Inductive generators, what is their output compared to their horsepower rating

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To convert a 40 hp three-phase AC motor into a single-phase induction generator, understanding its efficiency is crucial, with a typical efficiency around 95%. The power output in kilowatts can be determined by converting horsepower to kilowatts using the appropriate conversion factor. If the induction generator operates at 80% efficiency, this indicates that the electrical output will be 80% of the mechanical input from the internal combustion engine (ICE). Consequently, to achieve the desired electrical output, the ICE must produce 20% more power than the generator's rated output. Additionally, determining the required torque from the ICE involves calculating the motor's specifications and operational parameters.
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Let say I have a 3 phase ac motor rated at 40 hp. Now I'd like to be able to convert this to act as a single phase induction generator (I hear is possible but I don't know much about it). driven by an ICE at around 1800 rpm. What would its power output be in kilowatts. Or more importantly how would I find that out? What other information from the motor do I need? I appreciate your time in looking into my question. Thanks.
 
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Welcome to PF! All you really need to know is the efficiency. A good 3 phase motor/generator can be 95% efficient. After that, it's just knowing the English-metric conversion factor for hp to kW...
 
Thanks for that Russ. I wasn't sure if it was as simple and converting HP to kW. A few more follow up questions. If we assume an induction generator is 80% efficient is that referring to mechanical input to electrical output? Meaning the electrical output will be 80% of the mechanical input? If so does that mean I will need an ICE that will produce 20% more power than the rated output or would I just produce 20% less electrical power? Also how do I find out how much torque is needed from the ICE?
 
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