A 12 pole, 50 Hz synchronous motor has saturated Xd = 0.8 pu and Xq = 0.5 pu. It is supplying full load at rated voltage of 0.8 pf leading. Draw the phasor diagram and calculate the excitation emf in pu. Calculate its pu synchronizing power per degree elect. and pu synchronizing torque per degree

Question

A 12 pole, 50 Hz synchronous motor has saturated [latex] X_{d} = 0.8 pu[/latex] and [latex] X_{q} = 0.5 pu[/latex]. It is supplying full load at rated voltage of 0.8 pf leading. Draw the phasor diagram and calculate the excitation emf in pu. Calculate its pu synchronizing power per degree elect. and pu synchronizing torque per degree mech.

Accepted Answer

The phasor diagram is drawn in Fig. 8.92.

[latex] \bar{V}_{t}=1 \angle 0^{\circ}, \bar{I}_{a}=1 \angle 36.9^{\circ}, \cos ^{-1} 0.8=36.9^{\circ}[/latex] leading

[latex] \bar{E}_{f}^{\prime}=\bar{V}_{t}-j \bar{I}_{a} X_{q}=1-j 1 imes 0.5 \angle 36.9^{\circ}=1+0.5 \angle-53.1^{\circ}[/latex]

[latex] \bar{E}_{f}^{\prime}=1.3-j 0.4=1.36 \angle-17^{\circ}[/latex]

[latex] \bar{E}_{f}^{\prime}=1.36, \quad \delta=17^{\circ}, \quad E_{f} ext { lags } V_{t}[/latex]

[latex] \psi=\phi+\delta=36.9^{\circ}+17^{\circ}=53.9^{\circ}[/latex]

[latex] \bar{I}_{d}=I_{a} \sin \psi=1 \sin 53.9^{\circ}=0.808[/latex]

As it lead [latex] E_{f}[/latex] by angle [latex] \phi[/latex] sign of [latex]I_{a}[/latex] is negative in Eq. (8.109), therefore

[latex] E_{f}=E_{f}^{\prime}+I_{d}\left(X_{d}-X_{q} ight)=1.36+0.808(0.8-0.5)[/latex]

[latex] E_{f}=1.6 pu , \delta=17^{\circ}[/latex]

[latex] P_{ syn }=\frac{E_{f} V_{t}}{X_{d}} \cos \delta+V_{t}^{2}\left(\frac{X_{d}-X_{q}}{X_{d} X_{q}} ight) \cos 2 \delta[/latex] pu/elect.rad

In pu system multiplier 3 is not needed

Substituting values

[latex] P_{ syn }=\frac{1.6 imes 1}{0.8} \cos 17^{\circ}+1^{2}\left(\frac{0.8-0.5}{0.8 imes 0.5} ight) \cos \left(2 imes 17^{\circ} ight)[/latex]

= 1.913 + 0.622 = 2.53 pu/elect. Rad

Or [latex] P_{ syn }=\frac{\pi}{180} imes 2.53=0.044 pu[/latex] /degree elect.

[latex] heta_{m}=\left(\frac{2}{P} ight) heta_{e}=\frac{1}{6} heta_{e}[/latex]

[latex] n_{s}^{\prime}=\frac{120 f}{P} imes \frac{2 \pi}{60}=\frac{120 imes 50}{12} imes \frac{2 \pi}{60}=\frac{\pi}{16.67}[/latex] elect rad/s

[latex] T_{ syn }=\frac{16.67}{\pi} imes 6 imes 0.044=1.4 Nm[/latex] m (pu)/degree mech.

Question 8.31: A 12 pole, 50 Hz synchronous motor has saturated Xd = 0.8 pu...

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