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Published byMervin Walsh Modified over 9 years ago
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Problem # 5-7 A 3-phase, 8-pole, induction motor, rated at 874rpm, 30hp, 60Hz, 460V, operating at reduced load, has a shaft speed of 880 rpm. The combined stray power loss, windage loss, and friction loss is 350W. The rotor parameters in Ω/phase are R1= R2=0.191 XM=14.18 X1=1.338 X2= Rfe=189.1 ECE 441
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Equivalent Circuit for an Induction Motor with all parameters referenced to the stator
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Need to determine the value of s.
Problem # 5-7 continued The motor is NEMA Design C and Y-connected. Determine the motor input impedance/phase Need to determine the value of s. ECE 441
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The Input Impedance per Phase
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Solution continued Determine the line current ECE 441
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3-Φ Delta and Wye Connections
(a) shows the sources (phases) connected in a wye (Y). Notice the fourth terminal, known as Neutral. (b) shows the sources (phases) connected in a delta (∆). Three terminals ECE 441
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Solution continued Determine the active, reactive, and apparent power and power factor ECE 441
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Solution continued Determine the equivalent rotor current ECE 441
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Solution continued Determine the stator copper loss ECE 441
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Solution continued Determine the rotor copper loss ECE 441
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Solution continued Determine the core loss ECE 441
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Solution continued Determine the air gap power ECE 441
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Solution continued Determine the mechanical power developed ECE 441
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Solution continued Determine the developed torque ECE 441
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Solution continued Determine the shaft horsepower ECE 441
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Solution continued Determine the shaft torque ECE 441
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Solution continued Determine the efficiency ECE 441
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Solution continued Sketch the power flow diagram and enter all data
24.33hp 631.5W W W W ECE 441
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Solution continued Determine the expected minimum locked-rotor torque, breakdown torque, and pull-up torque From Table 5-1, Tlr=200%Trated Tlr = (2)(180.27) = lb-ft From Table 5-3, Tbreakdown = 190%Trated Tbreakdown = (1.9)(180.27) = lb-ft From Table 5-6, Tpull-up = 140%Trated Tpull-up = (1.4)(180.27) = lb-ft ECE 441
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