Excessive KVAR from Public Power Supply

[18Bash]

We use a 1000KVA diesel generator to power our plant and at the peak of consumption we rarely get to 65% loading. Is it now possible for the Demand Meter (reactive loads alone)to read up to 840KVA (on public power suply)when from all indication, the total load obtainable from the generator(with pf of 0.8)consisting of both the real and reactive loads has never exceeded 700KVA. Granted that we have a very poor power factor ( around 0.52), would this be a possibility or perhaps the Meter is faulty. Note that the CT Coils are appropriately rated for 3-phase, 50HZ, 415V supply.

 

[DanDel]

You state that the demand meter reads up to 840kVA, but that the total load from the generator has never exceeded 700kVA. Where are you measuring the 700kVA?

 

[jghrist]

When you run the diesel generation, are you disconnected from the public power supply? If so, how do you supply a 0.52 pf load with generation running at 0.8 pf? If not, how much additional load is supplied by the public power supply.

If the generator supplied the total real load when operating at 0.8 pf and 700 kVA, the plant real load would be 700·0.8 = 560 kW. If the plant were running at 0.52 pf, the total plant apparent power would be 560÷0.52 = 1077 kVA. The total reactive plant load would be sqrt(1077²-560²) = 920 kvar. The generator would supply sqrt(700²-560²) = 420 kvar reactive load. The public power supply would have to supply the remaining 500 kvar.

Without the generation, the public power supply would supply the entire 1077 kVA plant apparent power, 560 kW real load and 920 kvar reactive load.

 

[18Bash]

Dandel: the generator and the public supply are used completely exclusively, one as a standby for the other. the demand meter is for the public supply consumption. The rating of 700Kva is based on the percentage of the total load rating of the generator that is ACTUALLY used up wheneer we run on generator. the two power sources are never used simultaneously.

jghrist: the 0.8pf is the rated pf of the generator, while our actual pf whether on generator or public supply (but easily obtained while on public supply) is 0.52. Using your calculation now, it is obvious that ther is no way the reactive load alone at 920KVA can surpass the total apparent load of 1077KVA which is exactly the question i asking. the pulic Utility company sems to have stopped reading the Demand Meter or reads it incorrectly.

 

[busbar]

 
It may be time to apply known-accurate portable instruments to the system and confirm or disprove the displayed numbers are correct, based on per-phase voltages, currents and phase angles.

For example, 700 kVA @ 0.52 PF is 364.0 kW and 597.9 kVAR at an angle of 58.7° at the power frequency. Three voltages, their corresponding currents and respective phase angles should “add up” or mesh with power readings from fixed equipment.

[If there are voltage/current harmonics present, regardless of “true RMS” quantities, phase-angle measurements can quickly become meaningless.]

 

[digitrex]

The best thing you can do should be measuring the voltages and currents using clamp-on meter and use the results to verify that the kVA meters are accurate. The meter errors can be at the generator or at the public utility or both.

If you can use a clamp-on power meter, you can measure kW, PF etc.

Please use the correct units appropriately for kVA, kW, kVAR etc. I am a bit confused with the messages posted.

 

[tinfoil]

1) Are you sure that the backup generator supplies supplies exactly the same loads as the utility supply? Are there 'non-essential loads' not in service when the generator is running? Is there power factor correction or some other extra equipment that is used when on generator and not on the utility, or vice-versa?

2) Are you sure that the generator is running during the actual peak period(s)? The utility meter tends to record the maximum value reached during all 720 hours or so of a billing period, not just the 3-4 hours (or whatever) that you would be using the generator.

 

[18Bash]

Thanks all, i agree with Digitrex, the best option is to take actual measurements especially of the Kw and PF. However to extend the case further , is it impossible for for a plant whose overall pF has been established at 0.52 to be run on a generator with a rated pf of 0.8? Also what would be the limitation to getting my real power by the product of the voltage and the highest amperage obtained on any of the three phases.

 

[jghrist]

18Bash,

You should get your generator's Capability Curve from the manufacturer to be sure, but I believe the following could safely apply to a generator set rated 1000 kVA at 0.8 pf:

The generator is limited to 1000 kVA
The prime mover is limited to 800 kW
The exciter may be limited to 600 kvar

You could generate 702 kVA, 365 kW at 0.52 pf, limited by the 600 kvar.

As to your second question, to get the real power, you would multiply the current of each phase times the line-to-neutral voltage of each phase times the power factor of each phase and sum them. If the phases are balanced, the real power would be 3·I·V_L-N·pf or sqrt(3)·I·V_L-L·pf

 

[digitrex]

jghrist seems to put the generator capability curve as a rectangular shape, which I don't agree. The generator should be able to produce higher than 600kVAR at 0.52pf. However, jghrist's illustration is very clear that the engine is limited to 800kW and exciter has a limitation(rotor heating) at very low pf.

Without the capability curve, I can safely estimate the max reactive power limit at 0.52pf as 90% of 854kVAR = 770kVAR. This is, at 0.52pf, the generator can produce 900kVA (470kW and 770kVAR).

Watch out for the setting the generator's overcurrent protection because it is measuring the current in ampere, which is equivalent to kVA loading. If you don't have kVA meter, you need to monitor the ammeter(draw a red mark at generator's max rating, 1400A).

I am curious, at a power factor of 0.52, you should be paying penalty to the utility. May be the amount of penalties you had paid for 1 year were more than enough to install the PF correction capacitors. In addition, with the capacitor banks, you generator will be able to supply higher kW power.

 

[busbar]

Basic tests for harmonic content should be conducted to be ruled out as a contributor to low PF.

 

[18Bash]

Digitrex: Yes we currently pay up to 20% of our total bill as Maximum KVAR demand and we are taking steps to apply pf correction.

Busbar: is it important to know how much of the pf is contributed by harmonic distortions in selecting appropriate capacitive bank for pf correction? Our p[lant is an auto assembly one and we have more of spot welding machines and others with high inductive capacities than electronics which can be responsible for high harmonic distortions.

 

[jghrist]

As Busbar said, harmonics might be contributing to the low pf. Harmonics are produced by welding machines and are a critical factor in the application of pf correction. Capacitors can produce resonant points where the harmonic currents can cause harmful harmonic voltages. See

http://www.schneider-electric.ca/

http://www/en/crns/html/toledo.htm

which discusses pf correction and harmonics at a Chrysler plant in Ohio.

 

[busbar]

 
18Bash, PF capacitors can be adversely affected by high-frequency voltage and currents on your 50Hz system—acting as a “trap” for harmonics, instead of just providing magnetizing current at 50Hz for loads like motors. Simply, if fuses used to protect capacitors operate, then plant power factor is again reduced from loosing the means to correct PF in the first place.

Spot welders can add further voltage-control problems on a very short-term but repetitive basis, requiring dynamic high-speed correction as described by jghrist's reference.