Q2Electrical Machines II

Question

Find the synchronous impedance, synchronous reactance and the terminal voltage when full load is thrown off, Of a $250 Amp$ , $6600 Volts$ , $0.8 p.f.$ alternator, in which a given field current produces an armature current of $250 Amp$ on short circuit and a generated e.m.f. of $1500 Volts$ on open circuit. The armature resistance is $2 Ω$ .

Answer 1

Assume the given $6600 V$ is the terminal line voltage of a 3-phase, star-connected alternator. Terminal Phase voltage $V_{p h} = 6600/ 3 = 3810.5 V$ . Given: Full load armature current $I_{a} = 250 A$ , Power factor $cos ϕ = 0.8$ lagging ( $sin ϕ = 0.6$ ). Armature resistance $R_{a} = 2 Ω$ .

For a specific field current, Open Circuit line voltage is $1500 V$ . So, Open Circuit Phase voltage $E_{oc (p h)} = 1500/ 3 = 866.02 V$ . The corresponding Short Circuit current $I_{sc} = 250 A$ .

1. Synchronous Impedance ( $Z_{s}$ ): $Z_{s} = \frac{E _{oc (p h)}}{I _{sc}} = \frac{866.02}{250} = 3.464 Ω$ .

2. Synchronous Reactance ( $X_{s}$ ): $X_{s} = Z_{s}^{2} - R_{a}^{2} = 3.46 4^{2} - 2^{2} = 12 - 4 = 8 = 2.828 Ω$ .

3. Terminal Voltage when full load is thrown off ( $E_{0}$ ): This requires finding the internal induced phase EMF, $E_{p h}$ , which becomes the terminal voltage on no-load. Using the EMF equation for a lagging power factor: $E_{p h} = (V_{p h} cos ϕ + I_{a} R_{a})^{2} + (V_{p h} sin ϕ + I_{a} X_{s})^{2}$

$V_{p h} cos ϕ = 3810.5 \times 0.8 = 3048.4 V$ $V_{p h} sin ϕ = 3810.5 \times 0.6 = 2286.3 V$ $I_{a} R_{a} = 250 \times 2 = 500 V$ $I_{a} X_{s} = 250 \times 2.828 = 707 V$

$E_{p h} = (3048.4 + 500)^{2} + (2286.3 + 707)^{2} = (3548.4)^{2} + (2993.3)^{2}$ $E_{p h} = 12591142 + 8959844 = 21550986 = 4642.3 V$ .

The line EMF when load is thrown off is $3 \times E_{p h} = 3 \times 4642.3 = 8040.7 V$ .

Q2Electrical Machines II

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