Final answer:
To calculate the internally generated voltage (E0) for a synchronous generator during a short circuit test, we use the base phase voltage and the current to determine the impedance and hence the voltage drop. Considering an impedance of 10 ohms, E0 is approximately 554V.
Step-by-step explanation:
The student is asking to calculate the internally generated voltage (E0) for one phase of a synchronous generator when the terminal voltage is 480V line-to-line and the line current with a short circuit is 27.7A. The situation presented is essentially a short circuit test on the synchronous generator to determine the internal voltage drop across its synchronous impedance.
To calculate the phase voltage (V_phase) from the line-to-line voltage (V_L-L), we use the formula V_phase = V_L-L / √3. Given that V_L-L is 480V, the V_phase is approximately 277V (480V / √3).
With the line current (I) at 27.7A, the impedance per phase (Z_phase) can be calculated using Ohm's Law: Z_phase = V_phase / I, which gives us a value of approximately 10 ohms (277V / 27.7A).
Now, the internally generated voltage (E0) can be found by summing the voltage drops across the synchronous impedance and internal resistance, but the question simplifies by stating to ignore the resistance, leaving only the voltage drop across the reactance to consider. Thus, E0 = V_phase + (I * jX) where jX is the synchronous reactance. Assuming that the voltage drop across the reactance is in phase with the current (which it is, in a short circuit test), the internally generated voltage E0 can be estimated as 277V + (27.7A * 10 ohms), giving a value around 554V.