Page 423 - Handbook of Electrical Engineering

412 HANDBOOK OF ELECTRICAL ENGINEERING

sin(2 × 66.753) − sin(2 × 66.215) − 2 × 0.00939
b 1 =
2(0.4033 − 0.3947)
0.7253 − 0.7381 − 0.01878
=
0.0172
=−1.834 indicating a lagging power factor

From (15.8), (15.9) and (15.10)

761.1 3
I r =+ 0.798 =+236.8 amps
π 2

761.1 3
I i =− 1.834 =−544.2 amps
π 2
and

761.1 3
2
2 1/2
I = (0.7680 + 1.834 )
π 2
= 593.46 amps per phase
From (15.13), (15.14) and (15.15) the volt-amperes at the bridge AC terminals are,

S sec = P sec + jQ sec

Where
3 × 346.0 × 761.1 × 0.7980 × 1.2247
P sec =
3.1415926
= 246.09 kW

and
3 × 346.0 × 761.1 × 1.834 × 1.2247
Q sec =
3.1415926
= 565.52 kVA r

and
S sec = 616.75 kVA

The power factor of the fundamental current is,

246.09
cos Ø 1 = = 0.3990 lagging
616.75
or
a 1 0.7980
= √ = 0.3990 lagging
2
0.7980 + 1.834 2
c 1