What is reactive power and how to deal with it

Power factor is the ratio of active power (watts, kilowatts) to apparent power (volt-amperes, kilovolt-amperes). The power factor in the general case is always less than unity. Only with a purely active load (lighting, heating devices) is it equal to unity. The value of the power factor determines the fraction of the apparent (full) power of the generator or transformer that they can give to the electrical receiver in the form of active power.

Thank you very much, really understandable information.

That's just the article forgot to add that most of the reactive power is returned back to the electrical system! If you explain on the fingers, the current flows through the wire in both sides at the same time if there are disagreements - from the generator to the load and from the load (it returns energy) to the generator. And naturally, this is only possible with AC. And the consumer PAYS for energy that he did not actually use! Therefore, some things (such as lowering the level of consumption) occur only virtually because of the idiotic principle that the meter considers the passing energy, and WHERE it goes on the drum. Compensation is a thing of course necessary, but for the most part to energy companies. Well, if you think logically - how the introduction of an ADDITIONAL element with losses in the circuit can increase its efficiency ???? But as a method of combating harmonics and subsidence (excesses) of voltage in a line, it is effective, because agrees generator and load. Naturally, thinner wires can be used (for theoretical cos = 0, the current in the wire will double, becausewill flow through the wire in both directions the same SIMULTANEOUSLY). The load on control and distribution devices will also decrease due to the same. And generators with reverse current transformers do not like. And these processes occur during ANY load change (if it is not purely active, which generally speaking doesn’t really happen, even an ordinary lamp has a negligible inductance). In the 70s in the United States, due to DISCONNECTING, the plant immediately under the line brought under a hundred distribution transformers in several states ...

Andrey, household meters are “active energy meters”. With all the ensuing. They do not take into account reactive energy.

AndrewFirstly, the plant is always powered by several power lines. And even if the plant is completely de-energized, which is impossible in principle, since there are always several independent sources of energy supply, this cannot serve as a reason for de-energizing distribution substations. The plant is operating - the load is at the substations, the plant has shut down - the load has decreased by some value. This is not an emergency mode for the power system. It can only vice versa - the plant is de-energized as a result of the de-energization of several substations.

Cosine phi (power factor) is the ratio of active power to total power consumption. In principle, it cannot be equal to zero. All transformers located at substations designed for a certain power, and this power is full, that is, taking into account the active and reactive component. The consumed electric power, although active, even reactive, always goes in one direction. The power can have a different direction on the transit lines of substations, in this case, depending on the state of a particular section of the power system, the active and reactive power can have a different direction (consumption or return of electrical energy).

Dear friends (the author of the article and commenting), I do not agree with you on everything, but I will not discuss this. I want to state my vision of the physics of the process. In general, in nature, such a type of energy (power) as "Reactive", of course, does not exist. But there is a concept: Reactive energy (power). This concept characterizes the phenomenon that occurs in electrical circuits of alternating current. The essence of the phenomenon is simple. Inductive and capacitive elements create (arise) magnetic and electric fields. In alternating current circuits, these fields are naturally also variable. Energy is expended on the creation of these fields. For example, when a current flows in an inductance, a magnetic field arises. Moreover, when the current increases, the energy from the electric network (i.e. from the generator) is consumed to create this field, and when the current decreases, the energy stored in the inductance is returned to the network. Obviously, for each period, the magnetic field doubles from zero to a maximum and twice decreases in the opposite direction. A similar phenomenon occurs in the tank. Only in the capacitance do electric fields oscillate and this happens synchronously with a change in voltage. The oscillation phases of electric fields in a capacitance and magnetic fields in an inductance are always in antiphase. Similar phenomena occur in mechanical systems: for example, when a spring is compressed, energy is expended, and when unclenched, the stored potential energy is released (why not the capacity?) Or, for example, to accelerate water to a steady speed in a closed water supply system, it takes some time for the pump to work, if after that the pump turn off then the water circulation will continue for some time by inertia due to the stored kinetic energy (this is an analog of inductance).

Conclusion: Reactive energy is not some special type of energy, it is electrical energy, which is periodically consumed in alternating current circuits and given up by reactive elements.

PS. - Reactive energy (power) can be measured, meaning it exists.

The only thing I agree with the author is that there are a lot of legends around the concept of "reactive energy" ... Apparently, the author put forward his own revenge ... Confused ... contradictory ... all kinds of abundance: "' comes, the energy goes ... "The result was generally shocking, the truth was turned upside down:" Conclusion - the reactive current causes the wires to heat up without doing any useful work "Sir, dear! heating is already work !!! My opinion, here people with a technical education without a vector diagram of a synchronous generator under load can’t stick together the process description correctly, and for those interested, I can offer a simple option, without any fancy.

So about reactive energy. 99% of electricity with a voltage of 220 volts or more is generated by synchronous generators. We use different electrical appliances in everyday life and work, most of them “warm the air”, give off heat to one degree or another ... Feel the TV, computer monitor, I don’t even talk about the kitchen electric oven, everywhere it feels warm. These are all consumers of active power in the power supply of a synchronous generator. The active power of the generator is the irretrievable loss of generated energy by heat in wires and devices. For a synchronous generator, the transfer of active energy is accompanied by mechanical resistance on the drive shaft. If you, dear reader, rotated the generator manually, you would immediately feel increased resistance to your efforts and that would mean this one, someone included an additional number of heaters in your network, that is, the active load increased. If you have diesel as a generator drive, be sure that fuel consumption increases at lightning speed, because it is the active load that consumes your fuel. With reactive energy, it’s different ... I’ll tell you, it’s incredible, but some consumers of electricity themselves are sources of electricity, albeit for a very short moment, but they are. And if we take into account that alternating current of industrial frequency changes its direction 50 times per second, then such (reactive) consumers transfer their energy to the network 50 times per second. You know how in life, if someone adds something to the original his without consequences it does not remain. So here, provided that there are a lot of reactive consumers, or they are powerful enough, the synchronous generator is excited. Returning to our previous analogy where you used your muscle power as a drive, you will notice that despite the fact that you did not change the rhythm by rotating the generator, or did not feel a surge of resistance on the shaft, the lights in your network suddenly went out. Paradoxically, we spend fuel, we rotate the generator with a nominal frequency, but there is no voltage in the network ... Dear reader, turn off reactive consumers in such a network and everything will be restored. Without going into theory, the excitation occurs when the magnetic fields inside the generator, the field of the excitation system rotating together with the shaft and the field of the stationary winding connected to the network rotate in the opposite direction, thereby weakening each other. Electricity generation decreases with decreasing magnetic field inside the generator. The technology has gone far ahead, and modern generators are equipped with automatic excitation regulators, and when reactive consumers “fail” the voltage in the network, the regulator will immediately increase the excitation current of the generator, the magnetic flux will return to normal and the voltage in the network will restore It is clear that the excitation current has active component, so please add fuel in the diesel ..In any case, the reactive load negatively affects the operation of the mains, especially when the reactive consumer is connected to the network, for example, an asynchronous electric motor ... With a significant power of the latter, everything can end badly, by accident. In conclusion, I can add for an inquisitive and advanced opponent that there are also reactive consumers with useful properties. These are all those that have electric capacity ... Connect such devices to the network and the electric company already owes you)). In pure form, these are capacitors. They also give off electricity 50 times per second, but at the same time the magnetic flux of the generator, on the contrary, increases, so that the regulator can even lower the excitation current, saving costs. Why didn’t we make a reservation about this before ... but why ... Dear reader, go around your house and look for a capacitive jet consumer ... you won’t find ... Unless you spoil a TV or a washing machine ... but it will not be useful .... <

Well, as if 50 Hz is a change in the direction of the current 100 times per second it took another 1 year ... So everyone is literate.

Eugene, in the first year of seminary or the Institute of Physical Education? Would not be dishonored! He who has a brain has learned even in a class that way in the 7th-8th that hertz is a full period of oscillation per second! Those. with a sinusoidal waveform with a frequency of 50 Hz, the sign changes to the opposite 50 times per second, but the half-wave will be already 100! You read here, the hell takes it: electrical engineering has now become like a pagan faith: all obscurantism and heresy ...

Friends, by decreasing the reactivity, you are reducing the active, it is a fact! The counter will show this too!

Remember elementary physics!

To find out the indicator of active power, it is necessary to know the total power, for its calculation the following formula is used: S = U \ I, where U is the voltage of the network, and I is the current strength of the network.

The calculation of active power takes into account the phase angle or coefficient (cos), then: S = U * I * cos

So take ticks, measure reagent, if less than 0.9, put Conders of the appropriate rating and you will be happy!

All this is correct, but if we put a diode bridge in the circuit with a capacitor (all losses of active power for heating the diode bridge and capacitor, of course, will be taken into account by the counter as active power), and after connecting the diode bridge, connect the electrolytic capacitor, then it will charge to the maximum mains voltage, after which, having no way for its discharge, it will begin to stand charged at the maximum network voltage. The charge time can be arbitrarily long, but the capacitor consumed only current from the network through the diode bridge, gradually accumulating its charge and increasing the voltage on its plates to the maximum voltage of the network, and the capacitor consumed only the current, which is 90 phase degrees ahead of phase voltage, i.e. reactive current from the network. Yes, the capacitor did not return its charge to the electric network in the next quarter of the period, as it should have done if it had been connected to the electric network without a diode bridge. And then the power of the capacitor without taking into account the active losses due to heating of its plates would be considered a purely reactive power. But the capacitor was charged with current from a current source in the form of a diode bridge, and this current was a reactive current with respect to the electric network, since there is another capacitor in the circuit to the diode bridge. That is, the meter did not take into account this electric power, because it was reactive power and the current was ahead of the voltage by almost an angle of 90 electrical degrees, and the meter as active power takes into account only the power that coincides in phase with the current. In this case, the electrolytic capacitor connected after the diode bridge can no longer be discharged to the network, after charging to the maximum voltage of the network, it will remain in a charged state.That is, some part of the electric energy not taken into account by the meter is selected from the electric network. If the capacitor is discharged quickly enough to some load, for example, a resistor, then the charge accumulated by the electrolytic capacitor is converted into thermal energy and it will heat the resistor. The capacitor will again be charged from the mains. If a current flows continuously across the resistor, then the capacitor will smooth out the ripples of the rectified voltage, recharging from the network with reactive current. But at the same time, a rectified reactive current will flow through the resistor itself. The magnitude of the voltage drop across the resistor will depend on the magnitude of its resistance. The constant component of the current through the resistor will not be able to affect the electrical angle between the current and voltage in the part of the circuit to the diode bridge, since the voltage after the diode bridge is 1.41 times higher than the voltage to the diode bridge. Of course, due to the fact that the load voltage on the diode bridge coincides in phase with the drain at the ripple current and the ripples of the rectified voltage are completely smoothed out, the meter will not take into account part of the load power as the active power in the alternating current network. For a large load power, such a circuit is unacceptable because of the size of the capacitors and high currents. But such a scheme is used in power supply schemes for LED lamps with a ballast capacitor. If a ballast resistor is installed instead of a ballast capacitor, then the power consumption of the LED lamp immediately increases by 20 - 25 times due to large losses on heating the ballast resistor. Such a scheme can be used only at low capacities and exclusively for converting electric energy into heat, for example, into warm energy on the internal resistance of LEDs with the emission of light.

All commentators are so smart, you write or copy comments from different sites or books. So tell me, what do we live in such an asshole that we have to study the types of energy ourselves and how it works and what we pay for. Respect to the author.

in comments it is written even worse than in the article - no one is clear

And what kind of trick is this kind. Active energy is 53435. Reactive consumed-7345, and reactive released-36456 and this is according to the meter. Why is there such a difference between reactive energies and is it right that we are forced to pay for it

Where did you get these formulas from ?! Gross power: S = root of (P * P + Q * Q), where P is active and Q is reactive power. To find the reactive one, you need to multiply the active one (which P) by a certain coefficient (tg f), which is located from cos f according to the passport data of the receiver (if you need it, you will easily find it). Arr ... Now, you are looking for information on the internet, and you come across nonsense ... Reducing reactive power in no way reduce active !!! On the contrary, full power should strive for active !!!

"...at theoretical cos = 0, the current in the wire will double"m ... yes!
Well, draw already, even for yourself, this damn unit circle and thisfucking Cartesian cross with arrows (one to the right, one to the top).