Oxidizer and the rocket equation

[mjl1966]

So we know dV = Ve ln R and that Ve refers to the velocity of the propellant as it escapes the rocket nozzle. We also know that R is the ratio of total mass to dry mass. What I have always wondered is how oxidizer fits into all this. Is it considered part of the propellent mass or the dry mass? Since it's not going out the back end, I keep wondering if it's just dead weight.

Thoughts?

[Kryten]

In a chemical rocket, oxidiser and fuel are both propellant, and both leave the rocket. For example, a hydrogen-powered rocket produces energy though the reaction of hydrogen fuel and oxygen oxidiser to form water, which is what actually forms the exhaust.

[KerikBalm]

" Is it considered part of the propellent mass or the dry mass?"

Propellant mass

"Since it's not going out the back end, I keep wondering if it's just dead weight."

It most certainly is going out the back end.

Thats why NERVA engines get higher specific impulse than Liquid hydrogen+ liquid oxygen.

If oxygen was not part of the propellant, then they would both just be shooting hydrogen out the back end. NERVA doesn't get the hydrogen any hotter, it would melt the rods if it could.

Its simply shooting H2 out the back (MW: 2), while the chemical rocket is shooting H2O out the back (MW: 18)

[Kryten]

Although, in practice, most hydrogen engines run extremely fuel rich, and most of the exhaust is hydrogen.

[mjl1966]

I love being around people who are way smarter than me. Thanks for clearing this up!

[K^2]

Although, in practice, most hydrogen engines run extremely fuel rich, and most of the exhaust is hydrogen.

Not by mass it isn't.

[lajoswinkler]

Fuel and oxidizer are meaningless terms, actually. It's just a convention. Both kerosene and LOX are fuels. Kerosene reduces oxygen, oxygen oxidizes carbon inside kerosene. It's a transfer of electrons. One gets free of them, one gets stuffed with them.

The only reason this disinformation lingers is because we look at it from out antropocentric view, living in abundant oxidizing atmosphere, with reducers scattered throughout the world.

"Propane is burning" - yes. So does oxygen in propane. There's no difference. You either have a reaction or you don't and that's what matters.

[cantab]

Yeah, maybe it is Earth-centric, but it's still reasonable. It makes it fairly obvious that you need a fuel and an oxidizer - two fuels or two oxidizers generally won't work. (Notwithstanding a few compounds that can be either, or monopropellants). You also know that the oxidizer is liable to oxidize your metal components if you aren't careful.

[lajoswinkler]

It's reasonable as a convention in particular fields for the sake of simplicity (although that can be tackled, too), but basic meanings of such terms have nothing to do with it. You can have chlorine burning in a room full of hydrogen, and vice versa - hydrogen burning in chlorine room.

When you mix them in a chamber, it literally loses its every meaning outside pure chemical reaction which concernes electron transfer.

[KerikBalm]

Yea... it comes down to terminology.

In effect, the consumable(s) providing the energy are the fuel, and the reaction mass is the propellant.

In a NTR, the fissible material is the fuel, and the H2 (or Methane, or ammonia, or whatever...) is the propellant.

In a chemical rocket, the fuel and the propellant are the same thing (unless, as is the cas with liquid hydrogen rockets, the unburnt hydrogen is simply propellant)

I wonder though... how extremely fuel rich are we talking? is it done intentionally to increase Isp, or is it to get a more consistent burn?

[PlonioFludrasco]

AFAIK, molar weight is somewhat related to Isp, in particular if the molar weight is lower the Isp is higher. If all the hydrogen reacts with all the oxigen, you have only water (MW=18) escaping the nozzle. If part of the hydrogen doesn't react, you have a mixture of water and hydrogen (MW=2) leaving the nozzle, so the average molecular weight is lower than 18, thus increasing Isp.

EDIT: Ovviously, correct me if I'm wrong!

[shynung]

That's one reason LH2/LOX engines generally run rich. The other reason is that the extra hydrogen acts as coolant, reducing the combustion temperature, prolonging engine service life.

[KerikBalm]

AFAIK, molar weight is somewhat related to Isp, in particular if the molar weight is lower the Isp is higher.

If the MW is lower, the speed of the particles *at a given temperature* is higher.

If the speed of the exhaust is higher, the Isp is higher.

Expelling hydrogen at 500 kelvin is not going to get you a higher Isp than expelling methane at 2,500 kelvin, for instance.

[StrandedonEarth]

I think another reason why rocket engines run fuel rich is to prevent metal parts from oxidizing (burning) in the excess hot oxygen. After all, that's how a cutting torch works: First, the metal is heated to red-hot with a normal, neutral (neither oxygen-rich nor fuel-rich) flame, then a blast of oxygen does the actual cutting by burning the metal. It wouldn't take long to burn through an engine's combustion chamber if the mixture was too oxygen-rich

[Iskierka]

Running fuel rich actually reduces Isp, at least in basic theory - molecular weight may be lower, but the unreacted particles have to take energy from the reacted ones, rather than providing their own. Tsiolkovsky originally never thought rockets would have to be so humongous, as he assumed LH2/LOX rockets would be capable of around 5700 m/s Ve.

In practice, this isn't quite true - if LH2/LOX could fully react, it would perform better, but a small irony is that this fuel gets so hot that only so much of the fuels can react. If another water particle is formed, the energy released will break apart one that's already formed. This means that in the end, this fuel type actually gets almost the same Ve across a large range, from 3-4 times excess hydrogen as is typically used, down to theoretically perfect ratio. The actual reasons for running hydrogen-rich are that it means you have unreacted hydrogen in the engine, rather than both hydrogen and oxygen, and it means that the engine operates at a lower temperature than if you tried to push it to its true limit, helping with material limits. Tsiolkovsky's 5700 m/s engines would have had to somehow contain temperatures on the order of 6200 K or more.

Lower MW doesn't improve LH2/LOx performance when running rich, but what it does is keep the performance up, allowing us to make the engines operate way off-peak in order for them to actually be useful and not consume themselves in excessive heat and oxygen.

[OhioBob]

Running fuel rich actually reduces Isp, at least in basic theory - molecular weight may be lower, but the unreacted particles have to take energy from the reacted ones, rather than providing their own.

Running fuel rich DOES increase ISP, to a point, then it starts to decrease. There is an optimum mixture ratio at which ISP is maximized, and that optimum is on the fuel-rich side of a stoichiometric mixture.

Generally speaking, exhaust velocity is proportional to SQRT(T/M). In other words, the higher the temperature and the lower the molecular weight, the higher the exhaust velocity. Maximum temperature occurs when the mixture is stoichiometric, that is, there is just enough oxygen present to react with all the fuel. Unfortunately this also results in a high molecular weight. Reducing the proportion of oxygen to fuel lowers both temperature and molecular weight. Initially the molecular weight decreases faster than temperature, so exhaust velocity goes up. Eventually the trend reverses and temperature decreases more rapidly than molecular weight, so exhaust velocity goes down. Maximizing exhaust velocity, thus ISP, involves finding the optimum point where there is the right balance between temperature and molecular weight.

[OhioBob]

To better illustrate the point I made in my previous post, below is a table showing the temperature and average molecular weight resulting from the combustion of liquid oxygen and liquid hydrogen at different mixture ratios. Mixture ratio is the mass of the oxidizer divided by the mass of the fuel. A mixture ratio of 8 is stoichiometric, while lower mixture ratios are fuel rich. Also provided is (T/M)^1/2, which gives us a comparison of how the different mixture ratios will perform in regard to specific impulse. These values have been computed for a combustion chamber pressure of 68 atmospheres (1000 PSI).

[TABLE=class: grid, width: 500, align: center]

[TR]

[TD=align: center]Mixture Ratio[/TD]

[TD=align: center]Temperature

(Kelvin)[/TD]

[TD=align: center]Molecular Weight

(g/mol)[/TD]

[TD=align: center](T/M)^1/2[/TD]

[/TR]

[TR]

[TD=align: center]8[/TD]

[TD=align: center]3605[/TD]

[TD=align: center]16.01[/TD]

[TD=align: center]15.01[/TD]

[/TR]

[TR]

[TD=align: center]7[/TD]

[TD=align: center]3587[/TD]

[TD=align: center]14.79[/TD]

[TD=align: center]15.57[/TD]

[/TR]

[TR]

[TD=align: center]6[/TD]

[TD=align: center]3495[/TD]

[TD=align: center]13.36[/TD]

[TD=align: center]16.17[/TD]

[/TR]

[TR]

[TD=align: center]5[/TD]

[TD=align: center]3293[/TD]

[TD=align: center]11.74[/TD]

[TD=align: center]16.75[/TD]

[/TR]

[TR]

[TD=align: center]4[/TD]

[TD=align: center]2954[/TD]

[TD=align: center]9.94[/TD]

[TD=align: center]17.24[/TD]

[/TR]

[TR]

[TD=align: center]3[/TD]

[TD=align: center]2455[/TD]

[TD=align: center]8.01[/TD]

[TD=align: center]17.51[/TD]

[/TR]

[TR]

[TD=align: center]2[/TD]

[TD=align: center]1802[/TD]

[TD=align: center]6.02[/TD]

[TD=align: center]17.30[/TD]

[/TR]

[TR]

[TD=align: center]1[/TD]

[TD=align: center]986[/TD]

[TD=align: center]4.02[/TD]

[TD=align: center]15.66[/TD]

[/TR]

[/TABLE]

Note that, as described, both temperature and molecular weight decrease as the mixture becomes more fuel rich. As Iskierka explained, the reason the temperature is lower is because some of the heat generated by the reacted particles must heat up the unreacted particles. As the mixture ratio decreases there are fewer reacted particles and more unreacted particles. The reason the molecular weight goes down is because, as the propellant becomes more fuel rich, there is a greater abundance of lightweight unreacted hydrogen molecules.

Also note that at a mixture ratio of 8 you might expect the molecular weight to be 18, i.e. that of water. What is happening is the high heat of combustion is causing some of the large molecules to break down into smaller ones in a process known as dissociation. Although water is most abundant, there is also some H₂, HO, H, O₂ and O present in the mixture, thus lowering the average molecular weight.

If we look at the value of (T/M)^1/2 we see that it's greatest when the mixture ratio is 3. This means that the exhaust velocity, thereby the specific impulse, will be greatest at about this mixture ratio. (I say "about" because the specific heat ratio is also a factor in the exhaust velocity equation and has not been considered here.)

In practice we see that no LOX/LH2 engines actually use mixture ratios that are this fuel rich. Most LOX/LH2 engines operate at mixture ratios in the range of 5 to 6. The reason for this is because of hydrogen's extremely low density. At a mixture ratio of 3, the combined propellant density is very low, resulting in the need for very large propellant tanks. Increasing the proportion of oxygen to hydrogen significantly increases the density and lowers the propellant volume. Having smaller propellant tanks provides benefits such as lower mass and less drag. These benefits offset the decrease in ISP that results from operating at a less than optimal mixture ratio. In engineering practice it is found that mixture ratios of 5 to 6 strike the optimum balance between ISP and propellant density.

Edited January 18, 2015 by OhioBob