--- title: "Reactance diagram: two generators, two motors, j50 Ω line" subject: "Power System Analysis I" chapter: 3 key: "semester_4/power_system_analysis_i:reactance-diagram-two-gens-two-motors" source: https://www.bctengineer.com/q/semester_4/power_system_analysis_i/reactance-diagram-two-gens-two-motors site: BCT Engineering --- # Reactance diagram: two generators, two motors, j50 Ω line Power System Analysis I · Chapter 3: Computational Technique - Asked in: 2071 Bhadra, 6 marks - Video: [Impedance and Reactance diagram Problem-4 | Power System Analysis | PSA | per unit Reactance diagram](https://www.youtube.com/watch?v=keLDCqmhDiQ) — 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-two-gens-two-motors - Whole chapter: https://www.bctengineer.com/md/semester_4/power_system_analysis_i/chapter-3 ## Question Draw the per-unit reactance diagram for the power system shown below, choosing suitable base values. G1: 25 kV, 20 MVA, 20%; G2: 25 kV, 30 MVA, 30%; T1: 50 MVA, 33/220 kV, 15%; T2: 50 MVA, 220/11 kV; M1: 20 MVA, 11 kV, 30%; M1: 30 MVA, 11 kV, 20% (second motor, printed as M1). [Figure: G1 and G2 in parallel on a bus → T1 → line j50 Ω/phase → T2 → bus feeding motors M1 and M2.] ## Short answer On a 50 MVA base with 33 kV generator, 220 kV line, and 11 kV motor bases, per-unit reactances are: X_G1 = X_G2 = 0.287 pu, X_T1 = 0.15 pu, X_line = 0.0517 pu, X_T2 = 0.15 pu (assumed), X_M1 = 0.75 pu, X_M2 = 0.333 pu. ## Answer **Choice of base:** 50 MVA (transformer rating) and 220 kV on the line. Then the generator side base is 33 kV (T1 33/220 kV) and the motor side base is 11 kV (T2 220/11 kV). The reactance of T2 is not printed; it is assumed to be **15 %** on its own rating (same as T1). $$ Z_{b,line} = \frac{220^2}{50} = 968\ \Omega $$ #### Per-unit reactances (50 MVA base) $$ \begin{aligned} X_{G1} &= 0.20\times\frac{50}{20}\times\left(\frac{25}{33}\right)^2 = 0.2870\ \text{pu}\\ X_{G2} &= 0.30\times\frac{50}{30}\times\left(\frac{25}{33}\right)^2 = 0.2870\ \text{pu}\\ X_{T1} &= 0.15\ \text{pu}\\ X_{line} &= \frac{50}{968} = 0.0517\ \text{pu}\\ X_{T2} &= 0.15\ \text{pu (assumed)}\\ X_{M1} &= 0.30\times\frac{50}{20} = 0.75\ \text{pu}\\ X_{M2} &= 0.20\times\frac{50}{30} = 0.3333\ \text{pu} \end{aligned} $$ | Element | Base kV | X (pu) | |---|---|---| | G1 (20 MVA, 25 kV) | 33 | 0.2870 | | G2 (30 MVA, 25 kV) | 33 | 0.2870 | | T1 | 33/220 | 0.1500 | | Line (j50 Ω) | 220 | 0.0517 | | T2 (assumed 15 %) | 220/11 | 0.1500 | | M1 (20 MVA) | 11 | 0.7500 | | M2 (30 MVA) | 11 | 0.3333 | #### Reactance diagram ```text j0.15 j0.0517 j0.15 +--+---/\/\/---/\/\/-----/\/\/---+------+ | | T1 line T2 | | j0.287 j0.287 j0.75 j0.3333 | | | | (E1)(E2) (Em1) (Em2) | | | | +--+-----------------------------+------+ reference bus ``` **Answer:** On 50 MVA with 33 / 220 / 11 kV bases: $X_{G1} = X_{G2} = 0.287$, $X_{T1} = 0.15$, $X_{line} = 0.0517$, $X_{T2} = 0.15$ (assumed), $X_{M1} = 0.75$, $X_{M2} = 0.333$ pu. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Which bases are chosen? A. 50 MVA and 25 kV everywhere B. 50 MVA and 220 kV on the line, giving 33 kV (generator side) and 11 kV (motor side) C. 20 MVA and 11 kV on the line D. 30 MVA and 33 kV on the line Answer: B. Voltage bases follow the transformer ratios from the line base. (Key point: Choice of base) 2. Why is G1's X multiplied by (25/33)^2? A. Its rated voltage 25 kV differs from the 33 kV zone base, and pu reactance scales with (kV_old/kV_new)^2 B. Because the line is at 220 kV C. Because pu reactance scales with the first power of voltage D. Because of the transformer MVA Answer: A. Xnew = Xold x (MVAnew/MVAold) x (kVold/kVnew)^2. (Key point: Generator reactance conversion) 3. How is the line reactance of j50 Ω converted to pu? A. 50/220 = 0.227 pu B. 50 x 968 pu C. 50/50 = 1 pu D. 50/968 = 0.0517 pu, with Zb = 220^2/50 Answer: D. Zb on the line = 220^2/50 = 968 Ω. (Key point: Line reactance) 4. T2 reactance is not given. What is assumed? A. 0 pu B. 0.5 pu C. 15% on its own 50 MVA rating, equal to T1, so 0.15 pu on the base D. The value of G1 Answer: C. T2 has the same rating as T1, and its rating equals the base so no conversion is needed. 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