Author Topic: Starship On-orbit refueling - Options and Discussion  (Read 815427 times)

Offline InterestedEngineer

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Re: Starship On-orbit refueling - Options and Discussion
« Reply #2700 on: 12/07/2024 12:18 am »
Anyone suggesting to use drag to settle the propellant might want to take out an envelope and do a quick estimate of the associated heating...

Sure, why not.  The deltaV is 7m/sec over a period of 2 hours.

KE of say 500t combined Starships + fuel at 7,814 m/sec is 15,264,649,000,000 joules (3.6kt of TNT)

KE of say 500t of combined Starships + fuel at (7814 - 7) mm/sec is 15,237,312,250,000 joules (3.6kt of TNT)

That's a net change of  -27,336,750,000 over 7200 seconds, or 3.8e6 joules/second (3.8MW)

Like re-entry heating, 90% of the heat is dissipated in the passing particles, not on the heat shield tiles themselves, so that leaves us with 380kW of heat flux for the whole Starship.

Assuming the tiles have the same heat capacity as the Space shuttle (628 J/kg-K), and there are 10,000 kg of tiles on a Starship, that's a heating rate of 380kJ/sec / 10,000kg / 628 J/kg-K = 0.06degK/sec.

Now the surface area of the heat shield tiles on Starship is roughly 400m2, which gives us a heat flux of 380kW/400 = 950 watts/m2

We now have to calculate the stefan-boltzmann equilibrium of the tiles that gets us to a heat flux of -950W/m2 and thus equilibrium.

Let's say they start at 100K,  How much do they have to heat up for them to emit 950w/m2?

The answer is they must be heated to 360K.  That's going to be hardly any heat at all coming through the tiles to the tank, they are below the boiling point of water and their thermal conductivity is very low. (there's also an insulation blanket and back up heat shield behind them).

It will take 260K / .06K/sec  = 4333 seconds to engage the full heat capacity of the tiles.  This suggests that they will never reach the 360K equilibrium, but the calculation for the actual final temperature is a 4th order polynomial and I really don't feel like solving one of those right now.

TL;DR - no problemo on heating from the drag.




Offline InterestedEngineer

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Re: Starship On-orbit refueling - Options and Discussion
« Reply #2701 on: 12/07/2024 12:22 am »
In October 2020 up-thread, the best guess was that the necessary settling acceleration is on the order of 1e-4 m/s2.

Cool, thanks for the numbers!  That means much less deltaV will be needed, so any heating from drag will be negligible, and the refuel can take place at a much higher altitude than 100km.

Quote
Also, keeping the depot pointing toward the sun to minimize illuminated surface area and therefore boil-off is probably important. Either way, using drag for the acceleration doesn't really make sense, because the ship would have to be brought up to altitude again using propulsion. Besides, boil-off will create ullage gas that will have to be vented anyway; they might as well use it for settling.

I'm not quite sure how you conclude that it doesn't make sense, you just made the problem 10 times easier than my calculations above show.   thanks for the 10x margin... but why won't it work?

As far as pointing heat shield at the sun, if we are going to belly to belly one heat shield is pointed in the direction of travel and the other heat shield is pointed the opposite, so the sun will never hit the large unshielded surface area, so good news there.

Note that the heat flux for the sun was similar to what I calculated for the drag - you probably won't hit the heat shield thermal capacity in two orbits.


Offline InterestedEngineer

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Re: Starship On-orbit refueling - Options and Discussion
« Reply #2702 on: 12/07/2024 12:25 am »

I agree, logically VLEO is the best place for refilling, at least for the lowest "rung" of any refilling ladder. However I'm not convinced about using drag for propellant settling.

Thanks for the numbers I'll do the calcs in a minute, but given that settling acceleration is 1/10 the numbers bandied about here, why won't drag work for propellant settling?

Offline InterestedEngineer

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Re: Starship On-orbit refueling - Options and Discussion
« Reply #2703 on: 12/07/2024 12:47 am »

You may find this graph handy: https://space.stackexchange.com/questions/18223/where-can-i-find-data-for-atmospheric-density-vs-altitude

At hypersonic velocities you can assume a Cd of 2.2, so for a 9 m diameter 100 tonne vehicle you achieve 1 mm/s2 at just over 100 km altitude, and for a 2,000 tonne vehicle it's almost exactly 100 km.

thanks for the info!

For the newly found requirement of 0.1mm/sec2 to settle the propellant, the air density must be on the order of 1e-9 which from your chart implies over 150km altitude.  tons of margin.

plugging that into kinematic equations over a one hour fuel transfer, that's a deltaV of 0.36m/sec, so really rounding error.

What I find interesting is that the more refuels, the lower the ship needs to be to get that .1mm/sec2 of drag acceleration, so this follows the normal course of events, so by the time a Starship V3 is full, it'll be down in the 100-120km range, perfectly fine, all with air drag setting the altitude, no low Isp thrusters.

The only problem I see is mating the starships with the differential drag on the 350t refill ship with the 2300t almost full StarshipV3, as their drag acceleration will be different, the heavier one at 0.1mm/s2 and the lighter one at .66mm/s2.  The difference being .56mm/2.

F=ma, so the differential force on the lighter one will be 200N.  I would assume a mating connector can handle 200N.

Perhaps this is solved by the lighter one approaching in the heavier one's wake.

Who knew surfing the thermosphere could be so fun.

Offline Brigantine

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Re: Starship On-orbit refueling - Options and Discussion
« Reply #2704 on: 12/07/2024 10:10 am »
differential drag on the 350t refill ship with the 2300t almost full StarshipV3, as their drag acceleration will be different, the heavier one at 0.1mm/s2 and the lighter one at .66mm/s2.  The difference being .56mm/2.

F=ma, so the differential force on the lighter one will be 200N.  I would assume a mating connector can handle 200N.
I had assumed that, short of a total redesign of the plumbing, the settling acceleration should be axial. If so, then that differential drag creates a torque ≈2kNm. (assuming 1m of docking hardware, or else to 1sf)

Carefully placed cold gas thrusters can counteract this with at least 3x greater lever arm, so <70N of thrust.
Depending on details, the receiver ship's tanks need to be vented anyway to maintain the pressure differential. (how much?)

I'm curious whether this VLEO co-incidentally (depending on solar weather) is close to the altitude corresponding to 16 complete orbits each node (<24h per sidereal rotation and precession), and therefore daily launch opportunities with zero phasing. (Not at all, it seems that's ≈208km for 28.6⁰ inclination)

Offline InterestedEngineer

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Re: Starship On-orbit refueling - Options and Discussion
« Reply #2705 on: 12/07/2024 05:21 pm »
differential drag on the 350t refill ship with the 2300t almost full StarshipV3, as their drag acceleration will be different, the heavier one at 0.1mm/s2 and the lighter one at .66mm/s2.  The difference being .56mm/2.

F=ma, so the differential force on the lighter one will be 200N.  I would assume a mating connector can handle 200N.
I had assumed that, short of a total redesign of the plumbing, the settling acceleration should be axial. If so, then that differential drag creates a torque ≈2kNm. (assuming 1m of docking hardware, or else to 1sf)


If axial, it's mostly nose first (Tail-Tail), and the altitude is somewhere around 140km, given the 400/64 reduction in surface area.

Quote
I'm curious whether this VLEO co-incidentally (depending on solar weather) is close to the altitude corresponding to 16 complete orbits each node (<24h per sidereal rotation and precession), and therefore daily launch opportunities with zero phasing. (Not at all, it seems that's ≈208km for 28.6⁰ inclination)

Interesting, good point. And yes solar weather is going to have some effect, but it's probably calculable, so there will be slight yield differences as fuel is boosted higher or lower depending on space weather.

Online DanClemmensen

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Re: Starship On-orbit refueling - Options and Discussion
« Reply #2706 on: 12/07/2024 06:12 pm »
I'm sure this idea has been mentioned before, but I am not good at searching The NSF forums:

Should Depot be designed for nose-to-tail connection to other Ships?

Depot would have a nose that looks a whole lot like the top of Booster. It would be covered with a disposable fairing for launch. For docking, Ship does the rendezvous and then orients itself pointing away from Depot at a distance of perhaps 50 m. Depot then maneuvers to dock to Ship's tail.

All of Depot's propellant transfer machinery is near the forward end. This includes an SQD that swings out and up to connect to Ship's QD plate. The advantage of this configuration is that all of the Depot-specific pieces are in what would otherwise be the payload area, above the tanks, and there are no docking modifications to Ship. The basic mating interface is "merely" the same surfaces as those between Booster and Ship. Depot may need some active latches, since Ship depends on gravity to stay connected to Booster. Depot may need a 6DOF conformal active latching ring system similar to the one IDSS uses (but 9 m instead of 1 m, what could possibly go wrong?) Depot is the active docking partner and will need fairly powerful 6DOF RCS thrusters.

Since Ship needs no additions for propellant transfer (no funny docking ports) the Ship design is unconstrained by this scheme.


Offline eriblo

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Re: Starship On-orbit refueling - Options and Discussion
« Reply #2707 on: 12/07/2024 06:29 pm »
Anyone suggesting to use drag to settle the propellant might want to take out an envelope and do a quick estimate of the associated heating...

Sure, why not.  The deltaV is 7m/sec over a period of 2 hours.

KE of say 500t combined Starships + fuel at 7,814 m/sec is 15,264,649,000,000 joules (3.6kt of TNT)

KE of say 500t of combined Starships + fuel at (7814 - 7) mm/sec is 15,237,312,250,000 joules (3.6kt of TNT)

That's a net change of  -27,336,750,000 over 7200 seconds, or 3.8e6 joules/second (3.8MW)

Like re-entry heating, 90% of the heat is dissipated in the passing particles, not on the heat shield tiles themselves, so that leaves us with 380kW of heat flux for the whole Starship.

Assuming the tiles have the same heat capacity as the Space shuttle (628 J/kg-K), and there are 10,000 kg of tiles on a Starship, that's a heating rate of 380kJ/sec / 10,000kg / 628 J/kg-K = 0.06degK/sec.

Now the surface area of the heat shield tiles on Starship is roughly 400m2, which gives us a heat flux of 380kW/400 = 950 watts/m2

We now have to calculate the stefan-boltzmann equilibrium of the tiles that gets us to a heat flux of -950W/m2 and thus equilibrium.

Let's say they start at 100K,  How much do they have to heat up for them to emit 950w/m2?

The answer is they must be heated to 360K.  That's going to be hardly any heat at all coming through the tiles to the tank, they are below the boiling point of water and their thermal conductivity is very low. (there's also an insulation blanket and back up heat shield behind them).

It will take 260K / .06K/sec  = 4333 seconds to engage the full heat capacity of the tiles.  This suggests that they will never reach the 360K equilibrium, but the calculation for the actual final temperature is a 4th order polynomial and I really don't feel like solving one of those right now.

TL;DR - no problemo on heating from the drag.
I think those are some optimistic assumptions. For aerobraking it's typically 90% of the energy that goes into the vehicle skin rather than into the gas. For Starship the ballistic coefficient is high enough that you might get a decent shock but it will still be a higher fraction than on reentry (and if you have shocks you have hot spots).

There is also the problem of relying on the tanker heat shield to take all the heating. That orientation requires extra internal piping and is likely to be unstable with partially filled tanks. If you go tail first there is a lot less area and likely less insulation.

Any part of the depot tankage exposed to the flow will be hampered by having low emissivity and will get significantly hotter.


The biggest problem with using atmospheric drag is of course that you don't gain anything because you'll lose more propellant compensating than you save.
« Last Edit: 12/07/2024 06:31 pm by eriblo »

Offline InterestedEngineer

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Re: Starship On-orbit refueling - Options and Discussion
« Reply #2708 on: 12/07/2024 06:56 pm »
Anyone suggesting to use drag to settle the propellant might want to take out an envelope and do a quick estimate of the associated heating...

Sure, why not.  The deltaV is 7m/sec over a period of 2 hours.

KE of say 500t combined Starships + fuel at 7,814 m/sec is 15,264,649,000,000 joules (3.6kt of TNT)

KE of say 500t of combined Starships + fuel at (7814 - 7) mm/sec is 15,237,312,250,000 joules (3.6kt of TNT)

That's a net change of  -27,336,750,000 over 7200 seconds, or 3.8e6 joules/second (3.8MW)

Like re-entry heating, 90% of the heat is dissipated in the passing particles, not on the heat shield tiles themselves, so that leaves us with 380kW of heat flux for the whole Starship.

Assuming the tiles have the same heat capacity as the Space shuttle (628 J/kg-K), and there are 10,000 kg of tiles on a Starship, that's a heating rate of 380kJ/sec / 10,000kg / 628 J/kg-K = 0.06degK/sec.

Now the surface area of the heat shield tiles on Starship is roughly 400m2, which gives us a heat flux of 380kW/400 = 950 watts/m2

We now have to calculate the stefan-boltzmann equilibrium of the tiles that gets us to a heat flux of -950W/m2 and thus equilibrium.

Let's say they start at 100K,  How much do they have to heat up for them to emit 950w/m2?

The answer is they must be heated to 360K.  That's going to be hardly any heat at all coming through the tiles to the tank, they are below the boiling point of water and their thermal conductivity is very low. (there's also an insulation blanket and back up heat shield behind them).

It will take 260K / .06K/sec  = 4333 seconds to engage the full heat capacity of the tiles.  This suggests that they will never reach the 360K equilibrium, but the calculation for the actual final temperature is a 4th order polynomial and I really don't feel like solving one of those right now.

TL;DR - no problemo on heating from the drag.
I think those are some optimistic assumptions. For aerobraking it's typically 90% of the energy that goes into the vehicle skin rather than into the gas. For Starship the ballistic coefficient is high enough that you might get a decent shock but it will still be a higher fraction than on reentry (and if you have shocks you have hot spots).

Space Shuttle and Starship heat shields would melt on re-entry if 90% of the energy went into the heat shield.  But that might be because there's a bow shock and stagnation zone, at 150km there might well not be, and 80-90% might be correct. I"ll do more research.  Stefan Boltzmann being to the 4th power, I doubt it will change the equilibrium temperature much.

Quote
There is also the problem of relying on the tanker heat shield to take all the heating. That orientation requires extra internal piping and is likely to be unstable with partially filled tanks. If you go tail first there is a lot less area and likely less insulation.

You don't go tail first, you go nose first if we are doing axial cross loading.  The tails are docked.  The nose is completely covered in tiles.
Quote
Any part of the depot tankage exposed to the flow will be hampered by having low emissivity and will get significantly hotter.

Yes, there are certain orientations where this won't work. Fortunately none of those are likely (nose-nose, or docking on the heat shield side)


Quote
The biggest problem with using atmospheric drag is of course that you don't gain anything because you'll lose more propellant compensating than you save.

We need 0.1mm/sec2 of acceleration no matter what.  Which is cheaper in terms of propellant mass?  Maintaining that with 150Isp cold gas thrusters, or recovering the lost altitude using 365 Isp main engines?  Conservation of energy and momentum suggest it's the same problem, therefore, highest Isp wins (and that's not counting Oberth effect, you go faster at lower altitudes).

We are either refueling above 200km and using cold gas thrusters, or in the 110-160km range surfing the thermosphere.  The lower, the altitude, the better the Oberth effect, both from less altitude to launch tankers to, and firing propellant out the back at lower altitude (higher velocity) when launching to Mars or the Moon.

Offline InterestedEngineer

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Re: Starship On-orbit refueling - Options and Discussion
« Reply #2709 on: 12/07/2024 07:18 pm »
Anyone suggesting to use drag to settle the propellant might want to take out an envelope and do a quick estimate of the associated heating...

Sure, why not.  The deltaV is 7m/sec over a period of 2 hours.

KE of say 500t combined Starships + fuel at 7,814 m/sec is 15,264,649,000,000 joules (3.6kt of TNT)

KE of say 500t of combined Starships + fuel at (7814 - 7) mm/sec is 15,237,312,250,000 joules (3.6kt of TNT)

That's a net change of  -27,336,750,000 over 7200 seconds, or 3.8e6 joules/second (3.8MW)

Like re-entry heating, 90% of the heat is dissipated in the passing particles, not on the heat shield tiles themselves, so that leaves us with 380kW of heat flux for the whole Starship.

Assuming the tiles have the same heat capacity as the Space shuttle (628 J/kg-K), and there are 10,000 kg of tiles on a Starship, that's a heating rate of 380kJ/sec / 10,000kg / 628 J/kg-K = 0.06degK/sec.

Now the surface area of the heat shield tiles on Starship is roughly 400m2, which gives us a heat flux of 380kW/400 = 950 watts/m2

We now have to calculate the stefan-boltzmann equilibrium of the tiles that gets us to a heat flux of -950W/m2 and thus equilibrium.

Let's say they start at 100K,  How much do they have to heat up for them to emit 950w/m2?

The answer is they must be heated to 360K.  That's going to be hardly any heat at all coming through the tiles to the tank, they are below the boiling point of water and their thermal conductivity is very low. (there's also an insulation blanket and back up heat shield behind them).

It will take 260K / .06K/sec  = 4333 seconds to engage the full heat capacity of the tiles.  This suggests that they will never reach the 360K equilibrium, but the calculation for the actual final temperature is a 4th order polynomial and I really don't feel like solving one of those right now.

TL;DR - no problemo on heating from the drag.
I think those are some optimistic assumptions. For aerobraking it's typically 90% of the energy that goes into the vehicle skin rather than into the gas. For Starship the ballistic coefficient is high enough that you might get a decent shock but it will still be a higher fraction than on reentry (and if you have shocks you have hot spots).

Your criticism is correct, further research shows there's no bow shock above what I think the broadcast commenters call the "entry interface", or about 100km or so, so 90% of the deceleration energy is going to go into the heat shield above 100km.  It switches to 10% below the entry interface as the bow shock forms.

You know what they say about assumptions...

Since it appears we should be surfing the thermosphere in the 120-160km range to get the desired acceleration of 0.1mm/s2, I shall redo the calculations.

Let's stick to the 2 hour refuel, which is probably pessimistic (27kg/sec seems really slow).  We'll go axial, nose first, so the bare steel is protected.  We don't have to worry about non-impinging surfaces heating, there's no bow shock.

That's a deltaV of 7200s * 0.0001m/s2 = 0.72m/sec.  Thanks for that order magnitude of margin!

KE of say 500t combined Starships + fuel at 7,814 m/sec is 15,264,649,000,000 joules (3.6kt of TNT)
KE of say 500t of combined Starships + fuel at (7814 - .721) mm/sec is 15,261,832,182,960 joules (3.6kt of TNT)

That's a net change of  -2,816,817,040 joules over 7200 seconds, or 391 kj/second (391kW).

90% of that goes to heat shield, 10% to the passing particles, leaving us with a net heat flux of 350kW.

I note if we are not refueling axial but rather heat-shield-facing-forward, the result is nearly identical to above, about 360K equilibrium temperature.  Thanks for that order of magnitude smaller settling acceleration.

Let's stick with axial tail-to-tail, since that seems to be the most likely without significant plumbing changes.  The nose has surface area of 64m2, so the heat flux is 5.5kw/m2

The equilibrium temperature is now 560K.  This is well within the range of the tiles.  furthermore the heat is in the nose cone, with vacuum separation from the header tanks, and tiles have a layer of insulation underneath them.  The tiles will do their job, and the tanks won't heat up at all.

TL;DR - heat build up is still not a problem when accounting for the non-bowshock behavior of surfing the thermosphere, because the requirements dropped by an order of magnitude, saving my bacon (which alas will be undercooked)

Offline Keldor

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Re: Starship On-orbit refueling - Options and Discussion
« Reply #2710 on: 12/07/2024 08:20 pm »
Anyone suggesting to use drag to settle the propellant might want to take out an envelope and do a quick estimate of the associated heating...

Sure, why not.  The deltaV is 7m/sec over a period of 2 hours.

KE of say 500t combined Starships + fuel at 7,814 m/sec is 15,264,649,000,000 joules (3.6kt of TNT)

KE of say 500t of combined Starships + fuel at (7814 - 7) mm/sec is 15,237,312,250,000 joules (3.6kt of TNT)

That's a net change of  -27,336,750,000 over 7200 seconds, or 3.8e6 joules/second (3.8MW)

Like re-entry heating, 90% of the heat is dissipated in the passing particles, not on the heat shield tiles themselves, so that leaves us with 380kW of heat flux for the whole Starship.

Assuming the tiles have the same heat capacity as the Space shuttle (628 J/kg-K), and there are 10,000 kg of tiles on a Starship, that's a heating rate of 380kJ/sec / 10,000kg / 628 J/kg-K = 0.06degK/sec.

Now the surface area of the heat shield tiles on Starship is roughly 400m2, which gives us a heat flux of 380kW/400 = 950 watts/m2

We now have to calculate the stefan-boltzmann equilibrium of the tiles that gets us to a heat flux of -950W/m2 and thus equilibrium.

Let's say they start at 100K,  How much do they have to heat up for them to emit 950w/m2?

The answer is they must be heated to 360K.  That's going to be hardly any heat at all coming through the tiles to the tank, they are below the boiling point of water and their thermal conductivity is very low. (there's also an insulation blanket and back up heat shield behind them).

It will take 260K / .06K/sec  = 4333 seconds to engage the full heat capacity of the tiles.  This suggests that they will never reach the 360K equilibrium, but the calculation for the actual final temperature is a 4th order polynomial and I really don't feel like solving one of those right now.

TL;DR - no problemo on heating from the drag.
I think those are some optimistic assumptions. For aerobraking it's typically 90% of the energy that goes into the vehicle skin rather than into the gas. For Starship the ballistic coefficient is high enough that you might get a decent shock but it will still be a higher fraction than on reentry (and if you have shocks you have hot spots).

Your criticism is correct, further research shows there's no bow shock above what I think the broadcast commenters call the "entry interface", or about 100km or so, so 90% of the deceleration energy is going to go into the heat shield above 100km.  It switches to 10% below the entry interface as the bow shock forms.

You know what they say about assumptions...

Since it appears we should be surfing the thermosphere in the 120-160km range to get the desired acceleration of 0.1mm/s2, I shall redo the calculations.

Let's stick to the 2 hour refuel, which is probably pessimistic (27kg/sec seems really slow).  We'll go axial, nose first, so the bare steel is protected.  We don't have to worry about non-impinging surfaces heating, there's no bow shock.

That's a deltaV of 7200s * 0.0001m/s2 = 0.72m/sec.  Thanks for that order magnitude of margin!

KE of say 500t combined Starships + fuel at 7,814 m/sec is 15,264,649,000,000 joules (3.6kt of TNT)
KE of say 500t of combined Starships + fuel at (7814 - .721) mm/sec is 15,261,832,182,960 joules (3.6kt of TNT)

That's a net change of  -2,816,817,040 joules over 7200 seconds, or 391 kj/second (391kW).

90% of that goes to heat shield, 10% to the passing particles, leaving us with a net heat flux of 350kW.

I note if we are not refueling axial but rather heat-shield-facing-forward, the result is nearly identical to above, about 360K equilibrium temperature.  Thanks for that order of magnitude smaller settling acceleration.

Let's stick with axial tail-to-tail, since that seems to be the most likely without significant plumbing changes.  The nose has surface area of 64m2, so the heat flux is 5.5kw/m2

The equilibrium temperature is now 560K.  This is well within the range of the tiles.  furthermore the heat is in the nose cone, with vacuum separation from the header tanks, and tiles have a layer of insulation underneath them.  The tiles will do their job, and the tanks won't heat up at all.

TL;DR - heat build up is still not a problem when accounting for the non-bowshock behavior of surfing the thermosphere, because the requirements dropped by an order of magnitude, saving my bacon (which alas will be undercooked)

You know, even if you can refuel using the atmosphere to settle fuel, you'll have to pay it all back when you do a burn to get back to a stable orbit.  Using thrusters for settling, on the other hand, will give you a small boost towards your destination.

This is to say nothing of exciting problems such as aerodynamic occlusion resulting in one vehicle seeing more drag than the other.  And of course, since the vehicles will be different masses due to the weight of the fuel, they'd be slowed at different rates even if the drag was identical.
« Last Edit: 12/07/2024 08:23 pm by Keldor »

Offline InterestedEngineer

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Re: Starship On-orbit refueling - Options and Discussion
« Reply #2711 on: 12/07/2024 10:01 pm »
You know, even if you can refuel using the atmosphere to settle fuel, you'll have to pay it all back when you do a burn to get back to a stable orbit.  Using thrusters for settling, on the other hand, will give you a small boost towards your destination.

This is to say nothing of exciting problems such as aerodynamic occlusion resulting in one vehicle seeing more drag than the other.  And of course, since the vehicles will be different masses due to the weight of the fuel, they'd be slowed at different rates even if the drag was identical.

We are paying back 0.72m/s2 of deltaV, rounding error, with an ISP that is double that of cold gas thrusters, plus we don't have to worry about losing too much pressure from using cold gas thrusters.

Plus that atmospheric drag deltaV is going into dropping the orbit altitude, not to dropping the velocity.  It's roughly a 1km drop per 2 orbits if the starting orbit is 150km.

If our last orbit is at 120km (60 orbits, 1 refuel every 6 orbits), our net velocity gain is 18m/sec.  We have traded height for kinetic energy, which is a great tradeoff to make, due to the average launch height of our refuelers being 135km instead of 200km, safely out of noticeable aerodrag range.

Let's see how much fuel we save by refueling at an average altitude of 135km instead of 200km.

It's best to use C3 when doing these calculations.

If the average launch for each 200t refill is 135km, the C3 at that altitude is -6.106E+07 m2/s2.

This compares to 200km, where it is -5.985E+07 m2/s2.  A difference of 4.630E+05 m2/s2

Launching all that wet mass (320t) of Starship up to 200km on each trip thus costs an additional  kinetic energy of 3.5e+10 joules, which at an exhaust velocity of 3600m/sec is 11.4t of additional fuel.

If we are doing 10 refuels on a Starship V3 then we've wasted 114t of fuel refueling at the higher altitude, and that's not counting the fuel needed to run 20 hours of cold gas thrusters during this process, that's just boosting dry mass to a higher altitude.

Now that we have a 300t dry mass Starship with 2000t or 1886t of fuel, let's calculate the deltaV for the two scenarios. I'm using the standard C3 equations to calculate Vinf:

Scenario 1:  Boosting at 120km, MR is 7.67, deltaV is thus 7.333km/sec, Vinf is 10.360 km/sec

Scenario 2:  Boosting at 200km, MR is 7.287, deltaV is 7.150 km/sec, Vinf is 10.092 km/sec.

Thus we got a net gain of 268m/sec in Vinf by not boosting all that fuel up to 200km, or another way of putting it is we got 6% more fuel by refueling at a lower altitude.

Refueling at a lower altitude is more efficient.  You are going to get aerodrag there no matter what you do, you might as well use it for settling the propellants, and eliminate the need for continuous settling thrusters.


Possible questions:

Wait, you say, doesn't all that 2000t of fuel need to get boosted up to from 135km to 200km on the wait out to Mars?

Nope, thanks to Mr Oberth, you threw that all out the back before you get to 200km. It's dropping into the atmosphere silently and invisibly.  Your Starship is continuing on its merry way, a much lighter 300t and getting to Mars faster.

Why do you keep calling this surfing?
Because you are doing something very similar to what surfers do, trading potential energy for kinetic energy and using the water as a source of friction.  "surfing the thermosphere" sounds cool too.
« Last Edit: 12/07/2024 10:09 pm by InterestedEngineer »

Offline Twark_Main

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Re: Starship On-orbit refueling - Options and Discussion
« Reply #2712 on: 12/07/2024 10:25 pm »

I agree, logically VLEO is the best place for refilling, at least for the lowest "rung" of any refilling ladder. However I'm not convinced about using drag for propellant settling.

why won't drag work for propellant settling?

I didn't say it won't work. I'm just unconvinced that it's the best solution.

By flying so low you're incurring a lot of unnecessary drag whenever you're not currently in the process of refilling. However in all your calculations you're only calculating the loss during that 2 hour duration ("paying back 0.72m/s2 of deltaV"). You actually need to use the total duration of flight, integrated over the density of the atmosphere.

As for Oberth, you can still get all those same benefits without using drag to settle your propellant. You can simply wait for your orbit to decay down to 120 km (or whatever) and then perform the TMI burn. If you counter that the wait is too long, then that's actually good news because it means you can lower your original parking orbit. :)


Offline Greg Hullender

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Re: Starship On-orbit refueling - Options and Discussion
« Reply #2713 on: 12/07/2024 11:27 pm »
There are a few things I wonder about here:

1) I see how this might work for fueling up the HLS, but I'm not seeing how it would work for the (say) 10 refueling flights needed to fill the depot. I know Elon wants a fast turnaround time, but I don't think it's going to be so fast that the next tanker arrives just as the current one finishes! In that case, I'm not sure it's a great idea to use one method to fuel the depot and a different one to fuel the HLS.

2) For fueling, I thought the acceleration vector needed to be parallel to the starship and pointed at the base. I.e. the same as it is on the ground, only much smaller. Otherwise, I don't see how the existing plumbing will work. If you use air resistance, I think that means the engines have to take the brunt of the friction.

3) It would be nice to think that 0.1 mm/sē would work, but, unless I'm mistaken, all those numbers come from studies of what sort of ullage burns were required to relight upper stages. But that operation only needs to run for a few seconds--just enough to be sure the bottom of the tank is wet--because once the engines ignite, all the propellant will be pressed hard against the bottom of the tank. Refueling, on the other hand, has to run for a long time and the ullage burn needs to counter the force of the pumps. Figuring out how much force is really needed is likely to be a key thing learning from the early experiments.

Apologies if I missed something. If there are papers that address these issues, I'd love to see them.

Offline InterestedEngineer

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Re: Starship On-orbit refueling - Options and Discussion
« Reply #2714 on: 12/08/2024 01:55 am »
maybe a picture will sort a few things out.

The Depot/HLS will have to take the heat on the nose cone portion of the stainless steel, or install a heat shield on just the nose.

TBD, but even with a 0.35 emissivity stainless steel on the nose cone will only get to 725K, which 100K below its annealing temperature.  Paint it black, and it'll drop 130K.

Offline InterestedEngineer

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Re: Starship On-orbit refueling - Options and Discussion
« Reply #2715 on: 12/08/2024 02:03 am »

I agree, logically VLEO is the best place for refilling, at least for the lowest "rung" of any refilling ladder. However I'm not convinced about using drag for propellant settling.

why won't drag work for propellant settling?

I didn't say it won't work. I'm just unconvinced that it's the best solution.

By flying so low you're incurring a lot of unnecessary drag whenever you're not currently in the process of refilling. However in all your calculations you're only calculating the loss during that 2 hour duration ("paying back 0.72m/s2 of deltaV"). You actually need to use the total duration of flight, integrated over the density of the atmosphere.

I'm pretty sure I multiplied out the drag by calculating a drop of 1km per two orbits, then guessing at one refill every 6 orbits, so 60 orbits to fill Starship V3 to 2000t of prop and a total of 30km drop.  Did I do the math wrong?  Is 60 orbits, or over 4 days too optimistic?

for station keeping one could burp an engine for a second and throw away a half a ton of propellant to get a few m/s to raise the orbit, it's still cheaper to do that at double the Isp of cold gas thrusters.

Quote
As for Oberth, you can still get all those same benefits without using drag to settle your propellant. You can simply wait for your orbit to decay down to 120 km (or whatever) and then perform the TMI burn. If you counter that the wait is too long, then that's actually good news because it means you can lower your original parking orbit. :)

Once you've paid the expense to get the fuel up over 200km, there's no Oberth benefit.  Make sure you use a C3 calculation, I got fooled when I didn't.

If you want to refill below 200km, you are back in in the regime of "you have atmospheric drag, you might as well use it".

The greatest benefit turns out to be the not having the lift the fuel from the surface of the earth to over 200km, but to an average of 135km instead.  The second benefit is not wasting tons of prop on cold gas thrusters.
« Last Edit: 12/08/2024 02:06 am by InterestedEngineer »

Offline OTV Booster

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Re: Starship On-orbit refueling - Options and Discussion
« Reply #2716 on: 12/08/2024 02:20 am »
In October 2020 up-thread, the best guess was that the necessary settling acceleration is on the order of 1e-4 m/s2.

Cool, thanks for the numbers!  That means much less deltaV will be needed, so any heating from drag will be negligible, and the refuel can take place at a much higher altitude than 100km.

Quote
Also, keeping the depot pointing toward the sun to minimize illuminated surface area and therefore boil-off is probably important. Either way, using drag for the acceleration doesn't really make sense, because the ship would have to be brought up to altitude again using propulsion. Besides, boil-off will create ullage gas that will have to be vented anyway; they might as well use it for settling.

I'm not quite sure how you conclude that it doesn't make sense, you just made the problem 10 times easier than my calculations above show.   thanks for the 10x margin... but why won't it work?

As far as pointing heat shield at the sun, if we are going to belly to belly one heat shield is pointed in the direction of travel and the other heat shield is pointed the opposite, so the sun will never hit the large unshielded surface area, so good news there.

Note that the heat flux for the sun was similar to what I calculated for the drag - you probably won't hit the heat shield thermal capacity in two orbits.
I was just about to point out the solar flux is 1.3-1.4 kW/m^2 but then realized that was for visible light used by PVs and decided the post was pointless. Wouldn't the IR and UV add quite a bit more flux?
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Offline Twark_Main

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Re: Starship On-orbit refueling - Options and Discussion
« Reply #2717 on: 12/08/2024 02:36 am »

I agree, logically VLEO is the best place for refilling, at least for the lowest "rung" of any refilling ladder. However I'm not convinced about using drag for propellant settling.

why won't drag work for propellant settling?

I didn't say it won't work. I'm just unconvinced that it's the best solution.

By flying so low you're incurring a lot of unnecessary drag whenever you're not currently in the process of refilling. However in all your calculations you're only calculating the loss during that 2 hour duration ("paying back 0.72m/s2 of deltaV"). You actually need to use the total duration of flight, integrated over the density of the atmosphere.

I'm pretty sure I multiplied out the drag by calculating a drop of 1km per two orbits, then guessing at one refill every 6 orbits, so 60 orbits to fill Starship V3 to 2000t of prop and a total of 30km drop.  Did I do the math wrong?

Yes, you only multiplied by 7200 s (2 hours), not 6 hours.

Is 60 orbits, or over 4 days too optimistic?

For sending thousands of Starships to Mars per launch window, yes it's hilariously optimistic.

for station keeping one could burp an engine for a second and throw away a half a ton of propellant to get a few m/s to raise the orbit, it's still cheaper to do that at double the Isp of cold gas thrusters.

This is a very "spherical cow" picture of engines. In reality your Isp is going to suffer greatly for such a short burp, which is dominated by the startup and shutdown transients.

As for Oberth, you can still get all those same benefits without using drag to settle your propellant. You can simply wait for your orbit to decay down to 120 km (or whatever) and then perform the TMI burn. If you counter that the wait is too long, then that's actually good news because it means you can lower your original parking orbit. :)

Once you've paid the expense to get the fuel up over 200km, there's no Oberth benefit.

No, the Oberth benefit is still there. You just don't also get the benefit of launching to a lower orbit, but instead you trade that off for higher reboost costs (which may or may not be worth it).

Also I didn't say 200 km, I said as low as possible so the decay time to your desired TMI altitude isn't too long.

  Make sure you use a C3 calculation, I got fooled when I didn't.

I know, I'm the one who pointed that out.  ;)

If you want to refill below 200km, you are back in in the regime of "you have atmospheric drag, you might as well use it".

Not necessarily. Your mathematical approach starts out by assuming you must have 0.1 mm/s2, which isn't necessarily the optimal answer.

The greatest benefit turns out to be the not having the lift the fuel from the surface of the earth to over 200km, but to an average of 135km instead.

And the greatest downside is the high drag at that altitude. If your refilling takes too long, the necessary reboosts will swamp your savings.

  The second benefit is not wasting tons of prop on cold gas thrusters.

Arcjets aren't exactly groundbreaking technology, so I don't see that as too big an advantage. If it's really such a downside you just use an arcjet.

I expect the "real" optimization will be similar to what's used on ISS: you reboost just high enough that after orbital decay you meet your next tanker at an altitude that strikes the optimal balance between reboost and payload-to-orbit. If drag happens to be sufficient for ullage at that altitude then sure you use that (this is the right way to use your "might as well use it" reasoning), but otherwise you just combine your reboost and ullage burns.

You're letting your new favorite "one clever trick" wag the entire dog. In reality you want to optimize for the greatest overall system efficiency, balancing all sources of loss (payload-to-orbit, reboost, Oberth, boiloff, schedule risks, etc).
« Last Edit: 12/08/2024 02:58 am by Twark_Main »

Offline InterestedEngineer

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Re: Starship On-orbit refueling - Options and Discussion
« Reply #2718 on: 12/08/2024 03:20 am »

I'm pretty sure I multiplied out the drag by calculating a drop of 1km per two orbits, then guessing at one refill every 6 orbits, so 60 orbits to fill Starship V3 to 2000t of prop and a total of 30km drop.  Did I do the math wrong?

Yes, you only multiplied by 7200 s (2 hours), not 6 hours.

No, I converted a constant drag of 0.1mm/sec to altitude drop per orbit, which is about 500m/orbit.  Sorry if that was confusing.  I'll do that math again just to be clear.

nominal orbit is 7814.09 m/sec at 150km
Orbital period is 1.458 hours = 5249 seconds. (16 orbits per day)
deltaV = 5259 * .0001 = 0.5m/sec

0.5m/sec in the retrograde direction drops the orbit to 149.34 km, or a 660 meters per orbit.

Keep in mind this can be modulated by adjusting the angle of incidence to pretty much any surface area from 64m2 to 800m2 when docked.  When not actively fueling, it'll be adjusted to the minimum, so when not actually refueling the drag will be less than 0.1mm/s2.

Roughly average that out to 500meters drop per orbit.  There's 60 orbits before dropping to the "get it done fast" zone of 120km with another 5-10 orbits of margin left.

Is 60 orbits, or over 4 days too optimistic?

For sending thousands of Starships to Mars per launch window, yes it's hilariously optimistic.

Only passenger ships need a full fuel drop.  Cargo ships need half of that, they take the slow route, so it's only 1/10 ships that need the full load.

You think it'll take well over 4 days to fill up one passenger ship's tanks? Passengers aren't going to wait around that long anyways, the passengers will hate it.  If it's 100 ships that's 1000 refuelings over 4 days or 250/day, which will keep 250 starships busy at refueling, assuming they have a 24 hour cycle, and about 32 boosters and launch mount pairs if cycle time is 3 hours.  That doesn't sound completely insane by Elon standards.

With a launch window of 3 weeks (21 days), that gives 5 iterations of this, and because the fuel for cargo is a half load taking the slow route to Mars, there's your 10x.    32 booster/launch mount pairs, 250 fuelers, and add 10% for inevitable snafus.

Shorter is also much better for boil off reasons.  I don't think we want things sticking around in VLEO for very long due to that alone, so an orbital drop of 8km/day seems reasonable.
« Last Edit: 12/08/2024 03:21 am by InterestedEngineer »

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Re: Starship On-orbit refueling - Options and Discussion
« Reply #2719 on: 12/08/2024 03:53 am »

I'm pretty sure I multiplied out the drag by calculating a drop of 1km per two orbits, then guessing at one refill every 6 orbits, so 60 orbits to fill Starship V3 to 2000t of prop and a total of 30km drop.  Did I do the math wrong?

Yes, you only multiplied by 7200 s (2 hours), not 6 hours.

No, I converted a constant drag of 0.1mm/sec to altitude drop per orbit, which is about 500m/orbit.  Sorry if that was confusing.

The only confusing part is why you're then still quoting a penalty of only 0.72 m/s.  ???

That's a deltaV of 7200s * 0.0001m/s2 = 0.72m/sec.

We are paying back 0.72m/s2 of deltaV, rounding error




Is 60 orbits, or over 4 days too optimistic?

For sending thousands of Starships to Mars per launch window, yes it's hilariously optimistic.

Only passenger ships need a full fuel drop.  Cargo ships need half of that, they take the slow route, so it's only 1/10 ships that need the full load.

You're still not doing that over four days out of the total of 780 days within the synodic cycle. That is not highly efficient fleet and launch site utilization. If 900 ships are partly filled (or more likely, more heavily laden with payload) it doesn't really change that fact.



You think it'll take well over 4 days to fill up one passenger ship's tanks? Passengers aren't going to wait around that long anyways, the passengers will hate it.

Two and a half year journey, and you think four days is a show-stopper?  ???


If it's 100 ships that's 1000 refuelings over 4 days or 250/day, which will keep 250 starships busy at refueling, assuming they have a 24 hour cycle, and about 32 boosters and launch mount pairs if cycle time is 3 hours.  That doesn't sound completely insane by Elon standards.

So if you wait 8 days instead of four, you can halve the entire refilling fleet and ground support costs? Leaving such a massive cost efficiency gain on the table doesn't sound like Elon either.

With a launch window of 3 weeks (21 days), that gives 5 iterations of this, and because the fuel for cargo is a half load taking the slow route to Mars, there's your 10x.    32 booster/launch mount pairs, 250 fuelers, and add 10% for inevitable snafus.

21 days out of 780 days? Same argument. You're way better off stretching out your campaign (and therefore reducing your fixed costs) by staging propellant in orbit over a longer period of time. Flatten the curve.

Shorter is also much better for boil off reasons.  I don't think we want things sticking around in VLEO for very long due to that alone, so an orbital drop of 8km/day seems reasonable.

Squinting and saying "maybe that's right" isn't really the route to a proper system optimization.

The correct algorithm is to start with your optimization goal and then work backwards from there, not try to shoehorn it in.
« Last Edit: 12/08/2024 03:58 pm by Twark_Main »

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