Consider the three-phase transformer shown in Fig. 13.120 ....

Consider the three-phase transformer shown in Fig. $13.120 .$ The primary is fed by a three-phase source with line voltage of $2.4 \mathrm{kV} \mathrm{rms}$, while the secondary supplies a three-phase 120 -kW balanced load at pf of $0.8 .$ Determine: (a) the type of transformer connections, (b) the values of $I_{L S}$ and $I_{P S}$ (c) the values of $I_{L P}$ and $I_{P P}$ (d) the $\mathrm{kVA}$ rating of each phase of the transformer.

Consider the three-phase transformer shown in Fig. $13.120 .$ The primary is fed by a three-phase source with line voltage of $2.4 \mathrm{kV} \mathrm{rms}$, while the secondary supplies a three-phase 120 -kW balanced load at pf of $0.8 .$ Determine:
(a) the type of transformer connections,
(b) the values of $I_{L S}$ and $I_{P S}$
(c) the values of $I_{L P}$ and $I_{P P}$
(d) the $\mathrm{kVA}$ rating of each phase of the transformer.

Key Concepts

Transformer kVA Rating Calculation

The kVA rating of a transformer reflects its apparent power capacity, which is the product of voltage and current per phase multiplied by ?3 for a three-phase system. This rating is essential for ensuring that the transformer can handle the load without overheating or incurring losses, making the proper calculation of apparent power a key step in transformer design and analysis.

Transformer Connection Configurations

This concept covers the different ways that transformer windings in a three-phase transformer can be interconnected, such as delta (?) and wye (Y) arrangements. The connection type influences how voltages and currents are distributed between the transformer’s primary and secondary sides, thereby affecting the relationships between line and phase quantities. Understanding these configurations is crucial when analyzing performance, determining necessary ratings, and selecting connection-specific formulas.

Line-to-Phase Voltage and Current Relationships in Three-Phase Systems

In three-phase systems, the relationships between line and phase quantities differ according to the type of connection. For a wye connection, the line voltage is ?3 times the phase voltage and the line current equals the phase current; for a delta connection, the line voltage equals the phase voltage while the line current is ?3 times the phase current. These relationships are fundamental in transforming measured values from one side of the transformer to the other and in designing proper load and equipment sizing.

Balanced Load Analysis in Three-Phase Circuits

Balanced load analysis assumes that all three phases of a system have identical loads in magnitude and phase angle. This symmetry simplifies calculations, as the analysis can be performed on a per-phase basis rather than dealing with unbalanced conditions. It is critical for correctly applying line and phase relationships and ensuring that transformer ratings and current calculations accurately reflect the system's performance.

Transcript

0:00 Hello everyone.

00:01 Let us do the following question.

00:02 We have a three -phase transformer shown in figure.

00:06 The primary is fed by a three -phase source with the line voltage of 2 .4 -kilovolt rms while the scandry supplies a three -phase while a scand -rifice rating of each transformer.

00:32 Firstly, we will say that this is a delta -data connection.

00:38 This is delta -data connection.

00:41 Firstly, we will say this is a delta -data connection.

00:45 Now we will simplify this structure.

00:49 The easy phase, now we will solve second diagram, we will say that the easy way, the easy way to solve it is to consider the easy way to solve it to consider just one phase.

01:09 Consider just one face.

01:11 It can be written as one ratio n.

01:13 It can be written as one ratio and here it is 4 ratio 1 or n can be written as 1 by 4.

01:21 As value of n is v2 by v1 which lets as value of n is v2 by v1 which lets v2 by v1 which lets v2 is equal to n into v1 it will put value of n this is 0 .25 and value of v1 is equal to 2400 value of v2 is given as 600 volt that is it will write value of vlp value of vlp this is 2400 volt and value of bels this is 600 volt it will write, if we'll draw this is value vls.

02:04 Now it will draw the simplify, it will see the complex power factor that can be written as p by kose theta.

02:13 Value of this is 120 and this can be written as 0 .8 kilo volt ampere.

02:20 Value of this can be written as 150 kilo volt ampere.

02:24 Now it will write value of pl.

02:27 This is p by 3.

02:28 This we will show in the folding above diagram.

02:31 This is 120 by 3 value of this is 40 kilo watt...

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