A 208-V Y-connected synchronous motor is drawing 40 A at unity power factor from a 208-V power system. The field current flowing under these conditions is 2.7 A. Its synchronous reactance is 0.8 Ω. Assume a linear open-circuit characteristic.
The answer is given as:
However, I do not understand where they got that equation for EA. I was under the impression that the equation for the phase voltage is:
Wouldn't this mean that the equation for the internal generated voltage is EA = VΦ + jXsIA?
EDIT:
My bad. That's the equation for a synchronous generator, not a synchronous motor.
This is the equation for a synchronous motor:
Best Answer
If you supply mechanical power to a synchronous machine, it becomes a synchronous generator generating electrical power.
Electrical power converts the synchronous machine into a motor providing mechanical power.
From Electric Machinery Fundamentals. As a motor:
$$V_{\phi} = E_A + j I_A X_S + I_A R_A$$ As a generator:
$$V_{\phi} = E_A - j I_A X_S - I_A R_A$$
Note direction of current. The voltage seen at the terminals \$V_{\phi}\$ will depend on whether the synchronous machine takes current or generates current, hence the change of signs. Just Kirchhoff's Voltage Law.