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Regulation of 100 kVA alternator from tests

Short answer

DC resistance per phase is 0.5 Ω, effective Ra=0.8335 Ω. From OC and SC tests at same field current: Zs=4.62 Ω, Xs=4.544 Ω. At full load 0.8 pf lagging, generated emf per phase E=829.7 V, voltage regulation=30.64%.

The data obtained on 100 kVA, 1100 V, Y-connected, 3-phase alternators are: DC resistance Test: Voltage between lines 6 V DC, current in lines = 6 A Open Circuit Test: Field Current = 12.5 A DC, line voltage = 420 V AC Short Circuit Test: Field Current = 12.5 A DC, line current = rated current Calculate the voltage regulation of alternator at 0.8 p.f. lagging. Take effective resistance to be 1.667 times of DC resistance.

Regulation is found by the synchronous impedance (EMF) method.

Armature resistance

DC test between two lines (star: two phases in series):

Rdc,ph=12×66=0.5 ΩRa=1.667×0.5=0.8335 Ω\begin{aligned} R_{dc,ph} &= \frac{1}{2}\times\frac{6}{6} = 0.5\ \Omega \\ R_a &= 1.667\times0.5 = 0.8335\ \Omega \end{aligned}

Synchronous impedance (same field current 12.5 A)

Irated=100×1033×1100=52.49 AEoc,ph=4203=242.49 VZs=242.4952.49=4.620 ΩXs=4.6202−0.83352=4.544 Ω\begin{aligned} I_{rated} &= \frac{100\times10^3}{\sqrt3\times1100} = 52.49\ \text{A} \\ E_{oc,ph} &= \frac{420}{\sqrt3} = 242.49\ \text{V} \\ Z_s &= \frac{242.49}{52.49} = 4.620\ \Omega \\ X_s &= \sqrt{4.620^2 - 0.8335^2} = 4.544\ \Omega \end{aligned}

Regulation at full load, 0.8 pf lagging

Vph=11003=635.09 VVcos⁡ϕ+IRa=508.07+52.49×0.8335=551.82 VVsin⁡ϕ+IXs=381.05+52.49×4.544=619.56 VE=551.822+619.562=829.67 V per phase%Reg=829.67−635.09635.09×100=30.64%\begin{aligned} V_{ph} &= \frac{1100}{\sqrt3} = 635.09\ \text{V} \\ V\cos\phi + IR_a &= 508.07 + 52.49\times0.8335 = 551.82\ \text{V} \\ V\sin\phi + IX_s &= 381.05 + 52.49\times4.544 = 619.56\ \text{V} \\ E &= \sqrt{551.82^2 + 619.56^2} = 829.67\ \text{V per phase} \\ \%\text{Reg} &= \frac{829.67 - 635.09}{635.09}\times100 = 30.64\% \end{aligned}

Answer: ZsZ_s = 4.62 Ω, XsX_s = 4.544 Ω, regulation at 0.8 pf lagging = 30.64% (full load assumed).

✓ Answer checked by Claude Sonnet 5.5

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