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Back emf and new current of 5 kVA motor

Short answer

With unity pf, per-phase V=120.09 V, I=8.327 A; back emf Eb=137.33 V/phase (237.9 V line), power angle δ=29.02°. Keeping excitation constant and increasing δ by 20% to 34.82° gives I=9.85 A and pf=0.996 lagging.

A 3-phase, 5 kVA, 208 V, 4-pole, 50 Hz star connected synchronous motor has negligible armature winding resistance and synchronous reactance of 8 ohm per phase. It is operated from the 3-phase, 208 V, 50 Hz power supply and field excitation is adjusted so that the power factor is unity and the motor draws a power of 3 kW from the supply. (i) Find the back emf (or excitation) voltage and power angle. (ii) Keeping the excitation voltage constant, the power angle is increased by 20% due to increase in load on the shaft. Calculate the new armature current and power factor.

Given: star, 5 kVA, 208 V, 50 Hz, Xs=8 ΩX_s = 8\ \Omega, Ra≈0R_a \approx 0; unity pf, input 3 kW.

(i) Back emf and power angle

V=2083=120.09 VI=30003×208×1=8.327 A(in phase with V)Eˉb=Vˉ−jXsIˉ=120.09−j(8×8.327)=120.09−j66.62Eb=120.092+66.622=137.33 V per phaseδ=tan⁡−166.62120.09=29.02∘ (Eb lags V)\begin{aligned} V &= \frac{208}{\sqrt3} = 120.09\ \text{V} \\ I &= \frac{3000}{\sqrt3\times208\times1} = 8.327\ \text{A}\quad(\text{in phase with } V) \\ \bar E_b &= \bar V - jX_s\bar I = 120.09 - j(8\times8.327) = 120.09 - j66.62 \\ E_b &= \sqrt{120.09^2 + 66.62^2} = 137.33\ \text{V per phase} \\ \delta &= \tan^{-1}\frac{66.62}{120.09} = 29.02^\circ\ (E_b \text{ lags } V) \end{aligned}

Line value of back emf =3×137.33=237.9= \sqrt3\times137.33 = 237.9 V.

   O-------------------> V = 120.09
   O--------> I (upf)  |
     `-.               | jIXs = 66.62
        `-.            |
           `-. Eb = 137.33, delta = 29.02 deg

(ii) Power angle increased by 20%, excitation constant

δ′=1.2×29.02∘=34.82∘Eˉb′=137.33∠−34.82∘=112.74−j78.42Iˉ′=Vˉ−Eˉb′jXs=(120.09−112.74)+j78.42j8=7.35+j78.42j8=9.802−j0.919 AI′=9.845 A,∠−5.36∘pf=cos⁡5.36∘=0.996 lagging\begin{aligned} \delta' &= 1.2\times29.02^\circ = 34.82^\circ \\ \bar E_b' &= 137.33\angle -34.82^\circ = 112.74 - j78.42 \\ \bar I' &= \frac{\bar V - \bar E_b'}{jX_s} = \frac{(120.09 - 112.74) + j78.42}{j8} = \frac{7.35 + j78.42}{j8} \\ &= 9.802 - j0.919\ \text{A} \\ I' &= 9.845\ \text{A},\quad \angle -5.36^\circ \\ \text{pf} &= \cos 5.36^\circ = 0.996\ \text{lagging} \end{aligned}

New input power =3×120.09×9.802=3531= 3\times120.09\times9.802 = 3531 W.

Answer: (i) E_b = 137.33 V/phase (237.9 V line), δ = 29.02°; (ii) I = 9.85 A, pf = 0.996 lagging (input ≈ 3.53 kW).

✓ Answer checked by Claude Sonnet 5.5

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