Rocket Fuel Ejection: Intuitive & Math Explained

[bob012345]

OK. For a finite exhaust pellet, this $v_e$ is still not completely determined by the fuel energy density, but by the fuel energy density in conjunction with the mass ratio of the packet to the payload.

Right, which is why in my corrected model the bump in rocket speed is
$$ v_f = \frac {m_f v_e}{M}$$ where $M$ is the mass of rocket plus remaining fuel.

This assumes the energy is supplied in proportion to the ejected mass so I don't have to worry about it.

 

[bob012345]

I think it would be instructive to find real rocket engine data and extrapolate from that.

 

[Filip Larsen]

For this problem I am not convinced we need to concern ourselves with the energy.

I set out with an aim to find a way to compare a single-chunk ejection with continuous ejection via a model that makes sense in both situation in order to answer the original question about which of the two is yielding most rocket speed. In order for this to make sense, there must be some "physical similarity" between the two situations, otherwise we are just comparing apples and oranges. Constraining conservation of momentum by just requiring a fixed relative speed of $v_e$ is simple but also means the energy used is not proportional to the amount of spent fuel, potentially giving either situation an "unfair advantage" in the comparison.

So a natural next step is to constrain the energy to be "equal" in both cases and see what happens. And I do feel that including energy gives a more correct or interesting model for the comparison, even if it also do seem indicate that single-chunk ejection with such energy constrain is more speed-efficient than a continuous rocket, that is, the opposite conclusion the original question posed.

But that said, all in all, I still think the question is not well-defined enough for a conclusive answer, but that does of course not exclude us from having interesting discussions about it.

 

[bob012345]

I don't why you think the energy used is not proportional to the ejected fuel? I think it is exactly that. The energy used ideally is exactly that which makes the exhaust velocity $v_e$ relative to the rocket and is linear with ejected mass whether it is one lump or small lumps.

Instead of rocket engines just think of mass ejectors that eject mass with a fixed energy ratio per unit mass at a fixed relative velocity $v_e$.

I agree this is an idealized problem because there is no practical way to do this with real rockets. But in that spirit we can imagine such a one shot rocket.

I envision a spring loaded device that uses one large ideal spring vs. $n$ small ideal springs. Each gives the rocket a relative boost of $v_e$.

 

[jbriggs444]

I don't why you think the energy used is not proportional to the ejected fuel? I think it is exactly that. The energy used ideally is exactly that which makes the exhaust velocity $v_e$ relative to the rocket and is linear with ejected mass whether it is one lump or small lumps.

Instead of rocket engines just think of mass ejectors that eject mass with a fixed energy ratio per unit mass at a fixed relative velocity $v_e$.

I agree this is an idealized problem because there is no practical way to do this with real rockets. But in that spirit we can imagine such a one shot rocket.

I envision a spring loaded device that uses one large ideal spring vs. $n$ small ideal springs. Each gives the rocket a relative boost of $v_e$.

You are ignoring the recoil of the rocket. If you push off from a finite sized packet, the rocket recoil involves non-zero energy. By contrast, the energy from an infinitesimal packet winds up 100% in the packet, 0% in the rocket.

This is just a [time-reversed] inelastic collision. The energy in such a collision as measured in the center of mass frame is divvied out in inverse proportion to the mass ratio. The smaller ejected piece gets the larger share of the energy. If you eject something infinitesimal, all of the energy goes there. If you eject something finite, not all of the energy goes that way.

You can change your reference frame to the post-burn payload frame and do the energy accounting there, but you still have the same behavior. Energy usage and delta v are invariants. They do not change depending on the chosen reference frame. Choice of reference frame can change where the kinetic energy winds up. But it will not change the total amount by which kinetic energy increments or the delta v that results.

 

[bob012345]

You are ignoring the recoil of the rocket. If you push off from a finite sized packet, the rocket recoil involves non-zero energy. By contrast, the energy from an infinitesimal packet winds up 100% in the packet, 0% in the rocket.

This is just a [time-reversed] inelastic collision. The energy in such a collision as measured in the center of mass frame is divvied out in inverse proportion to the mass ratio. The smaller ejected piece gets the larger share of the energy. If you eject something infinitesimal, all of the energy goes there. If you eject something finite, not all of the energy goes that way.

You can change your reference frame to the post-burn payload frame and do the energy accounting there, but you still have the same behavior. Energy usage and delta v are invariants. They do not change depending on the chosen reference frame. Choice of reference frame can change where the kinetic energy winds up. But it will not change the total amount by which kinetic energy increments or the delta v that results.

I am not ignoring the recoil of the rocket! At least not knowingly. It has been in all my models. That it can be ignored in the infinitesimal limit is exactly why each of these models approach the rocket equation in that limit.

 

[Filip Larsen]

I don't why you think the energy used is not proportional to the ejected fuel?

For the model with $v_e$ as fixed relative speed when ejecting the fuel mass $m_f$ with $m_r$ remaining rocket mass, I get the total kinetic energy to be $\frac{1}{2}m_f v_e ^2 \frac{m-m_f}{m}$. Notice the trailing factor making this non-linear in $m_f$, especially for $m_f \gg m_r$ where the total kinetic energy goes towards zero even though nearly all of the initial mass has been ejected as fuel.

 

[bob012345]

For the model with $v_e$ as fixed relative speed when ejecting the fuel mass $m_f$ with $m_r$ remaining rocket mass, I get the total kinetic energy to be $\frac{1}{2}m_f v_e ^2 \frac{m-m_f}{m}$. Notice the trailing factor making this non-linear in $m_f$, especially for $m_f \gg m_r$ where the total kinetic energy goes towards zero even though nearly all of the initial mass has been ejected as fuel.

Is this the sum of both rocket and ejected mass? In the instantaneous CM frame? You have to be specific as to where you are referencing the kinetic energy from. Also, did you mean to write;

$$\frac{1}{2}m_f v_e ^2 \large \frac{m_r-m_f}{m_r}$$

This goes highly negative for $m_f \gg m_r$!

 

[Filip Larsen]

The total kinetic energy of what with respect to what reference frame?

I was referring to the single-chunk equations so there is really only one interesting frame. Before ejection the total mass sits at rest, after ejection the two masses move in opposite directions with some total kinetic energy that, as mentioned, does not vary linearly with the mass of the ejected fuel for the model in question. Please calculate the energy yourself and you will see.

 

[bob012345]

I was referring to the single-chunk equations so there is really only one interesting frame. Before ejection the total mass sits at rest, after ejection the two masses move in opposite directions with some total kinetic energy that, as mentioned, does not vary linearly with the mass of the ejected fuel for the model in question. Please calculate the energy yourself and you will see.

I'll happily do that. Whether we totally agree or not I find the discussion interesting.

First though, what about the negative kinetic energy I pointed out above and are we still forcing $v_e$ as the relative velocity in all mass ratio's? Thanks.

 

[Filip Larsen]

did you mean to write;

$$\frac{1}{2}m_f v_e ^2 \large \frac{m_r-m_f}{m_r}$$

No, you read it wrong.

I wrote the kinetic energy out like I did in terms of $m_f$ and the total (constant) mass $m = m_f + m_r$ to clearly show that it varies quadraticly, and not linearly, in $m_f$ and with the first part (i.e. $\frac{1}{2} m_f v_e^2$) separate because that is the equivalent energy released by the the fuel in the continuous case.

 

[bob012345]

No, you read it wrong.

I wrote the kinetic energy out like I did in terms of $m_f$ and the total (constant) mass $m = m_f + m_r$ to clearly show that it varies quadraticly, and not linearly, in $m_f$ and with the first part (i.e. $\frac{1}{2} m_f v_e^2$) separate because that is the equivalent energy released by the the fuel in the continuous case

Thanks!

I think I see what's happening here. There are different reference frames to consider and they are not the same nor is the kinetic energy required to be equivalent.

First, in the rocket's frame $R$, an observer sees the mass ejecta at $v_e$ as specified by the rocket engine and combustion chemistry or just by whatever is ejecting the mass. That is always $m_f$ moving at $v_e$ and its kinetic energy in that frame is always $K_r = \frac{1}{2}m_f \large v_e ^2$ and linearly proportional to $m_f$. The rocket velocity is by definition zero so that is the total kinetic energy. Also, one could look in the reference frame of the recoiling fuel mass where the total energy is $K_f = \frac{1}{2}m_r \large v_e ^2$ since $v_f$ is zero in that frame.

Next, we look at the velocities in the instantaneous center of mass reference frame. That is where you calculated;
$$K_{cm} = \frac{1}{2}m_f \large v_e ^2 \frac{m-m_f}{m}$$ or
$$K_{cm} = \frac{1}{2}\large v_e ^2 \frac{m_rm_f}{m_r + m_f}$$ I agree that is the total kinetic energy in the center of mass reference frame.

for $m_r \gg m_f$ we are essentially in the rocket frame $R$
$$K_{cm} ≈ \frac{1}{2}\large m_f \large v_e ^2$$
and for $m_r = m_f$
$$K_{cm} = \frac{1}{2}\frac{1}{2}\large m_f \large v_e ^2$$

The point is that in general $K_r ≠ K_f ≠ K_{cm}$ but all are valid.It is in the rocket frame however where the fuel is burned or just ejected that we measure the chemical or stored energy released. A real rocket is all about constant thrust or power which is mass flow times velocity or $\large m_f' ⋅ v_e$ where the energy per unit mass is ≈constant and $v_e$ is fixed by reaction chemistry.

The original OP question was simply;

Why does ejecting fuel over time cause rocket to go faster than ejecting all at once?

I think as long as we fix the relative velocity of the fuel mass wrt the rocket, as rockets do, we are keeping with the spirit of the question and we both saw that the rocket does better with a slow burn than throwing off a single lump.

 

[Filip Larsen]

There are different reference frames to consider

I am not sure why you want to bring in more than one frame of reference, but I hope we agree that for the process of ejecting a single chunk of mass there is only need for one, namely the frame where the center of mass is at rest.

There is actually a ejection mechanism that sensible will cover both cases (constrainted ejection speed vs constrained ejection energy) if we imagine the rocket being equipped with a linear accelerator that can accelerate a chunk of mass put on it.

In constrained ejection speed mode the accelerator will accelerate any chunk of mass put on it until the chunk has exactly the set speed $v_e$ relative to the accelerator (and rest of the rocket) itself. The energy to do so comes from some other internal or external energy source that can provide up to at least some maximum of energy. Note that we don't care how long the accelerator take to do its work, conservation of momentum works regardless of the time taken. The important part (for this to be a "discrete" rocket) is only that the ejected mass is ejected as a single chunk, that is, the particles of the ejected mass is all at rest relative to each other after ejection. In this mode, the conclusion then is, that you will get more total speed of the rocket if you eject the same mass in smaller chunks than if you do it in one big chunk.

In constrained energy mode the accelerator instead works by slurping in an amount of fuel (and oxidizer), combusting it to get energy, but collecting all the combustion products into a small (mass-less) tank, and then use exactly the released energy to accelerate the combustion products out the back of the rocket. In this mode, however silly it sounds, the conclusion is that the rocket will get less final speed if the same chunk is broken up into smaller chunks.

Since the original question does not specify details that can discern between the two models and since the two models give opposite conclusions I will stil join the consensus in this thread that the question is ill-posed.

One may argue that since the constrained speed model is nice and simple and since it seems to affirm the original question then this is the type of rocket that whoever posed the original question must have had in mind. But I would rather answer the original question with "yes, there are models of ejections mechanisms where this will be true, but there are also other models where it is not true. Which one are you thinking off?"

 

[bob012345]

I am not sure why you want to bring in more than one frame of reference, but I hope we agree that for the process of ejecting a single chunk of mass there is only need for one, namely the frame where the center of mass is at rest.In constrained ejection speed mode the accelerator will accelerate any chunk of mass put on it until the chunk has exactly the set speed $v_e$ relative to the accelerator (and rest of the rocket) itself. The energy to do so comes from some other internal or external energy source that can provide up to at least some maximum of energy. Note that we don't care how long the accelerator take to do its work, conservation of momentum works regardless of the time taken. The important part (for this to be a "discrete" rocket) is only that the ejected mass is ejected as a single chunk, that is, the particles of the ejected mass is all at rest relative to each other after ejection. In this mode, the conclusion then is, that you will get more total speed of the rocket if you eject the same mass in smaller chunks than if you do it in one big chunk.

In constrained energy mode the accelerator instead works by slurping in an amount of fuel (and oxidizer), combusting it to get energy, but collecting all the combustion products into a small (mass-less) tank, and then use exactly the released energy to accelerate the combustion products out the back of the rocket. In this mode, however silly it sounds, the conclusion is that the rocket will get less final speed if the same chunk is broken up into smaller chunks.

Rockets combust propellent and expel exhaust gases from within the rocket's frame of reference not the instantaneous CM frame although in the usual case where $m_r >> m_f$ discussed above, the rockets frame and the CM frame are essentially the same. As I showed above in that case the total kinetic energy in the CM frame is almost the same as in the rocket frame.

Thus the natural frame to think about energy is the rocket frame. Only in that frame are both cases of constrained $v_e$ and constrained energy simultaneously true. Constraining the kinetic energy in the CM frame forces the $v_e$ to be different for each case the mass ratio is different because it is dependent on both masses. Focusing on the CM is the original mistake I made in which you corrected me where I took $v_e$ relative to the CM and not the rocket. It is a good approximation in the continuous case but not the discreet case.

I think an answer to the OP regarding why is simply momentum conservation. Since delta $v_r$ is inversely proportional to total mass, the summation of single, discreet or continuous dictates that behavour.

 

[jbriggs444]

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

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

although in the usual case where $m_r >> m_f$ discussed above, the rockets frame and the CM frame are essentially the same.

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.

 

[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