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Given the mass ratio of 3, we know that the Polaris is 66% propellant and 33% everything else. Give the total mass of 1188.9 tons means 792.6 tons of propellant and 396.3 tons of everything else. Since each GC engine is 30 tons, that means 150 tons of engine and 246.3 of everything else.
"Everything else" includes a spacecraft's structure, propellant tankage, lifesystem, crewmembers, consumables, hydroponics tanks, cargo, atomic missiles, and other ship systems.
For some great notes, read Rick Robinson's Rocketpunk Manifesto essay on Spaceship Design 101. Also worth reading are Rick's essays on constructing things in space and the price of a spaceship.
What is the structure of the ship going to be composed of? The strongest yet least massive of elements. This means Titanium, Magnesium, Aluminum, and those fancy composite materials. And all the interior girders are going to have a series of circular holes in them to reduce mass (the technical term is "lightening holes").
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As an interesting side note, rockets constructed of aluminum are extremely vulnerable to splashes of metallic mercury or dustings of mercury salts. On aluminum, mercury is an "oxidizing catalyst", which means the blasted stuff can corrode through an aluminum beam in a matter of hours (in an atmosphere containing oxygen, of course). This is why mercury thermometers are forbidden on commercial aircraft. Why? Ordinarily aluminum would corrode much faster than iron. However, iron oxide, i.e., "rust", flakes off, exposing more iron to be attacked. But aluminum oxide, i.e., "sapphire", sticks tight, protecting the remaining aluminum with a gem-hard barrier. Except mercury washes the protective layer away, allowing the aluminum to be consumed by galloping rust. Alkalis will have a similar effect on aluminum, and acids have a similar effect on magnesium (you can dissolve magnesium with vinegar). As far as I know nothing really touches titanium, its corrosion-resistance is second only to platinum. |
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If you want a World War II flavor for your rocket, any interior spaces that are exposed to rain and other corrosive planetary weather should be painted with a zinc chromate primer. Depending on what is mixed into the paint, this will give a paint color ranging from yellowish-green to greenish-yellow. In WWII aircraft it is found in wheel-wells and the interior of bomb bays. In your rocket it might be found on landing jacks and inside airlock doors. Naturally this does not apply to strict orbit-to-orbit rockets, or rockets that only land on airless moons and planets. |
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When laying out the floor plan, you want the spacecraft to balance. That is, if you draw a line straight through the exhaust bell (in the direction that thrust is applied), it had better pass through the spacecraft's center of gravity, and if the ship is intended for atmospheric flight, it should also go through the spacecraft's nose. Otherwise your ship is going to loop-the-loop or tumble like a cheap Fourth of July skyrocket (Heinlein calls this a rocket "falling off its tail"). This also means that each deck should be "radially symmetric". That's a fancy way of saying that if you have something massive in the north-west corner of "D" deck, you'd better have something equally massive in the south-east corner. This is another reason to strap down the crew during a burn. Walking around could upset the ship's balance, resulting in the dreaded rocket tumble. This will be more of a problem with tiny ships than with huge cruisers, of course. Small ships might have "trim tanks", small tanks into which water can be pumped in order to adjust the balance. The ship will also have heavy gyroscopes that will help prevent the ship from falling off its tail, but there is a limit to how much imbalance that they can compensate for. |
A cursory look at the rocket's mass ratio will reveal that most of the rocket's mass is going to be propellant tanks.
Nuclear thermal rockets generally use hydrogen (if it is a Gas Core NTR, the fissionable fuel will probably be less than 1% of the total propellant load) since you want propellant with the lowest molecular mass. Liquid hydrogen has a density of 0.07 grams per cubic centimeter. 792.6 tons of propellant = 792,600,000 grams / 0.07 = 11,323,000,000 cubic centimeters = 11,323 cubic meters . The volume of a sphere is 4/3πr3 so you can fit 11,323 cubic meters in a sphere about 14 meters in radius . Almost 92 feet in diameter, egad! It is a pity hydrogen isn't a bit denser. If this offends your aesthetic sense, you'll have to go back and change a few parameters. Maybe a 2nd generation GC rocket, and a mission from Terra to Mars but not back. Maybe use methane instead of hydrogen. It only has an exhaust velocity of 6318 m/s instead of hydrogen's superior 8800 m/s, but it has a density of 0.42 g/cm3, which would only require a 1.7 meter radius tank. (Methane has a higher exhaust velocity than one would expect from its molecular weight, due to the fact that the GC engine is hot enough to turn methane into carbon and hydrogen. Note that in a NERVA style engine the reactor might become clogged with carbon deposits.)
Robert Zubrin says that as a rule of thumb, the mass of a fuel tank loaded with liquid hydrogen will be about 87% hydrogen and 13% tank. In other words, multiply the mass of the liquid hydrogen by 0.15 to get the mass of the empty tank (0.13 / 0.87 = 0.15). So our 792.6 tons of hydrogen will need a tank that masses 792.6 * 0.15 = 119 tons.
87% propellant and 13% tank is for a rocket designed to land on a planet. An orbit-to-orbit rocket could get by with more hydrogen and less tank. This is because the tanks can be more flimsy since they will not have to endure the stress of landing (A landing-capable rocket that uses a propellant denser than hydrogen can also get away with a smaller tank percentage). Zubrin gives the following ballpark estimates of the tank percentage:
If you are going to use aerobraking to land your rocket, Zubrin says mass of the heat shield and thermal structure will be about 15% of the total mass being braked. As a wild guess, aerobraking will be limited to killing a velocity of no more than 15 to 30 kilometers per second. The rule of thumb is that aerobraking can kill a velocity approximately equal to the escape velocity of the planet where the aerobraking is performed (10 km/s for Venus, 11 km/s for Terra, 5 km/s for Mars, 60 km/s for Jupiter). This will mostly be used for our purposes designing a emergency re-entry life pod, not a Solar Guard patrol ship. With a sufficiently advanced engine it is more effective just to carry more fuel, so our atomic cruiser will not need to waste mass on such a primitive device.
In the movie 2010, the good ship Leonov had a one-lung propulsion system, so they needed an aerobraking "ballute" to slow them into Jovian orbit. If you are thinking about aerobraking, keep in mind that many worlds in the Solar System do not have atmospheres.
If you cannot tap your propulsion system for electrical power, you will need a separate power plant (or it's going to be real dark inside your spacecraft). Since rocket designers are concerned with the mass of various components, they use a term called "alpha". This is the ratio of power plant mass (kilograms) to the electrical output (kilowatts). In other words, divide the plant mass by the kilowatts to get Alpha. So if a solar power array had an alpha of 90, and you needed 150 kilowatts of output, the array would mass 90 * 150 = 13,500 kg or 13.5 metric tons.
Occasionally you'll see a power plant with a rating in "Specific Power." This is kilowatts divided by plant mass. In other words it is 1/alpha (though sometimes specific power uses watts instead of kilowatts).
Radioisotope thermoelectric generators (RTG) are slugs of radioisotopes (usually plutonium-238 in the form of plutonium oxide) that heat up due to nuclear decay, and surrounded by thermocouples to turn the heat into electricity. There are engineering reasons that make it impractical to design an individual RTG that produces more than one kilowatt. However nothing is stopping you from using several RTGs in your power room.
Nuclear weapons-grade plutonium cannot be used in RTGs. Plutonium-238 has a half life of 85 years, i.e., the power output will drop to one half after 85 years. To calculate power decay:
P1 = P0 * 0.9919^Y
where:Example: If a new RTG outputs 470 watts, in 23 years it will output 470 x 0.9919^23 = 470 x 0.83 = 390 watts
Wolfgang Weisselberg points out that this equation just measures the drop in the power output of the slug of plutonium. In the real world, the thermocouples will deteriorate under the constant radioactive bombardment, which will reduce the actual electrical power output even further. Looking at the RTGs on NASA's Voyager space probe, it appears that the thermocouples deteriorate at roughly the same rate as the plutonium.
Plutonium-238 has a specific power of 0.56 watts/gm or 560 watts per kilogram, so in theory all you would need is 470 / 560 = 0.84 kilograms. Alas, the thermoelectric generator which converts the thermal energy to electric energy has an efficiency of only a few percent. If the thermoelectric efficiency is 5%, the plutonium RTG has an effective specific power of 560 x 0.05 = 28 watts per kilogram (0.036 kilogram per watt or 36 kg/kW). This means you will need an entire 17 kilos of plutonium to produce 470 watts.
Currently RTGs have an alpha of about 200 kg/kW (though there is a design on the drawing board that should get about 100 kg/kW). So an RTG with the theoretical maximum output of 1 kilowatt would obviously mass 200 kilograms.
Many RTG fuels would require less than 25 mm of lead shielding to control unwanted radiation. Americium-241 would need about 18 mm worth of lead shielding. And Plutonium-238 needs less than 2.5 mm, and in many cases no shielding is needed as the casing itself is adequate.
Solar Power ArraysAt Terra's distance to the sun, solar energy is about 1366 watts per square meter. This energy can be converted into electricity by photovoltaics. Solar power arrays have an alpha ranging from 20 to 100 kg/kW. The current state of the art is about 45 kg/kW. Of course their power level goes down the farther from the Sun they travel. The International Space Station uses 14.5% efficient large-area silicon cells. Each of the Solar Array Wings are 34 m (112 ft) long by 12 m (39 ft) wide, and are capable of generating nearly 32.8 kW of DC power. 19% efficiency is available with gallium arsenide (GaAs) cells, and efficiencies as high as 30% have been demonstrated in the laboratory. |
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 Mir space station
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Obviously the array works best when oriented face-on to the sun, and unshadowed. As the angle increases the available power decreases in proportion to the cosine of the angle (e.g., if the array was 75° away from face-on, its power output would be Cos(75°) = 0.2588 or 26% of maximum). Solar cells also gradually degrade due to radiation exposure (say, from 8% to 17% power loss over a five year period if the panel is inhabiting the deadly Van Allen radiation belt, much less if it is in free space). |
Typically solar power arrays are used to charge batteries (so you have power when in the shadow of a planet). You should have an array output of 20% higher voltage than the battery voltage or the batteries will not reliably charge up. Sometimes the array is used instead to run a regenerative fuel cell.
Like all non-coherent light, solar energy is subject to the inverse square law. If you double the distance to the light source, the intensity drops by 1/4. As a rule of thumb:
Es = 1366 * (1 / Ds2)
where:Remember that you divide distance in meters by 1.49e11 in order to obtain astronomical units.
Example: What is the available solar energy at the orbit of Mars?
Mars orbits the sun at a distance of 2.28e11 meters. That is 2.28e11 / 1.49e11 = 1.53 astronomical units. So the available solar energy is:
Es = 1366 * (1 / Ds2)This means that the available solar energy around Saturn is a pitiful 15 W/m
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 Back in the 1950's, on artist conceptions of space stations and space craft, one would sometimes see what looked like mirrored troughs. These were "mercury boilers", a crude method of harnessing solar energy in the days before photovoltaics. The troughs had a parabolic cross section and focused the sunlight on tubes that heated streams of mercury. The hot mercury was then used in turbines to generate electricity. These gradually vanished from artist conceptions and were replaced by nuclear reactors. Such systems are generally useful for power needs between 20 kW and 100 kW. Below 20 kW a solar cell panel is better. Above 100 kW a nuclear fission reactor is better. They typically have an alpha of 250 to 170, a collector size of 130 to 150 watts per square meter, and a radiator size of 140 to 200 watts per square meter.  |
Fuel CellsFuel cells basically consume hydrogen and oxygen to produce low voltage electricity and water. They are quite popular in NASA manned spacecraft designs. Each PC17C fuel-cell stack in the Shuttle Orbiter has an alpha of about 13 kg/kW, have a total mass of 91 kg, have an output of 7 kW, and produces about 2.7 kilowatt-hours per kilogram of hydrogen+oxygen consumed (about 70% efficient). The water output can be used in the life support system. A "regenerative" fuel cell saves the water output, and uses a secondary power source (such as a solar power array) to run an electrolysers to split the water back into oxygen and hydrogen. This is only worth while if the mass of the secondary power source is low compared to the mass of the water. But it is attractive since most life support systems are already going to include electrolysers anyway. |
 Fuel cell from Apollo Service Module. Red rectangle is a one foot ruler.
Image from the Scott Schneeweis Space Artifacts Collection.
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 SNAP-10A NASA spacecraft nuclear reactor
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Nuclear Fission ReactorsFor a great in-depth analysis of nuclear power for space applications, I refer you to Andrew Presby's engineer degree thesis: Thermophotovoltaic Energy Conversion in Space Nuclear Reactor Power Systems (PDF file). There is a much older document with some interesting designs here (PDF file). Nuclear fission reactors are about 18 kg/kW. However, Los Alamos labs had an amazing one megawatt Heat Pipe reactor that was only 493 kg (alpha of 0.493 kg/kW):
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Nuclear Thermal Rockets are basically nuclear reactors with a thrust nozzle on the bottom. A concept called Bimodal NTR allows one to tap the reactor for power. This has other advantages. Since the reactor is running warm at a low level all the time (instead of just while thrusting) it doesn't have to be pre-heated if you have a burn coming up. This reduces thermal stress, and reduces the number of thermal cyclings the reactor will have to endure over the mission. It also allows for a quick engine start in case of emergency. In the real world, during times of disaster, US Navy submarines have plugged their nuclear reactors into the local utility grid. This supplies emergency electricity when the municipal power plant is out. In the science fiction world, a grounded spacecraft with a bimodal NTR could provide the same service. |
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 SNAP-10A NASA spacecraft
nuclear reactor. Click for a larger image.
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Here is a commentary on figuring the mass of the reactor of a nuclear thermal rocket by somebody who goes by the handle Tremolo: Now, onto a more practical means for generation 1 MW of power using a Plutonium fission reaction. To calculate the mass required to obtain a certain power level, we have to know the neutron flux and the fission cross-section. Let’s assume the flux is 1E14 neutron/cm2/sec, the cross section for fast fission of Pu-239 is about 2 barns (2E-24 cm2), the energy release per fission is 204 MeV, and the Pu-239 number density is 4.939E22 atoms/cm3. Then the power is P = flux * number density * cross section * Mev per fission * 1.602E-13 Watt/MeV P = 1E14 * 4.939E22 * 2E-24 * 204 * 1.602E-13 = 323 W/cm3 So, for 1 MW, we need 1E6/323 = 3100 cm3. Given a density of 19.6 gm/cm3, this is 19.6*3100 = 60,760 gm or 60.76 kg. |
The next question to ask is: how long do you want to sustain this reaction? In other words, what is the total energy output?
For example, a Watt is one Joule per second. So, to sustain a 1 MW reaction for 1 year, the total energy is 1E6 J/s * 3.15E7 s/year = 3.15E13 J
For Pu-239, we have 204 Mev per fission and we have 6.023E23./239 = 2.52E21 atoms/gm. So, the energy release per gram is 2.52E21 * 204 Mev/fission * 1.602E-13 J/Mev = 8.24E10 J/gm.
Therefore, to sustain 1 MW for 1 year, we will use 3.15E13 J / 8.24E10 J/gm = 382 gm of Pu-239 or 0.382 kg. This is only a small fraction of the total 60.76 kg needed for the fission reaction.
Finally, this is thermal energy. Our current light water reactors have about a 35% efficiency for conversion to electric power. So, you can take these numbers and essentially multiply by 3 to get a rough answer for the total Pu-239 needed: 3 x 60.76 = 182 kg. Rounding up, you would need roughly 200 kg for a long term sustained 1 MW fission reaction with a 35% conversion efficieny.
These calculations assume quite a bit and I wouldn’t use these numbers to design a real reactor, but they should give you a ballpark idea of the masses involved.
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New reactors that have never been activated are not particularly radioactive. Of course, once they are turned on, they are intensely radioactive while generating electricity. And after they are turned off, there is some residual radiation due to neutron activation of the reactor structure. How much deadly radiation does an operating reactor spew out? That is complicated, but Anthony Jackson has a quick-and-dirty first order approximation: r = (0.5*kW) / (d2) where:r = radiation dose (Sieverts) kW = power production of the reactor core, which will be greater than the power output of the reactor due to reactor inefficiency (kilowatts) d = distance from the reactor (meters) This equation assumes that a 1 kW reactor puts out an additional 1.26 kW in penetrating radiation (mostly neutrons) with an average penetration (1/e) of 20 g/cm2. |
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As a side note, in 1950's era SF novels, nuclear fission reactors are commonly referred to as "atomic piles." This is because the very first reactor ever made was basically a precision assembled brick-by-brick pile of graphite blocks, uranium fuel elements, and cadmium control rods.
 Note that in reality one would not more than four radiator fins,
let alone eight. If the fins are spaced closer than 90 degrees, they intefere by radiating
heat into each other.
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HEAT RADIATORSPower plants and some propulsion systems are going to require heat radiators to avoid system meltdown. There are only three ways of getting rid of heat: convection, conduction, and radiation; and the first two do not work at all in the vacuum of space. So the ship designer is stuck with heat radiators. See Thermophotovoltaic Energy Conversion in Space Nuclear Reactor Power Systems and HIGH TRADER for details. Ken Burnside also noted that radiators are large, flimsy, and impossible to armor (except perhaps for the droplet radiator). A liability on a warship. If you want to calculate this for yourself: ∂Q/∂t = Re * (5.67x10e-8) * Ra * Rt4 where∂Q/∂t = amount of waste heat to get rid of (watts) 5.67x10e-8 = Stefan's Constant Re = emissivity of radiator (theoretical maximum is 1.0) Ra = area of radiator (m2) Rt = temperature of radiator (degrees K) |
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Ken Burnside says that if one examine the equation carefully one will notice that the radiator effectiveness goes up at the fourth power of the heat of the radiator. The higher the temperature, the lower the surface area can be, which lowers the required mass of radiator fins. This is why most radiator designs use liquid sodium or lithium (or things more exotic, still). 1600K radiators mean that you need a lot less mass than 273 K radiators. Propulsion systems like nuclear thermal rockets do not need heat radiators because the waste heat is carried away by the exhaust plume. In effect, the exhaust is their radiator (the technical term is "Open-Cycle Cooling"). Electrical powered drives like ion drives will require radiators on their power plants. Fusion drives may or may not require radiators, depending upon whether the design can dump the waste heat into the exhaust or not. |
 Artwork by Tero Niemi.
Click for larger image.
In reality a radiator glowing brighter than
dull red would melt, but it does make
for a dramatic image.
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My source (Matthew DeBell) says that if ∂Q/∂t = 150 gigawatts and Rt = 3000° K, Ra would be 34,941m2. Actually it could be half that if you have a two-sided radiator, which would make the radiator a square 90-odd meters on a side. I have no idea how to estimate how much mass a radiator of a given size will be. At a rough guess, you are looking at 0.01 to 0.05 kilograms per kilowatt dissipated. (The table I was looking at said that a flimsy radiator operating at 1100° K would be 0.01 kg/kW and an armored meteor proof radiator operating at 2000° K would be 0.05 kg/kW)
An advanced design is the liquid droplet radiator. Dr. John Schilling had a few comments:
Liquid-droplet radiators are also a possibility. There do exist liquids which have extremely low vapor pressure at high temperatures - certain organics up to ~600K, liquid metals (esp. lithium) to ~1500K. Using a carefully-designed nozzle to create a fan-shaped spray of fine droplets towards a linear collector results in a very efficient radiator, with minimal weight per unit radiating surface, high temperature, and high throughput.
The radiator would be essentially triangular when "deployed", with the spray nozzle at one vertex and the collector along the opposite side. If the nozzle-vertes is adjacent to the ship body, the collector "arm" will have to extend outwards. Alternately, the collector can be run along the side of the ship, and the spray nozzle extended on a boom and aimed inwards. A series of closely-spaced, narrow-angle nozzles would approximate a rectangular array.
There is always some loss of coolant due to evaporation in vacuum, hence use of liquids with extremely low vapor pressure. You also lose coolant if such a radiator is run under acceleration, unless the collector is over-long and aligned parallel to the thrust axis, which imposes a constraint on system geometry. You also lose coolant if the radiator "panel" is hit by enemy weapons fire; on the other hand there is no mechanical damage unless the much smaller nozzle or collector arms are hit. Bottom line - you'll need a small surplus of coolant, unless you are running a warship, in which case you'll need a large surplus.
If liquid metal is used as the coolant, MHD pumping can be used at the collector arms, resulting in a simplified design with no moving parts. Indeed, in such a case the coolant could also be used as the working fluid in an MHD generator, resulting in a single-fluid, single-cycle power system from primary energy generation to waste heat radiation. Again, a simple, efficient design with no moving parts.
Eric Rozier has an online calculator for droplet radiators here, and for coolant systems in general here. He had this analysis:
Given that the main thing we want to determine is the surface area of the lithium droplets to calculate the heat it can radiate, I decided to build a model of the surface area.
Since no such radiator has been built we have to work with some plausible model data. To model the lithium drops themselves I dug into some meteorological data and found that raindrops typically range in size from 1mm to 3mm, sounds pretty reasonable. Assuming droplets are spherical (a reasonable assumption in zero gravity) then the surface area of any given droplet is of course 4*π*r2.
Working off the wedge based idea you cited here. We then model the full radiating body of the droplets as a triangle, reducing the emitter to a point source for simplification. I'm not sure how space out the droplets should be, but I figure if the distance between any two droplets is roughly twice the radius, the model is probably pretty conservative. Thus for an emitter with distance h from the emitter to the collector, and a collector plate of length h, we get the number of droplets suspended between them to be:
(0.5 * b * h)/(16r2)
We can then model the surface area of the lithium droplets as:
(0.5 * b * h)/(16r2) * 4*π*r2
If you want to modify the spacing of the drops, you can change the inter-droplet gap to q instead of r, rendering the following equation:
(0.5 * b * h)/(4r2 + 4r*q + q2) * 4*π*r2
So the equations are:
a = (0.5*b*h) / (16*r2) * 4*π*r2
a = (0.5*b*h) / (4*r2 + 4*r*q + q2) * 4*π*r2
where:What color will the radiators glow? A practical one will only glow dull red. You can use the Blackbody Spectrum Viewer to see what temperature corresponds to what color. If it was glowing white hot, the temperature would be around 6000 Kelvin. This would be difficult for a solid radiator, since even diamond melts at 4300 degrees K.
Technically you also need radiators to keep the life-system habitable. Human bodies produce an amazing amount of heat. Even so, the life-system radiator should be small enough to be placed over part of the hull.
I had initially thought that the heat from the life-system could be simply dumped by the same radiator system dealing with the multi-gigawatt waste heat from the propulsion system. Richard Bell pointed out that I had not thought the problem through. Due to the difference in the temperatures of the waste heat from life-system and propulsion, unreasonably large amounts of energy will be required to get the low-level life-system heat into a radiator designed to handle high-level propulsion heat. The bottom line is that there will be two separate radiator systems.
The life-system radiators on the Space Shuttle are inside the cargo bay doors, which is why the doors are always open while the shuttle is in space.
Troy Campbell pointed me at a fascinating NASA report about spacecraft design (warning, 2 MB PDF file). In the sample design given in the report, the spacecraft habitat module carried six crew members, and needed life-system heat radiators capable of collecting and rejecting 15 kilowatts of heat (15 kW is the power consumption for all the systems included in the example habitat module). The radiator was one-sided (basically layered over the hull). It required a radiating surface area of 78 m2, had a mass of 243.8 kg, and a volume of 1.742 m3. It used 34.4 kg of propylene glycol/water coolant as a working fluid. In addition to the radiator proper, there was the internal and external plumbing. The Internal Temperature Control System (coldplates, heat exchangers, and plumbing located inside the habitat module) had a mass of 111 kg and a volume of 0.158 m3. The External Temperature Control System had a mass of 131 kg, a volume of 0.129 m3, and consumes 1.109 kilowatts.
Simple math tells me the radiator has a density of about 140 kg/m3, and needs a radiating surface area of about 5.2 m2 per per kilowatt of heat handled. The entire system requires about 35 kg per kilowatt of heat handled, and 0.13 m3 per kilowatt of heat. But treat these numbers with suspicion, I am making the assumption that these things scale linearly.
Here is some scary math about radiators from Tony Valle, along with some interesting conclusion:
It is surprising but there is an optimum temperature ratio at which to run a starship heat exchanger (or similar power source) to achieve maximum free power with a minimum of radiator area. The only assumptions necesary are that the power source obeys the laws of thermodynamics and that the starship may only get rid of waste heat by radiating.
Let us assume that we have a heat engine as a power source with a relative efficiency of η (its absolute efficiency is η times the Carnot efficiency ε). We can write the available free power, F, as:
F = Qηε = Qη(1 - T1/T2)where Q is the rate of heat flow into the exchanger and T1 and T2 are the temperatures of the cold and hot sides of the engine, respectively. The waste heat, H, released into the starship is Q - F, or:
H = Q(1 - η + ηT1/T2)For simplicity, we will measure temperature in units of T2 and let T1 be called just T. The amount of waste heat associated with a given free power is then (after dividing through by η):
H = F (η-1 - 1 + T) / (1 -T)Now this waste heat must be radiated away from the ship. The power radiated by a black body at temperature T and with area A is given by the Stephan-Boltzmann Law:
P = σAT4with σ a constant depending on the choice of units. Setting these equal to each other gives:
A = F (η-1 - 1 + T) / σ(T4 - T5)Now we can ask what value of T will give the minimum radiator area. Taking the derivative of A with respect to T and setting it equal to zero gives:
(T4 - T5) - (4T3 - 5T4)(η-1 - 1 + T) = 0Or, dividing by T3 and expanding:
T - T2 - 4η-1 + 4 - 4T + 5Tη-1 - 5T + 5T2 = 0After collecting terms, we have:
4T2 + (5η-1 - 8)T + 4(1 - η-1) = 0or, dividing through by 4:
T2 + (5/4η-1 - 2)T + (1 - η-1) = 0We write η-1 as γ then the solution to the above quadratic can be written:
T = 1 - 5/8γ + 1/8 sqrt(25γ2 - 16γ)In the special case where the exchanger runs at maximum theoretical efficiency, η = γ = 1 and the equation above gives T = 3/4. This means that the cold side of the heat engine is at 75% of the temperature of the hot side. Of course, this is not very efficient as a heat engine goes, but if the total waste heat is dropped by lowering the temperature of the cold side, the total radiator area must be increased! This is because of the T4 behavior of the radiation law - as the temperature of the radiator drops, it dumps heat much less efficiently. This function is fairly flat - the value of T only goes from 0.75 to 0.80 as the relative efficiency goes from 1 to 0.1.
From John Gwinner
Take a classic space opera warship. Onboard power is generated by one or more fusion reactors. If the overall power is 2 gigawatts, and the efficiency is 90% (a pretty generous estimate, since projections I've seen for MHD power generation are around 60%) then at full power, the reactors create 200MW of waste heat. At these sorts of power levels the waste heat of the crew, computers, coffee makers, etc. can be ignored. If there are energy weapons, assume they too are 90% efficient and use 500MW of power when fired, generating another 50MW waste heat. Lump in a lot of other minor systems and you get something like 300MW total waste heat that has to be gotten rid of at peak.
Where it gets complex, AFAIK, is the question of how hot you can allow the ship's interior to get. Let's assume that the environmental areas have heat pumps that allow them to stay a fair amount cooler than the engineering areas (since they're not generating the majority of the heat to begin with) and there are no low-temperature superconductors and so forth to worry about. If the engines and weapons can operate happily at 150 degrees Celsius, that's 423 degrees Kelvin. So that's our starting point -- the coolant (probably liquid sodium or lithium at that temp) gets that hot before it's pumped through the radiators to cool off again, at which point:
Heat lost [watts] = 5.67e-8[Stefan's Constant] * area [m^2] * emissivity * T^4 [degrees Kelvin].
(ed note: same equation as above)
If the radiators are perfectly black (emissivity of 1), and the coolant temperature is 423 degrees K, then in order to radiate away 275MW of heat, the radiator needs be about 150,000 square meters in area (of course it's double-sided, so the actual fin(s) only need to be 75,000 m^ 2). That's a square 275 meters on a side, or roughly a large city block, simply to deal with the ship's own waste heat at full power. If the ship needs to radiate away additional heat due to taking in, say, 400MW of energy from an enemy ship's lasers, you'd probably have to double or triple that figure (and make darn sure to keep your fins edge-on to the enemy ship firing at you! :). Of course all this is very crude and assumes perfect efficiency of a number of things (some of which I'm probably unaware of :). In reality you might get 80% of that theoretical performance. Or perhaps less. And the first thing damaged in a battle would probably be the radiators (big, hard to protect).
The structural mass of a large radiator fin could be a substantial fraction of the entire ship's mass, and that slows down the acceleration of the ship, which needs more power for thrust, which gives off more waste heat, and so on... So the idea of using spray wands and droplet coolants is attractive.
OTOH, if you need to keep the whole ship at a comfy temperature like 20C, then it's almost hopeless. The radiating area required is so enormous that high acceleration isn't practical at all (something like half a million m^2).
Another alternative is to design the ship to only radiate away normal, routine power levels, and to boil off propellant to deal with peak loads. But that goes through a lot of propellant pretty fast at high power levels. Dreadnaughts become like modern jet fighters -- only good for a few minutes of intense combat before the fuel runs out. Once it's gone, you can't crash, but you have to surrender or be boiled...
(ed note: He is assuming that the radiator temperature is 423K. Some of the other estimates were for radiator temperatures of 1600K to 3000K, which would drastically lower the radiator surface area.)
From 2001 A Space Odyssey by Sir Arthur C. Clarke (1969):
The spherical pressure hull formed the head of a flimsy, arrow-shaped structure more than a hundred yards long. Discovery, like all vehicles intended for deep space penetration, was too fragile and unstreamlined ever to enter an atmosphere, or to defy the full gravitational field of any planet. She had been assembled in orbit around the Earth, tested on a translunar maiden flight, and finally checked out in orbit above the Moon.
She was a creature of pure space - and she looked it. Immediately behind the pressure hull was grouped a cluster of four large liquid hydrogen tanks - and beyond them, forming a long, slender V, were the radiating fins that dissipated the waste heat of the nuclear reactor. Veined with a delicate tracery of pipes for the cooling fluid, they looked like the wings of some vast dragonfly, and from certain angles gave Discovery a fleeting resemblance to an old-time sailing ship,
At the very end of the V, three hundred feet from the crew-compartment, was the shielded inferno of the reactor, and the complex of focusing electrodes through which emerged the incandescent star-stuff of the plasma drive. This had done its work weeks ago, forcing Discovery out of her parking orbit round the Moon. Now the reactor was merely ticking over as it generated electrical power for the ship's services, and the great radiating fins, that would glow cherry red when Discovery was accelerating under maximum thrust, were dark and cool.
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Radiators are not the only thing spoiling the Polaris' sleek external lines. Roger Manning's radio needs large dish antennas. They might not be as large as the monsters on 2001's Discovery, but they won't be much smaller than the ones on the Apollo service module. It might also be a good idea to have landing radar on an outrigger or boom, so it won't be blinded by the exhaust. Both of these will be retracted during atmospheric re-entry, with the landing radar deployed when the air speed drops low enough so it won't be ripped off. |
A morbid but necessary fixture that nobody talks about will be the "C-Chute" (from the Isaac Asimov story with the same name). "C" is short for "Casualty". A dead body will quickly contaminate the air of the lifesystem, so there has to be a way to jettison the dear departed. Also of concern is the effect on crew morale. Personnel will be prone to morbid thoughts while their crewmate(s) mortal remains are lying in the next cabin. There will probably be a tradition of laying the dead to rest within twenty-four hours of death.
It will be important to have an already established protocol for laying the dead to rest. In the movie Conquest of Space they did not have such an established protocol, and the results were ugly. During an EVA astronaut Andre Fodor is killed by a meteor. Not knowing what to do, they leave the body out there still on the safety line.
Big mistake.
You can see the surviving crew start to freak out as they try to ignore their dead friend floating outside the porthole. Finally one of them cracks and starts to scream at the body. That's when the captain suddenly wakes up to the vital necessity of laying to rest the dear departed. Say a few words, and push the body off into space. Don't bother trying to push it into collision course with the Sun, it takes far too much delta V and if the course is only a tiny bit off the body will just sling-shot around and head off to the Oort cloud.
Somebody suggested using the spacecraft's rocket exhaust to cremate the body. Tuyu explains why this is not a good idea: EWWW! Can you say, "partially-burned semi-intact corpse flying off into the depths of space"? Unless you tether it, of course. Then you need to imagine a hot dog on a wire in the flame of a jet's afterburner. While ignoring the little flaming bits flying off in the jetwash.
Before you start designing the rest of the spacecraft, you should decide what sort of functions it will have to perform. Is it a cargo vessel? A tanker? A warship? The outer space version of a Coast Guard vessel? A blockade runner? This will give a rough idea of what most of the payload mass will be (cargo holds, remarkably large mass ratio, weapons, deep space rescue gear, stealth technology and oversized engines).
Speaking of cargo, present-day cargo ships are rated in "Net register tonnage", where each "ton" actually indicates 100 cubic feet of volume (2.83 cubic meters). The average cargo they carry has a density of 350 kg/m3. If the cargo has a wildly different density, some math will be needed, but for most cargo the net tonnage gives a good idea of the ship's cargo capacity. In practice, while filling the cargo hold it will either "mass-out" or "bulk-out", depending on which it runs out of first: lifting capacity or cargo space. In MANNA by Lee Correy (AKA G. Harry Stine) a surface-to-orbit shuttle bulked-out because it was carrying a cargo of fluffy non-dense cotton underwear. While the shuttle could have theoretically lifted more cargo mass, there wasn't any more room in the cargo hold.
In international shipping, a standard cargo container is 33 cubic meters and can have a maximum mass of 24 metric tons (2.2 tons of container and up to 21.8 tons of cargo). An extra large cargo container is 67.5 cubic meters with a max mass of 30.5 metric tons (3.8 ton container with up to 26.7 tons cargo). Thanks to Karl Hauber for pointing out an error in the the old figures posted here.
As you are beginning to discover, mass is limited on a spacecraft. Many Heinlein novels have passengers given strict limits on their combined body+luggage mass. Officials would look disapprovingly at the passenger's waistlines and wonder out loud how they can stand to carry around all that "penalty weight". There are quite a few scenes in various Heinlein novels of the agony of packing for a rocket flight, throwing away stuff left and right in a desperate attempt to get the mass of your luggage below your mass allowance.
From SPACE CADET by Robert Heinlein (1948)
Tex hauled out his luggage and hefted it. "It's a problem. I've got about fifty pounds here. Do you suppose if I rolled it up real small I could get it down to twenty pounds?"
"An interesting theory," Matt said. "Let's have a look at it -- you've got to eliminate thirty pounds of penalty-weight."
Jarman spread his stuff out on the floor. "Well," Matt said at once, "you don't need all those photographs." He pointed to a dozen large stereos, each weighing a pound or more.
Tex looked horrified. "Leave my harem behind?" He picked up one. "There is the sweetest redhead in the entire Rio Grande Valley." He picked up another. "And Smitty -- I couldn't get along without Smitty. She thinks I'm wonderful."...
...Matt studied the pile. "You know what I'd suggest? Keep that harmonica -- I like harmonica music. Have those photos copied in micro. Feed the rest to the cat."
"That's easy for you to say."
"I've got the same problem." He went to his room. The class had the day free, for the purpose of getting ready to leave Earth. Matt spread his possessions out to look them over. His civilian clothes he would ship home, of course, and his telephone as well, since it was limited by its short range to the neighborhood of an earth-side relay office...
..He called home, spoke with his parents and kid brother, and then put the telephone with things to be shipped. He was scratching his head over what remained when Burke came in. He grinned. "Trying to swallow your penalty- weight?"
"I'll figure it out."
"You don't have to leave that junk behind, you know."
"Huh?"
"Ship it up to Terra Station, rent a locker, and store it. Then, when you go on liberty to the Station, you can bring back what you want. Sneak it aboard, if it's that sort of thing." Matt made no comment; Burke went on, "What's the matter, Galahad? Shocked at the notion of running contraband?"
"No. But I don't have a locker at Terra Station."
"Well, if you're too cheap to rent one, you can ship the stuff to mine. You scratch me and I'll scratch you."
"No, thanks." He thought about expressing some things to the Terra Station post office, then discarded the idea -- the rates were too high. He went on sorting. He would keep his camera, but his micro kit would have to go, and his chessmen. Presently he had cut the list to what he hoped was twenty pounds; he took the stuff away to weigh it.
From BETWEEN PLANETS by Robert Heinlein (1951)
Long as he had been earthbound he approached packing with a true spaceman's spirit. He knew that his passage would entitle him to only fifty pounds of free lift; he started discarding right and left. Shortly he had two piles, a very small one on his own bed -- indispensable clothing, a few capsules of microfilm, his slide rule, a stylus, and a vreetha, a flutelike Martian instrument which he had not played in a long time as his schoolmates had objected. On his roommate's bed was a much larger pile of discards.
He picked up the vreetha, tried a couple of runs, and put it on the larger pile. Taking a Martian product to Mars was coal to Newcastle.
 From The Queen of Outer Space
Sorry ladies, I said No Skirts!
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Rocketeers would tend to be short and wiry. In Sir Arthur C. Clarke's classic THE OTHER SIDE OF THE SKY, space station construction crews got a pay bonus if they kept their weight below 150 pounds. Note that this would also be a good argument for rocketeers being [a] Oriental, [b] Female, or [c] Both. |
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Other innovations are possible. Perhaps boxes of food where the boxes are edible as well. The corridor floors will probably be metal gratings to save mass (This is the second reason why female cadet shipboard uniforms will not have skirts. The first reason is the impossibility of keeping a skirt in a modest position while in free-fall). In Frank Herbert's DUNE, spacemen had books the size of a thumb-tip, with a tiny magnifying glass. Keep in mind that every gram of equipment or supplies takes several grams of propellant. Try to make every gram do double duty. |
 Edible box made of soy biscuit. Artwork by Jerry Robinson
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With regards to low mass floors, the lady known as Akima had an interesting idea:
Unless the deck is also a pressure bulkhead, how about omitting deck plates and beams entirely, and making the floor a metal-mesh version of the trampoline decks used on sailing catamarans? That way, "weights" bearing on the decks would be transmitted into the tubular structure of the hull as an inward tension.
David Chiasson expands upon Akima's idea. There is an outfit called Metal Textiles which produces knitted wire mesh.
The meshes are knitted, as opposed to woven like a screen door. They are manufactured in densities (% metal by volume) from 10% to 70%. There are a wide variety of materials that the mesh can be made from, including aluminum, steels, Teflon, Nylon, even tungsten. Unfortunately, titanium is not on that list, I can only suspect that it must be difficult to get into a wire form suitable for making a knitted mesh.
Direct quote from site's main page: "In compressed form, knitted metal can handle shock loadings up to the yield strength of the material itself. The load may be applied from any direction-up, down or in from all sides."
I can speculate that with some kind of structural forming breakthrough, the mesh could be heated over a (ceramic?) mold to a near-melting point and simply pressed into place, compressing the mesh into a solid.
Michael Garrels begs to differ:
I need to point out some issues with the idea of mesh floors.
First off there's the idea that bulkheads have to be bulky. In nautical settings, bulkheads have to be bulky to withstand the large pressure of water, to mount things like hatches on, and to provide overall rigidity to the ship during turning and impact. Most partitions in a spaceship would be a thin pressure membrane sandwiched between a mesh to avoid punctures. The skin on the Apollo lander module was thinner than common aluminum foil. If all you're trying to do is partition, pull up pictures of Skylab - you'll see curtains and isogrid all over the place.
Next is your distinction between floors and walls. Unless there is spin or thrust, there will be no such distinction.
Which brings us to the most important point - the floor that you're currently standing on isn't made out of mesh for a reason. Remember that classic description of a gravity well with a weight on a rubber sheet? Many building codes don't limit the weight allowed on floors but instead the amount of deflection allowed. Floors have to be bulky with occasional beams - otherwise you'll never be able to wheel a torpedo or a gurney, and debris will roll toward where you're standing. It might work in a hallway, or as on your boat for stowage of light items, but not for spans more than a couple meters at 1 g using real materials - especially if you want to mount something like a chair and a console in the center of the cabin.
From DUNE by Frank Herbert.
"If it's economically feasible," Yueh said. "Arrakis has many costly perils." He smoothed his drooping mustache. "Your father will be here soon. Before I go, I've a gift for you, something I came across in packing." He put an object on the table between them-black, oblong, no larger than the end of Paul's thumb.
Paul looked at it. Yueh noted how the boy did not reach for it, and thought: How cautious he is.
"It's a very old Orange Catholic Bible made for space travelers. Not a filmbook, but actually printed on filament paper. It has its own magnifier and electrostatic charge system." He picked it up, demonstrated. "The book is held closed by the charge, which forces against spring-locked covers. You press the edge-thus, and the pages you've selected repel each other and the book opens."
"It's so small."
"But it has eighteen hundred pages. You press the edge-thus, and so . . . and the charge moves ahead one page at a time as you read. Never touch the actual pages with your fingers. The filament tissue is too delicate." He closed the book, handed it t o Paul. "Try it."
