I'm looking for a mathematician..
Hello..
This is about a theory:
I have 10 ultracaps of 2.7v and 3000f each.
I would like to build these 10 capacitors in series for a 24v system.
How long can I operate my 1000 watt water pump with the capacitors?
Thank you
(1 votes)
Loading...I want to connect a new battery, but yesterday I noticed that the positive and negative cables are missing? What can I do to connect a battery anyway? I own a Peugeot Ludix 50 scooter.
Hello. My socket is sparking. Should I take it out of service? Regards :3
I have this 13-year-old MacBook charging cable. Is it dangerous?
Hello How do I know if we have high voltage?
Hello, I can write as an answer: You attach the crocodile cables with the two sockets under the bulb and the beginning of the cables with the minus and plus pole cables I don't know how to phrase it, can you help me?
Hello. I'm taking the Energy Engineering module and preparing for the exam. I can't answer the following questions. What is the maximum power that can be delivered by the machine at the grid connection point in continuous operation? How can the synchronous reactance effective during nominal operation and during the operation mentioned above be calculated?
In order to calculate this, I assume that the 1000W applies at a voltage of 24V and the pump has a constant resistance.
How fast a capacitor is discharged depends on the time constant τ (Tau). Calculate by multiplying the capacitance C with the load resistor R.
1000W / 24V = 41,67A
24V / 41,7A = 0.576 Ohm
So your pump has about 0.58 ohm resistance.
The capacitance of capacitors (also Supercaps) is reduced in series connection:
3000F / 10 = 300F
Multiplies with the resistance of the pump:
τ = R * C = 0.576Ohm * 300F = 172,8s (This is not the final result, but only the time constant)
With the time constant, the voltage can then be calculated after a certain time. The pump will keep running slower and stop at some point. When that happens, you can only try. Depending on the model, this can be at 20V or even at 5V.
If I didn’t get involved, the formula is for the time after which the voltage has fallen to a certain value:
t = -ln(U/U0) *
“ln” is the natural logarithm, which can be expected by any reasonable calculator. U is the voltage to be reached. U0 is the voltage at the beginning, in your case 27V. τ is the time constant, in your case 172,8s.
For a voltage of 20V you get a time of 51.9 seconds. 10V are reached after 171.6s, and 5V after 291.4s.
WOW!!! You’re really great!
Thank you!
What would change if I doubled the supercap and then mount it in parallel?
10 caps 24v with 10 caps 24 in parallel.
Please answer me.
No problem;)
During parallel switching, the capacitances are added. Then, as a new capacity, you have the double, so 2 * C from before. Thus, the double time constant and the voltage decreases half as fast. For example, 10V are not already reached after 171.6s but only 343.2s.
Thank you
Sometimes school and interest in something useful is really beneficial ^w^
Thank you. You’re top.
In my eyes just genius!
3000F correspond to 3000 coloumb or there to a storage quantity of 1A × 1V × 1s in this unit as 1C.
At 2,7V × 3000F, the electrical storage capacity per cap is m.W. 8100 Vas
Stacked to 27 volts on a 24V/1000 watt pump should therefore m.W. after approx. 8 seconds to be under load.
https://youtube.com/shorts/coK8T8AZYhs?si=c4U6aNAoPXDtjwW
This should not go mathematically.
Right?
Although capacity and charge are connected, it is not that way.
Q = C * U
The charge is the product of capacity and voltage. At 2,7V and 3000F, 8100C.
Coulomb is by the way synonymous with As (amperesis customers) and technically also VF (Voltfarad). You can’t mix it. VAs (Voltamperesiscus) would be equivalent to Ws (Wattsecus), i.e. already an energy unit, but the energy in a capacitor is calculated differently.
E = (C*U^2)/2
This results in 10935Ws per condenser, i.e. 109350Ws for all ten.
You divide energy through power here to calculate the time. Would also work – if the performance would remain the same all the time. But the voltage of a capacitor decreases during deading, as a result the power becomes smaller and smaller over time. Apart from the questioner, there is a DC-DC converter that keeps the voltage constant, it could be expected. If this is the case, he would have to explain himself.