confused about reactive power

[Special K]

I understand the mathematics of it, but the concept is still kind of puzzling. Is reactive power just power that is absorbed by inductors and capacitors in a circuit, stored for awhile, and then released? Basically power that is consumed but doesn't do any useful work, but is necessary for the circuit's correct operation?

Apparent power is just taking the product of the current and voltage magnitudes to calculate power, without regard to the phase angle, correct? Shouldn't the apparent power be lower than the true power (real + reactive) since it neglects the phase shift between the voltage and current, and therefore ignores the reactive power?

 

[Xyo II]

Einstein Always Knows


Edit: crap link

 

[MrDudeMan]

apparent power is the squareroot of the square of the magnitudes of average and reactive power. in other words, S (apparent power) = sqrt(P^2 + Q^2) where P is average power and Q is reactive power. good design minimizes reactive power since the average power is what matters.

if there is a load of say 10kVAR at a warehouse, the power company will come out and install a -10kVAR capacitor bank to get the reactive component equal to 0 so all of the delivered power is real. at least thats what i think happens...i am pretty sure but not 100%. edit: i confirmed that this is what happens by reading my circuits book...so there you have it.

here is some stuff that might help you if you need to do some calculations. i had to memorize all of this stuff for circuits.

S = apparent power (VA - volt-amps)
Vm = Voltage max
V(r) = Voltage RMS
V* = V conjugate
Im = Current max
I(r) = Current RMS
I* = I conjugate
P = average power (watts)
Q = reactive power (VAR (volt-amps reactive))
Z = impedance
Z* = impedance conjugate


S = V(r)I(r)*
S = (.5)(Vm)(Im*)
S = (I(r))^2(Z)
S = (V(r))^2/Z
S = P + jQ
S = (P^2 + Q^2)^(1/2)

P = (.5)(Vm)(Im)cos(theta(v) - theta(i))
Q = (.5)(Vm)(Im)sin(theta(v) - theta(i))

hope that helps

 

[PowerEngineer]

Reactive power is one of the more difficult concepts to explain, and engineers are generally poor at explaining even simple things. But I will try...

Starting with DC circuits, we know that power (P) is equal to the voltage (V) times the current (I). The nice thing about DC is that the magnitudes of the voltage and current are constant (i.e. NOT a function of time); the power delivered at every instant in time is the same.

AC circuits are different because both voltage and current vary as a function of time. (Ideally) both are 60 Hz sinusoids. The power being delivered at any point in time is equal to the voltage at that time multiplied by the current at that time. In AC, the power delivered is not constant; it varys as a function of time. If you do the math (assume for now that both sinusoids are in phase), you'll see that the power delivered is an offset (i.e. not centered around zero) 60 Hz sinusoid. Delivered power is at a maximum when the voltage and current waves simultaneously reach their peaks and zero then both are zero.

We routinely overlook these fluctuations in delivered power because we're really only interested in the average power delivered (over any integer number of complete cycles). The average is clearly a function of the voltage and current peak magnitudes, but what we'd really like is a measure for the size of these sinusoids that give us the average power when multiplied together. It turns out that you can arrive at this desireable measure of magnitude by taking the square root of the mean of the square of the wave -- much more commonly called RMS (for root-mean-square). It's nearly always true that any AC magnitudes you are ever given will be RMS values. So, for AC: V(rms)*I(rms)=P(ave).

No reactive power so far...

But we've simplified our discussion to this point by assuming that the voltage and current sinusoids are in phase with each other. This is most often not true because the very nature of AC brings the capacitive and inductive natures of electrical lines and loads into play. Most often current lags voltage because lines, transformers, and motors all tend to have an inductive nature.

Let's consider an interesting special case, where the current lags behind the voltage by 90 degress. If you do the math (again), you will see that the power sinusoid is now centered around zero. What this means is that over each cycle, the average power delivered is zero! This situation (i.e. 90 degree lag) only occurs when the "load" is a pure inductor. You'll get the same zero average power result when current leads voltage by 90 degrees, which only happens when your "load" is a pure capacitor.

You can do more math to see how the magnitude of the power sinusoid stays constant while the its centerline moves toward zero as the angle between voltage and current (called the phase angle) increases (up to 90 degrees).

So, obviously V(rms)*I(rms)=P(ave) does not work so well. It needs to take phase angle into account. You can verify (with more math) that P(ave)=V(rms)*I(rms)*cosine(phase angle). The cosine of the phase angle is also called the "power factor".

We could be finished... but we engineers are not completely comfortable. We attach our metering devices and measure the voltage and the current, and we see that the so-called "apparant" power (A) is usually greater than what the load is actually consuming. If "real" power isn't being delivered, then what is? Well, we engineers started talking about "reactive" power (R) -- sometimes I wonder if it was just to confuse everyone else.

Reactive power R(ave)=V(rms)*I(rms)*sine(phase angle), which means that P(ave) and R(ave) form the legs of a right triangle with a hypotenuse of A (P^2+R^2=A^2).

The amount of reactive power compared against the real power gives you information about the phase angle between voltage and current.

So, to answer your questions...

Yes, reactive power is related to the energy stored and returned by capacitive and reactive devices through every cycle (and netting to zero). It is not "consumed" and does not (directly) increase real power losses (although reactine power increases the magnitude of currents to deliver the same real power and therefore increase RI^2 losses). Reactive power is an unavoidable characteristic of an AC power system, but utilities design their systems to minimize reactive power flows because it doing so reduces losses and voltage drops. For reasons described above, the "apparant" power is the maximum possible value for either real or reactive power. Unless the phase angle is zero or +/-90 degrees, both real and reactive power will be less than "apparant" power.

Hope this helps... If not take two of these ( :beer: ) and read again.

 

[Xyo II]

Go power engineer! :beer: for you!

 

[Special K]

Thanks for the explanation, that helped a lot.

I have a related question - I read that power is delivered over high voltage lines to minimize I^2*R losses. But since power also = V^2/R, won't the losses be the same even if you sent it using a high voltage?

 

[MrDudeMan]

Originally posted by: Special K
Thanks for the explanation, that helped a lot.

I have a related question - I read that power is delivered over high voltage lines to minimize I^2*R losses. But since power also = V^2/R, won't the losses be the same even if you sent it using a high voltage?

P=I^2*R is still a valid way to describe the loss due to the resistance in the wire, and since we know P=VI, if V goes way up, I goes way down. therefore, if I goes way down, I^2*R does not equate to a very big number, meaning less loss.

 

[CTho9305]

But since power also = V^2/R, won't the losses be the same even if you sent it using a high voltage?

You have to remember where the voltage drop is occuring. If you just had a circuit with a single power source and resistor, then you could simply compute V^2/R for the power. However, in the case we're interested in, there are two resistors: the power line, and the load. Now we have to first figure out what the voltage drop across each of them is.

Ideally, no voltage drop occurs over your wires, and all of it is over the load - you'd be dissipating no power at all from your wires. Unfortunately, the wires act like small resistors. Let's work out loss in the wires if you have a 10000v line and a 1000000v line, both supplying the same power - 1MW to a load.

I'm simplifying a little bit and assuming a load with nice properties (specifically, that it will draw the right amount of current regardless of the voltage actually delivered to it).

The 10000v line needs to have 100 amps flowing through it to supply the 1MW. The 1000000v line needs 1 amp. If we assume your wire has a resistance of 1 ohm, we can calculate the voltage drop across each wire:
10000V line: V=I*R V = 100*1 = 100 volts;
1000000V line: V=I*R V=1*1 = 1 volt;

Now calculate the power loss in each wire:
10000V line: P=I*V P=100*100 = 10000 watts
1000000V line: P=I*V P=1*1 = 1 watt

So, in order to deliver 1 megawatt, if your power lines are 1 ohm resistors, using 10kV will cost you an extra 10kW in the lines, while a 1MV wire will lose you almost no power at all.

edit: fixed a small 2-orders-of-magnitude error

 

[MrDudeMan]

good explanation ctho