A 1000 kVA, 6.6 kV, 50 Hz, Y-connected synchronous generator has a no-load voltage of 11.4 kV at a certain field current. The generator gives rated terminal voltage at full load 0.75 lagging power factor at the same field current. Calculate: (a) The synchronous reactance (armature resistance

Question

A 1000 kVA, 6.6 kV, 50 Hz, Y-connected synchronous generator has a no-load voltage of 11.4 kV at a certain field current. The generator gives rated terminal voltage at full load 0.75 lagging power factor at the same field current. Calculate:

(a) The synchronous reactance (armature resistance being negligible);

(b) the voltage regulation;

(c) the torque angle;

(d) the electrical power developed; and (e) the voltage and kVA rating, if the generator is reconnected in delta.

Accepted Answer

[latex] I_{a}(f l)=\frac{1000}{\sqrt{3} imes 6.6 imes 1000}=0.0875 kA[/latex]

[latex] \phi=-\cos ^{-1} 0.75=-41.4^{\circ}[/latex]

[latex] \bar{V}_{t}=6.6 / \sqrt{3} \angle 0^{\circ}=3.81 \angle 0^{\circ} kV[/latex]

[latex] E_{f}=V_{o c}=11.4 / \sqrt{3}=6.58 kV[/latex]

[latex] \bar{E}_{f}=6.58 \angle \delta kV[/latex]

Refer to circuit diagram of Fig. 8.111 from which we can write

[latex] 6.58 \angle \delta^{\circ}=3.81 \angle 0^{\circ}+j X_{ s } 0.0875 \angle-41.4^{\circ}[/latex] (i)

Equating real and imaginary parts of this equation yields

[latex] 6.58 \cos \delta=3.81+0.058 X_{s}[/latex] (ii)

[latex] 6.58 \sin \delta=0.0656 X_{s} [/latex] (iii)

Eliminating [latex] X_{s}[/latex] between Eqs (ii) and (iii), we get

[latex] 6.58 \cos \delta-5.82 \sin \delta=3.81[/latex] (iv)

Or [latex] \cos \left(\delta+41.4^{\circ} ight)=0.434 \Rightarrow \delta=22.8^{\circ}[/latex]

(a) From Eq. (iii) using the value of [latex] \delta [/latex] as obtained above

[latex] X_{s}=\frac{6.58 \sin 22.8^{\circ}}{0.0656}=38.9 \Omega[/latex]

(b) Voltage regulation [latex] =\frac{11.4-6.6}{6.6} imes 100=72.7 \% [/latex]

(c) Torque (power) angle [latex] \delta=22.8^{\circ}[/latex]

(d) Electrical power developed [latex] =3 E_{f} I_{a} \cos (\delta-\phi)[/latex]

[latex] =3 imes 6.58 imes 0.0875 \cos \left(22.4^{\circ}+41.4^{\circ} ight)=763 kW[/latex]

Also directly

electrical power developed [latex] =3 V_{a} I_{a} \cos \phi=3 imes 3.81 imes 0.0875 imes 0.75=750 kW[/latex]

This checks the result as there are no losses in armature resistance [latex] \left(R_{a} \approx 0 ight)[/latex]

(e) Machine reconnected in delta

Voltage rating [latex] =6.6 / \sqrt{3}=3.81 kV[/latex]

Current rating [latex] =0.0875 \sqrt{3} kA[/latex]

[latex] kVA ext { rating }=\sqrt{3} imes 3.81 imes 0.0875 \sqrt{3} imes 1000=1000 kVA[/latex] (same as in star connection)

Question 8.39: A 1000 kVA, 6.6 kV, 50 Hz, Y-connected synchronous generator...

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