can wattage can be increased?

[chandra2sekhar2000]

is it possible to increase the wattage source.
i.e for example if a consider a battery with 12v,6A, then the watts it can be produce is maximum of 72w. is it possible to extract more than that i.e 200w .by connecting any circuits.

[Wizpic]

Not quite sure wht you mean but i will try and answer it

Basicily the answer would be no from the battery of 6AH how ever it will supply the the 200watts for a very short time. to supply 200w of a 12volt and to last you would need a bigger battery.

What is it your trying to do ?

what's your load  and is it continuos load ?

if it's a transformer then it woudl have to be a biggger transformer to supply 200W because a 200W load at 12V is 16A and things would get very hot

explain a little more detail

wizpic

[chandra2sekhar2000]

actually my question is that is it possible to increase the wattage
that means a 12v supply can be increased to 230v or higher using a transformer.in the same way can we get wattage multipier in the sense by using 72w/hour of battery can we increase it to 200w/hour.

my intension is to drive the home power supply of 3kva using batteries while taking a 300w of power from mains.
 

[Wizpic]

oh you talking about an invertor ?

12V dc to 220V AC ?

Then I would say no, the invertors draw a lot of current of the batteries and would not last long or if this is hte case then I would buy one they are cheap enough to buy and a lot easier

I think what you are expecting is that if you use a transformer with 12V in you get 220V volt out the other side this is not the case it will not wotk like that

[SONSiVRi]

Its not possible to convert 100w/hour to 200w/hour.
Cause "energy cannot be created or destroyed".
If you say 10w/hour can be converted to 600w/minute technically yes.
But you need to save that energy while you sourcing it,
if you try to suck all of that energy from battery I think it will corrupt.

[Parmin]

Yes you can increase the wattage!!

Watt = Joule per Second

To increase the wattage of a fixed source, you should consider to increase the amount of current flow per unit time.

Back onto your original post, 12V 6A battery (P= IV) will give 12 x 6 = 72 Watt which is true.
This figure is given by the internal resistance of the battery which is about 2 ohm (V = IR) which limits the current flow off the battery.

But if you charge a large capacitor which have internal resistance of much much less than a battery, lets for discussion sake pick 0.1 ohm (may be more or less depended on the type and arrangement of the capacitor bank) and we also assume a perfect components without losses.
Now what we have is I = V/R so the current flow is now = 12/0.1 = 120A
and thus, the wattage output is now 12 X 120 = 1440 WATTS !!!!! a far cry over the puny 72Watt direct drive isn't it?

Unfortunately, the real world have much resistance to pure mathematics, your peak power is limited over the sustainable time of peak power delivery (how big is your capacitor bank? How long do you want the extra wattage to flow?), electrical resistance of the power delivery vehicle (what type of cable do you have? gold? silver? copper? carbon?), Magnetic resistance of the delivery vehicle (pulses of energy may be severely limited by bad electromagnetic designs) and many more other factors.


So there is my 2c.

[localcrack]

You can increase the wattage but
your Watt/Hour becomes Watt/Minutes or Watt/Seconds

That means 72 Watt/Hour becomes 500 Watt/Min or 1500 Watt/Sec

[sfiga69]

With battery you must consider the Ah (ampere-hour)

With battery 12v 7Ah  -> 84 Wh discharge it 1 hour with 7A    84W life 1 hour
                                                            30 min with 14A  168W life 30 min
                                                            15 min with 28A  338W life 15 min

In example, UPS 5Kw (from APC), it has  16 battery 12V 7.2A arranged in 4 pack  in serial, any pack 4 battery in parallel.
in total 48V 28.8Ah 1382VAh
Time up for UPS 1 hour at 1000 W
                      30 min at 2000 W
                      15 min at 4000 W
                      10 min at 5000 W

For a battery the question is:
For how much time it must supply power?

[Parmin]

Somehow I think the last two post are the same iterations of my post.. Hmm.. wonder why..

[SONSiVRi]

Parmin in fact you right but I learned sfiga69's message what I didnt know before.

In example, UPS 5Kw
in total 1382VAh
Time up for UPS 1 hour at 1000 W
                      10 min at 5000 W

Shouldn't it be? 1 hour at 1382 W
                    10 min at 8292 W (6*1382)

And where is our 5Kw power from beginning? Is UPS's power measuring by 10 mins of your calculation.

[Parmin]

Shouldn't it be? 1 hour at 1382 W
                    10 min at 8292 W (6*1382)

And where is our 5Kw power from beginning? Is UPS's power measuring by 10 mins of your calculation.

Ah, here is an example of the loses in calculations.

As I said before, capacitance loses, the electrical resistance and magnetic resistance of power delivery vehicle (in layman term, your cable) causes losses in the final power output.

So, if say the power converter is operating at about 60% efficiency, then the 8292 Watt x 60% gives 4975.2 watt.. or rounding it  gives 5000 watts.

Remember, there is always costs in electronics.  That is the first trouble most graduates of electronics had, they forgot to calculate the losses and thought that they invented a perpetual engine

[SONSiVRi]

Ty Parmin,
You make 1 part clear but I have another 2 questions also

UPS is pronouncing as 5 kW, shouldnt it be 5 kWh?
And we have 1382VAh power, what is the calculation to the 5 kW or kWh?

[Parmin]

The announcement of 5 kW means that it is capable to supply a maximum of 5 kW at a given moment. This capability is usually for a very short time ("short time" is a relative term ). Remember the components inside the UPS is stressed out really bad when supplying this maximum power.

As to the potential stored energy on the UPS, that is the kWh rating.. it is a way to say that it is capable to supply so many watt in so long time. ie Watt Hour rating.

[SONSiVRi]

I am a little bit confused, so they are selling their product by telling "peak power at a time" (kW) instead of "stored total power" (kWh) ?
Relation between kW and kWh is depends to quality of product, there is no calculation right?

[Parmin]

It sounds better when you describe your product at the PEAK output, because it sounds much more than it really is and for some, the more the better...

Relation between W and Wh is depended on the efficiency of the conversion.  It could be calculated given enough variables.

[jorge7458]

Hello: 
They are really two different electric magnitudes: Kw is measured of power and Kwh it is measured of energy (power in the time). - 
One is the power that he/she gives us in the instant and the other one is the energy consumed during so much time. - 
Greetings 
Jorge

[looser]

Hi sonsivri,

Dont be confuse on the matter.W and Wh are two different measure. As detaily explained above Watt is the power the mean device will use or produce on a continuous basis. Peak power is another description usually ' clever ' producer use this to impress their customers. It is a measure how much the device can whitstand or produce in a short time (couple of milliseconds for instance ) so is not directly has a use for the people in everyday life.

What hour in the mean time is the time extension for the power figure. If you would consider you have a ups that have average say 500W of power producing capacity. But notice you are to draw this power from a battery. Since the chemically stored energy is to be converted into electricity the time you are continuously draw energy from battery is directly proportional to the size of battery. The bigger the longer. If you have a battery saying 500Wh it means you can obtain 500W of power continuously for one hour.

Regards.

[is_razi]

no , the power (or wattage) cant be changed because it's the inherent property of the source, say if you can lift a 500kg mass yourself, so if you want to take more power from than that specified for the source it certainly will damage the device.

[Parmin]

say if you can lift a 500kg mass yourself, so if you want to take more power from than that specified for the source it certainly will damage the device.

Dude, just to contradict you as you contradict us, have you ever heard of pulleys? winch? as the increase in wattage happened because of an electronic power converter, a child could lift heavy weight because of mechanical power converter.  Read the above and ponder about it!

[is_razi]

Dude, just to contradict you as you contradict us, have you ever heard of pulleys? winch? as the increase in wattage happened because of an electronic power converter, a child could lift heavy weight because of mechanical power converter.  Read the above and ponder about it!

the power of a certain source can not be increased as it is the inherent property of that source.in pully you don't increase the power of your muscles but you change the way that the power is applied to the load,ie you apply a small power for a long time and long distant , and in the power converters and smps's the power doesn't increase and you just decrease the losses of control element by switching method,(ie using the power control device such as mosfet's in the on , off states rather than using them in the linear state).so now you tell me can you extract 500watt power from a 100watt power converter ,

[donno]

Have we not considered that nobody has mentioned that UPS and AC supplies are not rated in
simple kw or kw hours but are rated in kva (kilo volt amperes) a better idea of what can be
supplied from the kva rating can be calculated.

[Parmin]

Have we not considered that nobody has mentioned that UPS and AC supplies are not rated in
simple kw or kw hours but are rated in kva (kilo volt amperes) a better idea of what can be
supplied from the kva rating can be calculated.

Heh.. have YOU not consider that                   Volt x Ampere = WATT ??
So what is your point?

Posted on: December 22, 2007, 11:58:06 23:58 - Automerged

the power of a certain source can not be increased as it is the inherent property of that source.in pully you don't increase the power of your muscles but you change the way that the power is applied to the load,ie you apply a small power for a long time and long distant , and in the power converters and smps's the power doesn't increase and you just decrease the losses of control element by switching method,(ie using the power control device such as mosfet's in the on , off states rather than using them in the linear state).so now you tell me can you extract 500watt power from a 100watt power converter ,

As I said razi - ponder about it.. read books, THEN, post your opinion.  Otherwise, you sound foolish

[twonuts]

Basic theory

Wattage drawn from a battery depends on the load attached to the battery. The lower the load resistance of the load, the more current is drawn from the battery.

Battery manufacturers design these for specific purposes, and usually specify both Ah capacity, as well as max current which should be drawn from  their products. There are a number of good reasons for this.
Batteries may be designed for low power, deep discharge cycling, for recreational use, or for high power short duration use, as in automobiles.
The maximum current drawn from any battery is when you short circuit the battery. Then current is then limited only by the internal resistance of the battery.  This also limits the maximum wattage which can be taken from a battery. Sound good? Unfortunately in short circuit condition the battery will self destruct in a very short time. Dry cell batteries gas internally and usually burst explosively. Lead acid cells act in similar manner, with the acid gassing due to electrolysis effects, with the gas possibly igniting and exploding.

Effectively, you draw more power from a battery than the maximum power specified by the manufacturer, at your own risk. If you want more power from a battery system than the specified limits, use parallel battery systems.

 

[donno]

Heh.. have YOU not consider that                   Volt x Ampere = WATT ??
So what is your point?

Posted on: December 22, 2007, 11:58:06 23:58 - Automerged

As I said razi - ponder about it.. read books, THEN, post your opinion.  Otherwise, you sound foolish

Ya got me!....engage cranium before opinion?...that's the late shift for ya!...never mind, I''l flog myself later...
anyhow... what's this crap about pulleys and winches?...ohm's law never mentioned pulleys or winches?..Amperes,Ohms,Power and
Resistance only. Yeah.... I know Volts x Amps x Efficiency = Output. So I guess I'm too used to the AC side of the story but in this instance
I believe twonuts has "nailed it" perfectly...just try shorting an SLA or your car battery and you will quickly understand his point!

...must have "pondered about it" and read lots of books?

For those who wish to Ponder....http://www.pondergoembel.com/ponderbooks.html

and for those who wish to understand the basics of electricity...http://www.rmcybernetics.com/science/cybernetics/electronics_volts_amps_watts.htm
also...http://www.eng.cam.ac.uk/DesignOffice/mdp/electric_web/DC/DC_10.html

[is_razi]

Heh.. have YOU not consider that                   Volt x Ampere = WATT ??
So what is your point?

Posted on: December 22, 2007, 11:58:06 23:58 - Automerged

As I said razi - ponder about it.. read books, THEN, post your opinion.  Otherwise, you sound foolish

hehehe, do you mean that you want teach an electronic engineer and mechatronic MSc about power and watts, i am working about 10 years arround these type of things and how about you?, did you design any power converter in your life, i must tell that i am a desginer of power converters and smps's for industrial applications, so if you have any comment please propose that and be respectful about other comments.
also it seems that you havn't studied your lessons well,my dear parmin the V*I=Watt is not true, instead the V*I=S either V*I*Cos (Phi)=watt & V*I*Sin(phi)=var are the true equations.

Posted on: December 23, 2007, 02:36:16 14:36 - Automerged

Ya got me!....engage cranium before opinion?...that's the late shift for ya!...never mind, I''l flog myself later...
anyhow... what's this crap about pulleys and winches?...ohm's law never mentioned pulleys or winches?..Amperes,Ohms,Power and
Resistance only. Yeah.... I know Volts x Amps x Efficiency = Output. So I guess I'm too used to the AC side of the story but in this instance
I believe twonuts has "nailed it" perfectly...just try shorting an SLA or your car battery and you will quickly understand his point!

...must have "pondered about it" and read lots of books?

For those who wish to Ponder....http://www.pondergoembel.com/ponderbooks.html

and for those who wish to understand the basics of electricity...http://www.rmcybernetics.com/science/cybernetics/electronics_volts_amps_watts.htm
also...http://www.eng.cam.ac.uk/DesignOffice/mdp/electric_web/DC/DC_10.html

i am sorry about that crabs about pulleys,but they are because of parmin and i only answers them i must say that in my first post i respond to this question :
"is it possible to increase the wattage source.
i.e for example if a consider a battery with 12v,6A, then the watts it can be produce is maximum of 72w. is it possible to extract more than that i.e 200w .by connecting any circuits."
as you can see clearly  the answer is no ,you can't extract 200w from a 75w source if  you do that and the source doesn't protected against overload it will be damaged.but parmin wanted to teach me that it's possible because a baby can lift a mass using a pully(ooh what a strong logic!!!),and so i forced to answer that these two case are not related and in pullyes we don't increase the power but we apply small forces(and so small powers) in longer distances and times.
and dear donno in response to your comment about KW and KVar i must say that yes it is true we can increase the KW by applying some capacitances for phase compensation(this is true about inductive loads which is usually the case ) and so the efficiency will increase.
[/quote]

[Digix]

power can be increased in expense of duty ratio.
you can charge capacitor using 1W for 1minute and discharge it using 60w during 1 sec.
or even 10kW if you discharge if you discharge it during 1ms like it is done in photo flash lapms.

however total energy will newer change

[mylogin]

Transformer is rated by VA, or KVA.  a 12VA transformer has output of 12V can supply max. of 1A of current.  if overloaded( ex. 1.5A of current) the voltage will drop.  the energy of the transformer can supply is limited by the VA rating. you can calculate how much current the other side of the transformer should be by dividing the VA by the voltage input. so if your primary input is 110V the input current should be at least 12/110 Amp.  I your question you use the transformer to step up the voltage (12v/6A to 110v) the 110V side current will be max. (12x6)/110 Amp.  the total energy it can transform is the same.   it will not amplify the energy.   Same as your Wattage increase question.