I don't misunderstand. There's heat flow loss out of the body of the reactor. How much is this loss, compared to the amount generated?

What I'm missing are numbers: how much power do they expect to generate from fusion, how much power do they think it will take to run the full prototype system, and what's the operating temperature; ideally as part of a Sankey diagram. I think this isn't available because they simply don't know.

Again I ask, was Chicago Pile #1 a nuclear power plant? If not, what was the first nuclear power plant, and what distinguishes a power plant from other sorts of nuclear reactors? I think it must generate at least enough power to be be self-sustaining to qualify as a power plant. I therefore believe this General Fusion system, when built, will be an experimental fusion reactor used to develop the principles leading to a fusion power plant, but will not itself be a fusion power plant, experimental or otherwise.

BTW, the Wikipedia page says "They hope to have a working reactor by 2020", which is of course 6 years after Asimov's prediction. The WP does not give a source for that information.

First I'll note that the remaining problems seems to be engineering more than plasma physics -- and most of the show stoppers seems to be solved. If that don't make you go "wow", well...

>>what distinguishes a power plant from other sorts of nuclear reactors?

Again: I argued the main difference here is connecting a heat exchanger, you had nothing to say?

>>What I'm missing are numbers: how much power do they expect to generate from fusion

Again: The target is 6X the used energy for the pistons, according to the presentation. (It is a private company, details for investors.)

(You know the number of pistons, their size/speed. That should be enough to make a good guess on the target energy from fusion. You can also guess quite well the volume of the lead and hence wuite a bit of the pumping needs.)

There are other GF sources on the internet over the last few years, but since you ignore what I write please check them yourself.

My apologies. I did overlook your mention of a heat exchanger as being the requirement for a power reactor.

To me a heat exchanger is something which moves heat from one medium to the other, and is not necessarily part of power production. The radiator in a car is a heat exchanger, and not a power generator. Thus, I did not understand your viewpoint.

Quoting from the first link you gave: "Pumped through a heat exchanger, that hot lead will help generate steam just like a conventional thermal power plant". Thus, I consider it as not a power plant, because it only produces steam. While I see that you consider the production of steam to be sufficient, likely because it reduces the issue to a previously solved problem.

I am satisfied that the General Fusion design, which hasn't even started, is not a nuclear fusion power plant of the sort that Asimov predicted would be available by 2014. That is my primary interest in this discussion.

FWIW, after more looking around, I found this quote from http://www.technologyreview.com/news/414559/a-new-approach-t... :

> However, if the company can pull off its test reactor, it hopes to attract enough attention to easily raise the $500 million for a demonstration power plant.

Note how that magazine author also distinguishes between "test reactor" and "demonstration power plant"?

The information I found are not sufficient to make the estimates that you mentioned. For example, while I found pictures of the small sphere experimental setup, and a mention that the final velocity is 50 m/s, I did not find learn the mass of the piston, or equivalently the amount of power used to drive it. Nor did I find out how much power goes into the plasma injectors. There are images of some pretty hefty capacitor banks, but I couldn't tell if they are discharged at once, or if they are rotated through in order to lengthen the charge time for any one bank of capacitors.

Estimating the pumping needs is very complicated. I found no reference of what the internal surface of the sphere looks like. If liquid extends into each piston then there well be turbulence at each interface, causing power loss. Even if it's laminar, I still need to know the rotational speed.

The depictions I've seen show that the lead surface is a near cylinder, which means a very high speed, as the natural shape of a rotating fluid is a parabola. Assuming 0.25 meter radius for the upper part of the cylinder and a height of 3 meters gives 400-500 Hz. That seems rather a lot, given that they're talking about some 50 cubic meters of lead, or 500 tons, moving at some 2,000 m/s.

(I used the equation for a liquid mirror telescope: h = 1/(2g) * (omega * r) ^ 2. Hopefully I didn't calculate it incorrectly.)

Since the information I've found doesn't make sense, either my math is wrong, or the information I've found is incomplete. Since you are the one who believes that this is a near-term power plant, then surely you must have worked out these details already, or read them some place. That is, how fast is the molten lead mixture supposed to rotate, and how much energy is needed to keep it at that speed?

Again: I used the term "full sized prototype" in the first comment. Which is not exactly what Asimov wrote. The remaining possible show stoppers are less plasma physics and more mechanical, which is certainly better than what Asimov predicted.

(I wrote repeatedly that the main thing missing from the planned prototype to be a prototype power plant is a heat exchanger, like most every other power plant that heats a medium. It boggles incredibility to assume you failed reading that.)

Your link is from 2009, when they were new to building experimental hardware. Hardly an authoritative reference.

Edit: A simulation of the vortex. It was hard to find, took me 3 minutes with Google: http://www.cs.ubc.ca/~jgregson/images/JamesGregsonMAScThesis...

Edit 2: The old papers, before much experiments: http://www.generalfusion.com/wp-content/uploads/2013/08/GF_S...

Edit 3: The recent TED talk is (claimed) to be up today, but the ted.com site is down as I write.

Thank you for the find for #1. I am pleased and surprised to see that my back-of-the-envelope calculation on the speed is correct. I am still gob-smacked to think of that much lead spinning a 2km/sec. I don't have the knowledge of supersonic fluid flows to be able to evaluate that. The paper itself points out various difficulties, including delamination, decavitation, jet formation with speeds up to 6km/sec, and:

> practically generating shocks mechanically at Mach numbers greater than 1.5 without destroying the machine that creates them (as is the focus of this work) seems challenging. For example, a steel piston impacting liquid lead to produce an approximately Mach 1.1 shock would see a compressive stress of approximately 2 GPa, well above the yield stress of most steels.

The paper talks about pressures up to 400 GPa. Another back-of-the-envelope calculation suggests 20 GPa of centrifugal force at the equator, which of course means the pumps need all the more force. This is well beyond a regime where I can make any mechanical estimates.

At this point I suggest that the biggest problem is the engineering to bring the lead up to speed, keeping the sphere in shape with that much force on it, generating a plasma collapse through all of that, and connecting everything to the heat exchanger. That will be a huge challenge in its own right, and must be solved before I would call it an experimental power plant.

BTW, this vortex simulation is from 2008, ... "hardly an authoritative reference" by your own criterion. ;) Indeed, I note that the simulation says 100 kg pistons impacting at 100 m/s, while this 2012 PDF http://fire.pppl.gov/FPA12_Richardson_GF.pdf says the target impact velocity is 50 m/s, which is 1/4th the total amount of energy.

Link #2 directs me to http://generalfusion.com/downloads/ICC2008_MGL.pdf, which shows that in 2008 they indeed planned on a 100 m/s impact, so parameters from 2008 are obviously no longer valid for 2014.

>> I am still gob-smacked to think of that much lead spinning a 2km/sec.

That is NOT the value for the spin speed. The paper says:

"The collapse of the cavity is accelerated by geometric focusing, resulting in cavity wall velocities over 2 km/s at the end stages of collapse"

The spin speed -- think centrifuge, as in a washing machine or in a chemical lab. Lead is high density, but you'd hardly need 2 km/s (around 12K RPMs!!). That is ridiculous.

(I do think I've seen articles on that the demands on the pistons might be smaller than the first calculations, after experiments. The self reinforcing shock wave might be involved.)

Edit: If you really need to estimate the spin speed, this should work? Consider how many G you'd need at the inner part of the evacuated tube. Then check that on some online calculation (there is a simple formula) to get the RPM (and hence speed) instead of trying the "back of the envelope" thing... :-)

Edit 2: I googled a page with the G formula for RPMs. 0.2 meter empty in the middle at 1000 RPMs gives 112 G which should be <cough> more than ample. :-) The formula use (RPM/1000)², so 12,000 RPMs... I don't even want to think about > 1400 G on tons of lead!! http://clinfield.com/2012/07/how-to-convert-centrifuge-rpm-t...

That is the value for the spin speed. Quoting from the paper, 4 paragraphs after the 2km/s you referenced:

> Probably the most signicant feature of the flow in the concept reactor is that it involves a compressible liquid. With flow velocities exceeding 2 km/s and pressures reaching 400 GPa, compressibility is unavoidable.

"Flow" refers to the spin speed, not the collapse.

Like I said earlier, I used the equation for a liquid mirror telescope -- h = 1/(2g) * (omega * r) ^ 2 -- for my '"back of the envelope" thing.' See http://en.wikipedia.org/wiki/Liquid_mirror_telescope for the full derivation.

Using h = 2.7 m, g = 9.8m/s, and r = 0.20m gives 40 Hz (or 2,400 rpm). Note though that this gives 20 cm at the top, and 0cm at the bottom. It's a paraboloid, so the center is probably 14-15 cm across. Thus, 40 Hz is a lower bound.

Using 40 Hz in (2 * pi * 1.5m) * 40/s implies a minimum equatorial speed of 380m/s, which is Mach 1. The equatorial pumps of course must be providing fluid at an even higher speed. The g-forces at the equator, 1.5 meters away, are even larger than the g-forces at the surface 20 cm away. a = r * omega^2 = 1.5m * 1600/s/s = 2500 G.

To get a more uniform evacuated center requires higher speeds still, but the formula I used (first discovered by Newton, btw), no longer holds. It will likely need to be several times faster. Flow speeds of 2km/s require only 6 times faster than the minimum possible speed, which sounds reasonable. Even if 2km/s doesn't sound reasonable.

I don't think General Fusion has done the engineering testing to show that they can actually construct one of these, much less use it to provide power.

Feel free to correct my calculations.

>> Quoting from the paper, 4 paragraphs after the 2km/s you referenced

The paper doesn't even discuss what you claim. This is the paragraph before the one you quoted:

"There are several important flows in the concept MTF reactor. However, this work focuses only on the compressible aspects. Consequently, issues such as the vortex formation and cycling lead through the reactor to generate steam are left to follow up work by General Fusion."

I do think you know you are misrepresenting the content of the paper.

Your calculations are derived from a mild parabola, I'm not going to bother looking up where the formula fails. We're talking about a globe, which must change this drastically. Which I frankly also think you're aware of.

AGAIN: 10+G in a centrifuge would certainly put any liquid against the wall. And already 1000 RPMs @ 20 cm is > 110G! (See reference in previous comment.)

After your misrepresentation of the paper above and forcing me to repeat trivial points repeatedly I'm not going to bother with the differing results from your math or how it might be applicable to a sphere.

Based on your responses, I can only conclude that you have neither an engineering nor physics background.

I quote from the abstract:

> An Eulerian compressible flow solver suitable for simulating liquid-lead flows involving fluid-structure interaction, cavitation and free surfaces was developed and applied to investigation of a magnetized target fusion reactor concept. The numerical methods used and results of common test cases are presented. Simulations were then performed to assess the smoothing properties of interacting mechanically generated shocks in liquid lead, as well as the early-time collapse behavior of cavities due to free surface reflection of such shocks.

That's three parts. 1) fluid-structure interaction, 2) effects of shocks, and 3) collapse behavior. I'm discussing part #1, which mentions 2 km/s fluid flows. There's also part #3, which includes jets up to 6 km/s.

It's a compressible flow solver because liquid lead compresses at those pressures. That's why the author talks about "Probably the most significant feature of the flow in the concept reactor is that it involves a compressible liquid" and why the author later goes on to develop the equation of state for liquid lead, including a term for cavitation effects.

You might have confused it with compressing the plasma. But plasma compression isn't covered until "cavity collapse" on page 71.

You wrote "Your calculations are derived from a mild parabola."

Yes, of course it's a parabola. The free surface of all bulk liquids in a system with constant rotation forms a paraboloid. That's what the physics says. Why do you think otherwise?

In this case it's also a "mild" parabola. That's why it's the minimum bound for a solution.

The fact that this is a rotating sphere instead of cylinder doesn't change anything. The free surface is a consequence of the centripetal force balancing out potential gravitational energy, not the shape of the container it's in.

For some reason you think using a sphere "must change this drastically", but you offer no explanation. The counter-proof to your supposition is very simple. For all but thin-films, the surface has no way to know the shape of the container. It can't tell if it's in a cylinder or a sphere. Indeed, if you replaced the lead that's more than 30cm away from the rotation axis with a solid, then the free surface wouldn't change, and now it's spinning in a cylinder.

You also believe that 10g "would certainly put any liquid against the wall." This is true, but the actual question is how high will it push the liquid? You offer your conclusion without explanation. The counter-example is easy to show.

Let's suppose that 10g is enough to be near-vertical. We know the angle of the free surface, relative to vertical, must be arctan(1g / 10g) = 5.7 degrees. Let's say the vortex surface is 2 meters long. In that case, if the top is 20 cm from the rotation axis then the bottom - descending at 5.7 degrees - must be tan(5.7 degrees) = d/(2 meters). Since tan(5.7) = 1/10, then d = 0.2 meters. But there isn't enough room for 200cm of inclination!

Obviously, with 100 g then the displacement across 2m will be about 20cm. This means the 20cm vortex surface must have well over 100 g in order to maintain a near-vertical surface. Yes, this requires more than 1,000 rpm.

I've now worked this out in two different ways, and presented the math. I get basically the same result each way. Those also seem to agree with the engineering analysis you point out.

All you've said is that I'm "misrepresenting" things, that you're "not going to bother looking up where the formula fails", and give seat-of-the-pants answers that are easily shown to be invalid.

Since you had to look up a method to compute the G formula for RPMs - something covered in introductory physics for physics/engineering majors - I therefore conclude that you don't actually understand the engineering physics involved.

So your claim is that when "2 km/s" is referenced on page 3 of the reference, the first which obviously is about the implosion speed after the shock is not related to the second -- which is the necessary speed for the lead.

So you repeat the claim of 2km/s for a vortex in the GF reactor.

A few decimeters irregularity in the bottom of the vortex for the lower plasma injector invalidates that, which you certainly know.

I have no physics background (cs/chemistry), but considering e.g. your insisting on the exact definition of Asimov's prediction above (which I acknowledged in my first comment), I believe you're trolling.

So I'd really like to see other's arguments on this. I'm not going to spend time to get the cobwebs out of the part of my head where old math studies are stored for what likely is a troll. I have lost too much time as it is.

Edit: I don't think you really care, but cavity compression for a 1 m sphere was done 2012. Google for a pdf called "Update on Progress at General Fusion". ("Fusion Power Associates, 2012 Annual Meeting")

Addendum:

General Fusion has gotten tens of millions of dollars from serious investors. The idea that their design would need to accelerate many tons of molten lead to 2 km/s every pulse (i.e. once per second!) is just too ridiculous.

I need to be less polite/trollable.

Addendum:

The TED talk was on Youtube (and ted.com is up now). It was for a non-technical audience, the message is that the GF target is to make economical power plants. A commercial venture, as I wrote.

The only real news is that the plasma injectors are probably done now, over the last month.