System Modelling

Sections: Example | Students' Corner

Example Problems

Example 01.
A 300 MVA, 20kV, three-phase generator has a reactance of 20%. The generator supplies a number of synchronous motors over a 64-km (40-mile) transmission line having transformers at both ends.

The motors, all rated at 13.2 kV are represented by two equivalent motors. The neutral of one motor, M1, is grounded through the reactance. The neutral of the second motor, M2, is not grounded through the reactance. Rated inputs to the motors are 200 MVA and 100 MVA for M1 and M2, respectively. Both motors have X" = 20%.

The three-phase transformer, T1, is rated 350 MVA, 230/20kV with a leakage reactance of 10%. Transformer, T2, is composed of three single-phase transformers each rated 127/13.2 kV, 100 MVA with a leakage reactance of 10%.

Series reactance of the transmission line is 0.5 ohms per km.

Draw the equivalent reactnace diagram with all the reactances marked in per unit. Select the generator rating as base on the generator circuit.

Solution.
The three-phase rating of Transformer, T2, is 3 x 100 = 300 MVA
The line-to-line voltage ratio is x 127 = 220
Thus the three-phase ratings of transformer, T2 is 300 MVA, 220/13.2kV

A base of 300 MVA, 20kV in the generator requires a 300 MVA base in all parts of the system, the voltage bases in the transmission line will be: 230 kV
In the motor circuit: 230 x 13.2/220 = 13.8 kV

The reactances of the transformers converted to the proper bases
Transformer No. 1, T1:
X_T1 = 0.1 x (300/350) = 0.0857 per unit
Transformer No. 2, T2:
X_T2 = 0.1 x (13.2/13.8)² = 0.09715 per unit

The base impedance of the Transmission Line is: (230)²/300 = 176.3 ohms
The reactance of the line is: X"_TL = (0.5 x 64) / 176.3 = 0.1815 per unit

The reactance of Motor, M1, X"_M1 = 0.2 x (300/200) x (13.2/13.8)² = 0.2745 p.u.
The reactance of Motor, M2, X"_M2 = 0.2 x (300/100) x (13.2/13.8)² = 0.5490 p.u.

The equivalent reactance diagram would be:

---

Example 02.
If the motors, M1 and M2 have inputs of 120 and 60 MW, respectively at 13.2 kV and both operate at unity power factor, find the voltage at the terminals of the generators?

Solution:
Together the motors take 180 MW, or 180/300 = 0.6 p.u. or |V|·|I| = 0.6 p.u.

Also, V = 13.2/13.8 = 0.9565/0° p.u.
I = 0.6/0.9565 = 0.6273/0° p.u.

At the generator V = 0.9565 + 0.6273 (j0.0915 + j0.1815 + j0.0857)
V = 0.9565 + j0.2250 = 0.9826/13.2° p.u.

Then Eg = 0.9826 x 20 = 19.65 kV

---

Sections: Example | Students' Corner

Students' Corner

Assignment.
Solve Problems No. 6.15 and 6.16 on page 164 of Stevenson's.