More Reactive Power

.."reactive power production can limit a generator’s real power capability."...

This can cause the voltage to collapse and the system to shut down...

https://electricalnotes.wordpress.com/2011/03/21/importance-of-reactive-power-for-system/

https://electronics.stackexchange.com/questions/117619/why-does-reactive-power-affect-voltage

I found both of those links disappointing!

Am I mistaken? I thought that reactive power depended almost entirely on the load. Is it common to "inject reactive power" at the source?

Yes to control voltage....

..."The reason why the reactive power is needed in the first place is because it accounts for the magnetization of the equipment. If there's no reactive power, transformers, generator rotors/stators, machines etc. have no magnetic field. With no magnetic field, there is no torque, no magnetic coupling in the transformer etc. So, a lot of equipment have to consume reactive power in order to work. If there's too little reactive power available the equipment will try to draw more current to compensate. This will lead to higher voltage drops, which in the end might cause voltage collapse.

...In a meshed grid it can also be used to control power flow. This works because the active power flow through a cable is mainly given by the voltage difference over it. If you inject reactive power, the voltage and currents angles will change, thus it will affect the power flow. If you inject the right amount at the right place you can redistribute the power flow the way you want (but only to a small extent)"...

https://electronics.stackexchange.com/questions/140552/why-is-it-desirable-to-inject-reactive-power-into-a-transmission-system

http://ieeexplore.ieee.org/document/7465837/

https://www.electrical4u.com/capacitor-bank-reactive-power-compensation/

Thanks. I am well aware of the need for reactive power at the consumption end, and in fact have been investigating the possibility of adding switched capacitors to improve the power factor of our facility, where our largest loads are saturable transformers, running much of the time at 0.5 PF or even less.

I hadn't given much thought to what must be done at the generator end...

Yes depending on cost analysis, you may indeed be a candidate for corrective power injection...

Apparently, I have a problem with semantics. I hardly think of capacitors as injecting power, but as affecting the timing thereof, and again, this would occur at the load end of the transmission line.

To me, injecting power would require a source of energy, and a capacitor is not a source of energy, except momentarily.

Likewise, I have great difficulty understanding the articles written by people whose first language is not English. Incorrect sentence structure and other gramatical errors really distract me!

Well it adds power by seeking to control the angle incidence of different legs...remember there are 3 legs and their relationship of wave angles are critical to performance...

http://www04.abb.com/global/seitp/seitp202.nsf/0/18aa8879b8cc0186c125761f005035b7/$file/Vol.8.pdf

Thanks again. Now I know it is semantics.

This illustration from the ABB link

makes it look like the capacitor bank is supplying power to the motor, without specifying where much of that power originates, which is in the collapsing magnetic fields inside the motor.

If a capacitor could truly supply power, then (perhaps with a larger capacitor bank) we should be able to disconnect from the transmission line and have the motor continue to run...

It would take an animated illustration to show the fact that at some times power is flowing from the motor to the capacitor, while at other times it is flowing from the capacitor to the motor.

Reactive power doesn't add energy, only real power does.

Real power is the product of voltage and current in phase with the voltage. Reactive power is the product of voltage and current that is 90 degrees out of phase with the voltage.

"Reactive power doesn't add energy, only real power does."

Thanks! That helps.

It has been almost 60 years since I first studied complex numbers, and I think I understand the concepts pretty well. I know trig better than I know the back of my hand, but it is still difficult to wrap my head around "imaginary power" (reactive power). Imaginary power would, of course, be imaginary energy divided by time, but I have great difficulty imagining imaginary energy...

Power is voltage times current. If the load is purely reactive (say a perfect inductor or capacitor), the voltage waveform and current waveform will be sinusoids that are 90 degrees out of phase (blue and red). Half of the time the product (yellow) is positive and half the time negative. So the power flows in and out twice each cycle, but the average is zero.

"but the average is zero." I agree with everything you said in this post, and that is precisely why I have difficulty with the concept of a capacitor "injecting reactive power".

If current isn't in the same phase as voltage (reactive load), you can separate the current vector into in-phase with voltage and 90 degrees out of phase. So real power is the product of voltage and in-phase current, and reactive power is the product of voltage and 90 degree out-of-phase current.

The 90 degrees out-of-phase is imaginary because of Euler's equation:

e^it = cos(t) + i sin(t), where you can represent a sinusoid of any phase with e^it .

The sin(t) is 90 degrees out of phase with the real part, cos(t). It allows you to do the math involving phase with complex numbers. But it's really just sines and cosines.

Again, I'm totally in agreement.

I am confused. If a generator happens to have no reactive load (1.0 power factor, purely resistive load) what happens?

If that is true, then this generator will likely not support load on the local grid due to excessive voltage drop.

If all the local load is purely resistive, there is no problem whatsoever, but that never really exists. It is purely a hypothetical.

The voltage and reactive power flow will rise across the system in proportion to the equivalent impedance between the point of injection and the node(s) of interest. Unless the system in question is a SMIB (Single Machine Infinite Bus) series equivalent system, determining the actual node voltage(s) and line flow(s) will require the use of an iterative loadflow program.

Dear Mr.Sunil mahto,

You have not mentioned about the type of reactive power.

Is it inductive or capcacitive nature. The reactive power depends upon the inductance and capacitance, and which one is predominant is decided by the impedance and you know the equation which is

Z² = R² + ( XL² - XC²) where Z = Impedence, XL = Inductance, XC² = Capacitance.

If Inductance is predominant, the terminal voltage of the generator will be less than the induced voltage. If capacitance is predominant, the terminal voltage will be more than the induced voltage.

You can draw the vector diagram and you can ubderstand. As regards the reactive power it generates heat and dissipated not available for useful work, and hence it is a loss, efficiency will be less.

Dhayanandhan.S

I believe the original question was about the injection of VARS onto the grid, in excess of the reactive load demand in the system. This means typically that the phase of current would now be opposite angle that required normally in said system, and I suspect the prime movers would trip due to dV/dt.

Does "the injection of VARS onto the grid, in excess of the reactive load demand in the system." mean having excess capacitance at the source, or perhaps having some form of thyristors firing at incorrect timing, or something else?

Your suspicions aside, the rate of change of the voltage (dV/dt) relay is there to protect against rapid changes of the voltage, not the phase angle difference. That falls to the OOS (Out Of Step) relay.

The total reactive power generated has to be equal to the total consumed by the load. If you increased the reactive power supplied by a generator beyond that required by the load, the remaining generators would supply negative reactive power, in other words, a leading power factor.

On a theoretical basis you are correct, however from a practical standpoint actual utility grids do not operate that way. Grid codes specify the actual operational limits, with many layers of protective relaying and control systems that prevent those limits from being breached.

In your example, if one machine's excitation system tried to to supply all the reactive power, then the OEL (Over Excitation Limiter) would attempt to stop the increase in field current to he machine's limit, while the (UEL) Under Excitation Limiter) on all the other machines would attempt to stop the decrease in their respective field currents. If that failed, the TOC (Time Over Current) relays on each machine would act to prevent anything from carrying more than its rated current for too long, etc., etc. And let's not forget all the over/under voltage/current relaying on the transmission/distribution network.

One possibility is that a circuit protective device will operate, thereby disconnecting it.

Oxymoron.

VoltAmperes Reactive (VAR) is not power.

The only benefit of adding reactance is if you are correcting the power factor. If the circuit has high inductive reactance, then add capacitance. If the circuit contains capacitive reactance, then add inductive reactance. The goal is to end up with a power factor of 1 in the circuit, particularly on the transmission and distribution lines.

Adding reactance to a circuit increases current levels, but since the current and voltage are out of phase and the power factor is lower, the circuit is not able to do an equivalent level of work for the same voltage and current if they were in phase and the power factor is high.

"...VoltAmperes Reactive (VAR) is not power..." Actually it is the shortcomings of language (and education) that cause the myriad of misunderstandings regarding Reactive power. VARs are the electromagnetic/electrostatic power that charge/discharge every reactive element in a circuit.

There are three usages of "power"; i.e., Real Power (P), Reactive Power (Q), and Apparent Power (S) which is the vector sum (aka Dot Product) of P and Q. When written correctly as S = P + Q or with a bar ¯ over S,P,Q, the characters are being identified as Phasors. Phasors/Vectors are a mathematical construct /shorthand for easily(?) analyzing quantities that have both Magnitude and Angle.

"...The goal is to end up with a power factor of 1 in the circuit..." True for RF transmission lines (aka Standing Wave Ratio = 1), but generally impractical for the power grid, where the goal is to provide the customer with whatever power factor their load demands within the constraints of the applicable local grid codes. If the customer doesn't mind paying the penalty for a poor power factor as allowed by the grid codes/local utility, then it is incumbent upon the utility to deliver the required current within the proper voltage limits, not to dictate to a customer how to utilize that (apparent) power, unless of course the loading is detrimental to other customers on the grid.