Lecture18

Lecture18 - Example M1 T ST = 1.5 MVA X = 3% NH/NL = 5.4586...

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Page 1 Fundamentals of Power Systems Lecture 18 1 Example T S T = 1.5 MVA X = 3% N H /N L = 5.4586 Bus V = 4.16 kV M 1 M 2 M2 Rated 100 kW at unity PF X S = 0.8 pu Efficiency 90% Rated field current 3 A M1 50 Hz, 6 pole, Y-connected P R = 75 kW, X m = 7.2 R 1 = 0.082 , R 2 = 0.07 X 1 = 0.19 , X 2 = 0.18 Friction and windage losses 1.3 kW Core losses 1.4 kW Miscellaneous losses 150 W If M1 operates at a speed of 960 rpm, what is the operating condition of M2 in order to establish unity power factor at the transformer? Calculate all relevant quantities such as line currents, voltages, and field current. Fundamentals of Power Systems Lecture 18 2 Example (Solution) Approach this problem in steps 1) What is the voltage on the motor side? 2) Find the current the induction motor M1 draws 3) Find the rated operating condition of M2 4) Adjust the field current of M2 to reach unity PF at the transformer 1) The turns ratio and the Y -connection yields the no-load secondary voltage at the transformer to be 1, 2, 4160 1 440 5.4586 33 LL L LL H V N VV N == = Since the voltage drop will only be 3% for 1.5MVA reactive (full load, 90°
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Lecture18 - Example M1 T ST = 1.5 MVA X = 3% NH/NL = 5.4586...

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