Author Topic: The LM maneuvres pre docking  (Read 50526 times)

Offline Allan F

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Re: The LM maneuvres pre docking
« Reply #15 on: September 18, 2014, 06:08:36 PM »
Well, in my previous math-attempt I discovered the stagnation pressure. Now I have the exhaust temperature at the same point. And the pressure and temperature drops rapidly after it exits the nozzle.
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Offline ka9q

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Re: The LM maneuvres pre docking
« Reply #16 on: September 18, 2014, 06:08:54 PM »
In pulse mode it's all about the transient, so if you were watching the thrusters live you could expect to see a very short visible burst.
If ice crystals were observed, perhaps it's unreacted propellant that has frozen by rapid evaporation. Aerozine-50 is a mix of UDMH and N2H4. The N2H4 has the higher freezing point (+2C) so it would be the first to freeze out. N2O4 freezes at -11C and a higher vapor pressure than N2H4 so it's another candidate.



Offline JayUtah

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Re: The LM maneuvres pre docking
« Reply #17 on: September 18, 2014, 06:30:54 PM »
...it should be at least a little hotter than at the nozzle exit.

Temperature at the surface should be substantially colder than at the nozzle exit plane.
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Offline Bob B.

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Re: The LM maneuvres pre docking
« Reply #18 on: September 18, 2014, 06:33:15 PM »
If ice crystals were observed, perhaps it's unreacted propellant that has frozen by rapid evaporation. Aerozine-50 is a mix of UDMH and N2H4. The N2H4 has the higher freezing point (+2C) so it would be the first to freeze out. N2O4 freezes at -11C and a higher vapor pressure than N2H4 so it's another candidate.

If there are in fact ice crystals, I would believe this explanation before I would believe it to be water ice.  But I still have a hard time believing anything would freeze.  Given the temperature at which the thrusters operate, I would think all of the expelled matter would be in a gaseous state, even unreacted propellant.  I'm not sure if it's even possible to go from gas to solid under these conditions.  Both pressure and temperature are dropping so, looking at a phase diagram, we'd be moving down and to the left, which I think will keep us in the gaseous state.

Offline Allan F

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Re: The LM maneuvres pre docking
« Reply #19 on: September 18, 2014, 06:44:52 PM »
A chamber pressure of only 7.13 bar? Is that in the range usually used for rockets?
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Offline Bob B.

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Re: The LM maneuvres pre docking
« Reply #20 on: September 18, 2014, 06:59:25 PM »
A chamber pressure of only 7.13 bar? Is that in the range usually used for rockets?

For pressure fed engines, yes.  For pump fed, no.

For a pressure fed system it is necessary to pressurized the entire system, including the propellant tanks.  This means the tanks walls have to withstand the pressure, which means they must be thick and heavy.  To keep the weight down, they operate the system at as low a pressure as practical.  This lowers the specific impulse of the engine, but it is made up for in weight savings.

For pumped systems, the high pressure is only downstream of the pumps; therefore the tanks can remain of light construction.  You can therefore ramp up the pressure in the engine without suffering a big weight penalty.  These systems operate at much higher pressures.

Offline ka9q

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Re: The LM maneuvres pre docking
« Reply #21 on: September 18, 2014, 07:36:58 PM »
Given the temperature at which the thrusters operate, I would think all of the expelled matter would be in a gaseous state, even unreacted propellant.  I'm not sure if it's even possible to go from gas to solid under these conditions. 
The thrusters are hot only when firing. When the valves are first opened, some propellant will escape unreacted before ignition; this is why the plume is momentarily visible.

Liquids cannot exist for long in vacuum. A liquid (water, hydrazine, N2O4) suddenly exposed to vacuum immediately starts to boil, removing enough heat that the rest of it may freeze. This certainly happens when water is dumped to space through a simple nozzle without adding enough heat to vaporize it all.

I recently had a long argument with some of Boy Blunder's sycophants about why Apollo water dumps produced ice streams while the sublimators did not.

Offline ka9q

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Re: The LM maneuvres pre docking
« Reply #22 on: September 18, 2014, 07:40:24 PM »
A chamber pressure of only 7.13 bar? Is that in the range usually used for rockets?
Yes, for pressure fed engines. I'm familiar with the 400N thruster used on the AMSAT Phase III spacecraft; it was made by MBB originally for Galileo. The tanks were pressurized with helium to, IIRC, 10 bar. Obviously this must be greater than the combustion chamber pressure.

Offline Bob B.

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Re: The LM maneuvres pre docking
« Reply #23 on: September 18, 2014, 08:41:08 PM »
When the valves are first opened, some propellant will escape unreacted before ignition

OK, that makes sense.

Offline JayUtah

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Re: The LM maneuvres pre docking
« Reply #24 on: September 18, 2014, 08:51:17 PM »
Optimal mix occurs only at steady-state.  During the transient the mix is not optimal and so some unreacted propellant will naturally escape.

In the larger motors (e.g., the SPS) the oxidizer is intentionally pre-injected to smooth the ignition.  This results in a visible, likely incandescent plume for up to 300 milliseconds.
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Offline Allan F

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Re: The LM maneuvres pre docking
« Reply #25 on: September 18, 2014, 09:06:33 PM »
So when talking about the temperature of the descent engines exhaust, would it be reasonable to say the temperature of the exhaust on the surface is (assuming a 45 degree spread after nozzle exit) 896 K x (area of nozzle/area of impact)? Where impact area is (nozzle diameter/2 + distance between nozzle and ground)^2 x pi?

Around 1.7/4.5 x 896K = 338 K? Or 65 C?
Well, it is like this: The truth doesn't need insults. Insults are the refuge of a darkened mind, a mind that refuses to open and see. Foul language can't outcompete knowledge. And knowledge is the result of education. Education is the result of the wish to know more, not less.

Offline ka9q

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Re: The LM maneuvres pre docking
« Reply #26 on: September 18, 2014, 09:34:58 PM »
I think there's more to it than that. When the plume hits the surface, at least some of its kinetic energy is turned back into heat. That's what stagnation temperature is about.

Offline Allan F

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Re: The LM maneuvres pre docking
« Reply #27 on: September 18, 2014, 09:44:24 PM »
Compression heating or frictional heating?
Well, it is like this: The truth doesn't need insults. Insults are the refuge of a darkened mind, a mind that refuses to open and see. Foul language can't outcompete knowledge. And knowledge is the result of education. Education is the result of the wish to know more, not less.

Offline JayUtah

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Re: The LM maneuvres pre docking
« Reply #28 on: September 18, 2014, 09:59:45 PM »
Compression.
"Facts are stubborn things." --John Adams

Offline Bob B.

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Re: The LM maneuvres pre docking
« Reply #29 on: September 18, 2014, 11:03:24 PM »
So when talking about the temperature of the descent engines exhaust, would it be reasonable to say the temperature of the exhaust on the surface is (assuming a 45 degree spread after nozzle exit) 896 K x (area of nozzle/area of impact)? Where impact area is (nozzle diameter/2 + distance between nozzle and ground)^2 x pi?

Around 1.7/4.5 x 896K = 338 K? Or 65 C?

Not exactly.  The effect of volume on pressure is described by the following equation,

(P2 / P1) = (V1 / V2) k

Combining this with the equation in Reply #10 and we get the effect of volume on temperature,

T2 / T1 = (V1 / V2) k – 1

So in your example, where we have a 1.7 to 4.5 change in volume, the new temperature becomes,

T2 = 896 x  (1.7 / 4.5) 1.23 – 1 = 716 K

(ETA) Also note that the change in cross-sectional area doesn't necessarily equate to a proportional change in volume.  Area and volume are proportional only if the velocity remains the same.  As the gas cools its enthalpy (internal energy) decreases.  That decrease in enthalpy is converted to kinetic energy, thus the velocity increases.
« Last Edit: September 18, 2014, 11:13:10 PM by Bob B. »