A single phase 150 kva transformer with a primary rating of...

A single-phase 150 kVA transformer with a primary rating of 2.5 kV supplies an inductive load. The load voltage is equal with the secondary rated voltage of 240 V. The transformer parameters are determined by open- and short-circuit tests. The test results are:

Transcript

00:01 So, now we need to calculate the impedance parameters referred to the high voltage.

00:13 We need to use the test result and the formulas for calculating the equivalent circuit parameters of the transformer.

00:23 So here are the steps.

00:24 So, let's first calculate the open circuit test in which the voltage on hv side is let's take it as v2 is given to us as 6360 volts.

00:49 The current i is given to us as 1 ampere and the power p is given to us as 10 kv.

01:01 The open circuit test is used to determine the parameters for magnetizing branch from the test result.

01:10 We can find the equivalent impedance of the magnetizing branch as impedance on hv side which we can write as zoc is equal to v square by i which would be 6360 volts by 1.

01:45 So, therefore the answer is 6360 ohms.

01:53 From the power p and the rated apparent power s of the transformer, we can calculate the core loss of the component.

02:05 Therefore, the core loss of the component which can be written as rc is equal to power divided by i square.

02:34 So therefore, it would be 2000 divided by 1 square which is equal to 2000 ohms.

02:45 So, this is the core loss of the component and therefore the magnetizing branch is given by which can be written as zoc minus rc is equal to 6360 minus 2000 which is equal to 4360 ohms.

03:19 This is the magnetizing branch impedance.

03:33 So for second short circuit, let's calculate the same things for short circuit.

03:42 So for short circuit test, the short circuit voltage on hv side which is vsc is given to us as 180 volts.

04:08 The short circuit current isc is given to us as 66 amperes...

Question

A single-phase 150 kVA transformer with a primary rating of 2.5 kV supplies an inductive load. The load voltage is equal with the secondary rated voltage of 240 V. The transformer parameters are determined by open- and short-circuit tests. The test results are:

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