Transformers
Consider a single-phase transformer in which the total series impedance of the two windings referred to the primary is Z_{2} (Figure 2.18).
Zj
Then the p.u. impedance Zā_{u} = , where I_{2}, and V_{2}, are the base values of the
Vj/Ij
primary circuit.
Figure 2.18 Equivalent circuit of single-phase transformer
The ohmic impedance referred to the secondary is and this in p.u. notation is
V_{2} and I_{2} are base voltage and current of the secondary circuit. If they are related to the base voltage and current of the primary by the turns ratio of the transformer then.
Hence provided the base voltages on each side of a transformer are related by the turns ratio, the p.u. impedance of a transformer is the same whether considered from the primary or the secondary side. The winding does not appear in the equivalent circuit (Figure 2.18b), the transformer impedance in per unit is only calculated once and Equation (2.7) is not required.
Example 2.4
In the network of Figure 2.19, two single-phase transformers supply a 10 kVA resistance load at 200 V. Show that the p.u. load is the same for each part of the circuit and calculate the voltage at point D.
Figure 2.19 Network with two transformers-p.u. approach
Solution
The load resistance is (200^{2}/10 x 10^{3}), that is, 4 V.
In each of the circuits A, B, and C a different voltage exists, so that each circuit will have its own base voltage, that is, 100 V in A, 400 V in B, and 200 V in C.
Although it is not essential for rated voltages to be used as bases, it is essential that the voltage bases used be related by the turns ratios of the transformers. If this is not so the simple p.u. framework breaks down. The same volt-ampere base is used for all the circuits as V_{1}I_{1} ā V_{2}I_{2} on each side of a transformer and is taken in this case as 10 kVA. The per unit impedances of the transformers are already on their individual equipment bases of 10 kVA and so remain unchanged.
The base impedance in C
The load resistance (p.u.) in C
In B the base impedance is
and the load resistance (in ohms) referred to B is Hence the p.u. load referred to B
Similarly, the p.u. load resistance referred to A is also 1 p.u. Hence, if the voltage bases are related by the turns ratios the load p.u. value is the same for all circuits.
Figure 2.20 Equivalent circuit with p.u. values, of network in Figure 2.19
An equivalent circuit may be used as shown in Figure 2.20. Let the volt-ampere base be 10kVA; the voltage across the load (V_{R}) is 1 p.u. (as the base voltage in C is 200 V).
The base current at voltage level C of this single phase circuit
The corresponding base currents in the other circuits are 25 A in B, and 100 A in A.
The actual load current is 200V/4 V = 50 A = 1 p.u.
Hence the supply voltage VS
The voltage at point D in Figure 2.17
It is a useful exercise to repeat this example using ohms, volts and amperes.
A summary table of the transformation of the circuit of Figure 2.19 into per unit is shown.
Section of network |
^{S}base (common for network) |
V_{base} (chosen as transformer turns ratio) |
Ibase (calculated ^{from S}basel^{V}base^{)} |
Z_{base} (calculated ^{from V}Le^{/}Sbase^{)} |
A |
10kVA |
100 V |
100 A |
1V |
B |
10kVA |
400 V |
25 A |
16 V |
C |
10kVA |
200 V |
50 A |
4 V |
Example 2.5
Figure 2.21 shows the schematic diagram of a radial transmission system. The ratings and reactances of the various components are shown. A load of 50 MW at 0.8 p.f. lagging is taken from the 33 kV substation which is to be maintained at 30 kV. It is required to calculate the terminal voltage of the synchronous machine. The line and transformers may be represented by series reactances. The system is three-phase.
Vs |
|||
11 kV 132 kV |
132 kV 33 kV |
||
i ? ^ |
*"50 MW |
||
eyā |
_3C_ 50 MVA |
^{]100Š} 50MVA |
0.8 p.f. -?lagging |
X = 10% |
X = 12% |
30 kV |
Figure 2.21 Line diagram of system for Example 2.5
Solution
It will be noted that the line reactance is given in ohms; this is usual practice. The voltage bases of the various circuits are decided by the nominal transformer voltages, that is, 11, 132, and 33 kV. A base of 100 MVA will be used for all circuits. The reactances (resistance is neglected) are expressed on the appropriate voltage and MVA bases.
Base impedance for the line
Hence the p.u. reactance
Per unit reactance of the sending-end transformer Per unit reactance of the receiving-end transformer
Base current for 33 kV, 100 MVA
Hence the p.u. load current
Voltage of the load busbar
The equivalent circuit is shown in Figure 2.22. Also,
Figure 2.22 Equivalent circuit for Example 2.5