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 >