Newbie needs some help kw/rpm relation

[alibaba2]

theory check: i have a hydro site that has generates 2000Nm of torque at 100rpm. this has been calculated at 20.95 kw. i am looking to attach a low-rpm PMG making 20kw at 250 rpm. if i use a gear box to achieve the 250 rpm (leading to a gecrease in torque 2,5 times=800Nm) will i be making the 20kw that is rated at rpm according to manufacturer specifications? the generator has 93% efficiency, so in the end i will be making 93% of the 20kw , or 18,6 kw.

please let me know if my calculations are correct

big 10q

 

[russ_watters]

Minus losses in the gearbox (which could be substantial), your logic about the power sounds correct. Note, though, the rpm gets divided, but the torque multiplied by the gear ratio. That's how the power stays the same (p=rpm*torque).

 

[oswaler]

Your calculations appear to be correct. In order to achieve the desired 20kW output at 250rpm, a gear box will need to be used to decrease the torque to 800Nm. This decrease in torque will result in a decrease in power output, which can be compensated for by the efficiency of the generator. With a 93% efficiency, the final power output will be 18.6kW, which is close to the desired 20kW. It is important to also consider other factors such as the efficiency of the gear box and any losses in the system, but your overall approach seems reasonable. It may also be helpful to consult with a professional in the field for further confirmation and guidance.