Solution Assignment_1_for_Chapter_3 Fall 2010

Solution Assignment_1_for_Chapter_3 Fall 2010 - Assignment...

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1 Assignment #1 – Chapter 3 Fall 2010- EE 362Q /EE 394.9 Power Quality M1. Redo Example 10.5 (of R1) and sketch the RMS current profile seen by the recloser. We have the following information: Current through recloser R before the incidence of the fault: I R = 250A Current through 100T fuse on Tap 2 before the incidence of the fault: I 100T, 2 = 95A Three phase fault current on the load side of Tap 2: I f = 1500A During the fault, current through 100T fuse on Tap 2 = 1500A (ignoring load currents) We also have the following information about the recloser: (a) Two fast and two delayed operations (b) First reclose interval = 0.5 s (c) Second reclose interval = 2 s (d) Reclose intervals after a delayed trip = 5 to 10 s From the TCC in Fig. 10.19, we can read the operating times of each protective device for a known fault current magnitude. 100T Fuse: 0.45 - 0.6 s to clear a fault current of 1500A. Let us assume, it takes 0.5 s to melt. Recloser: Fast tripping for 560A phase recloser causes it to trip in 0.05 s. Delayed tripping causes it to trip in 3 s. Case (a) The fault is assumed to be temporary and a fuse saving scheme is employed The three phase fault at Tap 2 is temporary. The sequence of operation is presented below: t = 0 s: Fault appears, both recloser and 100T fuse see the fault current, I f = 1500A. t = 0.05 s: The recloser trips (first fast tripping). The 100T fuse requires 0.5 s to melt at the given fault current level. Since the recloser trips in 0.05, the fuse does not melt. The recloser remains open during its first reclose interval period of 0.5 s. Hence the recloser closes at t = 0.5 s + 0.55 s = 0.55 s. t = 0.55 s: The recloser closes at the end of the first reclose interval period. Both recloser and fuse see the fault current again. From the TCC curve, the recloser again trips in 0.05 s or at t = 0.55 s + 0.05 s = 0.6 s. t = 0.6 s: Recloser trips (second fast tripping). The fuse does not have sufficient time to clear the fault. It now enters the second reclose interval and stays open for 2 s. The recloser closes at the end of the second reclose interval at t = 0.6 s + 2 s = 2.6 s. t = 2.6 s: The recloser closes at the end of the second reclose interval. The fault is assumed to have extinguished at t = 2.0 s. The recloser sees the normal load current of 250A at t = 2.6 s and will not continue with another cycle. The system returns to its normal operation after t = 2.6 s. The rms current profile seen by the recloser is shown in Fig. 1
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2 0 Recloser Current (A) Time (s) 0.05 0.55 0.6 2.6 Fault clears itself 1500 250 OPEN OPEN CLOSED CLOSED Figure 1 – Current profile seen by the recloser for temporary fault Case (b) The three phase fault at Tap 2 is assumed to be permanent The sequence of operation is as follows: t = 0 to 2.6 s – same as Case A, except that fault does not clear in 2 s, but is a permanent fault. t = 2.6 s: The recloser closes at the end of the second reclose interval. Since the fault is
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Solution Assignment_1_for_Chapter_3 Fall 2010 - Assignment...

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