🚀 go-pugleaf

In article <slrnc00slc.213.isaac@latveria.castledoom.org>, 
isaac@latveria.castledoom.org says...
> On Sat, 10 Jan 2004 12:53:18 -0500, Keith R Williams <krw@attglobal.net> wrote:
> > In article <3fffee5b@news.rivernet.com.au>, 
> > noemail@noemail.noemail.com says...
> > 
> > Sure it is.  It's useless current in the transmission system that 
> > causes heating.  *That* is wasted energy.
> >> 
> >> Reactive power  includes an imaginary component that is truely imaginary.
> > 
> > I know what reactive power is. I *are* an EE, though not a power 
> > jock (thus asking questions here).  
> > 
> >> Maybe the better word is "imaginarily increased power". or "peak volts peak
> >> amps power".. the power  you get when you use peak voltage and peak amps to
> >> calculate power.
> > 
> > Imaginary increased power?  Doesn't that imply that you're 
> > wasting something.  But wait!  You said there is nothing wasted. 
> > Peak amps?  Peak volts?  
> >  
> >> Eg The actual power is 10 kW, the 'reactive power' is 14 kW, then there is
> >> 4kW *  time in second joules of imaginary power, and its perfectly imaginary
> >> because there was no transfer of those joules from supply to consumer !
> > 
> > No, it was dissipated by the transmission system.  You generated 
> > amps (4kW worth) that the load didn't consume.  That must be 
> > dissipated somewhere, no?  (wasted). 
> 
> You are confusing some concepts here.  If the transmission lines were 
> perfect zero ohm conductors, then the reactive power would not cause any 
> losses in the transmission lines.  Yet the reactive power would still
> exist.  

No, the *current* has to be dissipated *somewhere*.  The 
generator is producing VA and the load is dissipating W.  The 
difference *must* be dissipated *somewhere*.  ...and that is in 
the transmission/distribution system.
> 
> The reactive power comes from energy that is alternately stored in 
> electrostatic and magnetic fields and then returned to the system when 
> those fields collapse.  In an ideal system the energy stored and returned 
> each cycle adds up to zero and thus no net energy is required from the
> source of supply.

Sure, I fully understand reactive power.  However, the generator 
is producing VA. The load is consuming W.  If there is a 
reactance inbetween, there is a loss in the wires caused by the 
reactance.
> 
> The reactive power does increase the current transmitted over the system
> and the increased current results in increased dissipation in the
> transmission system.  But that dissipated power is real not imaginary.

Good grief!  That's my point!  It may be "imaginary", but the 
loss is *REAL*.

> The extra power being dissipated is not numerically equal to the 
> reactive power (except by a sheer coincidence).  Its size can be greater 
> or smaller than the size of the reactive power.

If the generator is producing 14kVA and the load consuming 10kW, 
exactly where is the rest going?

 
> In short, you are confusing the system losses associated with reactive 
> power with the reactive power itself.

Well, they're sorta related, no?  

Ok, a generator *can* be used to assume a MVAr regulation role, 
but it isn't putting out it's maximum power then either.  Someone 
else is getting a ride...
 
> I'm sure you knew this stuff at one time, but apparently the knowledge
> is fading from non use.

No, I'm not a power engineer (as I've said, I do high-end 
microprocessor development) and I never did study power systems.  
However, your arguments haven't convinced me yet.  Please 
continue to try though.  ;-)

-- 
 Keith