--- title: "Per unit circuit of three-generator system, 30 MVA base" subject: "Power System Analysis I" chapter: 3 key: "semester_4/power_system_analysis_i:pu-circuit-30mva-g1-g2-g3" source: https://www.bctengineer.com/q/semester_4/power_system_analysis_i/pu-circuit-30mva-g1-g2-g3 site: BCT Engineering --- # Per unit circuit of three-generator system, 30 MVA base Power System Analysis I · Chapter 3: Computational Technique - Asked in: 2072 Magh, 6 marks - Video: [PER UNIT REACTANCE - HOW TO CHANGE BASE VALUE /KTU/ POWER SYSTEM ANALYSIS](https://www.youtube.com/watch?v=Q4rL15xWPGs) — praveen raj - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_4/power_system_analysis_i/pu-circuit-30mva-g1-g2-g3 - Whole chapter: https://www.bctengineer.com/md/semester_4/power_system_analysis_i/chapter-3 ## Question Figure below shows a single line diagram of a power system. The ratings of the generators and transformers are given below: G1: 25 MVA, 6.6 kV, XG1 = 0.20 pu; G2: 15 MVA, 6.6 kV, XG2 = 0.15 pu; G3: 30 MVA, 13.2 kV, XG3 = 0.15 pu; T1: 30 MVA, 6.6Δ–115Y kV, XT1 = 0.10 pu; T2: 15 MVA, 6.6Δ–115Y kV, XT2 = 0.10 pu; T3: single phase units each rated 10 MVA, 6.9/69 kV, XT3 = 0.10 pu. Draw the per unit circuit diagram using base values of 30 MVA and 6.6 kV in the circuit of generator 1. [Figure: G1 (25 MVA, 6.6 kV, Y grounded) → T1 (Δ–Y grounded, 6.6/115 kV) → Line-1 j120 Ω → junction bus; from the junction T2 (Δ–Y grounded, 6.6/115 kV) up to G2 (15 MVA, 6.6 kV, Y grounded); and Line-2 j90 Ω → T3 (Y–Y grounded, √3×69/√3×6.9 kV) → G3 (30 MVA, 13.2 kV, Y grounded).] ## Short answer On 30 MVA base, XG1=0.24, XG2=0.30, XG3=0.1976, XT1=0.10, XT2=0.20, XT3=0.108, XL1=0.2722, XL2=0.2042 pu. Base voltages are 6.6, 115, 6.6, and 11.5 kV. Connect these reactances in series branches with grounded generator buses on the per-unit circuit. ## Answer Base: 30 MVA, 6.6 kV in the G1 circuit. Base voltages change by the line-to-line turns ratio of each transformer. #### Base voltages - G1 zone: 6.6 kV - Line zone (T1 6.6/115 kV): 115 kV, $Z_b = \frac{115^2}{30} = 440.83\ \Omega$ - G2 zone (T2 115/6.6 kV): 6.6 kV - G3 zone: T3 is a Y–Y bank of 6.9/69 kV single-phase units, so its line-to-line ratio is $\sqrt3\times69/\sqrt3\times6.9 = 119.5/11.95$ kV, i.e. 10 : 1: $$ V_{b,G3} = 115\times\frac{6.9}{69} = 11.5\ \text{kV} $$ #### Per-unit values (30 MVA base) $$ \begin{aligned} X_{G1} &= 0.20\times\frac{30}{25} = 0.24\ \text{pu}\\ X_{G2} &= 0.15\times\frac{30}{15} = 0.30\ \text{pu}\\ X_{G3} &= 0.15\times\frac{30}{30}\times\left(\frac{13.2}{11.5}\right)^2 = 0.1976\ \text{pu}\\ X_{T1} &= 0.10\ \text{pu}\\ X_{T2} &= 0.10\times\frac{30}{15} = 0.20\ \text{pu}\\ X_{T3} &= 0.10\times\frac{30}{3\times10}\times\left(\frac{11.95}{11.5}\right)^2 = 0.1080\ \text{pu}\\ X_{L1} &= \frac{120}{440.83} = 0.2722\ \text{pu}\\ X_{L2} &= \frac{90}{440.83} = 0.2042\ \text{pu} \end{aligned} $$ | Element | Base kV | X (pu) | |---|---|---| | G1 | 6.6 | 0.2400 | | G2 | 6.6 | 0.3000 | | G3 | 11.5 | 0.1976 | | T1 | 6.6/115 | 0.1000 | | T2 | 6.6/115 | 0.2000 | | T3 | 115/11.5 | 0.1080 | | Line-1 | 115 | 0.2722 | | Line-2 | 115 | 0.2042 | #### Per-unit circuit ```text j0.10 j0.2722 j0.2042 j0.108 +--/\/\/---/\/\/----+----/\/\/-----/\/\/---+ | T1 Line-1 | Line-2 T3 | j0.24 j0.20 (T2) j0.1976 | | | (E1) j0.30 (G2) (E3) | | | | (E2) | +-------------------+----------------------+ reference (neutral) bus ``` **Answer:** $X_{G1}=0.24$, $X_{G2}=0.30$, $X_{G3}=0.1976$, $X_{T1}=0.10$, $X_{T2}=0.20$, $X_{T3}=0.108$, $X_{L1}=0.2722$, $X_{L2}=0.2042$ pu. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. How is the base voltage in the line zone obtained from the 6.6 kV base in the G1 circuit? A. Keep 6.6 kV everywhere in the system B. Multiply by the line-to-line turns ratio of T1: 6.6 x 115/6.6 = 115 kV C. Multiply by the generator rating ratio 30/25 D. Use the T1 MVA rating as the base voltage Answer: B. Base voltages change across each transformer by its line-to-line turns ratio, so the line zone is 115 kV. (Key point: Base voltage change) 2. What is the base impedance in the 115 kV line zone for a 30 MVA base? A. 115^2/30 = 440.83 Ω B. 115/30 = 3.83 Ω C. 30^2/115 = 7.83 Ω D. 6.6^2/30 = 1.45 Ω Answer: A. Zb = (kV_base)^2 / MVA_base. (Key point: Base impedance) 3. How is X of G1 (25 MVA, 0.20 pu) converted to the 30 MVA base? A. 0.20 x 25/30 = 0.167 pu B. 0.20 x (30/25)^2 = 0.288 pu C. It stays 0.20 pu D. 0.20 x 30/25 = 0.24 pu Answer: D. New pu = old pu x (new MVA / old MVA) when the voltage base is unchanged. (Key point: Change of base for G1) 4. Why is the G3 zone base 11.5 kV when T3 is made of 6.9/69 kV single-phase units in Y-Y? A. Because G3 is rated 13.2 kV, so the base is 13.2 kV B. Because Y-Y banks give a sqrt3 change, 115/sqrt3 C. The line-to-line ratio is 10:1 (same as 69/6.9), so 115/10 = 11.5 kV D. Because the base is always 6.6 kV on the generator side Answer: C. Y-Y connection keeps the phase ratio 10:1 as the line ratio, so the zone base is 115/10 = 11.5 kV. 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