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Unformatted text preview: 3410 Power Systems
(Power Factor & Compensation)
H_02 Solution 1. A load that at voltage RMS value U = 480 V has the active power P = 10 kW, can operate at two dlferent values of the power
factor, A, = 0.5 and A; = 0.9. The shortcircuit power of the supply bus is SE = 0.25 M VA, and the reactancetoresistance
ratio at the bus X/R, = if, = 2. Calculate the needed distribution voltage RMS value, E, in both cases and the loss of the active power APJ in the supply. AAA/\AAAAAAAAMAA Since S  E2 :4 U2 hance the su l ' ' ' U2 " 4802 "
5c _ _ —, pp y 1mpedance 1s approxrmately equal to Z5 kt — — —— — 0.92 Q.
25 2S Ssc 0.25 x 106 The supply impedance depends on the supply resistance and reactance 2
ZS: R3+X§ =12s l1+’—;§=R§,/1+§3 hence, the supply resistance is approximately equal to
5 — 0'92 = 0.419 Rs=_z_____
4H4“? xll+22 Xs =§SRS =2x4.1=0.820
The complex impedance of the bus is approximately equal to
2s = zsel't’: = Rs+ jXS = 0.41 + j0.82 = and“ (2
Since the load active power, P, is equal to
P P=UIcos¢z=UIL hence I=m and the supply reactance Thus, the load current RMS value depends on theload power factor, 2..
1. For 2 = ’11 = cos Q), = 0.5, q), = cos“(0.5) = 60°, the load current is _ _ P _ 104 _
1—1,_———Wi_————480x0.5_41.7A. Assuming that the load voltage CRMS value is U = U, then the load cmrentis lagging by (2)1 = 60° and its CRMS value is
1,: 41.7 at“ A.
The needed distribution voltage CRMS value is E1=U+Z.1. =480+ 0.92e1'63" x 41.7 eJ'“=480+ 38.46213" = 518ef°2" v. 2. For A = 1/2 = cos (p; = 0.9, q}; = cos"(0.9) = 25.80, the load current is _ _ P _ 104 _
1‘12‘U12‘480xo.9"23'1A' Assuming that the load voltage CRMS value is U = U, then the load current is lagging by to, = 60° and its CRMS value is 12= 23.1 at?” A.
The needed distribution voltage CRMS value is 152 = U+ZSIZ=48O+ 0.92e1'63"x 23.1 e725‘8=480 + 21.2e1'37° = 491e1'15" V.
Power 1055 inthe supply AP=RSI2
Therefore, at 2:2,, 1:11 = 41.7A, AP} =R5112= 0.41><41.72 = 713 w
atl=1Q,I=12=23.lA, APz=R5122=041x23JZ=219W Conclusion: Lower distribution voltage RMS value is needed for the load with higher power factor 2.. Power loss in the supply
is higher at low power factor. Symbols E 1 — Distribution voltage CRMS value at power factor 2.1, I 1 — Supply current CRMS value value at power factor ’11,
E2  Distribution voltage CRMS value at power factor ha, I; — Supply current CRMS value value at power factor 24,
U  Load voltage CRMS value, U — Load voltage RMS value, P — Load active power Z,— Complex impedance of the supply, 28 Impedance of the supply, o5  Supply impedance phase angle
Rs — Supply resistance X,  Supply reactance, é — Supply Xs/Rs ratio APl — Active power loss in the supply at power factor ’11, APZ — Active powerless in the supply at power factor 1Q, 2. A resistiveinductive, 25 kW at 2.4 kV load with power factor A = 0.5, is supplied ﬁ'om 2.4 kV bus with the shortcircuit power
S“ = 0.5 MVA and the reactanceto—resistance ratio X/RJ = if, = 3, A capacitive compensator is installed at the bus to
improve the power factor to unity. Calculate the capacitance C of the compensator, the distribution voltage, E, before and after compensation, the change in the
power loss in the distribution system and draw diagrams of CRMS values before and after compensation. Calculate how
much the distribution voltage can be reduced and haw much power could be saved owing to the load compensation. AAMAAAMAMAAM
. . . U2 (2.4x103)2
The bus impedance 1s approxrmately equal to 2,, w —— = ——6— = 11.5 Q,
Ssc 0.5 x10 The supply bus reactance is
S _ l 5 _ Q Rs=_z______
Jug: J1+32 Xs =i§sRs =3x3.6=10.8Q
The complex impedance of the supply bus is approximately equal to
zS = 25 em = 125+ jX, = 3.6 + j10.8 = 1 1.42171" 9 and the supply reactance is The load current RMS value before compensation, A = ll= 0.5, and the current lagging angle, (a P 25x103 _ .1 _ .1 0
1:1 =——=——=20.8A, ail—cos {/1}—cos 0.5 =60
1 U1: 2.4x103x0.5 ‘ { } Assuming that U = Ue’go, then load current CRMS value is I = 1,: 11ef(9°°‘¢’1) = 20.8ef3°° A
Thus, the distribution voltage CRMS value before compensation has to be
equal to
E, = U + 2,11 = 240021900 +11.4ef710 X 20.8.21300 = 263321“0 V
The active power loss before compensation
A1351: R, 1,2 = 3.6x 20.82 ‘= 1557 w The load current RMS value after compensation, A = A; = l, and the current
lagging angle, a); = 0. 1:12 P 25"“? 10.4A, ¢=cos'l{/i¢}=0 . o . o
I = 12 = 12 e190 ' W) =10.4e19° A
Thus, the distribution voltage CRMS value after compensation has to be equal to E2 = U + 2,12 = 2400ej9°0+11.4ej710x10.4ej900 = 2441 ei916° v
The active power loss before compensation
A112 = R3122 = 3.6x 10.42 = 389 w
The distribution voltage after compensation can be reduced by AE=E1 E2 = 2633 — 2441 = 192 V.
The saved active power
AP,=AP,.—AP,1= 1577—389= 1188 W. Conclusions: Reactive current compensation enables us to reduce the voltage, currents and power loss in distribution system Symbols Z,— Complex impedance of the supply, 23 Impedance of the supply, (a, — Supply impedance phase angle Rs — Supply resistance X, — Supply reactance, 4’, — Supply X,/R, ratio U — Load voltage CRMS value, U— Load voltage RMS value, P — Load active power ,1, — Power factor before compensation, x14 — Power factor after compensation E, — Distribution voltage CRMS value before compensation, E2 — Distribution voltage CRMS value aﬁer compensation,
I 1 — Supply current CRMS value before compensation, I; — Supply current CRMS value after compensation AP“ — Active power loss before compensation, APg — Active power loss after compensation, AE — Reduction of distribution voltage owing to compensation AP,~ Reduction of power loss owing to compensation ...
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This note was uploaded on 02/06/2012 for the course EE 3410 taught by Professor Staff during the Fall '08 term at LSU.
 Fall '08
 Staff

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