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    Electrical Energy Conversion: Synchronous Generator

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    Synchronous Generator
    A 250 kVA, 280 V, three-phase, four-pole, 60 Hz, synchronous generator with a synchronous reactance of 0.99 ohms per phase is operating at rated conditions and a power factor of 0.832 lagging. The magnetization curve for the generator is shown in Figure 1.
    a. Sketch a phasor diagram for the generator.
    b. Name the 2 curves shown on the graph, and indicate briefly how they are found. Why is it that one is a straight line, but the other exhibits a curve?
    c. Determine:
    - the excitation voltage (magnitude and angle)
    - the open-circuit phase voltage
    - the voltage regulation
    - the no-load voltage if the field current is reduced to 60% of its value at rated load.

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    Solution Preview

    a. A phasor diagram for this generator

    - Power factor = 0.832 lagging, the corresponding angle phi = 37.4 degrees.
    - Resistance of armature Ra is not given, so it is assumed that Ra = 0 Ohm
    - Rated voltage line-to-line (vector): UL-L = 480 at 0 degrees Volts
    U on the above vector diagram is rated voltage per phase. Assume the winding is wye connected, U = 277 Volts
    - Rated current I:
    Magnitude: I = S /(1.732*U) = (250*1000)/(1.732*480) = 300.7 Amps
    Vector: I = 300.7 at -37.4 degrees Amps
    - Synchronous reactance Xt = 0.99 Ohm
    - E(delta) is the induced voltage at the air-gap or known as excitation voltage.

    b. The two curves

    - The curly curve is the ...

    Solution Summary

    This solution is attached in a Word document and includes a diagram for (a), calculations and explanations for each step. The method used in this assignment can be applied to any similar synchronous machines related questions.