Figure Q2B shows a 22-kV radial system with CO-7 over-current relays. The system
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Question
Figure Q2B shows a 22-kV radial system with CO-7 over-current relays. The system is operated at the frequency of 50 Hz The characteristics of CO-7 relay are shown in Figure Q2B (ii)Loads at bus 1, bus 2 and bus 3 are 6.4 MW, 4M and 10.8 MW respectively and all have the same power factor of 0.8 lagging The transient reactance for generator is 3.968 , and reactances for lines from bus 2 to 1 and bus 3 to are 0.846 and 0.266 , respectively. ?.?.R-400/5 CTR. 5 ?.?.R-2005 Figure 22B (i) (a) Calculate maximum fault (or short-circuit) currents at bus 1, bus 2 and bus 3. Neglect load currents for fault calculations. (3 marks) (b) Calculate the fault currents in the relay for bolted three-phase faults at bus 1, time. coils of the relays bus 2 and bus 3, considering one at (3 marks) (c) Calculate normal load currents, source current, and line currents for normal operation of the system. (4 marks) (d) Calculate thormal currents in the relays, and the current tap setting (CTS) for all CTs to protect the system from faults (4 marks)Explanation / Answer
(a) Using the Z-impedance formation approach, Z-bus matrix for given network is
bus-3 3.968 3.968 3.968
bus-2 3.968 4.234 4.234
bus-1 3.968 4.234 5.08
Z-bus is formed with following rule:
bus-3 is added to reference bus with impedance of 3.968 ohm. Bus-2 is added to bus -3 with impedance of 0.266 ohm. so diagnol entry of bus -2 is 3.968+0.266=4.234 ohm. off-diagnol entry is copied from bus-3 entry. similarly bus-1 impedance is formed too.
Now,Short circuit current is given by voltage(22kV) divided by diagnol entry of each bus as loading is neglected. So, for bus-1, short circuit current is 22k/5.08 equals 4.33kA current.
S.C current for bus-2 is 22k/4.234 equals 5.196kA.
S.C current for bus-3 is 22k/3.968 equals 5.544kA.
(b) As C.T ratio of relay at bus-3 is 400/5, Relay current is 5.544*(5/400)=69.3A.
As C.T ratio of relay at bus-2 is 200/5, Relay current is 5.196*(5/200)=129.9A.
As C.T ratio of relay at bus-1 is 200/5, Relay current is 4.33*(5/200)=108.25A.
(c) Applying KCL to each of buses, we get system of three equations. Solving them,
Voltage at bus-3 is 15.258kV.
Voltage at bus-2 is 15.041kV.
Voltage at bus-1 is 14.577kV.
Load currents at bus-3,2,1 are (10.8 M/ (15.258k*0.8)), (4 M/ (15.041k*0.8)), (6.4M/(14.577k*0.8)) equals 884.78, 332.42, 548.8 A respectively.
Line current between bus-2 and bus-1 is V2-V1/0.846 equals 548.46A.
Line current between bus-3 and bus-2 is V3-V2/0.266 equals 1830.82A.
source current is (22k-15.528k)/3.968 equals1631.04A.
(d)
As C.T ratio of relay at bus-3 is 400/5, Relay current is 1631.04*(5/400)=20.38A.
As C.T ratio of relay at bus-2 is 200/5, Relay current is 1830.82*(5/200)=45.77A.
As C.T ratio of relay at bus-1 is 200/5, Relay current is 548.46*(5/200)=13.71A.
Current tap setting is normal current to fault current.
For relay 3, 20.38/69.3 equals 0.294.
For relay 2, 45.77/129.9 equals 0.352.
For relay 1, 13.71/108.25 equals 0.126.
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