Well, not really ... If you are going to load it with elements such as a light bulb, then 10kVA = 10kW. However, when you think about electric motors, there is a slightly different conversion factor ... If the motor is 10kW and its cos? = 0.8, you will overload the generator ... Such a simple formula that will allow you to calculate the apparent power consumed by the device (i.e. power in VA): P (active power in Watts) * 1 / cos?
PS with engines and such things as washing machine, refrigerator, you must bear in mind the starting power (it can never be greater than the instantaneous power of the generator)
Welcome! The power in watts is the reactive power P = U * J * cos? The power in VA is the active power S = U * J (cos? not equal to 0) (maybe this will help to solve the problem) Depends on the type of load - motor or resistive (cooker, bulb), you need to consider starting with the motor.
I absolutely disagree with the last post. Active power is given in watts, while apparent power is given in volt-amps. P = U * I * cos (phi) S = U * I In short: apparent power is the sum of active power (converted into work) and reactive power. The generator power given in VA specifies the maximum current consumption from it at rated voltage, regardless of the type of load. Paweł1968: maybe it's time to refresh the news from school?
Welcome ! Buddy "menek" and you are absolutely right that you do not agree with me because YOU are RIGHT !!! Inattention and I mixed the powers. Thanks for the correction, a to the rest of my colleagues, SORRY !!!
I am going to connect various types of drills, sanders, balls, etc. to the aggregate.
Already have something specific in your eye? In single-phase generators, KVA can be compared with KW, because they are rarely used for motors or other large inductive loads. In most cases, power tools with commutator motors, the power of which rarely exceed 2kW, are connected to them. If, however, you want to use a three-phase and connect welding motors to it, etc., then you have to take into account the cos (phi) factor. As a rule, such devices have 0.8 and you multiply it by the power given on the gregat in KVA and you get the power in KW. But, there is also a very important "but". The starting current of such devices must also be taken into account. During start-up, the engines consume a few times more power than indicated on the plate, so you need to buy a suitably strong aggregate. More on this topic here: http://www.agregaty.elem.com.pl/index.php?page=dobor There are still different types of generators for different applications with different degrees of protection. Synchronous, asynchronous, inverter, cycloconverters, etc. Each has its own advantages and disadvantages. For your needs, I would choose synchronous or cheaper asynchronous. More about these generators here: http://www.bazafirm.pl/newbaza/edytorwww/pane...a=2373C0&id_zakladki=11981&l=1&id_grupy=11136 Also, before choosing an aggregate, it is worth considering this choice. There is also a question of the company, durability, reliability, popularity, spare parts, etc.
The discussion revolves around the conversion of kVA (kilovolt-amperes) to kW (kilowatts) when selecting a generator. It is clarified that while kVA can equal kW for resistive loads (like light bulbs), the conversion differs for inductive loads (like electric motors) due to the power factor (cos φ). The formula P (active power in Watts) = S (apparent power in VA) * cos φ is emphasized, particularly for motors with a typical power factor of 0.8. The importance of considering starting power for motors is also highlighted, as they may require significantly more power during startup. The conversation includes advice on selecting generators based on load types and the necessity of accounting for starting currents, especially for three-phase systems. Summary generated by the language model.
