Calculation of the energy from a pulse

AI Thread Summary
The discussion focuses on calculating the energy released by two capacitors in parallel during a discharge process. The energy stored in the capacitors can be calculated using the formula E = 1/2 CV², where C is the capacitance and V is the voltage (18.5 kV). To find the energy of each pulse, the area under the discharge curve must be determined, which requires plotting the voltage over time and integrating the resulting curve. The discharge characteristics depend on the circuit configuration, particularly the resistance and the behavior of the spark gap. Accurate calculations necessitate a clear understanding of the circuit and the parameters involved, including the resistance value, to ensure reliable energy estimates.
Sara87
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Hello
I need help for a basic question (i am awful in physics) :confused:
I have two capacitors in parallel when they are charged up (18.5 kV) they release the energy to a discharge device. The discharge device releases the energy in the form of electric pulses through two electrodes in a chamber (the pulses will go through the liquid in the chamber for pulsed electric field treatment)
An exponential decay waveform is generated at the applied voltage.
What I need is to calculate the area under the discharge curve of the capacitor to have the energy of each pulse.
But i don't know which parameters, equations and ... do i need.

I would appreciate any help.

Regards,
Sara
 
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Are you asking us how to calculate the area underneath a given curve?
 
I don't have the curve actually, I guess I need to draw it and yes have the area under.
 
The energy stored in the capacitor will be CV2/2 and that is the maximum amount of energy that will be delivered, irrespective of the time of the discharge. I would say that you are very unlikely to get a simple 'exponential' discharge because the load could well not be an Ohmic Resistance (hence no RC)

If you can measure and record the Volts across the Capacitor at a high enough sample rate, you could plot the discharge curve and calculate the energy supplied in any short instant by the difference between the energy at the start and end of the interval. That could be expensive if the total discharge time is very short. What are the time intervals involved?
 
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This seems like a homework question to me.
 
Let me to explain it better.
I have these devices as you can see in the picture attached.
The capacitor become charged up to 18.5 kV and releases the energy to the discharge switch and then the energy in form of pulses go through the electrodes in the chamber.
I am asked to calculate the energy which goes from the electrodes to the liquid in the chamber.
I guess for that I need to draw the exponential decay curve and calculate the area underneath.
I don't know how and where to start.
Please help me from the scratch (I have forgotten all the basic physics from high school)

P.S. This is not a homework. I am a master student working on my final project :sorry::cry:
 

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Sara87 said:
energy in form of pulses

It isn't clear what is actually happening in this circuit.
What forms the pulses? Are you referring to some sort of high voltage oscillation due to the spark gap and some resonance in the circuit or does each pulse get generated by the 'Grey box' in the picture? What initiates the discharge? Is it when the spark gap breaks down?
Can we have a proper circuit schematic of the setup?
 
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The owner of the device told me they are exponential decay pulses. yes it gives high voltage pulsed electrical field (20 kV/cm) within the liquid material in a treatment chamber in a very short period of time (4 µs).
Yes the discharge will happen at the breakdown voltage of the spark gap.
Unfortunately I don't have any better picture.
 
Sara87 said:
The owner of the device told me they are exponential decay pulses. yes it gives high voltage pulsed electrical field (20 kV/cm) within the liquid material in a treatment chamber in a very short period of time (4 µs).
Yes the discharge will happen at the breakdown voltage of the spark gap.
Unfortunately I don't have any better picture.
Unless the discharge is through an Ohmic resistance, I don't see how the discharge can he Exponential - unless there is a large resistive component in the supply and if you treat the arc as a short circuit. But in that case, it would be very hard to estimate the energy supplied to your chamber. If this project is important to you then you should make sure that you get the theory right - unless your equipment is calibrated to tell you the correct value for the Energy transferred.
I don't want a "picture" I (you and we) want / need a proper schematic diagram to show the function of the circuit. Failing that, you just have to rely on what the 'owner' has told you. There is no simple answer to this.
 
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Maximum energy from the power supplier unit is 200 J/s which goes to the capacitor. there will be some losses in the wires but it can be neglected. the spark gap device (PerkinElmer’s OGP ceramic-metal overvoltage spark gaps) have a resistor inside which I don't know the R value for that. When they reach the breakdown voltage they release the pulses.
I was told to calculate the
rc14.gif
and then I will be able to calculate the area under the curve in the picture attached.
 

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  • #11
You have the functions V(t), I(t) and τ=RC=1 μs-1.
But, energy is not the area under the curves. The axis is V(t), not V(I).
We calculate energy by:
$$ E = \frac{1}{2}CV_0^2 $$
To calculate C we must know the value of R.

If we have no this information we can see that the charge currency is the same to uncharge. So energy must be:$$ E=\int_0^{5\tau}V(t)I(t)\,dt = V_0I_0\int_0^{5\tau}(1-e^{-t/\tau})e^{-t/\tau}\,dt \approx \frac{1}{2}CV_0^2 $$ If we can assume that:
$$ V_0I_0 \approx 200 (w) $$ approx the source power then we have:$$ E \approx \frac{1}{2}V_0I_0\tau \approx 100\tau = 10^{-4} (J)$$per cycle.
Integral calculated from 0 to infinity and is τ/2.
 
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  • #12
Cycle time is 10τ so the power is:
$$ P = \frac{10^{-4}}{10^{-5}} = 10 (watts) = 5\% \text{ of power supply or less} $$
 
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