Step 4 - Power Production

Fuel Cells

The prices quoted above are for air breathing fuel cells. Multiply this by 2.5 for non-air breathing cells.

"Realistic" Fuel Consumption

There are a number of problems with Fuel Cells as presented by 2300AD. I discovered that reverse engineering of the fuel consumption rates for a number of Ground Vehicles often yielded efficiencies in excess of 100%. Also, I noted that the prices for Fuel Cells in the Naval Architects' Manual was far too high compared with the total price of many vehicles in the Ground Vehicle Guide and elsewhere.

Given these inconsistencies, I decided to research modern day fuel cells and extrapolate more "realistic" models for 2300AD. The following is the result of this research.

A standard Fuel Cell uses aqueous Potassium Hydroxide as an electrolyte and has an operating temperature of about 50° to 100° Celsius. The modern day version of this fuel cell has an efficiency of 70% ; I assume that by 2300AD this will be about 85% ( I assume that there will always be some waste heat produced ). NASA currently employs this type of fuel cell on the Space Shuttle, where it produces 12 kW of power.

To calculate the maximum possible ( i.e. 100% efficiency ) output of such a device requires us to understand the process involved. We are in effect combining oxygen with hydrogen by reverse electrolysis and generating power.

The electrochemical reaction at the cell's anode is 2H₂ + 4OH^- » 4H₂O + 4e^-. At the cathode, the reaction is O₂ + 2H₂O + 4e^- » 4OH^-. Thus, we can see that for every 2H₂ we cause 4e^- ( four electrons ) to flow around our circuit. Each electron is equivalent to 1.6022 x 10^-19 Joules of energy ( 1 electron Volt, or 1ev ). We can see from the electrochemical reactions that for every mole of Hydrogen we cause one mole of electrons to flow around our circuit. I mole of electrons is equivalent to 6.023 x 10²³ x 1.6022 x 10^-19 or 96500 Joules of energy. This is Faraday's constant, the energy equivalent of 1 mole of electrons, which is more accurately quoted as 96487 Joules. Now, one mole of Hydrogen is approximately 1 gram, so for 70 kg of Hydrogen that is 6,754,090,000 Joules. ( Why 70 kg ? Because 70 kg of liquid Hydrogen represents one cubic metre of volume, which is important for later calculations ). This is 1.876 Megajoules for 1 hour.

At the 70% efficiency of modern ( 1990s ) systems, that is 1.3 Megajoules for 1 hour ( 4,727,863,000 J ). With our assumed 85% efficiency for 2300AD, that is 1.59 Megajoules for 1 hour ( 5,740,976,500 J ).

This means that 44kg or 0.63 cubic metres of liquid hydrogen will produce 1 megajoule per second ( 1 megawatt ) for one hour. This is an important result. It also differs from the figures quoted in the Ground Vehicle Guide and in the Naval Architects' Manual, though it is closer to the former.

The product of a fuel cell reaction is ordinary water. Water occupies 1 cubic metre for 1000 kg. Of this, 111.1 kg is Hydrogen and 888.8 kg is Oxygen. Bearing this 1:9 ratio in mind, we can determine that 70 kg of Hydrogen will react with 560 kg of Oxygen in our fuel cell, or ( for 1MW ) 44 kg will react with 396 kg of Oxygen. For an air-breathing fuel cell operating in a standard atmosphere, this oxygen need not be carried. For a starship or vehicle operating without external oxygen, it must be carried as well.

1142 kg of Liquid Oxygen occupies 1 cubic metre. Therefore 396 kg of Oxygen occupies 0.35 cubic metres. When added to the Liquid Hydrogen volume for 44 kg ( 0.63 cubic metres ), we derive that for a 1 MW fuel cell, 440 kg or 0.98 cubic metres of fuel is consumed every hour. From this we can extrapolate that 1.65 cubic metres of fuel ( the standard volume listed in the Naval Architects' Manual ) is 740 kg, not 1000 kg as listed. The designer of Star Cruiser has made a simple error - he has taken the volume of 70 kg Hydrogen and 1142 kg of Oxygen ( 2 cubic metres ) and derived the volume for 1000 kg as 1.65 cubic metres, without realising that not of all of the Oxygen in this volume will be combined with the Hydrogen in the same volume. The Star Cruiser ratio of Hydrogen to Oxygen is therefore incorrect and should be 1:9 not 1:16.

For non-air breathing fuel cells this is 0.44 tons or 0.98 cubic metres per hour per megawatt of output, or 73.92 tons or 164.64 cubic metres per week per megawatt of output.

In the Ground Vehicle Guide, it suggests ~3 kg of Hydrogen powers a 0.1 MW fuel cell for an hour. From the above, we can see that this would be 4.4 kg of Hydrogen. This reduces the endurances of fuel cell powered vehicles in the GVG to ( 3 / 4.4 = ) 68% of their listed values.

Fuel Cell - Fuel Summary

Table Notes -

1. Figures in brackets are per week for the purposes of Star Cruiser.

Ice Refuelling

A number of starships can use ice found in space as a source of fuel. There are few extra factors to consider here.

• First of all, ice is less dense than water ( 1000 kg of ice = 1.09 m ³ thus 1 m³ of ice = 917.4 kg ) and so filling your fuel tanks full of ice is not the same as filling them full of water.
• Secondly, ice must be melted before electrolysis can take place. This requires about 334 Joules for 18g of ice ( this is the energy needed to turn ice at 0° Celsius to water at the same temperature ), which is 18,555,556 Joules to melt 1 m³. This is about 0.005MW of power per hour for 1 m³. Our average 1 MW solar panel could thus melt 200 m³ of ice per hour if all power was dedicated to this activity. This would yield 183.4 m³ of water. Bearing in mind that to fill our fuel tanks we need only 45% of their capacity as water ( see Closed System Fuel Cells, above ), we can use these figures to determine the amount of ice we will need and how long it will take to get the appropriate amount of water prior to Electrolysis. See below.

Closed System Fuel Cells

As mentioned before, 1000 kg of water occupies 1 cubic metre. Some ships store the water generated by Fuel Cells so that they can "crack" it back into Hydrogen and Oxygen with solar cells upon arrival in a system. This is known as a "closed system".

440 kg of Liquid Hydrogen and Liquid Oxygen occupies 0.98 cubic metres. The same mass of water occupies 0.44 cubic metres. Thus, fuel tanks are actually 0.44 / 0.98 = 45% full when all fuel is exhausted. In addition, the mass of a closed fuel cell system starship does not vary with power plant fuel expenditure, which is useful for accurate warp speed / travel time calculations.

Upon arrival in a system, the starship unfurls its solar array whilst discharging its stutterwarp drive, and begins electrolysis to convert the water back into Hydrogen and Oxygen. The target is to completely process all of the water within the 40 hours it takes to discharge the stutterwarp drive.

Electrolysis

One Faraday ( 96485 Coulombs per mole, and effectively 96485 Joules ) represents one mole of electrons. Therefore, one Faraday will join with one mole of hydrogen ions in solution to provide 0.5 moles of H₂ gas at the electrode in electrolysis.

Taking the molar mass of one mole of H₂ to be 2.016 grams, this means that 96485 Joules liberates 1.008 grams of Hydrogen gas.

1 MW for one hour is 3,600,000,000 Joules. That is 37311.5 Faradays, and thus 37311.5 x 1.008 = 37610.0 grams of Hydrogen. 1 m³ of Liquid Hydrogen is 70 kg, so 1 MW for one hour is 0.5373 m³ of Liquid Hydrogen.

Hydrogen is ¹/₉^th of water by mass, so for this 37610.0 grams there must be 8 parts Oxygen or 300880 grams. This is 0.2635 m³ given that 1 m³ of Liquid Oxygen is 1142 kg in mass.

The amount of water processed then must be 37610 + 300880 = 338490 grams, or 0.33849 m³.

Therefore, to process 1 m³ of Water would take 2.9543 hours with a 1 MW power plant. It produces 2.9543 x 0.5373 = 1.5873 m³ of Liquid Hydrogen and 2.9543 * 0.2635 = 0.7785 m³ of Liquid Oxygen for a total of 2.3658 m³ of products.

We can work out from this how many solar cells a starship needs to crack fuel in the 40 hours taken to discharge its stutterwarp drive. We know that around a star like our sun at a range of 1 AU, 100 m² of solar panels provides 0.1 MW. We also know that to fill the fuel tanks with Liquid Hydrogen and Liquid Oxygen requires 45% of their combined capacity as water.

• Let t = Total fuel volume required ( m³ of Hydrogen and Oxygen )
• Volume of water = 0.45t ( i.e. 45% of the total capacity )
• Number of hours to process at 1 MW = 2.95 * 0.45t
• Therefore, number of hours to process at 1 MW = 1.3275t
• Number of 40 hour periods to process = 1.3275t / 40
• Therefore, number of 40 hour periods to process = 0.0331875t
• Each 40 hour period requires 10 standard 100 m² solar arrays to generate 1 MW at 1AU from a Sol type star.
• Therefore, number of solar panels required = 10 * 0.0331875t
• Or, Number of solar panels required = 0.331875t

Compression & Liquefaction

Of course, once you have got Hydrogen and Oxygen gas, you need to cool and liquefy them for storage purposes. The easiest method to achieve this is to take advantage of the Joule-Thomson effect. Basically, if a gas is allowed to expand very rapidly, it loses heat energy. Do this enough times in a slow compression-rapid expansion cycle, and the gas temperature will drop below the critical point where the gas can be compressed into a liquid.

Now, while the expansion of the gas is effectively free ( the gas itself does the work, which is why it loses energy and cools ), the compression is not, and there is an energy overhead for the compression.

In order to derive some sort of power consumption figure for this process I attempted to calculate the energy required for each cycle in the process. I defined my initial environment to be Hydrogen gas at 1 atmosphere pressure and a temperature of 293 Kelvin, with a volume of 1 m³. I envisaged adiabatic ( i.e. with no change in the heat energy of the gas ) expansion, followed by isothermal ( i.e. without change in temperature ) compression back to the original volume of 1 m³.

Given that T represents temperature in Kelvin, p represents pressure in Pascals, and V is volume in cubic metres, then -

• For adiabatic expansion, the following relationship holds true - T₁V₁^(y-1) = T₂V₂^(y-1). For a monatomic gas, y = 1.67. For a diatomic gas such as hydrogen, y = 1.4. For polyatomic gases, y = 1.33.
• For isothermal compression, the following relationship holds true - p₁V₁ = p₂V₂.

Using these two relationships, we can attempt to calculate the work done in repeatedly compressing the gas isothermally until the gas will liquify. I have performed the calculations for the first cycle and then omitted then from subsequent cycles for clarity.

1. Initial conditions 101,325 Pascals, 293 Kelvin, 1 m³.

Section Incomplete
Thanks are due to Nick Munn and others on the Traveller Mailing List, who provided most of this information during a rather lengthy discussion on fuel related issues.

Graphite Nanofibre Cartridges

This is a relatively new technology developed for hydrogen cars in the 1990s. I first uncovered it sometime ago amongst the pages of New Scientist magazine. Since then, I understand that the at least one major car manufacturer has put the technology into practice.

Graphite nanofibres are narrow "stacks" of graphite layers, each layer being one carbon atom thick. A typical fibre is between 5 and 100 millimetres in "height" and between 5 and 100 nanometres (nm) in diameter. The layers are formed from hexagonal arrangements of Carbon atoms. Between each layer there is a gap of 0.335 nm.

When a hydrogen atom is pumped into an array of such fibres at room temperature and under a pressure of 120 atmospheres, it interacts with the electrons in the graphite and "shrinks" from a radius of 0.26 nm to 0.064 nm. As a result of this interaction, it becomes possible to store vast amounts of hydrogen between the layers of graphite in the nanofibre. In order to prevent the hydrogen from escaping, the pressure is maintained at 40 atmospheres.

In practice, upto a maximum of 75% of the mass of a saturated graphite nanofibre cartridge is hydrogen. This means that it is possible to achieve storage densities 95 times that of liquid hydrogen at room temperature !!! ( Consider that graphite has a density of 2.22 g/cc. 1 m³ of graphite is therefore 2220 kg. If this accounts for 25% of the mass of a saturated cartridge, then the remaining 75% or 6660 kg must be hydrogen. 6660 kg of hydrogen per cubic metre of graphite is 95 times the density of liquid hydrogen, which is only 70 kg per cubic metre ).

As an added bonus, the gaps between the graphite layers are too narrow to permit oxygen atoms to react with the hydrogen, thus vastly reducing the risk of explosion.

This technology has dramatic consequences. It becomes possible to store vast amounts of hydrogen safely in a small volume, with no requirement for liquefaction or cooling, although obviously the hydrogen has the same mass. Please note that this technology is hydrogen specific and does not work for oxygen, which must still be liquefied in the normal way.

In deriving the figures below I have assumed that a cartridge has a small overhead for protection, compressors, and so on. This reduces the density of the hydrogen in the cartridge to ~86 times that of liquid hydrogen. In addition, these cartridges have a power requirement of ???? when they are charged with hydrogen, a process which takes 20 minutes per cubic metre.

These figures may be scaled as desired. For example, 1.50 cubic metres would be 3750 kg unloaded and so forth.

Magnetohydrodynamic (MHD) and Magnetoplasmadynamic (MPD) turbines

MHD Turbines generate power by taking advantage of a scientific phenomenon called the Hall Effect. In this, a high temperature plasma ( Ionised Hydrogen gas at about 10,000 Kelvin ) is forced between two electrodes in the presence of a strong magnetic field ( about 4.5 to 6 Tesla ). As the plasma interacts with the magnetic field, an electrical potential develops between the two electrodes, generating power.

It is assumed that in 2300AD hydrogen gas is burnt in the presence of oxygen to produce the necessary ionized plasma, which is then accelerated through the magnetic field.

MHD Fuel Requirements

I have based my MHD Fuel Requirements on the fuel requirements for Fuel Cells, the only real difference being the efficiency of operation.

You may notice that the derived figures for Fuel Cells are pretty much the same as those quoted in the Naval Architect's Manual. Alternatively, you may have applied Errata from GDW to your copy of NAM, and swapped the fuel consumption of MHDs with Fuel Cells, and now be puzzled as to why the above results differ from your modified NAM.

I think that my calculations go some way to disprove the GDW Errata ; NAM was right to start with, and the Errata is itself incorrect. Given that this is the case, the original NAM figures of 0.60 tons of fuel per hour ( 100 tons per week ) must apply to MHDs and not Fuel Cells, and the original Fuel Cell consumption rates of 0.45 tons per hour ( 75 tons per week ) can be reinstated.

This makes sense to me ; MHDs are less efficient than fuel cells ( normally about 60% efficient compared to 85% for our proposed fuel cells ), and the difference in fuel consumption can be explained away nicely by this difference in efficiency. ( 85% / 60% * 0.44 tons/MW/hr = 0.6233, which is very close to the 0.60 quoted in uncorrected NAM ).

Magnetohydrodynamic Power Plants - Fuel Summary

Fission

The amount of fuel consumed by a fission reactor is actually very small. 1.498 x 10^-5 grams s^-1 of Uranium-235 produces 1 Mj per second ( 1 Mw ) of thermal power ; this means 1.294 grams per day produces 0.33 Mw of electrical power if we assume 1/3 efficiency converting thermal energy to electrical power. Assuming 3.3% of fuel is Uranium-235 and the rest is Uranium-238, then this is 39.21 grams of fuel per day. Even over a year, the total consumption is only 42.97 kg ( about 0.002 cubic metres, given a density of 19 tons per cubic metre for Uranium ).

This fuel is considered to be built into the reactor, which is assumed to have a life-span of perhaps 5 to 10 years.

Fusion

Basically, fusion reactors generate power by colliding atomic nuclei together so that they form heavier elements with a release of energy. Unfortunately, nuclei have strong positive electromagnetic charges, and so they naturally repel one another. To get around this, you have to give the nuclei enough kinetic energy to slam into one another, or increase the density of the medium by implosion to get the nuclei close enough to fuse. The former method is usually implemented by using very high temperature plasmas contained in magnetic bottles generated by superconducting electromagnets. The latter method involves small pellets of fuel compressed to enormous densities by high energy lasers.

In both cases, the fuel used is usually a combination of isotopes of hydrogen. Hydrogen has three known isotopes -

• Protium, the most common isotope, is also the most abundant element in the universe. It consists of a single proton orbited by a single electron. Protium is stable and does not decay radioactively.
• Deuterium, which occurs naturally, accounts for ¹/₆₇₀₀th of all the Hydrogen in the universe. Deuterium consists of one proton, one neutron and one electron. Deuterium is also stable and does not decay radioactively.
• Tritium, which does not usually occur naturally, consists of one proton, two neutrons and one electron. Tritium is radioactive, and decays into Helium-3 with a half life of 12.3 years. Because of this short half life, Tritium is almost never found in nature.

Below are some of the more common fusion reactions, culled from the sci.physics Fusion FAQ. Here, D stands for Deuterium, T stands for Tritium, p for proton and n for neutron. Reaction products are listed with their relative ratios and energies in mega-electionvolts ( MeV ).

---

D+D   -> T (1.01 MeV) + p (3.02 MeV) (50%)   
      -> He3 (0.82 MeV) + n (2.45 MeV) (50%)  <- most abundant fuel
      -> He4 + about 20 MeV of gamma rays (about 0.0001%; depends
                                           somewhat on temperature.)
      (most other low-probability branches are omitted below)
D+T   -> He4 (3.5 MeV) + n (14.1 MeV)  <-easiest to achieve
D+He3 -> He4 (3.6 MeV) + p (14.7 MeV)  <-easiest aneutronic reaction
                                     "aneutronic" is explained below.
T+T   -> He4 + 2n + 11.3 MeV
He3+T -> He4 + p + n + 12.1 MeV (51%)
      -> He4 (4.8) + D (9.5) (43%)
      -> He4 (0.5) + n (1.9) + p (11.9) (6%)  <- via He5 decay
                                    
p+Li6 -> He4 (1.7) + He3 (2.3)      <- another aneutronic reaction
p+Li7 -> 2 He4 + 17.3 MeV (20%)
      -> Be7 + n -1.6 MeV (80%)     <- endothermic, not good.
D+Li6 -> 2He4 + 22.4 MeV            <- also aneutronic, but you 
                                              get D-D reactions too.
p+B11 -> 3 He4 + 8.7 MeV <- harder to do, but more energy than p+Li6
n+Li6 -> He4 (2.1) + T (2.7)        <- this can convert n's to T's
n+Li7 -> He4 + T + n - some energy

---

The most promising cycle for conventional fusion is the Deuterium-Tritium reaction, since it is the easiest reaction to achieve.

Fusion Plant Summary

The following is based upon the table in NAM. Note that the smallest Fusion plant in NAM is the 150 MW plant, the largest is 300MW. However, a number of people have reversed engineered the starship designs in Ships of the French Arm and other sources, and there is evidence to suggest that plants outside of this range exist. I would suggest a range of 30 MW to 500 MW is acceptable.

Tech Level	Volume in m3/ MW output	Mass in tons / m3	Price in MLv / MW output3
Any	32.9	1	0.52

Muonic Fusion

Until recently, I always assumed that all fusion reactors in 2300AD would use one of these traditional methods, neither of which seemed particularly elegant. Then, whilst reading New Scientist magazine, I discovered a third option - Muon Catalysed Fusion.

Muons are exotic short-lived subatomic particles with the same electrical charge as an electron but over 200 times the mass. They can form "muonic" atoms with nuclei in much the same way as electrons combine with nuclei. However, since a muon is so much more massive than an electron, the radius of its orbit about the nucleus is much smaller. The result is that a muonic atom is electrically neutral but tiny compared to the more normal atom with electrons. That reduction in size, combined with electrical neutrality, means that muonic atoms can be used to bring nuclei exceedingly close together, close enough to fuse.

A muon has a life time of about 2.2 microseconds. In that time, the muon must catalyse enough fusion events to break even. Creation a muon is expensive in terms of energy.

HPG Design

Section not complete

---

dµ + p -> pdµ -> 3He + µ + gamma rays + 5.5 MeV
dµ + d -> ddµ -> 3He + n + µ + 3.3 MeV
tµ + p -> ptµ -> 3He + n + µ + 5 MeV
tµ + d -> dtµ -> 4He + n + µ + 17.6 MeV
tµ + t -> ttµ -> 4He + 2n + µ + 5 MeV

---

Solar Panels

Note that the price is per square metre, not cubic metre ! In addition, note that the volume is based on a folding array which can be deployed when needed. For a fixed array, the volume is 0.007 cubic metres per square metre and can be external to the hull. In this case, the solar panels are supported on arms or pylons away from the hull.

Solar Panel Performance

Solar panel performance depends upon the luminosity of the local star, the distance of the panel from the star, ,the area of the solar panel and its energy conversion efficiency.

Take Sol for an example. At the surface, Sol emits the equivalent of 3.9 x 10²⁶ Watts per m². At 1 AU, this has decreased to about 1400 W / m². Therefore, if we had a 1000 m² solar cell array working at 100% efficiency, we could generate 1.4 MW at 1AU. Unfortunately, nothing is ever 100% efficient, and so 1 MW per 1000 m² ( or about 70% efficiency ) seems reasonable.

The following table summarises the incident radiation per m² at various distances from Sol. It is derived from the equation L / ( 4 r² ) where L = 3.9 x 10²⁶ Watts per m² and r is the distance from Sol in metres. ( 1AU is taken to be 149,597,870 kilometres ).

This table is a little more accurate and realistic than the usual "Inner-Habitable-Outer Zone" rule applied to Solar Cells. Some notes apply as follows -

1. The first entry ( 0.07 AU ) is the stutterwarp discharge distance for Sol of 11 million kilometres. Of course, the stutterwarp distance will vary for other stars.
2. The final column in the above table gives the Megawatt output per 100 m² array of solar panels. This is correct assuming about 70% efficiency.
3. For stars other than Sol, multiply the output at a given distance by the ratio of the star's luminosity to that of Sol - i.e. a star with a luminosity twice that of Sol would empower a solar panel to 2 x 1.08 = 2.16 MW at 0.30 AU.
4. Please note that some browsers do not render superscript text properly in the above table. If your browser can't render ^... inside a <font size="X"> ... <font> block, then you will see x1011 instead of x10¹¹.

Batteries

Batteries provide auxiliary power for most vehicles, but are often the primary source for unmanned remotes / drones. They are also used extensively by personal computers and cyberdecks.

There was a brief treatment of batteries in an article in Challenge Magazine entitled "Star Cruiser Power" by C.W.Hess, and personal laser weapons are assumed to be powered by "fast discharge liquid metallic suspension (FD-LMS)" cells. Other than that, there is little mention of batteries in 2300AD, presumably because of the widespread use of fuel cells and equivalent technology.

According to New Scientist magazine (14th June 1997 pg22) the latest zinc-air powered cells store 0.65 MJ per kg. I have assumed that by 2300AD, cells will have storage densities of perhaps 0.80 MJ per kg. This increase is in line with my general assumptions of improvements over the modern era as applied to Fuel Cells and other power sources. By way of comparison, primitive Lead-Acid cells store about 0.10 MJ per kg.

Assuming that such cells are based on Zinc-Air cells, and given a density for Zinc of about 7000 kg per cubic metre, then we have a volume of about 142 cm³ or so as a minimum and probably 160 cm³ in reality.

Fast Discharge Liquid Metallic Suspension (FD-LMS) Cells

The disposable power cells for small arm lasers in the Adventurers' Guide are 1 kg in mass, have no defined volume and cost Lv5. Two types of cell are mentioned, one which stores 7 Mj and one which stores 5 Mj. These cells are designed to be used to provide a set number of pulses in one hundredth of a second, rather than sustained power output. Typically, they can provide a pulse of upto 0.4 Mj over this timescale.

Small arm lasers seem to be remarkably efficient. The P3 pistol for example has a 7 Mj cell, and a delivered pulse of 0.2 Mj. The "magazine capacity" is listed as 35 shots, so the input energy per shot is likewise 0.2 Mj. In other words, the laser is 100% efficient at converting the electrical energy from the FD-LMS cell into the excited photons of the laser beam.

I find this unlikely. Personally, I rule that the delivered output energy from a laser is one third of the input energy (the same rule applies to ship mounted lasers too), and multiply the stored energy in FD-LMS cells by three. Thus the P-3 now uses a 21 Mj cell, requires 0.6 Mj per shot to fire the laser, and delivers 0.2 Mj of energy per shot. 21 / 0.6 is still 35 shots, so there is no difference to the statistics or operation of the weapon.

The consequence of all this is that I have 15 Mj and 21 Mj FD-LMS cells, for small arm lasers, rather than 5 Mj and 7 Mj. Note that these energy densities are much higher than the 0.80 MJ per kg of Zinc-Air cells ( 15 Mj is 18.75 times 0.8 ) but this is consistent with the fact that FD-LMS are designed to provide pulses and Zinc-Air cells are designed to provide a sustained output. In addition, FD-LMS are limited to one use whereas I believe that Zinc-Air cells can be recharged / reused.

Storage	Volume	Mass in kg	Price
15 Mj	??? cm3	1	5
21 Mj	??? cm3	1	5

C. W. Hess' Batteries

In an article in Challenge Magazine on Starship Power systems, C. W. Hess described two further sorts of battery. These are a "Quick Charge / Quick Drain" type and a "Slow Charge / Slow Drain" type. These are both very low power when compared with the Zinc Air cells above or either of the FD-LMS cells ( the original 2300AD type or the x3 capacity type proposed by me ), and I have not included them in this article for this reason.

Radioisotope Thermal Generator ( RTG )

RTGs are a very low yield power source which use a radioactive isotope to produce heat which is converted into electrical power. They have been typically employed on space probes such as NASA's Voyager I and II, because of their reliability and long lifespan. The following device has 2.6 kg of Plutonium-238 embedded within it and will generate an initial output of just 75 watts. Since this isotope has a half life of 87.7 years, power output will be reduced to 37.5 watts after this time has elapsed, then 18.75 watts after another 87.7 years and so on.

Tech Level	Volume in m3/ MW output	Mass in tons / m3	Price / m3
Any	0.2	0.02	Lv 25000

Homopolar Generators

A homopolar generator (HPG) consists of a rapid spinning flywheel in a magnetic field. They can be used to provide continuous electrical power ; However, the usual application is to accumulate energy over a short period of time and then discharge it in a single, very powerful pulse.

To generate such a pulse, the flywheel is "pumped" to many thousands of revolutions per minute with the magnetic field disabled. Then the magnetic field is switched on, acting like a brake on the flywheel. The energy stored in the rapidly rotating flywheel is converted into a pulse of electrical energy.

This effect was first observed by Michael Faraday, hence the name Faraday Disk is also used sometimes to refer to the HPG.

Modern HPGs can produce instantaneous outputs of upto 500 Mj - that's half a gigajoule of energy !

This kind of pulse generation is obviously suited to laser weapons and other high power devices. A homopolar generator is implied in the descriptions of small arm laser weapons in 2300AD, and is assumed to be present in vehicle and starship mounted laser weapons and particle accelerators.

HPG Design

Section not complete

---

---

This page was last updated on the 24^th July 1998.

This page is © 1997,1998 Andy Brick except where components are already copyright / trademarked by others in which case their use is not intended as a challenge to such ownership.