angle will be 0° (zero degrees), the calculation is simple. The formula required to
calculate Power Factor is,
COS ∠0° = 1.000 Power Factor (V1 ∠0° ,I1∠0°)
Example: The default voltage is 120.00 Volts and current is 5 Amperes AC, and the user
inputs a Power Factor of ± 0.3 as LEAD and LAG Values. The angles required for full
scale output from the transducer is,
0.3 Power Factor = COS 72.5° or,
+ 72.5° LEAD and -72.5° LAG
When the test is Started, the measured voltage and current outputs are displayed and the
calculated Power Factor is based on the measured phase angle bewteen the voltage and
current outputs. This is the value that gets displayed in the Transducer Test Screen
under the MPRT Output, next to the Label of Power Factor 1 Element.
Another value of Power Factor gets calculated using the measured dc Volts or dc
milliampere output as displayed in the Transducer Output section. Let us assume that
in our next example transducer, the output is in dc milliamperes. For this example, let us
say that ±1 milliampere of dc current is equal to the full scale Power Factor of ± 0.5.
Therefore, the theoretical range of output from the transducer would be - 0.5 Power
Factor, if the output current is -1 milliampere, to +0.5 Power Factor, if the output current
is +1 milliampere. For this example, let's say that the measured output voltage is 120.0
Volts, at 0°, and the measured output current is 5.000 Amperes, at a lagging angle of -
30°. The calculated Power Factor (displayed next to Power Factor 1 Element) would
be,
COS -30° = - 0.866
For this example, let's assume the measured output current from the transducer is -
0.489 mA dc. Based on a Lead/Lag value of ±1 mA equals ±0.5 PF, the scaling would be
equal to
0.5 PF = COS 60°
1 mA / 60° or 0.016666 mA per degree.
Therefore, the displayed PF in the Transducer Output section of the Transducer Test
Screen should read
- 0.489 mA/ 0.016666 mA/Degree = - 29.34 Degrees
COS - 29.34° = - 0.871 PF
Power factor transducer accuracies are stated in units of Power Factor, not in % error.
Therefor, the Accuarcy window for Power factor transducers needs to change from %
error to ± 0.000 PF. The Accuracy displayed in the Transducer Output section would
be equal to the following,
0.871 - 0.866 = + 0.005 PF
-
143
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