Say that your space ship has mass 32 676 kg excluding fuel and that you must slow down from 2 400 to 1 500 m/s velocity to insert into Moon orbit. Your space ship has a P-22 KS rocket engine with 97 400 N thrust (at full blast). How much fuel do you require to carry out the braking maneuver?
If you suggest, e.g. 10 898 kg, you must support your answer with proper calculations to win the prize (€1M). I have a feeling you need >80 000 kg.
Heiwa, putting aside the fact that your methods and calculations are complete buffoonery, let’s at least get your numbers and facts straight.
Say that your space ship has mass 32 676 kg excluding fuel32,676 kg was not the mass “excluding fuel”. It was the CSM/LM mass at LOI cutoff, which still included a large amount of unused propellant. I must congratulate, however, in that this is one of the few numbers you got right. According to
this source, the mass was 72,037.6 lbm, or 32,675.7 kg.
you must slow down from 2 400 to 1 500 m/s velocity to insert into Moon orbitNo, those velocities are incorrect. From
this source we see that Apollo 11’s velocity at LOI ignition was 8,250.0 ft/s (2,514.6 m/s) and at LOI cutoff was 5,479.0 ft/s (1,670.0 m/s).
The cutoff velocity can also be confirmed using orbital mechanics, as it is simply the orbital velocity at that point. We see that Apollo 11’s orbit at LOI cutoff was 60 n.mi. x 169.7 n.mi, and that the altitude at LOI cutoff was 60.1 n.mi. I’m not going to show the math, but the velocity at an altitude of 60.1 n.mi. in a 60 x 169.7 n.mi. orbit is in fact 1,670 m/s.
Furthermore, your quoted numbers suggest that the delta-v was 2400 – 1500 = 900 m/s. This again is incorrect. From the same source was see that the velocity change was 2,917.5 ft/s, or 889.25 m/s, which is the delta-v imparted by the SPS.
Why is the delta-v not simply the initial velocity minus final velocity, i.e. 2,514.6 – 1,670.0 = 844.6 m/s? This is because the spacecraft was also changing altitude during the burn – Apollo 11’s altitude at LOI ignition was 86.7 n.mi. The drop in altitude caused a positive change in velocity (resulting from potential energy converting to kinetic energy) at the same time the propulsion system was slowing Apollo down. For this reason, the amount of delta-v delivered by the propulsion system is more than the actual apparent change in velocity.
This is another reason, Heiwa, that your attempted energy balance fails so miserably, you’ve completely ignored changes in potential energy resulting the spacecraft’s change in altitude. An energy balance must account for all energy, kinetic + potential, and must account for all system elements, spacecraft dry mass + fuel/exhaust mass.
Fortunately, the entire problem is more easily and correctly solved in terms of momentum using Tsiolkovsky's rocket equation. In this case we need not even know the initial and final velocities. All we need to know is the delta-v, which is 889.25 m/s
Your space ship has a P-22 KS rocket engine.No it doesn’t; the SPS engine was an AJ10-137.
with 97 400 N thrust (at full blast).The rated thrust was 91,189 N (20,500 lbf), though that’s not necessarily the actual on the job performance. Things as simple as the initial temperature of the propellants will have an effect of thrust. In reality the thrust varied depending on factors such as temperature, upstream pressure at the injector, ablative liner wear, and probably other things I’m not thinking of. However, it’s possible to back calculate the actual thrust based on the known and recorded performance.
If you suggest, e.g. 10 898 kgThat number is close to accurate. Referring back to
this source we see that the CSM/LM mass at LOI ignition was 96,061.6 lbm, or 43,572.8 kg. The change in mass, therefore, was 43,572.8 – 32,675.7 = 10,897.1 kg. Although the vast majority of that change is due to the consumption of SPS propellant, I suspect a small part was RCS propellant.
How much fuel do you require to carry out the braking maneuver? We’ll get to this later.
Say that your space ship has mass 12 153 kg excluding fuel and that you must speed up from 1 500 to 2 400 m/s velocity to get out of Moon orbit to carry out a so called trans-Earth injection. Your space ship still has a P-22 KS rocket engine with 97 400 N thrust (at full blast). How much fuel do you require to carry out the acceleration maneuver?
If you suggest, e.g. 4 676 kg, you must support your answer with proper calculations to win the prize (€1M). I have a feeling you need > 20 000 kg.
In order to proceed with the discussion, I suggest you try to clarify above basic questions of fuel consumption.
Again, many of your facts are wrong, though it looks like you managed to get a few right.
Say that your space ship has mass 12 153 kg excluding fuelI cannot confirm the above number from my current source, though it does appear to be approximately the mass I’d expect at TEI cutoff.
If you suggest, e.g. 4 676 kgAdding these two number together we get, 12,153 + 4,676 = 16,829 kg, which should represent the CSM mass at LEI ignition. Referring back to
this source we see the CSM mass after LM jettison was 37,100.5 lbm (16,828.5 kg). This appears to confirm that Heiwa is using legitimate mass numbers, so I’ll concede these two numbers are probably correct.
you must speed up from 1 500 to 2 400 m/s velocity to get out of Moon orbit to carry out a so called trans-Earth injection.Wrong again on the velocities. From
this source we see that the pre- and post-TEI velocities were 5,376.0 ft/s (1,638.6 m/s) and 8,589.0 ft/s (2,617.9 m/s) respectively.
However, as before, we don’t need to know the velocities to solve the problem. All that’s important is the delta-v, which we see is 3,279.0 ft/s, or 999.44 m/s.
Your space ship still has a P-22 KS rocket engineOnly the most obtuse person on the planet or a troll would still be calling the engine a P-22 KS.
97 400 N thrust (at full blast)As previously stated, 91,189 N rated, actual to be determined.
How much fuel do you require to carry out the acceleration maneuver? To come.
In order to proceed with the discussion, I suggest you try to clarify above basic questions of fuel consumption.Already done ad nauseum.
When I started this post my intent was only to point out may of Heiwa’s factual errors – I had no intent to actually perform the calculations. However, even though much of it has already been done, why not recap?
Lunar Orbit InsertionLet’s start by recognizing that the rated specific impulse of the AJ10-137 is 314 seconds, though like the thrust, actual performance varies depending on a multitude of factors. The first calculation assumes actual Isp = rated Isp.
Final mass = 32,675.7 kg
Delta-v = 889.25 m/s
Effective exhaust gas velocity, C = Isp * g
o = 314 * 9.80665 = 3,079.3 m/s
Using Tsiolkovsky's rocket equation,
Initial mass = Final mass * EXP[ delta-v / C ]
Initial mass = 32675.7 * EXP[ 889.25 / 3079.3 ] = 43,615.6 kg
Propellant used = Initial mass – Final mass
Propellant used = 43615.6 – 32675.7 = 10,939.9 kg
Knowing the actual initial and final masses, and the actual delta-v, we can also solve the problem to determine the actual effective exhaust gas velocity, and thus the actual specific impulse. In this case we see that the actual propellant consumption was less than that calculated above, indicating we got above nominal performance. Let's determine,
Initial mass = 43,572.8 kg
Rearranging Tsiolkovsky's rocket equation,
C = Delta-v / LN[ Initial mass / Final mass ]
C = 889.25 / LN[ 43572.8 / 32675.7 ] = 3,089.8 m/s
Therefore,
Isp = C / g
o = 3089.8 / 9.80665 = 315.07 s
We can also calculate the actual thrust of the engine, thus
Time of burn = 357.53 s
Propellant flow rate, q = (43572.8 – 32675.7) / 357.53 = 30.479 kg/s
Thrust = C * q = 3,089.8 * 30.479 = 94,174 N
(These calculations assume that the entire change in mass is due to the consumption of SPS propellant, ignoring the possibility some may be RCS propellant.)
Transearth InjectionFinal mass = 12,153 kg
Delta-v = 999.44 m/s
Again using the rated Isp of 314 s, we obtain
Initial mass = 12153 * EXP[ 999.44 / 3079.3 ] = 16,812.8 kg
Propellant used = 16812.8 – 12153 = 4,659.8 kg
This time the actual propellant consumption appears to be greater than our calculation, therefore the engine performance was a little below nominal. Let’s find out,
Initial mass = 16,828.5 kg
C = 999.44 / LN[ 16828.5 / 12153 ] = 3,070.5 m/s
Isp = C / g
o = 3070.5 / 9.80665 = 313.10 s
Let’s finish up by calculating the thrust during the TEI burn,
Time of burn = 151.41 s
Propellant flow rate, q = (16828.5 – 12153) / 151.41 = 30.880 kg/s
Thrust = C * q = 3,070.5 * 30.880 = 94,817 N
As we can see, the flow rate and thrust of the second burn was a little higher than the first, though the specific impulse went down a bit. As described earlier, these variations are not unexpected as conditions change. Say for instance, the Isp might have dropped because the propellants were colder, and the flow rate might have increased because of erosion of the ablative nozzle throat. There are many reason that can account for the difference.
Heiwa, your challenge has been met time and time again. Only your ignorance prevents you from seeing it.
EDIT: I had to revise several of the above LOI calculations because in the middle of the computations I inadvertently started using 886.25 for the delta-v instead of the correct number of 889.25 m/s. It was just a simple typo that is now corrected.
