The design of power transmitters
In contrast to the design of standard transmitters for electrical or physical quantities, the design of power transmitters when ordering requires some additional consideration, which will be outlined below.
The relationship between input and output of a transmitter can generally be determined using a simple value assignment
e.g. AC Transmitter 0 to 1A, 0 to 20mA.
In the case of a power transmitter, the indication of the power measuring range (W, KW, MW or VAR) is necessary but unfortunately not sufficient. In addition to the final value of the power, additional information on the rated values of the current and voltage inputs is required. If the transmitter is to be calibrated to primary values (normal, i.e. for the power of the converter primary side), the transmission ratios of the current and voltage transformers must also be known.
Example:
In a three-phase system with current transformers of 1500 / 5A and voltage transformers of 6000 / 100V, an active power measurement should be performed. The following design data are available:
Mains type: 3-wire 3-phase mains (the load type is not important here)
Input: 5 A and 100 V
Output: 4 to 20 mA
Entrance: 5 A and 100 V
Exit: 4 to 20 mA
Although the specification of the power is missing, this information would be the power transmitterCalibratable on the secondary side, but the power that results when applying 5 A and 100 V is only in exceptional cases the one that should be displayed.
Why? - From the equation for the active power (Pw) in three-phase systems is calculated using this information:
Pw = U x I x √3 100 x 5 x 1,732 = 866 W
This active power is set on the converter secondary side for the special case cos phi = 1, if 100 V are present and 5 A are picked up by the load.
In order to obtain the primary power, the secondary power (the power behind the converters, in the example = 866 W) must be multiplied by the transformer ratios:
Pprim = Psek x Ü(u) x Ü(i) = 866 W x 6000/100 x 1500/5 = 15,588 MW
However, sizing the transmitter to this final value (15.588 MW) results in a number of disadvantages: in practice
1) The scales of downstream display or recording devices are difficult to read because of the "crooked" final value (division!).
2) The span is not fully utilized because the cos phi is always less than 1 in practice, or because the current transformers in the grid have been (over-) calculated for future requirements.
Precise calculation is the case if, for the example given above, it is assumed that the current transformers are calculated according to the performance and the system works with an assumed cos phi of less than or equal to 0.9. Then the maximum primary power is calculated as:
P w = 15.588 MVA x 0.9 = 14.03 MW
The full-scale value is assumed to be an integer for the reasons mentioned and set to 14 MW. (This information is usually provided by the operator on the basis of his system knowledge.) However, it is still necessary to check whether the full-scale value can also be covered by the power transmitter. There are borders, which can be identified by the so-called calibration or calibration factor ("c ").
The calibration factor (calibration factor) is calculated as the quotient of the two values P s and P w
calibration factor = Pw / Ps
The apparent power P s is calculated in three-phase networks as follows:
Ps = U x I x √3
For the example, the following applies: calibration factor = 14 / 15,588 = 0,898
The calibration factor is therefore within the limits. This concludes the calibration of the power transmitter.
Review / Calibration:
When checking or recalibrating power transformers, the calibration factor must be observed! It would not be correct in the above example to apply only 100 V and 5 A to obtain an output signal of 20 mA (by the way, a common error).
Rather, one of the input values (regardless of whether the 100 V or the 5 A) must be multiplied by the calibration factor and only then switched. So, for example 5 A * 0.898 = 4.49 A.
When applying 100V and 4.49A, the transmitter must supply 20mA.