Question

A single-phase electrical motor draws a current of 20 A from 240, 60 Hz line. A Fluke wattmeter is connected into line gives an active power (P) reading of 3,792 W. with line, neutral, appropriate fuse, switch, and motor itself. (DO NOT forget chassis ground of the motor for the safety of people working in/around the motor). Show a simple wiring diagram (a) Calculate the power factor (cost) of the motor and reactive power () it absorbs. (b) If we place a capacitor bank of 2,500 VAr in parallel with the motor terminals, calculate the following. (0) The total active power absorbed by the system. Show a Cap bank across the motor (il) The net reactive power (Qnet) and new apparent (complex) power of the system. (ii) New motor line current I after Cap bank added; is the new motor current is less or more than rated current of 20A? Explain (ii) The new power factor of the system afer capacitor is connected. Is it improved to a better value? Make sure you indicate if it is still leading or lagging power factor (iv) Calculate total approximate capacitance (uF) needed for a 2,500 VAr Cap bank.

Answer 1

1 - phi motor: I = 20 A: v = 240 V: f = 60 Hz p = 3792 w Calculate p.f & reactive power of motor apparent power s = v times I Therefore s = 240 times 20 = 4800 vA but s = p/cos theta Therefore p.f cos theta = 3792/4800 = 0.79 lag Q = squareroot s^2 - p^2 = squareroot 4800^2 - 3792^2 = 2942.912 VAR If we place a 2500 VAR capacitor in parallel total active power absorbed by system show a cap bank across motor \r\n when a pare cap bank is placed the value of active power does not change Therefore p_active = 3792 w net reactive power Q = Q_motor + Q_cap = (j2942.913 - j2500) Q_new = j442.913 VAR new complex power: s = 3792 + j 442.913 = 3817.779 angle 6.662 vA I_L = (s/v)* = (3817.779 angle 6.662/240)* = 15.907 angle -6.66 A
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