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Electric propuslion thrust to weight ratio Q&A by aero on 26 Apr, 2012 20:27
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What type of electric propulsion has or promises the highest thrust to weight ratio. Not counting the power source but counting the fuel density. Is it scalable to higher thrust?
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#1 by Nomadd on 26 Apr, 2012 20:46
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What does fuel density have to do with figuring thrust to weight ratio?
The Air Force had a pretty good rundown of technologies.
http://www.dtic.mil/cgi-bin/GetTRDoc?AD=ADA513936 -
#2 by ChileVerde on 26 Apr, 2012 20:47
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What type of electric propulsion has or promises the highest thrust to weight ratio. Not counting the power source but counting the fuel density. Is it scalable to higher thrust?
MPD seems to have held the "most promising" place for decades.
http://en.wikipedia.org/wiki/Magnetoplasmadynamic_thruster -
#3 by strangequark on 26 Apr, 2012 20:50
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MPDT at high power levels (>1 MW)
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#4 by Robotbeat on 27 Apr, 2012 04:12
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MPDT at high power levels (>1 MW)
Yes, definitely. Anything the size of a shoebox that can take 1MW is going to be pretty hot, though, even if you can make it relatively efficient. It'll glow like a lightbulb or at least a toaster oven heating element (better be made of something like Tungsten, maybe titanium for the slightly cooler parts to save mass). -
#5 by Nomadd on 27 Apr, 2012 19:12
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I think "has or promises" is a little broad. Are you talking about real hardware or something from Star Trek?
Did they ever get anywhere with the ELF-375? That was going to be a 50kg 200kw engine that could do 18 Newtons. -
#6 by strangequark on 27 Apr, 2012 19:52
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MPDT at high power levels (>1 MW)
Yes, definitely. Anything the size of a shoebox that can take 1MW is going to be pretty hot, though, even if you can make it relatively efficient. It'll glow like a lightbulb or at least a toaster oven heating element (better be made of something like Tungsten, maybe titanium for the slightly cooler parts to save mass).
Rhenium. -
#7 by aero on 05 May, 2012 02:49
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I think "has or promises" is a little broad. Are you talking about real hardware or something from Star Trek?
Promises - like within the next 10 years, or lacking only funded development. That makes it real or near term hardware.
Did they ever get anywhere with the ELF-375? That was going to be a 50kg 200kw engine that could do 18 Newtons. -
#8 by Robotbeat on 06 May, 2012 04:54
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Thrust/weight is the wrong metric for electric propulsion. People generally talk about total electrical efficiency, specific power (how much power can be handled per kilogram), and exhaust velocity (i.e. Isp). As well as total propellant throughput (i.e. thruster life).
(Of course, you can derive thrust/weight from these things, but it's generally not what you're directly interested in for calculations.) -
#9 by QuantumG on 06 May, 2012 05:04
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That's just cause the thrust/weight is so notoriously low that it's not worth discussing.. if that wasn't the case it'd be more relevant.
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#10 by Robotbeat on 06 May, 2012 05:12
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That's just cause the thrust/weight is so notoriously low that it's not worth discussing.. if that wasn't the case it'd be more relevant.
Right, and if we had flying saucers, we wouldn't need jet airplanes. It's just a different kind of technology with its own set of relevant metrics. -
#11 by QuantumG on 06 May, 2012 11:23
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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.
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#12 by Robotbeat on 06 May, 2012 15:57
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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 thrust. -
#13 by aero on 08 May, 2012 22:08
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Thanks to All who replied. I have achieved my objective and posted on Advanced Concepts, here:
http://forum.nasaspaceflight.com/index.php?topic=28733.msg894897#new
Feel free to comment without flaming. -
#14 by strangequark on 09 May, 2012 00:33
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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
! However, that is a very fair point, and Quantum, it's not just because it's low. An electric thruster is a low thrust, continuous operation, in space device. Your acceleration rate in this regime isn't the most important factor. We are willing to accept low acceleration rates, because we'll be thrusting for a long period of time. What matters is power to weight. This is the same reason that military jet engines give thrust to weight as a metric (rapid acceleration is important), whereas civil jet engines use power to weight (sustained performance is important). Additionally, the solar arrays (or reactor + radiators) and thruster both will scale as a function of power (reflecting the contributions of both Isp and thrust). Therefore, specific mass (kg/W) is the important metric. For any kind of rocket, what you really care about is payload mass fraction, and you can write that equation either in terms of specific power or thrust/weight (See attachment). Notice in the first equation that what matters is the ratio of vehicle T/W to engine T/W. We can make this ratio the same for an electric stage as it is for a chemical stage, because we can make T/W for the vehicle about as low as we want, so T/W for the engine flat out doesn't matter. All the while, we can raise Isp/u_e to drop R very low and raise mass fraction (at the expense of a longer burn time).
With all of that said, MPDT is still probably your best near term technology for low specific mass (high power to weight) as well, but the real driver is going to be specific mass of your power processing hardware and photovoltaic array (or nuclear reactor + radiators if you swing that way)
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#15 by aero on 09 May, 2012 19:14
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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! However, that is a very fair point, and Quantum, it's not just because it's low. An electric thruster is a low thrust, continuous operation, in space device. Your acceleration rate in this regime isn't the most important factor. We are willing to accept low acceleration rates, because we'll be thrusting for a long period of time. What matters is power to weight. This is the same reason that military jet engines give thrust to weight as a metric (rapid acceleration is important), whereas civil jet engines use power to weight (sustained performance is important). Additionally, the solar arrays (or reactor + radiators) and thruster both will scale as a function of power (reflecting the contributions of both Isp and thrust). Therefore, specific mass (kg/W) is the important metric. For any kind of rocket, what you really care about is payload mass fraction, and you can write that equation either in terms of specific power or thrust/weight (See attachment). Notice in the first equation that what matters is the ratio of vehicle T/W to engine T/W. We can make this ratio the same for an electric stage as it is for a chemical stage, because we can make T/W for the vehicle about as low as we want, so T/W for the engine flat out doesn't matter. All the while, we can raise Isp/u_e to drop R very low and raise mass fraction (at the expense of a longer burn time).
With all of that said, MPDT is still probably your best near term technology for low specific mass (high power to weight) as well, but the real driver is going to be specific mass of your power processing hardware and photovoltaic array (or nuclear reactor + radiators if you swing that way)
Quote from wikipedia
http://en.wikipedia.org/wiki/Magnetoplasmadynamic_thrusterQuoteIn theory, MPD thrusters could produce extremely high specific impulses (Isp) with an exhaust velocity of up to and beyond 110,000 m/s, triple the value of current xenon-based ion thrusters, and about 20 times better than liquid rockets. MPD technology also has the potential for thrust levels of up to 200 newtons (N) (45 lbf), by far the highest for any form of electric propulsion, and nearly as high as many interplanetary chemical rockets. This would allow use of electric propulsion on missions which require quick delta-v maneuvers (such as capturing into orbit around another planet), but with many times greater fuel efficiency. [1]
Unfortunately the article goes from there into the problems of low power available in space and difficulty of testing high power MPD thrusters on Earth, none of which concerns me at this time.
I need a ball park estimate of mass scaling for the thruster/PPU in the 100 MW range. So far, I have used 200 kW devices and scaled using the given value of kg/kW. I'm sure that this approach gives a very pessimistic mass estimate when scaled up to MW/tonne. Does anyone have a reasonable approach to scaling the mass of high power electronics like used in the PPUs? What is it? -
#16 by strangequark on 09 May, 2012 20:08
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Unfortunately the article goes from there into the problems of low power available in space and difficulty of testing high power MPD thrusters on Earth, none of which concerns me at this time.
I need a ball park estimate of mass scaling for the thruster/PPU in the 100 MW range. So far, I have used 200 kW devices and scaled using the given value of kg/kW. I'm sure that this approach gives a very pessimistic mass estimate when scaled up to MW/tonne. Does anyone have a reasonable approach to scaling the mass of high power electronics like used in the PPUs? What is it?
No one has done any serious work in that regime for in space power. I have seen scaling done for devices up to about 1 MW, but 100 MW is huge.
One the other thread, you give 35 hours to lunar orbit. How are you calculating that?
I get much more optimistic numbers than yours, even with a straight linear scaling with a 35 hour burn time (which can be very different from transit time). -
#17 by Robotbeat on 09 May, 2012 20:58
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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!
..., 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.
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#18 by strangequark on 09 May, 2012 23:33
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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!
..., 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. -
#19 by aero on 09 May, 2012 23:56
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Unfortunately the article goes from there into the problems of low power available in space and difficulty of testing high power MPD thrusters on Earth, none of which concerns me at this time.
I need a ball park estimate of mass scaling for the thruster/PPU in the 100 MW range. So far, I have used 200 kW devices and scaled using the given value of kg/kW. I'm sure that this approach gives a very pessimistic mass estimate when scaled up to MW/tonne. Does anyone have a reasonable approach to scaling the mass of high power electronics like used in the PPUs? What is it?
No one has done any serious work in that regime for in space power. I have seen scaling done for devices up to about 1 MW, but 100 MW is huge.
One the other thread, you give 35 hours to lunar orbit. How are you calculating that?
I get much more optimistic numbers than yours, even with a straight linear scaling with a 35 hour burn time (which can be very different from transit time).
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. ELF-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.
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.
Wikipedia 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.
Edit 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.
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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!
..., 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?
! However, that is a very fair point, and Quantum, it's not just because it's low. An electric thruster is a low thrust, continuous operation, in space device. Your acceleration rate in this regime isn't the most important factor. We are willing to accept low acceleration rates, because we'll be thrusting for a long period of time. What matters is power to weight. This is the same reason that military jet engines give thrust to weight as a metric (rapid acceleration is important), whereas civil jet engines use power to weight (sustained performance is important). Additionally, the solar arrays (or reactor + radiators) and thruster both will scale as a function of power (reflecting the contributions of both Isp and thrust). Therefore, specific mass (kg/W) is the important metric. For any kind of rocket, what you really care about is payload mass fraction, and you can write that equation either in terms of specific power or thrust/weight (See attachment). Notice in the first equation that what matters is the ratio of vehicle T/W to engine T/W. We can make this ratio the same for an electric stage as it is for a chemical stage, because we can make T/W for the vehicle about as low as we want, so T/W for the engine flat out doesn't matter. All the while, we can raise Isp/u_e to drop R very low and raise mass fraction (at the expense of a longer burn time).