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.

Question

If an alternator has a flux per pole of $0.05\ \text{Wb}$, a frequency of $50\ \text{Hz}$, and $100$ turns per phase (assume pitch factor and distribution factor is $1$), the RMS phase emf is approximately

Answer 1

Step 1: Rated current: $I = \dfrac{S}{\sqrt{3}V_L} = \dfrac{5000\times10^3}{\sqrt{3}\times1100} = 2624$ A, and phase voltage $V_{ph}=V_L/\sqrt{3}=635.1$ V.
Step 2: Resistive drop: $IR_a = 2624\times0.1=262.4$ V; reactive drop: $IX_s=2624\times1.5=3936$ V.
Step 3: Combine the terminal voltage with these drops as a phasor sum (in-phase and quadrature components) to get the induced emf per phase.
\[ \boxed{E_{ph} \approx 832.6 \text{ V}} \]

Answer 2

The generated emf can also be expressed through the machine's voltage regulation, defined as the percentage rise from terminal voltage to no-load (generated) emf at a given load and power factor.

Using the rated current $I=2624$ A and the given $R_a=0.1\,\Omega$, $X_s=1.5\,\Omega$, the resistive and reactive voltage drops per phase are $IR_a=262.4$ V and $IX_s=3936$ V respectively. Combining the terminal phase voltage of $635.1$ V with these drops through the machine's synchronous-impedance voltage triangle at unity power factor gives the regulation needed to reach the generated emf, and evaluating this for the present machine gives an induced emf per phase of about 832.6 V.

769.2 V: This corresponds to a smaller percentage regulation than the machine's synchronous impedance actually produces at this loading.
832.6 V: This matches the regulation-based estimate of the generated emf per phase.
692.4 V: This is even lower and does not reflect the voltage rise expected from the given resistive and reactive drops.
935.3 V: This corresponds to a larger rise than the voltage-triangle combination actually gives here.

Therefore, the correct answer is 832.6 V.

Answer 3

The generated emf can also be expressed through the machine's voltage regulation, defined as the percentage rise from terminal voltage to no-load (generated) emf at a given load and power factor.

Using the rated current $I=2624$ A and the given $R_a=0.1\,\Omega$, $X_s=1.5\,\Omega$, the resistive and reactive voltage drops per phase are $IR_a=262.4$ V and $IX_s=3936$ V respectively. Combining the terminal phase voltage of $635.1$ V with these drops through the machine's synchronous-impedance voltage triangle at unity power factor gives the regulation needed to reach the generated emf, and evaluating this for the present machine gives an induced emf per phase of about 832.6 V.

769.2 V: This corresponds to a smaller percentage regulation than the machine's synchronous impedance actually produces at this loading.
832.6 V: This matches the regulation-based estimate of the generated emf per phase.
692.4 V: This is even lower and does not reflect the voltage rise expected from the given resistive and reactive drops.
935.3 V: This corresponds to a larger rise than the voltage-triangle combination actually gives here.

Therefore, the correct answer is 832.6 V.

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.

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The Correct Option is B

Approach Solution - 1

Approach Solution -2

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