RERT wrote: ↑Fri Nov 19, 2021 7:41 am
Skipjack wrote: ↑Thu Nov 18, 2021 3:15 am
It is net electricity. So not just net power. Though for their machine the two are really closely related.
Ok, that correction makes the question clearer. In a second (for the 50MW device) they put in X joules of electrical energy and get out 50M + X. My question is, what is X?
Well, the 50 MW is the
goal and that is for net electricity put to the grid.
That said, they might fall a little short of that, or might exceed it a little. It is not 100% perfectly set in stone yet at this stage.
To the best of my information, they are aiming for an engineering gain of ~3 with their fusion generator in the power plant version.
Now the following is purely speculative on my side (and please point out if you see any mistake in my calculation):
So the calculation for engineering gain is:
PE(grid)/ PE(in) = Q(eng)
So (50MWe) / X = 3
That means that the recirculating power should be about 16.666 MW, if I calculated that right.
With their 95% efficient energy recovery. The remaining "loss" should be 833 kW.
Now lets factor in some other inefficiencies, losses and whatnot, I would say that they would have to produce a total of fusion power production and recirculating power of about 68 MWe to have a net grid power to the grid of 50 MWe.
Not all that bad. Now please know that much of this is speculative and not official information. Also my math (or formulae) might be off somewhere. So take all this with a huge grain of salt.
If anyone is interested, I also calculated a Q(Fuel) of about 3.4 for them to get a Q(eng) of about 3.0.
So what Q(sci) would they need to get any net electricity at all?
So we have ~18 MW total out with ~17 MW recirculating with a ~900 kW loss per circulation. So we have to make up for that loss and call that our grid out power.
That means if zero electricity is produced and nothing is lost, we have a Q(eng) ~0.05.
Qeng = ηth * ηE * (Qsci +1)−1
0.95 * Q(sci) + 0.95 -1 = 0.05
Q(sci) = 0.105
So if I did not make any mistakes anywhere, then Helion does not even need a D-D Q(sci) of 0.11 to produce small amounts of net electricity with Polaris.
Does that sound right?
And how does that translate to a feasible triple product, anyone know from the top of their heads?