do the analysis is completely arbitrary. All choices will yield identical results for all invariants of the problem.

The CM frame is a nice choice because of the symmetry it offers and the fact the the pre-burn momentum and kinetic energy are both zero.

I know everything you're saying and do not dispute it but that is not at issue here. Of course there are invariants. Of course one can look at the problem in the CM frame. It is very useful.

The issue is whether @Filip Larsen's constant energy scenario is equivalent to the constant relative exhaust velocity scenario or different and if the CE scenario is necessary to use to be most equivalent to a rocket behavior for the one shot scenario.

 

[bob012345]

As jbriggs444 already mentioned it doesn't really matter which frame you choose, as long as you then stay in that frame for both the pre- and post-ejection momentum and energy equations. But if you for the single-chunk ejection choose to go with the rocket frame (i.e. the frame in which the rocket is at rest after the ejection) that means you will have to include the pre-ejection speed full rocket, as its speed will not be zero in that frame.

And correcting for relative speed is very simple even in the CM frame: $m_r v_r = m_f (v_e - v_r)$. There, that's all you need for the speed constrained model.

If you are riding the rocket it is always at rest with respect to you. That's the point of being on a rocket.

The thrust a rocket produces is always acting on the rocket no matter where you look at it from. It is always proportional to mass flow times exhaust velocity (in our simplified model). Rockets are capable of constant thrust which creates increasing acceleration as the rocket mass decreases.

 

[Filip Larsen]

The thrust a rocket produces is always acting on the rocket no matter where you look at it from. It is always proportional to mass flow times exhaust velocity (in our simplified model). Rockets are capable of constant thrust which creates increasing acceleration as the rocket mass decreases.

I am not sure what argument you are trying make here bringing forces into it, or the point of your last few posts actually. We all here know how this stuff works and you say you do too so I fail to see why you bring it up. If you prefer to calculate in whatever frame you choose then its fine by me. I'm just pointing out the CG frame is a very convenient choice, both with and without energy constraints.

 

[bob012345]

I am not sure what argument you are trying make here bringing forces into it, or the point of your last few posts actually. We all here know how this stuff works and you say you do too so I fail to see why you bring it up. If you prefer to calculate in whatever frame you choose then its fine by me. I'm just pointing out the CG frame is a very convenient choice, both with and without energy constraints.

We completely agree on the constrained $v_e$ model I believe. It is better to do the slow burn than the single shot.

I understand why you proposed the constrained energy model I'm just not yet convinced it is a fair comparison. Give me more time to work through calculations please.

 

[Filip Larsen]

Give me more time to work through calculations please.

No problem.

In the mean time I took some time calculating the total kinetic energy in the CM frame of the continuous rocket. During the burn the speed of the rocket is $$v_r = -v_e \ln r ,$$ where $r = m_r / m$ is the fraction of the mass left in the rocket. This means the kinetic energy of the rocket is $$E_r = \frac{1}{2} r m v_r^2 = \frac{1}{2}m v_e^2 r \ln^2 r $$ and the kinetic energy of a small fuel element $mdr$ at a specific value for $r$ where the fuel is ejected with the speed $v_e - v_r$ relative to the CM frame is $$dE_f(r) = \frac{1}{2} m dr (v_e - v_r)^2 = \frac{1}{2} m v_e^2 (1 + \ln r)^2 dr , $$ which integrated gives the total kinetic energy of the ejected fuel cloud as $$ E_f(r) = \frac{1}{2} m v_e^2 \int_r^1(1+\ln r)^2 dr = \frac{1}{2}m v_e^2 (1-r - r\ln^2 r). $$ The sum of the two is then the total kinetic energy $$E_k(r) = \frac{1}{2} m v_e^2(r \ln^2r - r \ln^2r + 1-r) = \frac{1}{2} (1-r)m v_e^2, $$ which is then the total usable mechanical energy released from combusting the fuel mass $(1-r)m$. This is also exactly the energy I used earlier for the discrete energy constrained model, so the kinetic energy coming out of that model match the continuous case exactly.

 

[bob012345]

No problem.

In the mean time I took some time calculating the total kinetic energy in the CM frame of the continuous rocket. During the burn the speed of the rocket is $$v_r = -v_e \ln r ,$$ where $r = m_r / m$ is the fraction of the mass left in the rocket. This means the kinetic energy of the rocket is $$E_r = \frac{1}{2} r m v_r^2 = \frac{1}{2}m v_e^2 r \ln^2 r $$ and the kinetic energy of a small fuel element $mdr$ at a specific value for $r$ where the fuel is ejected with the speed $v_e - v_r$ relative to the CM frame is $$dE_f(r) = \frac{1}{2} m dr (v_e - v_r)^2 = \frac{1}{2} m v_e^2 (1 + \ln r)^2 dr , $$ which integrated gives the total kinetic energy of the ejected fuel cloud as $$ E_f(r) = \frac{1}{2} m v_e^2 \int_r^1(1+\ln r)^2 dr = \frac{1}{2}m v_e^2 (1-r - r\ln^2 r). $$ The sum of the two is then the total kinetic energy $$E_k(r) = \frac{1}{2} m v_e^2(r \ln^2r - r \ln^2r + 1-r) = \frac{1}{2} (1-r)m v_e^2, $$ which is then the total usable mechanical energy released from combusting the fuel mass $(1-r)m$. This is also exactly the energy I used earlier for the discrete energy constrained model, so the kinetic energy coming out of that model match the continuous case exactly.

Nice integration!

We agree that the slow burn rocket's total energy conversion yields a total amount of energy of $\frac{1}{2} m_f v_e^2$. Now if we combust that same fuel simultaneously, assuming we can do that with some super strong engine without adding mass, it should also have an exhaust velocity of $v_e$ since that depends on the chemistry and with a total mass of $m_f$, the released energy chemical energy is also $\frac{1}{2} m_f v_e^2,$. Yet, if it has an relative velocity of $v_e$, the total kinetic energy in the CM is computed to be $\frac{1}{4} m_f v_e^2$ not $\frac{1}{2} m_f v_e^2$ . This leads to some questions.

1) How can we explode the total amount of fuel with the same chemistry and not release the same total energy?

2) It is clear the total KE in the CM computes to half what you want with the $v_e$ of this fuel so how can you make it twice as explosive without either using twice as much fuel or using a different fuel that has a higher $v_e'$ or a different engine design that if possible, raises the $v_e$ of the same fuel? You don't have an independent knob to tweak.

3) I assume you might say use a different fuel that has a higher $v_e'$. In any case, how is this anything like the same rocket for comparison?

 

[jbriggs444]

If we combust that same fuel simultaneously, assuming we can do that with some super strong engine without adding mass, it should also have an exhaust velocity of $v_e$ since that depends on the chemistry

The exhaust velocity is a fixed constant based on energy density for a continuous burn only.

For an impulsive burn, measuring exhaust velocity relative to the post-burn rocket, the result will also depend, in part, on the delta v imparted to the rocket by the burn. That delta v counts in the energy analysis. The payload delta v is not a constant, but depends on the mass fraction being ejected.

For an impulsive burn, measuring exhaust velocity relative to the pre-burn center of mass, the result will still depend, in part, on the delta v imparted to the rocket by the burn. That delta v counts in the energy analysis. The payload delta v is not a constant, but depends on the mass fraction being ejected.

 

[hutchphd]

With respect @jbriggs444 answered this in #4 (that would be 124 answers ago). I see he beat me to it.

 

[bob012345]

The exhaust velocity is a fixed constant based on energy density for a continuous burn only.

For an impulsive burn, measuring exhaust velocity relative to the post-burn rocket, the result will also depend, in part, on the delta v imparted to the rocket by the burn. That delta v counts in the energy analysis. The payload delta v is not a constant, but depends on the mass fraction being ejected.

For an impulsive burn, measuring exhaust velocity relative to the pre-burn center of mass, the result will still depend, in part, on the delta v imparted to the rocket by the burn. That delta v counts in the energy analysis. The payload delta v is not a constant, but depends on the mass fraction being ejected.

Thanks. How would you then judge some kind of equivalent situation as the OP asked and @Filip Larsen has proposed? I think you're saying there is no realistic equivalent.

 

[hutchphd]

The exhaust velocity of a given engine depends upon more than just the energy content of the fuel. The specific impulse for a given propellent mix is for a given engine (at a particular burn rate and pressure differential). Excursions into extreme rates will not be recognizable and turbulent flow from an engine will be at high pressure and not well directed.
But the exercise so far has been useful.

 

[jbriggs444]

Thanks. How would you then judge some kind of equivalent situation as the OP asked and @Filip Larsen has proposed? I think you're saying there is no realistic equivalent.

What @Filip Larsen has used is what I would use. Idealize the situation, conserve energy and momentum and let the chips fall where they may.

Not very realistic, but it's a toy problem in the first place. There are burn procedures that will follow this model quite accurately.

What ever approach you take, state your assumptions first and do the calculations after.

 

[bob012345]

The exhaust velocity of a given engine depends upon more than just the energy content of the fuel. The specific impulse for a given propellent mix is for a given engine (at a particular burn rate and pressure differential). Excursions into extreme rates will not be recognizable and turbulent flow from an engine will be at high pressure and not well directed.
But the exercise so far has been useful.

I have been looking at rocket motor design and formulas recently and yes they are very complex. Of course this discussion wasn't about real rockets but idealized toy models and more about how to think about the rocket equation in the discreet limit rather than the continuous limit. It was fun to see how the discreet sums for different model assumptions all approached the ideal rocket equation as the number of steps became very large.

 

[bob012345]

What @Filip Larsen has used is what I would use. Idealize the situation, conserve energy and momentum and let the chips fall where they may.

Not very realistic, but it's a toy problem in the first place. There are burn procedures that will follow this model quite accurately.

What ever approach you take, state your assumptions first and do the calculations after.

Can you elaborate? What burn procedures do you mean. This model proposed one big energy release in the CM equivalent to the entire slow burn energy. How does one do that? Some kind of directed energy explosion? The only difference between the two models is mine had half the total energy and assumed all mass had one velocity.

 

[Filip Larsen]

1) How can we explode the total amount of fuel with the same chemistry and not release the same total energy?

2) It is clear the total KE in the CM computes to half what you want with the $v_e$ of this fuel so how can you make it twice as explosive without either using twice as much fuel or using a different fuel that has a higher $v_e'$ or a different engine design that if possible, raises the $v_e$ of the same fuel? You don't have an independent knob to tweak.

3) I assume you might say use a different fuel that has a higher $v_e'$. In any case, how is this anything like the same rocket for comparison?

I assume you pose these questions in order to "fix" the speed constrained model to somehow match the energy output in the totals even if it doesn't match along the way? If so this approach has the problem that the model will only work if we burn exactly the "planned" amount of fuel. If we only burn half of it it or if we continue to burn passed the planned amount the energy won't match. Also note that the equation for the total kinetic energy of the speed constrained model mentioned so far is only for a single chunk and I fully expect that if we calculate the total energy for a multi-chunk variant of this model it will show the energy also varies with $n$ such that the total energy relative to the continuous case is half for $n = 1$, increasing to one in the limit to the continuous case. So, no matter what is done, keeping the speed constrain will simply make it impossible to also match the energy of the ejected fuel, making this model suited for situations where the actual speed of the ejected mass is fixed because its controlled to be that, like with a mass accelerator.

If you want the energy to match there is really no way around an energy constrained model where the energy always matches no matter how much fuel you use and no matter how many chunks you eject the fuel in.

 

[jbriggs444]

Can you elaborate? What burn procedures do you mean. This model proposed one big energy release in the CM equivalent to the entire slow burn energy. How does one do that? Some kind of directed energy explosion? The only difference between the two models is mine had half the total energy and assumed all mass had one velocity.

What I have in mind, I've already described more than once in this thread.

1. You burn all of the fuel for the pulse, storing the released chemical energy.
2. You package up the expended fuel into a capsule for ejection.
3. You eject the capsule using all of the stored energy to do so.
4. You tune your efficiencies so that you are capturing the same percentage of fuel energy into total kinetic energy as a particular rocket does. i.e. you don't cheat and harvest extra thermal energy.

Alternately, you can imagine a continuous burn scheme where you burn the first bit of fuel and eject it at low velocity, saving some of the kinetic energy that could have gone into the exhaust stream. You continue burning bits of fuel and ejecting the exhaust at higher and higher relative velocities, saving up less and less energy. At some point you will be expending more energy on each bit of exhaust than the fuel possesses. So you dip into the energy you'd saved up previously. At the end of the burn you get this nifty exhaust stream, all of which is moving rearward at the same velocity.

Neither scheme is practical. But neither is physically impossible.

 

[bob012345]

I assume you pose these questions in order to "fix" the speed constrained model to somehow match the energy output in the totals even if it doesn't match along the way? If so this approach has the problem that the model will only work if we burn exactly the "planned" amount of fuel. If we only burn half of it it or if we continue to burn passed the planned amount the energy won't match. Also note that the equation for the total kinetic energy of the speed constrained model mentioned so far is only for a single chunk and I fully expect that if we calculate the total energy for a multi-chunk variant of this model it will show the energy also varies with $n$ such that the total energy relative to the continuous case is half for $n = 1$, increasing to one in the limit to the continuous case. So, no matter what is done, keeping the speed constrain will simply make it impossible to also match the energy of the ejected fuel, making this model suited for situations where the actual speed of the ejected mass is fixed because its controlled to be that, like with a mass accelerator.

If you want the energy to match there is really no way around an energy constrained model where the energy always matches no matter how much fuel you use and no matter how many chunks you eject the fuel in.

I agree. I was merely asking how you would implement the energy constrained model. @jbriggs444 has a good approach. I also agree the $v_e$ constrained model energy goes from half for $n$=1 to 1 as it approaches the continuous case. I'm not trying to force the constrained $v_e$ model to work better. I was just trying to understand how both models fit with each other and with the ideal rocket equation. I think I do.In the continuous limit both models are consistent with each other. We break that consistency by forcing discreet lumps and are forced to choose. It is easy to see why low $n$ the constrained $v_e$ model underestimates the rocket velocity while the constrained energy model overestimates the rocket velocity as compared to the the rocket equation but both approach the rocket equation in the continuous limit as we have seen.

In both models, as $\Large \frac{m_f}{m_r}$ → $0$, $\large v_r → \large v_e \Large \frac{m_f}{m_r},$ and $\large v_f → \large v_e$ and $E_{cm} → \large \frac{1}{2} v_e^2 \large m_f$,

also, the relative velocity becomes;
$$ \large v_e' = v_e \large \sqrt{ 1 + \large \frac{m_f}{m_r}} → v_e$$

 

[bob012345]

Filling in a little data for the two models. These are all one-shots except the rocket eq.;

Fixing $\large v_e$ gives less $v_r$
$$\large v_r = \large v_e \frac{m_f}{m_r + m_f} < \large v_e \sqrt{ \frac{m_f^2}{m_r(m_r + m_f)} }$$

and gives less total CM energy;
$$ E_{cm} = \large \frac{1}{2} m_f v_e^2 \large \frac{m_r}{m_r + m_f} < \large \frac{1}{2} m_f v_e^2 $$

$m_f/m_r$	$v_r/v_e$@fixed $v_e$	rocket eq.	fixed energy
0.01	0.009901	0.0099503	0.0099504
0.1	0.0909	0.09531	0.09535
1	0.5	0.6931	0.7071
2	0.66	1.0986	1.155
3	0.75	1.3863	1.50
4	0.80	1.6094	1.789
5	0.83	1.7917	2.041
6	0.857	1.9459	2.268
7	0.875	2.0794	2.457
8	0.888	2.1972	2.666
9	0.90	2.3026	2.847
10	0.909	2.3979	3.015
25	0.961	3.258	4.903
100	0.990	4.615	9.95