Compact Fluorescent Lamp Load Calculations

Question

I help build/wire up large illuminated carnival/parade floats, and in the past we have used around 11,000 40w/60w GLS BC light bulbs in festoon sockets, powered by a 800KVA 3 phase diesel generator. However, these are becoming more and more difficult to buy, and so we have started to use energy saving compact fluorescent lamps.

The float uses normally around 80% of the capacity of the generator. My question is, as I start to replace more and more of the old incandesant bulbs with compact fluorescent lamps, what considerations do I need to apply when calculating the load that I'm putting on the generator? I'm concerned that as the load is becoming more inductive and less resistive. I've read that at some point power factor correction will be needed, but I'm not sure how to calculate this, and how to apply this on this scale.

For example, before I would have had:

4000 x 60w = 240kW
7000 x 40w = 280kW

but now I have:

3000 x 60w = 180kW
6000 x 40w = 240kW
2000 x 11w = 22kW   (11w energy saver bulbs)

All this runs ok, the generator isn't overloaded, everything runs smoothly, the phases are near perfectly balanced, and nothing gets warm!

As I increase the ratio of inductive to resistive load, what will happen? How do I calculate how much inductive load the generator can take?

UPDATE: I found some information about a Synchronous Condenser on Wikipedia, which appears to do what I need by just connecting an unloaded motor to the supply, but it lacks details about how to do the calculations. Can anyone help with that?

Answer 1

That's an impressive piece of kit - far larger than most of us will ever get to play with.

Alternators can and do behave differently with reactive (in this case inductive) load components, with exact results varying with the design.

SO, I'd say that the "right" answer is to ask the alternator's manufacturer for specifications of its characteristics with varying amounts of inductive load.
But as a starter ...

• Mechanical load wise the alternator will not mind inductive load component. VAR (Volt-Amps reactive) are "Wattless". The energy source (diesel motor?) will not know they are there overall. You will still however have to provide for the IR distribution losses for the reactive current so they will contribute some real Watts, but it will be small compared to if they had been pure-resistive-load Watts.

• Inductive load component may quite possibly affect the ability of the alternator to regulate well. I assume that you are producing a 3 phase supply. If so, you'll wan't to keep the reactive component spread reasonably evenly over all 3 phases.

You'll be able to find better references than this, but a quick search turned up this Southern Illinois University Laboratory experiment instruction sheet which seems to do a good job of covering the basics. Dr Louis Youn, whose web page this is on, should be able to advise further. They note:

• The output voltage of an alternator depends essentially upon the total flux in the air gap. At no load, this flux is established and determined exclusively by the dc field excitation.

  Under load, however, the air gap flux is determined by the ampere-turns of the rotor and the ampere-turns of the stator. The latter may aid or oppose the MMF (magneto motive force) of the rotor depending upon the power factor of the load. Leading power factors assist the rotor, and lagging power factors oppose it.

  Because the stator MMF has such an important effect upon the magnetic flux, the voltage regulation of alternators is quite poor, and the dc field current must continuously be adjusted to keep the voltage constant under variable load conditions.

  Related: If one phase of a three-phase alternator is heavily loaded, its voltage will decrease due to the IR and IXL drops in the stator winding. This voltage drop cannot be compensated by modifying the dc field current because the voltages of the other two phases will also be changed. Therefore, it is essential that three-phase alternators do not have loads that are badly unbalanced.

I suggest that you try isolating the CFL bulbs (or enough of them to make a difference) in a switchable bank and observe effects on regulation (if any) with them in an out of circuit. This may work best if you test load with a limited number of incandescent bulbs and a large percentage of CFLs. Or, "best" of all, a purely CFL load. Observe carefully / no responsibility taken / Do not bend fold staple or mutilate / YMMV / Caveat Emptor ... - but it will probably be OK.

Another possible issue is CFL startup, but it is unlikely that the load on the alternator can be worse than the horrendous cold filament surge load of incandescents.

If you do find that the inductive component is causing eg regulation problems you can add capacitive reactance to balance. At the load capacity you have it may be attractive to use a 3 phase motor adjusted to present a capacitive reactive load to the system. I think these need to be wound-rotor slip ring induction motors - Mr Google is not too forthcoming on this but I know he knows. The technique is mentioned here- but using a rather large and special induction machine to do it.