--- title: "Reactance diagram: three generators, 30 MVA 6.6 kV base" subject: "Power System Analysis I" chapter: 3 key: "semester_4/power_system_analysis_i:reactance-diagram-30mva-three-generators" source: https://www.bctengineer.com/q/semester_4/power_system_analysis_i/reactance-diagram-30mva-three-generators site: BCT Engineering --- # Reactance diagram: three generators, 30 MVA 6.6 kV base Power System Analysis I · Chapter 3: Computational Technique - Asked in: 2077 Chaitra, 8 marks - Video: [Impedance and Reactance diagram Problem-3 | Power System Analysis | PSA | per unit Reactance diagram](https://www.youtube.com/watch?v=TQ8W4-l9S2I) — Simple n Easy Learning - 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/reactance-diagram-30mva-three-generators - Whole chapter: https://www.bctengineer.com/md/semester_4/power_system_analysis_i/chapter-3 ## Question Draw the reactance diagram with all values in p.u. on a base of 30 MVA, 6.6 kV in the circuit of generator G1. Required data are as follows: G1: 25 MVA, 6.6 kV, j0.2 p.u.; G2: 15 MVA, 6.6 kV, j0.15 p.u.; G3: 30 MVA, 13.2 kV, j0.15 p.u.; T1: 30 MVA, 6.6 (delta) kV/115 (star) kV, j0.1 p.u.; T2: 15 MVA, 6.6 (delta) kV/115 (star) kV, j0.1 p.u.; T3: single phase units each rated 10 MVA, 69/6.9 kV, j0.1 p.u. [Figure: G1 (Y, grounded) → T1 (Δ–Y grounded) → line j120 Ω → junction bus; from the junction, T2 (Y grounded on HV side, Δ on LV side) up to G2 (Y, grounded); and a line j90 Ω → T3 (Y–Y, grounded) → G3 (Y, grounded).] ## Short answer On 30 MVA base, Vb=6.6 kV at G1, 115 kV in line, 11.5 kV at G3. Reactances: G1 j0.24, T1 j0.10, line (j120 Ω) j0.272, T2 j0.20, G2 j0.30, line (j90 Ω) j0.204, T3 j0.108, G3 j0.1976. ## Answer #### Base values - $S_b = 30$ MVA, and $V_b = 6.6$ kV in the G1 circuit. - Line section (through T1, 6.6/115 kV): $V_b = 115$ kV, so $Z_b = 115^2/30 = 440.83\ \Omega$. - G2 circuit (through T2, 115/6.6 kV): $V_b = 6.6$ kV. - T3 is a Y–Y bank of single-phase 69/6.9 kV units. Its line ratio is $\sqrt3\times69 = 119.5$ kV / $\sqrt3\times6.9 = 11.95$ kV, and its rating is $3\times10 = 30$ MVA. - G3 circuit: $V_b = 115\times\dfrac{11.95}{119.5} = 11.5$ kV. #### Per-unit reactances $$ \begin{aligned} G_1&: 0.2\times\frac{30}{25} = 0.24\\ G_2&: 0.15\times\frac{30}{15} = 0.30\\ G_3&: 0.15\times\frac{30}{30}\left(\frac{13.2}{11.5}\right)^2 = 0.1976\\ T_1&: 0.1\times\frac{30}{30} = 0.10\\ T_2&: 0.1\times\frac{30}{15} = 0.20\\ T_3&: 0.1\times\frac{30}{30}\left(\frac{119.5}{115}\right)^2 = 0.1080\\ \text{Line } j120&: \frac{120}{440.83} = 0.2722\\ \text{Line } j90&: \frac{90}{440.83} = 0.2042 \end{aligned} $$ | Element | p.u. (30 MVA base) | |---|---| | G1 | j0.24 | | T1 | j0.10 | | Line (j120 Ω) | j0.2722 | | T2 | j0.20 | | G2 | j0.30 | | Line (j90 Ω) | j0.2042 | | T3 | j0.108 | | G3 | j0.1976 | #### Reactance diagram ```text j0.24 j0.10 j0.2722 j0.2042 j0.108 j0.1976 +-/\/\--/\/\----/\/\----+(J)+---/\/\----/\/\----/\/\--+ | | | (E1) j0.20 (T2) (E3) | | | | j0.30 (G2) | | | | | (E2) | +------------------- reference -----------------------+ ``` Here J is the junction bus. Winding connections and earthing do not affect these positive-sequence reactances. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. With 6.6 kV base at G1 and T1 6.6/115 kV, what is the base voltage of the line section, and Zb at 30 MVA? A. 6.6 kV; 1.45 Ω B. 115 kV; 440.83 Ω C. 115 kV; 13225 Ω D. 69 kV; 158.7 Ω Answer: B. Vb = 115 kV and Zb = 115^2/30 = 440.83 Ω. (Key point: Base voltage in line zone) 2. T3 consists of single-phase 69/6.9 kV, 10 MVA units in Y-Y. What are its three-phase rating and line ratio? A. 10 MVA, 69/6.9 kV B. 30 MVA, 119.5/11.95 kV C. 30 MVA, 69/6.9 kV D. 10 MVA, 119.5/11.95 kV Answer: B. Rating is 3x10 = 30 MVA and the line ratio is sqrt(3)x69 / sqrt(3)x6.9 = 119.5/11.95 kV. (Key point: Y-Y bank of single-phase units) 3. What is the base voltage in the G3 circuit? A. 13.2 kV B. 6.6 kV C. 11.5 kV D. 6.9 kV Answer: C. Vb = 115 x 11.95/119.5 = 11.5 kV, so G3 (13.2 kV rated) needs (13.2/11.5)^2 correction. (Key point: Base of G3 circuit) 4. Do the Δ/Y windings and grounding change the positive-sequence reactance diagram? A. Yes, they add zero-sequence impedances B. Yes, Δ doubles the reactance C. No, they do not affect these positive-sequence reactances D. Only the grounding changes it Answer: C. Connections and earthing only matter for zero-sequence; the diagram uses positive-sequence reactances. 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