A Plumber’s Guide to Starships, Part 3 – Thermal Properties of Materials

Materials are the bedrock upon which we can build starship designs. Outside of science fiction, there is no use postulating a starship drive that requires materials that do not, or cannot, exist. So this article is about materials: what they can do, what we hope future materials can do, and what we expect materials to never be able to do.

Figure 1- Roman clay pipes. Modern vitrified clay pipes. Alumina ceramic pipes and linings. 2000 years of progress in materials.

For a Starship plumber, the question boils down to: “What’s the best material for my pipes?” And the answer is the lightest, the strongest, the toughest, and the most corrosion, heat and radiation resistant. Plus some contradictory requirements according to application: high or low thermal resistivity, high or low transparency.

Let’s concentrate on application, because it’s easy to get lost in the mass of information about materials. But first, a few cautionary words on material properties and behavior.

Material Properties

There is a huge difference between the theoretical properties of materials and their actual properties in practical applications. Take aluminum, for example. Its theoretical yield strength is about 6,900 MPa(1), but in existing aluminum products, the actual yield strength varies from 35 to 500 MPa. Thus, after a century of research, optimization and hard work, the strongest real aluminum alloys have about 7% of their theoretical yield strength, and most of them have much less. This holds true for pretty-much all materials, from metals to ceramics — their actual strength properties are usually between 3% and 10% of their theoretical limits. It is therefore important to take the values published for relatively new materials such as graphene and carbon nanotubes with much skepticism. Graphene may have a theoretical yield strength of 130,000 MPa, but planning for anything more than 13,000 MPa in real products is probably highly optimistic.

On the other hand, some properties are not affected in the same way — they are inherent properties largely unaffected by manufacturing processes. Melting points, vaporization points, and rigidity numbers such as Young’s modulus are examples of such properties.

It is also helpful to introduce a few definitions, because these terms are often confusing:

Tensile strength: The maximum stress that a material can withstand without breaking. Sometimes called ultimate tensile strength or ultimate strength.

Yield strength: The stress that causes permanent deformation. Metals will stretch and yield without breaking, while ceramics will often have practically the same yield strength as tensile strength, and break without any stretching.

Allowable stress: The maximum stress recommended for practical use. For most metals, it is about a quarter of the tensile strength.

Creep, Deformation, and Fatigue

Materials’ yield strengths and melting / boiling points still don’t tell the whole story. Before they break down, materials will stretch. As time passes, cyclical loading will reduce their strength. As the temperature goes up, under the influence of the stresses upon them, they will flow and stretch. We call this “creep”, and it can cause a turbine blade tip to stretch, hit the turbine walls and self destruct, or it can cause a pipe wall to thin, balloon out and fail catastrophically. Because of this, we must reduce even more the loads planned for pipes and structural elements.

For example, a fairly common stainless steel pipe, ASTM 304, has a tensile strength of 525 MPa, and a yield strength of 205 MPa. However, it will never be used at more than 125 MPa of stress, even at lower temperatures, to ensure good long term performance. As the usage temperature gets higher, the allowable stress goes down. At 800 K, the allowable stress is down under 100 MPa, and it then declines rapidly, down to 25 MPa at 980K. That’s still a long way below the melting point of steel, at 1773 K.

Inquisitive minds might ask, “Just how precise are the equations and tables for materials?” The truthful answer is, just OK. Experimental results have errors, contaminants will change the results, pipes have manufacturing tolerances, pipes expand with heat, measuring instruments are affected by the heat. For high temperature materials, real world applications are limited, so the values are not all available. So take the results and tables with a grain of salt, and don’t hesitate to round off results for clarity and add margins of safety to your designs.

And do try to round down, rather than up.

Radiators and Heat Exchangers

Figures 2 – A pipe and pump assembly, part of a large radiator array for a future Starship. The Sabre Engine Pro-cooler, an essential element of the Skylon Spaceplane.

Starship radiator pipes will be subject to high temperatures, corrosion and erosion, as well as strain from high internal pressures. Nevertheless, except for the initial startup period, they should not be subject to much fatigue and vibration; their environment should be fairly static. This can be compared to the exceptionally difficult environment in an aircraft engine, where quickly varying conditions are the norm. Beyond simply choosing the highest ranking material in a table, there are other aspects of material selection that need to be considered, notably thermal conductivity. After having touched on Radiation and Convection in earlier texts, we can here address the final form of heat transfer: conduction, the flow of heat directly through a material. The equation for thermal conductivity is:

Q = k*(A/x)*dt (1)

Where:

Q = Power (W)

k = Thermal conductivity (W/mK)

dt = temperature difference (Tin-Tout) (K)

A = Area (m2)

x = Thickness of material (m)

If we refer back to the calculations we did for radiators, a surface temperature of 2200 K (Tout), with an emissivity of 0.9, corresponds to a power of about 1000 kW per m2.

If we suppose a thickness (x) of 1cm, or 0.01m, for a power (Q) of 1000 kW per m2, then we can find the temperature difference through the material required to ‘drive’ the conduction in the radiator.

The following table lists the results for various materials:

Table 1 – Thermal conductivity in various materials

So, for example, with a radiator made from tungsten pipe with a hot liquid metal coolant circulating inside, the inner surface temperature would be 2200K + 58K = 2258K. That’s a significant difference, but not a deal breaker. If we tried to push that much power through a carbon fiber reinforced epoxy wall, it would melt long before it reached the required temperature difference.

Looking again at Table 1, we can see that for a thickness of a few mm, Tungsten or Zirconium Diboride do not increase thermal resistance by more than a few dozen degrees. So their use as corrosion protection from liquid metals or hot hydrogen for a carbon composite radiator might be possible.

We will be looking in a bit more detail at corrosion and thermal coatings in the next sections.

Conduction Alone as Heat Transfer Mechanism

Figure 3 -Solid radiators. 40 000m2 of graphene radiators, glowing at 2200K on average, with a much hotter core.

What would the results be if we tried to cool the radiation shield of a ship like Icarus by conduction alone? Can we do without the pumps, compressors and pipes and just use conduction? It would certainly be simpler. Figure 3 shows such an arrangement, with a central core, surrounded by tapering radiators. Let’s suppose we’re using graphene at its highest possible value, 5000 W/mK (though as mentioned above, that’s not a safe design basis!). The radiators in the image above are about 250m in diameter, or 125m in radius, for a total surface area of roughly 80,000 m2, counting the two sides. The shield at the center captures 100 GW of radiation and turns it into heat. We can imagine the heat would have to transfer through at least 60m of graphene (half the distance to the edge), on average, before reaching the surface.

The radiators would be about 2200K average, and 3000K at the core, for a temperature difference of 800K. Plugging these numbers into equation (1) and then solving for the area (A):

Q = k*(A/x)*dt becomes

A = Q*x/k*dt A = 100x109W * 60/5000*800 = 1,500,000 m2

Thus, the area at the base of the radiators (where they glow bright orange in the picture) would have to be 1,500,000 / 80,000 = 18 times larger than pictured. Simply making the radiator bigger might help, but this would require the input energy to be spread over a much larger area, meaning that the shield itself is larger and more massive.

The unfortunate conclusion is that solid radiators are a rather inefficient way to radiate large amounts of heat.

Turbine Application

Figure 4 – A 274 MW Gas turbine assembly for power generation, showing the compressor on the left and the turbine on the right. The turbine has an inlet temperature of 1870K , an exit of 900K and turns at 3600 rpm. The whole turbine assembly with casing and accessories weighs 440 tonnes. (Siemens model SGT6-8000H)

Turbines for pumps and power generation are probably the hardest applications for materials on a starship, as on Earth. High stress from rotation, high temperatures, cyclical loading and fatigue, corrosion, and erosion from very high fluid velocities all take their toll. Metals do not behave very well at very high temperatures and under high stress. They can fracture, distort, and creep. But at lower temperatures, they remain very strong. Ceramics, on the other hand, are not very tough, and can break easily, but they keep their properties at much higher temperatures than metals. The solution found by engine makers has been to blend the two together: a metal blade with a ceramic coating. The trick is to use a very refractory ceramic and to cool the metal on the interior of the blade using either gas or liquid coolants. The heat flow out of the blade is equal to the heat flow into the blade through the coating.

Figure 5 – Turbine blade schematic (Original Sketchup model by Rainman)

For example, suppose we have a small 0.05 m x 0.3 m x 0.001 m thick turbine blade in one of our ship’s power systems (Figure 5). The hot gas going over the blade is at 2200K, but our metal works best at 2000K; higher than that, it starts to fail after a few months of operation.

So we need to bring down the temperature in the blade by 200K (dt). As mentioned above, a ceramic coating can be used for this. Consider a thin coat of Zirconia, for instance, since it has very low thermal conductivity of 3 (k). Can a mere 1mm coat be useful?

Here are the calculations:

The total blade area (A) is about 0.05 x 0.3 x 2 = 0.3 m2.

Applying the equation for conduction:

Q = k*(A/x)*dt so Q = 3*(0.3/0.001)*200 = 18 000 W = 18 kW (2)

So to keep the metal at 2000K we need to remove 18 kW from the turbine blade.

Consider how much more energy would have to be removed without the coating; metal has a much higher thermal conductivity, so it would have to be substantially thicker to achieve the same result.

Incidentally, one of the smart ways to get rid of this excess heat is to use the cool gas from the outlet of the radiator and to pipe it into the blade. In the gases article, we supposed an outlet temperature of 1800K for the main radiator. That’s a little too hot, so we’ll divert some of the gas through a small secondary radiator to bring its temperature down even more, to 1600K.

We can then feed this cooler gas into the blade with a high pressure pump and let it flow through openings in the blade back into the hot gas stream after having cooled the blade.

Going back to the mass heat flow equation, we can calculate the amount of gas needed:

Q = m*Cp*dt (3)

Where:

Q = Power we want to remove = 18 (kW)

M = mass flow rate (kg/s)

Cp = specific heat of helium, 5.4 (KJ/kgK)

dt = Temperature gain of the cooling helium = 2000K – 1600K = 400K

Putting the numbers in equation (3) we find :

m = Q/Cp*dt = 18/5.4*400 = 0.01 kg/s

Now, we are looking for the area of the pipe required for the coolant. Assuming that we’re using helium gas as the coolant, its density at high pressure (3500 kPa) is 0.75 kg/m3. We find that:

Volume flow rate m/d = 0.01/0.75 = 0.011 m3/s (4)

Choosing a reasonable velocity for the helium, 100 m/s for example, we can calculate the area of the pipe required :

0.011 m3/s / 100 m/s = 0.00011 m2

This is about 30% of the cross area of the turbine blade. The blade will be partly hollow, on top of being made with tiny interior piping with multiple holes in the blade and covered with a super tough ceramic. Just like this, in fact:

Figure 6 – An element of a modern turbofan engine, with laser drilled holes and interior air channels (Wikicommons)

Upcoming

The next article in the Plumber’s Guide will discuss materials in high radiation environments.

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. The materials table in spreadsheet form.

Acknowledgements

Many thanks to David Weiss, VP Engineering/R&D Eck Industries, Inc. for his great input for this article. He has added to it immeasurably.

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

 Older articles

Part 1 Liquids in pipe

Part 2 Gases in pipes

Radiators

Deuterium production