Apparent and Real Power 3

[gladkenn]

HI guys,

Can you explain to me the said subject. I seem to have forgetten it already. Am a mechanical engineer by profession.

Thanks.

 

[Senselessticker]

To explain it simply: Apparent power is measured in Volt-Amps (usually kVA) Apparent Power = Voltage x Current

Real Power is measured in Watts and takes into account power factor. Real Power = Apparent Power x power factor.

 

[jghrist]

See

http://www.kilowattclassroom.com/Archive/PWR.pdf

for a basic explanation.

 

[cuky2000]

To explain the physical meaning of real (active) and apparent (reactive) power, consider an engine couple to a generator:

- The power delivered in the shaft of the engine can be converted to active power by the generator witch is consumed by the load (ex. Motors, heating, lighting, etc).
- The engine does not produce reactive power; however, the generator does. This power is used to magnetize the inductive components of circuits (ex. Motors, capacitor, line, ballast & transformers, etc.).This power keeps circulating in the system without requiring extra active power from the engine.

The additional current produced by the demand of reactive power will increase the size and capacity of the electrical components such as cable, generators, transformer, etc.

Fro additioanl reference see the enclose site

http://www.ibiblio.org/obp/electricCircuits/AC/AC_11.html

 

[unclebob]

Here is a nice exemple:

Suppose you have a glass of beer.
The liquid represent real power because it quench your thirst!
The suds or brew (or wathever you call it) on the top
is reactive power, it takes places and does nothing to you.

So, the glass contains both the liquid and the foam, that is
apparent power!! )

 

[rmw]

I love unclebob's analogy. LOL. However, I have a question (also asked as an ME for whom reactive power is black art).

Since the beer foam occupies space in the glass, doesn't that also require a larger glass?

Asked on topic, since reactive power produces heat in the conductors rather than output in the driven motors, isn't the energy required to produce that heating of the conductors produced by the engine driving the generator?

So, if you had a zero PF system (if there is such a thing) wouldn't the engine still have to produce the same HP to drive the generator to put all the power into heating conductors? More realistically asked, a system that had a highly inductive load of 50% PF wouldn't require half the engine to pull it as compared to a system with a purely resistive load at a PF of 1.0 would it?

rmw

 

[jghrist]

The amount of beer left in the glass after downing it (losses) is small in comparison to the total volume of the glass, even though it is real beer, not useless foam.

 

[rbulsara]

rmw:

On a sober note, the energy used in the heating will constitute part of the 'real' power part of the total (apparent power). I^2R losses. Reactive current in 'ideal' (zero resistance conductor) will not create heating.

And yes, the engine only provides real (kW) power and no reactive power. So a system with 0.5 pf system and 100kVA of load, for example, will only require a 50kW (plus losses) rated engine.

That is why, you can start a motor of kW rating same as an engine (after accounting for efficiecny losses), as long the alternater kVA rating is adequate to accomodate the starting kVA of the motor within acceptable voltage dip.

 

[abcd3286]

OK
supposedly there is no cost for reactive power ie what is required for magnetizing fields (motors etc).
Several people said this is not developed by the engine (or other prime mover of an alternator).

WHY then does
Seattle city light (a utility with a few alternators) tell me they have to open the throttle if reactive load changes.
ALSO if it is free why do they go thru the expense of power factor correction?
AND why do they bill you for it if is free.

To the bear and foam
granted the foam is useless
BUT are you able to make it for free?

 

[itsmoked]

Everything in the magnetic realm has to be built larger to carry the magnetizing currents. Hence the adder.

 

[edison123]

Reactive power is a necessary evil - like beer froth


* Why is the time of day with the slowest traffic called "rush hour"? *

 

[ijl]

One more attempt to explain the apparent power:

The momentary power at a given time in an electric circuit is equal to the momentary voltage times the momentary current. This is not a definition, but follows from the definition of power, voltage and current. The momentary power is of little practical use in an AC-network, because the current and voltage are functions of time. It is better to calculate the average power over one cycle. It is equal to the rms voltage times the rms current times the cosinus of the phase difference between the voltage and current. That is, P = U I cos(fii), where P is the power, U is the rms voltage and I is the rms current, and fii is the phase difference. The cos(fii) is also called the power factor, pf.

Because cos(fii) appears in the expression of the power P, it is tempting to define another quantity, Q = U I sin(fii). This is the reactive power. The apparent power S is then the vector sum of the power P and the reactive power Q, S = sqrt(P*P + Q*Q).

The reactive power and the apparent power do not have a direct physical interpretation. (I am afraid that the colleagues here might not agree with this view?) A load can generally be described as a resistance and an inductance in parallel. The power P is then the real, physical or “proper” power consumed in the resistance. It can be seen and felt as heat, mechanical power, etc. The reactive power is the “powerlike” quantity “consumed” in the inductance. It is related to the power that is oscillating between the load and the source.

The apparent power can be seen as the requirement for the maximum capacity of a transmission line. This can be seen in the following way: The voltage U in a network is more or less constant. Thus, if the power factor is low, then the current must be high in order to get the same power as with load having a higher power factor. The transmission lines must be capable of passing this current. The losses in the transmission lines are equal to the resistance times the current squared. (Again ,this is not a definition, but can be derived using Ohm’s law and the expression for power.) Because the losses depend on the resistance of the wires, not on the reactance, they do not “know” about the power factor. This means that the wires can be considered (and dimensioned for) to transmit a real power equal to the apparent power, but with a power factor equal to one.

The losses are apparently one of the reasons why the utilities do not like a low power factor, and why the power of the source must go up, when the power factor goes down.

 

[rmw]

ijl,

As a ME, I am hanging on by my fingernails trying to stay with you, but I think I understand what you said in your post.

Here is what I specifically remember from an EE class (for non EE majors) during my university days.

First, here is some background as to why I might have remembered this comment.

I returned to college after a stint in the US Navy where had worked in the machinery spaces of warships. While I was obviously more interested as a future ME (a lot of what propelled me into the power side of Mechanical Engineering) in the boilers, pumps, and turbines, we did generate and distribute the ships electrical power, too.

Upon returning to the University, I worked in the campus power house as a student utility worker, and while I had no significant operating responsibility, I was around the daily operations of the central power station, which included interactions with other utilities on the grid to which parts of the campus was connected.

I said all that to say that I had some small limited amount of working knowledge of power generation, transmission, and distribution systems when I later attended my obligatory EE courses.

I assume that when we are discussing reactive power, we are talking about the term I learned as VAR's?????

I still see VAR meters on power company power panels.

The way the professor described VARs to us, and yes, this was complete with vector diagrams, was to use the analogy of a freezing rainy winter day (in our part of the world, we get a lot of ice hanging on electrical lines during freezing rain events-often causing major system wide damage and outages) of the power company dispatcher calling up such and such a plant across the system and saying 'ship me some VARs.' Purpose being to get the ice off of the transmission wires.

Based on your detailed and excellent post, I assume that this analogy he used would be a manner of speaking that would really mean 'lower the power factor so that the line losses pick up, heating the wires, and melting the ice off of them'.

Am I anywhere near close in my understanding?

I was about to dig into my personal library to see if I couldn't clear myself up on this, as I am in and out of a lot of power stations, and while power factor and VARs rarely interact with my work, often, such as in measuring pump work or generator output, it enters in. So, it is a peripheral issue to me.

Thanks for the time you took in your above post. I give it a star.

rmw

 

[itsmoked]

I think you have it.

With respect to ijl's post which is nice. I would replace "momentary" with "instantaneous" but that may just be a USA semantic difference.

 

[ITSTOAST]

Itsmoked,I agree, the utility has to size its distribution system according to amperage and low P.F. costs them money. In return they charge my facilities $100K's per year for any percentage above or below unity. Yes, any. Its "Coming soon to your local provider" --start planning a strategy now.

 

[ijl]

itsmoked,
My instantaneous reaction to your comment was a momentary
confusion (or the other way round, or neither?

 

[RudyL54]

In the example, the glass is the cable capacity. So the glass should be taller to accomodate the beer suds.

Easy to remember. Apparent power equals real power if the unit is perfect. No losses due to reactive impedance.

However, in real life, there is no perfect system. So, it is well accepted that the apparent power should be slightly higher than real power. The factor added in the equation is known as the power factor (PF).

Power (VA) = Power (watts) x PF
Sources of power are rated in KVA or VA (Volts-Ampere)( ex. transformers, generators) while machines or equipment (ex. motors) that use power are rated in WATTS.

Some times they are in HP (Horse Power)(=746 watts).

 

[coconutalley]

Can't resist -- already lot's of good answers at different views:
Seattle City light wants you to get your Power Factor up because the more VA [apparent power] means to them that they have to generate more current [Amps] capacity--physically this means bigger generator and bigger wires.

 

[itsmoked]

Rudy's addition to the beer analogy is outstandingly correct! Wow... I am humbled...

 

[RudyL54]

My apology. I mistyped X instead of /.

So it should be Apparent Power (VA) = Real Power (in watts)/PF(power factor).

Power factor is normally less than one.

Thanks itsmoked for your comment.