Question

Question: A three-phase 34.5 KV feeder supplying a load 3.5 MVA is protected by 80E power fuses in each phase, in series with a recloser. The time-current characteristic of the fuse is shown in Figure 10.43. Analysis yields maximum and minimum fault currents of 1000 and 500 A, respectively, (a) To have the recloser clear the fault, find the maximum clearing time necessary for recloser operation. (b) To have the fuses clear the fault, find the minimum recloser clearing time. Assume that the recloser operating time is independent of fault current magnitude.

Short answer

Answer

(a) The maximum clearing time to clear the fault is 0.137 s.

(b) Theminimum clearing time to clear the fault of the recloser is 0.063 s .

Step-by-step solution

Write the given data from the question.

The three-phase voltage, V= 34.5 KV

The load S = 3.5 MVA.

The maximum fault current I_max =1000 A.

The minimum fault currentI_{mIN =}500 A .

Determine the maximum time required to clear the fault by the recloser.

If the maximum fault current flows through the fuse for a long time, then the material of the fuse get melts. Therefore, the maximum time of clearing the fault should be less than the minimum melting time for the fuse.

The minimum melting time can be obtained corresponding to the minimum fault current, i.e. 500 A .

From the time-current characteristics of the fuse , the maximum time corresponding to the 500 A is 0.135 s.

Hence, the maximum clearing time to clear the fault is 0.137 s .

Determine the minimum time required to clear the fault by the recloser.

For clearing the fault by the fuse, the fuse must melt before the recloser opens. The minimum clearing time for the recloser should be greater than the maximum time for the fuse. The fuse's maximum time can be calculated corresponding to the maximum fault current, i.e.,1000 A .

From the time-current characteristics, the maximum clearing for the recloser is 0.063 s.

Hence, the minimum clearing time to clear the fault of the recloser is 0.063 s.