--- title: "Reactance diagram: 30 MVA, 6.6 kV base, three lines" subject: "Power System Analysis I" chapter: 3 key: "semester_4/power_system_analysis_i:reactance-diagram-30mva-g1-motor" source: https://www.bctengineer.com/q/semester_4/power_system_analysis_i/reactance-diagram-30mva-g1-motor site: BCT Engineering --- # Reactance diagram: 30 MVA, 6.6 kV base, three lines Power System Analysis I · Chapter 3: Computational Technique - Asked in: 2073 Magh, 10 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-g1-motor - Whole chapter: https://www.bctengineer.com/md/semester_4/power_system_analysis_i/chapter-3 ## Question Compute the per unit values taking power base of 30 MVA and 6.6 kV in the circuit of generator G1 and draw the reactance diagram of a power system shown in figure below. The system has the following data: G1: 25 MVA, 6.6 kV, X = 0.2 p.u.; G2: 15 MVA, 6.6 kV, X = 0.15 p.u.; M: 30 MVA, 13.2 kV, X = 0.15 p.u.; T1: 30 MVA, 6.6/115 kV, X = 0.1 p.u.; T2: 15 MVA, 6.6/115 kV, X = 0.1 p.u. [Figure: G1 (Y, grounded) → T1 (Y grounded–Y grounded) → bus; Line-1 from this bus to a junction; Line-3 from the junction to a bus feeding T3 (Y grounded on line side, Δ on motor side) → motor M; Line-2 from the junction down to a bus feeding T2 (grounded windings) → G2 (Y, grounded). No line impedances or T3 rating are printed.] ## Short answer Using 30 MVA base, XG1=0.24, XG2=0.30, XM=0.15, XT1=0.10, XT2=0.20 pu; assume T3 is 30 MVA, 115/13.2 kV, X=0.10 pu. Line zone base is 115 kV, Zb=440.83 ohm, so each line reactance in pu is X(ohm)/440.83. Draw the reactance diagram with these series reactances and grounded generator references. ## Answer Base: 30 MVA and 6.6 kV in the G1 circuit. The line impedances and the T3 rating are not printed, so T3 is assumed to be **30 MVA, 115/13.2 kV, X = 0.1 pu**, and the line reactances are kept as symbols $X_1, X_2, X_3$ (Ω). #### 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 - Motor zone (T3 115/13.2 kV): 13.2 kV #### Per-unit values (30 MVA base) $$ \begin{aligned} X_{G1} &= 0.2\times\frac{30}{25} = 0.24\ \text{pu}\\ X_{G2} &= 0.15\times\frac{30}{15} = 0.30\ \text{pu}\\ X_{M} &= 0.15\times\frac{30}{30}\times\left(\frac{13.2}{13.2}\right)^2 = 0.15\ \text{pu}\\ X_{T1} &= 0.1\times\frac{30}{30} = 0.10\ \text{pu}\\ X_{T2} &= 0.1\times\frac{30}{15} = 0.20\ \text{pu}\\ X_{T3} &= 0.10\ \text{pu (assumed rating)}\\ X_{Line\,k} &= \frac{X_k}{440.83}\ \text{pu} \end{aligned} $$ For example, a line reactance of 100 Ω would be $100/440.83 = 0.2268$ pu. | Element | X (pu, 30 MVA) | |---|---| | G1 | 0.24 | | G2 | 0.30 | | Motor M | 0.15 | | T1 | 0.10 | | T2 | 0.20 | | T3 (assumed) | 0.10 | | Lines 1, 2, 3 | $X_k/440.83$ | #### Reactance diagram ```text j0.10 jXL1 jXL3 j0.10 +---/\/\/---/\/\/--+--/\/\/---/\/\/---+ | T1 Line1 | Line3 T3 | j0.24 | j0.15 | jXL2 (Line2) | (E1) | (Em) | j0.20 (T2) | | | | | j0.30 (G2) | | | | | (E2) | +------------------+------------------+ reference (neutral) bus ``` **Answer:** $X_{G1}=0.24$, $X_{G2}=0.30$, $X_M=0.15$, $X_{T1}=0.10$, $X_{T2}=0.20$, $X_{T3}=0.10$ pu, and each line $= X(\Omega)/440.83$ pu on the 30 MVA, 115 kV base. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. T3 is assumed 115/13.2 kV. What is the motor-zone base voltage? A. 6.6 kV B. 115 kV C. 13.2 kV D. 11.5 kV Answer: C. 115 x 13.2/115 = 13.2 kV. (Key point: Motor zone base) 2. What is X_G2 on the 30 MVA base? A. 0.15 B. 0.30 C. 0.20 D. 0.075 Answer: B. 0.15 x 30/15 = 0.30, no voltage correction as 6.6 kV equals base. (Key point: G2 reactance) 3. How is each line reactance converted? A. X/440.83 B. X x 440.83 C. X/115 D. X/30 Answer: A. Zb = 115^2/30 = 440.83 Ω. 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