So, who wants to win 1 million Euro?

[Jason Thompson]

Pls return to topic So, who wants to win 1 million Euro?

Stop telling us to 'return to the topic'. We ARE on topic by discussing where your calculations are wrong. It's part of it.

In order to win you have to understand basic space travel physics

We do. You don't. And no-one can win because you have set the bet to be unwinnable. When the conditions include convincing you your numbers are wrong, you have to be willing to accept the corrections given. You are not, ergo no-one can win. Since we are all aware of this, no-one gives a damn about your fictitious money.

Questions still outstanding:

Do you acknowledge you have the fuel wrong (it's Aerozine-50, not pure hydrazine)? Do you acknowledge you have the LM fuel loads wrong? How did you calculate the mass of fuel based on the quantity in liters. Give us the calculations you used.

[Jason Thompson]

The energy (fuel mass) used up to brake the space craft (the mass of the fuel 'burnt') is evidently not part of the space craft after braking but has been transmitted to the surrounding space through the rocket exhaust and cannot be used by the space craft. It is gone. For ever.

Yes, and you have to account for that in your calculations. You have not. The mass of exhaust, and the kinetic energy it has, are not things you can simply ignore. You don't find it remotely odd that when you include it suddenly all the numbers balance out OK? You don't think that maybe you're the one who misunderstands the whole issue rather than the thousands of qualified people around the world who have had access to this data all the time? Conservation of momentum is an alien concept to you?

[Bob B.]

Thanks for agreeing to the kinetic energy values of the space craft before/after the braking maneuver due to burning fuel in the rocket engine producing a brake force.

Based on the numbers used in the example calculation, I agree that the kinetic energy of the 43574 kg spacecraft prior to the burn is 125.4 GJ.  I also agree that the kinetic energy of that same 43574 kg mass after the burn is about 174.6 GJ.

The energy (fuel mass) used up to brake the space craft (the mass of the fuel 'burnt') is evidently not part of the space craft after braking but has been transmitted to the surrounding space through the rocket exhaust and cannot be used by the space craft. It is gone. For ever. Unless you can produce a method to recycle energy in space.

The expelled propellent is a constituent part of the system of particles under analysis.

Pls return to topic

We're discussing the topic that you introduced.

So, who wants to win 1 million Euro? In order to win you have to understand basic space travel physics, e.g. that a mass of fuel transformed into a force to brake the space ship in the voyage is gone.

Who have you designated as the judge?  Surely you do not intend to judge the winner yourself as that would be a clear conflict of interest.

[gwiz]

The energy (fuel mass) used up to brake the space craft (the mass of the fuel 'burnt') is evidently not part of the space craft after braking but has been transmitted to the surrounding space through the rocket exhaust and cannot be used by the space craft. It is gone. For ever. Unless you can produce a method to recycle energy in space.

Even though the exhaust is no longer part of the spacecraft, it still has kinetic energy.  This energy has to be included in the energy equation, otherwise the system has less energy after the burn than before it.  This would also be a disaster for your arguments, since you claim that there is not sufficient energy in the propellants.  If the system is losing energy, you don't require the propellant to provide any.

[Bob B.]

Even though the exhaust is no longer part of the spacecraft, it still has kinetic energy.  This energy has to be included in the energy equation...

Heiwa apparently doesn't understand the concept of a "system".  A system is a set of interacting or interdependent components forming an integrated whole.  Heiwa defined the system when he calculated the initial kinetic of the 43574 kg spacecraft.  When the final kinetic energy is calculated, we must include the energy of all the components that add up to that original 43574 kg.  It does not matter whether it is one solid mass or individual gas molecules flying through space.

[Glom]

I concede that the kinetic energy before the burn is 43574*2400²/2 = 125.4 GJ.  I'll also concede that the kinetic energy of the spacecraft and remaining propellant after the burn is 32676*1500²/2 = 36.76 GJ.  But you must recognize that the expelled mass also has kinetic energy, thus the total kinetic energy after the burn is that of the spacecraft plus that of the mass expelled during the burn in the form of exhaust gas.

Thanks for agreeing to the kinetic energy values of the space craft before/after the braking maneuver due to burning fuel in the rocket engine producing a brake force.
The difference in kinetic energy of the space craft before/after the braking maneuver is solely due to burning fuel aboard and causing the brake force to be applied to the space craft during the braking time.
The energy (fuel mass) used up to brake the space craft (the mass of the fuel 'burnt') is evidently not part of the space craft after braking but has been transmitted to the surrounding space through the rocket exhaust and cannot be used by the space craft. It is gone. For ever. Unless you can produce a method to recycle energy in space.

Pls return to topic So, who wants to win 1 million Euro? In order to win you have to understand basic space travel physics, e.g. that a mass of fuel transformed into a force to brake the space ship in the voyage is gone. Same applies to fuel used during travel at sea? Compare a car running out of fuel, etc, etc. 

You are attempting (unsuccessfully) to look at energy balance. That requires looking at where energy goes when the spacecraft loses it.

The fact you do not recognise this reveals (as if we didn't already know) your extremely faulty understanding of basic mechanics.

[ka9q]

If the system is losing energy, you don't require the propellant to provide any.

Indeed. I'm actually surprised he doesn't claim that the tanks should fill up during the lunar orbit insertion burn, since the spacecraft is losing kinetic energy.

[JayUtah]

Heiwa apparently doesn't understand the concept of a "system".

Indeed.  Energy-balance methods require one to define the system in terms of hard boundaries that unequivocally include ("the system") or exclude ("the environment") components under study, and to keep consistent frames of reference for the values.  These are basic prerequisites in order for the method to work.  Heiwa has done neither, and has intentionally omitted key parts of the problem to "simplify" it.  But then he wishes to attribute the resulting error to someone else rather than to his own incompetence.

[JayUtah]

Who have you designated as the judge?  Surely you do not intend to judge the winner yourself as that would be a clear conflict of interest.

That is exactly what he proposes to do.  As if anyone believes he has a million euros anyway.

[Jason Thompson]

Same applies to fuel used during travel at sea? Compare a car running out of fuel, etc, etc.

The concept of energy balance is indeed the same, but the application is not. In a car or a ship the fuel is burned on board and energy transferred to moving components which then transmit it to other moving parts to drive the vehicle forward. That's the system in that case. In a rocket the fuel is burned and blasted out the back at high speed. It's the 'blasted out the back at high speed' you seem to be having trouble with. It's the reaction of that mass being thrown out in one direction pushing the ship in the other that makes your attempt at balancing the energy wrong. That mass of exhaust is still part of the system that needs to be accounted for. You can't ignore it just because it is no longer aboard the spacecraft when it is the very act of dumping it overboard that gives you the change in momentum you are trying to describe! If you applied your energy balance equations to ANY rocket, even the ones used just to put things into orbit (which you say is evidently possible), you would find the same problem of apparent impossibility because you just are not doing the right equations.

[ka9q]

I'm computing the theoretical energy released by the ideal stoichiometric combustion of Aerozine 50 and N₂O₄. I'm doing it by summing the enthalpies of formation for the propellants and then subtracting the enthalpies of formation of their ideal combustion products N₂, H₂O and CO₂. I know the actual figure for a real rocket engine will be lower because of the rich mixture ratio and the presence of many other products of combustion, but I'm just trying to get a theoretical upper bound.

Aerozine 50 is said to be a 50-50 mixture of UDMH, (CH₃)₂N₂H₂, and straight hydrazine, N₂H₄, but is this 50-50 by volume, by mass or by moles?

BTW, I see that hydrazine has a hazmat diamond rating of 4-4-3, that is, the highest toxicity rating, the highest fire rating and the next-to-highest reactivity rating. Gee, I wonder what could be worse. Is there anything with a 4-4-4 rating?

[Jason Thompson]

Aerozine-50 is a 50/50 mix by weight of hydrazine and UDMH, according to Wikipedia. Since I assume both components are weighed out on Earth that makes it 50/50 by mass as well, of course...

[Jason Thompson]

BTW, I see that hydrazine has a hazmat diamond rating of 4-4-3, that is, the highest toxicity rating, the highest fire rating and the next-to-highest reactivity rating. Gee, I wonder what could be worse. Is there anything with a 4-4-4 rating?

This stuff, apparently:

http://en.wikipedia.org/wiki/T-butyl_hydroperoxide

[BazBear]

BTW, I see that hydrazine has a hazmat diamond rating of 4-4-3, that is, the highest toxicity rating, the highest fire rating and the next-to-highest reactivity rating. Gee, I wonder what could be worse. Is there anything with a 4-4-4 rating?

tert-Butyl hydroperoxide.

ETA-Ack! I was ninjaed by Jason!

[Bob B.]

Aerozine 50 is said to be a 50-50 mixture of UDMH, (CH₃)₂N₂H₂, and straight hydrazine, N₂H₄, but is this 50-50 by volume, by mass or by moles?

Echoing what Jason said, it's by mass.  Since the SPS had a mixture ratio of 1.6:1, the reactants on a mole basis are:

2.09 N₂O₄ + 1 C₂H₈N₂ + 1.875 N₂H₄

I happen to know that because I worked out the same problem yesterday.  I'm interested in seeing what you come up with.

What I did was to calculate the reaction at the chamber pressure of 100 PSI (that of the AJ10-137 engine) and recorded the enthalpy (which is the same as the reactants since the reaction is adiabatic).  I then expanded the gases isentropically to the pressure at the nozzle exit and recorded the enthalpy after expansion.  Taking the difference in enthalpy, I got 5.16 MJ/kg.

Of course, I had to know the pressure at the nozzle exit.  I couldn't find this anywhere, but I did find that the nozzle expansion ratio was 62.5.  Knowing the chamber pressure and the expansion ratio, I could calculate the theoretical nozzle exit pressure, which came to 0.1033 PSI, or 712 Pa.