7 OR CORRECTION.IN THREE.»
POWER AN USING A BANK OF CAPACT acts:
i ected to incr
factor load is cor case
A low. f as by connecting a bank of capacitor a
power fai bank of capacitors is either wye.
load. This
nected or delta-connected
Fo) Xe g| xX ,
ee
a Xe
Xe
Qc =P, (tan®, - tan Onew)
a = reactive power of the capacitor, (VAR)
a = old reactive power of the load, (VAR)
| new = New reactive power of the load, (VAR)
Scanned with CamScanner _ pid active power of the load (W)
AO ew active power of the load (W)
"gid apparent power of the load, (VA)
“> new apparent power of the load, (VA)
e II: After correction the pfis unity
te: Since the final pf is unity then, Qc =Q, -
E III: After correction the pf is leading.
Qc =F, (tan, + tan ®new )
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'47 KV, 60-Hz, 3-phase SUPPIY; Loa 1 Inductiya”™
ai2 ‘60 kVAR; Load 2: Capacitive load”
load, 60 kW and 66 ‘
240 kW at 0.8 pf; Load 3: Resistive load of 60 kW. 4
wye-connected capaitor bank is connected in Paral
with the loads. What is the capacitance per phase t,
improve the overall pf to 0.8 lagging?
EXAMPLE: Three |
Let: S, = total apparent power of the loads
Pp
5, #5 +5243 75" |
= (60+ j660)+( 240 2-cos0.8) +60
S, = 360 + j480 = 600.253.13°
Ooig = 53.13°
®new = COS! 0.8 = 36.87°
Qc =P, (tan6, - tan new) — if final pf is lagging
= 360(tan53.13° - tan36.87°)
Qc = 210 KVAR ,
ae ™*,
Ep = 7199.56 V
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a, 219-70 EVAR
3
2 p _nenset = 740.48 ohms
= "70,000
= 3.58 iF
if final pf is lagg)
41,91(tan31,788° - tan18.194°)
=12.2 KVAR ;
E\ Xe Xe
Scanned with CamScanner QA _ 12.198 _ 4,066 kVAR
3.3
2 oe se
cet = 4 = 66.85 pF
WXe _ (2n)(50)(47.614)
A three-phase 12-kVA load A operating at 0.70 pf
lagging and a three-phase 10-kVA load B operating a
0.8 pf lagging are connected in parallel and supplied ty
a three-phase 440-V, 60 Hz generator. A wye-connectel
capacitor bank is connected in parallel to correct the
power factor. Determine the capacitance per phase to
make the overall.pf equal to (a) unity (b) 0.9 lagging
ANSWERS: (a) 199-65 uF (b) 90-86 UF.
TRY SOLVING THIS! M,
A500 KVA transformer is at full load and 0.60 lagging
pf. A capacitor bank is added improving the power
factor to 0.90 lagging, After improvement, what pe
of rated kVA is the transformer carrying?
ANSWER; 66'67%
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|. goLvING THIS!
| ary. a8; 6-pole synchronous motor has a per phase
aaen nce of (0:5 +J9.987) ohm, When running on
impede? 35-hz supply mains, its field excitation is such
2000 M) cmmf induced in the machine is 1600 V.
ane the maximum torque developed.
OG WER: 5867 newton-meter
A
POWER FACTOR CORRECTION USING
ASYNCHRONOUS MOTOR
= reactive power of the load, (VAR) ;
Quy = reactive power of the synchronous motor, (VAR)
ht = total reactive powet, (VAR)
Scanned with CamScanner = active power
a ae Dowel of the synchronous motor (Ww)
= rent power of the load, (VA)
a i OO vont pOWer of the synchronous motor (vA
Srovar = total apparent power, (VA)
of the load (W)
52 = P2+Q
P
Pia Q
==} =—; tand=—
cos 0 Siu S!
Q
EXAMPLE: A factory load of 900 kW at 0.6 pf lag
to be increased to 0.9 by the addition of a synchr
motor that takes 450 kW. At what power factor
this motor operate?
P.=900 'Py=450!
Prot =1350 ;
@, = cos”! 0.6 = 53.13°
Ototal = COS”! 0.9 = 25.84°
Q=Ptane . ,
Qem = Q ~ Qtotal = 900 tan53.13 - 1350 tan 25:
Qcm = 546.21 KVAR
Scanned with CamScanner 546.21
Q_, tan9sm = Jen
wep? SM ~ "450
250.516"
4 = 008 50.516 = 0.636 — leading
pLE: An overexcited synchronous motor is
nected across 2 150-kVA inductive load of 0.7
gain power factor. The motor takes 12 kW while
running on no-load. Calculate the kVA rating of the
motor if it is desired to bring the overall power factor of
the motor-inductive load combination to unity.
Note: An over-excited synchronous motor operates at a
leading pf.
Protas =P, + Ps
6, =cos™! 0.7 = 45.57° -
P =Scos@ A, =150(0.7) = 105 kW
Q= P tand > Q, =105tan45.57 =107-11 KVAR
s =p? +Q2
Sou = fiz? +107.112 = 107.78 KVA
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MPLE: An electrical system has a load of
A tagging power factor of 0.67. If a 3000 wa ky
eynchron ‘ous condenser Is installed for power ¢,
‘a
correction purposes, calculate the overall Dower fy,
P cr,
ndenser is a synchronous
An 3 chron
pee like a capacitor.
motor operating
P.=Prta=5000 *
6, = cos! 0.67 = 47.93°
0, =P, tang, = 5000 tan 47.93 = 5539.437 KVAR
Onrtq) = 5539-437 - 3000 = 2539.437 KVAR
tano=2
Pp
2539.437
5000
Orotal = 26.925°
Pfiotat = 005 26.925 = 0.891 lag:
TAN Gigtal =
RY SOLVING THIS! in
Brnchronous Motor absorbing 60 KW Is conte
paral a @ load of 240 kW having a lagging Pa 4
0.9, at ae If the combined load has a power
231 at what power factor is it working?
ANSWER: 0-865 lagging
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