Real Power v. Apparent Power

[tallgirl]

I need some jargon for a patent application I'm trying to get finished today.

What are the precise definitions of such things as "Real Power" and "Apparent Power", and how does "Power Factor" affect such things as generation and transmission.

The point I'm trying to explain in this application is how power factor affects generation requirements. I understand the concepts, but lawyers aren't electrical engineers and if I don't explain it, they won't know what the heck to tell the PTO when they do the final write-up.

Sorry I can't give more information -- it's a patent application so everything except vague questions is secret

 

[crossman]

Looks as if some google searches would give far more precise info than might be obtained on the forum. It would take a small book to adequately sum it up.

Edit: I say this because of your use of "precise". If you want the mad ramblings of us pseudo-scinetists, then you are at the right place. If you want precise, well....

 

[charlie b]

Welcome back. We haven?t heard from you in a while.

Here are definitions from an old textbook of mine. If you really want ?precise definitions,? then these will do as well as any others. But I don?t know if they will help. ?Active power? is the more precise term for what is more commonly called ?real power.?

Power, active: The product of the rms value of the voltage and the rms value of the in-phase component of the current.

Power, apparent: The product of the rms value of the voltage and the rms value of the current.

Power, reactive: The product of the rms value of the voltage and the rms value of the quadrature component of the current.

Power factor: The ratio of active power to apparent power.

The impact of power factor on generation and transmission is simply that the further it is from a value of 1.0, the more energy must be generated and transmitted down the lines to do the same basic work at the other end.

 

[zog]

I need some jargon for a patent application I'm trying to get finished today.

What are the precise definitions of such things as "Real Power" and "Apparent Power", and how does "Power Factor" affect such things as generation and transmission.

The point I'm trying to explain in this application is how power factor affects generation requirements. I understand the concepts, but lawyers aren't electrical engineers and if I don't explain it, they won't know what the heck to tell the PTO when they do the final write-up.

Sorry I can't give more information -- it's a patent application so everything except vague questions is secret

If you want to explain it to lawyers use the beer and foam analogy.

Or rmaybe:
Apparant power: Billable hours
Real power: Actual hours worked on a case
Reactive power: Hours billed to a client actually spent in bars or on the golf course
PF: Ratio of actual hours to billed hours to a client

 

[engy]

...the more energy must be generated and transmitted...[/SIZE][/FONT]

By more energy do you mean watt's?

or is the real issue Amps?, since bad pf means more amps without doing any more real work(W), in effect - stealing capacity.

 

[charlie b]

By more energy do you mean watt's?

No, I meant Joules. But over any given period of time, that's what's watts anyway.

Oh, and since we are not changing the voltage, then an increase in amps is the same thing as an increase in power, apparently. :roll:

 

[winnie]

I would suggest defining _instantaneous_ voltage, current, and power.

Then describe what is meant by RMS voltage and current.

rms voltage and current are _averages_ taken over time, where the instantaneous voltage and current is constantly changing.

Power factor, crest factor, THD, and so on are all ways to describe the relationship between the RMS voltage, the RMS current, and the average power.

Just throwing an idea or 2 out

-Jon

 

[engy]

...that's what's watts anyway.
...same thing as an increase in power, apparently. :roll:

Hmmm, I've been thinking about the beer analogy...just think I'll have a beer:smile:

 

[breaker12]

AC Power: real vs apparent power

AC Power: real vs apparent power

A great resource I found was wikipedia. Search AC power. They also have the associated techincal terms highlited and link another page explaining the term.

Good luck

 

[billsnuff]

beer analogy

beer analogy

the amount i drink when wife is present.....the amount of beer i drink when wife is not present..........and the amount of electricity generated by beer drank without wife present........... sparks fly!!

 

[zog]

Hmmm, I've been thinking about the beer analogy...just think I'll have a beer:smile:

It really does work well to explain the concept while pouring 2 different beers, I have had 30 year electricians say "Dang, it took me 30 years to understand that, now I get it"

 

[engy]

It really does work well to explain the concept while pouring 2 different beers, I have had 30 year electricians say "Dang, it took me 30 years to understand that, now I get it"

I'll be sure to tell my wife I'm doing research ...I'll have her dad help, I hate to research alone:smile:

 

[kfenn22]

Apparant power: Billable hours
Real power: Actual hours worked on a case
Reactive power: Hours billed to a client actually spent in bars or on the golf course
PF: Ratio of actual hours to billed hours to a client

Would you say the laywers have a PF of 90% (comerical) or 60% (Industrial).

LOL

 

[crossman]

Hey Tallgirl:

Looks as if some google searches would give far more precise info than might be obtained on the forum.

If you want the mad ramblings of us pseudo-scinetists, then you are at the right place. If you want precise, well....

See?

 

[LarryFine]

Would you say the laywers have a PF of 90% (comerical) or 60% (Industrial).

I guess that depends on whether they're corporate lawyers.

 

[Jraef]

I have expanded the beer analogy when discussing PF with power providers as follows.

In the classic mug-o-beer transaction, there is a bar owner (barkeep) and a customer (sot). If you are the sot, you are paying for a full glass of beer, not foam. If the barkeep hands you a glass that is half foam, you will send it back to be topped off. In doing so, the foam goes out of the top of the mug, down the sides and down the drain. But as the sot, you couldn't care less about the losses incurred by the barkeep, that is his problem. (or so it seems).

If you are the barkeep, you bought a keg of beer; that is your source of supply (analogous to generating power). The foam which ends up down the drain represents lost revenue; it is inherent in the act of delivering beer to sots, but you get nothing for it because the sots refuse to pay for foam and the beer manufacturer does not give you a discount for the foam. So if you can require all of your customers to use slanted beer mugs that reduce the amount of foam created in the pour, more of the beer from the keg generates revenue. The sot gets the same amount of beer, you get A better return on your investment. You are happy, but the sot is not as happy because you made him buy a special slanted mug (analogous to pf correction equipment), and he sees no immediate tangible benefit from that investment.

But wait. Still from the point of view of the barkeep, if a new sot comes in wanting a mug of beer but refuses to use the slanted mug, then you can make an argument that he must pay a "foam penalty" (analogous to a pf penalty) by charging him more for each mug of beer. That puts him at par with the revenue you can get from the more cooperative sots. You recoup your losses, but he pays more for a weekend bender.

Back to the point of view of the sot then, if you come in with your own slanted mug, you end up paying less for a bender than the sot who refuses to. So indirectly, you DO benefit from improving your barkeep's pour effectiveness.

 

[boboelectric]

Hmmm, I've been thinking about the beer analogy...just think I'll have a beer:smile:

Didn't that law about drinking and surfing the Net pass?

 

[zog]

Bravo!

Bravo!

Jraef, I saw the first sentence of your post and thought, "Oh man, I cant wait to read this, this is going to be great!"

I have expanded the beer analogy when discussing PF with power providers as follows.

You did not let me down, that is hands down the best post I have ever read on this forum.

 

[gar]

080523-2041 EST

tallgirl:

From your profile it appears you have some electrical background, but not an EE degree, and you are probably not a patent attorney.

If you have a black box that converts all incoming steady state electrical energy to heat and there are no energy storage devices in the black box, then the integral over some reasonable averaging time of the product of the instantaneous voltage and current at the input is equal to the dissipated average power. Waveform is of no importance. This implies a resistive load or the equivalent. Also the product of the RMS voltage and current is also equal to the dissipated power.

Now introduce some reactive components, capacitors and/or inductors, and the integral calculation above is still valid because this measures the average power. Power is what relates to heat. But the product of the RMS values of the input voltage and current does not equal the heat dissipated in the black box unless the capactive and inductive components cancel each other.

Here is the problem in a transmission system. A major factor is wire resistance and power dissipated in the wire is Irms^2 * R and has nothing to do with the phase of the source voltage and its relationship to the current in the distribution wire.

Distribution system losses are defined by the current in the distribution system and its resistive component. Power at the destination which is what the power company gets paid for is determined by the above integral calculation. Suppose a large capacitor is the only load at the end of the distribution system and it causes a line current equal to normal full load. The power company gets paid nothing because there is no "real power" delivered to the end load. But all the losses in the distribution system from the capacitive current flowing in the distribution system resistance are costing the power company energy. Therefore in this case the power company actually looses money.

.

 

[LarryFine]

Suppose a large capacitor is the only load at the end of the distribution system and it causes a line current equal to normal full load. The power company gets paid nothing because there is no "real power" delivered to the end load. But all the losses in the distribution system from the capacitive current flowing in the distribution system resistance are costing the power company energy. Therefore in this case the power company actually looses money.

So does the customer, the owner of the electrical installation, because he has to pay for a system capable of carrying the useless power along with the useable power.

That's why it pays to install PF correction equipment as close as practicable wink to the offending load. Less of the system has to be sized for the total apparent power.