[AR] Re: OT Nozzle shapes
- From: Carlo Vaccari <airplaniac2002@xxxxxxxxx>
- To: arocket@xxxxxxxxxxxxx
- Date: Wed, 13 Nov 2013 16:47:37 -0500
You can't have twice the surface area without reducing thickness. The 100
meter diameter pipe had 2x-soda-can-thickness walls; if you double the
surface area it inherently needs either twice the metal, or half the
thickness.
On Wed, Nov 13, 2013 at 4:44 PM, Ian Woollard <ian.woollard@xxxxxxxxx>wrote:
> Aren't thin radiators better than thick radiators? Why would you want a
> pipe 100 metre in diameter? For example 4 pipes 50 metres in diameter have
> the same volume, and PV value and metal weight is the same, but you have
> twice the surface area.
>
> What am I missing?
>
>
> On 13 November 2013 21:25, Keith Henson <hkeithhenson@xxxxxxxxx> wrote:
>
>> Spinning was my first thought, even have some artwork based on the idea.
>>
>> But, if it will work, it would be less complicated to have a rocket
>> type expansion into the low pressure condensing section of ten tubes
>> 100 meters in diameter and 2.7 km long. The question: is the momentum
>> from the low pressure steam entering the tube enough to sweep the
>> condensed water to the far end of the tube where pumps can return it
>> for reuse? Sub question, what's the best dimensions for the nozzle?
>> Can you get gas this thin to choke in the nozzle?
>>
>> On Wed, Nov 13, 2013 at 1:08 PM, Carlo Vaccari <airplaniac2002@xxxxxxxxx>
>> wrote:
>> > 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
>> >>
>> >>
>> >
>>
>>
>
>
> --
> -Ian Woollard
>
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