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Power electronic control in electrical systems 395
I a I ab I ca 47:41 j11:40 48:76e j13:52 A
I b I bc I ab 50:24 j10:94 51:42e j167:72 A
I c I ca I bc 2:83 j22:33 22:51e j82:78 A
P 1 RefV ac I g 415e j60 48:76e j13:52 5741W
a
P 2 RefV bc I g 415e j120 51:42e j167:72 6496W
b
Check: P 1 P 2 12 237 W
2
2
2
2
2
jI ab j R ab jI bc j R bc jI ca j R ca 27:6667 15 25:085 1:2 12 237 W:
22. (i) Whatarethemaindisadvantagesofsingle-phasedistributionofACelectricpower?
(ii) Explain the methods used to reduce or eliminate the following harmonics in
AC power systems:
(i) odd triplen harmonics, i.e. 3rd, 9th, 15th etc.
(ii) odd non-triplen harmonics, i.e. 5th, 7th, 11th, 13th etc.
(iii) A star-connected, three-phase AC load is supplied with 415 V three-phase
sinewave AC power at a frequency f Hz, and the star point is solidly con-
nected to the supply neutral. The load impedances at 50 Hz (in ohms) are as
follows:
Z a 9:184 j0 Z b 3 j17 Z c 3 j17:
Calculate the rms values and phase angles of the line currents and the neutral
current when the frequency f is
(a) 50 Hz
(b) 150 Hz
(i) Unbalance; 100-Hz oscillation in the power flow causes vibration, noise, and
lamp flicker. Single phase cannot by itself produce a rotating ampere-conductor
distribution in electric motors (it needs a capacitor and a split-phase winding).
(ii) (a) Triplen harmonics are suppressed by star connection or trapped by delta
connection; or taken by neutral wire.
(b) Non-triplen harmonics. In electric machines, they are minimized by wind-
ing design (harmonic winding factors) and by having a sine-distributed
magnetic flux around the airgap. In transformers, they are minimized by
limiting the flux to a level below saturation. In non-linear power-electronic
loads such as rectifiers, the 5th and 7th can be cancelled by using a 12-pulse
rather than 6-pulse circuit. This requires a transformer with two sec-
ondaries, one wye and the other delta. Otherwise, the 5th and 7th can be
filtered by damped or tuned filters built up from L, C and R elements.
p
(iii) (a) I a 415/ 3/9:184 26:088 A
p p
I b 415/ 3e j120 /(3 j17) 415/ 3e j120 /17:263e j80
13:880e j200:0 13:044 j4:745 A
p p
I c 415/ 3e j120 /(3 j17) 415/ 3e j120 /17:263e j80
13:880e j200:0 13:044 j4:745 A
I n I a I b I c 0