HW05 EE6560 Power System Protection 1For HW05 assume that the ground overcuent Device 51N provides...

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HW05  EE6560 Power System Protection 1For HW05 assume that the ground overcuent Device 51N provides adequate and sufficient single line to ground fault coverage (perhaps you'll do 51N in a future homework.)  Concentrate on the phase overcuent Device 51 protection of the primary zone of this feeder for each of the following cases.  Line length is from eaker to recloser,R.  This utility uses a minimum Detection Margin of 1.5 and a minimum Load Margin of 1.15 1. What EOL (End Of Line) bolted fault type yields the lowest phase amps? 2. Determine the primary phase amps for a.  bolted EOL fault,  . highest phase overcuent Device 51 pickup amps primary that satisfy the Detection Margin,  c. the maximum load amps primary that meet the Load Margin, and  d. the resulting maximum load MVA allowed. e. Show hand calculations for 4.16kV case then can use EXCEL or ASPEN for the rest of cases.  Note: To simplify your calculations I'm providing reactance only, use 1.0 pu nominal source voltage.  If you think you need Zo assume it's 3Z+.  These line reactances are for 954 ACSR Cardinal conductor rated 1009 A for summer normal conditions.Read Blackburn Chapter 12 pages XXXXXXXXXXAs a minimum page turn to assure you are familiar with this material.) Paul Nauert 2/25/2023

Banasree · Accepted Answer

1.Ans.
To determine the EOL (End Of Line) bolted fault type that yields the lowest phase amps for the primary zone of the feeder protected by a phase overcurrent Device 51,  consider the fault impedance seen from the location of the protective device. The lowest phase amps would correspond to the fault type that presents the lowest impedance to the protective device. 
For an EOL bolted fault, the fault impedance is essentially the impedance of the entire line length from the location of the protective device to the end of the line. Therefore, the fault impedance will be lowest for the fault that occurs closest to the location of the protective device. Let say that the protective device is located at the breaker, the fault closest to the breaker will yield the lowest phase amps. This fault will be a bolted fault at the breaker.
2.a. Ans.
With give data, calculation of  the positive sequence impedance of the line as follows:
Zpos = Rpos + jXpos
Where,
 Rpos = 0.064 ohm/mile (for 954 ACSR Cardinal conductor)
and 	Xpos = 0.412 ohm/mile (for 954 ACSR Cardinal conductor)
Assuming a line length of L miles, the total positive sequence impedance of the line would be:
Zline = L * (Rpos + jXpos)
For a bolted EOL fault, the fault impedance is negligible, so the fault current would be limited by the positive sequence impedance of the line and the source impedance.
Assuming a nominal source voltage of 1.0 pu and a fault location at the end of the line, the fault current would be:
Ifault = Vsource / (Zsource + Zline)
Where,
 Zsource =  the source impedance, Assume as 3Zline for simplicity.
Substituting the values for Zline and Zsource:
Ifault = 1.0 pu / (3 * Zline)
Ifault = 1.0 pu / (3 * L * (0.064 + j0.412) ohm/mile)
Assuming a line length of 10 miles
Ifault = 1.0 pu / (3 * 10 * (0.064 + j0.412) ohm/mile)
Ifault = 982.6 A
Therefore, the primary phase amps for a bolted EOL fault would be approximately 982.6 A.
b.Ans.
Assuming a nominal source voltage of 1.0 pu and a line length of 10 miles, the positive sequence impedance of the line would be:

HW05 EE6560 Power System Protection 1 For HW05 assume that the ground overcurrent Device 51N provides adequate and sufficient single line to ground fault coverage (perhaps you'll do 51N in a...

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