A Plumber’s Guide to Starships Part 2: Gases in Pipes

posted by Michel Lamontagne on April 15, 2014

Figure 1

Figure 1. An array of 100 4 MW compressors, driving helium at 2 200 Kelvin through the radiators, glowing bright orange from the heat. The large cylinder holds the heat exchangers and includes the radiation shielding.


In this section, we will mainly be looking at gases in pipes from the heat transfer point of view.  Although there are plenty of other reasons to move gases around in a starship, starting with fuel feed, for example, heat transfer covers the most complicated cases. The information will be applicable to simpler problems.

Gases are trickier than liquids.  They compress and change volume.  They turn back into liquids under pressure.  They leak.   But kg for kg, they have higher specific heat than most liquids, so for a mass driven machine like a starship, they are an interesting choice of heat transfer fluid. 

The equations for gases and liquids are remarkably similar from the point of view of a starship plumber.  We will basically run through the same steps as for “Part 1: Liquids in Pipes”.  Pumps will change for compressors, but the rules for these two types of equipment are the same. 

The main differences are geometric, stemming from the large difference in density between the two fluids at the same temperature.   Liquid Beryllium, for example, has a density of 1,615 kg/m3 at 2200 K.  Helium, at this temperature, even if it is compressed up to 3,500 kPa (500 psi) has a density of only 0.75 kg/m3; almost 2,000 times less!  Therefore the pipes will be much larger.  However, gases can flow much faster than liquids, since they are less viscous.  So the cycle time will be shorter than for liquid metals, giving us a much lower coolant weight.  But the pipework will be larger, possible leading to heavier radiator weight. 

A bit of, this, a bit of that, it’s the classic iterative design process.

Another important parameter is the low thermal conductivity of gases compared to liquids.  Taking Beryllium again as our reference, its thermal conductivity is 200 W/mK, while the thermal conductivity of helium is only 0.6 W/mK at the temperatures our system will be using.  This means that because of the internal resistance of the gas to heat transfer, its low thermal conductivity, the radiator walls will be significantly cooler than the gas.  This results in larger radiators, and more weight for our starships.

The high pressure and low density of the coolant will affect the radiation shield design as well.  The coolant will probably run through pipes built into the shield, to limit pipe diameter and reduce hoop stress to manageable levels.  And, to avoid catastrophic failures, the system should probably be divided into a multitude of independent circuits that can be maintained individually.   

Again, we are interested in limiting the power required to make the gas circulate through the circuit, and in keeping the pipes as small as possible, so the pressure will create less strain in the walls. 


About Gases

Before we start our design work, we should get a little more familiar with the behavior of gases and the equations that govern this behavior.  Our starting point is the Ideal Gas Law:

P*V = n*R*T                                                                                                                          (1)

Delightfully simple, isn’t it?  But a bit abstract, so we will change right away to a more useful version:

ρ=Pw/RT                                                                                                                               (2)



ρ = Gas density (kg/m3)

P = Gas Pressure (kPa)

W = Molecular Weight (g/mole)

R = Ideal Gas Constant = 8.314 J/kg mole

T = Absolute Gas Temperature (K)


The interesting thing about the molecular weight expressed in g/mole is that it is exactly the same number as the atomic weight (or relative atomic mass) of the substance, that can be found in the periodic tables. We will round off the fractions, because real gases aren’t perfect anyway, so who needs so much precision?



molecular weight















Table 1, gas molecular weight


In this form we can easily find the density of a gas at various temperatures and pressures, and from the density determine the weight of the coolant.  The density will also be required later for our pressure drop calculations.

Another property we will want to determine about the coolant gas is its viscosity, since this changes with the temperature of the gas and we need to know it to determine the pressure drop, and therefore the size and power of the compressors.  Mr. Sutherland did the work for us here and his findings are compiled in the following equation and table:

μ = μ0 x ((T0+C)/(T+C)) x (T/To)3/2                                                                                              (3)



T = actual gas temperature (K)

T0 = reference gas temperature (K)

μ = actual gas viscosity (Ns/m2)

μ0 = reference gas viscosity (Ns/m2)

and C and To can be found in the following table:



Sutherland Constant

Reference Temperature

Reference Viscosity

Viscosity at System Temperature

Specific Heat Capacity





μ at 2200 K


























carbon dioxide






carbon monoxide
























sulfur dioxide



















Table 2, gas properties and factors for viscosity calculations, beryllium is included for comparison purposes



The Parameters

There are many possible arrangements for this system of a hot gas circulating in pipes.  The one illustrated here is simply one of the simplest to calculate, but probably not the one that will be eventually chosen.

Figure 2

Figure 2. The ship is illuminated by the pinch fusion drive, pushing the large robot probe up to 6 or 7% of the speed of light. Icarus Firefly, designed by Robert Freeland.


Here are the cooling process parameters(1):


The power to be handled and radiated out is 100 GW total, 50 GW per side.

The coolant is helium gas at 3,500 kPa (500 psi).

It is heated, and then cooled, from 1,800K to 2,600K.

The mass flow is 24,000 kg/s (24 tonnes per second), or 32,000 m3/s in volume flow.

The average density of the gas is 0.75 kg/m3.


Figure 3

Figure 3. Detail of the radiator assembly under load at full temperature, with one section separated to show the structure and the walls of the heat exchangers in transparency to show the radiation absorbing core.


The Geometry

The system is divided into 100 sections.  Each section is made of two elements: a 30m diameter heat exchanger, about 2m thick, which holds the radiation absorbing material; and a long pipe loop, stretching 300m out and returning, for a total length of 600m, that serves as the radiator.  There are 50 loops of pipe per side, 1 GW per loop, and 240 kg/s, or 320 m3/s of helium flowing in each loop.


Figure 4

Image 4. Closer detail of  a large circular heat exchanger, pump and the long pair of black pipes that serve as radiators. Stack 100 of these one atop the other, and you get the arrangement discussed in this section. For scale, the pipes are 2m in diameter and the exchanger is 30m in diameter.


The Calculations

The equation linking compression power to gas flow:

W=(Q*dp)/n                                                                                                                           (4)



W = compressor power (kW)

Dp = pressure change (kPa)

Q = flow = 32,000 (m3/s)

n = compressor efficiency (usually between 0.6 and 0.8)

The pressure drop in the pipe is calculated by the Darcy–Weisbach equation:

dp = f*(L/D)*((ρ*v2)/2)                                                                                                             (5)



dp = pressure drop (Pa)

f = friction factor

L = length of pipe (m)

D = Diameter of pipe (m)

ρ = density = 0.75 (kg/m3)

v = average gas velocity (m/s)


To choose the velocity, we can apply the upper limit of industrial design velocities for gases of about 100 m/s.  We find that the cycle time will be 600/100 = 6 seconds in the pipes, plus about 2 seconds in the radiator.  8 seconds x 24 tonnes per second = 192 tonnes.  That’s very good (not too heavy), especially if we compare this to the 1,425 tonnes we calculated for liquid metals.


With the fluid velocity known, we can determine the pipe size required for each of the 100 cooling circuits.  First the area of each pipe:

Q = v*A                                                                                                                                 (6)

Q = Flow = 320 m3/s

v = velocity = 100 m/s

A = Area = 3.2 m2 


and then the diameter:

D = 2 x (A/π)1/2 (m)  where  D = 2 x (3.2/π)1/2 = 2m in diameter                                                          (7)

All we are missing now to determine the pressure drop is (f), the friction factor.  The Haaland friction equation is used to find it:

f = (-1.8*log(((e/D)/3.7)1.11 + (6.9/Re)))-2                                                                                    (8)



e = pipe roughness factor (m)

Re = Reynolds number (dimensionless)

D = pipe diameter (m)


Roughness factors (e) is about 0.0000015 for the smooth pipe we should be using.

And the Reynolds number (Re):

Re = ρvD/μ                                                                                                                            (9)



ρ = density = 0.75 (kg/m3)

v = average fluid velocity = 100 (m/s)

D = Diameter of pipe = 2 (m)

μ = dynamic viscosity = 0.000067 (kg/m*s)(Pa*s) , From equation 3, and table 2.


Putting it all together, for helium at about 2200K

9) Re= 0.75 * 100 * 2/0.000067 = 2,238,800

8) f = (-1.8*log(((0.0000015/1.1)/3.7)1.11 + (6.9/2 238 800)))-2 = 0.0102

5) dp = 0.0102*(600/2)*((0.75*1002)/2) = 11,400 Pa  = 11.4 kPa

4) W = (320*11.4)/0.7 = 5,240 kW = 5.2 MW

For 100 circuits, the power is 520 MW.

We can also do the hoop stress right away:

σ = Pr/t                                                                                                                               (10)



σ = Hoop stress (Pa)

P = Internal pressure 3,500,000 (Pa) + pressure from the compressor 11,400 (Pa)

r = Interior pipe radius (m)

t = Pipe wall thickness (m)


And the relationship with the material of the pipe is:

S/SF = σ                                                                                                                             (11)

S = Material tensile strength (Pa)

SF = Safety Factor

σ = the allowable hoop stress (Pa)


The pipes are very large; if we chose the same middle of the range carbon-carbon pipe as we had for liquid metals, at 360 MPa, the radiators would weight 14,000 tonnes!  So we will move up to stronger, more speculative stuff; carbon nanotube reinforced ceramics, with a tensile strength of 3,600 MPa.

S/SF = σ = 3,600  MPa / 3 = 1,200 MPa or 1,200,000 kPa

For our radiator, 2 m in diameter with a 1 m radius, the wall thickness will be:

t = Pr/σ = 3511*1 /1,200,000 = 0.003m, or about 3 mm thick

So each pipe wall will have a volume of 0.003 x 600 x 2 x pi = 11 m3

With a density of 1500 kg/m3, each pipe will weigh 16 tonnes.

For 100 circuits, the total weight will be 1,600 tonnes.  Adding the 192 tonnes of coolant we calculated for Equation #5, we find a total weight of about 1,800 tonnes.


The Analysis

The pressure drop for gases in this arrangement is very low, compared to liquid metals.  However the volume flow rate is huge.  The coolant races through the system in just a few seconds and everything is moving very fast.  There is very little energy accumulated in the system, so it has very little buffer.  If the flow stops for as little as a minute, the radiation shield will melt, probably with explosive results.  So we must plan for the drive to be able to shut down quickly.

The poor heat transfer characteristics of gases tend to negate their excellent heat capacity and low weight.  This leaves them pretty much equal to liquid metals.  The high pressures required are hard on the pipes, however.  The large diameter of the radiator pipes forces us to move to more speculative materials to keep the weight reasonable.

We haven’t worked out the details of the heat exchanger yet.  That’s for a future essay. The pressure loss in a heat exchanger is usually the main equipment pressure loss in a piping system, just below the piping itself.  So we can guess that the total pressure drop of the system might be twice what we have calculated, for a total power of 540 MW x 2 = 1080 MW.  So the power require to circulate the gas may be up to twice the power required to circulate an equivalent liquid metal, at least for the present model.

It’s pretty obvious that we haven’t optimized the circuit.  We should look into reducing the pipe diameter, either by increasing the gas velocity over standard values (as we did for liquid metals) or by increasing the radiator size and increasing the number of circuits.  If we can determine materials that will resist corrosion from hot gases, we might switch to hydrogen cooling for its much higher specific heat.  Deuterium does not offer this advantage, since its specific heat is lower than the one of helium, so its use as a substitute for helium is problematic.

Alternatively, a compound gas might be interesting; a mixture of nitrogen and helium, or of neon and helium, might provide better overall performances.

This guide provides the equations.  The work is up to the starship designer.


Figure 5

Are we done yet?(2)


Not quite.  We’ve mentioned several times that the surface temperature of a radiator will be lower than the cooling fluid’s temperature, but we have not given the details yet.  Here they are.


Surface Temperature of Radiators

In our example for radiators with liquid metal coolants, we assumed the temperature of the pipe walls, and so of the radiator, would be the same as the temperature of the liquid metal.  In reality, this is not the case, but the error is not very large since liquid metals are very conductive and the pipes are small. For gases, we really need to take the thermal conductivity of the gas into account, the convection rate, as well as the size of the pipes and the other physical characteristics of the gas.  Generations of experimenters have labored tirelessly to produce the following equation, linking all these elements together, the Gnielinski correlation(3):

h = ((f/8) x (Re – 1000) x Pr) / (1+(12.7 x (f/8)1/2  x (Pr 2/3 -1))) x k/D                                             (12)



h = convective heat transfer rate for a gas in a pipe (W/m2K)

f = friction factor

Re = Reynolds number

Pr = Prandtl number

D = Pipe diameter (m)

k = Thermal conductivity of the gas


Admittedly, it’s a lot less elegant than the Ideal Gas Law. 

The Reynolds number, the friction factor (f) and the pipe diameter we already know.

Pr is the Prandtl number, a dimensionless number, standing in for the following:

Pr = cpμ/k                                                                                                                            (13)



cp =specific heat = 5,400 (J/kgK)

μ = dynamic viscosity = 0.00067 (Ns/m2) from Equation #3

k = thermal conductivity (W/mK) from Equation #4


We need to determine the thermal conductivity (k) for both Equations #12 and #13.  It varies with the temperature and the pressure.  Again we turn to experimental results and to the equations that were worked out during the great age of nuclear experimentation, in the fifties and sixties(4)


14a) Helium:  k=2.684×10-3 x T0.71   (mW/cmK)

14b) Deuterium: k= 0.7073+0.02368×103 x T+0.1048×106 x T2 (mW/cmK)

14c) Hydrogen: k=0.7897+0.03623×103 x T+0.01809×106 x T2 (mW/cmK)

14d) Nitrogen: k=0.051+0.7438×103 x T-0.1573×106 x T2 (mW/cmK)

14e) Oxygen: k=0.07979+0.6671×103 x T-0.05479×106 x T2 (mW/cmK)


Putting the numbers into equations 13) and 14) we find:

14) k = 2.684×10-3 x 22000.71= 0.6

13) Pr = 5,400 x 0.000067/0.6 = 0.57

12) h= ((0.0102/8) x (2,238,800-1000)x0.57) / (1+(12.7 x (0.0102/8)1/2 x(0.57 2/3 -1)))x0.6/2 = 600


As our final step, we want to determine the wall surface temperature. To solve this we will use the fact that the surface temperature for the radiation equation is the same number as the wall temperature for the convection in the pipes, and that the power that radiates out from the radiator has the same value as the power that moves out from the gas and to the radiator walls.


Qr = eBA(Twall4-Tspace4).                                                                                                           (15)

Qt = hA(Tfluid-Twall)                                                                                                                  (16)

We can combine these two equations into a single one, since Qr = Qt and find our last equation:

eB(Twall4-Tspace4) – h(Tfluid-Twall) = 0                                                                                              (17)



e = emissivity

B = Stephen-Boltzman constant

Tspace = 10K

Tfluid = 2200K

h = thermal conductivity = 600 (W/m2K)


We know all the numbers here, except Twall.  To find this temperature, you can either embark on a long and complex search for analytical solutions, or use the brute force approach available in most spreadsheet programs, using the solve function in Excel, for example.

In this case, the solution is 570 K of difference. So for an average gas temperature of 2200 K, the average radiator surface temperature will be 1630 K.

It is left as an exercise to the reader to determine the increase in radiator area this represents.

Tools of the Trade

You will find the tables and equations mentioned in this article in the following spreadsheet:

Plumbers guide to starships-Gases



(1) These are the parameters used in the Starship radiator post, and are repeated in the Tools of the Trade spreadsheet.

(2) Kitty was found here on the Internet: http://bit.ly/1kNw1aJ

(3) The Gnielinski correlation is actually expressed as a function of the adimensional Nusselt number, not of convective heat transfer.  But since its only use for us here is to combine the equation with the heat transfer equation h=Nu D / k, I did it right away and saved us all a step. The Dittus Boelter correlation is also often used for heat transfer in pipes, but it goes way off in the ranges of temperature we are working in, so we will not use it here.  And since we have the Haaland equation to determine ‘f’ directly, there is no longer a problem with using the Gnielinski correlation for simple calculations.

(4) The actual equation for helium is more complex and takes pressure into account

k = (2.682×10-3)(1 + 1.23×10-3 x P) x T x 0.71(1-0.0002 x P)  it can be found here: http://1.usa.gov/1lBsDgu.

For the other gases, the equations were taken from ‘Transport properties of gases and gaseous mixtures at high temperatures, S.C.Saxena, 1970’.  The curve fitting was only tested by the authors up to 1100 K, so the results of these equations should be taken with a grain of salt.

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27 Responses to A Plumber’s Guide to Starships Part 2: Gases in Pipes

  1. Asher Syed says:

    Excellent work! Michel

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  3. Charles quarra says:

    very good and interesting series of posts.

    an interesting possibility is to analyse the tradeoff between splitting the gas pipe between multiple ones, increasing the net gas-pipe contact area, and partially compensate the relatively low thermal conductivity of the gases. Although I’ve seen this applied only for small hydraulics, so I’m not sure how that would scale to a configuration with pipes running hundreds of meters. You made a mention about multiple circuits on the post, so it is clear that you already knew this though

    • Hi Charles,

      I did the simulations: for a 2m pipe the temperture difference is 570K, as mentionned in the paper. for a 10cm pipe, the temperature difference is 435K and for a small 1cm pipe the temperature difference is 325K. This corresponds to increases in efficiency of 37% and 75%. So it is a very effective solution, from the thermal point of view. The physical arrangement is probably heavier though. The numbers are here: http://bit.ly/RoG7mz.
      Michel Lamontagne

  4. yes Indeed a great set of articles,
    my math skills are poor however i still struggle on with my very small project,


    A ULA engineer who is interested in my idea provided this data for the Centaur in space stage;

    begin quote

    “LH2 volume ~1,700 ft3
    LO2 volume ~600 ft3

    The Centaur tank sidewall is ~25 ft long and 10 ft in diameter. You can calculate the area from this.

    We fly SOFI on the sidewall on all missions for insulation. Some missions also have MLI.”
    end quote
    The Engineer did not mention the alloy that makes up the Centaur,s tanks

    So I assume a small fission reactor radiator in the spent LO2 with gasses Xenon at 600 ft3 and the Effective radiator temperature 475 K
    Radiator size required 5.035 m2
    Radiator length required 240 cm

    This data is from the decadel survey small fission reactor study

    I assume gas law and the surface area losses through the fuel tank walls and friction in the Stirling engines
    because the SFR radiator has displaced LO2 in the tank needed for the chemical burn we need to increase the size of the O2 tank thus altering the tank volume needed to compute the gas law and surface area equation!

  5. I was inspired by the ISS radiator NASA paper as it is the largest in space radiator to date!
    if you would like to see this paper I have posted it on my blog with my take on how to recover waste heat from the ISS radiator.
    would there be some starship designs where radiators could be re purposed?


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  9. N/A says:

    Would supercritical CO2 be a middle ground of sorts? Ease of gas pumping, but thermal properties more like a liquid?

    • It might be interesting. Its molecular weight is 44, or 10 times denser than helium. However its specific heat is only 0,84 compared to 5 for helium, so my guess is that you would need a lot more weight, but that the overall volume would be smaller, so lighter pipes; so not a clear advantage.

  10. Oups! I have updated table 1, there was a mistake in the molecular weight table of the non inert gases. I will also be providing a more exact equation for compressor power in my upcoming pumps and compressor post.

  11. Martin Kuttner says:

    Regarding deuterium viscosity:
    Assael M.J.; Mixafendi S.; Wakeham W.A. The Viscosity of Normal Deuterium in the Limit of Zero Density J. Phys. Chem. Ref. Data 1987 Vol. 2, No. 2

    A pdf reprint is available on the NIST website http://www.nist.gov/data/PDFfiles/jpcrd318.pdf

  12. Devon says:

    Michel, I admire your work on this. Have you considered fractal metamaterials for heat transfer between the shield and the radiator? Link: http://nextbigfuture.com/2013/05/metamaterials-remove-heat-and-control.html#links

    This is a near field effect so it won’t replace the need for radiators. I realize that the fractals are a bit speculative and one can’t easily put numbers to it, but what are your thoughts on this?

    • Michel Lamontagne says:

      Hi Devon,
      Thanks for the link.
      Frankly, it looks like a series of cavities transmitting infra red radiation, added to closed convection cells. It might be a good replacement for pipe walls, as long as the overall product was gas tight. I’m a bit skeptical about ‘special’ electromagnetic waves, whatever they are.

      • Devon says:

        Yes, the word “special” is not too far from “magic”, and a poor descriptor of evanescent wave near-field effects. But it looks like the rate of heat transfer exceeds Stefan-Boltzmann by orders of magnitude, certainly much higher than convection and conduction also.

        Correct me if I’m wrong, but I don’t think there’s necessarily a gas medium? Seems like a fine vacuum effect unless I’m missing something.

        The only bottleneck may be conduction of the heat out of the neutron shield into the metamaterial, and then out of the metamaterial into the radiator wings, but this should be no worse than traditional conductive transfer. The advantage would be mass savings, no moving parts, no power or pumps, etc with regard to the plumbing.

  13. Bobby Baum says:

    Here we go again…
    Some of the equations use * for multiplication; others use x. This is unnecessarily inconsistent and confusing. I recommend using x as in equation 3 since many people aren’t familiar with using * for multiplication (did this come from early computer languages like fortran?).
    The explanation of Equation #2 lists P twice.
    Eable 2 has 2 as a superscript in the units for Reference Viscosity but not Viscosity at System Temperature, definitely confusing. Viscosity at System Temperature has commas as does Specific Heat Capacity for Be. Shouldn’t the data for deuterium from the earlier comment be added to the table?
    The numbers for the footnotes are shown as subscripts; in the other articles superscripts are used
    Image 4 – “of a the”
    The Darcy-Weisbach equation has p while the explanation has rho.
    Equation 7 – what is the funny a between the equations? Is there any reason 3.2 is in parentheses?
    Equation 8 – since it is raised to the -2 power (1 over the square) why have the – sign at the beginning? It cancels out when you square it.
    Equation 9 – mu has a comma. Doesn’t s cancel out in the units leaving just (kg/m)(Pa)?
    9) has a comma.
    Wall thickness – this time there’s a decimal point where there should be a comma (1,200,000 KPa)!
    “The pressure loss in a heat exchanger is usually the main pressure loss of a piping system, just below the piping itself.” If it’s just BELOW the piping itself the piping would be the main pressure loss.
    Equation 12 would be clearer if the k/D was on the same line as the rest of the equation.
    Equation 14a – should this have mW/cmK?
    The exponents definitely should be clearly indicated as in Equation 12; Equations 14a-12 and footnote 4 are hopelessly unclear.
    Footnote 4 has an extra (.
    If you’re using nanotube-reinforced ceramics how about adding long nanotubes perpendicular to the wall extending from inside the pipe far out into space? I would think this would significantly improve the thermal conduction (at the expense of a higher friction factor)?

    • Thanks again Bobby for your careful reading. I do hope the substance of the post is coming through despite these errors. The * comes from Excel (or google sheets, in this case), where I work out these formulas and test the numbers. The , and . problems mostly come from the fact that I do these in a french version of google sheets, and then translate them to american notation. I changed my setting in later posts and I hope to have eliminated most of these.
      Extending nanotubes might help if they were very long. However they would then stick out from the shadow shield, and I’m afraid they might gain as much power through neutron and x ray interactions as they lose. But this would need to be checked.

      • Bobby says:

        Table 2 still has a comma in the Specific Heat Capacity for Beryllium (and the 2 really should be a superscript in Ns/m2 (and why are the first 2 but not last 2 units in []?))
        In (13) k = thermal conductivity (W/mK) is not from Equation #4.
        In 14) there should be x between 2200 & 0.71 but the math doesn’t work; I get 4.2 rather than .6.
        The math doesn’t work for 13) either; I get .603 not .57.
        For 12) 1/2 & 2/3 are exponents.
        Equation #17 is quartic in Twall; there is an equation for solving quartic equations (I’m not sure where you’d find it). Note that quartic equations have 4 solutions; I suspect 2 of them are complex numbers. The other one is presumably the irrelevant case with the wall temperature higher than the fluid temperature.
        To totally change the subject, has Icarus Interstellar considered exhibiting at science fairs such as the gigantic USA Science and Engineering Festival in Washington, D.C. in April?

        • Hi Bobby, I’ve corrected as best I can. The editor keeps dumping superscripts, for some unknown reason. 13) is 2200^0.71 which explains the difference. For science fairs, it’s part of the educational charter of Icarus, but I must refer you to the Icarus administrators for specifics. I guess it depends if there is a member in the area, as budgets are quite short! Right now the big push is for the Starship congress in September.

          • Bobby says:

            Shouldn’t footnote 4 also be changed from T x .71 to T^.71? (and why is the equation so different from the others? Because He is monatomic?)
            For 13) I still get 5,400 x .000067/.6 = .60 not .57.
            Incidentally, I now think the other solution to the quartic equation is the even less relevant negative T.

          • Michel Lamontagne says:

            Yes, footnote 4 will be changed this evening. The .57 vs .6 is because the result of 14) is .63 and not .6 I rounded down the result, leading to the erroneous result in 13). If you actually solve the quartic equation, I will need to send you a voucher for a beer, or whatever your favorite beverage is 😉
            I don’t know why He is so different, your explanation makes sense. I’ll try to look this up in my references this evening as well.

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