I'm not sure what you're asking - is DG distributed generation? Full load would be provided by the DG; if it's a 40MW wind farm, it will be able to provide you with turbine data; each turbine will be able to provide so much kVA. I would use this number (instead of kW output) for sizing of equipment as this would give a maximum. You may also need to take into account any D-VAR and capacitor bank effects on this cumulative KVA.
To calculate the DG set current, you need the load current and the load voltage. To convert it into kilowatts it is divided by 1000.
Rating for DG set and any of electrical machines is calculated in KVA. KVA is calculated as KW/pf. One can calculate the required KVA for DG set with this formulation: (KW/pf)/load rate. For example KW=110, pf=0.8 and one loads the DG at 75%, so KVA= (110/0.8)/0.75=185 KVA.
I=Kva*1000/v*1.732 =500*1000/415*1.732 =500,000/718.78 =695.62 Amps. So max.load of 500kva DG is 695.62 Amps
There are several ways to calculate working load limit. One of these includes Minimum Breaking Load (MBL) divided by Working Load Limit (WLL) equals Working Load Limit (WLL).
Full load starting current is typically in the region of 5or 6 times the full load motor current;.
To calculate the DG set current, you need the load current and the load voltage. To convert it into kilowatts it is divided by 1000.
To calculate the DG set current, you need the load current and the load voltage. To convert it into kilowatts it is divided by 1000.
no load voltage - full load voltage by full load voltage
Rating for DG set and any of electrical machines is calculated in KVA. KVA is calculated as KW/pf. One can calculate the required KVA for DG set with this formulation: (KW/pf)/load rate. For example KW=110, pf=0.8 and one loads the DG at 75%, so KVA= (110/0.8)/0.75=185 KVA.
The oil pressure for the DG sets depends with the load.
The question has to be more specific. Full load amps, watts or voltage. Please restate your question.
To calculate the fuse rating level, you typically want to choose a fuse that is slightly higher than the full-load current to avoid nuisance tripping. For a full load of 4 amps, you could choose a fuse rating of 5 or 6 amps.
The fuel usage will be based on the load applied to the generator and how many KW the generator outputs. The more load the higher the fuel consumption. You do not list the output of the generator nor the load so all I do is point you to the diesel fuel generator consumption chart. Click the link to see the chart where you can figure it yourself.
max.load that can run on 62 kva dg is of 86 amperes.
Full load current ofthe motor x 0.58
I assume you are asking about AC generators. The controller is wired up with suitable transducers and other control wiring to obtain required signals from the electrical bus. Three DG sets are feeding the common bus bar through isolating breakers. Conditions for a successful synchronization are equal voltage, equal frequency (Hertz) defined by the speed, same phase sequence. (say R Y B phases must match with R Y B of the incoming DG set) One DG is started with some load (at least 60 to 70 % of load) and the common bus is energized. The second DG to be brought in by closing the breaker either manually or automatically. Start the second DG. The controller reads the bus parameter and incoming DG parameter in real time and moment the incoming DG parameters match with running DG parameters, controller allows the incoming breaker to close. Both DG are now synchronized and load is shared. Same way put on third DG set and synchronize.
I=Kva*1000/v*1.732 =500*1000/415*1.732 =500,000/718.78 =695.62 Amps. So max.load of 500kva DG is 695.62 Amps