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Capacitor calculations?

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James1986

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Hi Guys,

A quick couple of questions.

t=rc thus T=(V/I)xC = (24/27)x58= 51.5s

Can you tell me what this means, by your understanding? This is calculation I have done, and I just want to check it proves what I think it does!

Also, if an electric generator gives off 24v, and a capacitor is rated at 15v supply, can it accept 24v? or will I need to step the voltage down?

Thanks!
 
Hi,

Your calculation looks right but what are you running that requires 27 amps?
Also, the RC time constant is for one time constant, and does not hold
over all time unless you are using a constant current source.

And no sorry, a 15v capacitor could blow up if used at 24v.
You need to lower the voltage or get a higher rated cap.
 
Thanks,

the 27 amps comes from a 24v motor, used as a generator; and from a resulting calculation based on the P.E. of an 85kg load atop a 100m high hill.

This leads me on to a confusing contradicting result I just got...

The capacitor storage can be calculated by E=1/2 CV²

So with that in mind, E= 0.5 x 58 x 24² = 16704joules is the total energy able to be stored in a 58F capacitor, correct?



Now, the P.E. of an 85kg mass atop a 100m hill can be calculated by

P.E.=M x g x h = 85 x 9.81 x 100 = 83385joules

so for 1m 833.85j can be recaptured on the way down, assuming 100%efficiency, for every 1m/s. 1W is 1j/s and so there is 833.86watts of E available.

So a 24v motor turned generator at 78% efficiency will give W=VxI, 650=24xI

I = 650/24 = 27Amps!

Now bear with me,

T=RC and R=V/I therefore we can say T=(V/I)xC = (24/27)x58 = 51.5s

So it takes 51.5 seconds to charge the capacitor to 63% full!

Going back to the top, the full amount of energy able to be stored is 16704j, and based on the assumption of 78% efficiency, 650j is given from the motor every second (at 1m/s) then the capacitor will be charged in:

16704/650 seconds time?!?! which is = 25.69s?


Can somebody unravel that mess and help me figure out why one way says the cap will be full after 25 seconds and the other 51 seconds?

Thanks for your patience with this one!
 
Hello again,


There are a couple problems here.

For the one you are talking about, the energy is not transferred at a constant
rate therefore you cannot calculate the time to charge based on the
incremental charge transferred between two points in time (which you
are using 650j for). The incremental charge transferred over time
varies with time to a great degree, so it's not even possible to approximate
it this way. For example, in the first 10 seconds approx 630j are transferred,
but for the same time span near the end of charge about 10 times that
amount is transferred. This of course means that your first calculation is
the correct one.
 
Last edited:
Ah thanks Mr Al,

Sorry about the double post, not sure why that happened!:confused:

Another quick question...

The voltage output by the motor; as a 24v motor will it only ever put out 24v? or if I slow down my turning does the voltage drop? Conversely is it the current which fluctuates with speed?

Thanks
 
Hi again,


If the speed decreases then the area of the field swept out by a length
of wire in a time dt decreases which in turn causes a decrease in the voltage.

Also, the voltage and resistance is a better way of looking at this
problem anyway. This is the other problem i hadnt talked about in the
previous thread :)

I've included a graph to compare the two different views on charging
the capacitor. The blue indicates the voltage charged by a constant
current charging source, and the green is charging by a generator
that puts out a max of 24v and has an internal resistance of
24/27 ohms. The respective energies are shown in violet and orange.
We can see it takes a lot longer to charge than with a constant
current charging source as you were presuming. This will be more
realistic, although a good test would be in order before buying parts.
It takes roughly 3 to 4 times longer to reach the same energy level.

Get a smaller cap and test the time it takes to charge to 24v (or whatever
that generator puts out) and we can do a more precise calculation
after getting the results of that test.

For example, how long does it take to charge a 1F cap under the
exact conditions that will be used for the final design.
 

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Last edited:
Thanks MrAl,

I'm still a little confused though... the output of the generator will be variable, as is the nature of its use. Are you suggesting that I come up with some way to stabalise the current output, as the fastest means of charging? Or is the current output always steady, and the voltage is all that variates?

I will try and get hold of a 1F capacitor! :)
 
Thanks MrAl,

I'm still a little confused though... the output of the generator will be variable, as is the nature of its use. Are you suggesting that I come up with some way to stabalise the current output, as the fastest means of charging? Or is the current output always steady, and the voltage is all that variates?

I will try and get hold of a 1F capacitor! :)

hi,
I have never seen a 1F 25Vdc cap advertised.:)
 
Yeah, I'm having trouble finding a capacitor at 24V =( Plenty in America though..... anybody know any decent distibutor websites in the UK?
 
Yeah, I'm having trouble finding a capacitor at 24V =( Plenty in America though..... anybody know any decent distibutor websites in the UK?

hi,
You could make it up with 10 * 100,000uF in parallel, its getting expensive.!

**broken link removed**
 
:eek:

Not the perfect project for a student anymore:(

Could you take a few minutes to complete this survey for me; it would help my research to get a balanced opinion! I kindly have half my results from e-bike owners already!

**broken link removed**

Thanks!
 
Hi again,


Sorry, i assumed that if the OP could get a 58F cap then he would be able to
get a 1F cap even easier. Does this sound like an unreasonable assumption?
I also assumed that the OP had the other stuff like the generator.

Just a note on the speed again...

If the speed decreases then the base voltage decreases, which because
of the series resistance will force the current to decrease. This of course
means that both current and voltage will decrease when the speed
decreases for the same load.
If you want to charge the cap up to say 24v then the generator has to
be able to put out at least 24v and preferably even more than that.

A good estimate would be to use the familiar equation:

V=Vm*(1-e^(-t/RC))

where

Vm is the motor voltage (assuming open circuit and constant speed)
t is the time in seconds
R is the resistance of the motor in ohms
C is the capacitance
V is the desired charge voltage (and total energy = C*V^2/2 as before)
e is the natural log base, e=2.71828182845904523536
 
Hi again,

Thanks MrAl! Not an incorrect assumption, but as a student most of what I have to do is purely theoretical... like if I could have this stuff, then Y will happen. Which is why it is so difficult for me to understand; I'm much more of a practical person.

Blueroom, my other topic discussed different capacitor matters, and I apologise if you feel like you have to keep reading the same thing :(
Its easy when you are able to understand everything, but sadly I don't, so bear with me whilst I try. My course was in the middle of a syllabus change during my year in industry, and when I came back, they had practically cut electronics out. I'm doing as much reading as I can, but some things are still unclear. Its a shame that any examples I CAN find, never have what I am looking for.
 
Different application for your project

Hi

I want to apply this project to a smal ICE car. The car is front wheel drive and I want to use the back wheels to drive a motor on regen braking and storing the energie in UltraCap's. And then releasing the energie back the wheels, via the motor, to assist with pull-away.

Can you 'upgrade' this project to my 'idea'?

Cheers
 
Here a link to a 1F capacitor rated at 20V. For 40V operation you'll need 4 which will be expensive.

The capacitor probably needs to be rated for much more than 25V, I'd say at least 30V.

Passives | Capacitors | Capacitors, Aluminium
 
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