A Plumber’s Guide to Starships- Part 4 – Materials in High Radiation Environments

A Plumber’s Guide to Starships- Part 4 – Materials in High Radiation Environments

Materials near a fusion reaction are subject to very high levels of both X-ray and Neutron radiation.  How do they stand up to this?

swelling
Figure 7 – Material swelling due to radiation damage

Even with a nominally “aneutronic” fusion reaction – and much more so with D-D fusion — there will necessarily be a large radiation flux coming from the fusion drive, mostly in the form of high energy neutrons and X-rays.  Radiation flux alters materials, but, surprisingly, not necessarily in a negative way.  And the heat from the absorbed radiation needs to be managed, since we do not want to have our ship melt in its first few minutes of operation.  In some cases, we will want our materials to be as transparent to radiation as possible, in other cases, such as in radiation shields, we will want the opposite, maximum opacity. Fortunately, there is quite a lot of experimental data on this subject, thanks to research on classical fission reactors.  

The Deuterium-Deuterium fusion reaction produces primarily ions, but also many fast neutrons in the 2.45 MeV range, along with some 14.1 MeV Neutrons from tritium recombination and a wide range of X-ray energies. At these levels of energy, most materials are at least partly transparent to both neutrons and X-rays.

For an Icarus drive with 35,000 GW of power, the total neutron power is about 15,000 GW, or 15,000 GJ per second, and the X-ray power is about half of that, about 7,000 GW.  Dividing the neutron power by the neutron energy gives us a neutron production rate of of about 4 x1025 neutrons per second. 

For the pulse propulsion drive, the explosion is a point source, and the radiation level goes down as an inverse square law of distance from the source:

Pf = P/4*pi*r2 (5)

Where:

Pf = power or neutron flux (watts/m2) or (neutrons/m2)

P = Power (watts) or Neutrons

r = distance from the source (m)

For the Z-pinch drive, the equation is a little more complex, since we have a line source rather than a point source:

Pf = P/(4*pi*r2 + l*2*r*pi)  

Where ‘l’ is the length of the line, the fusion area of the pinch.

However, the results converge at distance, as shown in the following table:

 

   

Z-pinch

   

Pulse

 

Distance (m)

Area

(m2)

Neutron Power

(MW/m2)

Fluence/

m2

Area

(m2)

Power (MW/m2)

Fluence/ m2

0

0

15,000,000

4.07E+25

0

15,000,000

4.07E+25

2

553

27,129

7.36E+22

50

298,416

8.10E+23

5

1,571

9,549

2.59E+22

314

47,746

1.30E+23

10

3,770

3,979

1.08E+22

1,257

11,937

3.24E+22

20

10,053

1,492

4.05E+21

5,027

2,984

8.10E+21

50

43,982

341

9.25E+20

31,416

477

1.30E+21

100

150,796

99

2.70E+20

125,664

119

3.24E+20

200

552,920

27

7.36E+19

502,655

30

8.10E+19

500

3,267,256

4.6

1.25E+19

3,141,593

4.8

1.30E+19

1000

12,817,698

1.2

3.17E+18

12,566,371

1.2

3.24E+18

1600

32,572,033

0.5

1.25E+18

32,169,909

0.5

1.27E+18

Table 2 – Radiation power flux at various distances

 

Note that since the x-ray power is about 50% of the neutron power, the x-ray flux will be about 50% of the neutron flux at the same distance for this particular ship design.  Other reactions would have different ratios.

For comparison, the neutron fluences over the life of fuel pins  through the zirconium cladding of fuel pellets in a typical fission reactor can reach 1×1021 n/cm2, or 1×1025 neutrons/m2(4).  That’s about the fluence  near the fusion reaction as seen in the above table, but for much slower neutrons and over a longer period.  So the radiation levels are somewhat comparable to one with which we are already familiar, and for which existing materials can provide adequate resistance and durability.

Despite the extreme radiation environment, it is interesting to note that the physical strength of materials near the fusion drive of a starship do not need to be intrinsically high.  The actual forces from the ship’s thrust are quite modest.  For Icarus, thrust is about 30 times lower than those from a Saturn 5 rocket, or about 100 tonnes versus 3000 tonnes. We want highly transparent materials, but they will not necessarily have to be all that strong.

 

Transparency

 

zpinch  explosion 

Figures 8 –  A Z-pinch drive (left) and a pellet explosion drive (right). For the Z-pinch, Deuterium gas fuel feed (blue) is being compressed into a z-pinch electromagnetically confined fusion plasma, purple. For the pellet drive, the laser driven explosion happens up to 100m behind the ship, and the explosion is magnetically confined. All material elements are subject to tremendous neutron flux.  The human figure is for scale only, because you really don’t want to be there when these things are operating.

The secret to radiation resistance is twofold: transparency and controlled defects. In his Firefly (Z-pinch) design, Robert Freeland demonstrated that the better approach to handling radiation from the drive isn’t to shield the drive and handle the resulting waste heat, but to instead allow as much of the radiation as possible to escape directly into space. For a Z-pinch or magnetic confinement type of drive, materials are required right next to the fusion reaction, to carry the required currents and create the magnetic fields (Figure 8 left).  For pellet explosion driven ships, the requirements are slightly less extreme since there are no physical components near the explosion (Figure 8 right), but the radiation flux will still be very high.  

Most materials are quite transparent to neutrons and X-rays, but not perfectly so.  If the material is transparent enough and thin enough, most high energy radiation will go right through. The behavior of this at the atomic level is complex, but at the macroscopic level, the simple equation for attenuation of high energy neutrons and X-rays is the Beer-Lambert law:

I = Io*e-(μ/ρen)ρl       (X-rays) (6a)

I = Io*e-(μ/ρ)ρl     (Neutrons) (6b)

Where:

I = intensity at depth ‘l’ into the material

Io = Original intensity

μ = attenuation coefficient (m2/kg)

ρ = density (kg/m3)

l = thickness into the material (m)

μ/ρen= mass energy attenuation coefficient (m2/kg)

 

The attenuation coefficient ‘μ’ is an experimental value that varies from material to material and according to the type of radiation.  The mass energy attenuation coefficient ‘μ/ρen’ is a form used to express the attenuation coefficient when it is modified to take into account the generation of secondary radiation in the material.  This coefficient has been tabulated by the NIST for X-rays.  For neutrons, a slightly less precise mass attenuation coefficient is available from various sources (5). These are summarized in the following table for a power level of 15 MeV:

Mass attenuation coefficient

Neutron

at 15 MeV

X-rays

at 15 MeV

 

μ/ρ (m2/kg)

μ/ρen(m2/kg)

Hydrogen

0.06580

0.00225

Lithium

0.01570

0.00114

Beryllium

0.01140

0.00123

Boron

0.00836

 

Carbon

0.00803

0.001959

Oxygen

0.00451

0.001483

Iron

0.00346

0.002108

Nickel

0.00355

0.002234

Copper

0.00347

0.002174

Tungsten

0.00160

0.003072

Lead

0.00150

0.004972

Bismuth

0.00147

0.003123

Uranium

0.00154

0.003259

Table 3 – Mass and mass energy attenuation coefficients (3)(4)(5)

 

From Table 3, we can see that X-rays rays are best stopped by dense materials since μ/ρ gets larger with denser materials, while neutrons are best stopped by light materials. In a very general way, X-rays interact with electrons, while neutrons interact with nuclei.  So X-ray opaque materials must have the maximum number of electrons, while the best neutron absorbers must have the most tightly packed nuclei.

X-ray Example

 

Beryllium is often recommended as a material with good X-ray transparency. Its density is 1848 kg/m3 (ρ). From the NIST tables, at an X-ray energy of 10 Mev, the mass energy attenuation coefficient is 0.014 cm2/g, or 0.0014 m2/kg (μ/ρ).

 

If we suppose a thickness of 1cm of beryllium, then,from equation (6a):

 

I/Io = e-(μ/ρen)ρl = e-0.0014*1848*0.01  = 0.98

 

At this thickness, the beryllium has absorbed 2% of the radiation intensity, allowing the remainder to escape. That’s pretty good, but how much energy is that 2% absorbed?  Well, if the beryllium pipe wall is 2m away from the Z-pinch, for example, the X-ray radiation power (from Table 2) is 0.5 x 27,000 MW per m2.  So 2% of that is about 280 MW per m2.  That’s still a lot of waste energy (heat) so it is very important to minimize the mass in the Z-pinch area.

Neutron Example

 

The neutron mass attenuation coefficient for high energy neutrons in Beryllium is 0.0157 m2/kg.  That’s almost ten times higher than for X-rays. For the same 1cm thickness and equation (6b):

 

I/Io = e-(μ/ρ)ρl= e-0.0157*1848*0.01  = 0.75; therefore the beryllium has absorbed 25% of the radiation intensity.  Very intense cooling will be required, even for this semi-transparent material.

However, it is important to note that this description doesn’t quite tell the whole story.  In fact, many of the neutrons are not absorbed at all but are re-emited as secondary neutron radiation in a process called neutron scattering.  So the actual energy absorbed is less than what the Beer-Lambert equation calculates. Consult your local nuclear physicist for details!

 

Controlled Defects

 

The use of controlled defects is the other important aspect of radiation resistance, since radiation can actually “cure” defects in certain types of materials (radiation hardening phenomena and radiation annealing effects). If we choose materials that can only be manufactured with a large number of defects in the first place, the basic properties of the material are known and the design can be done.  Radiation then interacts with these defects and, if the materials are well chosen, these interactions can actually reduce the influence of the defects, increasing, rather than decreasing, the strength of the materials.  This process is called radiation annealing.  With such materials, we can start with values that are perhaps 1 to 2% of the ultimate values, and expect these values to increase with time.  Fortunately, for strong materials with radiation annealing properties, such as carbon-carbon composites(2), 1% of 60,000 kPa (yield strength) is still 600 kPa — a very respectable number.

If we instead started from perfect materials, the radiation would introduce defects that break down this perfection, making the materials weaker.  The defects would be random, and might lead to catastrophic failure.  By controlling the flaws in our materials and in designing for them, we have controlled and reliable properties: reliable mid range materials, rather than unreliable high strength materials.

Materials with radiation annealing properties are identified in the tables in the references.

 

Radiation Shields

 

shield2

Figure 9 – Radiation shield at 50m from a fusion reaction

 

Radiation shields are at the top of the heat chain in the cooling system of a spaceship.  They are the hottest elements of the ship, except for the plasma in the fusion drive.  In practice, they are very similar to a fission reactor’s core, except that the radiation comes from the outside, from the fusion drive, rather than from the inside, from nuclear fuel.  Thus, the thermal and radiation environment will be similar to the one that exists today in fuel elements for nuclear reactors. The neutron flux will be lower, but the temperature will probably be more extreme.

Because the intent of the radiation shield is to intercept radiation in the form of X-rays and neutrons, it needs materials that are as opaque to neutrons and X-rays as possible.  Single-material shields (like beryllium) could work, but composite shields can work better.

Consider, for instance, a shield comprised of high temperature ceramic boron carbide (for neutron capture), and tungsten (for X-ray capture). Note that the boron should be closer to the radiation source, because its neutron-capture reaction spawns gamma rays that in turn must be shielded.  (Gamma rays are generally attenuated by the same materials as X-rays, though less easily.) If we want to stop 99.9% of the radiation, a bit of experimentation with equation (6) gives us a thickness of about 200 mm for the boron and 5mm for the tungsten.

If the shield is 50m from the fusion reaction, and is 15m in diameter, then the energy flux from the neutrons will be 341 MW/m2 (from Table 2) and the energy flux from the X-rays will be about 171 MW/m2  (50% of the energy flux of the neutrons), for a total waste energy load of 512 MW/m2. Multiply that by the area of the shield to get the total power dissipation requirement:

 

P= 512 MW/m2 * 177 m2 = 90,000 MW = 90 GW

 

This gives the ~100 GW heat dissipation requirement that we’ve been using until now in these articles.

 

As far as the physical arrangement of the 200mm of boron carbide and the 5mm of tungsten shield is concerned, we must remember our example on conduction.  The shield cannot be a flat thin plate as in the left side of Figure 9.  The heat cannot leave the shield by conduction from the sides, since the rate is much too low and the middle would melt.  It cannot leave by radiation alone either, since the surface temperature required is over 8,100K.  Therefore ample space is required for a coolant to circulate in the shield, and the design should increase as much as possible the surface area of the neutron and x-ray capture materials in order to increase heat transfer to the coolant.  

 

Interestingly, if liquid metals such as aluminum or beryllium are used as coolants, these have good neutron capture properties, and it might be possible to simplify the shield into some kind of piping loop, or perhaps simply a tank, in which the coolant circulates and absorbs the heat generated by the radiation.

 

Fuel Lines and Tanks

 

Fuel lines and tanks offer the other extreme of material applications, cryogenics.

Tanks themselves are really just big pipes with end caps, plus a lot of fiddly details where the pipes connect to the tanks.  From the materials point of view, this isn’t new territory. Cryogenics systems are common — often quite large and already used for space applications.  As explained in the Saturn comparison earlier, the Icarus tanks are subject to much lower stresses than those found in a typical liquid fuel rocket, so this makes the job somewhat easier.

Composite fiber tanks may be used, perhaps with an inner liner of metal (or possibly graphene) to prevent leaks due to permeability of many materials to the small atoms of helium and hydrogen.  This is a good application for low-density carbon reinforced plastics.  

An important consideration for cryogenic tanks is revaporization. Particularly while the ship is in operation, there will be heat gain in the tanks due to radiation coming from the drive, from space, and from the radiators. Some kind of refrigeration system may be required for the ship’s fuel tanks.

The tanks must absolutely be shielded from the drive by a radiation shield, since liquid deuterium is one of the best-known absorbers of neutrons, and the tank walls themselves – depending on composition — might absorb significant quantities of X-rays (and, to a lesser extent, neutrons) even with very small thicknesses.

For example, we can calculate the stopping power of the skin of a tank with a thin 0.5mm liner of beryllium, located at 100m from a fusion drive with no shield:

I/Io = e-(μ/ρ)ρl= e-0.0117*1848*0.0005  = 0.001;

therefore the beryllium has absorbed only 0.1% of the neutron radiation intensity. However, the radiation flux at that distance is still 100 MW/m2, so the tank coating will absorb about 100 kW/ m2. And that’s only from the neutrons.

To put the 100 kW/ m2 into perspective, a standard baseboard heater has a power of about 1 kW per meter of length.  So each and every square meter of our tank exposed to the radiation from the drive gets as much heating as from 100 meters of electric baseboards. That’s not very good for cryogenic liquids.

 

The Future of Materials

 

Below is a table of properties we might expect from materials available in 2100. To build the table, current yield strengths were doubled, and maximum use temperatures were increased from 50% to 66% of the melting point. Remember that the allowable stress at room temperature is about 50% of the yield strength, and that it goes down to about 25% of the yield strength at the maximum use temperature.

 

   

Melting

Max Use Temp

 

Yield Strength

   

Material

Density

Point

2014

2100

theoretical

2014

2100

 

Kg/m3

K

K

K

MPa

MPa

MPa

Steel alloys

7800

1773

800

 

20,000

250-5000

500-10000

Tungsten

19600

3695

1850

2400

     

Tantalum

16400

3290

1645

2170

     

Molybdenum

10188

2896

1450

1900

     

Carbon – Diamond

3500

3823

 

2500

1600

   

Carbon-carbon(graphite) composite

1600-2000

3800

1800

2500

 

600

1200

Carbon – Nanotube composite

1600-2000

3800

 

2500

60,000

 

6,000

Carbon – Graphene composite

1600-2000

3800

 

2500

130,000

 

13,000

Silicon carbide- silicon carbide composite

3200

3000

1500

1980

     

Glass (glass sheets)

2600

1600

   

35,000

7-800*

 

Titanium Diboride – TiB2

4520

3225

 

2100

 

370

740

Zirconium carbide – ZiC

6560

3400

 

2200

     

Boron carbide – CB4

2510

3036

 

2000

     

Table 4)  Material properties, all the yield numbers have huge spreads

 

*Glass is highly variable. The gorilla glass on tablets has a bending strength of about 800 MPa but only for the topmost 50 μm layer of the glass (6).

 

We should expect the strength of materials to double from their present values in the next century, but they will still be far below their theoretical strengths.  Melting points will remain the same, but materials will be able to be used closer to them.  

The most promising radiation and heat-resistant materials are carbon-carbon composites, and for most applications that do not require corrosion resistance they will not need to be much improved over today’s values. Ceramics, in particular those containing boron, may be the best radiation shields.  Fiber and nanoparticle reinforced ceramics may have much better properties than what we have today.

Apart from fuel, the quantities of materials required for Icarus are not large on an industrial scale, but for some of these materials they are orders of magnitudes larger than today’s production capacities.  As requirements for aircraft parts, power turbines and nuclear reactors increase during the next century, we can expect that industry will greatly increase the production rates, making today’s exotic materials much more commonplace and affordable.

Present day materials are not really suitable for operation beyond 1800K.  Substantial increases in existing material capacities and availabilities will be required to build Icarus and operate at temperatures in the 2000 – 2500 K range.  Diamond pipes and graphene reinforced mega materials are not required, however.  But if they are around, they’ll certainly be appreciated!

 

Tools of the Trade

 

The complete information about materials took up too much space, and is rather boring to read, so it can be found in the spreadsheet under the following link.  All the equations from the text are also in the spreadsheet. Materials tables in spreadsheet form.

 

Acknowledgements

 

Many thanks to David Weiss, VP Engineering/R&D Eck Industries, Inc. for his great input for the two articles on materials. He has added to them immeasurably.  Pavel V. Tsvetkov kindly reviewed the nuclear aspects of the text, helping to reduce some of the confusion.

Older articles

Part 1 Liquids in pipe

Part 2 Gases in pipes

Part 3 Thermal properties of Materials

Radiators

Deuterium production

 

 

References

 

(1) Look up Frenkel theoretical strength for details. Here is a table for some materials http://bit.ly/1lxe1Sq

 

(2) Juhasz, Albert J. NASA Glenn Research Center, OH, United States, High Conductivity Carbon-Carbon Heat Pipes for Light Weight Space Power System Radiators http://ntrs.nasa.gov/search.jsp?R=20080045532

 

(3) Mass energy attenuation coefficients at the NIST  http://www.nist.gov/pml/data/xraycoef/index.cfm

 

(4)Chang, G.S., 2004, INEEL, Integrated fast neutron flux   http://www.inl.gov/technicalpublications/documents/3310878.pdf

 

(5) Go T. Chapman, Co Lo Storrs, National Energy Comission, 1955, Effective neutron removal cross sections for shielding  http://web.ornl.gov/info/reports/1955/3445603498123.pdf

 

(6) Gorilla glass competitor http://www.us.schott.com/xensation/english/products/xcover/technicaldata.htmlà

 

Further reading:

Thermophysical Properties of Materials for Nuclear Engineering: A Tutorial and Collection of Data , INTERNATIONAL ATOMIC ENERGY AGENCY,  VIENNA 2008  http://bit.ly/1gzzVO2