ece524-s09

# ece524-s09 - m .0: ZOHmmmm m_>_m_l_.m>m ~E>>On_...

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Unformatted text preview: m .0: ZOHmmmm m_>_m_l_.m>m ~E>>On_ ZH mHZmHmZ/EH .nmm mUm , , osmutobﬁazg , «a? a“. As 3 .ﬁ 3E 3: + a} “£3, ocmutoztﬂm>m9W. 5.} .5 . ocmvtobﬁgmzct .9 ELQL osmviobﬁgmch ECE 524: Session 7; Page 1/8 Transients in Power Systems Spring 2012 \$5 Transient R—L Circuit Example Given a 230kV:3‘4kV, A—Ygrounded, lOOMVA transformer supplied supplied by a source with a short circuit MVA of 2000. The transformer has a per unit reactance of 0. lpu on its ratings base. Both the transformer and the source impedance have an X/R ratio of 12. Calculate the complete phase A fault current if a fault occurs on the LV side of the transformer at 70 degrees past the voltage peak. GM Deﬁne Units: MVA :2 lOOOkW SBASE :2 IOOMVA pu :2 l Transformer parameters: Suited 2: lOOMVA XoverR :2 12 WW := 230w VIV :2 34.5w Find the transformer impedances referred to the LV side since the fault occurs on that side. 2 Vlv ZBLV :2 ZBLV = 11.99 Srated RtransLV Z: Rtrans'ZBLV RtransLV = 0.09929 XtransLV 3: Xtrans'ZBLV XtransLV : 1‘199 r- XtransLV *9 L’tranSLV m 1-4131}va : 2-7r-60Hz ECE 524: 7 Session 7; Page 2/8 Transients in Power Systems Spring 2012 \$.23, {am Source impedance MVA := ZOOOMVA V 2: 1.0pu / SC pu MVA MVAscju :2 SEAS: MVASCJDU = 20 pu 2 MVA :M- II“! Vpu stource :: m stource : 0-05 pu 1;“ g n :2 atan(X0verR) n = 85.24deg Zsource 3: mzsource‘eyrn Zsource : (0-0042 + 0049801711 Source impedance in Ohms referred to LV side: Zsource_0hm 5: Zsource'ZBLV Zsource_Ohm : (0-049 + 0-5930 Q In1(Zsource_0hm) L . = 1.57mH (2.“.60HZ) source_LV Lsource_LV 3: If we wanted to ﬁnd source impedance referred to LV side in 1 step: 2 (V1V> ' atan( XoverR) MVASC = (0.049 + 0.593i) 0 Equivalent circuit (since this is a three phase fault we can use per phase analysis: Rsrc H, Rtrans rns : Fault VLN__LV ECE 524: Transients in Power Systems Transient Current Solution: Requiv :: Re<zsource_0hm) + RtransLV Lequiv -— Decay constant: >\ :: Requiv Lequiv l T 2: —— >\ Driving point voltage: 2 Vm 3: “3"Vlv Fault inception angle: source_LV + LtransLV If we deﬁne our driving voltage as: v(t) = Vm-sin(w-t+ (b) w :2 2-71'60Hz >\ 2 31.42-1— s T = 0.0325 Vm = 28.17kV rad o.) = 376.99— s Requiv = 0.159 Lequiv = 4.73 mH “a. Session 7; Page 3/8 Spring 2012 Since the fault occurs 70 degrees phase the peak of the voltage (for analytical solution let fault occur at time t = 0): d) := 90deg + 70deg ( Initial condition: 10 ;= 0A Equiavlent impedance: ZequivﬂLV 3: Requiv + (W'Lequiv) (i) = l60deg Switch is open 2 ZequivALV : 1-790 avg 97 ECE 524: Transients in Power Systems Create time vector t2: Osee,1-10_ see. 2x104 1.5x104 itrans( t) 5x103 1x104 4 4 V(t) :2 vmsinw-t + (1)) 60Hz . Vm r...) Zequiv_LV . Vm 1trans“) 3: equiv_LV ‘9 sin(w>t + (1) — n) jsmm) —n)-e_ M 0.02 0.04 0.06 0.02 0.04 0.06 Session 7; Page 4/8 Spring 2012 ECE 524: Transients in Power Systems \JLA/ 000473[H] MVW 0449phm] Timed Fault Logic 7 0 7V l_Fault ‘7 , w MW.“ WM “MM ,oum amo'cem,rmno name '0. ka = 27.122kA a? amm_.ambv 0mm amp Session 7; Page 9/9 Spring 2012 m m L1. 1" ll ‘1 I ll '8 f a / ‘ L / a l :1 1 g 1 ,1 _’ l l 2 )} W1 {A 1L x 1‘ if 1 If 1K / \H‘r goes up a little with smaller time step \EQ ii? «acidmg hquD gogidc All Osmvtobﬁngphﬁ a} S ,, _, {a s g ,3 3 £3 ... .. .5938 30 3.. mi; 53 g: 720 30 E gt 0 . , CIGU-kobﬁhwxwmm‘m: 3335 3926 L6 5+3 SdgchU 23m 319: _ ,E 2 ga s3 6 a», g ozmniﬁtﬁgmcg «3d . 053103?ng ...
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## This note was uploaded on 02/28/2012 for the course ECE 524 taught by Professor Staff during the Spring '08 term at Idaho.

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ece524-s09 - m .0: ZOHmmmm m_>_m_l_.m>m ~E>>On_...

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