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Question

# A 3-phase, 100 MVA star connected, salient pole synchronous motor has following specification. Excitation voltage: Ef=1.5 p.u Terminal voltage: V = 1 p.u. Direct axis synchronous reactance,Xd=1 p.u. Quadrature axis synchronous reactance, Xq=0.5 p.u. Keeping that motor does not loose synchronism, the maximum mechanical power output supplied by motor is

A
1.74 p.u.
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B
3.55 p.u.
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C
1.55 p.u.
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D
2.14 p.u.
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Solution

## The correct option is A 1.74 p.u.Salient pole motor's output power is given by P=EfVXdsinδ+V2.(Xd−Xq)2XdXqsin2δ P=1.5sinδ+(1−0.5)2×1×0.5sin2δ P=1.5sinδ+0.5sin2δ For maximum power output, dPdδ=0 dPdδ=1.5cosδ+cos2δ=0 2cos2δ−1+1.5cosδ=0 Let cosδ=x Then, cos2δ=x2 2x2+1.5x−1=0 x=−1.5±√2.25+84 x=0.4253,−1.1757 (Not possible) Then cosδ=0.4253 δ=cos−1(0.4253) δ=64.84∘ ∴Pmax=1.5sin(64.84∘)+0.5sin(129.68∘) =1.74p.u.

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