Tuesday, December 05, 2006

Reactor Scaling

I have been doing a little back of the envelope calculations on the scaling laws for the Bussard Reactor.

Power scales as the 7th power of volume. Reactivity scales as the 5th power. If we assume unit radius and reactivity for a 1 MW unit what will be the relative radius and reactivity for other power outputs?


Power Radius Volume Reactivity
1,000MW 2.68 19.31 138.950
100MW 1.93 7.20 26.827
10MW 1.39 2.68 5.180
1MW 1.00 1.00 1.000
100KW .72 .37 .193
10KW .52 .14 .037



So control will be a problem with larger outputs due to high gain and sufficient reactivity will be a problem with smaller outputs due to low gain. It will be interesting to see what the actual reactivity is in the test reactor.

11 comments:

Tom Cuddihy said...

something's not quite right with that. In his google talk, Dr. Bussard said the first full-power model should be a 3-m 40MW reactor.

You have a 3-m reactor being over a GW.

I thought he said power scales as the 7th power of the size (p 26 of the IAC paper). (B^4R^3), with the B being proportional to the radius, thus P:r^7, not v^7?

M. Simon said...

Radius to the 7th power = power

Setting the radius for 1 MW = 1.00

1.39 to the 7th power = 10

1.39 to the 3rd power = 2.68

All the above are ratiometic numbers and not actual diameters or radii.

Tom Cuddihy said...

ah, ok.

Well, using Bussard's numbers, I did a 5 min BOE in excel.
V= (4/3)*pi*r^3 P ∞ V^7
psf = v^7*40/(113.097)^7
1.69003E-07
vol(m^3)rad(m) P (W) (MW)
0.52 0.50 142.89 0.00
1.77 0.75 2441.41 0.00
4.19 1.00 18289.90 0.02
8.18 1.25 87213.04 0.09
14.14 1.50 3.13E+05 0.31
23.27 1.77 1.00E+06 1.00
33.51 2.00 2.34E+06 2.34
65.45 2.50 1.12E+07 11.16
113.10 3.00 4.00E+07 40.00
268.08 4.00 3.00E+08 299.66
523.60 5.00 1.43E+09 1428.90
904.78 6.00 5.12E+09 5120.00
1436.76 7.00 1.51E+10 15062.52
2144.66 8.00 3.84E+10 38356.70
3053.63 9.00 8.75E+10 87480.02

M. Simon said...

Tom,

You are using actual values.

Mine are normalized to 1 MW.

Notice how everything on the 1 MW line is 1.00?

Tom Cuddihy said...

Yep, understand you're normalized.

You assumed a 1MW RX for 1 m. Just pointing out a 1 m reactor actually only gets you 20 kW using Bussard's reference: 3 m for 40 MW. A 1GW Rx is about 4.5 m.

Tom Cuddihy said...

so a nominal 1 MW RX (for thermal testing, etc) would be 1.77 m

M. Simon said...

You assumed a 1MW RX for 1 m..

Nope. Normalization removes the units.

100 MW radius/1 MW radius = 1.93

M. Simon said...

BTW Bussard says 100 MW for 3 m.

Power MW Radius m
1000 4.17
100 3.00
10 2.16
1 1.55
0.1 1.12
0.01 0.80
0.001 0.58


Radius m Vol. m^3
4.17 303.4057465
3.00 113.0973355
2.16 42.15809177
1.55 15.71482381
1.12 5.857847857
0.80 2.183567688
0.58 0.81394532

I'll send you my spread sheet if you will send me an e-mail.

M. Simon said...

3/1.55 = 1.93

Jeremy said...

Hmmm. A more important question, IMHO, is how to cool the reactor. There will be un upper limit to the power of the device, and this, from my understanding, won't be much bigger than break even sizing. Assuming a 100MW figure postulated at 95%alpha production, and ignoring anode losses, this leaves between 1 and approaching 5MW of heat encapsulated in an evacuated sphere no more than 5 metres across. Nasty. Even keeping the inevitable x-rays out of it. The happy reactor size between break even and meltdown I hereby designate the "M.Simon Region".

M. Simon said...

jeremy,

Cooling is going to be a very tough job.

Not so much the reactor vessel which will be hard enough.

Keeping the magnet coils at 20 deg K with a heat flux estimated at 5 to 10 MW is going to be very difficult.

I'm thinking a 3 coolant system.
Water - vacuum - LN2 - vacuum - LHe.

In a lot of ways the heat flux resembles what is required for the Space Shuttle Main engines.