Q1Power System - I
Question
Q.1. Two generators rated 10MVA, 13.2kV and 15MVA, 13.2kV are connected in parallel to a bus bar. They feed supply to 2 motors of inputs 8MVA and 12MVA respectively. The operating voltage of motors is 12.5kV. Assuming the base quantities as 50MVA, 13.8kV, draw the per unit reactance diagram. The percentage reactance for generators is 15% and that for motors is 20%.
Answer
Choosing a 50 MVA, 13.8 kV base, the per-unit reactances of G1 (0.15×50/10=0.75 p.u.), G2 (0.15×50/15=0.50 p.u.), M1 (0.20×50/8=1.25 p.u.) and M2 (0.20×50/12=0.833 p.u.) are computed by base-conversion and assembled into a per-unit reactance diagram with both generators feeding a common bus connected to both motors.
Given: G1: 10 MVA, 13.2 kV, X=15%; G2: 15 MVA, 13.2 kV, X=15%; M1: 8 MVA (input), 12.5 kV, X=20%; M2: 12 MVA (input), 12.5 kV, X=20%. Chosen base: Sbase=50 MVA, Vbase=13.8 kV (generator side).
Since the machine base voltages (13.2 kV, 12.5 kV) differ from the chosen system base voltage (13.8 kV), and assuming these are on the same voltage level (no intervening transformer per the single bus-bar arrangement shown), we convert each machine's given per-unit reactance to the new base using the standard base-conversion formula:
Generator G1: X_G1(new) = 0.15×(50/10)×(13.2/13.8)² = 0.15×5×0.9150 = 0.686 p.u. (using voltage correction) — or, if voltage correction is neglected as is common in introductory per-unit problems where machine-rated voltage is assumed equal to bus base voltage: X_G1(new) = 0.15×(50/10) = 0.75 p.u.
Generator G2: X_G2(new) = 0.15×(50/15) = 0.50 p.u. (with voltage correction: ×(13.2/13.8)² = 0.457 p.u.)
Motor M1: X_M1(new) = 0.20×(50/8)×(12.5/13.8)² = 0.20×6.25×0.8199 = 1.025 p.u. (without voltage correction: 0.20×6.25=1.25 p.u.)
Motor M2: X_M2(new) = 0.20×(50/12)×(12.5/13.8)² = 0.20×4.167×0.8199 = 0.683 p.u. (without voltage correction: 0.20×4.167=0.833 p.u.)
Per-unit reactance diagram: the two generators G1 (X=0.75 p.u., or 0.686 p.u. with voltage correction) and G2 (X=0.50 p.u., or 0.457 p.u.) are shown as EMF sources in series with their respective per-unit reactances, both connected in parallel to a common bus bar; from this common bus, two branches lead to motor M1 (X=1.25 p.u., or 1.025 p.u.) and motor M2 (X=0.833 p.u., or 0.683 p.u.), each motor represented as a reactance connected to its own internal EMF (since motors, like generators, have internal induced EMFs during operation).
In most introductory textbook treatments of this classic problem (as is standard in RTU's Power System-I syllabus), the machine-rated voltages (13.2 kV, 12.5 kV) are treated as sufficiently close to the chosen base voltage (13.8 kV) that only the MVA-ratio correction is applied and the voltage-ratio correction is neglected, giving the simpler, commonly quoted answer set: X_G1=0.75 p.u., X_G2=0.50 p.u., X_M1=1.25 p.u., X_M2=0.833 p.u.