[AR] Re: OT Nozzle shapes
- From: Nathan Mogk <nm8911@xxxxxxxxx>
- To: arocket@xxxxxxxxxxxxx
- Date: Wed, 13 Nov 2013 14:05:39 -0700
For a heat pipe, you don't need to have it spinning. The temperature
gradient will provide the circulation, but perhaps you can get more
circulation by spinning.
On Wed, Nov 13, 2013 at 1:13 PM, Ian Woollard <ian.woollard@xxxxxxxxx>wrote:
> Doesn't sound particularly difficult in principle, it's just a heat pipe.
>
> Maybe construct it as a round, flat disk with the lasers at the rim, and
> spin it around the centre axis.
>
> Make the disk double walled and have ties between the two faces to deal
> with the steam pressure. Or just spot weld it occasionally with dimples.
>
> So the water cycle goes, boils at the rim, steam rises up towards the
> axis, condenses against the wall and rains back down towards the rim again.
> No pumps. You'll need to score the interior surface of the metal with
> channels to let the water wet it and run back down again otherwise it will
> get blown about too much by the steam.
>
> ~300 watts/metre sounds about right; similarish power to that radiated by
> the Earth.
>
> You'll possibly also need a sunshade. You'll also have to precess it
> somehow for pointing or use mirrors.
>
>
> On 13 November 2013 05:59, Keith Henson <hkeithhenson@xxxxxxxxx> wrote:
>
>> 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
>>
>>
>
>
> --
> -Ian Woollard
>
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