Really, and I mean really really. It's not going to work.
It appears you are right that you couldn't achieve orbit with this design. Performance falls off too fast with increasing velocity. Still, I'm curious is there might be other areas where ion-craft could be useful.
Launching a nuclear engine is not a feasible solution. We can't even get people on board with building ground based ones. Even if you could get past the politics, submarine engines weigh over 100 tons just for the housing. That's not including the 80 tons of coolant or 800 tons of ancillary equipment. For reference, the empty shuttle weighs 80 tons give or take. An A380 weighs 650. So, assume you get something immense enough to lift all that off the ground. I'll estimate it weighs 10 to 20,000 tons. Nuclear power generates a slew of heat. For a 30 MW reactor, ill estimate 70 MW worth. In a sub, you dump that into the surrounding ocean. You're going to be swimming through superheated, ultra-rarefied plasma. The only way to get rid of that heat will be football field sized radiators.
The viability of nuclear fission in space is a whole new discussion all together. Suffice it to say I give it a better chance of working than you do.
You've posted a simplistic equation for calculating flow rate through an orifice using density, area, and speed; but did you calculate the power, current, or voltage that would be needed to achieve that kind of thrust with a EHD lifter? All I've seen is a 30MW number, but nothing to back it up...
I don't believe there's anything wrong with my equation for mass flow rate; I'm picturing a ion-craft where the high voltage is applied to the hull itself. That is, work is done on all particles in any vicinity of the hull. If you want current and voltage that's fairly simple application of electrostatics. For the applications we are discussing, air can be assumed to be 100% ionized, simplifying calculations greatly.
First, recall the change in kinetic energy of a particle is:
\DeltaKEIon = 1/2m(V
Exhaust2 - V
2)
and
P = .5 dm/dt V
Ex2-.5 dm/dt V
2, where P is the power (I*E) supplied to the craft's engines
we find that:
V
Exaust = sqrt( 2P/dm/dt + V
2 )
and thus,
\DeltaV = sqrt( 2P/dm/dt + V
2 ) - V
Now recalling the definition of electric field, and definition of voltage, and definition of work:
E = F / q;F = E * q
E = - dV/dr
W=\DeltaKE=integral(F,dr,0,r)
Integrating force over distance on a charged particle gives work done:
W = -1 * E * q * r
W = V * q
Changing the variable V(voltage) to \epsilon to avoid ambiguity with velocity:
W = \epsilon * q
Elementary
Setting work equal to delta KE; let n be a number representing the average ionization (avg # of missing electrons)
1/2m(V
Exhaust2 - V
2) = \epsilon * n * q
Elementary
And solving for \epsilon
1/2m
particleAvg(2P/dm/dt + V
2 - V
2) = \epsilon * n * q
Elementary
1/2m
particleAvg * 2P/dm/dt = \epsilon * n * q
Elementary
\epsilon = .5 * (mparticleAvg / ( n * qElementary)) * 2P/dm/dt
~~~~~~~~~~~~~~~~~~
To find current, realize that current is "charge in motion". 1 amp is a coulomb of charge passing a given point in 1 second. In other words, we can speak of current as a charge flow rate.
dq/dt = k dm/dt; where k is the ratio charge per unit mass
dq/dt = I = n * qelementary / mparticleAvg * dm/dt
~~~~~~~~~~~~~~~~~~~
Checking our work, we should find that P=I\epsilon
P = (.5 * (m
particleAvg / ( n * q
Elementary)) * 2P/dm/dt)(n * q
elementary / m
particleAvg * dm/dt)
and indeed we find that P=P, so our work checks
~~~~~~~~~~~~~~~~~~~
Now plugging in numbers; let's choose the 86km altitude again:
\rho=.000288kg/m
3
V=2000m/s
m
AvgParticle = the weighted average of the masses of nitrogen and oxygen molecules = .79 * 14 * 2 + .21 * 16 * 2 = 28.84amu = 4.789e-26kg
q
el=1.6e-19C
n = 1.2 (I'm taking a shot in the dark here)
\epsilon = (mparticleAvg / ( n * qElementary)) * P/(\rhoAV)
\epsilon = .2598V
I = 1.15e8A
As crazy as that sounds, I think it's right. The shear amount of charged particles flowing ac cross the spacecraft at those velocities means currents are astronomical.
At 2000m/s, a 30MW reactor only allows you to apply a potential of 1/4V. That's why this supplies so little thrust at high velocities; there just isn't enough power to sustain a strong enough electric field.
The advantage of operating at these higher velocities and altitudes however is that energy need not be spent ionizing the air. The supersonic shock-wave and solar radiation has already done that. I think ionizing the air is where 99+% of the energy is spent in a normal ion lifter.
Drag coefficients for typical satellites range from 2 to 4, but that's because there's virtually no effort to make them aerodynamic. I think it's reasonable to say you could decrease the drag coefficient if that was an issue, but 0.1 seems a little optimistic.
A subsonic wing can achieve .01 C
D, so .1 is in the range of sane numbers; not saying it's necessarily accurate for a satellite, but it's about what would be necessary for this to have any chance of working at all.
Also, thanks for the numbers; those look much more accurate.
It is possible to calculate the thrust put out by an EHD lifter, but some of it is a little "fuzzy" for less than 1 atm air. According to Wikipedia...
I have my doubts as to the usefulness of that page to this discussion. As stated above, the only reason that line needs to be at 30kV is to ionize the air, not to accelerate it. Because charge on an ion is very large in comparison to mass, a relatively low potential can cause meaningful thrust. That thrust statistic from Wikipedia assumes that you have to invest the ionization energy into every ion you want to ionize, whereas at speed and altitude, solar radiation and supersonic shock have done the ionizing for you; efficiencies should be orders of magnitude higher at altitude. My equations are derived from assumptions relevant to this discussion, while that may not necessarily be true of the equations on the wiki page.
Also in that article:
Further study in electrohydrodynamics, however, show that different classes and construction methods of EHD thrusters and hybrid technology (mixture with lighter than air techniques), can achieve much higher payload or thrust-to-power ratios than those achieved with the simple lifter design.
So the lifter is at the lower end of efficiency for the ionization reasons I already mentioned.
I also have another idea for a potential use of this technology. I'll crunch the numbers and post again.
