--- title: "Reactance diagram: 300 MVA, 20 kV generator, two motors" subject: "Power System Analysis I" chapter: 3 key: "semester_4/power_system_analysis_i:reactance-diagram-300mva-20kv" source: https://www.bctengineer.com/q/semester_4/power_system_analysis_i/reactance-diagram-300mva-20kv site: BCT Engineering --- # Reactance diagram: 300 MVA, 20 kV generator, two motors Power System Analysis I · Chapter 3: Computational Technique - Asked in: 2076 Bhadra, 10 marks - 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-300mva-20kv - Whole chapter: https://www.bctengineer.com/md/semester_4/power_system_analysis_i/chapter-3 ## Question A 300 MVA, 20 kV three-phase generator has a sub-transient reactance of 20%. The generator supplies two synchronous motors over a 65 kilometre line having transformers at both ends, as shown on the single line diagram. The neutral of one of the motors M1 is grounded while M2 is ungrounded. Rated inputs to the motors are 200 MVA at 13.2 kV and 100 MVA at 13.2 kV for M1 and M2 respectively. For both motors X"d = 20%. The three phase transformer T1 is rated 350 MVA, 230/20 kV with leakage reactance of 10%. Transformer T2 is composed of three single phase transformers, each rated 127/13.2 kV, 100 MVA with leakage reactance of 10%. Series reactance of the transmission line is 0.8 ohm/km. Draw the reactance diagram with all the reactances marked in p.u. Select the generator rating as base in the generator circuit. [Figure: G (Y, neutral grounded through an impedance) → T1 (Δ on generator side, Y grounded on line side) → transmission line TL → T2 (Y grounded on line side, Δ on motor side) → motors M1 (Y, grounded) and M2 (Y, ungrounded).] ## Short answer Using base 300 MVA, Vb=20 kV (gen), 230 kV (line), 13.8 kV (motors). Per-unit reactances: generator j0.2, T1 j0.0857, line (52 Ω) j0.2949, T2 j0.0915, M1 j0.2745, M2 j0.5490. ## Answer In a per-unit reactance diagram every element is expressed on **one common MVA base**, and the base voltage changes from zone to zone in the ratio of the transformer line-to-line voltages. Resistances, magnetising branches and the grounding impedances are neglected (they carry no current under balanced conditions). #### Step 1: Base values in each zone - Base MVA (all zones): $S_b = 300$ MVA - Generator zone: $V_{b1} = 20$ kV (given) - Line zone (T1 is 20/230 kV): $V_{b2} = 20 \times \frac{230}{20} = 230$ kV - T2 is a Y–Δ bank of single-phase units 127/13.2 kV, so its **line-to-line** ratio is $\sqrt{3}\times127 / 13.2 = 220/13.2$ kV. Motor zone: $$ V_{b3} = 230 \times \frac{13.2}{220} = 13.8\ \text{kV} $$ $$ Z_{b,\text{line}} = \frac{V_{b2}^2}{S_b} = \frac{230^2}{300} = 176.33\ \Omega $$ #### Step 2: Change of base $$ X_{pu,new} = X_{pu,old}\times\frac{S_{b,new}}{S_{b,old}}\times\left(\frac{V_{old}}{V_{b,new}}\right)^2 $$ #### Step 3: Per-unit reactances $$ \begin{aligned} X_G &= 0.20\ \text{pu (already on its own rating)}\\ X_{T1} &= 0.10\times\frac{300}{350} = 0.0857\ \text{pu}\\ X_{line} &= \frac{0.8\times 65}{176.33} = \frac{52}{176.33} = 0.2949\ \text{pu}\\ X_{T2} &= 0.10\times\frac{300}{300}\times\left(\frac{13.2}{13.8}\right)^2 = 0.0915\ \text{pu}\\ X_{M1} &= 0.20\times\frac{300}{200}\times\left(\frac{13.2}{13.8}\right)^2 = 0.2745\ \text{pu}\\ X_{M2} &= 0.20\times\frac{300}{100}\times\left(\frac{13.2}{13.8}\right)^2 = 0.5490\ \text{pu} \end{aligned} $$ (The three-phase rating of T2 is $3\times100 = 300$ MVA.) | Element | Rating used | X (pu on 300 MVA) | |---|---|---| | Generator G | 300 MVA, 20 kV | 0.2000 | | T1 | 350 MVA, 20/230 kV | 0.0857 | | Line | 52 Ω, base 176.33 Ω | 0.2949 | | T2 | 300 MVA, 220/13.2 kV | 0.0915 | | Motor M1 | 200 MVA, 13.2 kV | 0.2745 | | Motor M2 | 100 MVA, 13.2 kV | 0.5490 | #### Step 4: Reactance diagram ```text j0.0857 j0.2949 j0.0915 +---/\/\/-----/\/\/-----/\/\/---+--------+ | T1 line T2 | | j0.2 j0.2745 j0.5490 | | | (Eg) (Em1) (Em2) | | | +-------------------------------+--------+ reference (neutral) bus ``` **Answer:** On 300 MVA base, $X_G = 0.2$, $X_{T1} = 0.0857$, $X_{line} = 0.2949$, $X_{T2} = 0.0915$, $X_{M1} = 0.2745$ and $X_{M2} = 0.5490$ pu, with base voltages 20 kV, 230 kV and 13.8 kV in the three zones. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. T1 is 20/230 kV. What is the base voltage in the line zone? A. 20 kV B. 230 kV C. 13.8 kV D. 132.8 kV Answer: B. The base voltage follows the line-to-line ratio: 20 x 230/20 = 230 kV. (Key point: Base voltage zones) 2. T2 uses three single-phase 127/13.2 kV units (Y-Δ). What is its line-to-line ratio? A. 127/13.2 kV B. sqrt(3)x127/13.2 = 220/13.2 kV C. 127/(sqrt(3)x13.2) kV D. 3x127/13.2 kV Answer: B. The Y side line voltage is sqrt(3)x127 = 220 kV, the Δ side is 13.2 kV. (Key point: Single-phase bank ratio) 3. What is the motor-zone base voltage and why is X_T2 = 0.10 x (13.2/13.8)^2? A. 13.2 kV; rated equals base so no correction B. 13.8 kV; rated 13.2 kV differs from the base so voltage correction is needed C. 13.8 kV; since the motor is delta D. 230 kV; same as line Answer: B. Vb3 = 230 x 13.2/220 = 13.8 kV; rated 13.2 kV differs, so (13.2/13.8)^2 is applied. (Key point: Motor zone base and T2 reactance) 4. Which elements are neglected in the reactance diagram? A. Generator and motor reactances B. Line and transformer reactances C. Resistances, magnetising branches and grounding impedances D. Base values Answer: C. They carry no current under balanced conditions so are neglected. 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