0.0075 kw
75×1000
75000
a: there is no calculated load for the range. This 16 kw is the maximum that this particular range can safely use any thing less is OK.
starting current of 3 phase 75 KW induction motor
12 kw
Yes I could. How?
For a purely resistive load with a unity power factor, 9.41 kVA would equal 9.41 kW. However some equipment such as a motor will have a power factor less than 1. If the power factor is 0.8 then 9.41 kVA would equal 9.41 x 0.8 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.
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.
It depends on the total connected load (KW) of the house. If the total connected load is about 20 KW the alternator generator should be designed to meet peak 20 kw load
3phese 460 volt 75 kw a/c drivi full load current
It depends on the voltage. A load of 32 amps at 120 volts will be 3.84 kW. A load of 32 amps at 240 volts will be 7.68 kW. For any other voltage, multiply the voltage by 0.032. All these calculations assume a resistive (non-reactive) load.
No because there are losses along the way. Some of the needed electrical power will go to control circuitry and inefficiencies in the heating elements themselves.
a: there is no calculated load for the range. This 16 kw is the maximum that this particular range can safely use any thing less is OK.
12 kw
starting current of 3 phase 75 KW induction motor
The bungalow would probably have a supply rated at about 15 kW but the average load would be a lot less, maybe 1-2 kW.
Yes I could. How?
1.35*1.5 KW