## Energy Balance Formula for Polywell

Discuss how polywell fusion works; share theoretical questions and answers.

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Shubedobedubopbopbedo
Posts: 62
Joined: Wed Jun 27, 2007 9:38 pm

### Energy Balance Formula for Polywell

I posted this on the Nasa Spaceflight forum in June. Note that at the time the constants K1 and K2 were unknown for Bussard's scaling rule. If anyone would care to fill them in, it would show some progress here, and we could actually estimate the size of the reactor for a desired power output.

Hypothetical Solved Problem for Bussard Polywell Net Power Reactors:
---------------------------------------------------------------------------------------
The tentative formulas for power gain were given as:

Power Gain = FusionPower / DrivePowerLoss
Power Gain = (k2)*(B^4)*(R)

So the radius of the reactor volume would be:

R = PowerGain / (k2 * B^4)

Like I said, if anyone knows what the value of K2 is, it would be very useful here.

By inspection of the image linked to below of WB6 I estimated that the magnets would block 20 percent of the reaction products and so 20 percent of the Fusion Power would be dumped as heat into the magnets.

http://i194.photobucket.com/albums/z197 ... lywell.jpg

===================================
Here's a power balance for D-D reactions in a Polywell reactor:
===================================
If the Power Gain = 4 and the Fusion Power = 100 MW, then

- 35 MW of fusion power is converted to Bremmstrahlung radiation, of which 7 MW is dumped as heat into the ring magnets. The rest escapes to the walls.
- 20 MW of fusion power is dumped as heat into the ring magnets when reaction products including neutrons hit them.
- 73 MW of heat is dumped into the walls by reaction products and Bremms-radiation, of which 29 MW might be converted to electricity using a heat engine at 40 percent efficiency.
- 25 MW of electricity is used to drive the magnets and electron guns.
- 0.269 kW of electricity is used to drive the water pumps to replenish water flashed to steam in the water jackets surrrounding the magnets which absorb 27 MW of heat.
=======================================

Net Power = 3.731 MW.

Some of that power is needed for vacuum pumps, etc.

===================================
Here's a similar power balance for D-He3 reactions in a Polywell reactor:
===================================
If the Power Gain = 4 and the Fusion Power = 100 MW, then

- 19 MW of fusion power is converted to Bremmstrahlung radiation, of which 4 MW is dumped as heat into the ring magnets. The rest escapes to the walls.
- 20 MW of fusion power is dumped as heat into the ring magnets when reaction products including neutrons hit them.
- 76 MW of heat is dumped into the walls by reaction products and Bremms-radiation, of which 30 MW might be converted to electricity using a heat engine at 40 percent efficiency.
- 25 MW of electricity is used to drive the magnets and electron guns.
- 0.187 kW of electricity is used to the water pumps to replenish water flashed to steam in the water jackets surrounding the magnets.
=======================================

Net Power = 4.813 MW.

=====================================
Here's a similar power balance for D-T reactions in a Polywell reactor:
=====================================
If the Power Gain = 4 and the Fusion Power = 100 MW, then

- 0.7 MW of fusion power is converted to Bremmstrahlung radiation, of which 0.14 MW is dumped as heat into the ring magnets. The rest escapes to the walls.
- 20 MW of fusion power is dumped as heat into the ring magnets when reaction products including neutrons hit them.
- 79.86 MW of heat is dumped into the walls by reaction products and Bremms-radiation, of which 31.9 MW might be converted to electricity using a heat engine at 40 percent efficiency.
- 25 MW of electricity is used to drive the magnets and electron guns.
- 0.111 kW of electricity is used to drive the water pumps to replenish water flashed to steam in the water jackets surrounding the magnets.
=======================================

Net power = 6.789 MW

No other loss mechanisms are accounted for, such as synchrotron radiation, power needed to run the vacuum pumps, to process exhaust gases, to refine fuel, and transmission losses.

Can anyone tell me what the reactor radius would be to achieve these fusion power levels?

In every case that's less than 7 percent net power. Someone said that this is the technology that would power an industrial and capitalist society and that my Wind Ladderwouldn't work. But my inflatable wind powerplant gets 95 percent net power even with an integrated self-inflation device. Even if it only manages to generate 25 percent of rated power on average due to poor wind conditions, it still yields 80 percent net power over-and-above what it consumes to replenish its hydrogen inflation gas.

I hate to be inflammatory, but the energy balance doesn't look favorable for Polywell.

MSimon
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That was an estimation in which the numbers were just guesses.

Some of them are two orders of magnitude off.

First off the the 78 MW going to the walls is going either be used to make steam or it will be direct converted to DC. Direct conversion will be about 70% efficient.

Direct conversion = 55 MW power out (or better)
Stem plant conversion would give about half that - 30 MW

Reactor drive power is high in my estimation. Should be more like 5 MW.

However, despite the precision of the numbers their accuracy is in doubt.

jlumartinez
Posts: 143
Joined: Thu Jun 28, 2007 7:29 pm
Location: Spain
As I said you in June in NasaSpaceFlight forum you don´t know yet what the fusion gain is.

Fusion gain, Q = Energy output / Energy input

The real energy produced by Polywell is 100 MW: some wasted as heat and some useful as electricity

In your cases Q = 100 MW / 25 MW+ 0.000111 MW ( 0.111KW used for the pumps and 25 MW for the magnet) = 4.

25 MW to run a magnet and electron guns ??? Do you know what a MW is ??

Please first check you assumptions and later show us how you calculated those numbers. In June I asked you for the equations but no answers is still received. Again, what are the equations or references used to do your numbers?. Even the best nuclear engineer can still calculate the Polywell results. But you have done! Amazing! You keep on amazing me day by day: in June with your centrifugal pumps consuming 20 MW ( you haven´t seen any in your life, later you corrected to 0.111 Kw more normal but now too low almost like a bulb...not good calc yet ) . Yesterday with your helium condensers. Later mixing argon with electrolytic hydrogen. Have you ever seen a magnet of electron gun consuming 25 MW !!!. Do You know how much is a MW? 1 MW = Power for 200 houses

Please get concentrated in your wind mill. If you become a millonaire please advice me for being too stupid to not invest in such project. And a last recommendation: I think you are really a good boy, but very theorist. Please use more the experience

Shubedobedubopbopbedo
Posts: 62
Joined: Wed Jun 27, 2007 9:38 pm
25 MW is drive energy. I assume that means current x voltage for the 6 electromagnets. They drive the electrons magnetically toward the center reaction zone. That takes energy. For a gain = 4, that means drive power is 1/4th of the fusion power. Hence, 25 million watts (25 MW) drive power and 100 MW fusion power. The drive power is divided up so that 4.17 MW is consumed by each of six coils. Pretty simple. Obviously I have no reason to expect a gain = 4. But from my estimates of the losses, you would need a gain = 4 just to be respectable for net power production. Polywell wouldn't be very useful if it produced 100 MW of power but used it all up and left nothing to sell to consumers. It would be just another expensive research reactor like Z-Machine or the NIF.

Gain varies with reactor size. Most importantly in my estimates, and like MSimon says, I have no way of knowing if a reactor big enough to have a gain = 4 would also have a fusion power of 100 MW. Since the proportionality constants K1 and K2 were never provided by Dr. Bussard, there is no way to correlate the two estimates. They are therefore meaningless except as percentages. That is why I chose 100 MW fusion power, as a percentage reference. My estimates are really percentages of fusion power.

windmill
Posts: 29
Joined: Thu Jul 19, 2007 3:19 pm
We do have Bussard's own estimates of a 100Mw net power machine's gain, in his "Clean Energy Future" summary. He writes:

PROOF OF CLEAN FUSION POWER REQUIRES FULL SCALE TEST MACHINES (ca. 3 m DIAMETER, 40 MW DRIVE POWER, 15-25 kG MAGNETIC FIELDS, DRIVE ENERGIES UP TO 180 kV).

If this is accurate, (and why wouldn't it be?), the gain would be something on the order of 100+40/40, or about 3.5.

rexxam62
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Joined: Sun Jul 29, 2007 10:13 pm

### Re: Energy Balance Formula for Polywell

Shubedobedubopbopbedo wrote:Can anyone tell me what the reactor radius would be to achieve these fusion power levels?

In every case that's less than 7 percent net power. Someone said that this is the technology that would power an industrial and capitalist society and that my Wind Ladderwouldn't work. But my inflatable wind powerplant gets 95 percent net power even with an integrated self-inflation device. Even if it only manages to generate 25 percent of rated power on average due to poor wind conditions, it still yields 80 percent net power over-and-above what it consumes to replenish its hydrogen inflation gas.

I hate to be inflammatory, but the energy balance doesn't look favorable for Polywell.

I knew that is the reason you came here. To disscredit Nuclear Fusion and to disscredit ME you filthy... . NOBODY knows if Polywell Fusion will work or not. Everything is just educated guesses. Give someone the freakin chanse to test the concept first then we can judge if it works or not and start to talk about numbers. And dont even get me started with your balloons.

bcglorf
Posts: 436
Joined: Mon Jul 23, 2007 2:58 pm

### I think your missing something...

"Since the proportionality constants K1 and K2 were never provided by Dr. Bussard, there is no way to correlate the two estimates."

"The estimates for the k for power gain go as: n1*n2 *fusion cross section* colliding velocity. Where n1 and n2 are the densities (particles per unit volume) of the reactants. For D-D where both particles are the same it is (n^2)/2."
Short answer being the constants are dependent on configuration and Bussard left them out as mundane/standard physics.

As windmill kindly pointed out, Bussard's estimates for 1.5m radius give 100MW net power. Deriving the value of k2 for the described configuration is trivial.
gain=3.5
Magnetic Field ~ 25kG
k2=gain/(R*B^4)

As to getting that net power back out efficiently, even the worst cases in a Polywell design are a dream beside Tokomaks.

windmill
Posts: 29
Joined: Thu Jul 19, 2007 3:19 pm
...and if gain scales as the 5th power of the radius, just going from 1.5m to 2.0m radius will increase the gain by 1.33^5 or 4.2 times. Gain becomes 4.2 x 3.5 = 14.7!!!

I love this technology.

MSimon
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Location: Rockford, Illinois
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Shubedobedubopbopbedo wrote:25 MW is drive energy. I assume that means current x voltage for the 6 electromagnets. They drive the electrons magnetically toward the center reaction zone. That takes energy. For a gain = 4, that means drive power is 1/4th of the fusion power. Hence, 25 million watts (25 MW) drive power and 100 MW fusion power. The drive power is divided up so that 4.17 MW is consumed by each of six coils. Pretty simple. Obviously I have no reason to expect a gain = 4. But from my estimates of the losses, you would need a gain = 4 just to be respectable for net power production. Polywell wouldn't be very useful if it produced 100 MW of power but used it all up and left nothing to sell to consumers. It would be just another expensive research reactor like Z-Machine or the NIF.

Gain varies with reactor size. Most importantly in my estimates, and like MSimon says, I have no way of knowing if a reactor big enough to have a gain = 4 would also have a fusion power of 100 MW. Since the proportionality constants K1 and K2 were never provided by Dr. Bussard, there is no way to correlate the two estimates. They are therefore meaningless except as percentages. That is why I chose 100 MW fusion power, as a percentage reference. My estimates are really percentages of fusion power.

Update: I see where your error is. You don't understand Polywell. The magnets provide no drive. Zero. Nada. They provide shielding and confinement.

These are all theoretical, but we should come close in a real machine.

1. Magnet power is essentially zero even if LN2 cooled Cu magnet coils are used. Closer to zero with superconductors. For a 100 MWf output. (f standing for fusion)

2. Drive power - at 180 KV the energy gain is about a factor of 7. (1.2 Mev drive 8.4 Mev from fusion) If operated at the lower pBj resonance the gain goes up to around 22. At the cost of about 10X the volume or about 2X the linear size.

So let me ask you how did you determine coil power?

How did you determine anode power?

MSimon
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windmill wrote:...and if gain scales as the 5th power of the radius, just going from 1.5m to 2.0m radius will increase the gain by 1.33^5 or 4.2 times. Gain becomes 4.2 x 3.5 = 14.7!!!

I love this technology.

Actually there is an ultimate gain.

It is around 7 at 180 KV drive. (the 1.2 barn peak)
It is around 22 at 50 KV drive (the .1 barn peak)

The only thing that goes up in net power machines is the power output. The gain (power out/in) will not change much.

MSimon
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At

http://iecfusiontech.blogspot.com/

I explain how to calculate drive power. It is one of the first posts. Scroll down.

Shubedobedubopbopbedo
Posts: 62
Joined: Wed Jun 27, 2007 9:38 pm
I guess I'm not that interested. Fact is it's unproven technology, the physics is rather obscure, and Bussard isn't talking.

pstoller78
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Shubedobedubopbopbedo wrote:I guess I'm not that interested. Fact is it's unproven technology, the physics is rather obscure, and Bussard isn't talking.

So what your implying is Dr. Bussard is hiding something all because you aren't interested enough to take the time to understand the concepts presented to you by several members on this site. If your going to cast doubts please take the time to educate yourself on these concepts and provide well reasoned arguments as to why we should share your doubts.

JohnP
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Location: Chicago
- 20 MW of fusion power is dumped as heat into the ring magnets when reaction products including neutrons hit them.

Whatever else he's saying, this is an interesting point. Am I to assume that the magnets are getting a high pressure stream of LN2 to keep them cool while they're getting roasted with neutrons (in D-D)? I also assume that with p-B11 the helium nuclei will tend to fly off around the magnetic field lines and not at the magnets directly?

jlumartinez
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Joined: Thu Jun 28, 2007 7:29 pm
Location: Spain
it's unproven technology

Years ago laser, TV, chips, internet, telephones, cars, airplanes , even a simple bulb were also unproven technologies. Even a well-known scientific as Lord Kelvin stated that anything heavier than air will never fly ("Heavier-than-air flying machines are impossible". Lord Kelvin, ca. 1895) Some years later the technology evolved and airplanes were developed.

Now that they have changed the world nobody doubt about them. In the earlies XIX century the main energy source was coal, in the earlies years of XX century petrol was becoming the main energy source. In the middle years of XX century the fission energy was developed . Which is the next step?? I am relly expecting that in the earlies years of XXI century the fusion will become a reality. Polywell is a great candidate to make it real. Polywell is better than Tokamaks because it focus the ions to the center forcing them to fuse (as in the Sun). In the other side, Tokamaks is just a "bowl" for the plasma but this doesn´t force the ions to fuse , just keep them hot to let fusion occur.