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I am taking a course on AC circuits and so far I have seen that for the different types of power (read real, reactive and complex) different units are used.

For the real power, which is power in fact dissipated by the system, the SI compliant watt is used. I am very much in agreement with this.

However, for reactive and complex power the equivalent SI derived unit volt-ampere is used. In the special case of reactive power, a lowercase "r" is appended.

Now, I wonder: why should any other unit be used instead of watt, if the purpose of the unit is to specify the type of magnitude of a value? Moreover, why should a symbol be appended to the unit when its use is simply that of a reminder?

Also, is there a reason why using watts for reactive power would be considered wrong?

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    \$\begingroup\$ Why don't we use Joules for torque also? Like work, it's force times distance, just not in the same direction. \$\endgroup\$ – Kaz Apr 21 '13 at 20:31
  • \$\begingroup\$ It is different to compare energy and force, than to compare an energy transfer rate with another. One comparison deals with different types of quantities, the other does not. \$\endgroup\$ – Severo Raz Apr 22 '13 at 4:53
  • \$\begingroup\$ If you apply voltage V to a Resistor and current I flows then power dissipated is V x I Watts. If the load is "complex" with R + L + C components then multiplying V x I at any given time will NOT give you power. If the voltage source is a hand cranked alternator then you will be very clearly be able to distinguish between a pure resistive and pure reactive load even when V and I are the same in each case. So too can a grid power station. \$\endgroup\$ – Russell McMahon Apr 22 '13 at 5:37
  • \$\begingroup\$ Should the product of V·I be not a measure of power when dealing with reactive components, then why would reactive power be called power in the first place? The product yields the rate at which energy is being transferred to/from the reactive component, how is that not power? \$\endgroup\$ – Severo Raz Apr 22 '13 at 5:54
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Simply because a watt is a measure of work done. The real part of V*A is watts. There are exact equivalents in mechanical systems. \$W=\dfrac{J}{s}=\dfrac{N*m}{s}\$ so \${N*m}\$ here is Force that does work through a distance but it can also be a measure of Torque over a distance and that is static(non moving). One is the potential to do work, one is the work itself.

Using watts for reactive power would be wrong because reactive power is stored power and not capable of doing work.

Using VAr - r for reactive is just short hand for the purely imaginary part, instead of using i or j (for you physicists out there).

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  • \$\begingroup\$ I get the torque-energy unit duality, since each represents a different type of magnitude. However, I still don't get why two measures of power will need to be represented with different units. The watt is not a measure of work done, it is a measure of energy conversion or transfer rate. The reactive power is the energy transfer rate from and to a reactive load; this latter definition does not collide at all with the former. Energy dissipation is a form of energy transfer, but not vice versa. \$\endgroup\$ – Severo Raz Apr 22 '13 at 4:35
  • \$\begingroup\$ What you are calling energy dissipation is real work done by the VI. From the voltage/current point of view, if it moves something, creates heat, makes sound that is all work. Imaginary VA is just stored. It is called reactive power but that is misusing the term power. If you have an ideal reactive load (purely imaginary) there is no average energy transfer. There is instantaneous energy flow but the Expected value is zero. These terms are used are indiscriminately in EE. You see in more recent years the attempt to tighten usage up, don't know if this is intentional or not. \$\endgroup\$ – placeholder Apr 22 '13 at 4:43
  • \$\begingroup\$ I think we have a misunderstanding when it comes to the difference between the terms "work" and "power". While "work" (energy) can be stored, how could power? You cannot store a flow as itself, however you can store whatever flows in the flow. Sorry for the word game. I don't see how the reactive power is a misuse of the term "power", since it is the energy transfer rate from the active source to the reactive load, be it an inductor storing the energy in a magnetic field, or the capacitor in an electric field. \$\endgroup\$ – Severo Raz Apr 22 '13 at 4:51
  • \$\begingroup\$ Well it's a pedants question to begin with, with watts being real and also the real part. It makes sense to call the imaginary part it something different. Even for calculation reasons. \$\endgroup\$ – placeholder Apr 22 '13 at 4:56
  • \$\begingroup\$ My short answer is that if you consider a transformer (traditional copper iron for instance), the VA is what you get as useful output and the Watts is what comes out as heat. So a 120VA transformer with a 12V secondary will supply 10A, but may draw (say) 150W, which you could measure as heat. So that's why it's useful to make the distinction, because I'm far more interested in knowing I've got 10A at 12V coming out than in what load it presents to the upstream supply. In this context. \$\endgroup\$ – Ian Bland Nov 4 '16 at 0:09
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These are three different powers therefore they have different units:

P [W] - real power

Q [var] - reactive power

S [VA] - complex power

Complex power S is a sum of real and reactive powers:

$$S = P + jQ$$

where j is a an imaginary unit. This can be shown on a complex plane like this: enter image description here

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A power transformer is rated in VA. It is not rated in watts because that would be wrong. The transformer can supply a certain rated voltage at a certain rated current and if the load on the transformer is purely reactive then no-watts (net) are transferred BUT the transformer is at its limit.

To say a 50VA transformer is rated at 50W is nonsence - the load may not "lose", "transfer" or "dissipate" any watts but it sure will be pushing the 50VA transformer to its maximum ratings.

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  • \$\begingroup\$ The fact is that the transformer will be undoubtedly "transferring" energy, at a rate. This is why such phenomenon can be called "power". In the case of an inductor, it is known that power delivered to it will be converted to a magnetic field. It is also valid to say that power can be drawed from an inductor in a time-varying magnetic field. Since the watt is the unit for energy conversion or transfer rate, I don't see where is the nonsense in rating a transformer with watts. \$\endgroup\$ – Severo Raz Apr 22 '13 at 4:43
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Here's an example: I have a device that apparent power of 10 VA with \$cos \phi=0.7\$. It uses \$P=S*cos \phi=0.7*10 \mbox{ } W=7 \mbox{ }W\$. On the other hand it also uses \$Q=S*sin \phi=0.714*10=7.14 \mbox{ }W\$.

We could then add those two together since hey, it's both consuming 7 W and 7.14 W at the same time and get 14.14 W, which is obviously incorrect, since device will never use more than 10 W, even if its power factor somehow got corrected back to 1.

Basically var is used to signify that the reactive power is not in phase with active power and that you can't add them together. On paper, power is power but by using var (sometimes incorrectly referred to as VAr, so r isn't just appended, it's part of a different unit whose name is often misspelled), volt-ampere and watt, we (at least hope to) clearly signify the context in which such unit is used. So var not only signifies magnitude, it signifies phase as well.

This is actually nothing new and is common in other fields as well. For example we have radian and steradian as measurement units for angles which are actually 1. Same thing goes for bit as well. So while we could technically say that speed of a communication link is 1 MHz, it implies different context than 1 Mb/s.

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  • \$\begingroup\$ Why would you add together reactive power and the active power, units aside? This is what I meant with the "reminder" purpose of the var unit. \$\endgroup\$ – Severo Raz Apr 22 '13 at 4:44
  • \$\begingroup\$ @Wolter Hellmund Well I wouldn't, but from my experience many people would if they wanted to calculate total load on the system (which is needed when choosing dimensions of conductors, transformers etc.), producing incorrect result. If we use a different unit, hopefully people will remember that they can't be just added together. \$\endgroup\$ – AndrejaKo Apr 22 '13 at 8:07
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If a wall transformer is driving a reactive load, then during part of each AC cycle it will be transferring energy from the mains to the load, but during another part of each AC cycle it will be transferring energy from the load back to the mains. If during each 1/60sec cycle a load takes one joule and returns 0.25 joules, then the power taken by that load will be 60 watts and the power returned will be 15 watts, so the total power consumed will be 45 watts--the 60 to the load minus the 60 to the mains. On the other hand, the total power conveyed by the transformer would be 75VA--60 watts to the load plus the 15 watts to the mains.

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It is just a convention. It explicitly differentiates the different types of power when speaking or writing (which have different meanings mathematically, even though they have the same fundamental units).

If the different types of power are given "different-looking/sounding" units (even though VA = W), than it is very clear to a person familiar with the field what you are talking about, even if you don't tell them.

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