--- title: "Salient pole generator phasor diagram in per unit" subject: "Electric Machines II" chapter: 1 key: "semester_5/electric_machines_ii:salient-generator-phasor-pu" source: https://www.bctengineer.com/q/semester_5/electric_machines_ii/salient-generator-phasor-pu site: BCT Engineering --- # Salient pole generator phasor diagram in per unit Electric Machines II · Chapter 1: Three Phase Synchronous Generator - Asked in: 2076 Asoj, 7 marks - Video: [23=Problems in voltage regulationin salient pole synchronous machine Salient pole alternator](https://www.youtube.com/watch?v=07_KVnOMJrI) — CS Boopathi - Answer checked by Claude Sonnet 5.5 on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_5/electric_machines_ii/salient-generator-phasor-pu - Whole chapter: https://www.bctengineer.com/md/semester_5/electric_machines_ii/chapter-1 ## Question 3 phase star connected, 50 Hz synchronous generator has direct axis and quadrature axis reactance of 0.6 pu and 0.45 pu. Draw the phasor diagram at full load 0.8 lagging and hence calculate (i) load angle (ii) Id and Iq (iii) open circuit voltage (iv) voltage regulation. Resistance drop at full load is 0.015 pu. ## Short answer At full load 0.8 lagging, using two-reaction theory: ψ = arctan((V sinφ+IXq)/(V cosφ+IRa)) = 52.18°, load angle δ = ψ−φ = 15.31°. Id=0.79 pu, Iq=0.613 pu. Excitation emf E = V cosδ + Iq Ra + Id Xd = 1.448 pu, voltage regulation=44.8%. ## Answer Using the **two-reaction (Blondel) theory**: $X_d$ = 0.6 pu, $X_q$ = 0.45 pu, $R_a$ = 0.015 pu, $V$ = 1 pu, $I$ = 1 pu, pf = 0.8 lagging ($\phi = 36.87^\circ$). ```text E (on q-axis) /: / : j Id (Xd - Xq) / : / .-+ E' = V + I(Ra + jXq) /.' delta O-------------- V \ phi \ I (psi = phi + delta from q-axis) ``` #### (i) Load angle $\delta$ $$ \begin{aligned} \tan\psi &= \frac{V\sin\phi + IX_q}{V\cos\phi + IR_a} = \frac{0.6 + 0.45}{0.8 + 0.015} = \frac{1.05}{0.815} = 1.2883 \\ \psi &= 52.18^\circ \\ \delta &= \psi - \phi = 52.18^\circ - 36.87^\circ = 15.31^\circ \end{aligned} $$ #### (ii) $I_d$ and $I_q$ $$ \begin{aligned} I_d &= I\sin\psi = \sin 52.18^\circ = 0.790\ \text{pu} \\ I_q &= I\cos\psi = \cos 52.18^\circ = 0.613\ \text{pu} \end{aligned} $$ #### (iii) Open-circuit voltage (excitation emf) $$ \begin{aligned} E &= V\cos\delta + I_qR_a + I_dX_d \\ &= \cos 15.31^\circ + 0.613\times0.015 + 0.790\times0.6 \\ &= 0.9645 + 0.0092 + 0.4740 = 1.448\ \text{pu} \end{aligned} $$ #### (iv) Voltage regulation $$ \%\text{Reg} = \frac{E - V}{V}\times100 = \frac{1.448 - 1}{1}\times100 = 44.8\% $$ **Answer: δ = 15.31°, I_d = 0.790 pu, I_q = 0.613 pu, E = 1.448 pu, regulation = 44.8%.** ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. The salient-pole machine is analysed using A. a single Xs B. Blondel's two-reaction theory with Xd and Xq C. only the power angle D. negative sequence Answer: B. The armature mmf resolves into d and q components. (Key point: Two-reaction) 2. Load angle is found from tanψ = (V sinφ + I Xq)/(V cosφ + I Ra), giving A. δ = ψ + φ B. δ = ψ = 52.18° C. δ = ψ - φ = 15.31° D. δ = φ = 36.87° Answer: C. ψ = 52.18° is the angle between E and I. (Key point: Load angle) 3. Id and Iq are A. I cosψ = 0.790 and I sinψ = 0.613 B. I sinδ and I cosδ C. 0.8 and 0.6 pu D. I sinψ = 0.790 and I cosψ = 0.613 pu Answer: D. The d-axis current is the component of I along the d-axis. (Key point: Id and Iq) 4. E = V cosδ + Iq Ra + Id Xd = A. 1.448 pu; regulation 44.8 % B. 1.0 pu; 0 % C. 0.6 pu; 60 % D. 1.448 pu; 14.8 % Answer: A. Regulation = (1.448 - 1)/1. 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