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New member

Re: load bank testing

Dear Mr Shrivatsa,

 

I hope your are still with Caterpillar and I hope your a mechanical engineer and not electrical engineer. Any alternator which cannot give rated power at unity power factor cannot deliver the rated power at 0.8 power factor too. and kvar is not delivered by excitation of alternator.

 

lagging power factor means the demagnetisation effect of field and avr has to give more power to field to produce the same magnetic field and i remind here the avr will grab power from main supply only.

 

I had tested alternators on dynamo meters at both unity and 0.8 power factor. Prime movers has to deliver more torque to alternator at 0.8 PF rather than at Unity power factor.

 

Please visit CPRI labs to clarify if I am wrong.

 

In India genset manufacturers will never get either CE approval or ISO approval since you people take only ARAI approval for sound and poor customer has to wait for 7.5 hours to utilise 110% of load. i pity customers in India.

 

Please go through all the documents of IS/ISO 8528 series and no generator in India will get even G3 class performance.

 

Can you show your customer, the desired KVA, KVAR and KW at 0.8 Power factor by installing an inductive load instead of resistive load and zero KVAR.

Cat Dealer
Dealer

Re: load bank testing

Dear Srivatsa,

 

Nice to meet you. My name is Tuvshinjargal, I'm a technical communicator of Wagner Asia Equipment LLC. We would like to purchase the load bank for the 3400 and 3500 series engines. What type of load bank have you been using?  Could you please let us know the information for load bank?

 

Regards,

Tuvshinjargal

Regular Contributor
Employee

Re: load bank testing

Hi,

The real power in a system is the Voltage multiplied by the same pahse current and the power factor.. Power factor of a circuit is the Cosine of the angle the current lags (or leads) the voltage... Why do we have 2 components ?? let us take the case of a motor. It has a coil (winding) wound into a core of magnetic material. When voltage is applied to it the core has to be magnetised to maintain the voltage in the sustem.. This is also termed as emf... To maintain this excitation we need some current.. This is the reactive current that is drawn from the system by the load to maintain excitation... The greater the coil inductance, the greater the excitation current needed and hence more reactive power drawn and the power factor will be lower...

In an electrical system the capacity is defined by the KVA (i.e Kilo Volt Ampere) .. This KVA has to provide the motor both Real power (KW) needed to produce the torque and the Reactive Power (KVAr) to maintain the excitation across the motor. The equation is KVA sqr = KW sqr + KVAR sqr...

 

The KW is provided by the Prime Mover( the Engine) and the KVAr is provided by the Generator excitation system.

 

In the case of a resistive load bank there is no excitarion current required as there is no magnetic coil and there is no need to provide excitation currents and hence KVAr is zero... KVA = KW.. The KW is 292 in this case and hence the 422 amps for load bank...

The motor full load current capability will depend on the power factor of the motor.. If the motor full load power factor is >0.8 then the full load current will also change for the same KW rating. Example if the full load power factor of the motor is 0.9 the load current is 452 amps...

 

Hope this is clear...

Cat Dealer
Dealer

Re: load bank testing

Reactive power is a result of inductive or capacitive reactance in a circuit causing a phase offset between the voltage and current. The overall current supplied to the load can be split into two seperate waveforms -- one in phase with the voltage, and one out of phase by 90 degrees (either leading or lagging the voltage waveform, but typically lagging due to an inductive load). Multiplying the in-phase current waveform by the voltage results in the real power produced by the generator, which is always positive and thus requires torque from the prime mover. However, multiplying the out-of-phase waveform by the voltage results in a power waveform that alternates between positive and negative, with an average of 0. Each phase will absorb power from the generator for 1/4 cycle, and then return the power back to the generator the next 1/4 cycle. If this reactive power is balanced on all phases of a 3-phase system, the reactive power cancels out completely and only results in additional heat due to the increased current in the generator stator and transmission lines, as well as the increased excitation power required to maintain the same voltage.
Visitor

Re: load bank testing

thanks for your clarifications but if we have a system (one generator and one motor)  the same generator feeding a motor (inductive load) and it takes the full load current which is 525 amp (neglecting starting of the motor) so from where the current above the 422 amp come from

in other words from where the reactive power come from

does it come from the generator only (how come)

or from the engine also as the active power

 

thanks for your helping

Visitor

Re: load bank testing

thanks very much for all of you
Regular Contributor
Employee

Re: load bank testing

Hi, The KW equation is KW = sqrt 3 x V x I x cos phi.. Cos phi is the power factor. In a genset application the real KW load that the Engine can take is 292 KW in this case.. So if the load has a 1.0 pf the current is 292/ (1.732 x 0.4) which is 422 amps. Also please note that at 1.0 pf the KW = KVA from the power triangle and there is no KVAR component of power..

Hope this clarifies the reason of using lower current value when using a resistive load bank for testing.

Regards

Srivatsa

Contributor

Re: load bank testing

For resistive load test use the 292kW rating as the full load value as it should reflect the rated driver kW.

The 363kVA / 525 amp rating can only be realized with an inductive component to your load bank.

 

hope this helps

Visitor

load bank testing

we have 1 mw resistive load bank(in 100 , 50 ,10,5 kw steps) and we want to test 4306 gen set have the following 363 kva , 292 kw at 0.8 pf ,    525 amp rated , 400 volt

 

what is the referance in the load bank testing

is it the active power which means to load the gen set with 292 kw resistive load so it will draw  421 amp only (unity power factor)

or the total rated current 525 and that means to load the gen-set with 363 kw

 

 

thanks in advance