How is the load current found for 1.28 MW at 0.8 pf lagging, 13.5 kV?

I = 1.28e6/(sqrt3 x 13500 x 0.8) = 68.43 A

I = 1.28e6/(13500 x 0.8) = 118.5 A

I = 1.6e6/(sqrt3 x 13500) = 68.43 A

I = 1.28e6/(3 x 13500) = 31.6 A

Comparing 0.8 lagging and unity pf, what happens to regulation?

It is 18.60 % at 0.8 lagging and 3.23 % at unity

It is 3.23 % at 0.8 lagging and 18.60 % at unity

2076 Baisakh · 8 marks

1600 kVA alternator: emf, regulation, power angle

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.

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.

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 $V$ ).

✓ Answer checked by Claude Opus 5.5 (high effort)

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1600 kVA alternator: emf, regulation, power angle

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