A three-phase, star-connected, 10 kVA , 400 V 50 Hz alternator has armature resistance of 0.5 ohm / phase and synchronous reactance 10 ohm/phase. Determine its torque angle and voltage regulation when it supplies rated load at 0.8 pf lagging.

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Accepted Answer

[latex] ext { Here, } \quad ext { Rated power }=10 kVA ; R=0.5 \Omega ; X_{5}=10 \Omega ; \cos \phi=0.8 ext { lagging }[/latex]

[latex] ext { Rated load current, } I_{L}=\frac{10 imes 10^{3}}{\sqrt{3} imes 400}=14.4 A[/latex]

[latex] ext { Rated phase current, } I=I_{L}=14.4 A[/latex]

[latex]\bar{Z}_{S}=R+j X_{S}=0.5+j 10=10.012 \angle 87^{\circ} \Omega[/latex]

[latex] ext { Rated phase voltage, } V=\frac{V_{L}}{\sqrt{3}}=\frac{400}{\sqrt{3}}=230.9 V[/latex]

Taking phase voltage V as reference phasor,

[latex] herefore[/latex]

[latex] \bar{V}=V \angle 0^{\circ}=230.9 \angle 0^{\circ}=(230.9 \pm j 0) V[/latex]

At 0.8 lagging power factor

[latex] ext { Current, } \bar{I} =I \angle-\cos ^{-1} 0.8=14.4 \angle-36.87^{\prime \prime} A[/latex]

[latex] E_{0} =\bar{V}+\bar{I} \overline{Z_{s}}[/latex]

[latex] \left.=230.9+j 0+\left(14.4 \angle-36.87^{\circ} ight)(10.012) \angle 87^{\circ} ight) [/latex]

[latex] =230.9+144.2 \angle 50.13^{\circ}=230.9+92.4+j 110.6 [/latex]

[latex] =323.3+j 110.6=341.7 \angle 18.9^{\circ} V [/latex]

[latex] herefore E_{0} =341.7 V[/latex]

Torque angle between V and [latex] E_{0}[/latex] is [latex] \delta=18.9^{\circ}[/latex] (leading)

[latex] ext { Voltage regulation }=\frac{E_{0}-V}{V}=\frac{341.7-230.9}{230.9}=0.4798 ext { pu }[/latex]

Question 6.24: A three-phase, star-connected, 10 kVA , 400 V 50 Hz alternat...

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