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Generator on infinite bus with 15% excitation rise

Short answer

At rated kVA, 0.8 pf lagging, excitation is 231.7 V per phase, power angle 28.63°. After 15% field increase with constant turbine power, stator current is 16.51 A, pf 0.673 lagging, active power 4.0 kW, reactive power 4.40 kVAR.

A 3-phase, 5 kVA, 208 V, 4-pole, 60 Hz, star connected synchronous generator has negligible armature winding resistance and synchronous reactance of 10 ohms per phase. The generator is first connected to an infinite bus of 208 V, 60 Hz. i) Determine the excitation voltage and the power angle when the generator is delivering rated kVA at 0.8 pf lagging. ii) If the field excitation is now increased by 15% (keeping turbine power constant), find the stator current, power factor, active and reactive power constant.

Given: star, 5 kVA, 208 V, 60 Hz, Xs=10 ΩX_s = 10\ \Omega, Ra≈0R_a \approx 0, on a 208 V, 60 Hz infinite bus.

(i) Excitation voltage and power angle at rated kVA, 0.8 pf lagging

V=2083=120.09 VI=50003×208=13.88 A,Iˉ=13.88∠−36.87∘Eˉ=Vˉ+jXsIˉ=120.09+j10×13.88(0.8−j0.6)=(120.09+83.27)+j111.03=203.36+j111.03E=231.70 V per phase,δ=28.63∘\begin{aligned} V &= \frac{208}{\sqrt3} = 120.09\ \text{V} \\ I &= \frac{5000}{\sqrt3\times208} = 13.88\ \text{A},\quad \bar I = 13.88\angle -36.87^\circ \\ \bar E &= \bar V + jX_s\bar I = 120.09 + j10\times13.88(0.8 - j0.6) \\ &= (120.09 + 83.27) + j111.03 = 203.36 + j111.03 \\ E &= 231.70\ \text{V per phase},\quad \delta = 28.63^\circ \end{aligned}

P=3VIcos⁡ϕ=4000P = 3VI\cos\phi = 4000 W, Q=3VIsin⁡ϕ=3000Q = 3VI\sin\phi = 3000 VAR.

(ii) Excitation increased by 15%, turbine power constant

E′=1.15×231.70=266.45 Vsin⁡δ′=Esin⁡δE′=111.03266.45=0.4167,δ′=24.63∘Eˉ′=266.45∠24.63∘=242.22+j111.03Iˉ′=Eˉ′−VˉjXs=122.13+j111.03j10=11.10−j12.21I′=16.51 A,∠−47.73∘pf=cos⁡47.73∘=0.673 lagging\begin{aligned} E' &= 1.15\times231.70 = 266.45\ \text{V} \\ \sin\delta' &= \frac{E\sin\delta}{E'} = \frac{111.03}{266.45} = 0.4167,\quad \delta' = 24.63^\circ \\ \bar E' &= 266.45\angle 24.63^\circ = 242.22 + j111.03 \\ \bar I' &= \frac{\bar E' - \bar V}{jX_s} = \frac{122.13 + j111.03}{j10} = 11.10 - j12.21 \\ I' &= 16.51\ \text{A},\quad \angle -47.73^\circ \\ \text{pf} &= \cos 47.73^\circ = 0.673\ \text{lagging} \end{aligned}

Active and reactive power

P=3×120.09×11.10=4000 W (unchanged)Q=3×120.09×12.21=4400 VAR (lagging, delivered)\begin{aligned} P &= 3\times120.09\times11.10 = 4000\ \text{W (unchanged)} \\ Q &= 3\times120.09\times12.21 = 4400\ \text{VAR (lagging, delivered)} \end{aligned}

Increasing the field only raises the reactive power supplied to the bus; active power is set by the turbine.

Answer: (i) E = 231.7 V/phase (401.3 V line), δ = 28.63°; (ii) I = 16.51 A, pf = 0.673 lagging, P = 4.0 kW (constant), Q = 4.40 kVAR.

✓ Answer checked by Claude Sonnet 5.5

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