A 2-winding single-phase motor has the main and auxiliary winding currents Im = 15 A and Ia = 7.5 A at standstill. The auxiliary winding current leads the main winding current by α = 45° elect. The two windings are in space quadrature and the effective number of turns are Nm = 80 and Na = 100.

Question

A 2-winding single-phase motor has the main and auxiliary winding currents [latex] I_{m} = 15 A[/latex] and [latex] I_{a} = 7.5 A[/latex] at standstill. The auxiliary winding current leads the main winding current by [latex] \alpha=45^{\circ}[/latex] elect. The two windings are in space quadrature and the effective number of turns are [latex] N_{m} = 80[/latex] and [latex] N_{a} = 100[/latex]. Compute the amplitudes of the forward and backward stator mmf waves. Determine the magnitude of the auxiliary current and its phase angle difference [latex] \alpha [/latex] with the main winding current if only the backward field is to be present.

Accepted Answer

The mmf produced by the main winding

[latex] \bar{F}_{m}=N_{m} \bar{I}_{m}=80 imes 15 \angle 0^{\circ}=1200 \angle 0^{\circ}[/latex]

The mmf produced by the auxiliary winding

[latex] \bar{F}_{a}=N_{a} \bar{I}_{a}=100 imes 7.5 \angle 60^{\circ}=750 \angle 60^{\circ}[/latex]

From Eq. (10.17a), [latex] \bar{F}_{f}=\frac{1}{2}\left(\bar{F}_{m}-j \bar{F}_{a} ight)[/latex] the forward field is given by

[latex] \bar{F}_{f}=\frac{1}{2}\left(\bar{F}_{m}-j \bar{F}_{a} ight)=\frac{1}{2}\left(1200 \angle 0^{\circ}-j 750 \angle 45^{\circ} ight)=334.8-j 265.2[/latex]

[latex] F_{f}=427.1 AT[/latex]

Similarly from Eq. (10.17b) [latex] \bar{F}_{b}=\frac{1}{2}\left(\bar{F}_{m}+j \bar{F}_{a} ight)[/latex] the backward field is given by

[latex] \bar{F}_{b}=\frac{1}{2}\left(\bar{F}_{m}+j \bar{F}_{a} ight)=\frac{1}{2}\left(1200 \angle 0^{\circ}+j 750 \angle 45^{\circ} ight)=865.2+j 265.2[/latex]

[latex] F_{b}=904.9 AT[/latex]

Since the forward field is to be suppressed

[latex] \bar{F}_{b}=\frac{1}{2}\left(1200 \angle 0^{\circ}+j 100 I_{a} \angle \alpha ight)=0[/latex]

Or [latex] 1200+100 I_{a} \sin \alpha+j 100 I_{a} \cos \alpha=0[/latex]

Equating real and imaginary parts to zero

[latex] 100 I_{a} \cos \alpha=0[/latex]

Or [latex] \alpha=90^{\circ}[/latex]

[latex] 1200+100 I_{a} \sin 90^{\circ}=0[/latex]

Or [latex] I_{a}=-12 A[/latex]

Note: Minus sign only signifies a particular connection of the auxiliary winding with respect to the main winding

Question 10.3: A 2-winding single-phase motor has the main and auxiliary wi...

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