Part 1 : Liquids in Pipes
Moving liquids around is hardly new technology – even if it’s in a starship. Pretty much all of the design tools were worked out in the eighteenth century, and the most recent developments – magnetic pumps for liquid metals – date back to the 1960s. However, it is important to understand the limitations of pumps and pipes, since these can have a significant impact on the ship’s mass and power requirements.
We will not try to go all the way back to basics, but rather present practical formulae, as in a plumber’s guide. We will use equations and variables, with reasonable simplifications, since we want to be able to test our designs with a large array of fluids. Although the detailed engineering will eventually be done with software and simulations, the conceptual design can be done with these simple equations and tables. Solutions found for one situation can be generalized using rules such as the pumping affinity laws.
Piping design is an iterative process; the solution will be a compromise between the various factors that make up the system.
The simplest way to practice is to do an example. Let’s try to work out the best piping system for the radiation shield of the Icarus probe. The piping goes from the shield to the radiators, through the radiators and back again to the shield.
The Parameters
These are the cooling process parameters_{(1)}:
The power to be transported: 100 GW total, 50 GW per side
The fluid: liquid beryllium
from 1800K to 2600K
mass flow of 57,000 kg/s (57 tonnes), or
35 m^{3}/s in volume flow, 18 m^{3}/s per side
density of 1,615 kg/m^{3}
The geometry: 100 m of large pipe from the shield to the radiator and back.
400 m of small pipe in 50 loops in the radiator itself.
Our design objectives are to keep both the mass and the pumping power as low as possible. These are contradictory requirements, since to keep the mass low, the pipes need to be smaller for the fluid to circulate faster, but smaller pipes will mean more pressure loss, and bigger pumps will be needed.
Another point is that the nuclear reactors that are creating the electricity to drive the pumps need to be cooled themselves by radiators. So making the piping too small will eventually lead to added heat loss from the reactors, and will create an ever increasing need for larger radiators!
The Calculations
The equation linking pumping power to fluid flow is very simple:
W=(Q*dp)/n (1)
Where:
W= pumping power (kW)
dp = pressure drop (kPa)
Q = flow = 18 (m^{3}/s)
n = pump efficiency (usually between 0.6 and 0.8)
Unfortunately, it is not solvable at this point because we do not know the pressure drop. All we know is the flow. However, if we know the physical parameters of the system and of the coolant, it is possible to calculate the pressure drop and solve the equation.
The pressure drop in the pipe is calculated by the Darcy–Weisbach equation:
dp = f*(L/D)*((ρ*v^{2})/2) (2)
Where:
dp = pressure drop (Pa)
f = friction factor
L = length of pipe = 500 (m)
D = Diameter of pipe (m)
ρ = density = 1615 (kg/m^{3})
v = average fluid velocity (m/s)
Now we’re missing three numbers: f, D and v!
Here is the time to simplify things a bit, and apply a little intuition to the design process, by choosing the system velocity. To do this, let’s consider the length of the piping circuit: 500m. Applying the standard economical industrial design velocities of about 3m/s, we find that the cycle time will be 500/3 = 166 seconds. Not quite 3 minutes.
That may seem OK at first, but if we multiply it by the mass flow required by the cooling process, 57 tonnes per second, we find a coolant mass of almost 9500 tons! (57 x 166 = 9462)
This is way too heavy. We need to go faster. But how fast? We can use the pump affinity laws to help us decide:
v1/v2 = Q1/Q2 = N1/N2, velocity is proportional to flow and rotation speed (3a)
p1/p2 = (v1/v2)^{2}, pressure is proportional to the square of the velocity (3b)
W1/W2 = (v1/v2)^{3}, power is proportional to the cube of the velocity (3c)
Putting these into a table:
Velocity 
Pressure 
Power 
Coolant mass 
m/s 
ratio 
ratio 
tonnes 
3 
1x 
1x 
9,500 
6 
4x 
8x 
4,750 
9 
9x 
27x 
3,167 
12 
16x 
64x 
2,375 
15 
25x 
125x 
1,900 
20 
44x 
296x 
1,425 
25 
69x 
579x 
1,140 
30 
100x 
1,000x 
950 
40 
178x 
2,370x 
712.5 
50 
278x 
4,630x 
570 
100 
1,111x 
37,037x 
285 
Table 1, pump affinity laws
There are obviously diminishing returns for increasing fluid velocity past a certain point. A velocity of 20 m/s seems like an aggressive, but perhaps feasible, choice, so let’s try it.
At this velocity, 500m / 20 m/s = 25s x 57 tonnes/s = 1,425 tonnes. That’s heavy, but tolerable.
With the fluid velocity known, we can determine the pipe size required to feed one of the two radiators.
Q = v*A (4)
Where:
Q = Flow = 18 (m^{3}/s)
v = velocity = 20 (m/s)
A = Area (m^{2})
We can find the main pipe size via Equation #4:
π(D/2)^{2} = Q/v > D = 2*((18/20)/π)^1/2 = 1.1 m in diameter
We need to also determine the diameter of the smaller pipes. If we divide the flow into 50 smaller loops, the diameter of each small pipe will be about 0.15m.
All we are missing now is (f), the friction factor. Yes, there is an equation for that_{(2)}, specifically for pipe flows — the Haaland friction equation:
f = (1.8*log(((e/D)/3.7)^{1.11} + (6.9/R_{e})))^{2} (5)
Where:
e = pipe roughness factor (m)
R_{e} = Reynolds number (dimensionless)
D = pipe diameter (m)
Roughness factor (e) is about 0.0000015 for the smooth pipe we should be using.
The Reynolds number (R_{e}) is our last equation of the day:
R_{e} = ρvD/μ (6)
Where:
ρ = density = 1615 (kg/m3)
v = average fluid velocity = 20 (m/s)
D = Diameter of pipe = 1.1 (m)
μ = dynamic viscosity = (kg/m*s)(Pa*s)
The last variable, the dynamic viscosity (μ) characterizes the resistance to flow of the fluid. It changes with temperature, but not by much. Here are some values for prospective coolants:
Fluid 
Dynamic Viscosity 
Melting Point 
Boiling Point 
Heat Capacity 
Density 

kg/m*s 
K 
K 
kJ/kg*K 
kg/m3 
Beryllium 
0.0011 
1600 
2700 
1.92 
1616 
Aluminum 
0.0027 
933 
2773 
0.91 
2700 
Lithium 
0.00035 
453 
1573 
4.3 
500 
FLiBe 
0.003 
733 
1703 
2.4 
2300 
Tin 
0.0002 
500 
2875 
1 
6990 
Water 
0.00013 
0 
100 
4.18 
1000 
Silicon 
0.0005 
1687 
3538 
0.71 
2570 
Table 2, fluid properties
Putting it all together, for molten Beryllium at about 2000K, once for the big 1.1m pipe and once for the small 0.15m pipe:
Big Pipe:
6) R_{e} = 1616*20*1.1/0.0011 = 32,320,000
5) f = (1.8*log(((0.0000015/1.1)/3.7)^{1.11} + (6.9/32,320,000)))^{2} = 0.007
2) dp = 0,007*(100/1.1)*((1615*20^{2})/2) = 210,000 Pa = 211 kPa
1) W = (18*211)/0.7 = 5,400 kW = 5.4 MW
Small Pipe:
6) R_{e}= 1615*20*0.015/0.0011 = 4,407,000
5) f = (1.8*log(((0.0000015/0.15)/3.7)^{1.11} + (6.9/4,407,000)))^{2} =0.0096
2) dp = 0,0096*(400/0,15)*((1615*20^{2})/2) = 8,325,000 Pa = 8325 kPa
1) W = (18*8325)/0.7 = 214,000 kW = 214 MW
Adding the two, the pumping power is 219 MW for one radiator, or 430 MW for both. Put in a more familiar unit, it’s a bit less than 600,000 horsepower, or about 5000 car engines.
The Analysis
That’s a lot of power, but it’s far from an impossible number. If we search for a present time analog installation, we find the example of the Bath County Pumping Storage Station, in the United States, that is equipped with six 500 MW Francis pump turbines and has been operating since 1985.
Looking at the results for the two pipe sizes, it’s clear that most of the pumping power is required for the small pipes. It’s much harder to pump through the 50 small pipes than through the single large pipe, since there is much more surface for the same amount of liquid going through. So it might make sense to change the velocities in the pipes, making the fluid go faster in the large pipes and slower in the small pipes. It could also help to create an arrangement with the shortest length of pipe possible.
The pressure at the outlet of the pump will be 8,500 kPa. This is harder to put in familiar terms: to put it terms of usage, it’s in the range of the oil pressures that are used in hydraulic machinery, such as backhoes and mechanical shovels. It’s certainly a high pressure, but not outlandishly so.
Depending on the technology, the nuclear reactor will weigh from 1 to 4 kg/kW_{(4)}, so we can expect the reactor to drive these pumps to weigh somewhere between 450 tonnes and 1,800 tonnes. Reducing the speed from 20 m/s to 15 m/s would reduce the power requirements by almost half, for an added mass of about 500 tonnes of coolant. So for a reactor weighing only 1 kg/kw, the reduction would be about 225 tonnes, for a net gain in mass of 275 tonnes. That doesn’t help. But for a reactor at the higher end of the range, the reduction in the mass of the reactor might be 900 tonnes, for a net reduction of 400 tonnes – a significant improvement overall.
Not quite. As a bonus to anyone who read this far, a few more equations, concerning the strength of the pipes.
Hoop Stress in Pipes
Although a modern, detailed analysis of stresses in pipe walls requires sophisticated finite element modeling, it’s good to know that a mathematical model based on a beer keg can do almost as well.(5)
The hoop stress is the force exerted on the pipe wall by the pressure in the pipe. The relationship between the two is the following equation:
σ = Pr/t (7)
Where:
σ = Hoop stress (Pa)
P = Internal pressure (Pa)
R = Interior pipe radius (m)
t = Pipe wall thickness (m)
The relationship with the material of the pipe is :
S/SF = σ (8)
Where
S = Material tensile strength (Pa)
SF = Safety Factor
σ = the allowable hoop stress (Pa)
Here are some possible materials for the piping:
Material 
Tensile strength (MPa) 
Aluminium Alloys 
35 – 500 
Aluminium oxide crystals 
400 
Kevlar cloth (combine with matrix material) 
3,620 
Tungsten 
1,510 
Tantalum 
170 
Titanium 
830 
Steel alloys 
250 – 1,600 
Carbon fiber composite 
1600 – 4,000 
Carboncarbon composite 
300 – 400 
Carbon nanotubes 
7,000 
Graphene 
130,000 
Diamond 
1,600 
Table 3, Piping materials
So, choosing a middle of the range carbon carbon composite at 360 MPa, and a factor of safety of three, because… it divides nicely, we will find:
S/SF = σ = 360 MPa / 3 = 120 MPa or 120,000 kPa
For our big pipe, 1.1m in diameter with a 0.55 m radius, the wall thickness will be:
t = Pr/σ = 8536*0.55 / 120,000 = 0.04m, or about 4 cm thick
And for the small pipes:
t = Pr/σ = 8536*0.075 / 120,000 = 0.005m, or 5 mm thick
The volume of the big pipe will be about:
0.04m x 1.1m x π x 100 = 4.4m^{3}.
And the volume of the small pipes will be:
0.005m x 0.15m x π x 50 x 200 = 23.6 m^{3} for a total of 28 m^{3}.
For a density of 1,500 kg/m^{3}, the net piping mass will be 42,000 kg, or 42 tonnes.
All together, the total mass of the system might be about 1600 tonnes, excluding accessories such as pumps and the radiation shield. Fin space between the pipes and a corrosion coating might add some mass to this as well.
As a final thought: since all the pressure is lost in the small pipes, with thin walls, why have only a big central pump, forcing us to have very thick walls on the main pipes? Why not have a central pump sized for the pressure drop in the main pipe, 211 kPa, and then distributed pumps, at 8,300 kPa, located just before the small pipes? This arrangement is called distributed compound piping, and is just one of the refinements possible to reduce the mass of the ship.
Tools of the trade
You will find the tables and equations mentioned in this article in the following spreadsheet:
Plumbers Guide to Starships – Liquids
And here is a link to the calculations related to radiators:
Notes
(1) These are the parameters used in the Starship radiator post. They are in the “Tools of the Trade” spreadsheet.
(2) In fact, there are half a dozen or more equations for friction in pipes. Most of them require computers, or at least a lot of patience to solve. Some are only applicable to water. The Haaland equation was developed fairly recently, in 1983. It provides values that are as close to the experimental results as any other solution. After all, the experimental results themselves only have a precision of about 10%. It looks a bit nasty at first glance, but is really easy to put into a spreadsheet. It grows on you.
(3) http://www.redbubble.com/people/debbiechayes/works/6706441allwornout
(4) For 1 kW/kg we should think nuclear lightbulb or fission fragment reactors. For 4 kW/kg we are in the range of conventional reactors with turbines and cooler core.
(5) Valid for thin walled cylinders with no ends. For a cylinder with ends, you must add the axial stress, and for a thick walled cylinder, a radial stress. Since we are designing our ship to be as light as possible, thick walled cylinders are not a required refinement.