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#20 by Robotbeat on 10 May, 2012 02:12
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Well, no need to apologize! But I may just take you up on that offer... (It just so happens that I'll be in Phoenix next week on business.)
Yes, I'm a physics grad student right now*There's no need for that attitude.. arcjets have good thrust/weight compared to other electric propulsion technologies and have been proposed for ground launch systems before.
Because of their low Isp. For the same power, Isp is inversely proportional to the square root of thrust.
Now, now, you're a physicist IIRC. Be precise
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..., but I'm pretty sure (just derived it again to make sure) that if you keep power level the same (and allow mass rate to change), Isp is just inversely proportional to thrust.
Now, if you keep mass rate the same (and allow power to change), then Isp is proportional to thrust (and power is proportional to the square of Isp, or rather Isp is proportional to the square root of power).
The fly in the ointment of these calculations, as you know, is that they assume efficiency remains constant (100%). In reality, you tend to get higher efficiency with higher Isp (for ion thrusters, at least), since the ionization losses are less significant when you are putting a lot more kinetic energy into each ion.
(Everyone: It works better if you think "exhaust velocity" every time "Isp" is mentioned, here.)
*Yay for below-poverty-level stipends.
Well dammit if I didn't just put my foot in my mouth. You are absolutely correct. I was thinking of mass flow rate, and didn't actually revisit the equations myself. You have my apologies, and a free beer if you're ever in the Phoenix area.
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#21 by strangequark on 11 May, 2012 23:50
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Quote
I did it all on an Excel spreadsheet. I started with a range of mast heights, 500 to 3000 meters and settled on 1500 meters. Two right triangular solar arrays 1500 meters per side, per mast, and two masts, at 4.2% efficiency and with specific mass of 4 g/m^2 gives 257,229 kW at 18,000 kg for the solar power source.
Okay, this is strange, because you're using very, very aggressive values for specific mass, while being abysmally pessimistic on efficiency. A triple junction gallium arsenide cell will have 30% efficiency. However, state of the art specific mass for solar arrays is maybe 3-4 kg/kW, versus your 0.07 kg/kW.QuoteELF-375 thrusters mass 0.25kg/kW (total 64307.25 kg) and the PPU masses 0.45 kg/kw (total 115,753.05) so the solar arrays, thrusters and PPUs together mass 198,060.3 kg. Then I added 195,960 kg of propellant and an extra 100,000 kg for the balance of spacecraft to come up with the total mass of 494,020.3 at the start of the mission.
This set of assumptions is okay.
The ELF-375 thrusters at high thrust give 95 mN/kW for a total of 24436.755 N, and the mass flow for one 200 kW thruster is 1200 mg/s so it totals 1.543374 kg/s from 257,229 kW's worth of thrusters.QuoteWikipedia gives 8km/s as the delta V needed for a low thrust trip to LLO, so I broke the trip into 400 intervals and use the trapezoidal rule to integrate the spacecraft mass and acceleration. It’s a linear math problem that way so trapezoidal integration should be accurate. I adjusted the integration time step size to reach exactly 8000 m/s at step 400 and iterated on the fuel load until the remaining fuel was zero when the velocity reached 8000 m/s. I actually stopped with 0.89 kg of fuel in the tank and 0.23 m/s over speed and the final time step was 317.42 seconds per interval.
Okay, delta-V sounds a little high, but fair enough. However, you are going about this the hard way. The rocket equation is an exact analytical solution to what you are trying to do with the above.QuoteEdit added: So you see I didn't really get to the moon though I did travel 466,567 km along whatever path I was on. I need to find and learn to use a trajectory integrator program.
It's easy enough to do a poor man's version in Excel (or better yet, C++/FORTRAN if you can code). I have attached a plot from an Excel sheet I built to do just that (I leave it as an exercise to the reader, but I can give you a little more to go on if you like). The plot omits several of the inner loops of the spiral so that it isn't as crowded. Dimensions are thousands of kilometers.
It sounds like you started with a solar array size, and moved from there. A better approach would be:
1. Specify your inert mass. For now, leave out tanks, solar arrays, propulsion bus, and just use your desired payload mass
2. Specify maximum trip time
3a (Better Option). Identify optimum Isp that maximizes payload mass fraction for the given trip time, reality check to make sure that electric propulsion is still viable
3b (Okay Option). Use the demonstrated Isp from your favorite tech
4. Use the Tsiolkovsky equation (aka The Rocket Equation) to identify how much propellant you will need for your payload
5. Use propellant and trip time to determine mass flow rate (constant thrust assumption, fine for EP), which with Isp gives you thruster power
6. Taking into account thruster efficiency, determine solar array power
7. Size solar array, based on array efficiency
8. Update payload size with values for solar arrays, etc. based on calculated power level and propellant mass
9. Reiterate 4-8 until convergence
You can also get a closed form solution, which is easier to work with, once you've done the math. For 100 mT of desired payload, and under reasonable-ish assumptions, I get the following similar spacecraft:
Trip Time: 30 days
Isp: 2109 s
Thrust: 673 N
Power: 14 MW
PVA Array Area: 34000 square meters (Equivalent to 10 full ISS Arrays)
Mass Breakdown
Payload: 100 mT
Propellant: 84 mT of cryogenic Argon
Misc. Structure: 10 mT
Tank Mass: 6 mT
PV Array Mass: 56 mT
Propulsion Bus Mass: 7 mT
Total Mass: 263 mT
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#22 by aero on 12 May, 2012 03:47
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@ strangequarkQuoteOkay, this is strange, because you're using very, very aggressive values for specific mass, while being abysmally pessimistic on efficiency. A triple junction gallium arsenide cell will have 30% efficiency. However, state of the art specific mass for solar arrays is maybe 3-4 kg/kW, versus your 0.07 kg/kW.
I believe you looked at my reference for the solar cells. I copied the mass and efficiency values straight from the reference. I know that the mass does not include necessary conductors or power conditioners needed to make it a real solar power plant, but I also know that the efficiency given is for laboratory items. Maybe, if it ever reaches production, it will be at a higher efficiency. I could arbitrarily double the mass and efficiency, but it would be just that, arbitrary, as I don’t have any data to justify either one.
I chose this extremely low mass solar cell deliberately in support of my mission which I outlined in my first post on the other thread. The mission is to win the Yet To Be Conceived X-Prize Electric Rocket Circumlunar Race. The yet to be conceived prize is offered to encourage space tourism beyond LEO. It starts and ends in LEO with at least one complete LLO, hence it is a race, because tourists will not pay well to spend months in isolation in cramped quarters between Earth and Moon. With those ground rules, I believe I have won the race so far, but I have yet to complete the second leg, that is, from LLO back to LEO.QuoteA better approach would be: …
Would that really be a better approach for my task? The task is to carry a big enough payload to provide a comfortable trip for tourists, and to win the race by going faster than anyone else electric.
Yes, I can program but right now I am exploring NASA’s GMAT program for the mission analysis. It is complete with displays and ephemerides for the major bodies in the solar system. Using GMAT, I can plan my spacecraft and mission. Maybe I can actually go to the moon and know that I got there, numerically. The inputs to GMAT seem to be along the lines you suggest as a better approach.
Now, I have a question for you regarding your mass breakdown. What kind of thrusters did you use? I see 7 mT for the propulsion bus mass which must be thruster plus PPU mass. If so, that is 0.5 kg/kW which is better than the 0.7 kg/kW of the ELF-375. So what kind of engines are you using?
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#23 by aero on 12 May, 2012 17:49
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This morning while reading a NASA study paper about solar power station development, it occurs to me than maybe we could all buy electrical power beamed to our yachts from a Space Solar Power station. A 5 to 10 GW solar power station could surely spare a coupe hundred MW of power to energize our yachts beamed either by microwave or laser. It would sure cut the mass of our yachts and motivate development of MW class thrusters, too.
What would be some of the problems with that system? -
#24 by rds100 on 12 May, 2012 17:52
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This morning while reading a NASA study paper about solar power station development, it occurs to me than maybe we could all buy electrical power beamed to our yachts from a Space Solar Power station. A 5 to 10 GW solar power station could surely spare a coupe hundred MW of power to energize our yachts beamed either by microwave or laser. It would sure cut the mass of our yachts and motivate development of MW class thrusters, too.
What would be some of the problems with that system?
I for one wouldn't want to stand anywhere near the receiving end of that beam.
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#25 by aero on 12 May, 2012 18:03
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This morning while reading a NASA study paper about solar power station development, it occurs to me than maybe we could all buy electrical power beamed to our yachts from a Space Solar Power station. A 5 to 10 GW solar power station could surely spare a coupe hundred MW of power to energize our yachts beamed either by microwave or laser. It would sure cut the mass of our yachts and motivate development of MW class thrusters, too.
What would be some of the problems with that system?
I for one wouldn't want to stand anywhere near the receiving end of that beam.
Why not? You're in a spacecraft in transit between the earth and the moon, shielded against cosmic rays with receiving antenna between the spacecraft and the shields. Admittedly laser power receivers are only about 50% efficient but microwave receivers are much better. If the beam misses your antenna, it misses your spacecraft so no harm, no foul.
I'm going to start a thread over on Advanced Concepts to explore this idea a little more fully. But Later, its Saturday afternoon here on a beautiful, sunshiny day.
Edit: I looked into it a little further, so no I'm not going to start an SPS thread anywhere for powering spacecraft. If we could overcome the efficiency problems for an SPS, and the transmitter efficiency and range problems, then the receiver efficiency and size problems I still wouldn't use it on my space yacht.
If I were racing my yacht, doing a turn-over maneuver and lost the power beam, where would I go? No place good, I'm pretty sure. -
#26 by Robotbeat on 14 May, 2012 16:11
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This morning while reading a NASA study paper about solar power station development, it occurs to me than maybe we could all buy electrical power beamed to our yachts from a Space Solar Power station. A 5 to 10 GW solar power station could surely spare a coupe hundred MW of power to energize our yachts beamed either by microwave or laser. It would sure cut the mass of our yachts and motivate development of MW class thrusters, too.
Totally pointless when you have good access to an essentially limitless beamed power source (the Sun).
What would be some of the problems with that system?
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