I tired to connect a 60 W 240 V resistive load to a 1000 W generator which was running at 200 V immediately after the load was connected the voltage drops to 60 V and the generator stops spinning. By the way it is a three phase 415 V delta generator.
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asked Dec 23, 2014 at 3:12
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\$\begingroup\$ What was turning the generator? Power out = power in minus losses \$\endgroup\$– Andy akaCommented Dec 23, 2014 at 8:34
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\$\begingroup\$ Im using wind to turn the generator. As the voltage generated reaches a 180 V i tired to hook up a 10 W load. The rotor start to slow down so drastically till it literally stop spinning. \$\endgroup\$– 3.1415926535897932384626433832Commented Dec 23, 2014 at 9:25
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1\$\begingroup\$ How large a diameter WT?. What wind velocity? What did the 10W load consist of. eg a 10W light bulb may book like a 100W+ load until properly lit. | Power ~= 600 x A x V^3 x efficiency Watts. A m^2. V m^3. Efficiency 5%-40% deep-ending on design. \$\endgroup\$– Russell McMahon ♦Commented Dec 23, 2014 at 10:06
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1\$\begingroup\$ It's only a 1000 watt generator if there is 1000 watts (plus mechanical losses and electrical losses) mechanical power turning the rotor. \$\endgroup\$– Andy akaCommented Dec 23, 2014 at 10:10
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1\$\begingroup\$ Lots of good info on making wind turbines here scoraigwind.co.uk If your blades are less than 2.5m dia, they are probably too small to drive a 1kw generator. \$\endgroup\$– user16324Commented Dec 23, 2014 at 11:02
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Reason for the PMG getting locked:
(1) My wind Source is not strong enough. Needs a proper channeling to the blade. May be a reverse blower concept might be implemented.
(2) My Blade Design is needed to be improvised. Blower Fan is too heavy and not effecient enough.
Reasons: Tried to hook up a 60 W load and use extra force on the blade, by means of rotating it by hand. The light starts to turn ON. Concludes that at current state the blades requires additional torque in load condition. At no load it is generation 270 V 30 Hz, I would need to push up the no load condition up to 415 V 50 Hz then only try to load the generator and see.
answered Dec 24, 2014 at 9:35
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\$\begingroup\$ You gave us a windspeed but not a ducting diameter. Given both, we can get the max power available (before turbine and generator losses) \$\endgroup\$– user16324Commented Dec 31, 2014 at 12:52
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\$\begingroup\$ @BrianDrummond the diameter of ducting is not finalized yet as we haven't implement the reducer yet so far the ducting end is 1ft by 1ft and the wind speed is 20 m/s. \$\endgroup\$ Commented Jan 2, 2015 at 0:06
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1\$\begingroup\$ Near enough. Cross sectional area <0.1m^2 and 20m/s gives 2m^3/s air volume or 2.4kg/s. So there's <480W in the airflow to begin with. \$\endgroup\$– user16324Commented Jan 2, 2015 at 11:19