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I want to be able to move the windmill I am designing out of the wind when the power output is to much (high wind scenario). To do this I need to measure the output of the power. To set everything up I need to be able to select my equipment for max current/voltage/power/etc. The generator I am planning to use for now is this generator. The max output from the graph seems to be about 800V DC rectified. At max RPM however the max OC DC rectified voltage is about 675V. Some calculations later I figure about 500VAC LL output @ 675V DC rectified. This is assuming a balanced load and current flow but there is none at the open-circuit point. I am assuming the voltage will be lower under load. Being under load all the time (diversion unit in place) will allow me to "safely" use equipment rated for typical 3-phase AC voltage of 480V?

Also for testing and maintenance since this is a prototype I wanted a shut off switch to be able to work on the control section. I am thinking the best way is to have a 3 phase 480V knife switch (lock out tag out method) in series between the diverting modular and the control modular. Any other suggestions?

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    \$\begingroup\$ "Knife switch" sounds like parts you get from the supplier Dr. Frankenstein used for his gear. Isolators can't switch load - they can only safely be used when the load or supply has been switched off or isolated elsewhere. You may want to use a proper fully enclosed disconnect switch that can interrupt the load and provide finger-proof terminals. \$\endgroup\$
    – Transistor
    Commented Jun 25, 2017 at 23:17
  • \$\begingroup\$ There exist knife swithches with an integrated finger-proof cover. eg: dealdey.com/deals/change-over-switch-knife-switch \$\endgroup\$
    – Jasen Слава Україні
    Commented Jun 26, 2017 at 6:28

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Instead of trying to measure max power to decide whether to move the turbine out of the wind - measure the rpm of the input shaft - if too high shut down (brakes on) and / or move out of wind.

This will work no matter what the fail is : generator fail, equipment fail or just high wind - you may need to consider an external power source hydraulic pressure may be possible...

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  • \$\begingroup\$ This doesn't answer my question but it gets me to my final solution. I initially thought of using the RPM but figured the output power would be more useful for monitoring so that was stuck in my head and never thought of RPMs again. Now that I have it all laid out, I can see there will be more parts and cost using the output power then I wanted. Since the system is open ended in that it could be used to use/store power locally or connect to the grid the power can be calculated later on. I will probably calculate the power/RPM to calibrate the system once working. Thank you. \$\endgroup\$
    – K.Doe
    Commented Jun 26, 2017 at 0:06
  • \$\begingroup\$ Extracting the power is the "brake" that keeps the frequency within range - why convert to dc? Unless you are charging batteries? \$\endgroup\$
    – Solar Mike
    Commented Jun 26, 2017 at 5:46
  • \$\begingroup\$ could be a simple as a centrifugal governor moving the tail of the turbine. \$\endgroup\$
    – Jasen Слава Україні
    Commented Jun 26, 2017 at 6:32
  • \$\begingroup\$ @Jasen or a spring loaded tail even easier \$\endgroup\$
    – Solar Mike
    Commented Jun 26, 2017 at 6:35
  • \$\begingroup\$ @SolarMike I do not plan on converting to DC but that is what the bad documentation on the generator provides its results based on. The 15k model does not have a dc rectifier but the 20k has that option. \$\endgroup\$
    – K.Doe
    Commented Jun 27, 2017 at 16:17
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The motor voltage will tell you the speed. That is probably what you should use for overspeed protection.

The published specifications are really poor. Is the rectifier integrated into the motor housing? If not, they should publish the AC output voltage and current ratings. If so, they should publish the equivalent internal DC resistance, not the per-phase resistance. You shouldn't need to calculate ratings by reading curves.

You can calculate the full-load voltage based on the internal resistance, but the published specifications should make that easier to do. According to my calculations, the output voltage will drop about 10% from no-load to full-load.

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  • \$\begingroup\$ I agree they are really poor. Hard to figure stuff out when they give you nothing. I did not choose the generator. \$\endgroup\$
    – K.Doe
    Commented Jun 25, 2017 at 23:56

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