### Author Topic: Is my math correct? LM descent engine force at nozzle  (Read 14661 times)

#### Allan F

• Jupiter
• Posts: 919
##### Is my math correct? LM descent engine force at nozzle
« on: April 26, 2013, 05:30:14 PM »
I was wondering, what the force of the exhaust of the LM's descent engine was, when the LM was close to landing, and the descent fuel was almost used up.

So I used the formula F = m x a, with mass m 6.800 kg, acceleration a 1.62 m/s, to calculate what the force needed to balance the lunar gravity. and got 11000 N. That's the equivalent of lifting 1120 kg on earth. So far so good.

The nozzle area is pi x r^2, with r being the radius of the nozzle at the bottom - 0.75 meters. That's 1.766 m^2. That's 17660 cm^2.

That gives a force at the nozzle of 0.62 N/cm^2.

This force seems very very low, so I would like a reality check on my math, please.

Edit: That equates to a force equivalent to 62 grams/cm^2.
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.

#### ka9q

• Neptune
• Posts: 3010
##### Re: Is my math correct? LM descent engine force at nozzle
« Reply #1 on: April 26, 2013, 10:15:52 PM »
Yeah, your calculation sounds about right. But I'm not sure what the result actually means.

I'm still learning all this stuff myself, so maybe you can help me figure it out. The nozzle is a heat engine that turns the random thermal motions of the slow moving, hot, high pressure gas molecules in the combustion chamber into a cool, low pressure but rapidly moving exhaust gas. Ideally all of the heat is turned into kinetic energy and you have a rapidly moving exhaust stream at zero temperature and zero pressure.

You could do this in a vacuum with an infinitely long nozzle but that's not practical. The conversion can't be complete. As in any heat engine, some of the input heat is wasted as residual heat at a lower temperature. The exhaust is much cooler than in the combustion chamber, but still hot. So it has residual pressure, and in a vacuum the plume will immediately start to expand sideways as it leaves the nozzle, and those molecules don't contribute to thrust. Depending on why you're calculating it (e.g., to examine the effects on the lunar surface) you might want to take it into account.

The textbooks say that rocket engines produce two kinds of thrust: momentum thrust and pressure thrust. Momentum thrust is easy to understand; it's the actual exhaust velocity times the mass flow rate. Classic action-reaction. Momentum thrust is always positive, but it depends on the efficiency of the nozzle in converting heat to kinetic energy. It's generated when molecules bounce off the top of the combustion chamber, or the inside of the nozzle, transfer their momentum, and then leave without hitting any other part of the engine.

I'm still trying to grok pressure thrust. It's defined as the nozzle exit area times the difference between exhaust and ambient pressure. In an atmosphere this can be either positive or negative, but in a vacuum it's always positive. Optimum efficiency occurs when the exhaust pressure is exactly equal to ambient and pressure thrust is zero, because this maximizes both momentum thrust and overall thrust. There's an ideal nozzle length in an atmosphere. In a vacuum, a longer nozzle is always better (except for weight and size).

What bothers me is that "pressure thrust" sounds a lot like the misconception that rockets work by pushing on the air. Obviously that's not the case here, but I don't fully understand it.

It's easy to see that the random heat motions that happen to be sideways don't contribute to thrust. What I don't quite understand is the effect of those residual thermal motions that happen to be in the desired direction, i.e., parallel to the exhaust flow, but I think this is where pressure thrust comes from. I'm trying to visualize how these molecules transfer their momentum to the rocket. I think it's indirect; they bounce into each other until one eventually bounces into the engine nozzle.

What I think you've computed is yet another kind of pressure: stagnation pressure. If I understand it correctly, it's what you get when the plume impacts a large flat surface. Although the pressure in the plume may be very low (i.e., it doesn't tend to expand), when it hits the molecules will deliver their momentum in the process of being deflected sideways. Many will collide with each other and the plume will heat up as well.
« Last Edit: April 26, 2013, 10:32:19 PM by ka9q »

#### Allan F

• Jupiter
• Posts: 919
##### Re: Is my math correct? LM descent engine force at nozzle
« Reply #2 on: April 26, 2013, 10:37:56 PM »
Yes, that seems to be the stagnation pressure. I'm aware of the sideways expansion, once the exhaust leaves the nozzle. And you're correct, it is to get a feel for the impact it had on the surface.

I don't think the exhaust has much effect on the force on the spacecraft, once the gas has exited the nozzle.

And I seem to have forgot a "^2" after 1.6 m/s.

Edit: The rest of your post is beyond my current level, but I'll see if I can find some reading material to improve my understanding.
« Last Edit: April 26, 2013, 10:57:37 PM by Allan F »
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.

#### ka9q

• Neptune
• Posts: 3010
##### Re: Is my math correct? LM descent engine force at nozzle
« Reply #3 on: April 26, 2013, 11:12:30 PM »
For what it's worth, I've been playing with the equations for isentropic flow in an ideal rocket nozzle, so here are the numbers I calculated for the LM DPS at 20% throttle, which is about where it was at the Apollo 11 landing.

The chamber pressure in psi was almost exactly equal to the throttle setting in percent, so the chamber was 20 psi.

The expansion ratio is given as 47.5:1 and the exit diameter as 59.0 inch. From that I compute a throat area of  0.0372 m^2. (Caveat: the J-series LMs had longer nozzles, and I'm not sure where this 47.5:1 figure came from. I think it's for the pre-J LMs, including Apollo 11).

From Bob B's web pages I extracted the following information for Aerozine-50 & N2O4:

k = 1.228
mol wt = 21.8 g/mol
adiabatic flame temp = 3100 K

I may have done this wrong; these values are from graphs that vary with chamber pressure and mixture ratio, and I didn't know the latter offhand. I had to extrapolate his plots down to the LM's chamber pressure, and Bob himself gave the caveat that, unlike other propellants, his calculations seem to differ from practice for AZ-50/N2O4.

Anyway, I compute the following:

Mass rate = 3.09 kg/s
exit temp = 927 K
exit press = 207 Pa (.002 atm!; 25-50% Mars atmospheric pressure)
exit density = .000585 kg/m^3 (0.05% of sea level air)
exit velocity = 2988 m/s
Isp = 317 sec, effective velocity = 3106 m/s
Carnot efficiency = 70.1% (efficiency of nozzle as heat engine = 1 - exit_temp/chamber_temp)
Total thrust = 9590 N
Momentum thrust = 9225 N
Pressure thrust = 365 N

As a check, the Isp is specified to be 311 s at an unknown throttle setting, and actual landing thrust would be 20% * 45.04 kN = 9 kN. So I'm not too far off given the questionability of some of my source numbers. But it's rather striking to see the extremely low density, pressure and temperature of the exhaust, and of course these figures would be even lower by the time the plume impacted the surface.

Oh, and these low numbers underscore just how silly Ralph Rene's leaf blower "tests" really were.

« Last Edit: April 26, 2013, 11:24:13 PM by ka9q »

#### ka9q

• Neptune
• Posts: 3010
##### Re: Is my math correct? LM descent engine force at nozzle
« Reply #4 on: April 26, 2013, 11:27:38 PM »
Edit: The rest of your post is beyond my current level, but I'll see if I can find some reading material to improve my understanding.
It was beyond mine too until just recently, when I decided to finally try to learn the thermodynamics of rocket engines. I was prompted by an argument with yet another Youtube moron who claimed that rockets can't work in a vacuum because they push on air. As always, I learn something while they learn nothing.

#### Allan F

• Jupiter
• Posts: 919
##### Re: Is my math correct? LM descent engine force at nozzle
« Reply #5 on: April 26, 2013, 11:53:01 PM »
Your molar weight seem low. Nitrogen tetroxide should be 4x16 + 2x14 = 92 g/mol and aerozine 50, (60 + 32)/2 = 46.

Yes, the gain seems to be one-sided in these discussions.

Edit: The molar weight you quote, is that for the combustion products?
« Last Edit: April 26, 2013, 11:59:52 PM by Allan F »
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.

#### Allan F

• Jupiter
• Posts: 919
##### Re: Is my math correct? LM descent engine force at nozzle
« Reply #6 on: April 26, 2013, 11:58:54 PM »
What numbers do you use for the nozzle? I've been using 1.5 meter, but I'm willing to accept a superior source.
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.

#### ka9q

• Neptune
• Posts: 3010
##### Re: Is my math correct? LM descent engine force at nozzle
« Reply #7 on: April 27, 2013, 12:24:45 AM »
Edit: The molar weight you quote, is that for the combustion products?
Yes, it's the combustion products that count. For AZ-50 & N2O4 at the usual rich mixture I'd expect mostly CO, CO2, N2, H2O and H2.

What complicates the process is that the equilibrium among these compounds changes as the exhaust cools in the nozzle. This is called "shifting equilibrium" but it's currently beyond my level to analyze.

#### ka9q

• Neptune
• Posts: 3010
##### Re: Is my math correct? LM descent engine force at nozzle
« Reply #8 on: April 27, 2013, 12:47:37 AM »
What numbers do you use for the nozzle? I've been using 1.5 meter, but I'm willing to accept a superior source
I don't see a nozzle or throat diameter specifically listed, but as with the F-1 and J-2 engines I took the advertised engine diameter and assumed that was the diameter of the nozzle exit (as it appears to be in the pictures). The figure I found for the LM is 59.0 inch or 150 cm, but I don't know for sure if that's the pre-J or J version.

#### VQ

• Earth
• Posts: 146
##### Re: Is my math correct? LM descent engine force at nozzle
« Reply #9 on: April 27, 2013, 01:11:30 AM »
0.62 N/cm^2 isn't all that low of pressure, either. That is 6200 N/m^2=6200 Pa, or about 6% of atmospheric pressure at sea level.

#### ka9q

• Neptune
• Posts: 3010
##### Re: Is my math correct? LM descent engine force at nozzle
« Reply #10 on: April 27, 2013, 01:12:33 AM »
Another interesting thing I've noticed is if you compute an "equivalent thrust area" by dividing the total thrust by the chamber pressure, you get an area somewhere between 1.5 and 2 times the throat area. I take this to mean that somewhere between half and two-thirds of the total thrust is generated by hot, high pressure gas pushing on the top of the combustion chamber opposite the nozzle, and the remainder is generated by the expanding, lower pressure gas pressing on the inside of the much larger nozzle as it expands. You can visualize those molecules producing thrust as they bounce off the curved shape of the nozzle, deposit their momentum and are redirected toward the exit.

#### ka9q

• Neptune
• Posts: 3010
##### Re: Is my math correct? LM descent engine force at nozzle
« Reply #11 on: April 27, 2013, 01:15:32 AM »
0.62 N/cm^2 isn't all that low of pressure, either. That is 6200 N/m^2=6200 Pa, or about 6% of atmospheric pressure at sea level.
It's interesting to compare that to my calculated 207 Pa exit pressure. That's what you'd read on a gauge drilled through the edge of the nozzle right before the gas exits. Your 6200 Pa reflects the "stagnation" pressure, the total pressure you'd get as the fast-moving gas slams into a blunt obstacle.

The engine does a pretty good job of turning all that heat and pressure into kinetic energy, but you can turn it back to heat and pressure by putting a solid barrier in the flow.
« Last Edit: April 27, 2013, 01:21:12 AM by ka9q »

#### AtomicDog

• Mars
• Posts: 368
##### Re: Is my math correct? LM descent engine force at nozzle
« Reply #12 on: April 27, 2013, 12:18:50 PM »
Eh, it's all Rocket Science to me.
"There is no belief, however foolish, that will not gather its faithful adherents who will defend it to the death." - Isaac Asimov

#### Daniel Dravot

• Mercury
• Posts: 24
##### Re: Is my math correct? LM descent engine force at nozzle
« Reply #13 on: April 27, 2013, 01:18:58 PM »
I was prompted by an argument with yet another Youtube moron who claimed that rockets can't work in a vacuum because they push on air.

Seems an awful waste, carrying all that oxygen around, when the damn things only work in places where oxygen is available for free.

#### ka9q

• Neptune
• Posts: 3010
##### Re: Is my math correct? LM descent engine force at nozzle
« Reply #14 on: April 27, 2013, 02:17:49 PM »
Yeah, I keep pointing that out. The Saturn V was something like 2/3 LOX by weight at launch. They never seem to have an answer for that. In fact, they never seem to have an answer for a lot of questions that would back them into a corner.