Temperature/Voltage needed for Nuclear Fusion from Electrical Charge?

[CNBA3]

I was wondering, for those who know these kind of things, what is the temperature needed to achieve nuclear fusion with electrical fields?

This is based on the item called Fusor.

And by proxy, what would the voltage be for such a process?

Fusor - Wikipedia

en.wikipedia.org

Otherwise, would the temperature still be the same as regular nuclear fusion temperatures of 120 million degrees?

EDIT: I looked further into it and found something with "For fusion to occur the ions must be at a temperature of at least 4 keV (kiloelectronvolts), or about 45 million kelvins."

Thoughts?

 

[CNBA3]

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[CNBA3]

So I was doing a bit of calculation with how Fusors utilize Voltage for heating up ions to achieve nuclear fusion and this is what I found. I will be utilizing the Ohm's Law as well.

First we convert Electron-Volts (30000) or 30 KeV to that of Voltage with using an elementary charge as Coulombs (1.60217663e-19) during the process which 30000 Voltage.


Container = 97.04 px = .23 meters

Fusor = 30.97 px = .07 meters

The measurement of the container size is based on being roughly the length of an average human foot (Note the image is bad quality).

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Length = 75.63 px = .07 meters

Diameters = 40 px = .04 meters

Radius = .02 meters

The Cross Section Area = 0.0012566 m^2

We would need to find the Ohms from the Ohms meters of the material, in this case I will be using Tungsten 4.9e-8 Ohms meters which is suggested to be one of the preferred materials for Fusors.

(4.9e-8 / .07 length) * 0.0012566^2 cross section area = 1.1053305e-12 Ohms


Amps = 30000 / 1.1053305e-12 = 2.7141203e+16 Amps

Then we shall find the number of Watts from multiplying Amps and Voltage which is:

814236090000000024576.00 Watts

Thoughts?

@AbaddonTheDisappointment @Agnaa @DMUA

 

[CNBA3]

@Psychomaster35 @Damage3245

 

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@Flashlight237

 

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@CloverDragon03

 

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[DontTalkDT]

The temperature obviously depends on pressure.

What your calc is concerned... Why and how are you trying to convert an energy unit (keV) to a voltage?
I also don't think the container has much to do with how the electric field that's produced.
I also don't think ohms law is applicable since I think the voltage associated with the field, doesn't qualify as that of a current running through a conductor.
It is in any case wrong, as it's several orders of magnitude beyond what any real life nuclear fusion device uses or produces. You would need to detonate 12 million Little Boy nuclear weapons each second to cover the energy consumption you predicted.
For reference, ITER will use 300000000 watt to infuse 840 m^3 of plasma with 50000000 watt of power. Wendelstein 7X, a magnetic confinement reactor of another type, uses 14000000 watt to cause fusion in 30 m^3 of plasma. Under comparable pressure conditions, those can be taken as the upper limit of what the result would be. Sci-fi or magic devices could easily be much more efficient, as they likely solved the problem of keeping the energy in the plasma.


After all these years, are you still trying to get Wall's nuclear fusion reactor calculated to values far beyond any real life nuclear fusion reactor, while using the parameters equal or less than real life nuclear fusion reactors as parameters?

 

[CNBA3]

The temperature obviously depends on pressure.

What your calc is concerned... Why and how are you trying to convert an energy unit (keV) to a voltage?
I also don't think the container has much to do with how the electric field that's produced.
I also don't think ohms law is applicable since I think the voltage associated with the field, doesn't qualify as that of a current running through a conductor.
It is in any case wrong, as it's several orders of magnitude beyond what any real life nuclear fusion device uses or produces. You would need to detonate 12 million Little Boy nuclear weapons each second to cover the energy consumption you predicted.
For reference, ITER will use 300000000 watt to infuse 840 m^3 of plasma with 50000000 watt of power. Wendelstein 7X, a magnetic confinement reactor of another type, uses 14000000 watt to cause fusion in 30 m^3 of plasma. Under comparable pressure conditions, those can be taken as the upper limit of what the result would be. Sci-fi or magic devices could easily be much more efficient, as they likely solved the problem of keeping the energy in the plasma.


After all these years, are you still trying to get Wall's nuclear fusion reactor calculated to values far beyond any real life nuclear fusion reactor, while using the parameters equal or less than real life nuclear fusion reactors as parameters?

for the last part, from the images i posted would say no, This is different, but to clarify on that still one final time as to not go in too deep here, there is no mention of any reactor and the process is external of Wall's body as the material for it is drawn from the atmosphere. but that is besides the point.

now onto the rest of your post, the Conversation of the electron-volt to voltage is to convert it later to watts. I did this by using a calculator website with adding the Coulomb value for the conversion.

from what i tried to find with how to get Ohms, which would require doing some physical measurements of said fusor container with it's length and cross section and using the Ohms meter of said material.

For this case, it is a device created from a by a race that transcended conventional science and philosphy.

 

[CNBA3]

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[CNBA3]

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[CNBA3]

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[CNBA3]

I just realized that I made an error with dividing instead of multiplying the ohms meters by the length in meters