It is a bit of a theoretical thing, of little practical use, but I just want to understand the physics behind it. I am aware that I'm simplifying things quite a bit.

In electric power we differentiate Real, Reactive and Apparent power and of course we want the reactive part small, but with practical loads this is rarely the case.

electric power

The other day, a colleague of mine and I, were discussing a multi-MW Rotary Diesel UPS (demo takes a while to load) in one of our datacenters and the following question came to mind, which we weren't able to answer ourselves:

enter image description here

Say that the load on the UPS causes a non-ideal \$\text{cos(}\varphi\text{)}\$ on that UPS, causing Reactive Power \$Q > 0\$ being transported through the power lines back and forth. Would the Diesel engine still use fuel just for the Real power part or does Reactive power have its impact on fuel consumption too? Theoretically reactive power is not consumed, but it feels weird once grid power is replaced by a Diesel engine. Does Reactive power exist in the mechanical world?

  • 1
    \$\begingroup\$ A diesel engine has no constant torque during a full revolution. There is no constant force from each piston. Using a V8 or V12 engine and a large rotating mass minimizes variations of the torque, but there will be still some reactive power caused by the diesel engine. \$\endgroup\$
    – Uwe
    May 22, 2022 at 16:13

5 Answers 5


Reactive power would put no extra load on a generator shaft if everything were perfect. However, real generators have real losses, with some of those proportional to the square of the current. The reactive load causes more current in the wires than there would be with a purely resistive load of the same real power. The extra current causes additional real power to be lost.

So the answer is that the engine will see a somewhat higher load and therefore use a bit more fuel. This is because of more inefficiencies and losses in the system, not be reactive power itself makes the generator harder to turn.


I should have mentioned this before, but somehow it slipped thru my mind at the time.

A reactive load on a perfect generator does not require more shaft power averaged over a cycle, but it does add "bumps" to the torque. One attribute of a 3 phase AC generator is that the torque is constant over a cycle with a resistive load. However, with a reactive load parts of the cycle will require more power and other parts less. The average power is still the same, but constant pushing forwards and backwards relative to the average torque can cause undesirable mechanical stresses and vibrations.

You can think of this a bit like moving two magnets past each other. Let's say they are oriented to repell. At a distance there is little force. You have to apply force to move them close together, meaning you put energy into the system. The magnets push in the direction of motion as they move away, thereby giving you back the energy you put in earlier. The net energy spent is 0, but there was definitely energy flow back and forth. There is always some loss as energy is moved around or converted back and forth in real systems.

Again, the reactive power itself doesn't cause the problem, but real power is lost because energy can't be moved around and converted with perfect efficiency. This real power loss has to be made up with more real power input. In addition, the extra mechanical forces can decrease the life of the generator and engine driving it.

  • \$\begingroup\$ It's actually quite a significant effect in any sizable generator, especially those designed for local power generation without step down i.e. backup generators. \$\endgroup\$ May 31, 2012 at 20:40
  • \$\begingroup\$ I compared your permanent magnets to a shaft that has some (theoretically fully elastic) torsion. Is that a valid comparison in your opinion, @OlinLathrop? \$\endgroup\$
    – jippie
    Jun 1, 2012 at 16:33
  • 2
    \$\begingroup\$ @jippie: If you're likening it to a tortion spring in series with the shaft, then yes in the sense that the spring could momentarily cause higher or lower torque load, will average to the same without the spring, and won't dissipate any power. \$\endgroup\$ Jun 1, 2012 at 16:36
  • \$\begingroup\$ Coming to think of it, your magnets aren't such a bad analogy either. I remember a generator has something to do with rotating magnetic fields too... \$\endgroup\$
    – jippie
    Jun 1, 2012 at 16:45
  • \$\begingroup\$ With a Balanced Three Phase reactive load, the torque is constant over a cycle, just as with a balanced resistive load. The 'bumps' you refer to happen with any unbalanced load. (The 'constant' torque value is zero for a balanced reactive load). \$\endgroup\$
    – david
    Feb 26, 2013 at 7:31

Like Olin Lathrop answered to your first question.

Does Reactive power exist in the mechanical world?

In mechanical system the reactive power does exists. But there is no simple way to explain this without going into simple harmonic motion.

Simply imagine a object M was mounted to a string and you due to the centrifugal force it moving on a circular orbit. Suppose it's angular velocity is \$\omega\$ and you could describe it's x projected motion equation as \$R \cos (\omega)\$ and y projected motion as \$R \sin (\omega)\$.

\$ x= R \cos (\omega t)\$

\$y =R \sin (\omega t)\$

And you need to accelerate it , so you need to feed it some power.Suppose you seeding a force F to this such as F lags \$ \alpha \$ from it's current angle.

[see the figure] power factor in mechanical example So the F's component which is parallel to it's linear velocity will only do some active power. Which is \$F \cos(\alpha)\$ ,

so the active Power \$ = F \cos(\alpha ) v = F \cos (\alpha ) \omega R \$

reactive power \$ = F \sin (\alpha) 0 = 0\$

But a person who look at this will think that I'm applying a force 'F' and it's moving in the speed of 'v' so power should be Fv , but because of phrase difference it won't. This is happened to your watt meter too. Because it does not count the phrase difference between the current and the voltage , as well as in the above mechanical example it does not count the direction of the force vs direction of the motion.

  • \$\begingroup\$ Why did you make your post community wiki? With suggested edits now possible, there's little use for that option. \$\endgroup\$ May 31, 2012 at 20:39
  • \$\begingroup\$ I'm going to delete this post. Or can somebody edit this idea more, Add math latex to it and make this more clear please. \$\endgroup\$ May 31, 2012 at 20:44
  • \$\begingroup\$ Can I compare this with a (fully elastic) torsion of a shaft? \$\endgroup\$
    – jippie
    Jun 1, 2012 at 16:24

Pure reactive component of power will not consume extra fuel.

The energy flow of reactive component will keep changing direction maintaining averaged zero. When energy flow is directed backwards, the torque applied to generator shaft will reduce (for few milliseconds every few milliseconds), because generator will act a tiny bit like a motor, but staying mostly a generator.

The combustion part of machine will see averaged load equal to active component only. Say if the function of fuel supply route is to maintain constant velocity, then variations of torque (load) will be mirrored in amount of fuel. More torque, means more fuel, more consumed active power, with same velocity.

Small scale experiment is to turn permanent magnet AC motor shaft with fingers when it is disconnected. Then connect capacitor and compare.

  • \$\begingroup\$ Interesting experiment, but I haven't got this stuff available. Sounds like a cool classroom experiment (only couple of years late for me ...). I could ask if I can use one of the datacenter engines for this experiment ;o) \$\endgroup\$
    – jippie
    Jun 1, 2012 at 16:36

As noted above, the torque required for a balanced 3-phase reactive load is constant and zero. This hides the fact that for half each cycle, each reactive load is pushing power back into the phase/phases that is/are accepting energy.

If the reactive load is not balanced, the energy is fed back into the generator. You can't recover chemical energy, and some of that energy fed back into the generator is lost, but some of the energy is fed back into rotating kinetic energy of the generator. Which makes the generator turn faster-slower-faster-slower etc. A small generator does not have much rotating kinetic energy, so most of this energy is lost, and it just stresses the system.

Also hidden is the fact that if the generator spins faster, more energy goes into Capacitive Loads, and energy comes out Inductive loads.

For a very large generator set, with significant stored energy, the return of reactive energy from an Inductive network can cause the transmission frequency to increase, and eventually make the whole system unstable (higher frequency, more reactive return, higher frequency, more reactive return, generator spins out of control and self destructs). For this reason, power grids are designed to operate with a slighty capacitive load -- even though this increases the peak currents and reduces the grid efficiency.

Going back to your original question, as the generator set spins up, it pours energy into all the attached reactive loads, even the balanced ones, as the voltage comes up. Small it may be, but you can't really get that energy back. When you detach the generator, you don't get the chemical energy back again.


I thought generators generate electrical energy which is in KVA. Of this generated KVA energy first part kvar will be utilized by inductive load to keep the equipment magnetically charged and second part kw will be utilized to produce torque which would be dependent upon load. At higher loads kvar is negligible compared to kw. But still generator has to produce it. If a pure inductive coil is connected to load, generator will only generate kvar component in lbs & fuel consumption will be more than no load


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.