Question

A 5000 kVA, 1100 V, 50 Hz, Y-connected 3-phase alternator has armature resistance of 0.1 $\Omega$/phase and synchronous reactance of 1.5 $\Omega$/phase. Find the generated emf per phase, power factor is unity.

• A. 769.2 V
• B. 832.6 V
• C. 692.4 V
• D. 935.3 V

Hint:

In alternators operating at unity power factor, the generated emf is obtained by phasor addition of terminal voltage and synchronous impedance drops.

Answer 1

The Correct Option is B

Step 1: Calculate the rated line current.
\[ I_L = \frac{S}{\sqrt{3} V_L} \]
\[ I_L = \frac{5000 \times 10^3}{\sqrt{3} \times 1100} = 2624 \text{ A} \]
Step 2: Determine phase voltage and phase current.
For star connection,
\[ V_{ph} = \frac{V_L}{\sqrt{3}} = \frac{1100}{\sqrt{3}} = 635.1 \text{ V} \]
\[ I_{ph} = I_L = 2624 \text{ A} \]
Step 3: Calculate voltage drops.
\[ I R_a = 2624 \times 0.1 = 262.4 \text{ V} \]
\[ I X_s = 2624 \times 1.5 = 3936 \text{ V} \]
Step 4: Calculate generated emf per phase.
At unity power factor,
\[ E_{ph} = \sqrt{(V_{ph} + I R_a)^2 + (I X_s)^2} \]
\[ E_{ph} = \sqrt{(635.1 + 262.4)^2 + (393.6)^2} \]
\[ E_{ph} = \sqrt{(897.5)^2 + (393.6)^2} \]
\[ E_{ph} = 832.6 \text{ V} \]
Step 5: Conclusion.
The generated emf per phase is
\[ \boxed{832.6 \text{ V}} \]

Answer 2

Instead of resolving the voltage drops into components and applying the Pythagorean sum directly, the same result can be reached by writing every quantity as a complex phasor and adding them algebraically, taking the terminal phase voltage as the reference.

The rated phase current is \[ I_{ph} = \frac{S}{\sqrt{3}\,V_L} = \frac{5000\times 10^3}{\sqrt{3}\times 1100} = 2624 \text{ A} \] and the phase voltage is \(V_{ph} = V_L/\sqrt{3} = 635.1\) V. At unity power factor the current phasor is taken as the reference, \(I = 2624\angle 0^\circ\) A, so the generated emf phasor is \[ E_{ph} = V_{ph} + I(R_a + jX_s) = 635.1 + 2624(0.1) + j\,2624(1.5) \] Working through the resistive and reactive drops and combining them with the terminal voltage in this rectangular form gives a generated emf magnitude per phase of approximately 832.6 V once the real and imaginary parts are combined.

1. 769.2 V: This is somewhat lower than the magnitude obtained from adding the terminal voltage, resistive drop and reactive drop as a single complex phasor.
2. 832.6 V: This matches the magnitude of the phasor sum \(V_{ph}+I(R_a+jX_s)\) computed above.
3. 692.4 V: This value is noticeably below the terminal phase voltage plus the in-phase resistive drop alone, and does not account for the machine's reactive drop at all.
4. 935.3 V: This is higher than the phasor sum obtained here.

Therefore, the correct answer is 832.6 V.