--- title: "Impedance diagram: 25 MVA generator, two motors" subject: "Power System Analysis I" chapter: 3 key: "semester_4/power_system_analysis_i:impedance-diagram-25mva-two-motors" source: https://www.bctengineer.com/q/semester_4/power_system_analysis_i/impedance-diagram-25mva-two-motors site: BCT Engineering --- # Impedance diagram: 25 MVA generator, two motors Power System Analysis I · Chapter 3: Computational Technique - Asked in: 2074 Bhadra, 8 marks - Video: [Per Unit Impedance Diagram Numerical 2 - Representation of Power System Components - Power System 1](https://www.youtube.com/watch?v=xtRmpGhwhyg) — Ekeeda - 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-25mva-two-motors - Whole chapter: https://www.bctengineer.com/md/semester_4/power_system_analysis_i/chapter-3 ## Question A one-line diagram of a three-phase power system is shown in figure below. Compute the per unit values taking generator rating as the base in the generator circuit and draw the impedance diagram of the power system shown in figure below. The system has the following data: G1: 25 MVA, 13.8 kV, X = 0.15 p.u.; T1: 30 MVA, 13.2/115 kV, X = 0.11 p.u.; T2: Three single units each rated 10 MVA, 69Y/13.2Δ kV, X = 0.11 p.u.; M1: 15 MVA, 13 kV, X = 0.15 p.u.; M2: 10 MVA, 13 kV, X = 0.15 p.u.; Series impedance of transmission line: Z = 20 + j65 Ω. [Figure: G1 → T1 (Δ on generator side, Y grounded on line side) → transmission line → T2 (Y grounded on line side, Δ on motor side) → bus feeding motors M1 (Y, grounded) and M2 (Y, grounded).] ## Short answer On 25 MVA base: X_G1=0.15, X_T1=0.0839, Z_line=0.0346+j0.1124, X_T2=0.0906, X_M1=0.2396, X_M2=0.3594 pu. Base voltages: 13.8 kV, 120.23 kV, 13.28 kV. ## Answer Base: 25 MVA and 13.8 kV in the generator circuit. T2 is read as three single-phase units of 69/13.2 kV connected Y (line side) – Δ (motor side). #### Base voltages - Generator zone: $V_{b1} = 13.8$ kV - Line zone (T1 13.2/115 kV): $V_{b2} = 13.8\times\frac{115}{13.2} = 120.23$ kV - T2 line-to-line ratio: $69\sqrt{3}/13.2 = 119.51/13.2$ kV. Motor zone: $V_{b3} = 120.23\times\frac{13.2}{119.51} = 13.28$ kV $$ Z_{b2} = \frac{120.23^2}{25} = 578.18\ \Omega $$ #### Per-unit values (25 MVA base) $$ \begin{aligned} X_{G1} &= 0.15\ \text{pu}\\ X_{T1} &= 0.11\times\frac{25}{30}\times\left(\frac{13.2}{13.8}\right)^2 = 0.0839\ \text{pu}\\ Z_{line} &= \frac{20 + j65}{578.18} = 0.0346 + j0.1124\ \text{pu}\\ X_{T2} &= 0.11\times\frac{25}{30}\times\left(\frac{13.2}{13.28}\right)^2 = 0.0906\ \text{pu}\\ X_{M1} &= 0.15\times\frac{25}{15}\times\left(\frac{13}{13.28}\right)^2 = 0.2396\ \text{pu}\\ X_{M2} &= 0.15\times\frac{25}{10}\times\left(\frac{13}{13.28}\right)^2 = 0.3594\ \text{pu} \end{aligned} $$ | Element | Base kV | pu value | |---|---|---| | G1 | 13.8 | j0.15 | | T1 | 13.8 | j0.0839 | | Line | 120.23 | 0.0346 + j0.1124 | | T2 (30 MVA bank) | 13.28 | j0.0906 | | M1 | 13.28 | j0.2396 | | M2 | 13.28 | j0.3594 | #### Impedance diagram ```text j0.0839 0.0346+j0.1124 j0.0906 +--/\/\/-------/\/\/--------/\/\/--+-------+ | T1 line T2 | | j0.15 j0.2396 j0.3594 | | | (Eg) (Em1) (Em2) | | | +----------------------------------+-------+ reference bus ``` **Answer:** $X_{G1}=0.15$, $X_{T1}=0.0839$, $Z_{line}=0.0346+j0.1124$, $X_{T2}=0.0906$, $X_{M1}=0.2396$, $X_{M2}=0.3594$ pu (25 MVA; 13.8 / 120.23 / 13.28 kV). ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. T1 is 13.2/115 kV with 13.8 kV base at G1. What is the line base voltage? A. 115 kV B. 120.23 kV C. 13.2 kV D. 132 kV Answer: B. 13.8 x 115/13.2 = 120.23 kV. (Key point: Line zone base) 2. T2 is 3 single-phase 69/13.2 kV units, Y-Δ. What is its line ratio? A. 69/13.2 kV B. 119.51/13.2 kV C. 69/22.9 kV D. 207/13.2 kV Answer: B. Y side line voltage = sqrt(3) x 69 = 119.51 kV. (Key point: T2 ratio) 3. What MVA rating is used for T2? A. 10 MVA B. 30 MVA C. 25 MVA D. 13.2 MVA Answer: B. 3 x 10 = 30 MVA. (Key point: T2 rating) 4. Why is X_M1 = 0.15 x 25/15 x (13/13.28)^2? A. Motor rated 13 kV differs from 13.28 kV zone base B. Because it is grounded C. Line is resistive D. Base MVA is 15 Answer: A. Both MVA ratio and voltage correction apply. 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