--- title: "1600 kVA alternator: emf, regulation, power angle" subject: "Electrical Machines" chapter: 6 key: "semester_4/electrical_machines:alt-1600kva-regulation-power-angle" source: https://www.bctengineer.com/q/semester_4/electrical_machines/alt-1600kva-regulation-power-angle site: BCT Engineering --- # 1600 kVA alternator: emf, regulation, power angle Electrical Machines · Chapter 6: Three Phase Synchronous Machines - Asked in: 2076 Baisakh, 8 marks - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_4/electrical_machines/alt-1600kva-regulation-power-angle - Whole chapter: https://www.bctengineer.com/md/semester_4/electrical_machines/chapter-6 ## Question A 3 phase star connected alternator is rated at 1600 kVA, 13.5 kV having per phase armature effective resistance and synchronous reactance of $1.5\ \Omega$ and $30\ \Omega$ respectively. Calculate the line value of emf generated, voltage regulation and power angle for a load of 1.28 MW at, i) 0.8 pf lagging ii) unity power factor. ## Short answer Using Vph=7794.2 V, Ra=1.5 Ω, Xs=30 Ω: (i) 0.8 lagging, I=68.43 A, E_line=16011 V, regulation 18.60%, δ=9.85°; (ii) unity pf, I=54.74 A, E_line=13936 V, regulation 3.23%, δ=11.78°. Calculated via phasor addition. ## Answer **Given:** 1600 kVA, 13.5 kV, star, $R_a = 1.5\ \Omega$, $X_s = 30\ \Omega$ per phase. Load $P = 1.28$ MW at rated voltage. $$ V_{ph} = \frac{13500}{\sqrt3} = 7794.2\ \text{V}, \qquad I = \frac{1.28\times10^6}{\sqrt3\times13500\times\cos\phi} $$ $$ \vec E = \vec V + \vec I(R_a + jX_s), \qquad \text{Regulation} = \frac{E - V}{V}\times100 $$ #### (i) 0.8 pf lagging $$ I = \frac{1.28\times10^6}{\sqrt3\times13500\times0.8} = 68.43\ \text{A} $$ $$ IR_a = 102.64\ \text{V}, \quad IX_s = 2052.80\ \text{V} $$ $$ E = \sqrt{(7794.2 + 102.64\times0.8 + 2052.80\times0.6)^2 + (2052.80\times0.8 - 102.64\times0.6)^2} $$ $$ E_{ph} = 9244.2\ \text{V}, \qquad E_L = 16011\ \text{V} $$ $$ \text{Regulation} = \frac{9244.2 - 7794.2}{7794.2}\times100 = 18.60\% $$ $$ \delta = \tan^{-1}\frac{1642.2-61.6}{7794.2 + 82.1 + 1231.7}\approx 9.85^\circ $$ #### (ii) Unity pf $$ I = \frac{1.28\times10^6}{\sqrt3\times13500} = 54.74\ \text{A}, \quad IR_a = 82.11\ \text{V}, \quad IX_s = 1642.24\ \text{V} $$ $$ E = \sqrt{(7794.2 + 82.11)^2 + 1642.24^2} $$ $$ E_{ph} = 8045.7\ \text{V}, \qquad E_L = 13935.6\ \text{V} $$ $$ \text{Regulation} = \frac{8045.7-7794.2}{7794.2}\times100 = 3.23\%, \qquad \delta = \tan^{-1}\frac{1642.24}{7876.3} = 11.78^\circ $$ | Case | $I$ (A) | $E_L$ (V) | Regulation | Power angle $\delta$ | |---|---|---|---|---| | 0.8 lagging | 68.43 | 16011 | 18.60 % | 9.85° | | Unity | 54.74 | 13936 | 3.23 % | 11.78° | **Answer:** as in the table (emf is line value; power angle is the angle between $E$ and $V$). ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. How is the load current found for 1.28 MW at 0.8 pf lagging, 13.5 kV? A. I = 1.28e6/(sqrt3 x 13500 x 0.8) = 68.43 A B. I = 1.28e6/(13500 x 0.8) = 118.5 A C. I = 1.6e6/(sqrt3 x 13500) = 68.43 A D. I = 1.28e6/(3 x 13500) = 31.6 A Answer: A. P = sqrt3 VL I cos(phi), so I = 1.28e6/(sqrt3 x 13500 x 0.8) = 68.43 A. (Key point: Load current) 2. How is voltage regulation of the alternator defined here? A. (V - E)/E x 100 B. (E - V)/V x 100 C. (E - V)/E x 100 D. (V - E)/V x 100 Answer: B. Regulation = (no-load emf - rated terminal voltage)/rated terminal voltage x 100. (Key point: Regulation definition) 3. Comparing 0.8 lagging and unity pf, what happens to regulation? A. It is 3.23 % at 0.8 lagging and 18.60 % at unity B. It is 18.60 % at 0.8 lagging and 3.23 % at unity C. It is the same for both D. It is negative in both Answer: B. A lagging load needs more excitation (E_L = 16011 V, 18.60 %) than unity pf (E_L = 13936 V, 3.23 %). (Key point: Effect of power factor) 4. What is the power angle delta? A. Angle between I and V B. Angle between E and I C. Angle between E and V D. Angle between Ia and Ra Answer: C. The power (load) angle is the angle between the generated emf E and terminal voltage V; here 9.85 degrees and 11.78 degrees. 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