It's because the voltage is a specified value, and the current drawn has a limiting value. So multiplying those together gives the VA or kVA that it can supply.
2
To select the size of a cooling tower for a 1000 KVA diesel generator (DG) set, first determine the heat rejection requirement, which is typically around 10-15% of the DG's output power. For a 1000 KVA DG set, the heat rejection would be approximately 100-150 kW. Next, consider the cooling water flow rate and the temperature difference between the inlet and outlet water to determine the cooling tower capacity. Finally, consult manufacturer specifications and guidelines to choose a cooling tower that can handle the calculated heat load efficiently.
Presumably, you are asking what is the rated secondary current for a 45 kV.A (not 'kva') transformer? The answer depends on its rated secondary voltage. To obtain the rated secondary current, you divide the (apparent) power rating by its secondary rated voltage.
A gen set will be rated for a certain kVA at a certain power factor, or certain kW at a certain power factor. If rated in kVA, the power factor indicates the amount of real power you will get (ie if rated at 10,000kVA at .95 pf, the gen set can generate .95 x 10,000kW, or 9,500kW).
A transformer's capacity is rated in volt amperes(V.A). This is the product of the secondary winding's current rating and voltage rating.
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.
What is the minimum height required for chimney of 100 KVA DG Set which is kept in
2
To calculate the power factor (pf) of a 1010 kVA diesel generator (DG) set, you can use the formula: [ \text{Power Factor (pf)} = \frac{\text{Real Power (kW)}}{\text{Apparent Power (kVA)}} ] First, determine the real power (kW) output of the DG set, which can be obtained from the manufacturer's specifications or by measuring the actual load. Then, divide the real power by the apparent power (1010 kVA) to find the power factor. Typically, a power factor of 0.8 to 0.9 is common for DG sets.
The oil pressure for the DG sets depends with the load.
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 three main parts of dg sets one is avr and anotther one is coolent oil level
max.load that can run on 62 kva dg is of 86 amperes.
It is 7 .5 litre per hour
1- low lube oil safety 2-water safety 3-temprechar safety any athar safety
sump capacity means the oil capacity for the requirements of DG sets. Which is required for controlling, maintanance of DG sets
This is the rated output of the transformer, obtained by multiplying the rated secondary voltage by the rated secondary current. And it's 'kV.A', not 'kva'.