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 xray 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 DeuteriumDeuterium fusion for energy production, these drives produce huge amounts of waste power in the form of neutrons and Xrays (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 xrays, 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 xrays 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 * r^{2 }))
Where:
p=Incident power through surface A (W)
P=Source Power (W)
A=Area of the shield (m^{2})
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 = 700m^{2}. Using these factors, the power radiated into the shield is:
15 000 GW x 700m^{2} / 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.
Radiator Area
The next step is to determine the area required to radiate out the power. The equation for this is the StefanBoltzmann equation:
(3) Q = ε * B * A(T_{r}^{4} – T_{i}^{4})
Where:
Q =Power (W)
ε = Emissivity
A = Area (m^{2)}
B = StefanBoltzmann constant
= 5.67 x 10^{8} J/sm^{2}K^{4}
T_{r} = Radiator surface temperature (K)
T_{i} = Temperature of space around the radiator (K)
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 carboncarbon composite, such as the type used on the Space Shuttle’s wings[v] with an emissivity of 0.9, a density of 1800 kg/m^{3}, 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:
(100,000,000,000W) / (0.9 * 5.67×10^{8} J / sm^{2}K^{4} * 2200^{4}K) = 84,000m^{2}
Since the radiator has two sides, the area will be 42,000 m^{2}, 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 
Carboncarbon 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. If the radiators are 50 away, this adds another 100m to the path and 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 = 1,625 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 250 m away from the shield, rather than 50m, 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 m^{3}/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 m^{3}/s / 20 m/s x 2 = 3.8 m^{2}
Let’s divide this into 100 pipe loops, coming and going in the radiator. 3.8 / (100×2) = 0.02 m^{2} 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 m^{2} of pipe wall.
We can add another 34,000 m^{2} for the fin area between the pipes, for a total area of 54,000 m^{2}. The final step is to take this 54,000 m^{2} 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 m^{2} x 0.005m = 270 m^{3}. We multiply this result by the density of the carboncarbon composite, 1800 kg/m^{2}, 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 carboncarbon 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.
[i] Project Daedalus, p.65
[ii] 200 000 000 x 100 hp x 745 W/hp = 14 000 000 000 000 Watts
[iii] ‘Yes Virginia, there is a Santa Claus’, a famous 1897 editorial
[iv] 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.
[v] Wikipedia, reinforced carboncarbon