I have two questions about reactive power. Would you give me a hand, Please? - Page 1
Forum section: electronics / beginners

author: Michael George
author_rank: Contributor
timestamp: April 09, 2016, 01:24:14 pm
content: 1. I read on the internet that the reactive power is negative real power. It is a power that moves from the load to the source (with 90 deg. of phase shift). Why is reactive power measured in VAR not in watt?A purely inductive load store some watts (a positive power that moves from the supply to the load). After a certain amount of time, It releases them as VARs !! It stores Watts and I think it should releases watts also. Is this just two names of the same unit?2. For inductive load, I also read that I can correct power factor using a capacitor that is parallel to the supply. But I think the only way to correct power factor is to use a series capacitor with the load for the following reason:For the attached circuit that has a parallel inductor and capacitor, The total impedance (Z) has imaginary part (i). That means it has a reactants and it will consume reactive power.The total impedance of the second circuit does not contain an imaginary part so there is no reactance and it will not consume any reactive power. If there is a series resistance with the load, The circuit will be a pure resistive circuit.How Can I improve power factor by adding a capacitor in parallel with the supply?Thank you very much,

author: rstofer
author_rank: Super Contributor
timestamp: April 09, 2016, 02:36:11 pm
content: You should try this stuff with a vector diagram drawn in the first quadrant (for inductance). Along the X axis is the real power in Watts extending from the origin to the right on the graph. At the right end of the real power vector, draw the reactive power (up for inductive, down for capacitive) upward at 90 degrees. The units are Volt-Amperes Reactive (VARs). Now close up the triangle by drawing the hypotenuse and this vector is in Volt-Amperes. Note that VA is always equal to or greater than Watts.I like to think of a glass of beer. The liquid is Watts, it does the work. The foam is VARs and it's pretty much useless. The total volume is VA.Back to your power factor correction: Yes, the capacitor is placed in parallel and it causes you to add another VAR vector pointing down in opposition to the inductive VAR vector. As these two vectors cancel each other, the VA vector rotates down closer to the W vector and it gets shorter. If both reactive vectors have the same magnitude, there will be no reactive power left in the circuit and VA will equal W - a glass full of beer with no foam!In the real world of electrical power, power factor is an issue because the Amps in the circuit have to provide for both real and reactive power. More amps means more I^2*R losses (heat) and utilities begin to charge for this when it gets out of hand. They have to deliver Amps through the wires.

author: BobsURuncle
author_rank: Regular Contributor
timestamp: April 09, 2016, 04:55:07 pm
content: Quote from: Michael George on April 09, 2016, 01:24:14 pmTo keep track of the energy budget in a circuit you need to know how much of the energy you are pumping does work, that is energy that is lost to the circuit per second, i.e. watts. The energy that is not doing work is expressed as VAR.It is important to keep clear the difference between energy and power. Power (Watts) is energy (joules) consumed within a specified time frame(sec), i.e. 1W= 1 J/s. Inductors and capacitors don't store power, they store energy. The next thing to remember when talking about AC power systems is we are generally talking about averages, so not instantaneous power but average power."Real" power is power that does work, i.e. turns motors, drives speakers, heats up resistors, emits light..... Volts don't do work unless they are moving a charge in the direction of the EMF field and that moving charge is doing some work, like crashing into atoms generating heat or generating a magnetic field that turns a motor.So for example, if you had an ideal gasoline 100V AC voltage generator driving an ideal inductor with an impedance of 50 ohms, then the generator is delivering 2A. So how many watts is that? Answer is zero. Instead we say it has supplied 200 VAR. Because no net (average) work is being done. The inductor is storing the energy delivered to it in a magnetic field and when that field collapses it sends all that energy back to the generator -it is never consumed. So the generator would run forever without the need to provide it any more fuel. Actually I guess it would oscillate between being a motor and a generator - but never mind that. Now if you swap in a 50 ohm resistor for the inductor then 200 Watts are consumed as heat by the resistor, which means a gasoline energy equivalent of 200J/sec needs to be supplied to the generator to keep it going. So the guy supplying the gasoline needs to know whether he is supplying Real Power or Reactive Power.You know the power is 200VA reactive, in the first case, and not 200W because the voltage and current are 90 degrees out of phase. If you take the instantaneous power (watts) at each moment in a cycle and add them up over a complete cycle it adds up to zero; at some instants it is zero, sometimes positive and sometimes negative, but the sum is zero. Then why watts, when doing instantaneous calculations of power? Because when it is positive it is doing work building the magnetic field in the inductor and when it is negative the inductor is doing work generating magnetic field in the generator. Because in AC power systems we talk of averages - no net work is done - so no Watts.