WB6 Coil question

[tombo]

Has anyone here run the numbers on the heat flow through the copper and across the solid/liquid interface to the water/ln2 ?
I expect there to be a maximum copper thickness we can use.
i.e. a maximum distance we can expect to have the heat travel.
Even a ballpark number for now would be useful in coil design.

[MSimon]

Has anyone here run the numbers on the heat flow through the copper and across the solid/liquid interface to the water/ln2 ?
I expect there to be a maximum copper thickness we can use.
i.e. a maximum distance we can expect to have the heat travel.
Even a ballpark number for now would be useful in coil design.

The easiest way to work it out is to assume a power. Get your LN2 flow right - power vs delta T vs flow. then work back to thickness and see if it will support the delta Pressure and the current flow.

I'll see if I can dig up my Bitter Coil calculations and post them.

[eros]

Tony,

If we were to build net power deuterium-burning machines using copper magnets, magnet power would still be a major consideration. You might still be able to reach net power by some profitable margin, but ohmic magnet heating will never be trivial in copper. Dropping resistance dramatically using LN2 would be a huge benefit if it can be done practically, in which case a cooling jacket with something like borated (B10) water would probably be essential to minimize heat loading the copper with fusion energy. Silver conductors might marginally improve the situation.

I am quite sceptic that copper magnets are posible gain energy. 0.017ohm/mm² is totally too much.
Say 1 meter coils, 30cm thick copper coils, 0.13T field, give ~1W fusion power, but coil consumes ~400kw. and 30cm coil is maybe too thick..?

[MSimon]

Tony,

If we were to build net power deuterium-burning machines using copper magnets, magnet power would still be a major consideration. You might still be able to reach net power by some profitable margin, but ohmic magnet heating will never be trivial in copper. Dropping resistance dramatically using LN2 would be a huge benefit if it can be done practically, in which case a cooling jacket with something like borated (B10) water would probably be essential to minimize heat loading the copper with fusion energy. Silver conductors might marginally improve the situation.

I am quite sceptic that copper magnets are posible gain energy. 0.017ohm/mm² is totally too much.
Say 1 meter coils, 30cm thick copper coils, 0.13T field, give ~1W fusion power, but coil consumes ~400kw. and 30cm coil is maybe too thick..?

So you make the reactor bigger. Coil power goes up linearly with size. Output power goes up as the 7th power of linear size.

[eros]

So you make the reactor bigger. Coil power goes up linearly with size. Output power goes up as the 7th power of linear size.

I make calculation sheet for that purpose. Yes if go Big enough breakeven is posible. But it found at 2.5km core and 250m thick coils if use 10% thickness rule. Breakeven come ~5GW power level.
I am quite sure that 2.5km machine is imposible because it use so much copper.. And that is only breakeven. If put same coil 1T field coils take 500GW, but fusion power 50000GW.
In 10m machine 1m coil breakeven occur in 30T field power level ~2600GW

If want copper machine startup energy to be realistic then it needs to be super giant. Further throtling B give much gain.

And ewerything depend copper resistivity 0.017ohm/mm². I see breakeven is only posible with superconductors. 1m machine 5T superconduit give ~2MW, with 10T it give 34MW.

If want drop some values, I can put them in calulator.

But please check calculations. You find B field formula, copper resistivy, find WB6 size and power. I user scaling rule B^4 * r^3

[stevethetinker]

I was told once some years ago that if you published an idea, you had a year to patent it. Might be worth looking into. The only thing I see about your idea is that it has the same problems that anything has that does not use round conductors: the electrons tend to collect at the regions of highest curvature, and the extra current there causes uneven heating. But then you knew that. Just wanted to let you know about the patent thing.

[D Tibbets]

So you make the reactor bigger. Coil power goes up linearly with size. Output power goes up as the 7th power of linear size.

I make calculation sheet for that purpose. Yes if go Big enough breakeven is posible. But it found at 2.5km core and 250m thick coils if use 10% thickness rule. Breakeven come ~5GW power level.
I am quite sure that 2.5km machine is imposible because it use so much copper.. And that is only breakeven. If put same coil 1T field coils take 500GW, but fusion power 50000GW.
In 10m machine 1m coil breakeven occur in 30T field power level ~2600GW

If want copper machine startup energy to be realistic then it needs to be super giant. Further throtling B give much gain.

And ewerything depend copper resistivity 0.017ohm/mm². I see breakeven is only posible with superconductors. 1m machine 5T superconduit give ~2MW, with 10T it give 34MW.

If want drop some values, I can put them in calulator.

But please check calculations. You find B field formula, copper resistivy, find WB6 size and power. I user scaling rule B^4 * r^3

After just reading the above post and associated posts I am suspicious of the above calculations. Perhaps he multiplied diameter by 7 as opposed to using the 7th power of diameter. I'm too lazy to carefully follow the math, but using examples with the presumed power gain = diameter to the 7th power (ratio of diameter change ^7th power)
If size is increased from 2.5 meters to 2.5 kilometers, the power would go up by a facter of 1000 to the 7th power, or 10 ^ 24th . A LARGE number. If you start at 2.5 meters diameter, and increase to 5 meters then power output increase would be 2 ^7th power. = 128 x increase. Using the conservative net gain of diameter ^5th power ( uncetain of energy input costs. I've seen mention scaling of ~ 2nd power- from increased magnetic power and drive energy needed, giving a net gain of diameter to the 5th power?). So, to overcome the above ratio of 1 watt out / 100,000 watts in, the size would need to be increased from 2.5 meters to 25 meters (ratio of 10). 10^5= 100,000. Or, using only the linier increase in resitivity losses mentioned earlier in this thread the power gain would be diameter to the 6th power (even smaller size needed).
Also, the power output for the example 2.5 meter machine is given as 1 watt. If the 7th power gross gain scaling is applied to the almost 1 milliwatt output of WB6 at 30 cm diameter would give an output of
( 250cm/30cm = 8 x increase in diameter-> 8^ 7th power) ~2,000,000 gross watts out for a 2.5 meter machine. If the magnet input power is linier the input power would increase from 100,000 watts (assuming magnet energy is much larger than electron drive energy) to 800,000 watts. I don't know how much the electron input energy increases ( perhaps ~ same or mldly higher ratio than the magnetic energy based on above arguments, and estimates that ~ 3 meter diameter is needed for net power machines).

I'm uncertain where the copper resistance value used above comes from (diameter and length of wire x resistance in the example). Is this the resistance for the total length of copper wire ? I'm guessing that this is at room temerature. As mentioned by M. Simon earlier in this thread, the resitance of the copper decreases at colder temperatures. If cooled to near liquid nitrogen temperatures the resistance would be 4-5 (?) times less, decreasing the benifits from and the complications of superconductivity.
A nice calculater to show the resitivity vs temperature of coper is at:

http://circuitcalculator.com/wordpress/ ... alculator/

Forgive my wordy post, it could no dought be more percisly presented, but I hope it gets my points across. If my assumptions, or math are too far off please correct me.

Dan Tibbets

[MSimon]

At 83K (a good number for engineering purposes) Cu has 1/6th the resistance it does at 25 deg C.

[eros]

After just reading the above post and associated posts I am suspicious of the above calculations. Perhaps he multiplied diameter by 7 as opposed to using the 7th power of diameter. I'm too lazy to carefully follow the math, but using examples with the presumed power gain = diameter to the 7th power (ratio of diameter change ^7th power)
If size is increased from 2.5 meters to 2.5 kilometers, the power would go up by a facter of 1000 to the 7th power, or 10 ^ 24th . A LARGE number. If you start at 2.5 meters diameter, and increase to 5 meters then power output increase would be 2 ^7th power. = 128 x increase. Using the conservative net gain of diameter ^5th power ( uncetain of energy input costs. I've seen mention scaling of ~ 2nd power- from increased magnetic power and drive energy needed, giving a net gain of diameter to the 5th power?). So, to overcome the above ratio of 1 watt out / 100,000 watts in, the size would need to be increased from 2.5 meters to 25 meters (ratio of 10). 10^5= 100,000. Or, using only the linier increase in resitivity losses mentioned earlier in this thread the power gain would be diameter to the 6th power (even smaller size needed).
Also, the power output for the example 2.5 meter machine is given as 1 watt. If the 7th power gross gain scaling is applied to the almost 1 milliwatt output of WB6 at 30 cm diameter would give an output of
( 250cm/30cm = 8 x increase in diameter-> 8^ 7th power) ~2,000,000 gross watts out for a 2.5 meter machine. If the magnet input power is linier the input power would increase from 100,000 watts (assuming magnet energy is much larger than electron drive energy) to 800,000 watts. I don't know how much the electron input energy increases ( perhaps ~ same or mldly higher ratio than the magnetic energy based on above arguments, and estimates that ~ 3 meter diameter is needed for net power machines).

I'm uncertain where the copper resistance value used above comes from (diameter and length of wire x resistance in the example). Is this the resistance for the total length of copper wire ? I'm guessing that this is at room temerature. As mentioned by M. Simon earlier in this thread, the resitance of the copper decreases at colder temperatures. If cooled to near liquid nitrogen temperatures the resistance would be 4-5 (?) times less, decreasing the benifits from and the complications of superconductivity.

1st scaling rule is B^4 * r^3. So if size grow power grow at same rate as volume. If magnet field grow power grow ^4. Some losses grow same rate as field.

Broblem is copper resistivity, you can pack lot of copper and achive high field, but then you lost field gain totally in copper loss. I asking people to check my calculations.
There is maybe error pi() in field calculation and real field is 3.1415.. more than my calculator say so things go little nicer. And further LN2 cooling drops resistivity 5x.

If I add these optimistic corrections I got:
10m machine 1m coil, 3.5T 505MW coil power and 513MW fusion power.
3m machine, 0.3m coil 11.9T 1826MW coil power and 1851MW fusion power.

and research machine:
1m machine, 0.3m coils 0.13T, 7.72kw coil power and 0.98w fusion power.
And if pump coil to 11.9T 66.7MW it give 68MW fusion power. So pulse mode breakeven demo is maybe posible..?

Power is for six coils. No electron beam or other losses included.
However fusion power level can be higher than WB6 showed, but it is some value that we have data

[MSimon]

Let us take this a step at a time.

1. Coil volume goes up with the square of the linear size.
2. Coil resistance is a linear function of linear size

Thus coil resistance (we can assume a one turn coil for simple calculation purposes) declines linearly with size.

1. Field strength declines linearly with linear size (at constant current)

If we keep the heat load per unit length constant (increasing current as the square of linear size) the magnetic field increases linearly as the size increases. Coil power in that case goes up linearly with an increase in size.

Any disagreement so far?

[MSimon]

Let me try again.

I'm going to use the = sign to mean proportionate to.

R = linear size/coil area

R = linear size / (linear size)^2

So coil resistance is inversely proportionate to linear size.

R/unit length = 1/(linear size)^2

If we keep current per unit volume constant. Current goes up as the square of coil size. And the B field goes up linearly with coil size. Coil power would then go up as the cube of linear size while B field goes up linearly.

Fusion power then goes up as the 7th power of linear size.