I have 100 W vertical axis wind turbine which is mounted on PMSG. And its generated 3 phase voltage is converted to dc using H bridg rectifier. When I connect the load, the turbine speed will reduce. Why it happen so?
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1\$\begingroup\$ turbine loose coupling raises source impedance thus speed "load regulation" occurs. Also bridge cap charging presents more load \$\endgroup\$– D.A.S.Commented Mar 4, 2017 at 3:20
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\$\begingroup\$ Impedance matching relies on generator load Voltage/current profile to match with source impedance profile vs speed for MPT \$\endgroup\$– D.A.S.Commented Mar 4, 2017 at 3:30
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6 Answers
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It's a very basic behavior of any generator. Load creates torque reverse to one that is driven by the power source. Imagine a small generator, like your own, connected to an electrical locomotive. No chance it will move!
Another way to look is by power. Power it limited at some point. Mechanically it is rotation speed times torque, electrically it's voltage (speed times a factor) times current. When no load applied, mechanical power is completely wasted on mechanical losses. If load is applied, electrical power is no longer zero, so mechanical lisses must be reduced, and it means less speed.
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\$\begingroup\$ How can I control the same speed of the turbine, even when the battery is connected? \$\endgroup\$– soumyaCommented May 24, 2017 at 13:55
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\$\begingroup\$ How can I control this using matlab simulation tool? and which type of controller should I use to maintain the speed? \$\endgroup\$– soumyaCommented May 24, 2017 at 13:57
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\$\begingroup\$ Using MATLAB simulation how can I control the speed by using controller? Is there any other controlling methods? \$\endgroup\$– soumyaCommented May 24, 2017 at 14:00
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This is more an aerodynamics than electronics question.
When a turbine aerofoil is gliding through the air, it settles down to a speed where the 'propulsive' force, the lift * aerofoil angle, is equal to the total load, comprising aerodynamic drag, and frictional and load resistance from the output shaft.
If the load is increased, then the excess of load over the propulsive force will slow the turbine down initially. What this does is, for constant wind velocity, and a constant mechanical blade angle, it increases the angle of attack, increases the aerodynamic blade angle, which increases the propulsive force to match the new load. This is the stable operating region of the turbine, where the wind power input can cope with the load, and small load increases result in small speed decreases.
If you increase the load too much, then the blades will slow down so much that the angle of attack increases too much, and the blades enter an aerodynamic stall. This is an unstable operating region. The lift now decreases with angle of attack. A small increase in load results in a decrease in speed and a decrease in lift, which results in the turbine stopping.
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The turbine slows down because the electrical load on the generator results in torque that resists the rotation of the wind turbine.
Since your permanent-magnet AC generator drives a bridge rectifier, the complex behavior of a generator interacting with an AC line does not apply here -- this speed reduction is simply a matter of the increased load on the generator. Conservation of energy requires that if the generator's electrical power output increases (amps times volts), then the mechanical power input to the generator (torque times angular velocity) must also increase. This is why the torque at the generator's input shaft increases when you connect the load.
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The turbine slows down because you're increasing the load on it. This is the same basic effect as when a table-saw blade bites into wood and slows down a bit (you can hear it)... or when you're riding a bicycle and hit a hill and you slow down.
Part of good generator design is assuring that the generator doesn't put such a load on the turbine as to make it "bog". You want it to slow down enough to get maximum useful power (torque x revs or volts x amps), but not too much that it bogs.
This problem is not unlike the MPPT problem for solar panels.
answered Mar 4, 2017 at 6:22
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Other answers have answered the immediate question : why does it slow down? But I'm sensing an underlying question : what can I do about it?
The answer to that is, tune the load to get the best performance from the generator at all windspeeds.
At no load, the generator spins happily but delivers no power. As you add load, you extract power, slowing the generator down, until the turbine blades are stalled and produce very little power.
Somewhere in between, you will extract maximum power from the turbine - the ideal load will vary according to the windspeed. I did some working out for a specific generator in this answer which may help.
For initial experiments you can simply connect in different load resistances and measure voltage across them, to evaluate performance. But a practical way of automatically tuning the load may be similar to the "MPPT" tracking systems used in solar power.
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\$\begingroup\$ If I connect the diode in series to the load , It doesn't matter for the turbine to reduce its speed. \$\endgroup\$– soumyaCommented Mar 12, 2017 at 17:51
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For a motor or generator the general rule is: speed \$ \propto \$ voltage and torque \$ \propto \$ current. When current starts to flow torque will increase and the speed will drop.
For a specific wind speed there is a speed - torque line for the turbine. This becomes a voltage - current line on the electrical side.
See Figure 120 in this reference for a sample torque - speed diagram. Good inverter design can ensure that the generator runs a maximum power.
