Questions about psspy.dist_clear_fault

asked 2025-02-15 17:34:41 -0600

JQ
1 ●14 ●20 ●19

updated 2025-02-16 01:02:53 -0600

3452 ●1 ●9 ●20

Does psspy.distclearfault(1) put the tripped branch between bus 40809 and bus 41450 back into service after clearing the fault on the branch in the following code?

psspy.run(0,1.0,500,50,0)
psspy.dist_branch_fault(ibus=40809, jbus=41450, id = '1', units = 3, basekv = 0.0, values1 = 0, values2=0)
psspy.run(0,1.05,500,6,0)
psspy.dist_branch_trip(ibus = 40809, jbus =41450, id = '1')
psspy.run(0,1.10,500,6,0)
ierr = psspy.dist_clear_fault(1)
psspy.run(0,15,500,6,0)

answered 2025-02-16 01:02:36 -0600

perolofl

3452 ●1 ●9 ●20

updated 2025-02-16 01:03:52 -0600

Clear fault just removes the fault shunt inserted by dist_branch_fault.

Branch faults are modelled with a large line shunt on the faulted line. Tripping the line, means that the fault is isolated from the rest of the system, i.e. the fault is cleared at time 1.05 when the line is tripped.

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What is the difference between psspy.dist_branch_trip(ibus = 40809, jbus =41450, id = '1') and psspy.dist_clear_fault(1)? I think dist_branch_trip trips the branch between bus 40809 and 41450, while dist_clear_fault only removes the fault shunt from the branch.

JQ ( 2025-02-16 11:11:21 -0600 )edit

The trip disconnects the branch. The branch fault is a part if the branch data (line shunt at ibus side). Disconnecting the branch means that the fault is no longer connected to the system. Tripping branch means tripping line and shunt fault. There is no need for clear fault in this contingency.

perolofl ( 2025-02-16 14:18:08 -0600 )edit

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