How [Not] to burn for system leave.

[PB666]

Let me start with the following disclamer. The results I present are for an ideal ship weighing approximately 335 t with 39t of fuel (Xenon or Argon). The ship was designed to shuttle 200t to mars and bring 20t back. The fuel that is provided is ideal fuel and assumes that burns are instantaneous. In reality more fuel would need to be brought along. The drive is assumed to have a nuclear powersupply, no such power supply exists AND if it did the cooling systems for the power supply would be much greater in mass than the mass provided for engines, power-supply, and radiative cooling. The vessel produces in excess of 0.0105 acceleration and has a total dV in the 8.7 km/sec  range.

The first set up assumes that the burn angle occurs around the periapsis of an opposing exit vector. So if a burn span is 270' then the burn begins 45' after the exit vector and continues until 45' bfore the exit vector, resting as it travels through the apoapsis. The first set up gets the ship to Earths average hill sphere with zero velocity relative to the Earth.

SpanBurn	TBurn (sec)(cumm.)	T (to escape)	Dv (m/s)	Waste(m/s)
360 ' = 2π	604413	>7 days	6633	3422
270 = 3π/2	524607	>7 days	5726	2514
240 = 4π/3	452699	>13 days	4917	1705
180 = π	364657	>60 days	3937	725
120 = 2π/3	324279	>200 days	3491	280
90 = π/2	313283	>400 days	3370	159
60 = π/3	306224	~1000 days	3293	81

Burning to Mars. Adding the 6471769 J/kg required to reach Mars at its apogee.

SpanBurn	TBurn (sec)(cumm.)	T (to escape)	Dv (m/s)	Waste(m/s)
360	800005	>9 days	8902	5018
270	719818	>8 days	7964	4080
240	634915	>13 days	6863	3099
180	527567	>60 days	5759	1875
120	491702	>200 days	5355	1471
90	483796	>400 days	5265	1381
60	476635	~1000 days	5185	1301

Note that some waste in the final burn to mars is unavoidable, because once spefici orbital energy >=0 one cannot 'kick' past the periapsis and so once the final burn that has SME_>0 has also be a parto of the burn that goes to mars despite the fact that most of the burn occurs at low velocity (and thus low dE/dt during the burn).
There are several ways to improve this.

1. Burn down 5 to 10' more than prograde to keep Pe low (from 270 to to 200)
2. On the last jump, jump to 270 and also burn down at 30'
3. Switch the ION drive to burn at a lower ISP and higher thrust.
4. Have an auxillary chemical based engine on board. The dV difference between escape, and mars intercept is only about 400 at burn from 6531 km and so a reasonably small fuel tank with powerful engine would suffice, the fuel tank can be ditched after the burn, it will go into a circumsolar orbit.

Lessons to be learned:

In  a burn to Mars from Earth orbit, you do not want to use a spiral (360 burn span) orbit. It wastes both fuel and time.
Since a Mars mission would be at least 90 days the loss of 5 days getting a better fuel economy is worth it. This means that blocking off 120' of the orbit as a no burn zone is a reasonable move.
For a space tug that has no real cost of time except for capital cost of inventory. blocking the burn from a half to 2/3rds of the orbit is a reasonably good choice.

 

Edited December 19, 2017 by PB666

[Streetwind]

Well, yes. i suppose it's nice to have an example with numbers. But I don't think anyone flies spiral trajectories unless they absolutely have to. It's easily proven mathematically that they are always worse. Not just for going from Earth to Mars - for literally any and all applications. Worst case, you might end up paying as much as 2.4 times the dV. (We went over that a while ago on a whim of mine.)

Trajectories like that are a necessary evil if your thrust is absurdly low, and it only works out in your advantage if your engine Isp is so much higher that you can afford to waste dV like that. If you have a ship capable of higher acceleration, like the one in your example, then it is always better to split the burn into periapsis kicks. The problem here is that your example spacecraft is impossible to build within the laws of physics. A vessel with your specs running an electric engine would need to deliver a guesstimated 1.7 gigawatts of power to its engines, sustained. That sort of thing is impossible to generate within a 96 T mass envelope, from which you still need to subtract the not insubstantial mass of the engines themselves, as well as the structural dry mass of the craft. Even at a phenomenal 2,000 W/kg specific power - our best solar panels today manage close to 300 W/kg - you'd still be looking at over 860 metric tons just for the power source alone.

And thus, even the best Mars craft we could realistically build would still fly spiral trajectories when equipped with electric engines. Not because it is better, but because there is no way we can give it enough thrust for anything else.

Edited December 21, 2017 by Streetwind
Fixed a serious unit conversion mistake

[PB666]

43 minutes ago, Streetwind said:

Well, yes. i suppose it's nice to have an example with numbers. But I don't think anyone flies spiral trajectories unless they absolutely have to. It's easily proven mathematically that they are always worse. Not just for going from Earth to Mars - for literally any and all applications. Worst case, you might end up paying as much as 2.4 times the dV. (We went over that a while ago on a whim of mine.)

Trajectories like that are a necessary evil if your thrust is absurdly low, and it only works out in your advantage if your engine Isp is so much higher that you can afford to waste dV like that. If you have a ship capable of higher acceleration, like the one in your example, then it is always better to split the burn into periapsis kicks. The problem here is that your example spacecraft is impossible to build within the laws of physics. A vessel with your specs running an electric engine would need to deliver a guesstimated 1.7 terawatts of power to its engines, sustained. That sort of thing is impossible to generate within a 96 T mass envelope, from which you still need to subtract the not insubstantial mass of the engines themselves, as well as the structural dry mass of the craft. Even at a phenomenal 2,000 W/kg specific power - our best solar panels today manage close to 300 W/kg - you'd still be looking at over 860 metric tons just for the power source alone.

And thus, even the best Mars craft we could realistically build would still fly spiral trajectories when equipped with electric engines. Not because it is better, but because there is no way we can give it enough thrust for anything else.

Some folks here have a hard time understanding that. Aside from that, even if you had to spiral out, it provides very little harm in terms of time and provides alot of benefit if the first halfof the orbits are Pe kicks.

Originally I put up this example to show that if we had a Magic Wand fusion reactor, it would not grant us 39 days to mars, that the 39 days to Mars is a fantasy scenario. The point of the exercise is to show that player x is a space tug operator who carrys supplies to mars (200t to mars, 20t back), and these are the ways he has to get from LEO, where is picks up supplies from SpaceZ and carries them to Mars and back again. There is a huge connundrum for using ION Powered systems which you touched on but missed the Major point. In an ideal world you have a ship that burns in LEO all the energy it needs to get to Mars, but even through kicks during the last pass you need a system that can chug out 500 dV over a few minutes. KIcks can save most of the money by spiraling but not all. Given all this space tug operate has to choose amoungst the options for the most time economic and fuel economic way of get the cargo to Mars. Just to make the point that ventures that rely on ION drive systems (with no Magic wand power supply) are extremely wasteful and will not get the traveler back home.

As I stipulated at the beginning, I don't believe Fusion power is an answer at all, it creates 2 problems for every problem it solves (see below).
It does not violate the laws of physics per say, it disregards current human abilities to convert heat to energy. 1.7 TW is not a problem if the fusion reactor is 100% efficient in the power conversion, if it is only 99% efficient it generates 10 MW of power, as I have stated here recently that a fusion power reactor would need at least 100 times its mass in radiative cooling (see below).

The engines themselves are rated at 24 kw per 0.3731 meters so we can generate 64.32 kw of power to thrust conversion per meter. So for a system that weighs 380 t at 0.0105 acceleration requires 3933 N of thrust

These are the parameters, they are not as unrealistic as you think, save one (which itself is composed of many). 3933 x 81128.7 / 2 x 0.75 eff = 212,719,415 watts of power. This can be delivered by 3307.2 meters squared of ION drives (93 tons might be a credible weight). HiPEP with a little more investment could get the efficiency higher, its power density could increase, it did perform at 80% efficiency at ISP 9650 and 39 KW, it could be possible to get it to 85% efficiency at 8000 ISP at say a power density of 100 kw/meter. It is possible to get 93 tons of ION drive laid out on six spans that cover the desired area once they flip out. At the center could have a carriage that would insert a 50 ton reactor, all of that is credible, hard but credible.

What is incredible is the fusion reactor.
Scenario 1. we can get a fusion reactor that produces 212,719,415 watts of electric power at its destination. The best power conversion in space at the moment (idealistically) would be a three phase thermocouple that delivers 25% of the input heat as electric power. So thats ~ 1 Gigawatts of power. How do you start a fusion reactor, you need batteries that can generate 1 Gigawatts of power. Most of the power if not all of the power would go back into initiating the reactor. Suffice it to say we don't know how bad the waste problem will be with a fusion reactor because we do not currently have a working fusion reactor, but even in this situation a sizable fraction of the engines total mass will be tirgger batteries required to initiate the reaction. So if we started with a 50t reactor we can add another say 20 tons for batteries. Next we have to deal with waste heat. If all the above is true then there is going to be 0.7 GW of waste heat. If you set the maximum temperature of the radiator at 100'C then you can see (Stefan-Boltzman constant = 5.67067 x10⁻⁸)

Where T = temperature epsilon-sigma-temperature^4 is Q the heat flux the specific Power.  If we set emmissivity at 0.5 and temper at 373K we can calculate the area required to dissipate a give amount of heat to be about 50watts per meter.
So if you don't see the problem now, its very simple to make it visible. .7 GW/ .00000005 GW = 12,755,00 square meters (12 sq.km) this could be possible down by a factor of 3 to about 4 km this produces a sphere with a radius of 1 km.
Since in our sphere will be facing half its surface to the sun, that side cannot be used to transfer heat, then we would need 8 km of shell or a radius of 1.414 km in Assuming the cost is 1 kg per meter or radiative heating (very optimistic) then such a Fusion reactor would require approximate 8 million kg of radiative cooling. So now our fusion reactors power system dropped from 0.01 a to 0.00045 a.  This is the best case scenario. 

Scenario 2. A probable scenario is that Fusion will work but only at the Terawatt range as you say, in which spectacular amounts of heat will be generated again we are talking about heat shells 10 to 100s of kilometers in radius.

Scenario 3. Fusion electric power is never a thing. Replace Fusion reactor with a lower efficiency per mass Fission reactor and the cooling problem is the same as scenario 1.

So if we disregard all nuclear electric power systems over 10 kw and look at solar. . . . . . . . .
Alt. Scenario 1. Solar power at 212719415 / 0.4 KW = 531,798.537 kg Removing 50t for the reactor and adding 531t gives us acceleration of 0.0046 but 500,000 square meters is 0.5 square kilometers of panels. And so now we move to alt 2

Alt. Scenario 2. We split scenario 1 into say 20 ships of managable solar panel size and we link them in a long teather to carry the payload behind. (ION drives will need to point off angle).

 

 

 

[Starman4308]

A few issues with your analysis of the cooling required:

First, water is used as the coolant here on Earth because we can easily afford to have gigantic cooling systems. In space, you're more likely to see a liquid metal-cooled reactor; see the SNAP-10A as an example. Assuming an NaK-cooled reactor could be run at 973K (85 Kelvins below the NaK boiling point), cooling per unit area rises by a factor of 46. While I'm unsure how practical a lead-bismuth eutetic cooling system would be, that would increase cooling per unit area by a factor of 635 vs. a water-based cooling system.

Second, there are some strange unit issues, such as "watts per meter". I re-did the math, and came up with 25.4 kW/m^2 (assuming 973K temperature, 0.5 emissivity) for an NaK cooling system and 0.55 kW/m^2 for a water cooling system.

For Streetwind's figure of 1.7 GW and a 0.25 heat efficiency (thus 6.8 GW of heat that must be rejected), this would entail 12.4 km^2 for a water-cooled system, vs. 0.27 km^2 for an NaK system. Using the published specs of the ISS radiators (https://www.nasa.gov/pdf/473486main_iss_atcs_overview.pdf), I'd estimate perhaps 20 kg/m^2 for radiator mass; while it's probable that the ISS radiators can be improved upon, I suspect NaK is harder to handle than ammonia. This would mean 5400 tonnes of radiators for NaK, vs 250000 tonnes of water radiators.

Third... why a sphere? You can hugely reduce wasted area by having flat radiator panels edge-on to space, versus having a sphere where half is actually absorbing solar radiation heat.

Overall, while I'm still not confident that a high-thrust ion trajectory is practical, since any way you slice it you're going to need absurd quantities of radiators or solar panels, there are still baffling flaws in your analysis.

[PB666]

1 hour ago, Starman4308 said:

A few issues with your analysis of the cooling required:

First, water is used as the coolant here on Earth because we can easily afford to have gigantic cooling systems. In space, you're more likely to see a liquid metal-cooled reactor; see the SNAP-10A as an example. Assuming an NaK-cooled reactor could be run at 973K (85 Kelvins below the NaK boiling point), cooling per unit area rises by a factor of 46. While I'm unsure how practical a lead-bismuth eutetic cooling system would be, that would increase cooling per unit area by a factor of 635 vs. a water-based cooling system.

Second, there are some strange unit issues, such as "watts per meter". I re-did the math, and came up with 25.4 kW/m^2 (assuming 973K temperature, 0.5 emissivity) for an NaK cooling system and 0.55 kW/m^2 for a water cooling system.

For Streetwind's figure of 1.7 GW and a 0.25 heat efficiency (thus 6.8 GW of heat that must be rejected), this would entail 12.4 km^2 for a water-cooled system, vs. 0.27 km^2 for an NaK system. Using the published specs of the ISS radiators (https://www.nasa.gov/pdf/473486main_iss_atcs_overview.pdf), I'd estimate perhaps 20 kg/m^2 for radiator mass; while it's probable that the ISS radiators can be improved upon, I suspect NaK is harder to handle than ammonia. This would mean 5400 tonnes of radiators for NaK, vs 250000 tonnes of water radiators.

Third... why a sphere? You can hugely reduce wasted area by having flat radiator panels edge-on to space, versus having a sphere where half is actually absorbing solar radiation heat.

Overall, while I'm still not confident that a high-thrust ion trajectory is practical, since any way you slice it you're going to need absurd quantities of radiators or solar panels, there are still baffling flaws in your analysis.

Responses

1. Thermocouple efficiency is temperature dependent if you want to get away from mechanical systems then each phase needs to drop temperature. If you are going to use 973K the efficiency will be less, and it is highly unlikely that fusion would work at all. But if you had a 10% efficiency reactor you would need 2.5 times the reactor and many more kilometers of cooling. If you use a high temperature transfer temperature such as 973 k you will only benefit from 1 thermocouple and the power efficiency will be 10%, which means power needed it 2.5 times higher and waste heat.  In my system the first therocouple is silicone germanium type  with incoming T below 1380K and dropping out at 800K, the second thermocouple is lead alloy, and would drop the temperature to 450K the last thermocouple would be Bismuth/ lead telluride.  Since no single TC is more than 10% efficient the 25 % comes from layering from hot to cold side of the radiator (The first one could be close to the reactor for liquid sodium transfer to the radiator).

2a. (do you see your 2 errors. lol) My temperature and calculations are based on 373K, as I properly stated and assume a three phase temp drop thermocouple system, you changed that. 0.7 GW of power = 0.5 x E x A x T^4. A = 1.4E9 / (5.8E-8 x 373⁴) = 1,243,000 sq. meters.  . . .x 20 kg/meter = 24,000 t. This would decrease a from 0.0105 a to 0.000164, the dV would go from 9600 to 130. <--- Will not got to Mars, period. Conclusion is the same.

2b. However in your system you have a much lower power efficiency and since your efficiency is lower to higher radiator heat, you have 212,000,000 watts/ 0.10 = 2.12GW reactor output x 0.90 = 1.98GW. =---> 76,175 m² x 20 kg/m² = 1,523 t. This system would drop the a from 0.0105 to 0.002 and drop dV to 1828. <---- Will not go to Mars from LEO.  Conclusion is pretty much the same.

2c. Streetwind said 1.7 terrawatts not gigawatts. He may be right, it might be impossible to have fusion in space for anything smaller, which means Fusion is summarily off the table for power plant. The system I created was proportioning weight with power need to avoid impossible scenarios like launching at 1 kT fusion reactor into space. Conclusion is pretty much the same.
2d.  Agreed that current systems are too weighty, I was giving a best possible case scenario. All the more reason to reject Nuclear. Makes matters worse for proponents of Nuclear Electric propulsion.

3. A sphere is used as being the smallest possible structure that can support itself. Part of the weight issue for NASA's solar panels and radiators are the weight of the truss and the need to be extendable. A sphere could be support for much less the size, an accordian like truss would not work.

 

[Streetwind]

47 minutes ago, PB666 said:

2c. Streetwind said 1.7 terrawatts not gigawatts.

Apologies, I screwed that up royally and meant gigawatts from the beginning. Actually edited my post several hours ago to reflect that, too. But apparently you were already typing your response by the time i made my edit, so you didn't see it =/

Edited December 21, 2017 by Streetwind

[Crampman]

I don't want to say something stupid, but in such a scenario, with (relatively) short burns, is it really necessary to use such overpowered generators ? These generators would be useful only some a very short time, right ?

Could a system with solar / supercapacitors be sufficient ? If you can split the burns into short ones, for example ?

[wumpus]

1 hour ago, Crampman said:

Could a system with solar / supercapacitors be sufficient ? If you can split the burns into short ones, for example ?

Assuming an ISS-like orbit of 90 minutes, a capacitor would need to hold 15 minutes of energy.  Extremely expensive, but probably not compared to a hypothetical fusion power plant.  A larger issue would be the Van Allan belts.  Current solar panels are likely to be degraded each time they pass through the Van Allan belt.  To be honest, I'm not even sure crewed flight would work using these assumptions.  Perhaps some sort of flow-battery might work.  As far as I know, they aren't sufficiently efficient in the lab but the market is more than willing to justify the R&D for commercial uses.

I'd assume that spiral patterns are done because you simply don't have the time for Perapsis kicking.  Current ion drives are closer to a year than week to burn to escape to Mars (assuming they could survive the Van Allan belts, NASA launches them at escape velocity to avoid this issue), and multiplying that time by 100 simply isn't reasonable (it makes more sense to pack more Xenon, or perhaps switch to Argon if you need unreasonable amounts of fuel).

One final thing: assuming you are on the "year to Mars" plan (presumably bringing probes or cargo) and are worried about efficiency, you would probably head out to LM-2 and find a gravity trick to launch you toward Mars (probably not a strict Hohamm transfer, you probably want to do a more spiralish [burning fuel] transfer for ballistic capture at Mars).

Edited December 22, 2017 by wumpus
sorry, no time to deal with Squads stupid strikethrough bug.