[AR] OT Nozzle shapes
- From: Keith Henson <hkeithhenson@xxxxxxxxx>
- To: arocket@xxxxxxxxxxxxx
- Date: Tue, 12 Nov 2013 21:59:23 -0800
I have a fairly bizarre problem. Need to get rid of 3 GW of waste
heat from lasers in orbit. The pump diodes use a total flow of 60
cubic meters per second at 22 deg C inlet and 34 deg C outlet.
The heat removed from the local cooling loop for one laser diode, is
1.5 kW, which amount to lowering the temperature of 0.03 l/s by 12
degrees to get the water back to 22 deg C. Evaporation takes 2640
kJ/kg at that temperature. Injected tangentially into a low-pressure
"boiling drum," the pumps could pick up the water from scoops as it
swirled around the inside while vapor came off the surface in the
center. At that temperature, evaporating water takes 2460 kJ/kg.
Lowering the temperature 12 degrees for a kg of water would require
removing 4.2 kJ/kg-deg K x 12 deg K or 50.4 kJ. The fraction of the
water flow evaporated would be 50.4/2640 or about 2%.
Two% of 60,000 liters per second is 1200 liters per second or 1200
kg/s. 120 kg/s split into ten radiator sections.
It is not clear how to minimize the system mass, but large pipes do
not contribute much to the mass since the pressure over 22-degree
water is only 2644 Pa. If we allow a ten-degree temperature difference
to drive the low-pressure steam into the cooler radiators, they will
be at 12 deg C. At that temperature and an emissivity of 0.95, the
surface will radiate 355 W/m^2.
How to get an emissivity coating on aluminum (or something else) that
will radiate well at that temperature is a question that needs more
research.
Three GW at 355 W/m^2 would take 8.45 square km of area.
The original rough partitioning of mass for the LPS gave 6,000 tons to
the radiator. As a design-to-mass project, we initially allocate 4000
tons to the radiator pressure vessel 4,000 tons of aluminum at a
density of 2.7 tons per m^3 is 1481 cubic meters. Li-Al alloys can be
10% lighter and there are substances like graphite and other forms of
carbon that are much lighter for the same strength. 1481 cubic meters
spread over 8.45 million square meters is 0.175 mm. That is only twice
as thick as a soda can.
On a per meter basis, and using a 400 M Pa strong aluminum alloy, the
maximum hoop stress is 70129 N. The vapor pressure of water at 12 deg
C is 1400 Pa. Thus 70129 = 1/2 D x P or D = 2 x 70129/1400, which is
almost exactly 100 meters in diameter. Circumference is 314 m, each km
of tube would have 0.314 square km of area. For 8.45 square km, the
length is 26.85 km. A radiator with ten tubes would be almost 2.7 km
in one dimension and, with spacers and reflectors, about 4 km in the
other dimension. (See figure C on page 7 of the above URL.)
The internal volume of the radiator tubes would be ~212 million cubic
meters. Steam at 12 deg C has a density of 0.01067 kg/m^3 and would
mass 2262 tons. How much water would stick to the inside is unknown.
Only a tenth of a millimeter would increase the mass by 845 tons.
The question is how to size the nozzle between the relatively hot
steam coming off the evaporator at 22 deg c and 2644 Pa and the major
part of the radiator at 1400 Pa and 12 deg C main radiator tube?
Secondary to that question is: will the momentum from the steam carry
the liquid water to the far end of the radiator in the absence of
gravity.?
I realize this is way off topic for AR, but it is a physics problem
about nozzles and some of you may have some insight into how to solve
the problem.
Keith
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