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# Mass of Exhaust Products, Exhaust Velocity, and Efficiency

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I have been wondering about something I've read in books about rocket engines, specifically that exhaust products of a lower mass are better than ones of a higher mass, hence the suggested use of hydrogen in NTRs. However, I don't understand why a lighter exhaust product would be better than a heavier one.

This is the thought experiment I've used:

Imagine a ship that weighs 1 kg, and carries 2 particles of propellant. It magically applies 50 joules of energy to 1 of those particles of propellant at a time. All of that energy is magically converted into perfect backwards movement of the particle, imparting an equal momentum to the particle and to the ship.

E = 1/2 M Vwhere E is energy, M is mass, and V is velocity.

So if the first particle weighs 1 kg, it ends moving backwards with a velocity of 10 m/s.

Plugging those numbers into the rocket equation results in a forward velocity of the ship (with one remaining particle of fuel) of 4.05 m/s.

Then we repeat with with the second particle, which adds 6.93 m/s to our total Delta V, resulting in 10.98 m/s.

Now, there is an immediately apparent reason to use a lighter particle of propellant, some the energy gained from the first particle has to be spent pushing the second particle. So this is enough reason to use a lighter propellant, however I have read that the mass of the exhaust products themselves, including the ones that are formed in chemical engines, have an affect on the efficiency of the engine. Now, obviously exhaust velocity is one of the primary measures of engine efficiency, and one of the ways to increase it is by making the exhaust lighter, but I don't understand why exhaust velocity is a measure of engine efficiency.

Going back to the thought experiment, if we ignore the need to carry more than one particle, we can use a particle of any weight without worrying about anything but the amount of Delta V we can get from that one particle. If momentum is conserved between the ship and the particle, wouldn't an equal energy input to particles of different masses result in the same momentum, regardless of mass? And if equal momentum is imparted to the ship, wouldn't that equal the same amount of velocity, and thus Delta V, since the ship never changes weight?

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The short answer is: the core principle of all reaction engines is, as the name suggests, conservation of momentum. Momentum is proportional to both mass and velocity while energy, like you pointed out, is proportional to the square of velocity. Since energy is usually the limiting factor it's therefore better to have a low reaction mass with high velocity than a slow but high reaction mass with the same energy.

It's quite late so maybe I'm not making sense but ill try to answer your question in a more detailed way tomorrow.

Edited by Three_Pounds
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Limiting factor in NTR is temperature, not energy. Lighter propellant carries away more heat, requiring higher input from reactor, but that's usually not hard to achieve. But it also leaves at higher velocity. So for the same 1kg of propellant, you get more momentum.

For a conventional rocket, it doesn't matter. Your limiting factor is energy, and all you really want is consistent masses. Having some light and some heavy exhaust would actually reduce efficiency. But beyond that, you just care about how much chemical energy you have per total mass.

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1 hour ago, K^2 said:

Limiting factor in NTR is temperature, not energy. Lighter propellant carries away more heat, requiring higher input from reactor, but that's usually not hard to achieve. But it also leaves at higher velocity. So for the same 1kg of propellant, you get more momentum.﻿﻿

For a conventional rocket, it doesn't matter. Your limiting factor is energy, and all you really want is consistent masses. Having some light and some heavy exhaust would actually reduce efficiency. But beyond that, you just care about how much chemical energy you have per total mass.﻿﻿

This, for an high ISP engine as an NTR is by default you want hydrogen even if this reduce your TWR as TWR is not very important in space.
Saturn 5 used RP1 on first and second  stage then hydrogen on last. New Glenn plans of using an hydrogen upper stage. Same with using SRB who has low ISP but high trust.

Now its a few settings you want TWR above all, RPG (rocket propelled grenades) is most common, here you need to burn out the rocket before it leaves the tube. its however pretty easy to tune solid rockets for this.

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Lighter molecules have a higher average velocity for a given temperature. So there's that. For non thermal engines - like a mass driver - it doesn't really matter...

Exhaust velocity is a measure of efficiency in one regard, that being how much of your rocket needs to be propellant/reaction mass. Exhaust velocity is equivalent to specific impulse (when used as impulse - change in momentum - per unit mass, which is dimensionally equivalent to velocity), and specific impulse (velocity) tells us how much change in momentum we get out of 1 kg of propellant. Thus, a higher specific impulse means less mass needed for a given change in momentum, meaning a smaller mass ratio. That's how it measures efficiency. Planes get an advantage, they measure change in momentum using the air taken in by the engine, but only measure the used reaction mass as the fuel carried on board, and not the air. They get to use air as part of their reaction mass.

55 minutes ago, magnemoe said:

This, for an high ISP engine as an NTR is by default you want hydrogen even if this reduce your TWR as TWR is not very important in space.
Saturn 5 used RP1 on first and second  stage then hydrogen on last. New Glenn plans of using an hydrogen upper stage. Same with using SRB who has low ISP but high trust.

Saturn V used hydrolox on second stage.

Edited by Bill Phil
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Chemical and Nuclear rockets operate via the temperature of their propellant. Temperature correlates to velocity via the equation provided in this online caculator:

Hydrogen has a (root mean squared) velocity of 5909 m/s at 2800 K, Water though has a velocity of 1969 m/s at 2800 K. This means you can get a lot more impulse out of hydrogen than water per kg at the same temperature. Worse more complex molecules hold more energy in their chemical bonds (cause they have more bonds and more orientational energy states) this heat capacity is basically energy wasted.

Ion engines though are limited by the velocity they can accelerate particles via electrostatic or magnetic forces: the bigger the mass and lower the ionization energy of an ion, the better. the velocity at temperature is irrelevant because temperature is not how it is being accelerated.

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4 hours ago, Three_Pounds said:

The short answer is: the core principle of all reaction engines is, as the name suggests, conservation of momentum. Momentum is proportional to both mass and velocity while energy, like you pointed out, is proportional to the square of velocity. Since energy is usually the limiting factor it's therefore better to have a low reaction mass with high velocity than a slow but high reaction mass with the same energy.

It's quite late so maybe I'm not making sense but ill try to answer your question in a more detailed way tomorrow.

So does that mean that objects of different mass gain a different momentum from the same amount of energy applied, not just a different velocity?

18 minutes ago, Bill Phil said:

Lighter molecules have a higher average velocity for a given temperature. So there's that. For non thermal engines - like a mass driver - it doesn't really matter...

[...]

That is my question, I can't tell if it does matter (beyond the thing about the weight of the unused propellant). Ignition says "So, to get a good energy term, we need an exhaust molecule with a high heat of formation and a low molecular weight" and I think I have read similar things elsewhere. Note that that quote is talking about the exhaust molecules specifically, not propellant molecules.

38 minutes ago, RuBisCO said:

Chemical and Nuclear rockets operate via the temperature of their propellant. Temperature correlates to velocity via the equation provided in this online caculator:

Hydrogen has a (root mean squared) velocity of 5909 m/s at 2800 K, Water though has a velocity of 1969 m/s at 2800 K. This means you can get a lot more impulse out of hydrogen than water per kg at the same temperature. Worse more complex molecules hold more energy in their chemical bonds (cause they have more bonds and more orientational energy states) this heat capacity is basically energy wasted.

Ion engines though are limited by the velocity they can accelerate particles via electrostatic or magnetic forces: the bigger the mass and lower the ionization energy of an ion, the better. the velocity at temperature is irrelevant because temperature is not how it is being accelerated.

I understand that lighter exhaust products go faster for the same energy input, and that thrust = exhaust velocity * mass flow rate, but I don't understand why that is, since it seems like the momentum would be the same.

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The momentum isn't really the same, given the same energy. Solve for momentum in the kinetic energy equation. A higher velocity reduces momentum - as it is in the denominator beneath energy - and the lower mass being used for thrust reduces momentum as well, this also means that thrust, the change in momentum over time, would be less.

Momentum is still conserved in the whole system though. Assuming the energy output is equal, then, at least for thermal engines, the exhaust momentum will be higher in the lower isp engine. One way of looking at it, although it may not be totally correct, is that the rocket gains more momentum, per unit of used propellant, when using the high isp engine as compared to the low isp engine. Less momentum in the exhaust means more momentum in the rocket, but also a lower thrust.

Thrust is change in momentum over time, mv is momentum, the mass flow provides the m, exhaust velocity is v, and the mass flow also provides the time.

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54 minutes ago, Mad Rocket Scientist said:

So does that mean that objects of different mass gain a different momentum from the same amount of energy applied, not just a different velocity?

That is my question, I can't tell if it does matter (beyond the thing about the weight of the unused propellant). Ignition says "So, to get a good energy term, we need an exhaust molecule with a high heat of formation and a low molecular weight" and I think I have read similar things elsewhere. Note that that quote is talking about the exhaust molecules specifically, not propellant molecules.

I understand that lighter exhaust products go faster for the same energy input, and that thrust = exhaust velocity * mass flow rate, but I don't understand why that is, since it seems like the momentum would be the same.

Yes, that's exactly what it means. E = 1/2 M*V^2, but p = M*V. I want to maximize momentum for a given amount of energy, because that'll give the most efficient rocket. Conservation of Momentum is what moves the rocket. Energy is held constant in this case, so I can only vary the mass. If you want to, you can combine the two equations and solve for momentum. You'll get p = (2 * E / M)1/2, which shows pretty clearly I think that you want to decrease mass in order to increase the momentum.

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9 minutes ago, Xeorm said:

Yes, that's exactly what it means. E = 1/2 M*V^2, but p = M*V. I want to maximize momentum for a given amount of energy, because that'll give the most efficient rocket. Conservation of Momentum is what moves the rocket. Energy is held constant in this case, so I can only vary the mass. If you want to, you can combine the two equations and solve for momentum. You'll get p = (2 * E / M)1/2, which shows pretty clearly I think that you want to decrease mass in order to increase the momentum.

Shouldn't there be another M term in there? (2 * E / M)1/2 is equal to v, but p = Mv. Shouldn't it be p = M * (2 * E / M)1/2?

55 minutes ago, Bill Phil said:

The momentum isn't really the same, given the same energy. Solve for momentum in the kinetic energy equation. A higher velocity reduces momentum - as it is in the denominator beneath energy - and the lower mass being used for thrust reduces momentum as well, this also means that thrust, the change in momentum over time, would be less.

How are solving for momentum? I think I'm missing something.

55 minutes ago, Bill Phil said:

Momentum is still conserved in the whole system though. Assuming the energy output is equal, then, at least for thermal engines, the exhaust momentum will be higher in the lower isp engine. One way of looking at it, although it may not be totally correct, is that the rocket gains more momentum, per unit of used propellant, when using the high isp engine as compared to the low isp engine. Less momentum in the exhaust means more momentum in the rocket, but also a lower thrust.

By less, do you mean magnitude or value (i.e. momentum of exhaust is negative)?

55 minutes ago, Bill Phil said:

Thrust is change in momentum over time, mv is momentum, the mass flow provides the m, exhaust velocity is v, and the mass flow also provides the time.

That's a helpful way of looking at it, thanks.

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Rockets dont work that way. They heat up fuel by burning it so that the velocity of particles are high. That becomes the thrust.

By some high school physics, Velocity of a gas molecule = square root of 3/2RT, where R is a constant, Na is avogardo constant, and T is temperature. Or V square is proportional to T. Now momentum is mv, so you get where this is going. Higher mass also means more attractive forces between molecules, so it wont heat up that much.

Someone correct me if I am wrong.

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Ok ok if I weigh say 100 kg and I throw 1 kg at 5909 m/s how much velocity will I gain in the opposite direction?

Now if I throw 1 kg at 1969 m/s how much velocity will I gain in the opposite direction?

It seems to make perfect sense to me.

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Sir, thrle thing is that light stuff tends to heat up more, do more expansions, gain more velocity that outweights the lower mass.

14 hours ago, RuBisCO said:

Ok ok if I weigh say 100 kg and I throw 1 kg at 5909 m/s how much velocity will I gain in the opposite direction?

Now if I throw 1 kg at 1969 m/s how much velocity will I gain in the opposite direction?

It seems to make perfect sense to me.

The lighter fuel is the 5909m/s thingy.

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Yeah, I guess the most important point is the fact that these engines are reaction engines -it works by reactional forces, hence conservation of momentum and not exactly of energy, unlike steam or internal combustion or electric engines.

39 minutes ago, Xd the great said:

light stuff tends to heat up m﻿ore﻿

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By light stuff, i mean with lower molecular mass.

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53 minutes ago, Xd the great said:

By light stuff, i mean with lower molecular mass.

Ammonia and Xenon.

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10 minutes ago, YNM said:

Ammonia and Xenon.

About that , I am talking about chemical rockets. For ion thrusters, the max velocity is fixed, so the heavier the better.

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5 hours ago, Xd the great said:

... I am﻿﻿﻿﻿﻿﻿ talking ﻿about chemical﻿ rockets﻿﻿. ...

That's even worse to do. Consider the other thread and the propellants.

This is why you want to test your engines and not just ballpark it from paper.

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Okay, this is getting out of hand. Lets use nuclear pulse drive.

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4 hours ago, Xd the great said:

Okay, this is getting out of hand. Lets use nuclear pulse drive.

No that is outmoded, the most awesome semi-plausible engine idea ever is the continuous nuclear fission explosion/Zubrin drive/Nuclear salt-water rocket. Yes just spray uranium enriched water down a high neutron flux/neutron reflecting nozzle, ride the nuclear explosion and keep spraying water and uranium into it until you reach the velocity you want or you die, which ever comes first, either way: totally awesome!

Edited by RuBisCO
Added image of awesome
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On 7/13/2018 at 3:38 AM, RuBisCO said:

Ok ok if I weigh say 100 kg and I throw 1 kg at 5909 m/s how much velocity will I gain in the opposite direction?

Now if I throw 1 kg at 1969 m/s how much velocity will I gain in the opposite direction?

It seems to make perfect sense to me.

That makes sense to me too, I was imagining a situation where momentum stayed constant with the same "push" as mass varied, but I think my mistake was in how momentum worked. I think it has to do with assuming that impulse = energy, but I'm not sure.

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15 minutes ago, Mad Rocket Scientist said:

I think it has to do with assuming that impulse = energy, but I'm not sure﻿﻿.

Reaction engines don't 'use' energy. There's a reason why the metric used is always a form of fuel flow + thrust and not the power output, true for rocket engines and jet engines. The only exception I could imagine is ion engines.

And with chemical engines, there's always the combination between how many energy is released + how fast the exhaust will be. This is why you need to actually test your theories.

6 hours ago, Xd the great said:

Okay, this is getting out of hand. Lets use nuclear pulse drive.

Wrong thread then.

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29 minutes ago, Mad Rocket Scientist said:

That makes sense to me too, I was imagining a situation where momentum stayed constant with the same "push" as mass varied, but I think my mistake was in how momentum worked. I think it has to do with assuming that impulse = energy, but I'm not sure.

With less mass you generally get less of an increase in velocity, assuming energy is the same. One fourth as much mass and you only get twice as much velocity. This means you get half the momentum in the exhaust, but you get more change in momentum per unit of propellant.

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To make sure I understood what people were saying, I plugged the formulas into fxSolver, and got this:

Clearly, two particles of 0.5n mass given the same initial push have more momentum than one particle of n mass, given the same push. So ignoring everything else, lighter exhaust products are better. Thanks for the help.

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