A 440 V- 50 Hz- Y-connected synchronous generator has a direct axis reactance of 0-12-phase and a quadrature axis reactance of 0-075-phase and the armature resistance being negligible- The generator is supplying 1000 A at 0-8 lagging pf- The excitation after taking saliency into account is Volts phase-

V_t=440√(3)=254 V tanψ =V_tsinϕ+I_aX_qV_tcosϕ+I_aR_a 254×0.6+1000×0.025254×0.8=0.873 or, nbsp; nbsp;ϕ= 41.1^∘ δ=ϕ ϕ=41.1^∘ 36.9^∘=4.2^∘ E_f=V_tcosδ+I_d ...

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Standard XII

Physics

Kirchhoff's Junction Law

A 440 V, 50 H...

Question

A 440 V, 50 Hz, Y-connected synchronous generator has a direct axis reactance of $0.12 Ω / p h a s e$ and a quadrature axis reactance of $0.075 Ω / p h a s e$ and the armature resistance being negligible. The generator is supplying 1000 A at 0.8 lagging pf. The excitation after taking saliency into account is_______ Volts (phase).
256.25

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Solution

The correct option is A 256.25
$V_{t} = \frac{440}{\sqrt{3}} = 254 V$

$tan ψ = \frac{V_{t} sin ϕ + I_{a} X_{q}}{V_{t} cos ϕ + I_{a} R_{a}}$

$\frac{254 \times 0.6 + 1000 \times 0.025}{254 \times 0.8} = 0.873$

or, $ϕ = {41.1}^{\circ}$

$δ = ϕ - ϕ = {41.1}^{\circ} - {36.9}^{\circ} = {4.2}^{\circ}$

$E_{f} = V_{t} cos δ + I_{d} X_{d}$

$I_{d} = I_{a} s i n δ = 1000 sin {4.2}^{\circ} = 73.24$

$E_{f} = 254 cos {4.2}^{\circ} + 73.24 \times 0.04$

$= 256.25 V (o r) 444 V (l i n e)$

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The correct option is A 256.25Vt=440√3=254V tanψ=Vtsinϕ+IaXqVtcosϕ+IaRa 254×0.6+1000×0.025254×0.8=0.873 or, ϕ=41.1∘ δ=ϕ−ϕ=41.1∘−36.9∘=4.2∘ Ef=Vtcosδ+IdXd Id=Iasinδ=1000sin4.2∘=73.24 Ef=254cos4.2∘+73.24×0.04 =256.25V(or) 444V(line)