Section 2 Technology
Section 2.2 The Matter Antimatter Reaction Assembly
The warp engines are fuelled by plasmas, energised from energy created in matter antimatter reactions in the warp core. The fuel for these reactions is stored in the primary deuterium tank in the ship’s stardrive section. The tanks are loaded with deuterium at 13.8 K (Sternbach and Okuda 1991) which is 4.57 K below the freezing point of deuterium. (18.37 K). The supply is held in compartmentalised tanks within the main body tank to prevent loss due to damage. The total internal volume is some 63,200 m3 but is more normally loaded to a volume of 62,500 m3. These tanks have a rated loss below 0.0002 kg/day.
Though we know the volume, for matter antimatter reactions we need an approximation of the total mass. Given that the deuterium is in a slush form below its freezing point we can use an approximation of its density as:
0.1967 gram per cubic centimetre
For the creation of energies the matter reactant is forced in to the core through an injector from deck 30 (Galaxy Class) to meet in the reaction chamber with antimatter (anti-deuterium) from deck 42. The anti-deuterium is stored in thirty 100m3 pods giving a total storage capacity of 3,0000m3.
In the core dilithium filters anti-deuterium and deuterium to create a maximised reaction. When bringing the reaction chamber up to levels for warp flight a ratio of reactant (matter) to reactant (antimatter) is about 25:1. Though one would expect the annihilation ratio to be 1:1 it is likely that there is some degree of matter loss or matter destruction that is not involved in the reaction. When breaking the warp one limit this loss drops to 10:1. the matter antimatter ratio changes to increasingly more efficient values up to warp 8 where the ratio stabilises at 1:1 and remains at this level for all speeds above warp 8. Engine efficiency of the starship and the resultant energies from matter annihilation is very high of the order of 97%. In 2370 Chief Engineer Geordie LaForge increased the power conversion levels to 97.2%
Galaxy Class fuel storage for three year mission disregarding refueling, bussard collector usage or antimatter creation using onboard antimatter generators.
3000 m3 of antimatter and
62,500 m3 of deuterium
Given
a loss rate of 0.00002 kg/day of deuterium over the 3 year mission
365.256*3
= 1095.768
1095.768 * 0.00002 = 0.0219 kg
For or our hypothetical
situation we can assume anti-matter to be the limiting factor on mission length.
The Galaxy Class project was designed to have an operational life of three years
at warp six before fuel exhaustion. This means that after three years the fuel
will be consumed, we interpret this, given the nature of reactions, to mean that
the antimatter will be depleted first with the possibility of redundant
deuterium left.
To calculate the energy
released during the mission aiming for fuel exhaustion in three years we must
estimate fuel mass.
Total mass of deuterium:
62,500,000,000 * 0.1967 = 12,293,750,000 grams
12,293,750 Kg
minus the loss over the period:
12,293,750Kg – 0.0219Kg = 12,293,749.9781 Kg
Given that there are
86,164.0905 (seconds) * 1095.768 (days)
2,265,980,474.856096 seconds in the three-year period
Matter reacting is: 2 *(12293749.9781) = 24,587,499.9562 Kg
(Assuming 1 part matter 1 part antimatter. As we have discussed this is not necessarily so and there will be deuterium waste, additional matter that is not annihilated. However, given that we started with such an excess of deuterium and that the level of reactants must be lower than the ratio 10:1, we can ignore the imbalance, also as discussed below should deuterium run low, there exists a supply of liquid deuterium approximating, 297,600 Kg [9,300*32] of liquid fuel that can be fed into the reactor)
Also note: there are thirty two cryo tanks in the saucer section module. these function as auxiliary fuel supplies, however the fuel is not in the same physical state. The auxiliary tanks store deuterium in a liquid state. The Volume of each tank is 113 cubic metres and each is capable of storing 9.3 metric tonnes of fuel. This means that the density in these pods must be far lower than in the PDT.
Specifically
9,300 Kg / 113 =
82.3 kg/m3 or 0.0823 gram per cubic centimetre
Using the Matter from the Primary fuel supply only we can calculate the energy output of the warp core.
By
E=mc2
E(joules)
= 24,587,499.9562 (kg)
* (299 792 500 )2 (ms-1)
E(joules)
= 2,209,814,910,958,998,089,136,250
Over
the duration of the mission this gives
2,209,814,910,958,998,089,136,250
(joules) / 2,265,980,474.856096 (seconds)
975,213,571,113,994.3617
joules per second
975.214
Terrawatts
Given
the loss of energy of 2.8% (100-97.2 efficiency) this gives:
A
loss of 27.305992 terrawatts over the mission duration:
947.908
Terrawatts (3 d.p.) Total useable power created in the M/ARA.
It
could be argued that an integral is required to take account of the effect
reduced mass has on the ship, in that as the mission continues the ship will
become lighter and more easily propelled. We assume that the mass of fuel is
negligible compared to the overall space frame and that the SIF field reduces
apparent starship mass to avoid relativistic effects.
In
comparison to the energy released in the M/ARA the systems on the former
Cardassian station DS9 consists of six fusion reactors supplies an energised
plasma to the station. The maximum
output for all these working in concert is 790 terrawatts, a value
roughly comparable to the output of a M/ARA on a Galaxy Class starship. Though
the Deep Space Nine core is many times larger, the values represent the
efficiency of the two systems with matter energy conversion being far higher in
a Galaxy Class ship.
Energy
Requirements for Deep Space Operation
For
a standard three year mission, many other systems would need considerable energy
including environmental and replicator systems. However, in contrast the ship
would very likely drop from warp on several occasions for extended periods of
time for this operation. We assume that our vessel is running in a form of grey-mode for a three year flight where energy requirements are for propulsion only.
We hope that this in someone balances with the environmental system
requirements:
For
warp flight we must consider the energy required for the main deflector. Even if
all other systems are minimal the main deflector must continually be powered or
else the ship could suffer a catastrophic failure
80%
normal output = 614.4MW
During
normal space flight one field generator must be active in each hull. Each
generator consists of a cluster of twenty 12 MW graviton generators. Therefore
we must allow power for 40 *12MW generators. This is 480 MW of power. In
addition There are four subspace distortion amplifiers in use cooled by 300,000
mega joule per hour coolant systems. This is 4 * 83.34 Mgawatt = 333.34
Megawatts.
This
means 813.34 Megawatts must be tied into the SIF field.
For
normal cruise functions one generator in each of the three major areas must be
functioning, these include:
Primary
Hull, Secondary Hull and one each for the nacelles. The normal cruise mode
requires the combined generators to output 1152 MW.
To
keep the ship operating we need at least:
614.4
+ 813.34 + 1152 = 2579.74 MW
This
is 2.258 Giga Watts.
This
number is insignificant compared to the energy output of the core.
947,908
Giga watts – 2.258 Giga watts = 947,905.75 Giga watts
For
a three year mission averaging warp 6 velocities until fuel exhaustion 947.905
Terrajoules per second are required.
However:
By
the Warp speed power graph (Sternbach and Okuda 1991pg.55) warp 6 has a power
usage of approximately 3,000,000 mega joules per cochrane.
This
equates to:
392
* 3 terrajoules = 1176 terrajoules
which
matches quite closely with the assumptions of a warp 6 flight for 3 years:
975.211
(terra watts) / 1176 (terrajoules) = 80.6% of the three year mission.
Which
is 2 years 5 months 12 days and 4 hours 45 minutes and 24 seconds.