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/m^{3} 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/m^{3}; 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)
Where:
ρ = Gas density (kg/m^{3})
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?
Gas 
molecular weight 
Hydrogen 
2 
Deuterium 
4 
Helium 
4 
Nitrogen 
28 
Oxygen 
32 
Fluorine 
38 
Neon 
20 
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 = μ_{0} x ((T_{0}+C)/(T+C)) x (T/T_{o})^{3/2} (3)
Where:
T = actual gas temperature (K)
T_{0} = reference gas temperature (K)
μ = actual gas viscosity (Ns/m^{2})
μ_{0} = reference gas viscosity (Ns/m^{2})
and C and T_{o} can be found in the following table:
Gas 
Sutherland Constant 
Reference Temperature 
Reference Viscosity 
Viscosity at System Temperature 
Specific Heat Capacity 

C 
T_{0} 
μ_{0} 
μ at 2200 K 
C_{p} 


K 
μNs/m^{2} 
Ns/m^{2} 
kJ/kgK 
Air 
120 
291.15 
18.27 
0.000067 

Nitrogen 
111 
300.55 
17.81 
0.000063 
1.3 
Oxygen 
127 
292.25 
20.18 
0.000075 

carbon dioxide 
240 
293.15 
14.8 
0.000066 

carbon monoxide 
118 
288.15 
17.2 
0.000064 

Hydrogen 
72 
293.85 
8.76 
0.000029 
14 
Deuterium 
? 
? 
? 
? 
5.1 
Ammonia 
370 
293.15 
9.82 
0.000052 

sulfur dioxide 
416 
293.65 
12.54 
0.000070 

Helium 
79.4 
273 
19 
0.000067 
5.4 
Beryllium 
n/a 
n/a 
n/a 
0.0011 
1.92 
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.
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 m^{3}/s in volume flow.
The average density of the gas is 0.75 kg/m^{3}.
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 m^{3}/s of helium flowing in each loop.
The Calculations
The equation linking compression power to gas flow:
W=(Q*dp)/n (4)
Where:
W = compressor power (kW)
Dp = pressure change (kPa)
Q = flow = 32,000 (m^{3}/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*v^{2})/2) (5)
Where:
dp = pressure drop (Pa)
f = friction factor
L = length of pipe (m)
D = Diameter of pipe (m)
ρ = density = 0.75 (kg/m^{3})
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 m^{3}/s
v = velocity = 100 m/s
A = Area = 3.2 m^{2}
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)
Where:
e = pipe roughness factor (m)
R_{e} = 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 (R_{e}):
R_{e} = ρvD/μ (9)
Where:
ρ = density = 0.75 (kg/m^{3})
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) R_{e}= 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*100^{2})/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)
Where:
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 carboncarbon 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 m^{3}
With a density of 1500 kg/m^{3}, 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.
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 (R_{e} – 1000) x Pr) / (1+(12.7 x (f/8)^{1/2 }x (P_{r}^{2/3} 1))) x k/D (12)
Where:
h = convective heat transfer rate for a gas in a pipe (W/m2K)
f = friction factor
R_{e} = Reynolds number
P_{r} = 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.
P_{r} is the Prandtl number, a dimensionless number, standing in for the following:
P_{r} = c_{p}μ/k (13)
Where:
c_{p }=specific heat = 5,400 (J/kgK)
μ = dynamic viscosity = 0.00067 (Ns/m^{2}) 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 T^{0.71} (mW/cmK)
14b) Deuterium: k= 0.7073+0.02368×10^{3} x T+0.1048×10^{6} x T^{2} (mW/cmK)
14c) Hydrogen: k=0.7897+0.03623×10^{3} x T+0.01809×10^{6} x T^{2} (mW/cmK)
14d) Nitrogen: k=0.051+0.7438×10^{3} x T0.1573×10^{6} x T^{2} (mW/cmK)
14e) Oxygen: k=0.07979+0.6671×10^{3} x T0.05479×10^{6} x T^{2} (mW/cmK)
Putting the numbers into equations 13) and 14) we find:
14) k = 2.684×103 x 2200^{0.71}= 0.6
13) P_{r} = 5,400 x 0.000067/0.6 = 0.57
12) h= ((0.0102/8) x (2,238,8001000)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.
Q_{r} = eBA(T_{wall}^{4}T_{space}^{4}). (15)
Q_{t} = hA(T_{fluid}T_{wall}) (16)
We can combine these two equations into a single one, since Q_{r} = Q_{t} and find our last equation:
eB(T_{wall}^{4}T_{space}^{4}) – h(T_{fluid}T_{wall}) = 0 (17)
Where
e = emissivity
B = StephenBoltzman constant
T_{space} = 10K
T_{fluid} = 2200K
h = thermal conductivity = 600 (W/m^{2}K)
We know all the numbers here, except T_{wall}. 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 starshipsGases
Notes
(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(10.0002 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.