[AR] Re: OT Nozzle shapes
- From: Carlo Vaccari <airplaniac2002@xxxxxxxxx>
- To: arocket@xxxxxxxxxxxxx
- Date: Wed, 13 Nov 2013 16:08:05 -0500
Normally, heat pipes are filled with a porous material for capillary
action. I imagine that's less feasible weight-wise, and it would perhaps
require more liquid volume, so I would be inclined to agree that spinning
would be helpful.
On Wed, Nov 13, 2013 at 4:05 PM, Nathan Mogk <nm8911@xxxxxxxxx> wrote:
> 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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