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One- or two-variable speed drive solution

Which one is the best, using two variable speed drives (method B) or one variable speed drive (method A)? In method B the rating of the motor is the same, but in method A the load motor is bigger than the test motor, how can I calculate the rating of the load and test motor? If for example I have tested motor 11 KW, what is the rating of the load motor?

In answer to "one- or two-variable speed drive solution", I have to say: What is it you are trying to accomplish?

If all you want to do is load your variable speed drive/motor combination to 100% (or higher), the single variable speed drive solution will cost you the least in materials and will allow you to achieve your goal. If you need to load the motor/variable speed drive combination over a wide speed range, you need two variable speed drives. If you can get an identical motor and variable speed drive, and the variable speed drive manufacturer OK's the common-bus solution proposed above would be the least-cost option for that scenario.

If common-bus is not an option, the load variable speed drive sizing will be dependent on how much braking power the load variable speed drive will absorb over the duration you need it to. My application required 100% load for fifteen minute periods. Toshiba advised that a 37kW variable speed drive I had available could absorb 13kW in regen power continuously using their standard heavy-duty dynamic braking resistor. They also recommend that the motor connected to their variable speed drive not be less than 50% of the variable speed drive rating. I had a 30kW motor available of the same nameplate speed as the test motor, so that's what I went with. Testing proved that 15kW was all the 37kW variable speed drive would absorb. If I tried to regen above that, the variable speed drive would not accept a lower speed reference because the DC buss voltage could not be brought down by the DB resistor circuit.