Why do the primary values of current and voltage transformers and the final power value have to be specified when ordering power transducers?

The dimensioning of power transmitters

• In contrast to the sizing of standard transmitters for electrical or physical quantities, the sizing of power transmitters requires some additional considerations when ordering, which are outlined below
• The relationship between the input and output of a transducer can generally be defined using a simple value assignment

e.g. alternating current transducer 0 … 1A -→ 0 … 20mA

• In the case of a power transmitter, the specification of the power measuring range (W, KW, MW or VAr) is necessary but unfortunately not sufficient. In addition to the final power value, information on the rated values of the current and voltage inputs is also required. If the transducer is to be calibrated to primary values (normal case, i.e. for the power of the transformer primary side), the transformation ratios of the current and voltage transformers must also be known
  

Example: An active power measurement is to be carried out in a three-phase system with current transformers of 1500/5A and voltage transformers of 6000/100V. The following design data is available:

• Grid type: 3-wire three-phase grid (the load type does not matter here)
• Input: 5 A and 100 V
• Output: 4 … 20 mA
  

Although the power is not specified, the power transmitter could be calibrated secondary with this information, but the power that results when 5 A and 100 V are applied is only the power that should be displayed in exceptional cases.

Why?

• The equation for the active power (Pw) in three-phase systems can be 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 transformer secondary side for the special case cos phi = 1 when 100 V is applied and 5 A is taken up by the load.

To obtain the primary power, the secondary power (the power behind the transformers, 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

In practice, however, dimensioning the transducer to this final value (15.588 MW) leads to a number of disadvantages:

• The scales of downstream display or recording devices are difficult to read due to the "crooked" final value (division!)
• The measuring span is not fully utilized because the cos phi is always less than 1 in practice, or because the current transformers in the network have been (over)dimensioned for future requirements
  

However, sensible dimensioning is achieved if it is assumed for the above example that the current transformers are dimensioned to match the output and the system operates with an assumed cos phi of less than or equal to 0.9. The maximum primary power is then calculated as follows:

Pw = 15.588 MVA x 0.9 = 14.03 MW

For the reasons mentioned above, the final value of the measuring range is assumed to be an integer and set at 14 MW. (This information is normally provided by the operator based on his knowledge of the system) However, it must now be checked whether the power transducer can also cover the full scale value. There are limits that are defined by the so-called calibration or* calibration factor* ("c").

The calibration factor (calibration factor) is calculated as the quotient of the two values Ps and Pw and must (depending on the device, please refer to the data sheet) be within the limits, e.g. 0.75 …1.3:

Calibration factor = Pw / Ps

The apparent power Ps is calculated in three-phase networks as follows:

Ps = U x I x √3

The following therefore applies to the example: Calibration factor = 14 / 15.588 = 0.898

The calibration factor is therefore within the limits. This completes the dimensioning of the power transducer.

Inspection / calibration:
When checking or recalibrating power transmitters, the calibration factor must be observed! In the above example, it would therefore not be correct to apply only 100 V and 5 A in order to obtain an output signal of 20 mA (a common mistake, by the way).

Instead, one of the input values (whether the 100 V or the 5 A) must be multiplied by the calibration factor and only then connected. So e.g. 5 A * 0.898 = 4.49 A.

When 100 V and 4.49 A are applied, the measuring transducer must supply 20 mA.