thanks ,it should be 400amps.
Since the breaker that is installed on the generator set will be sized to the output of the 30 kW generator, the load will stay connected until the thermal trip of the breaker trips the load off line. This will be in the matter of seconds before it happens. To handle a 75 kW load and depending if it is an inductive or a resistive load you will need at least a 80 to 100 kW generator.
When using a resistive load bank to test a generator, it does not matter if you load the generator to its kW or kVA rating, because those two numbers are the same when considering a resistive load. Power factor, which is the difference between true and apparent power, only comes into play when there is a reactive (inductive or capacitative) load.
First you will need a three phase generator. Mathematically there are 746 Watts per horsepower, but I like to use 1000 Watts for ease of mental calculation. This would mean you would need a 30 kW generator. If using 746 Watts per HP, you would need 22380 Watts, or 23 kW. Make sure this 23 kW is the normal load rating of the generator, not the surge rating! 30 kW would provide more of a safety cushion.
If a load takes 50 kW at a power factor of 0.5 lagging calculate the apparent power and reactive power Answer: Apparent power = Active power / Power Factor In this case, Active power = 50 kW and power factor = 0.5 So Apparent power = 50/0.5 = 100 KVA
maximum demand is measured in kva because current drawn is dependent on power factor for the same load and current drawn is calculated with kva
The kW rating of a transformer can be calculated by multiplying the kVA rating by the power factor. For example, if the power factor is 0.8, then the kW rating of a 100 kVA transformer would be 80 kW. You can also use the formula: kW = kVA x power factor.
160 Amp. MCCB 45 KW Contactor (67.5-112.5) Thermal Over Load
To find the minimum kW service demand load for twenty 6.5 kW ranges in a multifamily dwelling, you would simply multiply the number of ranges by the kW rating of each range. In this case, 20 ranges x 6.5 kW = 130 kW minimum service demand load.
Since the breaker that is installed on the generator set will be sized to the output of the 30 kW generator, the load will stay connected until the thermal trip of the breaker trips the load off line. This will be in the matter of seconds before it happens. To handle a 75 kW load and depending if it is an inductive or a resistive load you will need at least a 80 to 100 kW generator.
Think about it. Assuming a total transformer capacity of 60 kV.A, how could it possibly supply a load of 100 kW?
When using a resistive load bank to test a generator, it does not matter if you load the generator to its kW or kVA rating, because those two numbers are the same when considering a resistive load. Power factor, which is the difference between true and apparent power, only comes into play when there is a reactive (inductive or capacitative) load.
Transformers are rated in VA or kVA. That is because the voltage is limited by the power loss in the magnetic core, and the current is limited by the power loss in the resistance of the windings. The rated voltage times the rated current gives the transformer's rating in kVA.
0.0075 kw
It does not matter, when testing a generator with a resistive load bank, if you load it to kVA or KW. For a resitive load, i.e. non-reactive load, the power factor is one, so kVA and kW are the same.
1. KW OR HP Rating of you load 2. VOLTAGE Rating or Range of your load 3. PHASE 4. Frequency Range 5. Application - say HVAC 6. Enclosure Type 7. With Filter and Ventilation Fan?
Because it is the current rating of the windings that determine the maxium load current, and the product of rated current and rated voltage, in a.c., is apparent power (in volt amperes), nottrue power (in watts).And, incidentally, the correct symbols are kV.A and kW, not kva and kw.
Motors are rated according to their output power, expressed in watts or kilowatts (or horsepower, in North America). This is because the motor's rating must be matched to the power requirements of its mechanical load.