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I would like to monitor the power produced by an AC generator in a wind turbine. The goal is to monitor how the produced power varies with wind conditions. I connected the power resistor to AC generator to burn out the power. The voltage sensor measures the voltage across the resistor, and the current sensor measures the current flowing into the load resistor.

For computing average power, I rectified the output from the AC generator and use the DC voltage and current to compute the average power using $$1/N\sum_{i=1}^{N}V_{DC}(i)I_{DC}(i)$$ I thought that if I rectified the output, I can compute the average power more reliably. However, I observed that there is lots of variability in the computed average power. In other words, even for the same wind condition, the computed average power changes for different experiments (wind tunnel).

As an alternative, can I use non-rectified AC voltage and the current to compute the average power? Just computing the root mean square of the power time series would results in an sufficiently accurate result? The RPM of the generator is about 500, and the sampling frequency of measuring the voltage and the current is about 1000Hz.

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  • \$\begingroup\$ How variable have your readings been? And during what length of time? \$\endgroup\$
    – user39962
    Commented Feb 18, 2015 at 7:40
  • \$\begingroup\$ I don't see why you cannot use RMS of the AC voltage. Note that even if you rectify it, you still need to compute RMS. Rectified AC is not really DC. If you add a filter capacitor after the rectifier, then I guess it will be "DC, with ripple," but the combination of rectifier + cap will present a different and somewhat nasty load to the generator. I think it is best to start by measuring a purely resistive load. \$\endgroup\$
    – user57037
    Commented Feb 18, 2015 at 8:05
  • \$\begingroup\$ The average power, computed using 60-sec power time series with 1000Hz sampling, fluctuates roughly 10 percent of its mean value. In other words, I obtained 10 average powers, they varies with the standard deviation of 10% mean. \$\endgroup\$
    – Jinkyoo Park
    Commented Feb 18, 2015 at 8:28
  • \$\begingroup\$ I don't exactly understand "measuring a purely resistive load". It that mean measuring the voltage and the current without processing them through rectifier? \$\endgroup\$
    – Jinkyoo Park
    Commented Feb 18, 2015 at 8:30

2 Answers 2

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As an alternative, can I use non-rectified AC voltage and the current to compute the average power?

If you want an accurate number for power this is the only sensible way to do it because any other method will be only accurate if the voltage supply is sinusoidal and the current drawn has no harmonics.

Take high-speed simultaneous samples of voltage and current waveforms. Multiply i and v samples. Average over a period to suit you (best and most accurate if taken over integer numbers of AC cycles). This is proper average power.

If the current or voltage has harmonics then you need to sample sufficiently high enough to be able to ignore aliasing effects.

EDIT showing voltage and current waveforms multiplied together. V is blue, I is red and magenta is power: -

enter image description here

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  • \$\begingroup\$ "Take high-speed simultaneous samples of voltage and current waveforms. Multiply i and v samples. Average over a period to suit you (best and most accurate if taken over integer numbers of AC cycles). This is proper average power" If I proceeds this procedure, would the result be zero? I thought that the AC current and the AC voltage has wave form whose means are zero. Thus, if we add the voltage*current over the time period, the + power and - power are canceled out. I think I lack of basic knowledge on this. Would you please elaborate more why we need to just take average instead of RMS? \$\endgroup\$
    – Jinkyoo Park
    Commented Feb 18, 2015 at 18:20
  • \$\begingroup\$ The average of two multiplied sinwaves is definitely NOT zero except when one is sin and the other is cosine. \$\endgroup\$
    – Andy aka
    Commented Feb 19, 2015 at 8:49
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There's another thing going on here that I don't know if your experiments have accounted for. The time constant of a wind turbine to level out at a new power is relatively long compared to the electrical transient caused by changing load conditions. In any other engine driven application, a governor would ramp to the perfect throttle to maintain a fairly exact speed, but a wind turbine cannot do this.

And i could be wrong, you could have accounted for this. In which case, if you have variable pitch blades, you need to check and see if they're hunting - that is, if they're turning back and forth trying to find a steady state. If using a fixed blade setup, it's possible that something else is not what you think it is. But i suspect you may have just had firsthand experience with the inherent variability of wind.

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  • \$\begingroup\$ Thanks for the insightful comments. Before averaging power, I wait about two minutes to stabilize the power. Only after the power output reaches "seemingly" steady state (i.e., without rising or decreasing ramp), I starts to compute the average power. Flapping blades would be one issues to although I try to fix and hold them during each experiment. \$\endgroup\$
    – Jinkyoo Park
    Commented Feb 18, 2015 at 18:27

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