King Saud University. College of Engineering. Electrical Engineering Dept.
EE449 –Power System Protection. Final Exam – 2nd Semester 1427-1428.
King Saud University College of Engineering Electrical Engineering Dept.
EE449 –Power System Protection Final Exam – 2nd Semester 1427-1428 Time Allowed: 3 Hours
Question#1 (20%) Complete the following sentences. 1- The circuit breaker interrupts the circuit if its ………….
…………… is energized.
2- The function of zone-3 in a distance relay is ………………………………...………… …………………………………………………………………………………. 3- To overcome the problem of inrush current for differential protection of transformers, one usually uses ………………………………………………………………………… 4- The main protection for stator winding of the generator is the ………………….………………………… 5- In a three- phase distance relay, ground relays are used for ………………………. Faults 6- The abbreviation CTS means ………………
…………………
………………
7- The abbreviation CCVT means ……………… …………… …………… …………… 8- In the differential protection of Y-Δ transformer, the CTs are connected as ……… …….. 9- The coordination process of overcurrent relays in a radial system must start with the relay in …………………………………………. 10- The standard rating for CT secondary is: American = ………… European = …………… 11- The generator side windings of the step-up transformer in a power plant are usually connected into DELTA to ………………………………………………………………………………..
12- The dependability requirement for a protective relay means that …………………………… ………………………………………………………….
King Saud University College of Engineering Electrical Engineering Dept.
EE449 –Power System Protection Final Exam – 2nd Semester 1427-1428 Time Allowed: 3 Hours
Question#2 (20%) A) Explain the causes of inrush current during energization of power transformer. Discuss its effect on the differential relay response. Propose methods to overcome this problem. B) Discuss the sources of errors for the application of differential relays to bus protection. C) A 50 MVA, 13.8 kV Δ/110 kV Y transformer is protected by a percentage differential relay with taps. i. Select CT ratios for primary and secondary sides of the transformer. ii. Draw the connection diagram of the three-phase differential relay showing the CT connections. iii. Select suitable relay tap settings. iv. Select suitable percentage mismatch value. Assumptions - Available CT ratios: 50, 100, 150, 200, 250, 300, 400, 450, 500, 600, 800, 900, 1000, 1200, 1500, 2000, 2400, 3000, 4000, 5000, 6000:5 A. - Available relay tap settings: 5:5, 5.5, 6.6, 7.3, 8, 9, and 5:10.
Question#3 (20%) A) The burden of a 1000:5, 5C300 CT is 10 Ohms. What will be the percentage error of the CT if the primary current is 9000 A. B) A three-phase distance protection unit is used to protect a transmission line. This protection unit is composed of SIX distance relays; 3 relays for phase faults (Rab, Rbc, Rca) and 3 relays for ground faults (Rag, Rbg, Rcg). Each of the 6 relays is a 3-zone directional impedance relay with the following settings: Zr1 = 0.3+j3 Ω (Instantaneous), Zr2 = 0.45+j4.5 Ω (0.25 s), Zr3 = 0.8+j8 Ω (0.5 s). The compensation factor for ground relays is m=2.0. A B-C fault has occurred on the line. The resulting fault currents and voltages seen by the distance protection unit are: Ia = 0, Ib = -Ic=15.3 A /-175o, Vag = 80 V /0o, Vbg = 74 V /237o, Vcg = 73 V /123o i)
Calculate the impedance seen by each of the six relays. Draw the 3-zone characteristics of the relay and locate the calculated impedances on the characteristics.
ii) What will be the response of each of the six relays (Block / Trip (delay=?)). iii) Comment on the performance of these distance relays. Question#4 (20%) A) A 100 MVA, 13.8 kV, Y-connected synchronous generator is operating at rated voltage at no load. Its per unit parameters are: X1 = 0.2, X2= 0.15 and X0 = 0.05 pu. i- Find the ratio of the fault line current for a LG fault to the fault line current for a symmetrical three-phase fault. Assume that the neutral is solidly grounded and the fault is occurring at the terminals of the generator. ii- Calculate the pu reactance to be inserted in the neutral connection (Xn) to limit the fault line current for a LG fault of part (i) to that of a symmetrical three-phase fault.
B) The motor of fig-2 is supplied from a generator through a step down transformer. A fault occurs at bus#2. The symmetrical component currents supplied by the motor to the fault are: I0m = 0, I1m =0.8 – j2.6 and I2m = j2; while that supplied by the transformer are: I0t = 0, I1t = -0.8 – j0.4 and I2t = j1 all in per unit. 2
1
Fig-1
M
G
i- Calculate the symmetrical component currents in the fault, from which identify the type of fault. Justify your answer. ii- Find the phase currents supplied by the generator during the fault. Consider the transformer phase shift. Question#5 A) Discuss the use of directional relay in the protection of lines including: its main function, for which type of networks it should be used? for which type of relay it can be added? B) The power network shown in Fig-2 is protected using 6 protection units as described in the table below. The system is subjected to an AG fault at one of its lines. The phase-a currents flowing during the fault are illustrated in the figure. Assume that the currents at phases b&c are equal to zero. Study the response of all the relays (phase and ground for all units), then answer the following questions i. Specify the relays that will block and that will start its operation. ii. Specify the relays that will issue a TRIP signal, and determine its time delay. determine the sequence of breaker opening (e.g, B2 will open after ..s then B1 after..). Assume that all the used circuit breakers have an interruption time of 5 cycles. iii. Comment on the performance of the protection sysem.
1 B3
3200 A
2000 A
B5 2
B1
B2
B4
B6
640 A
Fig-2 Protection Unit B1 B2
CT Ratio (A) 800:5 800:5
B3
400:5
B4
400:5
B5
400:5
B6
400:5
Relay Type Overcurrent (CO8) Overcurrent (CO8) Directional Overcurrent (CO8) Directional Overcurrent (CO8) Directional Overcurrent (CO8) Directional Overcurrent (CO8)
Sett. of Phase Relays CTS (A) TDS 6 4 6 4
Sett. of Ground Relay CTS (A) TDS 2 1 2 1
5
3
1
0.5
5
3
1
0.5
5
3
1
0.5
5
3
1
0.5
