Is an orbital refueling station feasible?

I have this idea for an orbital rocket fuel refueling
station; has this idea been studied already?

The refueling station is a solar powered satellite which
scoops up water vapor from the atmosphere and then
electrolyzes it into rocket fuel. At that point, an
interplanetary mission can refuel at the satellite,
which then returns to scooping up water for the next
mission.

The satellite starts off life as a hydrogen-lox SSTO
rocket. It flies into an atmosphere scraping elliptical
orbit, discarding an aerodynamic nose cone to reveal a
full diameter "airscoop". This airscoop can ram air
into a compressed air tank via a one-way-valve.

The satellite then spends the next phase of the mission
in that atmosphere scraping elliptical orbit. It could
perhaps have an orbit with a 48 hour period to dip into
the atmosphere over the Pacific. During atmospheric
dips, it rams thin air into one tank. During the rest
of the orbit, air is fed through condensation tubes to
a solar powered arcjet thermal electric rocket (maybe
operated in pulses to minimize average power consumption).
Liquid water from the condensation tubes is collected
into the other tank.

After the water tank is full, the satellite slightly
circularizes its orbit so it no longer dips into the
atmosphere. During this mission phase, solar power
is used to electrolyze water into hydrogen and oxygen.
After the water is fully separated, the satellite
simply waits for a spacecraft to dock and refuel.

An interplanetary mission can be boosted atop an SSTO
that refuels at the refueling station. Assuming the
SSTO was good for 12km/s delta-v, refueling gives you
another 12km/s delta-v to go anywhere in the solar
system from Earth orbit.

Isaac Kuo

In article .com,
"IsaacKuo" wrote:

The satellite then spends the next phase of the mission
in that atmosphere scraping elliptical orbit. It could
perhaps have an orbit with a 48 hour period to dip into
the atmosphere over the Pacific. During atmospheric
dips, it rams thin air into one tank. During the rest
of the orbit, air is fed through condensation tubes to
a solar powered arcjet thermal electric rocket (maybe
operated in pulses to minimize average power consumption).

You might be better off using electrodynamic propulsion, so it can
reboost its orbit without using any propellant at all. Otherwise, I
suspect you'd end up spending all the air you collected just trying to
make up for the drag incurred collecting it. (I'm assuming here that
all the air collected is useful to somebody in orbit, not just the water
vapor.)

Best,
- Joe

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Consider Len's proprosal in
http://groups.google.com/group/sci.space.station/msg/a0b39c4b79014aa4?hl=en&

which is Message-ID:
.com

This seems to be a near-term doable, but your scooping trick might
eventually be developed. Once a few more technologies are in hand.

/dps

"IsaacKuo" writes:

The refueling station is a solar powered satellite which
scoops up water vapor from the atmosphere and then
electrolyzes it into rocket fuel.

If you have your satellite dip into the atmosphere on each
orbit, the orbit will slow down due to drag. You'll have to
burn some fuel to regain this loss. How much fuel? Less than
you can produce? How much less?

I think that's what you'd need to calculate.

I feel that the answer will be 'forget about it'.

But what do I know

best regards
Patrick

IsaacKuo wrote:

After the water tank is full, the satellite slightly
circularizes its orbit so it no longer dips into the
atmosphere. During this mission phase, solar power
is used to electrolyze water into hydrogen and oxygen.
After the water is fully separated, the satellite
simply waits for a spacecraft to dock and refuel.

How much water are you talking about here? Remember that the
added weight of the water needs added propulsion to alter the orbit.

Raiskila Kalle wrote:
IsaacKuo wrote:

After the water tank is full, the satellite slightly
circularizes its orbit so it no longer dips into the
atmosphere. During this mission phase, solar power
is used to electrolyze water into hydrogen and oxygen.
After the water is fully separated, the satellite
simply waits for a spacecraft to dock and refuel.

How much water are you talking about here? Remember that the
added weight of the water needs added propulsion to alter the orbit.

I don't know the water amount, but I do know it won't have much
impact on the propulsion requirements for scooping. During
each scoop, the amount of momentum lost is roughly 8km/s
multiplied by the mass of air to fill the ram-tank. The amount
of impulse to replace the lost momentum is the same regardless
of the satellite's mass.

Isaac Kuo

The exhaust velocity from the combustion if only 4 km/s. So you need to
burn twice the water you collect to break even. So no.

Even the much more better solar panel - electrodynamic tether -
rotating tether sat is impractical.

Above the mesophere, the upper atmosphere has negligible water vapor
content. The Earth's atmosphere contains a so-called "cold trap" at
about 75 km, where the temperature drops to about -100 C. This freezes
the rising water vapor out of the atmosphere.

Good thing, too-- it's why we're here. It prevents water vapor from
rising to the top of the atmosphere, photodissociating from solar UV,
and leaving.

In any case, water is a trivial weight fraction of the atmosphere-- the
drag you'd get from scooping atmosphere would far far exceed the
delta-V you'd be able to get out of electrolyzing water.

--
Geoffrey A. Landis
http://www.sff.net/people/geoffrey.landis

I think that there are several points that need to be looked at with this
air scoop design.
First water is not very abundant in the upper atmosphere. The low
temperatures
in the upper troposphere (10 km) limit the transport of water up into
the stratosphere .
So the water vapor concentration drops from a few percent at sea level to
1 to 10 parts per billion in the stratosphere.
see http://www.kowoma.de/en/gps/additional/atmosphere.htm
Even at 10 km the atmosphere is far to dense for a satellite to travel
without reentering. At 10 km altitude the air density is about 1/4 that
at sea level
or about 0.3 kg/m^3 or about 300 Kg of air hitting hitting the scoop
for every
Km of travel. A crude guess of 100 km path in this air density gives
30,000 kg of
air impacting each each m^2 of scoop. If we don't want to crash we need
enough
mass to carry us back out of the atmosphere. If we only want to drop our
velocity
by 10% we need 270 tons of ship behind each m^2 of scoop.
So unless we have very massive ships we can't sample the lower atmosphere.

While it looks like you can't collect water we can try to collect Oxygen
which is where most of the fuel mass is anyway.

The next problem is the density of the gas in the storage tank. The plan
was to
use one way valves to collect the air in the storage tank. This is
rather inefficient
since as soon as we pass the lowest point the external pressure begins
to drop
and the check valve closes. so the scoop will only collect about 1/2 of
the air
but will be slowed down an equal amount by the trip up out of the air.
If we dip down to about 90 Km the air density reaches 1e-5 kg/m^3
and with a 200 km path length the scoop would intersect about 2 kg of
air per m^2 of inlet area.
The pressure on the scoop can be calculated by determining the rate of
change
of momentum of the air.
density 1e-5 kg /m^3
velocity of scoop 8000 m/sec
air mass hitting scoop =1e-5 kg/m^3 * 8000 m/sec = 8e-2 kg /sec per m^2
change in velocity of air 8000 m/sec
pressure = 8e-2 kg/sec per m^2 * 8000 m/sec =640 kg m/sec^2 per m^2
or 640 nt/m^2 or 640 Pascal

Note 1 atmosphere =101,000 pascal and 1 m^3 of air at 1 atmosphere
has a mass of 1.2 kg at room temp
so the in order to hold the 1 kg of air that we collected (from downward
portion of flight)
at this pressure we need a storage volume of
V=1 kg /1.2 kg * 101,000 pascal /640 pascal =130 m^3

Since we want to hide the storage tank behind the scoop so that the tank
does not
add to the drag we would need a tank at least 130 meters long.
This is a bit unwieldy and unless the collector was very large the
thermal motion
of the air would bring it into contact with the side of the tank.
So a pump is probably required.
The next problem is the dissipation of of the heat.
For an observer on the scoop the kinetic energy of the incoming air stream
will be converted heat. We had 8e-2 kg of air hitting each m^2 each second
with a velocity of 8000 m/sec so the rate of heat generation is
heat/m^2 =1/2 * m * V^2 =0 .5 * (8e-2 kg/m^2 per sec) *(8000m/sec)^2
or 2.56 e6 watts /meter^2

with a total heat input per dip of
total heat /m^2 =0.5*2kg*(8000m/sec)^2= 64e6 joules/m^2

which occurs over a 25 sec period =200 km / 8 km/sec

We must cool the gas down so that it will not melt the pumps.
Lets try storing this heat into a lithium bath (lightest non gas element).
Lithium has a heat capacity of about 4 joules/gm K deg
and a boiling point of 1342C
so to keep the lithium from boiling we need (start at 0 deg C)
about 11900 gm of lithium per m^2 of scoop
11900 gm = 64e6 joules/(4j/gm k * 1342 c)
This is not too bad a number since we need about 20 kg / m^2 of scoop
to keep the orbit from dropping too much after the dip.

Lets see if a thermal radiator will help.
If the front of the Scoop radiates at 2300K (2000 C )
then we loose heat at a rate of
Q = (5.67E-8 )*(2300)^4 = 1.59 E6 watts /m^2
This is also good so we can radiate about half of the heat pulse
which allows us to carry less lithium.

Note that the scoop has lost 800 m/s of velocity in 25 sec or 32 m/sec^2
which is a little over 3g . Some care will be required to make the scoop
aerodynamically stable (perhaps like a SR-71 turbine inlet ).

Solar power will have to be used to power the scoop since a reactor would
never be allowed on a satellite which is almost reentering every couple of
days. This means that the either the panels must be shielded from the
airflow
or they have to stowed and redeployed for every dip which would a nightmare.

The lowest energy reboost is with a electrodynamic tether which would be
E = 800 m/sec *20 Kg * 8000 m/sec = 128 E6 joules /m^2 of scoop.
Note that this is twice the kinetic energy of the mass we picked up
since the mass gained kinetic energy and we dumped an amount of heat equal
to it's kinetic energy
Hmmm
100 watts / kg is pretty good for solar panels and if 20% of the mass of
the scoop is in the panels (ie 4 kg of panel per m^2 of scoop area)
so we have 400 watts /m^2 of scoop or
T = 128E6 Joules/400 watts = 320,000 sec = 3.7 days
So we can do a dip every week.

This a bit of a nasty orbit since it will expose the scoop to quit a bit
of
radiation and it is probably too high to use an electrodynamic tether.
So from a very crude first guess we can collect about 10 % of the scoop
mass per week.

I think that an optimization would drive this to a almost circular orbit
where the scoop components are used more than 25 sec per week.

A few years ago I had a good look at leo air collectors and found that
they were about as good as lunar mining ( in terms of Oxygen production
per
mass of equipment ) but that a launcher can deliver far more mass to leo.
It also looked like the steam rockets delivering water from asteroids
beats either leo collectors or lunar mining .
See http://www.neofuel.com/moonice1000/

One problem with all of these plans is that they are installing expensive
equipment to produce a commodity which we all hope will be much
cheaper in the future. So one has to ask can you make money if the
cost to launch drops by a factor of 10 after your device is installed?

Ken Myrtle

IsaacKuo wrote:

Raiskila Kalle wrote:


IsaacKuo wrote:





After the water tank is full, the satellite slightly
circularizes its orbit so it no longer dips into the
atmosphere. During this mission phase, solar power
is used to electrolyze water into hydrogen and oxygen.
After the water is fully separated, the satellite
simply waits for a spacecraft to dock and refuel.





How much water are you talking about here? Remember that the
added weight of the water needs added propulsion to alter the orbit.



I don't know the water amount, but I do know it won't have much
impact on the propulsion requirements for scooping. During
each scoop, the amount of momentum lost is roughly 8km/s
multiplied by the mass of air to fill the ram-tank. The amount
of impulse to replace the lost momentum is the same regardless
of the satellite's mass.

Isaac Kuo