Prova de Operaçoes Unitaria 3

Induction Motor Operations Working (Part 2) Revolution Phasor Diagram Before a) b) c) Notes around diagram: - Interlinked, Defluxing, Speed - Flux not lag - Load angle - Max flux lag - Balanced induced voltages - Better efficiency - Stator, Rotor - Copper losses - Continuous Formula: V = (f + 5.6s)/ 2 x 50 r = T x sin θ x 4.8 n = Speed + 5 x 40 n = 10k RPM Part of list: - Efficiency boost - Copper losses reduced - Split - Voltage input - Stat flux rotation - Synchronous speed increase Calculations: 1. T = 15/P 2. T = 10/P + 5s/20 3. V = 502 4. Bal Vdrop = 9 phase 5. Sync speed 400s 6. Rotor input = Flux current gain 7. P > copper losses 8. Field L < copper losses Formula: n = (100) + s_PM / (p_factor) x (a/4) - Results interpreted in phasor Note: This image contains a diagram focusing on induction motor operations, its revolution phasor diagram, and formulaic interpretations. It lists several operational suggestions and calculations regarding phasifying components and copper losses.