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What is the impedance voltage of a dry type cast resin transformer?


What is the impedance voltage of a dry type cast resin transformer? The impedance voltage of the transformer is to short-circuit the secondary winding of the transformer,  so that the voltage of the primary winding increases slowly.  When the short-circuit current of the secondary winding reaches the rated current, the ratio of the voltage (short-circuit voltage) applied by the primary winding to the rated voltage is the percentage.

Impedance voltage Uk (%) is an important economic index related to transformer cost, efficiency and operation, and one of the main parameters for transformer state diagnosis.

In the power system, transformer plays a core and decisive role in the whole process of power transmission, a lot of switching equipment, following the setting, cable selection and so on, are carried out around the specifications of the transformer. 

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This is a common dry transformer, it can be seen that its high voltage side is a triangle connection with three lines connected from the end to the end.

Impedance voltage: refers to the low voltage of rated frequency applied at the primary side when the secondary side is directly connected. When the secondary side flows through the rated current, the pressure value of the primary side is expressed in the form of a general percentage, such as Uz%. From this definition it follows that the percentage of short-circuit voltage is equal to the percentage of short-circuit impedance, so short-circuit voltage is generally called impedance voltage.

How does the impedance voltage come out, for example, the 630kVA transformer in the figure above, how to calculate the 4%, what is the use?

The transformer secondary side short circuit, in the primary side gradually applied voltage, when the secondary winding resistance through the rated current, a winding resistance applied voltage Uz and rated voltage Un ratio percentage, namely: Uz%=Uz/Un×100%.

Like the transformer up there, 630kVA as an example, the rated current In=630÷1.732÷0.41=887.17A,

Then his impedance voltage is to short-circuit the low-voltage side, and then apply voltage on the high-voltage side. When the low-voltage side reaches 887.17A, the ratio of the voltage Uz applied by the primary winding to the rated voltage Un. Into the formula

4%=Uz/10, therefore, Uz=0.04*10=0.4kV=400V, therefore, Uz=400V, and 6% is 600V.

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