Question

A three-phase synchronous motor with synchronous impedance of \(0.1 + j0.3\) per unit per phase has a static stability limit of 2.5 per unit. The corresponding excitation voltage in per unit is ................ (Round off to 2 decimal places).

Hint:

Static stability limit in synchronous machines is determined by excitation voltage and synchronous reactance: \(P_{max} = EV/X_s\).

Answer 1

Step 1: Formula for maximum power.
The steady-state power per phase is: \[ P = \frac{EV}{|Z_s|}\sin\delta \] Maximum power transfer occurs at \(\delta = 90^\circ\): \[ P_{max} = \frac{EV}{|Z_s|} \]

Step 2: Static stability limit.
Given static stability limit: \[ P_{max} = 2.5 \, \text{p.u.} \] Synchronous reactance magnitude: \[ |Z_s| = \sqrt{(0.1)^2 + (0.3)^2} = \sqrt{0.01 + 0.09} = \sqrt{0.10} = 0.316 \]

Step 3: Solve for \(E\).
Assume terminal voltage \(V = 1 \, \text{p.u.}\): \[ 2.5 = \frac{E(1)}{0.316} \Rightarrow E = 2.5 \times 0.316 = 0.79 \]

Final Answer:
\[ \boxed{0.79} \]