[vansig]

ground crews can retreat to safe distance when the craft is in operation. otherwise we’d have to space anodes as close as 3 or 4 cm apart, tesselating a sphere, to get shielding mass under 100t.

when it comes to space, propellant cost is king. it isn’t a fair comparison to say Falcon, etc. is cheaper, as those represent single-use vehicles and a mature technology, whereas the purpose of envisioning this is to have routine, possibly daily (hourly?), surface-to-orbit and return missions. the facilities cost is amortized over many launches and vehicles. final cost is more to do with day-to-day operation.

let’s see.. 600 kN nominal but probably 1800 kN max thrust at lift-off;
1800 kN / (100t x 9.8) = 1.84 gee.

[vansig]

zapkitty wrote:
… but energy density is still the problem. Even with the most optimistic shielding estimates you still have to have over 4.2 metric tons of water with each power module in addition to the FF cores, caps, onion etc… and both safety and cooling mean that you will not be able to squeeze all the cores into one module.

not quite. the power modules don’t need to be shielded from each other; so really there is just enough shielding to protect the payload and crew. i haven’t worked the numbers through, for EHD, as i’m not yet optimistic about it, for different reasons. but regardless, it seems the biggest problem to manage is heat, after all.

for 10 GWt of heat rejection,
keeping airfoil surfaces below glowing red hot, or around 800 kelvin, requires either a radiative surface area of 43 hectares,
or expulsion of coolant mass. ( 800 K radiates at most 23.226 kW/m²; and 10 GW / 23.226 kW/m² = 4.30e5 m² )
the situation is improved if very high temperatures can be tolerated. for 1600 K, 2 hectares would do; and for 3200 K, 1681 m² would do.
Ta4HfC5 melts at 4488 K, but is not exceptionally resistant to oxidation; layered hafnium carbide/silicon carbide seems the most resistant to ablation in air, (ablation rate of 5 µm/s measured at 595 kW/m², which corresponds to T=1800K).

it would be nice if all of the heat could somehow be carried away by air flowing over the wings and through the engine, but to accomplish that, i am told, we’d have to avoid stagnant air anywhere in the system.

[vansig]

here is an article detailing scaling parameters for EHD thrusters.

http://www.wbabin.net/physics/borg1.pdf

[vansig]

Some of these materials, boron nitride included, have intriguing semi-conductive properties. Cheap large-scale fabrication is the key, and the use of something simple like ultrasound could definitely help, quite a bit.

[vansig]

this will probably yield a particle fountain, at the tail end. but, the alphas will eventually collide with something that will carry the charge away

[vansig]

mjv1121 wrote: Is my grasp of expected efficiency more or less accurate then (see the top half of my first post)

There are some differences. Notably, there is residual charge remaining in the capacitors after the shot. If i recall, Eric mentioned that about 70% of the energy goes into the plasmoid; the other 30% is recovered in a second capacitor bank. The amount of x-rays is not fully worked out, and depends on the optimization of the magnetic field.

[vansig]

rickPS wrote: So the coil to extract energy from the beam is really an energy recovery device, it’s the X-ray PV cells that will collect the surplus energy from the fusion ?

to achieve the predicted system efficiency of 50%, you’ll want to use both.

[vansig]

ok, these numbers are a little better.. they do not yet deal with heat management, but i’ve looked at other flight parameters, such as lift, drag, rate of climb, shock waves; at high mach numbers, air intake needs to be a large fraction of the forward facing area, (about half). at hypersonic speeds, leading surfaces will be attached to the shock wave, to stabilize flight and prevent air from becoming stagnant.

let me know what looks impossible here.

assume 100t mass; (47t empty);
lift/drag ratio 4.5; glide ratio 4.5 (about the same ratios as the space shuttle)
122m² underwing area

calculate, density-versus speed, given approx constant air inflow of 80 kg/s, 600 kN thrust, 15500 m/s exhaust velocity, and drag from 218 kN to 300 kN. the engine possibly cannot operate at below mach 2.4, but i have not proved that. particularly, thrust may possibly be doubled in dense atmosphere, with extra energy.. which would enable vertical take-off.

mach 2.4+ at 36 km altitude (density ~.01 kg/m³; drag ~218 kN);
mach 8 at 45 km altitude (density ~.003 kg/m³);
mach 24 at 54 km altitude (density ~.001; drag ~300 kN);
net forward propulsion to >100km

disregarding efficiency, *any* sort of propulsion needs to add a lot of kinetic energy to the incoming air
E_k = .5mv² = .5(80 kg)(15500)² = 7 to 10 GJ each second

if system efficiency were terrible, say 1%, then
decaborane consumption rate would be 14.5 g /s, and this makes
Isp = 600,000 kg m/s² / ( .0145 kg/s ) / 9.8 m/s² = 4.2e6 s

efficiency is probably not that terrible. like VASIMR, traditional pressure from heating becomes inefficient at the needed exhaust velocity, so the engine will rapidly ionize air and then accelerate it magnetically, before it thermalizes. I’ve not seen a good argument as to why VASIMR is supposed to be operated in vacuum only. However, Ad Astra does discuss formula for efficiency of their engine. http://www.adastrarocket.com/AIAA-2010-6772-196_small.pdf

finally, thrust bottoms out at ~2 kN in space proper, where alpha particles go directly to exhaust

Oh, and here’s a nice article, mentioning DPF and nuclear thermal rockets
http://quantumg.blogspot.com/2011/02/making-fusion-rockets-relevant.html

[vansig]

and here i thought that a magnetic field line was a virtual path through the solution space of a system of differential equations, forming a hamiltonian cycle.
but i actually have very little knowledge of them.

[vansig]

jamesr wrote:
When the magnetic field collapses the rate of change in B-field creates the huge E-field which accelerates the ions in one direction at ~2.6±0.2MeV and the electrons in the other.

I thought the ions in the exit beam would average around 600 keV?

[vansig]

Mine, too. 🙂

One very significant approximation is the power required for hypersonic flight. I’m beginning to wonder whether my early guess of 400 MW is in the right order of magnitude.

My present best-case reasoning is leading to 800 MW, and worst-case 17.6 GW, so i need to debunk my assumptions.

Comparing again to solid rocket booster performance,
the GEM-40 burns HTPB 12%/ammonium perchlorate 68%/powdered aluminum 20%,
at a rate of 185.8 kg/s, thrust 499 kN, Isp 274 s, nozzle ratio of 16:1

Typical ramjet Isp is higher: 1400 s, i am told. Scramjets would be higher still, i assume.
I plugged in assumptions for two different configurations:

“Small mouth” assumes total mass = 100t, form drag 1250 kN at v=8 km/s, air intake 102 kg/s at p=0.01 bar;
and produces minimum needed exhaust velocity = 20236 m/s.

“Big mouth” assumes total mass = 100t, form drag 1250 kN at v=8 km/s, air intake 1020 kg/s at p=0.01 bar;
and produces minimum needed exhaust velocity = 9224 m/s.

If this exhaust velocity is obtained by non-thermal processes, then the entire mass is accelerated aft, and the kinetic energy required each second comes from the difference between the propellant’s intake and exhaust speeds, relative to the ship.

E = E_new – E_old = .5 m v_new² – .5 m v_old² = .5 m ( v_new² – v_old² ),
= .5 (102) (20236² – 8000²) = 17.62 GJ for small mouth,
or
.5 (1020) (9224² – 8000²) = 10.75 GJ for big mouth.

If this exhaust velocity is obtained by thermal processes, then temperature is given by
T = v_rms² M / ( 3R ), where M = mass of 1 mole of air particles in kg, and R = 8.314 (the gas constant)..
= 20236² (0.029) / (3 x 8.314) = 476120 K, for small mouth; or
T = 9224² (0.029) / (3 x 8.314) = 98925 K, for big mouth.

I’m not liking these numbers, at all. They make me hope i’ve overestimated something. (eg: drag? total mass?).

But, rockets can reduce drag by flying higher; and jets can reduce drag by flying slower, each with their own consequences

[vansig]

Rezwan wrote:
So the meta question is, “what is charge”?

i have yet to see a better answer than
“charge is one component of ‘spooky’ action at a distance”

[vansig]

I might not want to do mach 24 in the stratosphere, but it is a conversation point.

The engine needs to have enough air intake to give net forward thrust and lift, after overcoming drag, at any given altitude and velocity. Flying higher reduces drag by reducing density of air, but thrust also declines, requiring faster flight to feed the engine. There is a cross-over point from air-breathing to on-board propellant, as the min and max altitudes for hypothetical top speed converge.

At speeds much lower than mach-24, the craft cannot rely on centripetal motion, so engines and surfaces must provide for both drag and lift. it is interesting that there is a point where these are in the same magnitude. my hope is that the discussion will lead to a proper formula for a flight curve for both suborbital trans-continental transport and orbital insertion.

neglecting drag, perigee at 8 km/s and 36 km altitude would entail apogee near 400 km, and delta-v of ~70 m/s to complete the maneuver. finding the point along this curve where we must switch to on-board propellant will yield knowledge of how much propellant we need to carry.

[vansig]

The moon is unfortunately rather barren of easily extractable resources. Its advantage, for mining operations, is that it is fairly nearby. But it has almost no hydrogen, which is desperately needed as a resource for materials extraction. I would find asteroid mining to be more worthwhile, probably.. if-only we could park one into earth orbit.

What Focus Fusion is attempting, is aneutronic fusion. No neutrons means no long-lived radioactive waste, and no bomb-making.

And therefore there is no plan to breed tritium, and no good reason to want to, if you can just burn boron.

[vansig]

Quantitatively, what is the expected drag on a craft as it reaches orbital speed?

A 100 tonne craft, (roughly 3 m² area, and 16 m long), flying at ~36 km altitude (pressure=.01 bar) and mach-24 (8 km/s), needs 980 kN lift to remain climbing, but most of the lift is provided by centripetal motion, since 8 km/s is sufficient for orbital velocity.
( v²/R = 8² /(6371 + 36) = 0.0999 km/s² = 9.99 m/s² )

So only forward drag really matters, here. From the formulae at
http://www.grc.nasa.gov/WWW/K-12/airplane/drageq.html
http://www.grc.nasa.gov/WWW/K-12/airplane/dragco.html and
http://www.grc.nasa.gov/WWW/K-12/airplane/eqstat.html
as well as
http://www.lpi.usra.edu/meetings/lpsc2009/pdf/2059.pdf

We get that form drag dominates at high mach numbers. Assuming drag coefficient, Cd ~ 0.92; cross-sectional area, A ~ 3m², and density of air (at p=.01 bar and 244 K), rho = 0.0143 kg/m³,

then drag = Cd x Area x rho x v²/2
= .92 x 3 m² x 0.0143 kg/m³ x (8000 m/s)² /2 = 1263 kN

practical flight will have to exceed this by a good margin, for at least several minutes as these speeds and heights are obtained.
compare this needed performance to a gang of 3x GEM-40 boosters, (at 1000mm diameter and 492.9 kN each, 64 seconds burn time, only).

[Author]

Posts