https://www.onlineconversion.com/energy.htm
hhttp://www.edasolutions.com/Groups/Tech/ComputerElectricalCalcs.htm#ELEC-IMPttp://www.monachos.gr/eng/calculators/kvar_calculator.htm
According to what I have learned thus far the capacitance of a capacitor is determined by the following equation:
[DxAx(N-1)]/S,
Where D=dielectric constant of the capacitor
A=the surface area of plates in centimeters
C=the capacitance in picofarads
N=the number of plates
S=the spacing between the plates.
Now, I have been working with a concept, based on some recently declassified military capacitor designs, for creating an extremely powerful capacitor by increasing the surface area of the capacitor to extreme values while taking up relatively little volume. It is based on the so called Koch snowflake concept where the surface areas of a three dimensional solid can approach infinity as the volume remains constant by cutting the solid into smaller and smaller components, thus increasing the surface area exponentially.
Assume that one has a solid cube of metal with a surface area of
6x(1 meter)^2, corresponding to the six sides of a cube, that is 6 square surfaces composing the entire surface of the cube. Now if one cuts the cube directly in half, one adds two more sqaures to the surface area of the cube, so that the two solids now have a combined surface area of
8x(1 meter)^2. Without getting into the detailed mathematics, I found that you could feasibly cut the cube into 2^59 slices and get approximately 10^-18 meters thick slices. The cube can be sliced three ways to where you end up with an astronomical number of microscopic cubes each having a total surface area of around 6x10^-18 meter^2. The total number of plates contained is around 3x2^59 cubes total.
As it turns out the total surface area of all the cubes combined is
1,729,382,257,000,000,000 meters^2 of surface area all contained in around 3 meters^3 of volume. This is an immense amount of surface area to be contained in a box that is a little bigger then the average sized person.
This is around 172,938,225,700,000,000,000 centimeters^2 of surface area.
When plugged into the capacitance equation, where the dielectric constant is equal to 5, the number of plates is 3x2^59 plates total, and the plates are seperated by approximately 10^-20 centimeters (at at least one point, the cubes are a type of typical carbon-based liquid oil and assume the form of spherical, or ball shaped, dropplets when cut, which reduces the probability of a short circuit at close ranges and also helps out with "the real world collision" problem)
Then the capacitance is approximately 1.495381495x10^59 picofarads, which translates into 1.495381495x10^50 Farad Capacitor.
in expanded form, (written out), that is 14,953,814,950,000,000,000,000,000,000,000,000,000,000,000,000,000 Farads. The question I have is, how much energy in joules would this capacitor release? The initial voltage has an upper limit of approximately 1 volt, so this capacitor is rated as a 1 volt capacitor. Anybody want to take a wack at it? I've not found any reference that would allow one to calculate the energy a capacitor releases based solely on the farad rating of the capacitor. All references I've seen state that the power yield of the capacitor in watts is determined by the time constant of the capacitor in seconds by the following equation.
Vd=(1/E^Tc)*Vi
Where,
Vd= Discharge Voltage
Vi= Initial voltage, or terminal voltage
Tc=time constant = (Time T in seconds x Resistance R of Capacitor in ohms x Capacitance C of Capacitor in farads)
R= Resistance in ohms
C= capacitance in farads
How much energy would a 14,953,814,950,000,000,000,000,000,000,000,000,000,000,000,000,000 Farad capacitor release?
Inquisitively,
Rand
hhttp://www.edasolutions.com/Groups/Tech/ComputerElectricalCalcs.htm#ELEC-IMPttp://www.monachos.gr/eng/calculators/kvar_calculator.htm
According to what I have learned thus far the capacitance of a capacitor is determined by the following equation:
[DxAx(N-1)]/S,
Where D=dielectric constant of the capacitor
A=the surface area of plates in centimeters
C=the capacitance in picofarads
N=the number of plates
S=the spacing between the plates.
Now, I have been working with a concept, based on some recently declassified military capacitor designs, for creating an extremely powerful capacitor by increasing the surface area of the capacitor to extreme values while taking up relatively little volume. It is based on the so called Koch snowflake concept where the surface areas of a three dimensional solid can approach infinity as the volume remains constant by cutting the solid into smaller and smaller components, thus increasing the surface area exponentially.
Assume that one has a solid cube of metal with a surface area of
6x(1 meter)^2, corresponding to the six sides of a cube, that is 6 square surfaces composing the entire surface of the cube. Now if one cuts the cube directly in half, one adds two more sqaures to the surface area of the cube, so that the two solids now have a combined surface area of
8x(1 meter)^2. Without getting into the detailed mathematics, I found that you could feasibly cut the cube into 2^59 slices and get approximately 10^-18 meters thick slices. The cube can be sliced three ways to where you end up with an astronomical number of microscopic cubes each having a total surface area of around 6x10^-18 meter^2. The total number of plates contained is around 3x2^59 cubes total.
As it turns out the total surface area of all the cubes combined is
1,729,382,257,000,000,000 meters^2 of surface area all contained in around 3 meters^3 of volume. This is an immense amount of surface area to be contained in a box that is a little bigger then the average sized person.
This is around 172,938,225,700,000,000,000 centimeters^2 of surface area.
When plugged into the capacitance equation, where the dielectric constant is equal to 5, the number of plates is 3x2^59 plates total, and the plates are seperated by approximately 10^-20 centimeters (at at least one point, the cubes are a type of typical carbon-based liquid oil and assume the form of spherical, or ball shaped, dropplets when cut, which reduces the probability of a short circuit at close ranges and also helps out with "the real world collision" problem)
Then the capacitance is approximately 1.495381495x10^59 picofarads, which translates into 1.495381495x10^50 Farad Capacitor.
in expanded form, (written out), that is 14,953,814,950,000,000,000,000,000,000,000,000,000,000,000,000,000 Farads. The question I have is, how much energy in joules would this capacitor release? The initial voltage has an upper limit of approximately 1 volt, so this capacitor is rated as a 1 volt capacitor. Anybody want to take a wack at it? I've not found any reference that would allow one to calculate the energy a capacitor releases based solely on the farad rating of the capacitor. All references I've seen state that the power yield of the capacitor in watts is determined by the time constant of the capacitor in seconds by the following equation.
Vd=(1/E^Tc)*Vi
Where,
Vd= Discharge Voltage
Vi= Initial voltage, or terminal voltage
Tc=time constant = (Time T in seconds x Resistance R of Capacitor in ohms x Capacitance C of Capacitor in farads)
R= Resistance in ohms
C= capacitance in farads
How much energy would a 14,953,814,950,000,000,000,000,000,000,000,000,000,000,000,000,000 Farad capacitor release?
Inquisitively,
Rand