From 240 deg on, all three are contributing to the overall power in a circuit. Phase 1 begins at 0 deg, phase 2 begins at 120 deg and phase 3 begins at 240 deg. It shows the output of a generator wound in a manner that produces three separate single phase curves, each separated (or, out of phase) by 120 electrical degrees. If you thought that the single phase curve was complex, look at the three phase power curve in Figure 3. When a variable frequency drive reduces frequency to control speed, it also reduces voltage proportionally and maintains a constant V/hz ratio. This also occurs here in the United States in variable speed applications. All 50 Hz values are 5/6 of the 60 Hz value. In countries that use 50 Hz power, phase voltages are applied at lower intensities to keep V/hz constant. By lowering the peak voltage, we have decreased the RMS voltage of the 50 Hz curve to 380 V (5/6 of the 60 Hz RMS voltage) and balanced the V/hz ratios of the two frequencies. It is 7.6 for the 60 Hz curve but increases to 9.2 for the 50 Hz curve.įigure 2 shows the simple fix to this discrepancy. The box in the lower right of Figure 1 shows the Volts/Hz ratio for both curves. (We will cover resistive and inductive circuits next month.) To keep motor flux and the torque it produces constant, the volts per Hz (cycle) must remain constant at all frequencies. In the case of resistive circuits (heaters, incandescent lights, etc.), AC frequency does not have a major impact, but it is especially important with inductive devices (especially motors). I mentioned last month that RMS voltage is related to the heating equivalency of AC versus DC. Therefore, it must produce twice the torque. If a four pole motor is to do the same amount of work as a two pole motor, it has to accomplish twice as much work per rotation. This relationship is analogous to that of the torque produced by two pole (3,600 rpm) and four pole (1,800 rpm) motors of the same horsepower. During a period of one second, both will produce identical total voltages in the circuit, but the total voltage per cycle (Hz) will be greater for the 50 Hz curve since the number of cycles is just 5/6 that of the 60 Hz curve. The same is true for the negative portion although it is not as apparent. If you look at the positive section of the two curves, note that the area under the blue curve is greater that the area under the red curve. The 60 Hz curve finishes one cycle in about 16.66 milliseconds while the 50 Hz curve requires 20 milliseconds. Figure 1 shows one cycle of a 50 and 60 Hz, 460 V AC sine wave with peak voltages of 650 V.
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