It really is rocket science

[raven]

I knew I should have made a left turn at Albuquerque Station.

[ka9q]

Of course, the main propellant tanks can be a little smaller

Or the main engines can burn a little longer.

I'd also like to see how much power can be obtained from a kilogram of monopropellant versus a kilogram of fuel-rich bipropellant to see if there is an advantage to one versus the other.

Another idea: if the problem is to avoid burning up the turbine blades, why not burn a stoichiometric ratio of the main propellants along with something to cool the flame and produce even more gas -- like maybe water?

The efficiencies I got for gas generators seemed so low that almost anything ought to be an improvement, I'd think.

[cjameshuff]

Another idea: if the problem is to avoid burning up the turbine blades, why not burn a stoichiometric ratio of the main propellants along with something to cool the flame and produce even more gas -- like maybe water?

The efficiencies I got for gas generators seemed so low that almost anything ought to be an improvement, I'd think.

Reducing the temperature is exactly the opposite of what you want to do to increase efficiency. The maximum efficiency of a heat engine is 1 - Tcold/Thot, and you're talking about reducing Thot to something closer to Tcold. You're also adding a separate tank, plumbing, and pumps for water or some other fluid that you have to carry but which doesn't contribute any energy...if you must reduce operating temperature, you may as well just use some extra oxidizer or fuel.

[ka9q]

Reducing the temperature is exactly the opposite of what you want to do to increase efficiency.

Oh, I know that. The problem is that the turbine blades simply can't withstand the actual flame temperature, so you have no choice but to cool the hot gases before they reach the blades even though that reduces the thermodynamic efficiency of the heat engine.

Whether you mix water with the hot gases or combine the propellants in a highly non-stoichiometric ratio so they don't burn as hot in the first place, either way you are decreasing efficiency. But I don't know enough about turbine design to know which approach reduces the efficiency less.

[Bob B.]

I don't know if I'm going about this the right way or not, but I've attempted to compare the effectiveness of several propellants for driving turbines.  I've assumed in all cases that the pressure at the turbine inlet is 1000 PSI and and the pressure at the outlet is 50 PSI, for a pressure ratio of 20:1.  For each propellant, I calculated the enthalpy of the gases at the inlet and then at the outlet assuming isentropic expansion.  Taking the difference in enthalpy, I think, gives the amount of energy available to drive the turbine.  If the enthalpy difference is low, that means a larger mass of propellant is needed to produce a given turbine power.

In the first case I took 100% hydrogen peroxide, which when decomposed at 1000 PSI gives a temperature of 1279 K and an enthalpy of -5.5145 MJ/kg.  Expanding the gas until the pressure drops to 50 PSI reduces the temperature to 669 K and the enthalpy to -6.5573 MJ/kg.  The enthalpy difference is 1.0428 MJ/kg.  (Note that concentrations below 100% are less effective propellants.)

Next I did hydrazine.  The inlet and outlet temperatures are 911 K and 530 K, and the enthalpy difference is 1.5005 MJ/kg.

Next on my list is fuel-rich LOX and kerosene (for which I used C₁₂H₂₆).  I used a mixture ratio of 1.13 because this gives just the right amount of oxygen to oxidize all the carbon to CO, but not enough to oxidize the hydrogen.  This mixture yields inlet and outlet temperatures of 1599 K and 1090 K, and the enthalpy difference is 2.0502 MJ/kg.  (In practice, other mixture ratios might be common, which I'd have to do further research to determine.)

Finally, I used a stoichiometric mixture of LOX and kerosene with water added.  I added enough water to lower the inlet temperature to approximately the same as that of the fuel-rich LOX/kerosene mixture.  I ended up with about 59% water by mass.  The inlet and outlet temperatures are 1595 K and 925 K, and the enthalpy difference is 1.4233 MJ/kg.

Therefore, if my logic and math are correct, fuel-rich LOX/kerosene is best, requiring the least amount of propellant to drive the turbines.  Hydrogen peroxide is the worst, requiring double the mass of LOX/kerosene.  Hydrazine and watered-down LOX/kerosene are about equal and lie in the middle.

[ka9q]

Very interesting!

Now I have to go back to the books to make sure I know what "Enthalpy" is and how and why it changes... :-)

Oh, when you did hydrazine, what did you decompose it to? Ammonia and nitrogen, or nitrogen and hydrogen?

And what about the energy necessary to vaporize the LOX? The other propellants are all stored at room temperature.

[Bob B.]

Very interesting!

Now I have to go back to the books to make sure I know what "Enthalpy" is and how and why it changes... :-)

I've done some more calculations since yesterday.  I wanted to bring the LOX/kerosene combustion temperature down to something comparable to the hydrogen peroxide, so I lowered the mixture ratio from 1.13 to 0.5.  This changed the inlet and outlet temperatures to 1270 K and 958 K.  It also lowered the enthalpy change to 1.4902 MJ/kg.

Similarly, I increased the amount of water in the watered-down scenario to about 64%.  This changed the inlet and outlet temperatures to 1275 K and 707 K, and lowered the enthalpy change to 1.1460 MJ/kg.

With these revised numbers we see that LOX/kerosene doesn't look as good as it did before, but it's still better than H₂O₂ and about equal to hydrazine.  Of course if we lower the mixture ratio further, it gets closer and closer to H₂O₂.

Oh, when you did hydrazine, what did you decompose it to? Ammonia and nitrogen, or nitrogen and hydrogen?

I included H, H₂, N₂ and NH₃ in the products and let my computer program (STANJAN) sort out the equilibrium mixture.  On the inlet side I had almost complete dissociation of the ammonia (about 98%).  On the outlet side, because of the lower temperature, some of the ammonia reformed to where I had about 87% dissociation.

It's my understanding that it's possible to vary the design of the catalyst chamber to control the amount of ammonia dissociation.  About 30%-40% is the minimum that can be maintained, which results in a hotter temperature of about 1400 K.  This is desirable in thrusters because less dissociation produces a higher specific impulse.  However, I've read that a higher percentage dissociation is typically allowed in gas generators to keep the temperature lower.

Unfortunately, with the computer program I'm using, I have to control over how much ammonia dissociation there is.  I think that if I could reduce the dissociation and get a higher temperature, we'd find that the change in enthalpy would likely be considerably better than LOX/kerosene.

And what about the energy necessary to vaporize the LOX? The other propellants are all stored at room temperature.

That's been taken into account.  I reduced the heat in the gas by the amount needed to vaporize the LOX and bring it up to the same starting temperature as the other propellants.

[cjameshuff]

With these revised numbers we see that LOX/kerosene doesn't look as good as it did before, but it's still better than H₂O₂ and about equal to hydrazine.  Of course if we lower the mixture ratio further, it gets closer and closer to H₂O₂.

It also (in the case of LOX/RP1 rockets, anyway) uses the stuff you already have large, mass-efficient tanks full of and which you already need to pump around.

[Glom]

Very interesting!

Now I have to go back to the books to make sure I know what "Enthalpy" is and how and why it changes... :-)

Oh, when you did hydrazine, what did you decompose it to? Ammonia and nitrogen, or nitrogen and hydrogen?

And what about the energy necessary to vaporize the LOX? The other propellants are all stored at room temperature.

Enthalpy is the heat of reaction. Or maybe it's better termed the change in chemical energy of the system since an exothermic reaction has negative energy.

[Tedward]

Watching this thread develop is fascinating.

[Bob B.]

Watching this thread develop is fascinating.

It is interesting to talk about this stuff, but we're so far off topic now that we should have started a new thread.

[JayUtah]

Introduction to enthalpy as it relates to the fluid-cycle design of rocket engines.
http://exploration.grc.nasa.gov/education/rocket/enthalpy.html

This is the e-book for Sutton and Biblarz, the standard work.  The thermodynamics chapter is skimmable without having to buy the book.
http://books.google.com/books?id=pFktw0GYSX8C

Yes, this is an awesome thread, and I'm sure LunarOrbit wouldn't object to splitting it off into the Reality forum.

Bob, what program are you using?  Is it something I can rewrite to give you control over the ammonia dissociation?

[Bob B.]

Introduction to enthalpy as it relates to the fluid-cycle design of rocket engines.
http://exploration.grc.nasa.gov/education/rocket/enthalpy.html

Please allow me to add the following about isentropic compression/expansion.
http://www.grc.nasa.gov/WWW/K-12/airplane/compexp.html

If it will help, I'll provide a quick example of the calculations I've been doing. 

Let's do the calculation for atomic hydrogen, H.  As mentioned earlier, I assumed a turbine pressure ratio of 20:1, and let's assume our turbine inlet temperature is 1200 K.  Atomic hydrogen has a specific heat ratio, γ, of 1.667 and it's heat capacity, C_p, is 20.79 J/mol-K.  Therefore,

T₂/T₁ = (P₂/P₁)^[1-1/γ]
T₂/1200 = (1/20)^[1-1/1.667]
T₂ = 362 K

(h₂-h₁) = C_p(T₂-T₁)
(h₂-h₁) = 20.79*(362-1200)
(h₂-h₁) = -17420 J/mol

So each mole of atomic hydrogen has 17.42 kJ less energy at the turbine outlet than it had at the turbine inlet.

I used atomic hydrogen in this example because its heat capacity (or specific heat) and specific heat ratio is constant over all temperatures.  The heat capacity and specific heat ratio of molecular gases vary with temperature (here's an example), thus the calculations are more complicated.  The calculations also include a mixture of different gases, and the composition of those gases vary with temperature.  That is, the ratio of, say, H to H₂ changes as the temperature increases or decreases, with the molecular form dominating at low temperature and the atomic form at high temperature.

Because the calculations can become insanely complicated, a computer program is typically needed or else it would take forever to do just one calculation.

This is the e-book for Sutton and Biblarz, the standard work.  The thermodynamics chapter is skimmable without having to buy the book.
http://books.google.com/books?id=pFktw0GYSX8C

Thanks for the reference.  Although I haven't had a chance to study it yet, I'm sure I'll find it interesting.

Another e-book I've found helpful is this one
http://books.google.com/books?id=TKdIbLX51NQC&printsec=frontcover#v=onepage&q&f=false

The part about gas generators starts on page 116.  Note that page 118 states that the European Ariane uses one of the methods proposed by ka9q, i.e. using water as a diluent in a near-stoichiometric gas generator.

Also note that the gas generator example shown in Table 4-3 (page 119) indicates a LOX/RP-1 mixture ratio of 0.342, which is considerably lower than I was using.  Redoing my calculations at the lower mixture ratio gives inlet and outlet temperatures of 1179 K and 891 K, and an enthalpy change of 1.2788 MJ/kg.

I've also redone my calculations using hydrogen peroxide at 90% concentration rather than 100% (It's my understanding 100% is unattainable).  This changes the inlet and outlet temperatures to 1033 K and 524 K, with an enthalpy change of 0.8525 MJ/kg.

Yes, this is an awesome thread, and I'm sure LunarOrbit wouldn't object to splitting it off into the Reality forum.

I second that if Lunar Orbit doesn't mind doing it.  We'd have to split off anything having to do with gas generators, preburners, and staged combustion.

Bob, what program are you using?  Is it something I can rewrite to give you control over the ammonia dissociation?

I'm using freeware called STANJAN, written by Stanford University Mechanical Engineering Professor Bill Reynolds.  The version I'm using has a 1984 copywrite.  Being that old it's not very user friendly.  I also had to create/format for myself most of the JANAF data tables used by the program.  Nonetheless, STANJAN has served me well over the years and I wouldn't have been able to accomplish half of what I have without it.

I've often wished I could write a Windows version that's easier to interface with, but I can't read the programming to figure out how it works.  Just a few minutes ago a stumbled upon the following document, which might give me what I need to know.  However, it's probably more of a pain in the neck to write a new program than to continue using what I already have.
http://www.stanford.edu/~cantwell/AA283_Course_Material/STANJAN_write-up_by_Bill_Reynolds.pdf

Regarding the ammonia dissociation issue, I apparently figured out how to force STANJAN to give me what I want because I found old notes from an investigation I did several years ago regarding monopropellant engines.  At that time I calculated the temperature, molecular weight, and specific heat ratio of hydrazine at different percentages of dissociation.  For the life of me I don't remember how I did it.  I'll probably eventually figure it out again.

[Bob B.]

(h₂-h₁) = C_p(T₂-T₁)
(h₂-h₁) = 20.79*(362-1200)
(h₂-h₁) = -17420 J/mol

Wait a minute, is that right?  C_p is constant pressure heat capacity, but I'm not at constant pressure.  Can anyone verify whether or not I goofed here?  Nonetheless, the error, if it exists, is just in my one-time manual example calculation.  I was using STANJAN for all the other calculations, so hopefully those are correct.

EDIT:

I think I might be OK.  I plugged my example into STANJAN and I got a enthalpy change of -17270 J/g.  Multiply that by the atomic weight of hydrogen and I get, -17270 x 1.008 = -17408 J/mol.  I think that's close enough to my answer for verification.  Either my calculation is correct, or both STANJAN and I are wrong.

[JayUtah]

Wait a minute, is that right?  C_p is constant pressure heat capacity, but I'm not at constant pressure.  Can anyone verify whether or not I goofed here?

IIRC, the natural logarithm of the pressure ratio applies here somewhere, but I'd have to go back to references to determine where.

I'm using freeware called STANJAN, written by Stanford University Mechanical Engineering Professor Bill Reynolds.

[...]
I've often wished I could write a Windows version that's easier to interface with, but I can't read the programming to figure out how it works.

It's in Fortran 77.

[backs away slowly, not making eye contact]