A 25-kVA, 230-V, three-phase, four-pole, 60-Hz, wye-connected synchronous genera

Question

A 25-kVA, 230-V, three-phase, four-pole, 60-Hz, wye-connected synchronous generator has a synchronous reactance of 1.5 ohms per phase and negligible stator resistance. The generator is connected to an infinite bus of constant voltage magnitude and constant frequency at 230 V and 60 Hz.

Part A.

Determine the excitation voltage, line-to-line, EA [V], when the machine is delivering rated power at 0.8 lagging power factor.

Determine the power (torque) angle, , when the machine is delivering rated power at 0.8 lagging power factor. [degrees].

Part B

The field excitation current, If, is increased by 20% without changing the power input from the prime mover. Find the magnitude of the stator current IA [A].

For the operating conditions of Part B, find the power factor. For this question only, indicate a lagging power factor with a positive value and a leading power factor with a negative value. For unity power factor, enter 1.0.

For the operating conditions of Part B, find the reactive power supplied by the machine. [kVAR]

Part C

With the field excitation current, If, as in Part A, the input power from the prime mover is increased very slowly. What is the steady-state power limit [kW]?

With the prime mover input power at the steady-state power limit, determine the resulting stator current, IA [A]

With the prime mover input power at the steady-state power limit, determine the resulting power factor. For this case, indicate a lagging power factor by a positive value and a leading power factor by a negative value. Enter 1.0 for unity power factor.

With the prime mover input power at the steady-state power limit, determine the reactive power, Q [kVAR], that is delivered to the system. If the machine is absorbing reactive power, indicate by a negative value.

Explanation / Answer

Armature Current (Ia) = kva / voltage = 25000/230 = 108.69 A

excitation Voltage = (Armature Current Ia * synchoronous reactance ) + output Voltage = (108.69*j1.5)+230

line to line Voltage = root 3 * Vph = 1.732* 230 =398.37

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