--- title: "Impedance diagram: three generators, j100 and j130 Ω lines" subject: "Power System Analysis I" chapter: 3 key: "semester_4/power_system_analysis_i:impedance-diagram-three-gens-j100-j130" source: https://www.bctengineer.com/q/semester_4/power_system_analysis_i/impedance-diagram-three-gens-j100-j130 site: BCT Engineering --- # Impedance diagram: three generators, j100 and j130 Ω lines Power System Analysis I · Chapter 3: Computational Technique - Asked in: 2070 Magh, 8 marks - 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/impedance-diagram-three-gens-j100-j130 - Whole chapter: https://www.bctengineer.com/md/semester_4/power_system_analysis_i/chapter-3 ## Question Compute the per unit values taking a common base and draw the impedance diagram of a power system shown in figure below. The system has the following data: Generators: G1: 25 MVA, 6.6 kV, X = 0.2 p.u.; G2: 15 MVA, 6.6 kV, X = 0.15 p.u.; G3: 30 MVA, 13.2 kV, X = 0.15 p.u. Transformers: T1: 30 MVA, 6.6/115 kV, X = 0.1 p.u.; T2: 15 MVA, 6.6/115 kV, X = 0.1 p.u.; T3: three 1-ph units of 10 MVA, 69/6.9 kV, X = 0.4 p.u. [Figure: G1 (Y, grounded) → T1 (Δ–Y grounded) → line j100 Ω → junction bus; from the junction T2 (Y grounded on bus side, Δ on generator side) up to G2; and line j130 Ω → T3 (Y–Y, grounded) → G3 (Y, grounded).] ## Short answer Using a 30 MVA base with voltage bases 6.6/115/6.6/11.5 kV, per-unit reactances are: X_G1=0.24, X_G2=0.30, X_G3=0.1976, X_T1=0.10, X_T2=0.20, X_T3=0.432, X_L1=0.2268, X_L2=0.2949 pu. ## Answer Common base: **30 MVA, 6.6 kV in the G1 circuit**. Base voltages in other zones follow the line-to-line ratios. #### 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 69/6.9 kV units, so the line-to-line ratio is 119.5/11.95 kV (10 : 1). Then $V_b = 115/10 = 11.5$ kV #### Per-unit values (30 MVA) $$ \begin{aligned} X_{G1} &= 0.2\times\frac{30}{25} = 0.24\ \text{pu}\\ X_{G2} &= 0.15\times\frac{30}{15} = 0.30\ \text{pu}\\ X_{G3} &= 0.15\times\frac{30}{30}\times\left(\frac{13.2}{11.5}\right)^2 = 0.1976\ \text{pu}\\ X_{T1} &= 0.1\times\frac{30}{30} = 0.10\ \text{pu}\\ X_{T2} &= 0.1\times\frac{30}{15} = 0.20\ \text{pu}\\ X_{T3} &= 0.4\times\frac{30}{3\times10}\times\left(\frac{11.95}{11.5}\right)^2 = 0.432\ \text{pu}\\ X_{L1} &= \frac{100}{440.83} = 0.2268\ \text{pu}\\ X_{L2} &= \frac{130}{440.83} = 0.2949\ \text{pu} \end{aligned} $$ | Element | X (pu, 30 MVA) | |---|---| | G1 | 0.2400 | | G2 | 0.3000 | | G3 | 0.1976 | | T1 | 0.1000 | | T2 | 0.2000 | | T3 | 0.4320 | | Line j100 Ω | 0.2268 | | Line j130 Ω | 0.2949 | #### Impedance diagram ```text j0.10 j0.2268 j0.2949 j0.432 +--/\/\/---/\/\/----+----/\/\/-----/\/\/---+ | T1 line 1 | line 2 T3 | j0.24 j0.20 (T2) j0.1976 | | | (E1) j0.30 (G2) (E3) | | | | (E2) | +-------------------+----------------------+ reference (neutral) bus ``` **Answer:** $X_{G1}=0.24$, $X_{G2}=0.30$, $X_{G3}=0.1976$, $X_{T1}=0.10$, $X_{T2}=0.20$, $X_{T3}=0.432$, $X_{L1}=0.2268$, $X_{L2}=0.2949$ pu (base 30 MVA; 6.6 / 115 / 6.6 / 11.5 kV). ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Which base is chosen? A. 30 MVA, 13.2 kV everywhere B. 30 MVA, 6.6 kV in the G1 circuit; other zones follow transformer ratios C. 25 MVA, 115 kV everywhere D. 10 MVA, 6.9 kV Answer: B. One MVA base and voltage bases changing at each transformer. (Key point: Common base) 2. How is the G3 zone base found, and what is it? A. 115 x 6.9/69 = 11.5 kV (line-to-line ratio 10:1) B. 13.2 kV, G3 rating C. 6.6 kV D. 115 kV Answer: A. T3 ratio is 10:1, so the base is 115/10. (Key point: G3 zone base) 3. Why is X_T2 = 0.20 pu? A. 0.1 x 15/30 B. 0.1 unchanged C. 0.1 x (30/15)^2 D. 0.1 x 30/15, converting from its 15 MVA rating to the 30 MVA base Answer: D. Only the MVA ratio is used because its voltage rating equals the base. (Key point: Transformer T2 conversion) 4. How is T3's X = 0.432 pu obtained? A. 0.4 x 30/10 B. 0.4 x 10/30 C. 0.4 x 30/(3x10) x (11.95/11.5)^2, using the three-phase rating 30 MVA and the voltage mismatch D. 0.4 unchanged Answer: C. Three 10 MVA units give 30 MVA; the rated 11.95 kV vs 11.5 kV base gives the squared correction. 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