Introduction to Electric Circuits

5.5 Power in balanced three-phase circuits 119

Since the phase angle ~b is given by tan -~ (reactive power/real power) then

4~ = tan-' [V'3(P~- P2)/(PI + P2)] (5.14)

The power factor is then simply cos oh.
Summarizing, using the two-wattmeter method in any balanced star or delta
connected three-phase circuit in which the readings are PI and P2, we can
obtain the following information"
9 total real power
W = (P~ + P2) watts (5.12 bis)
9 total reactive power
Q = ~3(P~ - P2) Var (5.13 bis)
9 power factor
cos 4~ = cos {tan -~ [V'3(P~ - P2)/(P1 + P2)]} (5.15)

Example 5.5

Two wattmeters are connected to measure the power input to a 400 V, 50 Hz,

three-phase motor running on full load with an efficiency of 90 per cent. The

readings on the two wattmeters are 30 kW and 10 kW. Calculate (1) the input

power to the motor, (2) the reactive power, (3) the power factor, (4) the useful

output power from the motor.

Solution

1 Let the readings on the two wattmeters be P~ and P2. Then, from Equation

(5.12), the power to the motor is

PI + P2 = 30 + 10 = 40 kW

2 From Equation (5.13) the reactive power is

V/3(P~- P2) = V'3 • 20 = 34.64 Var

3 From Equation (5.14)

~b = tan -1 [V'3(P~- Pz)/(PI + P2)] = tan -1 (34.64/40) - 40.89 ~

so the power factor is

cos r = cos 40.89 ~ = 0.756 lagging

4 The efficiency (r/) of the motor is defined as the useful output power (Po)

divided by the total input power (Pi). Thus

Po = r/Pi = 0.9 • 40 = 36 kW