A Plumber’s Guide to Starships Part 1: Liquids

Part 1 : Liquids in Pipes

Figure 1 – A maintenance robot, hard at work on a leaky radiator section

 

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

Figure 2, a possible Icarus design, showing the main elements of the cooling system

Figure 2, a possible Icarus design, showing the main elements of the cooling system

 

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 m3/s in volume flow, 18 m3/s per side

density of 1,615 kg/m3

 

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 (m3/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*v2)/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/m3)

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 (m3/s)

v = velocity = 20 (m/s)

A = Area (m2)

 

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/Re)))-2                                                                           (5)

Where:

e = pipe roughness factor (m)

Re = 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 (Re) is our last equation of the day:

Re = 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) Re = 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*202)/2) = 210,000 Pa   = 211 kPa

1) W = (18*211)/0.7 = 5,400 kW = 5.4 MW

 

Small Pipe:

6) Re= 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*202)/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.

 

Are we done yet?(3)

Are we done yet?

 

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

Hoop stress in pipe

 

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

Carbon-carbon 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 = 4.4m3.

And the volume of the small pipes will be:

0.005m x 0.15m x π x 50 x 200 = 23.6 m3  for a total of 28 m3.

For a density of 1,500 kg/m3, 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:

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/6706441-all-worn-out

(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.