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Regarding this question, I will try to use the analogy of the differential equations $$F = m\frac{dv}{dt}$$ and $$V = L\frac{dI}{dt}$$ and $$w = \int Fv dt$$ with $$w = \int VI dt$$ to figure out the concept of reactive power.

If I move a mass back and forth, so that its velocity changes like $$v = v_0sin(\omega t)$$ my hands are doing a force similar to what a spring would do. But in this case, there is no spring and no friction also.

Formally, if the product Fv is integrated along a cycle, the result is zero because part of the time they have the same sign, and part opposite sign. But if I am the source of that accelerated movement, I am doing work along the cycle, in the meaning that I am spending energy.

Now compare with a current, also changing in a sinusoidal way in an inductor and without resistance. There is a changing voltage associated with that current, and without any capacitor, this voltage must be supplied by the source. In this case it is also true that the work, as the integral of the electric power during the time of the cycle is zero.

But the source of the electrical energy (say a diesel generator) is burning fuel to keep the AC running in the circuit.

It is different in the case of a spring-mass or inductor-capacitor system. Without friction (resistance) the oscillations don't need a source (except for initiate the harmonic motion).

Is it correct this mechanical analogy to understand reactive power?

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  • \$\begingroup\$ Perhaps you're looking for something that doesn't exist. There doesn't have to be a viable mechanical analogy for every electrical property. \$\endgroup\$
    – Chu
    Aug 25, 2021 at 10:20

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I'm not sure if this will help ...

Try pushing and pulling a friction load with a coil spring between your hands and the load. Hands and load will be out of phase. The real power will be applied to the load. The reactive to the spring which will take it and give it back every cycle.

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  • \$\begingroup\$ I think that with a friction load, the idea makes sense to me. Even in my example of a mass without spring. In part of the half cycle, I need to do the work necessary for the mass get a given kinetic energy. This energy is consumed by the friction load in the remaining half cycle, without I supply any force. The same for the second half cycle, but in the opposite direction. At the end of the day, the net work is against the friction load. \$\endgroup\$ Aug 25, 2021 at 15:16
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You are adding efficiency into the mix again, which is confusing you.

To start: if you have a force \$F(t) = F_0\cos \omega t\$, then the velocity will be \$v(t) = \frac{F_0}{\omega} \sin \omega t + C\$. If you assume that \$C = 0\$ it matches your equation above. The power at the mass is \$P(t) = F(t) v(t) = \left(F_0\cos \omega t\right)\left(\frac{F_0}{\omega} \sin \omega t\right)\$.

The power at the mass averages out to zero. It is purely reactive.

But if I am the source of that accelerated movement, I am doing work along the cycle, in the meaning that I am spending energy.

Yes. Yes you are expending energy to move that mass which on average is absorbing exactly zero energy. That's because muscles are not efficient movers of reactive objects. So when you attempt to say that power isn't reactive because the thing exerting the power is using real power, you are letting the inefficiency of the source of power confuse you about the character of the reactive power at the sink (the mass, in this case).

To reiterate -- on average, the mass is not absorbing any real power. Your muscles are wasting power, but the mass is not absorbing any real power. Because an alternating force is being exerted on the mass, and the mass is moving, there is "power" happening, but because the mass is not absorbing any energy on average, that means that the power at the mass averages out to zero, and is purely reactive.

But the source of the electrical energy (say a diesel generator) is burning fuel to keep the AC running in the circuit.

It is different in the case of a spring-mass or inductor-capacitor system.

Only in your mind. You are letting the larger picture confuse you from the definition of reactive power. The power at the inductor averages out to zero. It is purely reactive.

To determine if a load is drawing reactive power, you ignore what is happening inside the source. If the source is inefficient in the face of supplying reactive power you ignore that. You only care what's happening at the interface to the load.

So -- draw a boundary around the load. Ignore what the generator is doing. Only pay attention to the power transfer at the boundary. If no average power is being transferred at that boundary, then even if the generator is glowing white from heat generated by inefficiencies, the load is still drawing no real power -- it's power is entirely reactive.

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One of the subjects that constantly helps me form a good analogy in electricity is water or fluid dynamics, a very easy topic to learn and super useful in analogies.

I'm writing this without an answer, but I'm confident we will find a very simple analogy as we go.

First let's start with voltage which will be the velocity of water, then current which will be the amount of water that can move through a cross-sectional area per time, finally a pipe will be the wire which the water/electricity will pass through.

Now that we have set the equivalent for each one, let's start with a straight pipe with a random diameter of 1", you open a switch and electrons move through wire, you open the water tap and water moves through the pipe.

Now add a perpendicular hole to the pipe and glue a bigger diameter tube with a short length, it will have an area that can store some water like the capacitor.

Now cut the wire and solder a capacitor.

Now open the switch, the capacitor will charge and the electricity will lag until the capacitor is charged.

Now open the water tap the bigger diameter tube will fill and water will lag until the bigger diameter tube is full.

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