--- title: "Per unit values: 100 MVA generator, two motors" subject: "Power System Analysis I" chapter: 3 key: "semester_4/power_system_analysis_i:per-unit-100mva-6kv-two-motors" source: https://www.bctengineer.com/q/semester_4/power_system_analysis_i/per-unit-100mva-6kv-two-motors site: BCT Engineering --- # Per unit values: 100 MVA generator, two motors Power System Analysis I · Chapter 3: Computational Technique - Asked in: 2069 Poush, 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/per-unit-100mva-6kv-two-motors - Whole chapter: https://www.bctengineer.com/md/semester_4/power_system_analysis_i/chapter-3 ## Question Compute the per unit values of a power system shown in figure below taking the common base of the generator. [Figure: G (100 MVA, 6 kV, X = 6%, Y grounded) → T1 (Δ–Y grounded, 120 MVA, 6.6/132 kV, X = 4%) → transmission line (10 + j50) Ω → T2 (Y grounded on line side, Δ on motor side, 120 MVA, 6.6/132 kV; no reactance printed) → bus feeding motors M1 (50 MVA, 6 kV, X = 6%) and M2 (40 MVA, 6 kV, X = 6%).] ## Short answer On a 100 MVA, 6/120/6 kV base: generator j0.06 pu, each transformer j0.0403 pu (T2 assumed identical, 4%), line 0.0694+j0.3472 pu, motor M1 j0.12 pu, motor M2 j0.15 pu. ## Answer Common base: **100 MVA, 6 kV** in the generator circuit. The reactance of T2 is not printed; it is taken equal to T1, **X = 4 %** on 120 MVA, 6.6/132 kV (an identical transformer). #### Base voltages - Generator zone: $V_{b1} = 6$ kV - Line zone (T1 6.6/132 kV): $V_{b2} = 6\times\frac{132}{6.6} = 120$ kV, $Z_{b2} = \frac{120^2}{100} = 144\ \Omega$ - Motor zone (T2 132/6.6 kV): $V_{b3} = 120\times\frac{6.6}{132} = 6$ kV #### Per-unit values (100 MVA) $$ \begin{aligned} X_G &= 0.06\times\frac{100}{100} = 0.06\ \text{pu}\\ X_{T1} &= 0.04\times\frac{100}{120}\times\left(\frac{6.6}{6}\right)^2 = 0.0403\ \text{pu}\\ Z_{line} &= \frac{10 + j50}{144} = 0.0694 + j0.3472\ \text{pu}\\ X_{T2} &= 0.04\times\frac{100}{120}\times\left(\frac{6.6}{6}\right)^2 = 0.0403\ \text{pu}\\ X_{M1} &= 0.06\times\frac{100}{50}\times\left(\frac{6}{6}\right)^2 = 0.12\ \text{pu}\\ X_{M2} &= 0.06\times\frac{100}{40} = 0.15\ \text{pu} \end{aligned} $$ | Element | Base kV | pu value | |---|---|---| | G | 6 | j0.06 | | T1 | 6 / 120 | j0.0403 | | Line | 120 | 0.0694 + j0.3472 | | T2 (assumed 4 %) | 120 / 6 | j0.0403 | | M1 | 6 | j0.12 | | M2 | 6 | j0.15 | #### Impedance diagram ```text j0.0403 0.0694+j0.3472 j0.0403 +--/\/\/-------/\/\/--------/\/\/--+-------+ | T1 line T2 | | j0.06 j0.12 j0.15 | | | (Eg) (Em1) (Em2) | | | +----------------------------------+-------+ reference bus ``` **Answer:** $X_G = 0.06$, $X_{T1} = X_{T2} = 0.0403$, $Z_{line} = 0.0694 + j0.3472$, $X_{M1} = 0.12$, $X_{M2} = 0.15$ pu on 100 MVA (6 / 120 / 6 kV). ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. What is the base voltage in the line zone? A. 132 kV B. 6 x 132/6.6 = 120 kV C. 6 kV D. 6.6 kV Answer: B. Use the transformer ratio on the generator base. (Key point: Line zone base voltage) 2. How is the line (10 + j50) Ω converted? A. Divide by Zb = 120^2/100 = 144 Ω: 0.0694 + j0.3472 pu B. Divide by 132^2/100 C. Divide by 6^2/100 D. Multiply by 144 Answer: A. Zb = kV^2/MVA. (Key point: Line impedance) 3. Why is X_T1 = 0.0403 pu? A. 0.04 x 120/100 B. 0.04 unchanged C. 0.04 x (6/6.6)^2 D. 0.04 x 100/120 x (6.6/6)^2, correcting MVA and voltage Answer: D. Transformer rated 6.6 kV on a 6 kV base. (Key point: T1 conversion) 4. T2 reactance is missing. What is assumed? A. Zero B. Equal to the motors C. Equal to T1, 4% on 120 MVA, giving 0.0403 pu D. 10% Answer: C. It is an identical transformer. 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