DG Set rating at various power factor

The generator minimum amperage will be at Unity Power Factor with the generator connected to a load larger then generator kVA or synchronized to the grid, as the voltage regulator is adjusted the power factor is adjusted leading or lagging, the generator terminal voltage is fixed by the grid and the generator amperage changes with the change in voltage regulator setting.

Caution must be used when adjusting the generator voltage regulator to make sure you do not exceed rated power factor because the generator can slip poles causing mechanical damage.

You may find additional topics answering this question.

For the Diesel sets, the Prime mover namely the diesel Engine has got restricted power.In a 500 kVA set, the engine can deliver only 400 kW power.May be it can take 10 % overload.

The alternator is rated for 500 kVA,0.8 pf.This can give an output of 400kw.

Supposing the alternator has to deliver full load current of 696 Amps at pf of 0.9, the power required by the Alternator from the engine would be equal to 1.732X 415X696X0.9/1000, (assuming efficiency 100%) which is equal to 450 kW.Since the engine is rated for 400kW, it cannot deliver 450kW

Therefore, DG set cannot be loaded beyond rated current at rated power factor also pf cannot be improved to unity and at the same load the alternator to rated current.

For any more clarifications you can contact cbr@cuppu.com

Thank you for your good feed back

G,

Power factor has always been a baffling subject, I can remember in school sorting out the various formulas calculating the current, KVA, and power. I have sailed as an Instructor on a Maritime College Cadet training ship, instructing Diesel Engine, Refrigeration and Air Conditioning, and Electricity. I have attempted to simplify power factor to explaining it as the ratio of the power supplied by the source to a load, (True Power), and the amount of power that is reflected back to the source from the load (Reactive power). The ammeter will measure the sum of both currents, the current flowing from the source to the (Reactive) load, and the current flowing back to the source from the load, thus the KVA (Apparent Power) will be greater than the true power (KW).

Citing the example you presented an alternator operating at .8 power factor is receiving 20% of the power it produced and sent to the load reflected back to the alternator and there dissipated as heat.

Inductive loads reflect the power back to the source with the current lagging behind the voltage, Capacitive loads reflect the power back to the source with the current leading the voltage. The power factor can be also expressed as an angle, which is a vector of the true power and the apparent power.

A purely restive load, as a incandescent lamp or a heater will not reflect any power back to the source, and therefore the power factor is 1, or unity, all the power delivered to the load from the source is adsorbed by the load.. A network system is likely to have both capacitive and inductive loads on the system, and if the inductive current is equal to the capacitive current the system will appear to the source as a purely resistive load, and the power factor of the network will be unity.

The same phenomenon occurs in radio frequency circuits, only it is called standing wave ratio. In radio an antenna (Load) must be tuned to resonance so as not reflect any power back to the transmitter (Source). In radio the phenomenon is explained as a pulse of energy travels from the feed point of the antenna to the end of the antenna at the speed of light, and is reflected back to the feed point. if the antenna is too short the reflected energy arrives at the feed point before the next voltage pulse arrives at the feed point, and some of the energy travels back to the source ahead of the voltage pulse, the reflected current leads the voltage, the antenna is said to be capacitive. If the antenna is too long the reflected pulse arrives at the feed point after the next pulse arrives from the source, and some of the reflected energy travels back to the source after the voltage pulse, the current lags behind the voltage, the antenna is said to be inductive. When the antenna is just the right length, the reflected pulse arrives at the feed point at the same time as the next pulse from the source, and no energy is reflected back to the source, the antenna is resonant, and appears to the source to be purely resistive, the standing wave ratio is 1:1, the power factor is unity.

Various means are employed on a power grid and in radio frequency circuits to correct for power factor, capacitors are added to correct lagging power factor, inductors to correct leading power factor. Synchronous motor excitation can be adjusted to correct power factor, and Induction generators will correct lagging power factor.

Hope I didn't overload you with information, I hope perhaps I explained power factor without all the mathematical mumbo jumbo.

I have worked in the Middle East (Oman), and travelled in India, I have great respect for Indian Engineers.

Regards CEKM

G,

You can search "Power Factor" to get electrical and mathematical definitions of PF. This answer is a practical definition and how it changes your machinery selection and how it will affect your DG operation and life expectancy of the engine and alternator (generator).

Alternator (generator):

There are two types of loads that affect your generator rating and life. First: "Real Power" loads that convert energy to heat, light or motion. We measure these loads in 'watts'. Second: The other load is "Reactive Power". This reactive power is energy utilised to build and maintain an electromagnetic field found in capacitive loads (I can't think of too many capacitive loads in a building . . . maybe a large LCD panel ?) and inductive loads (like an induction motor). We measure these loads in 'volt-amps reactive' (VARs).

Watts do the real work (light - motion - heat) and VARS just recirculate from the load to the generator and back again as a magnetic field. The VARs do no 'work', but are necessary for the reactive loads to operate. The VARs arrive to / from the load in the form of extra amperes (current flow) that are not doing any 'work' but DO cause the generator windings to heat up more. So a 0.9 PF load will have less amps than a 0.6 PF load. Both loads might do the same work measured in watts, but the type of load that produces 0.6 PF will have substantial recirculating magnetism from the load to the generator and be practically measured by noticing the amps are much greater.

Both the watts and the reactive power have current demand (amps) on the generator. Watts tend to heat up the stater windings primarily and the rotor windings a bit less. VARs tend to heat up the rotor windings primarily and the stator winding a bit less.

Generator manufacturers have standardised on an world wide power factor average for a typical average load to be 0.8 PF. This gives a base line for manufacturing the average generator to give the customer good life from the stater winding and the rotor windings.

If you typically have very low power factor loads such as consistently 0.6 PF, then the generator maker will build your rotor differently (add some cooling tricks and some additional wire) and they will add more stater winding wire and different insulation to avoid the damage due to heat. heat from Amps. Amps from low power factor loads.

For high kilo-watt loads, this is real 'power' and take real engine power to turn the generator during high kilowatt loads. So they build the rotor shaft stronger, bearings larger, bearing cooling (oil or grease) differently, etc. So there are electrical and mechanical issues with watts and reactive power.

Lastly there is the temperature rise. This will relate to how hot the winding get during the rated load and the life of the winding before the insulation breaks down. There is an exponentially relationship to the decrease in insulation life when you exceed the temperature rise. Temperature rise can be exceeded if you specify a 0.8 PF generator at 400 kW but you operate it on loads with 0.6 PF. Amps = winding heat.

Black start emergency standby generators are typically rated for 130* C rise over 40* C ambient (they only operate a few hours a year in developed countries). Continuous 24/7 full load duty machines are rated for 80*C rise over 50*C. Therefore a 400 kW, 500 kVA unit rated at 80*C rise over 50* will have MUCH more copper than a 400 kW, 500 kVA rated for 130*C rise over 40*C ambient.

Engine Power and life:

The engine will only 'see' the real power watts load. Reactive loads recirculate magnetism and current but this is not 'seen' as watts by the engine. So a 400 kW load at 1.0 PF uses the same engine power (fuel) as a unit running at 400 kW at 0.6 PF. The engine does not 'see' the VARs. The generator will be MUCH hotter on the 0.6 machine, but the engine doesn't have any extra load.

So the engine people rate the engine power demand ONLY on the watt rating. A continuous duty 24/7 unit can typically be overloaded 20% for one hour in 24. A black start unit typically cannot be overloaded at all, as it is already rated very high to be competitive (small engine running at max power and small generator running at max heat).

Hope this helps.

Nice and educative one.

ramesh

Hi this is bala from Ethiopia,africa for this case can we use some external capacitors to make the power factor from 0.6 lagging to 0.8 lagging

The optimum pf is 0.8.therefore you can connect capacitors to improve the pf to 0.8 lag

dear sir,

normal DG designed at 0.8 PF if i am to add capacitor up to 0.9 to 0.99.

what happen DG set any issue the DG

please replay

vijay_ppppl@yahoo.com

sir

D G set power factor 0.8 standerd when D g trai 0.75 pf 100 Amp ups conect 413 .412.413 Voltage 51 hz 90 Kw 1525 Rpm what is Improve Pf 0.8