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Synchronous reactance and regulation from test data

Short answer

Effective resistance per phase = 0.12 Ω. Using O.C. 700 V line and S.C. 376 A at the same field current, Z_s = (700/√3)/376 = 1.0749 Ω, so X_s = √(1.0749² – 0.12²) = 1.068 Ω. At full-load 0.8 pf leading, generated emf = 1175.8 V/ph (2036.6 V line), regulation = –11.45%.

A 3-phase, 1500 kVA, star connected, 50 Hz, 2300 V alternator has a resistance between each pair of terminals as measured by direct current is 0.16 Ω. Assume that the effective resistance is 1.5 times the ohmic resistance. A field current of 70 A produces a short-circuit current equal to full-load current of 376 A in each line. The same field current produces an emf of 700 V on open circuit. Determine the synchronous reactance of the machine and its full load regulation at 0.8 pf leading.

Effective resistance

DC resistance between two terminals = 0.16 Ω, which is two phases in series (star).

Rdc,ph=0.162=0.08 ΩRa=1.5×0.08=0.12 Ω\begin{aligned} R_{dc,ph} &= \frac{0.16}{2} = 0.08\ \Omega \\ R_a &= 1.5 \times 0.08 = 0.12\ \Omega \end{aligned}

Synchronous impedance and reactance

The same field current (70 A) gives 700 V (line) on open circuit and 376 A on short circuit:

Zs=700/3376=404.15376=1.0749 ΩXs=Zs2−Ra2=1.07492−0.122=1.068 Ω\begin{aligned} Z_s &= \frac{700/\sqrt3}{376} = \frac{404.15}{376} = 1.0749\ \Omega \\ X_s &= \sqrt{Z_s^2 - R_a^2} = \sqrt{1.0749^2 - 0.12^2} = 1.068\ \Omega \end{aligned}

Regulation at full load, 0.8 pf leading

Vph=23003=1327.91 V,I=376 AVcos⁡ϕ+IRa=1062.32+376×0.12=1107.44 VVsin⁡ϕ−IXs=796.74−376×1.068=796.74−401.62=395.12 VE=1107.442+395.122=1175.82 V per phase%Reg=1175.82−1327.911327.91×100=−11.45%\begin{aligned} V_{ph} &= \frac{2300}{\sqrt3} = 1327.91\ \text{V},\quad I = 376\ \text{A} \\ V\cos\phi + IR_a &= 1062.32 + 376\times0.12 = 1107.44\ \text{V} \\ V\sin\phi - IX_s &= 796.74 - 376\times1.068 = 796.74 - 401.62 = 395.12\ \text{V} \\ E &= \sqrt{1107.44^2 + 395.12^2} = 1175.82\ \text{V per phase} \\ \%\text{Reg} &= \frac{1175.82 - 1327.91}{1327.91}\times100 = -11.45\% \end{aligned}

Answer: XsX_s = 1.068 Ω per phase; full-load regulation at 0.8 pf leading = −11.45% (E = 1175.8 V/phase, 2036.6 V line).

✓ Answer checked by Claude Sonnet 5.5

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