--- title: "Impedance diagram, 100 MVA 132 kV base, with load" subject: "Power System Analysis I" chapter: 3 key: "semester_4/power_system_analysis_i:impedance-diagram-100mva-132kv-load" source: https://www.bctengineer.com/q/semester_4/power_system_analysis_i/impedance-diagram-100mva-132kv-load site: BCT Engineering --- # Impedance diagram, 100 MVA 132 kV base, with load Power System Analysis I · Chapter 3: Computational Technique - Asked in: 2079 Chaitra, 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-100mva-132kv-load - Whole chapter: https://www.bctengineer.com/md/semester_4/power_system_analysis_i/chapter-3 ## Question Prepare an impedance diagram of the system shown in the figure below and show all impedances in per unit on a 100 MVA, 132 kV base in the transmission line circuit. [Figure: G1 (50 MVA, 13.8 kV, X = 0.15 p.u.) → T1 (80 MVA, Y-Y, 12.2/161 kV, Xt = 0.1 p.u.) → HV bus; from this bus a line of 40 + j160 Ω goes to the HV side of T2, and a parallel path of two sections 20 + j80 Ω and 20 + j80 Ω also joins the two buses, with a load of 50 MVA, cos φ = 0.8 lag, V = 154 kV tapped at the junction of the two sections; T2 (40 MVA, Y-Y, 161/13.8 kV, Xt2 = 0.1 p.u.) → G2 (20 MVA, 13.8 kV, X = 0.15 p.u.).] ## Short answer Using 100 MVA base and 132 kV in the transmission line, base impedances are derived from transformer ratios. Results: G1 j0.571, T1 j0.186, main line 0.2296+j0.9183, each parallel section 0.1148+j0.4591, load 2.1778+j1.6333, T2 j0.372, G2 j1.116. ## Answer #### Base values - $S_b = 100$ MVA, and $V_b = 132$ kV in the transmission circuit, so $Z_b = 132^2/100 = 174.24\ \Omega$. - G1 side, through T1 (12.2/161 kV): $V_b = 132\times\dfrac{12.2}{161} = 10.0025$ kV. - G2 side, through T2 (161/13.8 kV): $V_b = 132\times\dfrac{13.8}{161} = 11.3143$ kV. #### Per-unit impedances Use $Z_{new} = Z_{old}\dfrac{S_b}{S_{rated}}\left(\dfrac{V_{rated}}{V_b}\right)^2$. $$ \begin{aligned} G_1&: 0.15\times\frac{100}{50}\left(\frac{13.8}{10.0025}\right)^2 = j0.5710\\ T_1&: 0.1\times\frac{100}{80}\left(\frac{161}{132}\right)^2 = j0.1860\\ T_2&: 0.1\times\frac{100}{40}\left(\frac{161}{132}\right)^2 = j0.3719\\ G_2&: 0.15\times\frac{100}{20}\left(\frac{13.8}{11.3143}\right)^2 = j1.1157 \end{aligned} $$ Lines: $$ \begin{aligned} Z_{40+j160} &= \frac{40 + j160}{174.24} = 0.2296 + j0.9183\\ Z_{20+j80} &= \frac{20 + j80}{174.24} = 0.1148 + j0.4591\ \text{(each section)} \end{aligned} $$ Load (50 MVA, 0.8 lagging, 154 kV): $V = 154/132 = 1.1667$ p.u. and $S = 0.5$ p.u. $$ Z_L = \frac{|V|^2}{S^*} = \frac{1.1667^2}{0.5}\angle36.87^\circ = 2.7222\angle36.87^\circ = 2.1778 + j1.6333\ \text{p.u.} $$ #### Impedance diagram ```text j0.571 j0.186 0.2296+j0.9183 j0.3719 j1.1157 +-/\/\----/\/\--+-------/\/\/\/\-------+--/\/\----/\/\--+ | | | | (E1) +-0.1148+j0.4591-+-0.1148+j0.4591-+ (E2) | | | | 2.1778+j1.6333 (load) | | | | +---------------- reference -----+---------------------+ ``` | Element | p.u. (100 MVA, 132 kV line base) | |---|---| | G1 | j0.5710 | | T1 | j0.1860 | | Line (40 + j160 Ω) | 0.2296 + j0.9183 | | Each section (20 + j80 Ω) | 0.1148 + j0.4591 | | Load | 2.1778 + j1.6333 | | T2 | j0.3719 | | G2 | j1.1157 | ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Base impedance in the transmission circuit is A. 132 ohm B. 174.24 ohm C. 1.74 ohm D. 100 ohm Answer: B. 132^2/100 = 174.24 ohm. (Key point: Z_b) 2. The base voltage on G1 side through T1 (12.2/161) is A. 13.8 kV B. 10.0025 kV C. 11.31 kV D. 12.2 kV Answer: B. 132 x 12.2/161 = 10.0025 kV. (Key point: G1 side base) 3. The load 50 MVA, 0.8 lag at 154 kV has p.u. voltage and power A. V = 1.1667, S = 0.5 B. V = 1, S = 0.5 C. V = 1.1667, S = 50 D. V = 154, S = 0.5 Answer: A. 154/132 = 1.1667 and 50/100 = 0.5. (Key point: Load impedance) 4. Z_L is A. 2.1778 + j1.6333 p.u. B. 1.6333 + j2.1778 C. 2.7222 p.u. resistive D. 0.5 angle 36.87 Answer: A. Z = V^2/S* = 2.7222 angle 36.87 = 2.1778 + j1.6333. 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