10-12-2009 10:25 PM
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
10-13-2009 02:55 PM
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
10-13-2009 10:12 PM
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.
10-19-2009 12:23 PM
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
10-19-2009 07:00 PM
10-19-2009 09:42 PM
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...
09-06-2011 02:52 AM
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?