CURRENT TECH: ION THRUSTER
Current Leading Technology: Ion Thrusters
A typical ion thruster converts gas (i.e., argon, xenon, or hydrogen) into superheated plasma. To expel high speed ions out of exhaust, a magnetic nozzle directs the ion motion into useful linear momentum. Ions accelerate to speeds on the order of 50,000 meters per second (about .000167 c). While this exhaust velocity far exceeds speeds achieved by exhaust particles from chemical fueled vehicles/s, it is not nearly enough to achieve gforce for a interplanetary range.However, an ion thruster might be a good way to inject ions into Thought Experiment's (TE's) onboard particle accelerator propulsion system. The best ion thruster might be the VASIMR plasma drive, now under development.
Franklin ChangDiaz 
Dr. ChangDiaz explains the plasma drive: “...rocket engine of the future. As plasma is released through an exhaust nozzle, it creates the rocket effect and pushes the engine (in opposite direction). ... a plasma engine is more efficient and faster. In fact, a plasmadriven rocket could push a cargo vehicle from Earth to Mars in ninety days, about twice as fast as solid or liquid (i.e., chemicals) fueled rockets.”
Momentum is mass times velocity;
in a closed system, momentum exchanges are equal. EXAMPLE: a space borne vessel is a closed system. Left Side Momentum equals Right Side Momentum
IMPULSE: Momentum per time.To determine impulse, divide each side by one second. 
Redistribute terms
as shown.
Define New Terms
ff_{Sec} (fuel flow consumed per sec) ≈ exhaust mass per sec.
a (acceleration) = ship velocity increase per second
For simplicity, artificially assume:
negligible mass growth due to relativity
100% efficiency (all consumed particles contribute to thrust due to exhaust).
Given a fusion propulsion system,
assume exhaust particle velocity of 1 million m/sec.
Thus, a particle exhaust rate of .025 gram/sec will propel the 5 metric Tonne (mT) vessel 5 millimeters/sec faster for each second of powered flight. .025 gm is a small quantity of mass; how much fuel would be required for constant thrust throughout the 90 day voyage to Mars? First, compute vessel's daily fuel requirement.
Fuel requirement for entire voyage can be computed as a percentage of ship's initial gross weight.
Given above flight profile, determine what size ship can VASIMR propel with fuel consumption rate of one gram per second?
For every gram of ff_{Sec }, M_{Ship} equals 200 mT.
 For transport between planets, VASIMR proponents now call for a traditional chemical rocket to insert VASIMR vehicle into Earth orbit. Then, humans will enter the VASIMR powered interplanetary vehicle and travel to Mars with plasma drive. This would be much more efficient than a traditional orbital insertion via a chemical rocket.
For a trip to Mars, plasma drive is much faster than a chemical rocket. Plasma engine stays on throughout the entire trip; thus, velocity constantly increases. In fact, it increases so much that flight profile calls for a "slow down" at midway. At a carefully chosen position near midway, vessel flips around to propel in opposite direction and start decelerating. Otherwise, the vessel would arrive at Mars way too fast for orbital insertion.

While way too hot for material containers (>1,000,000°C), plasma's charged ions respond very well to magnetic forces.
Thus, VASIMR uses a magnetic "flux tube", an invisible sleeve shaped like a nozzle; a magnetic nozzle.
Technology of plasma containment originates from research in magnetic fusion, Dr. ChangDiaz's original field.
For magnetic fusion, one must confine the plasma away from any material walls; then, heat it and compress it to thermonuclear temperatures to ignite the plasma.
A persistent obstacle to fusion energy has been the difficulty of containing hot, electrically charged plasma. Ironically, this plasma confinement problem led ChangDiaz to an opportunity; use a controlled plasma leak to propel spacecraft.
Well arranged VASIMR components enable super heated, high speed plasma to exit in one direction to create thrust to propel the spacecraft in the opposite direction.
Thus, TE considers Three Ways of Traveling to Mars:  
1) Current Technology  Chemical Reactants:
Use short burn to leave Earth's orbit, enter a new Solar orbit from Earth's orbit to Mars's orbit; then, accomplish a 2nd short burn to enter orbit of Mars. With only chemical reactants, the shortest possible trip would be 6 months in a partial orbit. Most estimates call for a much longer trip. Most practical use for chemical reactants would be to move vehicle from Earth's surface to Low Earth Orbit (LEO). Even this dangerous manuever could be avoided once a space elevator is constructed.

2) Pending Technology  Plasma Drive:
Use plasma leaks for constant (though small) thrust throughout the trip. This propulsion provides a constant acceleration to midway; then, deceleration from midway to destination. The resulting trip could be as short as 3 months.
While 3 months is shorter than 6 months, it's still a long time to spend in near 0g conditions. A more practical use for a plasma drive engine (such as the VASIMR) would be as a plasma injector into a particle accelerator propulsion system.
 
VASIMR Diagram
High temps (>1,000,000°C) greatly expand fuel’s volume as it transforms from gas to plasma. However, relativistic effects are slight as mass increase is at most one thousandth.  
3) Thought Experiment: Particle Accelerator Propulsion
Thought Experiment (TE) introduces a third choice. Present day accelerators routinely take ions to near light speeds; thus, TE conservatively assumes exhaust particle velocity of .866c which introduces a relativistic growth factor (n) of 2 (particle is going so fast, it doubles in size).
Such acceleration provides gforce to simulate Earth surface gravity (gforce) during powered portions of flight. Furthermore, a trip time to Mars reduces from months to just days, a much more reasonable duration. Considering both this enormous velocity and relativistic growth of exhaust particles at this speed, TE uses simple momentum exchange equation to show that particle exhaust flow of .1 gm/sec will propel a 5 mT vehicle about 10 m/sec faster for every second of powered flight. 
Simple momentum exchange equation
shows fuel consumption of .1 gm/sec which grows to particle exhaust flow of .2 gm/sec which propels a 5 mT vehicle about 9.8 m/sec faster for every second of powered flight.  
1. Exhaust Particle Velocity (V_{Exh})
Exhaust speed of particles expressed as decimal light speed
V_{Exh} = d_{c }× c
d_{c}, decimal component of light speed, can be computed:
d_{c }= v_{Exh }/c
If v_{Exh} is already expressed as decimal c, then determine d_{c} by inspection.
(Example: if v_{Exh} = .866c; then, d_{c} = .866.)

 

2. Ship Acceleration (A_{Ship})
Step 1:
Determine each acceleration in meters per second per second (m/sec^{2}). Given constant acceleration, let initial velocity (V_{0}) be zero; then, determine final velocity(V_{1}) one second later.
Step 2:
Convert each value to decimal g as shown.
 
3. Ship Mass (M_{Ship})
Thrust depends on exhaust momentum.
M_{Ship} = V_{Exh} × ff_{Exh} / A_{Ship}
M_{Ship}'s potential mass depends on thrust.
Substitute values for light speed (c) & gravity (g) and recombine.M_{Ship} = (d_{c}×c) × (ff_{Exh}) / (d_{g}×g) Substitute terms from previous work. M_{Ship} = d_{c} × 30.57×10^{6}f_{Exh}
If ff_{sec }is in grams (kg),
M_{Ship }is in metric Tonnes (mT). 
 
Two kilograms/second of exhaust particles at .866c can gforce propel a vehicle of 52 kiloTonnes.
However, gforce fuel consumption must factor in Relativity and Efficiency.  


4. Growth Factor (n)
Relativistic Growth (n) factor comes from the Lorentz Transform which can be used to quantify mass growth due to relativity.
***
ff_{Sec}: fuel flow per second, mass per second consumed at rest. This quantity converts to plasma prior to injection into particle accelerator. Note: n can be any rational number >1.0, but choosing certain d_{c} values gives integer growth factors. Choosing particle exhaust speeds with corresponding growth factors as integers is a convenience but in no way a requirement.  
ff_{Exh}: exhaust mass per second
exits the spacecraft for propulsion.
Due to relativistic speeds via particle accelerator,
consumed fuel mass grows to become fuel flow exhaust.
ff_{Exh} = n × ff_{Sec}
 
5. Efficiency (E)
is a value between 0 and 100% which reflects inevitable design flaws and peripheral needs which inevitably decrease expected output. However, TE further assumes a learning curve in the design process. Thus, propulsion efficiency will start out low as we journey to nearby destinations such as Mars and Ceres. However, TE assumes efficiency will gradually improve as we learn to design better vessels to journey to further destinations.

Relativity: as initial fuel mass speeds up from relative rest (0 m/s) to .866c, initial 1.0 kgm per sec grows to an exhaust flow of 2.0 kgm per second. Theoretically, this should gforce propel a vehicle of 53,000 mTs.
M_{Ship} = (d_{c}×c) × (n×ff_{sec}) / g
However, above equations disregards inevitable inefficiencies. Like all other systems, Gforce spaceships will never achieve 100% efficiency; however, the conversion rate from at rest, consumed particles (ff_{sec}) to near light speed, exhaust particles (ff_{Exh}) will improve over time.M_{Ship} = d_{c} × 30.57×10^{6}× (n×ff_{sec}) M_{Ship} = .866 × 30.57×10^{6}× (2×1.0 kg) = 52,947 mT Since system inefficiencies are inevitable, actual thrust produced from theoretical fuel consumption requirement will be much less then needed for gforce. For this example, TE optimistically assumes system efficiency of 70%. Thus, system inefficiencies divert 30% of plasma particles away from the thrust stream. While relativity increases the consumed 1 kgm of fuel to 2 kgm, only 1.43 kgm would actually exit the vehicle in the exhaust flow.
*****
In this arbitrary scenario, an onboard weight scale would measure a 100 lb object at only 70 lbs. CONCLUSION: At 70% efficiency, fuel consumption of 1.0 kgm produces enough thrust for only .7gforce.
*****
Thus, TE uses an efficiency factor (ε) to determine added consumption required for gforce.  
Efficiency Factor (ε) is reciprocal of Efficiency (E).
ε = 1/E
To compensate for system inefficiencies, consumption rate must increase. In TE's fictional, arbitrary example, determine the Efficiency Factor (ε) by dividing 1 by E. Then, determine increased consumption rate by multiplying original rate by ε. Thus, consumption must increase from 1.0 kg/sec to 1.43 kgm sec as shown.
*****
In retrospect, an efficiency rate of 70% is very optimistic. System efficiency for initial gforce flights will likely be closer to 10%. Efficiency of 90% probably won't happen until lots of gforce flights push the learning curve for many years.Items contributing to inefficiencies might include:
Actual efficiency data can't be obtained until gforce operations actually begins. Thus, TE must make do with assumptions. For above items, TE assumes each requires 10% of propulsion power output: thus, TE arbitrarily assumes an efficiency of 70% (a very optimistic assumption.)
Actual efficiency (E) will need to be validated with empirical data.
ff_{Sec} = ε × M_{Ship }/ (d_{c} × 30.57×10^{6 }× n)
ff_{Sec} = 1.43 × 52,947 mT / (.866 × 30.57×10^{6}× 2 × 1.0) = 1.43 kg

 
6. Daily Difference (∇_{Day})
Ship's mass decreases due to powered flight's fuel consumption. TE uses term, ∇_{Day }, to describe daily decrease of ship's mass due to fuel consumption.
Recall that gforce ship's mass depends on quantity of fuel actually expelled, ff_{Exh}. Furthermore, gforce thrust particles are at relativistic velocities near light speed, d_{c }c. Consider following:

Daily decrease of ship's GW (∇_{Day }) depends on Efficiency Factor (ε), relativistic growth (n) and ship's acceleration (d_{g}, expressed as decimal g).
 
7. Typical Flight Duration (t)
 
8. TOTAL DIFFERENCE (_{∇}_{Ttl})
To determine total fuel consumption for trip duration, accummulate all the daily GW differences for entire trip.
(Also: difference between initial GW and final GW.)
Quick way to approximate:
Multiply daily diff (∇_{Day}) times total trip time (t).
∇_{Ttl} = ∇_{Day} × t
This works well for extremely small daily fuel consumptions
or for short trip durations.
Both these circumstances exist in following table.

EXPONENTIALS
However, many flight profiles have significant fuel consumption and/or lengthy trip duration. EXAMPLE: Let daily fuel consumption to be 1% ship's gross weight (GW)
∇_{Day} = 1% GW/day = .01 GW/day
EXAMPLE: Let total trip duration (t) be 40 days.
Initial approximation estimates total fuel consumption is 40% initial gross weight (GW_{0}). Thus, if GW_{0} = 100 mTs; then, final gross weight (GW_{40}) is about 60 mTs, and total fuel consumptions is about 40 mTs.
Δ_{Ttl} = .01 GW/day × 40 days ≈ .4 GW_{0}
GW_{Fin} = GW_{40 }≈ GW_{0 } Δ_{Ttl} = .6 GW_{0}
Easy to compute, but inaccurate!!!
A more accurate model of fuel consumption might involve exponents.
Axiomatic: If daily fuel consumption is 1% gross weight (_{∇}_{t} = .01 GW_{t}), then succeeding day's gross weight will be 99% of previous day (GW_{t} = .99 GW_{t1} ).
Example: If ship's initial gross weight GW_{0} is 100 metric Tonnes (mTs), then ship's GW can be computed for first, second and following days:
Exponentials can determine fuel consumption for longer duration flights.
 
SUMMARY: In succeeding days of powered flight, ship's Gross Weight (GW) decreases due to fuel consumption. Subsequently, lighter ship requires less fuel, and fuel consumption decreases. To model this, TE initially considers a convenient rate of fuel consumption (ff_{sec}=1.0 kg/sec); then, TE determines ship's GW.
Define Gross Weight (GW)
GW = Structure + Payload + Fuel
While structure and payload remain fixed throughout the voyage,
fuel decreases throughout powered flight.
Assume first day's consumption is .233% of 100,000 mT or 233 mT;
then, compute GW for end of day 1:
GW_{1 }= GW_{0}  (GW_{0}×_{∇Day}) = 100,000mT 233 mT = 99,767 mT
Second day's fuel consumption could be computed
as .233% of first day's GW:
F_{2}= _{∇Day }× GW_{1 }= .00233_{ }× 99,767 mT = 232.46 mT
Consider much longer flights. For an indeterminate flight duration,
F_{t}= _{∇Day }× GW_{t}= _{∇Day }× (1_{∇Day})^{t}× GW_{0} any given day's fuel consumption should use exponents as shown:
Example: 40 days. Someday, a gforce trip to Kuiper Belt could require 40 days. Thus, reconsider previous example of 40 days duration with ship size of 100 kiloTonnes.
F_{40}= .00233 × (.99767)^{40}×100,000 mT =212.74 mT
Decreased daily consumption translates into smaller fuel flow rates.
To determine fuel flow for Day 1, divide total fuel consumption by total seconds per day.
ff_{sec }= F_{t }/ 86,400 sec = 233×10^{6} gm/86,400sec
ff_{sec }= 2,696.8 gm/sec
Similarly, Day 2's 232.46 mTs translates into 2,690.5 gm/sec
and so on for succeeding days.
For the last day (Day 40), we get 2,462.3 gm/sec,
a significant 234.5 gm/sec less then the initial flow fuel for Day 1.
Thus, to maintain constant gforce throughout powered flight,
fuel flow must be constantly monitored and adjusted.
Gforce sensors (i.e., scales with 100 lbs. weights should indicate 100 lbs. throughout powered flight). Sensor servo connection could automatically adjust fuel flow. 
Sidebar: SPECIFIC IMPULSE(I_{sp})
Specific impulse is measured in seconds.
Specific Impulse (I_{sp}) is the length of time (usually “seconds”) that each unit weight of propellant propels its own weight. For ships in vacuum of space, it proves convenient to compute specific impulse as average particle exhaust speed divided by g, near Earth gravity:EXAMPLE: one pound of typical solid rocket motor fuel produces one pound of thrust for 250 seconds.
In contrast, Impulse (I) is the average propulsion Force (F) times the total duration (t) of firing.
I = F × t
Straight forward work rearranges terms for Impulse equation such that Impulse equals exhaust mass flow times velocity of exhaust particles.
F = ma = m × v/t
I = F × t = m × v (i.e., momentum)
For now, assume perfect efficiency;
thus, exhaust fuel flow equals input fuel flow:
I = ff_{Exh} × V_{Exh}
Effects of relativity on size of exhaust particles.
ff_{Exh} = n × ff_{sec}
For VASIMIR rocket engine, relativity is negligible and n=1
I = ff_{sec} × v_{Exh}
Recall ff_{Exh} is the amount of exhaust mass per time ejected out of the rocket (adjusted for relativity). Assuming constant exhaust velocity, we get: I = ff_{Exh} × v_{Exh} where m is the total mass of the propellant. We can divide this equation by the weight of the propellants to define the specific impulse. The word "specific" just means "divided by weight".
Thus, Specific Impulse (I_{sp}) shows Impulse from each unit weight of propellant. For example, typical solid rocket motor produces 250 pounds of thrust for every pound of fuel injected into the combustion chamber (per second).
Expressed mathematically:
If we substitute for the equivalent velocity in terms of the thrust:
I_{sp} = F / (mdot × g0)
Mathematically, the I_{sp} is a ratio of the thrust produced to the weight flow of the propellants.
A quick check of the units for I_{sp} shows that: = (m/sec) / (m/sec^{2}) = sec
Why specific impulse?
 
CONCLUSION: Gforce vessel must vary daily fuel consumption.
Fortunately, VASIMR's first name is "Variable"; thus, will vary plasma quantity injected into particle accelerator.

0 Comments:
Post a Comment
Links to this post:
Create a Link
<< Home