0195136047.pdf

4.3 MEASUREMENT OF POWER 211

tanφ=

√

3

WA−WC

WA+WC

=

√

3

WB−WA

WB+WA

=

√

3

WC−WB

WC+WB

(4.3.10)

The two-wattmeter method discussed here for measuring three-phase power makes use of
single-phase wattmeters. It may be noted, however, that three-phase wattmeters are also available,
which, when connected appropriately, indicate the total real power absorbed. The total reactive
power associated with the three-phasebalancedload is given by

Q=

√

3 VLILsinφ=

√

3 (WC−WA) (4.3.11)

based on the two wattmeter readings of the two-wattmeter method.
With the generator action of the source assumed,+Pfor the real power indicates that the
source is supplying real power to the load;+Qfor the reactive power shows that the source is
delivering inductive VARs while the current lags the voltage (i.e., the power factor is lagging); and
−Qfor the reactive power indicates that the source is delivering capacitive VARs or absorbing
inductive VARs, while the current leads the voltage (i.e., the power factor is leading).

EXAMPLE 4.3.1

Considering Figure 4.3.1, let balanced positive-sequence, three-phase voltages withV ̄AB =
100

√
3 0° V (rms) be applied to terminalsA, B, andC. The three-phase wye-connected balanced
load consists of a per-phase impedance of (10+j 10 ). Determine the wattmeter readings of
WAandWC. Then find the total three-phase real and reactive powers delivered to the load. Based
on the wattmeter readings ofWAandWC, compute the load power factor and check the sign
associated with the power factor angle.

Solution

V ̄AB= 100

√

3  0° V; V ̄BC= 100

√

3  −120°; V ̄CA= 100

√

3  120°

V ̄AN= 100  −30° V; V ̄BN= 100  −150°; V ̄CN= 100  90°

I ̄A=

V ̄AN

Z ̄

=

100  −30°

10

√

2  45°

= 5

√

2  −75° A(rms)

I ̄C=

V ̄CN

Z ̄

=

100  90°

10

√

2  45°

= 5

√

2  45° A(rms)

The load power factor angleφ=45°, and it is a case of lagging power factor with the inductive
load.

WA=VABIAcos(30°+φ)= 100

√

3

(

5

√

2

)

cos 75°=317 W

WC=VCBICcos(30°−φ)= 100

√

3

(

5

√

2

)

cos 15°=1183 W

The total three-phase real power delivered to the load

WA+WC= 317 + 1183 =1500 W

which checks with
√
3 VLILcosφ=

√

3 100

√

3

(

5

√

2

)

cos 45°=1500 W