ECE456 Power System Protection

Transcription

1 ECE456 Power System Protection Assignment : #5 (Solutions) 1. A phase b-c-g fault is experienced at F in the system shown in figure 1. Calculate the impedances seen by the b-c, b-g and c-g units of the distance relay installed at circuit breaker 2. Solution Figure 1 The equivalent sequence diagram for the given system and its connection for a B-C-G fault are shown below in figure 1a and 1b respectively. For a two phase to ground fault all the three impedance networks are connected in parallel at the fault point as shown in figure 1b. In addition since the transformer configuration is YnD, there will not be any zero sequence current at the relay location for ground faults on the star side of the transformer. 1 Of 11

2 Figure 1a Figure 1b The three phase transformer ratio is considered to be 1. Since we are only interested in finding the impedance at location 2, we have not considered the 30 shift in the sequence currents introduced by the transformer (this shift will affect the phase relationship between the currents and voltages across the primary and secondary of the transformer and will not affect the phase relationship between voltages and currents on any one side). We can calculate the sequence currents and voltages at the relay point looking at the figure 1b as follows (All values are given in per unit) Total equivalent impedance; Zeq = Zs1+ZL1+Zt1+Zt2+ZL2+Zs2 = j 1.14 pu. Sequence currents at the relay location ir1 = E1/Zeq = 1/( j1.14) = j pu ir2 = -ir1 = j pu ir0 = 0 Sequence voltages at the relay location vr1 = E1-iR1*Zs1 = j pu vr2 = -ir2*zs2 = j pu vr0 = 0 The phase currents and voltages can then be calculated as, ira=ir0+ir1+ir2 = 0 pu irb = ir0 + a 2 * ir1 + a * ir2 = j pu = <-179º irc = ir0 + a * ir1 + a 2 * ir2 = j pu m = <1.005 º irn = ira+irb+irc = 0 2 Of 11

3 vra=vr0+vr1+vr2 = i pu = < º vrb = vr0 + a 2 * vr1 + a * vr2 = j pu = < º vrc = vr0 + a * vr1 + a 2 * vr2 = j pu = < º The impedance measured by BC, B-G and C-G units are calculated as BC = (vrb RB-vRC)/(iRB vrc)/(irb-irc) irc) = j 0.37 pu ZBC BG = vrb/(irb+k*irn) = j pu CG = vrc/(irc+k*irn) = j pu ZBG ZCG Where k is the earthfault compensation setting of the impedance relay calculated as k = (z0 - z1)/(3*z1) 2. A three-phase line is connected to sources A, B and C (figure 2). A threephase fault occurs at the half way point between T and C. Calculate the impedance seen by the relays at circuit breaker 1 and 2 before any circuit breaker open. Assume that the voltages of all the three sources are 1.0 p.u. Solution Figure 2 The equivalent circuit of the above system can be drawn as shown in the figure 2a below; 3 Of 11

4 j0.2 j0.35 j0.05 j0.05 j0.25 Figure 2a Which can be re-drawn for reduction as shown in figure 2b; A j0.2 T j0.15 B j0.2 c j0.35 j0.05 j0.05 I T j0.25 E 1.0 pu Figure 2b From the above circuit we can find the fault current and current flowing in different branches as; Ic = E / j( ) = 1.0 / j( ) = -j 2.5 pu E 1.0 It = = = j j j If = It + Ib = -j ( ) = -j 6.5 pu 0.4 Ia = j 4 = j Of 11

5 Ic = It-Ia = -j(4-2) = -j 2 We have been asked to calculate the impedance seen by the relays at circuit breaker 1 (location A) and circuit breaker 3 (location C). Impedance seen by relay at A, Za = Va/Ia Where Va, is the voltage at the point A with respect to ground. This voltage can be found as Va = E Ia * j 0.20 = 1.0 (-j 2) * (j 0.20) = 0.6 pu Therefore, Za = Va/Ia = 0.6 / (-j ( 2) ) = j 0.3 p.u But from the system diagram we know that the relay at A should have measured only =0.25 p.u, but in actual case the relay measures higher impedance and thus will make it to underreach. This under reaching is due to the infeed at point T from the source B. Similarly the impedance seen by the relay at B, Zb = Vb/Ib Where Vb, is the voltage at the point B with respect to ground. This voltage can be calculated as Vb = E Ib * j 0.25 = 1.0 (-j 2) * (j 0.25) = 0.5 pu Therefore, Zc = Vb/Ib = 0.5 / (-j ( 2) ) = j 0.25 p.u. Again we know that the relay at C should have measured only =0.2 p.u, but due to the infeed at T from the source A, the relay measures the fault impedance as 0.25 p.u., thus causing it to underreach. 3. Determine the three zone settings for the relay Rab in the system shown in figure 3. The system nominal voltage is 138 kv and the positive sequence impedances for the various elements are given in the figure3. Assume that the maximum load at the relay site is 120MVA, and select a CT ratio accordingly. The available distance relay has zone 1 and zone 2 5 Of 11

6 settings from 0.2 to 10 ohms and zone 3 settings from 0.5 to 40 ohms in increments of 0.1 ohm. The angle of maximum torque can be adjusted to 75 or 90. Solution Figure 3 First we have to select an appropriate CT. The primary of the CT should be close to the maximum load current. In our case the maximum load current is given as 120MVA, therefore the load current can be calculated as I L = = 502A Thus we can select a CT ratio of 600:5 Since the system voltage is given as 138KV, the VT ratio would be 138KV/120V Thus the impedance transformation ratio (for transforming the impedance from system primary to relay secondary values) will be Tr = CTR / VTR = (600/5) / (138000/120) = Zone 1 of the relay has to be set to protect 80% of the line, thus Z1 = 0.8 * (3+j40) * = Ω Since the setting range available in the relay for impedance is in steps of 0.1Ω, we will set the zone 1 as 3.3Ω. 6 Of 11

7 The angle setting available in the relay is either 75 or 90 only. So select the angle setting as 90 Zone 2 of the relay has to be set to protect atleast 120% of the protected line. At the same time this setting should not reach beyond the zone 1 setting of the shortest downstream line relay, which is the relay on line BD. So we can write that the Z2 setting should be more than Z2min = 1.2 * (3+j40) * = Ω And less than Z2max = [(3+j40)+0.8*(1.5+j25)] * = Ω So we will set the Z2 to 5.5 Ω, which would cover 100% of the protected line and 50% of the shortest adjacent line Zone 3 element of the relay should provide back-up protection to the longest line from the remote bus, which in this problem is the line BE, with 5+j100 Ω impedance. Normally it s a practice to set the Z3 as 110% of the sum of the protected line plus the longest adjacent line. So we get; Z3 = 1.1*(3+j j100) * = Ω Select a Z3 setting of 14.6 Ω, which would be a value settable in the given realy. So the settings for the relay Rab would be Zone Ω Zone Ω Zone Ω Relay characteristic angle 90 4 Consider the system shown in figure 4, assume that F in that figure is a AG fault at the end of the line. Considering no load condition, calculate the impedance seen by AG unit of the relay at circuit breaker 2 for the two following cases: a. the current in the parallel circuit is available to the relay at breaker 2 7 Of 11

8 b. the current in the parallel circuit is not available to the relay at breaker 2 Figure 4 Solution In order to calculate the seen impedance, fault calculation should be performed to determine the fault currents and voltages at the relay 2 and 3 locations. Since there is AG fault, three sequence network are in series as shown in Figure 5. Figure 5 8 Of 11

9 In order to calculate the fault current simplified sequence network shown in right hand side of figure 5 is used to drive equivalent sequence impedances as follow: Zeqpos=ZL1pos*ZL2pos/(ZL1pos+ZL2pos)+Zs1= j Zeqneg=Zeqpos= j Zeqzero=(ZL1zero*ZL2zero-ZLm0^2)/( ZL1zero+ZL2zero-2*ZLm0)+Zs0= j Fault current is calculated as below Ifzero=230/sqrt(3)/(Zeqpos+Zeqneg+Zeqzero)= j Ifpos=Ifzero= j Ifneg=Ifzero= j Fault current seen by relay 2 IR2pos=Ifpos*ZL2pos/(ZL1pos+ZL2pos)= j IR2neg=Ifneg*ZL2neg/(ZL1neg+ZL2neg)= j IR2zero=Ifzero*(ZL2zero-ZLm0)/(ZL1zero+ZL2zero-2*ZLm0)= j IR2a=1.8208< IR2b= < IR2c= < Fault current seen by relay 3 IR3pos=Ifpos*ZL1pos/(ZL1pos+ZL2pos)= j IR3neg=Ifneg*ZL1neg/(ZL1neg+ZL2neg)= j IR3zero=Ifzero-IR2zero = j Voltage at relay point is calculated: VRpos=230/sqrt(3)-Zs1*Ifpos = j VRneg=-Zs2*Ifneg = j VRzero=-Zs0*Ifzero = j VRa= < VRb= < VRc= < k0=(zl1zero - ZL1pos)/(ZL1pos)= j k0m=(zlm0)/(zl1pos) = Of 11

10 a) For double circuit line having access to second line current, seen impedance is calculated from following equation: ZR2seen=VRa/(IR2a+k0*IR2zero+k0m*IR3zero)=4+j40 Ω b) if there is no access to second line current, IR3zero should be considered as zero in above equation. Hence we have: ZR2seen=VRa/(IR2a+k0*IR2zero) = j Ω As you see, relay2 under-reaches in this case. 5 A zone of a distance relay with mho characteristic is set at 10 ohms secondary. The CT ratio is 500:5 and the VT ratio is 20,000: What is the fault resistance, which this relay tolerates for a fault at a distance of 80% of the zone boundary? You may assume that the remote terminal of the line is out of service. The line impedance angle is equal to 80 while the relay maximum torque angle is equal to 70. Solution We have to calculate the available fault resistance coverage at 80% of the relay impedance setting for mho characteristics. For this we have to find the distance of the zone boundary from the relay characteristics line at 80% parallel to the R axis (0 ). We can either do this by constructing the characteristics on a graph paper and measuring the distance or by analytical geometry as follows; 10 Of 11

11 Relay Characteristic Line 8sin(80)= x Figure 5a Referring to figure 5a, we can write We have to find the point on the circle which has the same longitude as the 80 percent of the transmission line characteristic as below: (x-5cos(70)) 2 + (8*sin(80)-5sin(70)) 2 = 5 2 By solving above equation: X= & According to the figure 5a, is acceptable. Fault resistance of relay characteristics at 80% of setting can be faind as R = cos(80) = 4.18 Ω Impedance transformation ratio = (500/5)/(20000/69.3) = The relay will operate for a fault resistance upto 4.18/0.347 = Ω (primary) at 80% of its setting 11 Of 11