Regenerative Drives - Effect of Power Factor

err no - 100% of the true power is already there - the 10% reactive power is not useful in the sense it is not something you can phase shift from imaginary to reality and then use it to boil a kettle for tea.

In a traditional L-C world, on the wires between a load with a poor power factor and a phase shift network to restore it, energy is bouncing back and forth between the two reactances, but not much of it is  being dissipated except at the load.

To help with the visualization consider this analogy. In many ways there is much in common with the space in a laser cavity.

Consider the photons buzzing about trapped between 2 mirrors of perhaps 89% reflection, 10% transmission and 1% loss  - numbers for easy sums only . In equilibrium there may be light at a power of 1kW bouncing about in the cavity, and the light intensity, or no of photons, between the 2 mirrors matches that, but coming out of the mirrors bazooka-like is just 100W at each end, (10% transmission means that 10% of the stored energy gets out on each bounce, remember)  and the mirrors themselves heat to the tune of 10 watts each (1% loss) the rest, the 89% just joins the new power coming in and heads off to the other mirror to try and get out another time….

Now the real power that is coming into the system is only the 220watts - we only have to replenish just enough energy to replace those photons leaving as the wanted beams plus the energy dissipated on the mirrors. (plus any cavity loss, neglected here).

The apparently higher level of energy bouncing about reactively is only there as stored energy, built up over many cycles. Sure if you blew one of the mirrors off a few seconds after switch on you could get just the one cavity full of stored energy but that would be it. (such things as Q-switch cavity dumper lasers do actually exist by the way, and routinely fool the folk trying to look at the safety case from a peak power perspective...)

So much for the simple optical system. 

In terms of power factor,  you will usually have transmission coefficients (representing the ratio of the stored energy levels, as I2L/2 or  CV2/2  to the power dissipated or arriving from the supply ) nearer 50%-100%, but the science and the maths is un-altered - the stored energy in the inductors or capacitors is cycling in and out over the cycle period but as you end up where you started after one cycle, having to put back in as much as you took out, that energy is not actually available to burn. If anything the electric case is less extreme as the stored energy per cycle may be less than the dissipated, the ‘Q’ factor of stored to burnt can be less than 1. In a laser there is always a lot more stored then used so it is more obvious.

The equivalent of the laser cavity dump has you cut the wire to the PF correction caps at the very instant that their voltage peaks, and then walk off with them - you get one lot of CV2/2 to play with and then it's gone. (this is a thought experiment, do not try it ! )

Mike

err no - 100% of the true power is already there - the 10% reactive power is not useful in the sense it is not something you can phase shift from imaginary to reality and then use it to boil a kettle for tea.

In a traditional L-C world, on the wires between a load with a poor power factor and a phase shift network to restore it, energy is bouncing back and forth between the two reactances, but not much of it is  being dissipated except at the load.

To help with the visualization consider this analogy. In many ways there is much in common with the space in a laser cavity.

Consider the photons buzzing about trapped between 2 mirrors of perhaps 89% reflection, 10% transmission and 1% loss  - numbers for easy sums only . In equilibrium there may be light at a power of 1kW bouncing about in the cavity, and the light intensity, or no of photons, between the 2 mirrors matches that, but coming out of the mirrors bazooka-like is just 100W at each end, (10% transmission means that 10% of the stored energy gets out on each bounce, remember)  and the mirrors themselves heat to the tune of 10 watts each (1% loss) the rest, the 89% just joins the new power coming in and heads off to the other mirror to try and get out another time….

Now the real power that is coming into the system is only the 220watts - we only have to replenish just enough energy to replace those photons leaving as the wanted beams plus the energy dissipated on the mirrors. (plus any cavity loss, neglected here).

The apparently higher level of energy bouncing about reactively is only there as stored energy, built up over many cycles. Sure if you blew one of the mirrors off a few seconds after switch on you could get just the one cavity full of stored energy but that would be it. (such things as Q-switch cavity dumper lasers do actually exist by the way, and routinely fool the folk trying to look at the safety case from a peak power perspective...)

So much for the simple optical system. 

In terms of power factor,  you will usually have transmission coefficients (representing the ratio of the stored energy levels, as I2L/2 or  CV2/2  to the power dissipated or arriving from the supply ) nearer 50%-100%, but the science and the maths is un-altered - the stored energy in the inductors or capacitors is cycling in and out over the cycle period but as you end up where you started after one cycle, having to put back in as much as you took out, that energy is not actually available to burn. If anything the electric case is less extreme as the stored energy per cycle may be less than the dissipated, the ‘Q’ factor of stored to burnt can be less than 1. In a laser there is always a lot more stored then used so it is more obvious.

The equivalent of the laser cavity dump has you cut the wire to the PF correction caps at the very instant that their voltage peaks, and then walk off with them - you get one lot of CV2/2 to play with and then it's gone. (this is a thought experiment, do not try it ! )

Mike