Answer Happy

Power Factor: 0.85 Ztotal: 6.62<31.75 Stator Current:
19.19<-31.5 Motor speed: 1176rpm Pload: 5374.24 TLoad: 43.64
Problem 2 (Induction Motor) A three phase, Y connected, 220 V, 60 Hz, 6 pole induction motor has the following constants for its equivalent circuit (use circuit in slide 22): Xm = 13.25 12 R1 = 0.29 2 X1 = 0.5 12 R2 = 0.14 2 (referred to stator side) X2 = 0.21 1 (referred to stator side) =

Problem 3 (Power Factor Correction) In an industrial plant with a large number of induction motors, during certain loading conditions, the inductor motors have a large reactive component and the power factor is low. Capacitor banks and synchronous motors are two commonly used methods for power factor correction. a) Capacitor based power factor correction: A capacitor bank is connected in parallel with the induction motor in problem 2. Calculate the capacitance value needed to bring the total power factor to unity. (1 point) b) Synchronous motors can also be used for power factor correction. A synchronous motor with 5 kW power input is connected in parallel with the induction motor, what power factor should the synchronous motor operate at to bring the total power factor to unity? (1 point) c) Both capacitors and synchronous motors can achieve the goal of power factor correction. Discuss the implications of a low power factor. Compare these two methods and discuss the advantages and disadvantages of each method with considerations of social, environmental, economic factors. (2 points) d) For a MW level large industrial plant located in rural area, its low power factor has been causing high utility bills. This specific plant has enough budget to install a power factor correction system and they are looking into a solution with long service life and flexible power factor control, which solution should they choose, capacitor banks or synchronous motor? (1 point)