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#### MarkE

• Hero Member
• Posts: 6830
##### Re: Silly question about capacitors
« Reply #15 on: February 15, 2014, 03:23:26 AM »
Lets say we have 2 equal size air tanks, 1 filled to 300psi and the other with 0 psi.

We connect the 2 together with a hose and we open the valves on the tanks. We should end up with 150psi in each. But we lost 50% of the total energy stored the first 300psi tank by the simple transfer, expansion and equalization of the 2 tanks.  Now, each tank contains half of the total air pressure that was in the 300psi tank. How did we lose 50%? The total amount of air(electrical charge) is still there in the 2 tanks, but we lost pressure.

We lost it stupidly.    It wasnt lost due to resistance. It wasnt lost due to air leaking out. We lost it by reducing the pressure that was in the first tank and expanding it into 2 cambers, each the size of the first full tank.

Now, if we had a pump motor that uses air pressure to provide mechanical output(turbine may be a bit lossy due to blow by of the turbine fan) and we connect the air motor inline with the hose between the 2 tanks(1 full, 1 empty), then we are capturing that lost 50%(or so) of the total energy of the first tank during the transfer. So now we are not losing that energy stupidly.

And in the end, we still have 150 psi in each tank, just like when we transfer the pressure from 1 tank to the other directly without the air powered motor, but without the huge stupid loss.

So I dont believe that resistance is the issue with losing 50% of the initial energy of the full tank equalizing to the second tank.  Its just the fact that we reduced the energy of 300psi air pressure of 1 tank and expanded it into a tank that is twice as large, without doing anything with the actions of the transfer. And once the 2 tanks are equalized, there is no way that the 2 tanks of 150psi can be used to compress the total amount of air back into 1 tank at 300psi. But, if we mounted a large flywheel to the air motor, the energy that was lost in in the direct connect example is stored in the flywheel, and once the 2 tanks become equal(150psi), the flywheel is fully charged and would continue to pump the originally empty tank, pretty close to what was in the originally full tank, minus losses in the mechanical motor and resistances of the air being pushed through the hoses and fittings, etc.

Now with the caps, we can use an inductor as the air motor and get most all the energy of the full cap into the empty cap. Using a diode(check valve) to keep the previously empty cap, now nearly full, from going back to the first tank. A bit of loss in the diode due to voltage drop inherent in the diode.

Mags
Magluvin, what you propose will not fix the problem.  If we were to scale the capacitors or tanks so that they are not equal then we can greatly improve the efficiency going one way.  That is how charge pump power converters operate with relatively good efficiency for small loads.

#### Farmhand

• Hero Member
• Posts: 1583
##### Re: Silly question about capacitors
« Reply #16 on: February 15, 2014, 03:40:59 AM »
There is no loss of "charge".

Did no one look at the "Charge" in "Coulombs", If we take 2 x 10 000 uF capacitors and we have one at 16 volts the electronics assistant tells me it has 160 Millicoulombs of charge.
Now if I was to connect the two caps together and equalize the charge which results in both having 8 volts across them then the electronics assistant tells me they will both have 80 Millicoulombs of "charge" in them so added together there is still 160 Millicoulombs of charge, so no loss.. Where is the loss. The energy in a capacitor is really only "potential energy". I think the capacitor actually stores "charge", not "energy".

The voltage increase on a capacitor has a non linear effect on the "potential energy". meaning from 0 to 8 volts is less "potential energy" than from 8 volts to 16 volts, the voltage increase is the same but the potential energy increase is not.

Cheers

#### Magluvin

• Hero Member
• Posts: 5886
##### Re: Silly question about capacitors
« Reply #17 on: February 15, 2014, 03:47:11 AM »
Magluvin, what you propose will not fix the problem.  If we were to scale the capacitors or tanks so that they are not equal then we can greatly improve the efficiency going one way.  That is how charge pump power converters operate with relatively good efficiency for small loads.

Hey Mark

What problem are we talking about?  I thought the issue was described in the first post about the 50% loss when connecting a full cap to an empty cap.    From what I understand, the size difference in caps still have losses due to the same issues, just different ratios of losses depending on the size differences in capacitors used. Like if we had a 10uf cap with 1000v and dumped it into a 10,000uf cap, the loss would be huge and beyond 50%, but the other way around the loss would be less. What I am claiming is, I dont believe that the loss from cap to cap directly is related to resistance losses during direct transfer.  What hit me hard when I first encountered this problem was the fact that there is a 50% loss, no matter what the resistance is, whether it be high resistance or extremely low resistance, the loss is the same. It seems odd, no?  For some reason, resistance is the claimed cause of the loss, but there is no variance in loss with different values of resistance. Where in other circuitry, there is a difference when there is a difference.

For example, say we have 2 caps that had no resistance, zero resistance, zero losses of any kind. When we connect them, one charged and one empty, and lets get rid of inductance for this for now to eliminate oscillations,  do you believe that when we discharge a 100uf cap with 100v into an empty 100uf cap, that there would be more than 50v in each cap??

Gota go pick up my pizza.

Mags

#### MarkE

• Hero Member
• Posts: 6830
##### Re: Silly question about capacitors
« Reply #18 on: February 15, 2014, 03:57:18 AM »
Hey Mark

What problem are we talking about?  I thought the issue was described in the first post about the 50% loss when connecting a full cap to an empty cap.    From what I understand, the size difference in caps still have losses due to the same issues, just different ratios of losses depending on the size differences in capacitors used. Like if we had a 10uf cap with 1000v and dumped it into a 10,000uf cap, the loss would be huge and beyond 50%, but the other way around the loss would be less. What I am claiming is, I dont believe that the loss from cap to cap directly is related to resistance losses during direct transfer.  What hit me hard when I first encountered this problem was the fact that there is a 50% loss, no matter what the resistance is, whether it be high resistance or extremely low resistance, the loss is the same. It seems odd, no?  For some reason, resistance is the claimed cause of the loss, but there is no variance in loss with different values of resistance. Where in other circuitry, there is a difference when there is a difference.

For example, say we have 2 caps that had no resistance, zero resistance, zero losses of any kind. When we connect them, one charged and one empty, and lets get rid of inductance for this for now to eliminate oscillations,  do you believe that when we discharge a 100uf cap with 100v into an empty 100uf cap, that there would be more than 50v in each cap??

Gota go pick up my pizza.

Mags
Magluvin, take a 100uF cap.  Charge it to 10V.  Connect that to a 1uF cap.  Calculate the net energy.  It is close to the original.  The worst case is always when the capacitors are the same value.

#### Marsing

• Sr. Member
• Posts: 300
##### Re: Silly question about capacitors
« Reply #19 on: February 15, 2014, 04:02:22 AM »

Did no one look at the "Charge" in "Coulombs", If we take 2 x 10 000 uF capacitors and we have one at 16 volts the electronics assistant tells me it has 160 Millicoulombs of charge.
Now if I was to connect the two caps together and equalize the charge which results in both having 8 volts across them then the electronics assistant tells me they will both have 80 Millicoulombs of "charge" in them so added together there is still 160 Millicoulombs of charge, so no loss.. Where is the loss. The energy in a capacitor is really only "potential energy". I think the capacitor actually stores "charge", not "energy".

ok,  but you need energy to put back  only one cap  full of charge
the lose here is about to back to first condition.

btw, tesla have a great way to take advantages of this

#### Magluvin

• Hero Member
• Posts: 5886
##### Re: Silly question about capacitors
« Reply #20 on: February 15, 2014, 04:51:19 AM »
Magluvin, take a 100uF cap.  Charge it to 10V.  Connect that to a 1uF cap.  Calculate the net energy.  It is close to the original.  The worst case is always when the capacitors are the same value.

Ugh, just wrote a long post and my laptop went for a shutdown.

Your right. And I was, well, half right. lol

"From what I understand, the size difference in caps still have losses due to the same issues, just different ratios of losses depending on the size differences in capacitors used. Like if we had a 10uf cap with 1000v and dumped it into a 10,000uf cap, the loss would be huge and beyond 50%, but the other way around the loss would be less"

Im trying to figure out what I was thinking when I wrote that. Was running late to pick up the pizza. I had gone over these things with Poynt a while back. I get what you are saying.

Anyways, thanks for catching that.

Mags

#### Farmhand

• Hero Member
• Posts: 1583
##### Re: Silly question about capacitors
« Reply #21 on: February 15, 2014, 04:51:32 AM »
ok,  but you need energy to put back  only one cap  full of charge
the lose here is about to back to first condition.

btw, tesla have a great way to take advantages of this

I guess the question is why ? Why would you create the second condition if you want to keep the first ? You've already done the work to put the potential energy in the first capacitor, why not just keep it there if that's where you want it ?

As I said the "energy in a capacitor" thing is really just "potential energy" it can be wasted, or used wisely.

Same thing with two 10000 liter water tanks side by side, one empty one full, if you let the full one equalize with the empty one, the same amount of water is there. None wasted. If you want all the water in the first tank just leave it there.

With no reason to do a thing it is always a loss. It's like making a big high stack of bricks then leveling it out into two shorter stacks it takes energy to make the big stack into the two small stacks. It cannot do itself, something must do it.

Cheers

P.S. Imagine making one tall stack of bricks by hand then changing that to two shorter stacks, does that take no energy to do ? No you must expend energy to get from state one to state two. Then if you want to make the two stacks back into one tall one you must expend more energy again. It's really quite simple. The amount of brick remains the same.

Also the higher the stack gets the harder it becomes to add another brick to the top of the stack.  It depends on how you take them down and if you store the energy or not when you do it as to how much loss is involved.

...

#### Marsing

• Sr. Member
• Posts: 300
##### Re: Silly question about capacitors
« Reply #22 on: February 15, 2014, 04:57:58 AM »

I guess the question is why ? Why would you create the second condition if you want to keep the first ? You've already done the work to put the potential energy in the first capacitor, why not just keep it there if that's where you want it ?

hey...    PM , oscillation .. etc
this is not about why,    But What and how

Did you forget this  forum name ?

#### TinselKoala

• Hero Member
• Posts: 13968
##### Re: Silly question about capacitors
« Reply #23 on: February 15, 2014, 04:59:43 AM »
There is no loss of "charge".

Did no one look at the "Charge" in "Coulombs", If we take 2 x 10 000 uF capacitors and we have one at 16 volts the electronics assistant tells me it has 160 Millicoulombs of charge.
Now if I was to connect the two caps together and equalize the charge which results in both having 8 volts across them then the electronics assistant tells me they will both have 80 Millicoulombs of "charge" in them so added together there is still 160 Millicoulombs of charge, so no loss.. Where is the loss. The energy in a capacitor is really only "potential energy". I think the capacitor actually stores "charge", not "energy".

The voltage increase on a capacitor has a non linear effect on the "potential energy". meaning from 0 to 8 volts is less "potential energy" than from 8 volts to 16 volts, the voltage increase is the same but the potential energy increase is not.

Cheers

Charge is indeed the conserved quantity here, and capacitors do "store" charge... and just like springs or containers of compressed air, the more charge you put into a capacitor the more _energy_ it takes to do it and the more energy will be returned when you let the charge back out. To say that this is not storing energy is playing with words, I think.
Also, this is real energy, not potential energy we are talking about. The formula for the energy on the capacitor is, as we have seen, 1/2(CV2) in Joules, where C is capacitance in Farads and V is voltage in Volts. Note that this is the same form as the expression for Kinetic Energy of a moving mass: 1/2(mv2) where m is mass and v is velocity. It's not the "potential energy" form like mgh.
The relationship between energy on a cap and the voltage is, as the formula shows, quadratic, not "linear".  Voltage is "charge pressure". You can think of individual negative charges all trying to repel each other, in a fixed volume. The more charges you stuff in there the harder it is to stuff any individual charge in there. By the formula, if you take a 100 uF cap and charge it from 0 to 6 volts, you have to put 0.0018 Joules of work (energy) into the charging. If you then charge it from 6 to 12 volts ... the same number of volts increase as before ... you have to put 0.0054 Joules in to do it ... because you are pushing against the charge pressure of the charges already in there.
If I did the math right, that is.

#### Farmhand

• Hero Member
• Posts: 1583
##### Re: Silly question about capacitors
« Reply #24 on: February 15, 2014, 05:03:43 AM »
My undertanding is stored energy is potential energy and energy at work is kinetic energy.

..

I made a P.S. to my last post.

#### MileHigh

• Hero Member
• Posts: 7600
##### Re: Silly question about capacitors
« Reply #25 on: February 15, 2014, 05:38:23 AM »
Farmhand:

Quote
Same thing with two 10000 liter water tanks side by side, one empty one full, if you let the full one equalize with the empty one, the same amount of water is there. None wasted. If you want all the water in the first tank just leave it there.

No water is wasted but you have lost one half of the GPE of the water.  It has all turned into heat.  There is no point in challenging the premise of the experiment, it is a given.

MileHigh

#### TinselKoala

• Hero Member
• Posts: 13968
##### Re: Silly question about capacitors
« Reply #26 on: February 15, 2014, 06:01:26 AM »
My undertanding is stored energy is potential energy and energy at work is kinetic energy.

..

I made a P.S. to my last post.
Sure, OK. I was just trying to make the point that the equation has the same form (and for the same reason if you look at the full derivation) as the formula for KE of a moving mass. The energy on a capacitor can also be expressed as QV/2 where Q is the charge in Coulombs. This is more analogous to "potential energy", as it is linear in voltage.

#### Farmhand

• Hero Member
• Posts: 1583
##### Re: Silly question about capacitors
« Reply #27 on: February 15, 2014, 06:19:20 AM »
Farmhand:

No water is wasted but you have lost one half of the GPE of the water.  It has all turned into heat.  There is no point in challenging the premise of the experiment, it is a given.

MileHigh

WHo is challenging the premise of the experiment. I helping to explain in layman's terms. Why energy is expended and where. How it can be a more efficient expenditure or not.

Look what I also wrote.

Quote
With no reason to do a thing it is always a loss. It's like making a big high stack of bricks then leveling it out into two shorter stacks it takes energy to make the big stack into the two small stacks. It cannot do itself, something must do it.

Cheers

P.S. Imagine making one tall stack of bricks by hand then changing that to two shorter stacks, does that take no energy to do ? No you must expend energy to get from state one to state two. Then if you want to make the two stacks back into one tall one you must expend more energy again. It's really quite simple. The amount of brick remains the same.

Also the higher the stack gets the harder it becomes to add another brick to the top of the stack.  It depends on how you take them down and if you store the energy or not when you do it as to how much loss is involved.

So I'm not saying energy is not expended.

Tinsel, I get ya, energy is energy, I probably didn't word my post well. There is no assurance that the potential energy will be converted into "wanted work" at any specific efficiency.

Cheers

#### Farmhand

• Hero Member
• Posts: 1583
##### Re: Silly question about capacitors
« Reply #28 on: February 15, 2014, 06:26:19 AM »
Farmhand:

No water is wasted but you have lost one half of the GPE of the water.  It has all turned into heat.  There is no point in challenging the premise of the experiment, it is a given.

MileHigh

Well yes of course. It would take an additional input of energy to equalize the water "up" to a level with the same top height as the first one. In other words equalize "up" (rather than equalize down) and the GPE would then be increased.  That's just common sense.

..

#### MarkE

• Hero Member
• Posts: 6830
##### Re: Silly question about capacitors
« Reply #29 on: February 15, 2014, 06:32:28 AM »
Farmhand:

No water is wasted but you have lost one half of the GPE of the water.  It has all turned into heat.  There is no point in challenging the premise of the experiment, it is a given.

MileHigh
And this is of course the problem that our friend webby can't bring himself to see with the scenario he set up in defense of our favorite investor funds funeral director Mr. Wayne.