--- title: "Reactance diagram: 50 MVA, 13.8 kV base, three generators" subject: "Power System Analysis I" chapter: 3 key: "semester_4/power_system_analysis_i:reactance-diagram-50mva-13-8kv" source: https://www.bctengineer.com/q/semester_4/power_system_analysis_i/reactance-diagram-50mva-13-8kv site: BCT Engineering --- # Reactance diagram: 50 MVA, 13.8 kV base, three generators Power System Analysis I · Chapter 3: Computational Technique - Asked in: 2075 Baisakh, 6 marks - Video: [Problem #2 | Impedance & Reactance diagram | PSA](https://www.youtube.com/watch?v=Mq3pAkxuIjc) — FOR STUDENTS - 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-50mva-13-8kv - Whole chapter: https://www.bctengineer.com/md/semester_4/power_system_analysis_i/chapter-3 ## Question Draw the reactance diagram using a base of 50 MVA and 13.8 kV on generator G1 of a given single line diagram of a power system. The ratings of the generators and transformers are given below: G1: 20 MVA, 13.8 kV, X" = 20%; G2: 30 MVA, 18 kV, X" = 20%; G3: 30 MVA, 20 kV, X" = 20%; T1: 25 MVA, 220/13.8 kV, X = 10%; T2: 3 single phase units each rated 10 MVA, 127/18 kV, X = 10%; T3: 35 MVA, 220/22 kV, X = 10%. [Figure: G1 (Y, grounded) → T1 (Δ on generator side, Y grounded on HV side) → line Section 1, j80 Ω [?] → central bus; from the central bus T2 (Y on HV side, Δ on LV side) → G2; and line Section 2, j100 Ω → T3 (Y grounded on HV side, Δ on LV side) → G3.] ## Short answer On 50 MVA base, Vb=13.8 kV (G1), 220 kV (line, Zb=968 Ω), 18 kV (G2), 22 kV (G3). Per-unit reactances: G1 j0.5, G2 j0.3333, G3 j0.2755, T1 j0.2, T2 j0.1667, T3 j0.1429, line L1 (j80 Ω) j0.0826, L2 (j100 Ω) j0.1033. ## Answer All reactances are converted to a common base of 50 MVA; base voltages follow the transformer line-to-line ratios starting from 13.8 kV at G1. #### Base values - G1 zone: $V_b = 13.8$ kV - Line zone (T1 13.8/220 kV): $V_b = 220$ kV, $Z_b = \frac{220^2}{50} = 968\ \Omega$ - G2 zone: T2 is a Y–Δ bank of 127/18 kV units, line-to-line ratio $\sqrt{3}\times127/18 = 220/18$ kV, so $V_b = 220\times\frac{18}{220} = 18$ kV - G3 zone (T3 220/22 kV): $V_b = 22$ kV Formula: $X_{new} = X_{old}\,\frac{S_{b,new}}{S_{old}}\left(\frac{V_{old}}{V_{b,new}}\right)^2$ #### Per-unit values $$ \begin{aligned} X_{G1} &= 0.2\times\frac{50}{20} = 0.5\ \text{pu}\\ X_{G2} &= 0.2\times\frac{50}{30}\times\left(\frac{18}{18}\right)^2 = 0.3333\ \text{pu}\\ X_{G3} &= 0.2\times\frac{50}{30}\times\left(\frac{20}{22}\right)^2 = 0.2755\ \text{pu}\\ X_{T1} &= 0.1\times\frac{50}{25} = 0.2\ \text{pu}\\ X_{T2} &= 0.1\times\frac{50}{3\times10} = 0.1667\ \text{pu}\\ X_{T3} &= 0.1\times\frac{50}{35} = 0.1429\ \text{pu}\\ X_{L1} &= \frac{80}{968} = 0.0826\ \text{pu}\\ X_{L2} &= \frac{100}{968} = 0.1033\ \text{pu} \end{aligned} $$ | Element | X (pu, 50 MVA base) | |---|---| | G1 | 0.5000 | | G2 | 0.3333 | | G3 | 0.2755 | | T1 | 0.2000 | | T2 | 0.1667 | | T3 | 0.1429 | | Line section 1 (j80 Ω) | 0.0826 | | Line section 2 (j100 Ω) | 0.1033 | #### Reactance diagram ```text j0.2 j0.0826 j0.1033 j0.1429 +-/\/\/--/\/\/---+------/\/\/------/\/\/--+ | T1 L1 | L2 T3 | j0.5 j0.1667 (T2) j0.2755 | | | (E1) j0.3333 (E3) | | | | (E2) | | | | +----------------+------------------------+ reference (neutral) bus ``` **Answer:** $X_{G1}=0.5$, $X_{G2}=0.333$, $X_{G3}=0.276$, $X_{T1}=0.2$, $X_{T2}=0.167$, $X_{T3}=0.143$, $X_{L1}=0.0826$ and $X_{L2}=0.1033$ pu on 50 MVA, with base voltages 13.8, 220, 18 and 22 kV. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. T2 is three 10 MVA, 127/18 kV single-phase units in Y-Δ. What is the G2 zone base voltage? A. 18 kV B. 127 kV C. 31 kV D. 13.8 kV Answer: A. Line ratio is 220/18 kV, so Vb = 220 x 18/220 = 18 kV. (Key point: Base voltage G2 zone) 2. What MVA rating is used for T2? A. 10 MVA B. 30 MVA C. 50 MVA D. 127 MVA Answer: B. Three units of 10 MVA make 30 MVA, giving X = 0.1 x 50/30 = 0.1667. (Key point: T2 rating) 3. Why is X_G3 multiplied by (20/22)^2? A. G3 rating 20 kV differs from the 22 kV base of its zone B. T3 is Y-Δ C. The line is at 220 kV D. Because 50/30 is not 1 Answer: A. Rated 20 kV vs base 22 kV needs the voltage-squared correction: 0.2x50/30x(20/22)^2 = 0.2755. (Key point: G3 correction) 4. What is the p.u. reactance of the j80 Ω line? A. 80/968 = 0.0826 B. 80/484 = 0.165 C. 80/220 = 0.364 D. 80x968 Answer: A. Zb = 220^2/50 = 968 Ω. 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