In a power system different power equipment with different voltage and power levels are connected together through various step up or step down transformers. However the presence of various voltage and power levels causes problem in finding out the currents (or voltages) at different points in the network. To alleviate this problem, all the system quantities are converted into a uniform normalized platform. This is called the per unit system . In a per unit system each system variable or quantity is normalized with respect to its own base value. The units of these normalized values are per unit (abbreviated as pu) and not Volt, Ampere or Ohm. The base quantities chosen are:
- VA base ( Pbase ): This is the three-phase apparent power (Volt-Ampere) base that is common to the entire circuit.
- Voltage Base ( Vbase ): This is the line-to-line base voltage. This quantity is not uniform for the entire circuit but gets changed by the turns ratio of the transformer.
Fig. 1.20 Three balanced sources supplying two balanced load through balanced source impedances.
Fig. 1.21 Per phase equivalent circuit of the network of Fig. 1.20.
Based on the above two quantities the current and impedance bases can be defined as
 | (1.119) |
 | (1.120) |
Assume that an impedance Z is defined as Z1 per unit in a base impedance of Zbase _ old . Then we have
 | (1.121) |
The impedance now has to be represented in a new base value denoted as Z base_new . Therefore
 | (1.122) |
From (1.120) Z2 can be defined in terms of old and new values of VA base and voltage base as
Example 1.1:
Let us consider the circuit shown in Fig. 1.19 (a) which contains the equivalent circuit of a transformer. Let the transformer rating be
500 MVA, 220/22 kV with a leakage reactance of 10%.
The VA base of the transformer is 500 MVA and the voltage bases in the primary and secondary side are 200 kV and 22 kV respectively. Therefore the impedance bases of these two sides are
Ω and 
W
where the subscripts 1 and 2 refer to the primary (high tension) and secondary (low tension) sides respectively. Assume that the leakage reactance is referred to the primary side. Then for 10%, i.e., 0.1 per unit leakage reactance we have
Ω
The above reactance when referred to the secondary side is
Ω
Hence the per unit impedance in the secondary side is 0.0968/0.968 = 0.1. Therefore we see that the per unit leakage reactance is the same for both sides of the transformer and, as a consequence, the transformer can be represented by only its leakage reactance. The equivalent circuit of the transformer is then as shown in Fig. 1.22. Since this diagram only shows the reactance (or impedance) of the circuit, this is called the reactance (or impedance ) diagram .
Fig. 1.22 Per unit equivalent circuit of a transformer.
Example 1.2:
Consider the 50 Hz power system the single-line diagram of which is shown in Fig. 1.23. The system contains three generators, three transformers and three transmission lines. The system ratings are
| Generator G1 | 200 MVA, 20 kV, Xd = 15% |
| Generator G2 | 300 MVA, 18 kV, Xd = 20% |
| Generator G3 | 300 MVA, 20 kV, Xd = 20% |
| Transformer T 1 | 300 MVA, 220Y/22 kV, Xd = 10% |
| Transformer T 2 | Three single-phase units each rated 100 MVA, 130Y/25 kV, X = 10% |
| Transformer T 3 | 300 MVA, 220/22 kV, X = 10% |
The transmission line reactances are as indicated in the figure. We have to draw the reactance diagram choosing the Generator 3 circuit as the base.
Fig. 1.23 Single-line diagram of the power system of Example 1.2.
As we have chosen the circuit of Generator3 as the base, the base MVA for the circuit is 300. The high voltage side of transformer T2 is connected wye. Therefore its ratedline to line voltage is √3 X 130 = 225 kV. Since the low voltage side is connected in D , its line to line voltage is 25 kV. The base voltages are chosen as discussed below.
Since the base voltage of G3 is 20 kV, the base voltage between T3 and bus 1 will be 20 X 10 = 200 kV. Also as there is no transformer connected in bus 1, the base voltage of 200 kV must be chosen for both the lines that are connected to either side of bus 1. Then the base voltage for the circuit of G1 will also be 20 kV. Finally since the turns ratio of T2 is 9 (= 225 ÷ 25), the base voltage in the G 2 side is 200 ÷ 9 = 22.22 kV. The base voltages are also indicated in Fig. 1.23.
Once the base voltages for the various parts of the circuit are known, the per unit values for the various reactances of the circuit are calculated according to (1.123) for a base MVA of 300. These are listed below.
| Generator G1 |  |
| Generator G2 |  |
| Generator G3 |  |
| Transformer T 1 |  |
| Transformer T 2 |  |
| Transformer T 3 |  |
The base impedance of the transmission line is
Ω
Therefore the per unit values of the line impedances are
pu and 
pu
The impedance diagram is shown in Fig. 1.24.
Fig. 1.24 The impedance diagram of the system of Fig. 1.23.
Closure
This completes our discussion on the modeling of power system components. In the subsequent portion of this course we shall use these models to construct a power system and use the per unit notation and the impedance diagram to represent the system.
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