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Minimum current and emf of 10 HP motor

Short answer

At minimum current, pf = 1. Input power = 8776.5 W, so I_min = 8776.5/(√3×400) = 12.67 A. Induced emf Ef = √(V_ph² + (I Xs)²) = 263.4 V per phase (456 V line).

A 400 V, 10 HP, 3-phase synchronous motor has negligible armature resistance and synchronous reactance of 10 Ω/phase. Determine the minimum current and the corresponding induced emf for full load conditions. Assume an efficiency of 85%.

For a given load, armature current is minimum at unity power factor, because then the whole current is active current. Assumptions: star connection, 1 hp = 746 W.

Minimum current

Pout=10×746=7460 WPin=74600.85=8776.5 WImin=Pin3 VL×1=8776.53×400=12.67 A\begin{aligned} P_{out} &= 10 \times 746 = 7460\ \text{W} \\ P_{in} &= \frac{7460}{0.85} = 8776.5\ \text{W} \\ I_{min} &= \frac{P_{in}}{\sqrt3\,V_L \times 1} = \frac{8776.5}{\sqrt3 \times 400} = 12.67\ \text{A} \end{aligned}

Induced emf

At unity pf, E⃗f=V⃗−jXsI⃗a\vec E_f = \vec V - jX_s\vec I_a with IaI_a in phase with VV:

V=4003=230.94 V,IaXs=12.67×10=126.68 VEf=230.942+126.682=263.40 V per phaseδ=tan⁡−1126.68230.94=28.75∘\begin{aligned} V &= \frac{400}{\sqrt3} = 230.94\ \text{V},\quad I_aX_s = 12.67 \times 10 = 126.68\ \text{V} \\ E_f &= \sqrt{230.94^2 + 126.68^2} = 263.40\ \text{V per phase} \\ \delta &= \tan^{-1}\frac{126.68}{230.94} = 28.75^\circ \end{aligned} 
 V = 230.94 ------------------>|
   \                           | IXs = 126.68
    \ delta                    |
     '------------------------>'
          Ef = 263.40   (Ia along V)

Line value =3×263.40=456.2= \sqrt3 \times 263.40 = 456.2 V.

Answer: Minimum current =12.67= 12.67 A (at unity pf); induced emf =263.4= 263.4 V per phase =456= 456 V line.

✓ Answer checked by Claude Sonnet 5.5

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