--- title: "Reactance diagram: three generators, two 220 kV lines" subject: "Power System Analysis I" chapter: 3 key: "semester_4/power_system_analysis_i:reactance-diagram-three-gens-two-lines" source: https://www.bctengineer.com/q/semester_4/power_system_analysis_i/reactance-diagram-three-gens-two-lines site: BCT Engineering --- # Reactance diagram: three generators, two 220 kV lines Power System Analysis I · Chapter 3: Computational Technique - Asked in: 2076 Baisakh, 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-three-gens-two-lines - Whole chapter: https://www.bctengineer.com/md/semester_4/power_system_analysis_i/chapter-3 ## Question Draw the reactance diagram of the following figure with its equivalent per unit system. Take base MVA = 100 MVA and base voltage = 11 kV for generator. Compute the reactance diagram. [Figure: G1 → T1 and G2 → T2 feed a common 220 kV bus at the sending side; Line-1 and Line-2 run in parallel from there to a receiving bus, which feeds G3 through T3. Data: G1 100 MVA, 11 kV, X = 25%; G2 100 MVA, 11 kV, X = 20%; G3 100 MVA, 11 kV, X = 20%; T1 100 MVA, 11/220 kV, X = 6%; T2 100 MVA, 11/220 kV, X = 7%; T3 100 MVA, 220/11 kV, X = 7%; Line-1 100 MVA, 220 kV, X = 10%; Line-2 100 MVA, 220 kV, X = 10%.] ## Short answer Base 100 MVA, 11 kV at generators, 220 kV line. All components rated at base MVA and matching voltage, so per-unit reactances equal percent/100: G1 j0.25, G2 j0.20, G3 j0.20, T1 j0.06, T2 j0.07, T3 j0.07, each line j0.10 (parallel combination j0.05). ## Answer #### Base values - $S_b = 100$ MVA, and $V_b = 11$ kV in the generator circuits. - Line section (through 11/220 kV transformers): $V_b = 220$ kV, so $Z_b = 220^2/100 = 484\ \Omega$. - G3 circuit (through T3, 220/11 kV): $V_b = 11$ kV. #### Conversion to the common base $$ X_{pu,new} = X_{pu,old}\times\frac{S_{b,new}}{S_{b,old}}\times\left(\frac{V_{b,old}}{V_{b,new}}\right)^2 $$ Every component is rated at 100 MVA, and its rated voltage equals the base voltage of its section. Both correction factors are therefore 1, and each p.u. value equals its percentage reactance divided by 100. | Element | Rating | Calculation | p.u. reactance | |---|---|---|---| | G1 | 100 MVA, 11 kV, 25 % | $0.25\times1\times1$ | j0.25 | | G2 | 100 MVA, 11 kV, 20 % | $0.20\times1\times1$ | j0.20 | | G3 | 100 MVA, 11 kV, 20 % | $0.20\times1\times1$ | j0.20 | | T1 | 100 MVA, 11/220 kV, 6 % | $0.06\times1\times1$ | j0.06 | | T2 | 100 MVA, 11/220 kV, 7 % | $0.07\times1\times1$ | j0.07 | | T3 | 100 MVA, 220/11 kV, 7 % | $0.07\times1\times1$ | j0.07 | | Line 1 | 100 MVA, 220 kV, 10 % | $0.10\times1\times1$ | j0.10 | | Line 2 | 100 MVA, 220 kV, 10 % | $0.10\times1\times1$ | j0.10 | In ohms, each line is $0.10\times484 = 48.4\ \Omega$. #### Reactance diagram ```text j0.25 j0.06 j0.10 (L1) +-/\/\--/\/\--+------+--/\/\--+------+ j0.07 j0.20 | | | | +--/\/\--/\/\--+ (E1) | +--/\/\--+ | | | j0.10 (L2) (E3) | j0.20 j0.07| | +-/\/\--/\/\--+ | | | (E2) | +----------------------- reference -----------------+ ``` #### Simplified values (useful for fault studies) - The two lines in parallel: $j0.10\,\|\,j0.10 = j0.05$. - G1–T1 branch: $j0.31$. G2–T2 branch: $j0.27$. - These two branches in parallel: $\dfrac{0.31\times0.27}{0.58} = j0.1443$. - G3–T3 branch: $j0.27$. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Why is each p.u. value simply the percentage divided by 100? A. All reactances are already on 100 MVA and the rated voltage equals the section base voltage B. Percent and p.u. are always equal C. The lines have no reactance D. Transformers are ignored Answer: A. Both S_new/S_old and (V_old/V_new)^2 factors equal 1. (Key point: Correction factors) 2. What are the base voltage and Zb for the line section? A. 11 kV; 1.21 Ω B. 220 kV; 484 Ω C. 220 kV; 48.4 Ω D. 110 kV; 121 Ω Answer: B. Zb = 220^2/100 = 484 Ω. (Key point: Line zone base) 3. What is the combined reactance of two parallel j0.10 lines? A. j0.20 B. j0.10 C. j0.05 D. j0.025 Answer: C. 0.10 || 0.10 = 0.05 p.u. (Key point: Parallel lines) 4. The G1-T1 branch is j0.31 and G2-T2 is j0.27. What is their parallel value? A. j0.58 B. j0.1443 C. j0.04 D. j0.29 Answer: B. 0.31x0.27/0.58 = 0.1443. 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