A watt is simply the SI unit of measurement for power. So, the power of a mechanical load applied, say, to an electric motor is typically measured in kilowatts. In North America, a mechanical load is still generally measured in horsepower, which is the Imperial unit of measurement for power. The watt and the horsepower are simply two different units for measuring precisely the same thing!
As the motor is drawing 9.7×110 = 1,067 watts (or 1.067 kW), and delivering 1.25×746 watts (or .9325 kW) of mechanical energy, it should release 1,067-932.5 = 134.5 watts (or .1345 kW) of heat.
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.
The mechanical load of a motor determines the necessary output power rating of an electric motor. As mechanical loads are defined in terms of watts (or, in North America, horse power), then motor's output must be rated in watts, too.The so-called 'power' rating of a transformer is determined by the rated voltage and the rated current of its secondary winding. The product of these two quantities is the transformer's rated 'apparent power', expressed in volt amperes.Incidentally, the symbol for "kilowatts" is "kW", not"KW's"!
thanks ,it should be 400amps.
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.
Yes, But in this case the current will rise up by about 15% at full load. So if you will use the motor to drive any mechanical load take in your consideration the mechanical load Power (KW) must not exceed 85% of motor power (KW)
Assuming the generator converts 90% of the mechanical power into electrical power, it has an efficiency of 90%, which means it consumes 11 kW of mechanical power under a full electrical load of 10 kW. Under no load the frictional losses will still apply, but the resistive losses in the windings will not be present. Therefore the no-load losses can be estimated as 500 watts in this conditon.
Because the machines which use power i.e; load, are always rated in KW and the machines which produce power i.e generators, are rated in KVA.AnswerA motor's mechanical load is expressed in watts (horsepower in North America), so the output power of the motor must also be expressed in watts (horsepower in North America) so that the machine can be matched to its load. You cannot express a mechanical load in terms of volt amperes.
As the motor is drawing 9.7×110 = 1,067 watts (or 1.067 kW), and delivering 1.25×746 watts (or .9325 kW) of mechanical energy, it should release 1,067-932.5 = 134.5 watts (or .1345 kW) of heat.
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.
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.
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
Theoretically 22 kW is equivalent to 29.5 HP but there are some power losses in the motor and the probable maximum output of mechanical power is 25 HP.
It could overheat if the mechanical load exceeds the rated load (maybe due to poor lubrication) or because it is being run on an excessive voltage (in that case it would overheat even on no-load) or because of an internal fault such as a breakdown in insulation in one of the windings.