As opposed to designing standard measuring transducers for electrical or physical quantities, the dimensioning of power measuring transformers necessitates several additional considerations during order placement which are outlined below.
The relationship between the input and the output of a measuring transformer can generally be established by simply assigning the respective values, for example an AC measuring transducer: 0...1 A --> 4...20 mA.
In the case of a power measuring transducer, specification of the power measuring range (W, KW, MW or VAr) is necessary, although it is not enough. In addition to the upper power range value, rated values must also be specified for current and voltage inputs. If the measuring transducer is to be calibrated to primary values (power at the primary side of the transformer), the transformation ratios of the current and voltage transformers must be known as well.
Active power is to be measured in a 3-phase electrical system with 1500 / 5A current transformers and 6000 / 100V voltage transformers. The following data are known:
Mains: 3-wire, 3-phase system (load type is irrelevant in this case)
Input: 5 A and 100 V
Output: 4 ... 20 mA
Although power is not specified, the power measuring transformer could be calibrated at the secondary side with these data. However, the power value which results when 5 A and 100 V are applied only corresponds to the value which should actually be displayed in exceptional cases. Why?
Active power can be calculated using the equation for active power (Pa) in 3-phase systems with the help of these data:
Pa = U x I x 3 --> 100 x 5 x 1.732 = 866 W
This active power value settles in at the secondary side of the transformer in the special case where cos phi = 1, if 100 V are applied and 5 A of the load are consumed.
In order to determine primary power, secondary power (power downstream from the transformer, 866 W in our example) must be multiplied by the transformer ratios:
Pprim = Psec x R(u) x R(i) = 866 W x 6000/100 x 1500/5 = 15.588 MW
However, in actual practice, dimensioning of the measuring transducer to this final value (15.588 MW) leads to a number of disadvantages:
1) As a result of the three decimal places, scales at downstream indicators and recorders are not readily legible (scale division!).
2) Available measuring span is not fully exploited because, in actual practice, cos phi is always less than 1, or because the current transformer in the electrical system has been (over)dimensioned in anticipation of future requirements.
However, reasonable dimensioning results if, in the case of the above example, one assumes that the current transformers have been dimensioned appropriately with reference to power, and if the system operates with an assumed cos phi of less than or equal to 0.9.
In this case, maximum primary power is calculated as follows:
Pactive = 15.588 MVA x 0.9 = 14.03 MW
For the above mentioned reasons, the upper range limit is rounded off to a whole number and set to 14 MW. (This value is usually furnished by the operator based upon his knowledge of the system.)Testing must still be performed in order to assure that the power measuring transformer is capable of accommodating the upper range limit. Limits exist which can be determined by means of the so-called calibration factor (“c”).
The calibration factor is calculated from the two values for Papparent and Pactive, and must lie within a range of, for example, 0.75 to 1.3 (depending upon device type, refer to data sheet for details):
Calibration factor = Pactive / Papparent
Papparent is calculated as follows for 3-phase systems:
Apparent power = U x I x 3
The following applies in this example: calibration factor = 14 / 15.588 = 0.898
Thus the calibration factor does not exceed the specified limits. Dimensioning of the power measuring transducer is thereby concluded.
Testing / Calibration:
It is absolutely essential to take the calibration factor into consideration when testing and recalibrating power measuring transducers! In the example included above it would thus be incorrect to apply only 100 V and 5 A in order to obtain an output signal of 20 mA (which, incidentally, is a common error).
On the contrary, one of the input values (either 100 V or 5 A) must first be multiplied by the calibration factor, and then applied.
For example: 5 A * 0.898 = 4.49 A.
When 100 V and 4.49 A are applied, the measuring transducer must generate a 20 mA signal.