A 440 V, 50 Hz, Δ -connected synchronous generator has a direct-axis reactance of 0.12 Ω and a quadrature-axis reactance of 0.075 Ω /phase; the armature resistance being negligible. The generator is supplying 1000 A at 0.8 lagging pf. (a) Find the excitation emf neglecting saliency and assuming

Question

A 440 V, 50 Hz, [latex] \Delta[/latex]-connected synchronous generator has a direct-axis reactance of [latex] 0.12 \Omega [/latex] and a quadrature-axis reactance of [latex] 0.075 \Omega / ext { phase } [/latex] ; the armature resistance being negligible. The generator is supplying 1000 A at 0.8 lagging pf. (a) Find the excitation emf neglecting saliency and assuming [latex] X_{s} = X_{d}[/latex].

(b) Find the excitation emf accounting for saliency.

Compare and comment on the results of parts (a) and (b).

Accepted Answer

On equivalent star basis

[latex] X_{d}=0.12 / 3=0.04 \Omega, \quad X_{q}=0.075 / 3=0.025 \Omega[/latex]

(a) Saliency ignores; [latex] X_{s}=X_{d}[/latex]

[latex] \bar{E}_{f}=(440 / \sqrt{3}) \angle 0^{\circ}+j 0.04 imes 1000 \angle-36.9^{\circ}[/latex]

[latex] =279.8 \angle 6.6^{\circ}[/latex]

Or [latex] E_{f}=279.8 V[/latex] or 484.6 V (line)

(b) [latex] I_{a}=1000 A , \phi=+36.90^{\circ}[/latex] (lagging)

[latex] an \psi=\frac{V_{t} \sin \phi+I_{a} X_{q}}{V_{t} \cos \phi+I_{a} R_{a}}[/latex]

[latex] =\frac{254 imes 0.6+1000 imes 0.025}{254 imes 0.8} ; V_{t}=440 / \sqrt{3}=254 V[/latex]

= 0.873

Or [latex] \psi=41.1^{\circ}[/latex]

[latex] \delta=\psi-\phi=41.1^{\circ}-36.9^{\circ}=4.2^{\circ}[/latex]

[latex] E_{f}=V_{t} \cos \delta+I_{d} X_{d} ; \quad I_{d}=I_{d} \sin \psi=1000 \sin 4.2^{\circ}[/latex]

[latex] =254 \cos 4.2^{\circ}+73.2 imes 0.04[/latex]

= 256.3 V or 444 V (line)

Question 8.46: A 440 V, 50 Hz, Δ -connected synchronous generator has a dir...

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