Starship Radiators

posted by Michel Lamontagne on February 2, 2014

Looking at the Daedalus starship, it might appear at first glance that the ship has no radiators.  So why have these cumbersome appendages suddenly reappeared on the Icarus starships?  Are they absolutely required?  And if so, how do we design them, and how much of a mass penalty will they add to the ship?

 

Figure 1

Figure 1. Why do we need this?

 

Heat Sources

 

The Daedalus starship’s thermal control requirements depended on the assumption that the neutrons produced by side fusion reactions were absorbed directly in the fuel pellets[i].  It was also assumed that x-ray production was low. Therefore the radiation power was quite low, compared to the power produced by the fusion reaction (a fraction of a percent) and the ship’s reaction chamber was sufficient to act as a radiator. (That’s where the radiator was hidden, in plain sight!)  The infrared radiation emitted by the reaction chamber was deflected by a simple mirror, and therefore little work was required to keep the tanks cool, and no particular shielding was required to protect the payload from the radiation coming from the drive.

The situation for the designs that have emerged from the Icarus workshop sessions is dramatically different. Using Deuterium-Deuterium fusion for energy production, these drives produce huge amounts of waste power in the form of neutrons and X-rays (up to 60% of the reaction’s power) radiating out from the reaction area in all directions. How much total waste power are we talking about?  For a large probe with about the same payload capacity as Daedalus the waste power from the fusion reaction can be 15,000 GW or more. To put this number in perspective, it’s about the power of 200 million cars running at full throttle[ii].

The portion of the radiation that streams forwards, striking the tanks and the payload bay, will boil away the fuel and fry the payload in no time.  So some form of shielding is required between the drive and the fuel tanks and payload. It would be great if there was a substance that could bounce away neutrons and x-rays, chase away gamma rays and protect the equipment from radiation.  However, in the spirit of the season, but in a negative way: ‘No, Virginia, there is no neutron mirror’[iii].

Fortunately, there are x-rays absorbers, such as tungsten, and neutron absorbers, such hydrogen and other light elements. These absorbers stop the radiation, but by doing so they turn its power into heat – lots and lots of heat.  So much heat, in fact, that the shield’s temperature will quickly rise over its melting point, and it will be destroyed.  So now we need something to protect the shield.  The way to do this is to move the heat out of the shield as quickly as it is added by the radiation. and that’s what the radiators are for.  The radiators function to take the extremely concentrated heat present in the shields, spread it out over a larger area, and radiate it away into space. 

There are many possible shapes and modes of operations for radiators in space, but the classical solution is a large rectangular structure made from pipes, built at right angle from the radiation flow, with a circulating coolant. (See Figure 2.)

 

Figure 2

Figure 2. The discovery II, NASA report TM-2005-213559

 

Radiator Capacity

In order to size these radiators, the first number we need to know is: how much power do we have to radiate out?  Obviously, we want to intercept as little of the waste power from the drive as possible, and let as much of it as possible escape into space.  We thus want as small a shield as possible, and this leads to the very narrow and very long designs we see in the Icarus workshop (or Figure 2), rather than the short and stumpy Daedalus design. 

 

To determine the absorbed power we can use

Equation #1 (the inverse square law) that allows us to find the radiation flow through a certain area, when we know the power of the source.

(1) p = ((P * A) / (4 * π * r2 )) 

Where:

p=Incident power through surface A (W)

P=Source Power (W)

A=Area of the shield (m2)

r=radius(m)

 

From this formula, we see that the incident power to the shield goes down as ‘r’ gets longer and up as the shield gets bigger.  So we will want to position the shield as far as possible from the radiation source and make it as small as possible[iv]. The designs from the Icarus workshop all took this approach and arrived at similar dimensions.

Let’s suppose 100m distance and a 30m diameter shield, with (15m)2 * π = 700m2. Using these factors, the power radiated into the shield is:

15 000 GW x 700m2 / 4* π * (100m)2 =  83 GW

To put this number into perspective, it’s about the waste heat from 60 standard nuclear reactors.  Or if you prefer, we’re down to about 1,000,000 cars.  It would be a good idea to add some extra capacity for backup and for cooling structural elements that are between the shield and the drive, so let’s choose 100 GW as our radiator power.

 

Coolant

The next step is to find a fluid that will be able to carry the heat from the shield to the radiator.  The equation for this process is the following:

(2) Q = mf * Cp (Tin – Tout)

Where:

Q = Power (W)

mf = mass flow (kg/s)

Cp = Specific heat (kJ/kgK)

Tin = Fluid temperature as it enters the equipment it is cooling or heating (K)

Tout = Fluid temperature as it leaves (K)

 

We want a fluid with the lowest density and the highest specific heat.  We also want a fluid that boils at the highest temperature possible, so the radiator can be as hot as possible.  Here is a table of possible fluids.

 

Table 1, coolant fluids

Material

 

Aluminum

Beryllium

Lithium

FLiBe

Helium

Hydrogen

Deuterium

Water

Melting

 

933

1600

453

733

 

 

 

 

Boiling

 

2773

2700

1573

1703

 

 

 

 

Power

kW

100 000 000

Heat capacity

kJ/kgK

0,91

1,92

4,3

2,4

5,2

14

5,2

5,5

Supply temp

K

2600

2600

1500

1600

3000

2400

2400

2200

return temp

K

2000

1800

800

1000

2600

1800

1800

1800

Temperture difference

K

600

800

700

600

400

600

600

400

Average temp

K

2300

2200

1150

1300

2800

2100

2100

2000

Flow

kg/s

183 150

65 104

33 223

69 444

48 077

11 905

32 051

45 455

 

Tonne/s

183

65

33

69

48

12

32

45

Density

kg/m3

2700

1700

500

2300

3

1

2

54

Flow

M3/s

67,8

38,3

66

30,2

16 025

11 904

16 025

841,75

 

M3/hr

244 200

137 868

239 203

108 696

57×10^6

42×10^6

57×10^6

3×10^6

 

The liquid metals seem promising.  FLiBe, a salt used in nuclear reactors, a little less.  The gases are also interesting, in particular deuterium, which happens to be our fuel, suggesting an interesting possibility of double usage.  Water is also very interesting.  It must be noted though that the water in this table is still liquid at 2000K.  To achieve this, it has to be under very high pressure, and this will require much stronger piping.

Let’s choose Beryllium, as a compromise between the heavier Aluminum and the cooler Lithium.  We won’t use the gases in our radiators because, although they are very good heat carriers, they are not very good heat conductors.  On average, they are roughly 100 times less heat conductive than liquid metals.  So even though we might be able to have gases at 2000K and more, the radiator surface will be much cooler, easily 500K less, due to the difficulty the heat will have in moving from the gas to the wall or the radiator. For liquid metals, on the other hand, the temperature difference between the fluid and the radiator wall will be less than 20K.

Using equation 2, we can determine the flow of beryllium required to cool the shield:

100,000,000,000 W / 1.92 kJ/kgK * (2600 – 1800) =  65,000 kg/s, or 65 tonnes per second.

 

Radiator Area

The next step is to determine the area required to radiate out the power.  The equation for this is the Stefan-Boltzmann equation:

(3) Q = ε * β * A(Tr4 – Ti4)

Where:

Q =Power (W)

ε  = Emissivity

A = Area (m2)

B = Stefan-Boltzmann constant

    = 5.67 x 10-8 J/sm2K4

Tr = Radiator surface temperature (K)

Ti = Temperature of space around the radiator (K)

 

Figure 3

Figure 3. Temperature Scale

The second important factor is ε, the emissivity.  This is a factor of how well the radiator surface can emit heat.  As a rule of thumb, the darker the material the higher the emissivity will be.  Inversely, a very shiny surface will have low emissivity.

So the radiator must be as hot as possible and have high emissivity.  An excellent material combining these properties is carbon-carbon composite, such as the type used on the Space Shuttle’s wings[v] with an emissivity of 0.9, a density of 1800 kg/m3, a yield strength of 700 MPa and a working temperature of about 2200 K, the coolant average temperature from Table 1.  Let’s use it. 

Plugging the numbers in Equation #3:

 (1000000000W) / (0.9 * 5.67-8 J / sm2K4 * 22004K) = 84000m2

Since the radiator has two sides, the area will be 42,000 m2, or about 200m x 200m.

 

Table 2, Emissivity

Material

Emissivity

Aluminum Foil

0.04

Aluminum Anodized

0.77

Aluminium, molten

0.1

Beryllium

0.18

Beryllium, Anodized

0.9

Beryllium, molten

0.61

Black Epoxy Paint

0.89

Carbon-carbon composites

0.85 – 0.9

Glass smooth

0.92 – 0.94

Ice smooth

0.966

Inconel X Oxidized

0.71

Lithium, molten

0.035 – 0.05

Mercury liquid

0.1

Mild Steel

0.20 – 0.32

Molybdenum polished

0.05 – 0.18

Nickel, oxidized

0.59 – 0.86

Silicon Carbide

0.83 – 0.96

Stainless Steel, type 301

0.54 – 0.63

Titanium polished

0.19

Tungsten polished

0.04

Tungsten aged filament

0.032 – 0.35

Zirconium, molten

0.32

 

 

Coolant Velocity

A final number we need to think about is the velocity of the coolant in the pipes.  The pump affinity laws state that the pumping head of a fluid through a pipe rises to the square of the velocity, and that the pumping power rises to the cube of the velocity.  High velocity also causes turbulence, which can in turn cause vibration and wear.  So it’s best to keep the velocity low.  However, low velocities mean that a lot more coolant will be required, and that the mass of the ship will go up in proportion. 

The calculations required to establish the velocity are beyond the scope of this essay, but we will use 20m/s.  This is a high value – about 6 times the usual industry standard – but since mass is so important on a starship, we can afford to use bigger pumps.

 

Radiator Mass

With the above, we can finally find the mass of our cooling system.

First, let’s find the mass of the coolant.  Since the radiators are 200 m long and 200 m wide, the average distance the coolant has to cover to go all the way to the end of the radiator and return is 400m.  Adding another 100m to go from the shield to the radiator, the distance covered is 500m.  At 20 m/s it will take 25 seconds for the coolant to do a whole cycle.  So the coolant mass is the flow rate times the trip time:

65 tonnes/s x 25 seconds = 1625 tons.

Note that it is important to put the radiators as close as possible to the shield, due to the limited velocities possible for the coolant.  For example, if the radiator is 200 m away from the shield, the coolant will have to travel 400 m more distance.  This adds 20 seconds to the cycle time and therefore nearly 1300 tonnes to the ship’s mass!

Now let’s find the mass of the radiator panels themselves.  From Table 2, we see that the volume flow rate of the beryllium coolant is 38 m3/s.  Since the velocity is 20 m/s, and since the coolant has to both enter and leave the radiator, the total pipe section area is:

38 m3/s / 20 m/s x 2 = 3.8 m2

Let’s divide this into 100 pipe loops, coming and going in the radiator.  3.8 / (100×2) = 0.02 m2 per pipe, giving a diameter of 0.15m per pipe and a circumference of about 0.5m per pipe.  Multiplying this by the number and length of the pipes, we find:

0.5m x 200m x 200 pipes = 20,000 m2 of pipe wall. 

We can add another 34,000 m2 for the fin area between the pipes, for a total area of 54,000 m2. The final step is to take this 54,000 m2 and multiply it by the pipe wall thickness.  Rigorous pipe wall calculations are beyond the scope of this essay, but a thickness of 5mm is reasonable.  54,000 m2 x 0.005m = 270 m3.  We multiply this result by the density of the carbon-carbon composite, 1800 kg/m2, to get the final mass: 270 x 1800 = 486,000 kg, or about 500 tonnes.

Adding all this up, the final mass of the cooling system, excluding the coolant pumps and the shield itself, as well as any spares and reserve coolant, is about 2100 tonnes. That adds a lot of mass to the designs, though thankfully still within the bounds of practicality. No wonder Daedalus got rid of them!

 

Advanced Radiator Designs

Can we do better than this?  Perhaps, even though radiators glowing bright orange are already pretty far beyond what we can do today.  An alternative raised by several designs in the Icarus Workshop would be to do away with the radiator walls altogether, and use liquid droplet radiators or fountains.  Although interesting, this solution is handicapped by the low emissivity of liquid metals.  This lower emissivity leads to larger radiators, and longer cycle times for the coolant, increasing the coolant mass, and possibly offsetting the gains form the removal of the radiator walls.  It may also prove impossible to pump the very low pressure liquid metals used in liquid droplet radiators up to the velocities required to keep the mass down, since the pump action may cause cavitation due to a lack of suction head.

Another possibility is to dope lithium with nanoparticles of boron.  This might produce a coolant with properties half way between these materials, with the low density and high specific heat of lithium, and the high boiling point of boron.  We can also imagine using the lithium under high pressure, rather like high pressure water nuclear reactors, to increase its boiling point and run it at higher temperatures, reducing the weight of coolant considerably.  One of the limitations of lithium though is that it is highly corrosive to many materials, including carbon-carbon composites, so its use would require some kind of interior cladding for the pipes, or a thinner, high density corrosion resistant metal radiator, such as tungsten.

The Icarus team is actively evaluating all of these concepts, and they may form the subject of future blog posts, so stay tuned.

Figure 4

Figure 4. large radiators on a project Icarus type Starship

 
 

[1] Project Daedalus, p.65

[2] 200 000 000 x 100 hp x 745 W/hp = 14 000 000 000 000  Watts

[3] ‘Yes Virginia, there is a Santa Claus’, a famous 1897 editorial

[4] Footnote readers will have noticed that if you increase r a lot, the incident power goes down, eventually to practically 0.  This leads to the various tethers designs for starships (e.g. – Valkyrie).  The design of the tether, and of ways of preventing the radiation from destroying the tether, and of keeping tether weight below the radiator weight, might be the subject of another blog post, if there is any demand.

[5] Wikipedia, reinforced carbon-carbon


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35 Responses to Starship Radiators

  1. A very mass efficient design is to eliminate the pipes altogether and spray droplets of a low vapor pressure liquid into space, to then be collected and recirculated. This idea was worked out in considerable detail in the early 1980′s:

    http://arc.aiaa.org/doi/abs/10.2514/3.62557
    http://www.sciencedirect.com/science/article/pii/0094576585901304
    http://www.sciencedirect.com/science/article/pii/0094576582900844
    http://arc.aiaa.org/doi/abs/10.2514/3.25077

    Studies showed these systems to have one tenth the mass of the equivalent heat pipe radiators.

    • Michel Lamontagne says:

      Thank you for the links. This idea was raised in the conclusion of the post, and it’s possible we may decide in favor of droplet radiators as the design progresses. One difficulty I see is the low emissivity of liquid metals, that increases the weight. One tenth of the mass seems very optimistic though, since most of the mass at these temperatures is the coolant itself,and I don’t see how liquid droplet radiators could reduce the coolant weight.

  2. Antonio says:

    Hi,

    in Equation 1, the last “p” should be absent.

  3. Antonio says:

    Very nice article. Will there be similar articles about other parts of the starship?

    • Michel Lamontagne says:

      I don’t know about the others, but I’m working on ‘A plumbers guide to stardrives’ about pumps and pipes, and another post about deuterium production.

  4. Petr says:

    Using deuterium both for propulsion and radiative cooling would seem very efficient, if the low heat conductivity could be solved.
    There are interesting gas heat exchangers developed by REL, ltd for Skylon. They are meant to operate in atmosphere and with lower temperatures than above discussed, but it is surely worth a look…

    http://www.reactionengines.co.uk/heatex_rel.html

    • Michel Lamontagne says:

      This looks like a great design to use for the radiation shield, where the coolant will be moving through the radiation absorbing material at high pressure and high velocity. This is also encouraging regarding the radiators themselves, that will be rather like this exchanger, but rolled out flat, and painted high emissivity black. One challenge will be the circuit and valve design to isolate tubing in case of leaks. I hope the use of the maintenance robots for repairs, rather than depending on robust radiators to stand up to years of wear and impacts will allow us to reduce the weight of the system.

  5. Gerry says:

    For an interstellar vessel high exhaust velocity is paramount. But a design like this could be adapted for travel within our solar-system. The radiator fins could be replaced by a system that ejects the waste heat through a stream of rocket propellant; perhaps hydrogen or even water. This would entail carrying a greater mass of propellant, and would afford a higher acceleration in exchange for a lower exhaust-velocity, which would probably work pretty well for quick travel around our Solar System. And an interplanetary-scale economy is often considered to be one of the necessary conditions for the construction of large, high-power interstellar vehicles.

    • Michel Lamontagne says:

      Yes, the Icarus is rather overpowered for travel in the solar system. However this will mean it can get around in the target system with very little fuel, perhaps allowing us do do away with most of the sub probes (except landers, of course!)
      I agree water is an excellent add-in for a fusion drive, reducing power for more optimised ISP. However, there will not be enough water to compensate for the heat produced by the drive, since we will still be using a few kg per second. The water itself will dissociate and turn into plasma, and the neutrons will still be there, streaming through everything. We might be able to have much smaller radiators, though, and perhaps even get away with a ‘hot tank of gas’ type of shield between the drive and the more sensitive parts of the ship.

      • Gerry says:

        Michel, thanks for your response. What kind of range of performance could this drive produce in combination with a diluent (water, etc.)? I suppose it has to do with the heat capacity of the “diluting” propellant, and the resulting mass flow and velocity. Could the ISP and thrust be adjusted by, say, a factor of ten by adding more or less propellant to the fusion plasma, allowing the same drive system to function both in a “low” gear and a “high” gear mode, similar in principle to a VASIMR, but pulsed of course? That would be quite a versatile vehicle!

        • Hi Gerry; the Icarus is not well designed for travel in the solar system. It’s designed to fire it’s drive for 15 years, and the requirements for the solar system are more in the range of 15 days. So let’s say you want a 1500 ton ship, with 500 tons of cargo. And you have plenty of water, for a mass ratio of 4 and a total ship weight of 6000 tons. A 27 GW drive (about 1:500 the power of Icarus), burning a few grams per second of deuterium, and adding about 2,5 kg/s of water, will give you an ISP of 14 000 seconds, and deltaV of 200 km/s which gets you to Mars or Venus and back in a few weeks, or Jupiter in a month or so.
          You will need radiators to cool the radiation shield and the superconducting nozzle. You will also need something like a 50 MW nuclear reactor, to get the drive stated and run pumps and other stuff.
          Here is a link to a less powerful drive, not even fusion level,but it uses exactly the mixing method you’re thinking about. The description is near the end of the paper. Water gives an interesting purple glow to the drive plume…. http://bit.ly/1fQoLac

          • Gerry says:

            Michel, thanks for taking the time to provide such a thorough reply. I appreciate it very much. The “real world” physics and engineering considerations of space travel are quite a fascinating area, as I’m sure you’d agree…

          • Devon says:

            Hi Michel,

            I’m very late to this thread, but I was wondering if you’d be so kind as to explain how to calculate Isp and thrust for the case of injecting propellant (water, H2, whatever) into the exhaust flow.

          • Hi Devon,

            It’s fairly simple. 1) Ve= Isp*9,81
            2) P=1/2*mr*Ve^2
            where P is power, Ve is ejection velocity and mr is the mass flow rate.
            P is the power from the fusion reaction.

            Let’s suppose you know that your pure fusion reaction has an ISP of 1 million seconds (very good, you might want to be a bit less optimistic, like 500 000 s). You can use the above equations to find the power from a mass flow rate of fuel x (you will only need a tiny mass flows to get huge power numbers).

            This gives you your fusion power. You then substitute the mass flow rate you want to use, (that’s your added mass of water, or lithium, or whatever) and find the new ISP for the same power. From the ISP you can determine your thrust using: F=2P/Ve

            You need ta take a bit of care with the units but it should be fairly easy to do.

            Regards,

            Michel Lamontagne

          • Devon says:

            Hi Michel,

            Thanks for the reply. You’re right, that’s quite simple! I’m a little surprised, I thought it would also depend on things like molecular mass and temperature of the injected propellant. (Really, it makes no difference if you inject superheated H2 or an equal mass of cold lead? Weird.)

            Ah…is this because the power of the fusion reaction is so great that the propellant is approximated to have zero power of its own prior to injection?

          • Devon says:

            So, for example, it looks like according to that simple formula Gerry can achieve his 10x higher thrust by simply dumping 100x more mass into the exhaust flow?

          • Yes, the mass is the propellant, the energy comes from the fusion. But there’s a big caveat, it isn’t obvious how you will transfer the energy to the propellant. If you mix it too early, the fusion will fail; if you don’t mix it properly, you will lose some energy. Icf fusion using lithium jackets seems like the best bet these days. If you throw a block of lead at a fusion reaction, it won’t work ;-)

  6. CharlesJQuarra says:

    It seems the most interesting option regarding radiator was ignored. A graphene radiator sail of 10 square kilometers would weight approximately 100 kilograms. If we add some capillary veins to transport coolant we might increase the weight a factor of ten, but still almost a tiny fraction of what your other options imply. Also, graphene can stand 1000 Kelvin like it isn’t even a thing

  7. CharlesJQuarra says:

    It seems the most interesting option regarding radiator material was ignored. A graphene radiator sail of 10 square kilometers would weight approximately 100 kilograms. If we add some capillary veins to transport coolant we might increase the weight a factor of ten, but still almost a tiny fraction of what your other options imply. Also, graphene can stand 1000 Kelvin like it isn’t even a thing. Graphene stays robust well above 3000K

    • Michel Lamontagne says:

      Unfortunately for this clever idea, the thermal conduction of graphene is not high enough to compensate for the inherently bad thermal conduction characteristics of solids.The equation is Q=k*A/x*dt . With Q=100 GW, and k = 5300 for graphene, and x=about 2500m on average, you will find a gigantic cross section for the sail, many hundred m thick, in fact. Fins are ok in conductive spaces, not so good in space.

  8. Robert Myers says:

    I’ve always wondered about liquid metal cooling systems in a cold start situation. Does the metal have enough thermal conductivity to cool the reactor before the radiator system is fully thawed, or would you need double walled pipes to pump a lower melting material like potassium to preheat it?

    • Hi Robert,
      It’s a good question that will require careful thought if we go for liquid metals (‘I’m told the general favor is more for helium). My personal scenario goes like this: a small fission nuclear reactor is never shut off; it cools itself using just a small section of radiator. At start up, when we reach the target star, it passes a current through either the liquid metal itself, or through resistive windings in the radiator walls, to melt the coolant. It has to be sized to provide a bit more than the minimum energy that is required to keep the metal molten in the radiators. This will be simple for tin, that has a wide liquid phase, much harder for other metals with a narrow liquid range. We then start the second, larger reactor, the drive reactor. First, it loads up the superconducting magnets for the nozzle. Then it does the same melting work for the shield cooling radiators, bringing them up to about 1000K for aluminium, 1700 K for beryllium, etc. The main circulating pumps are started as well. Then we start the drive at part power. In less than 10 seconds, the whole circuit goes up to near the 1800 to 1900K average for regular operations. Then we push the drive to full throttle, the radiators glow bright red, or light orange, and we are off.

      • Andy P. says:

        You might consider going on the NASA TRS and finding the old SP-100 papers. They did a lot of work on starting up a liquid metal cooled fission reactor. These and the Russian papers on their space reactors *might* have some speculation on RESTARTABLE liquid metal cooled fission reactors.

  9. Antonio says:

    I’m wondering… will the final Icarus report be freely available? (I mean, the report with all the technical details about the ship.)

  10. Duncan Meyers says:

    Is there any way this abundance of heat could be turned into a form of electrical energy and utilized? Say for a secondary ion propulsion?

    • Hi Duncan; it does seem a waste, doesn’t it? But the heat is a side effect of a fusion process that creates a power gain in the order of 10-30 to 1. On the other hand, an ion drive, since there is no fusion involved, will always have a gain of less than 1:1, so it will always create extra waste heat. Another way of saying this is than any fuel used for the fusion drive will provide at least 20 times more thrust than fuel used for an ion drive. So there will never be any fuel available for an ion drive, it makes no sense to use it that way. The other way to use this heat would be to create electricity for some part of the fusion process. However, again, it will always be more efficient to use some form of direct energy production, such as induction from the drive stream, or direct conversion from the drive ions. The heat can be used as an alternative heat source for the auxiliary fission power drive, but only while the drive is working. So it can be used to save fission fuel for the auxiliary nuclear reactor. But the electricity from direct conversion can also be used this way, so with the drive working, the auxiliary reactor can be shut down, so there is no real savings. It really IS waste heat ;-)

      • Andy P. says:

        Umm…if you’re rejecting waste heat in space at 2200K…you should be able to apply parasitic thermophotovoltaics at least somewhere. There are papers out there which discuss using them for parasitic power (of course, in terrestrial power plants sans radiation and where they can be maintained) at temperatures much less than this. I think I shared this stuff with Dr. Long a few years back. If he doesn’t have it he knows how to contact me.

        • Hi Andy, At the very high power levels this thing is operating at, I expect the photovoltaics will be too heavy to be any use. Just the induction from the drive has to be 350 GigaWatts,basically induced by the drive flow in plain metal bars; that’s a whole lot of solar cells! The ships has way too much power, in fact; we just don’t need more, we need to find the best possible way to gain less!

        • An interesting thing we can do is turn the heat into mechanical energy. Put a turbine on the hot side, a compressor on the cold side, a shaft between the two and we have a pump with no electrical input. The sub-cooling shouldn’t be too bad, and the weight less than for an auxiliary reactor. We need to use helium of hydrogen for this to work though.

    • Oups, re reading myself, I see I made a mistake. Saving on the fission fuel is a net energy gain, since it saves us from inducing power away from the drive thrust. So some heat can be used to replace the auxiliary fusion reactor. About 500 MW, or about 0,5% of the loss, or 0,05% of the drive power. Better than nothing!

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