Starship Radiators

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?

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.)

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 = m f * C p (T in – T out )

Where:

Q = Power (W)

m f = mass flow (kg/s)

C p = Specific heat (kJ/kgK)

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

T ou t = 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 / (1,920 J/kgK * (2600 – 1800)) = 65,000 kg/s, or 65 tonnes per second.