Re: Is an orbital refueling station feasible?
- From: Ken Myrtle <myrtle@xxxxxx>
- Date: Tue, 06 Dec 2005 12:30:45 -0800
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
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