Determine the time of operation of a relay of rating 5 A, 2.2s IDMT and having a relay setting of 125%, TMS = 0.6. It is connected to the supply circuit through a CT 400: 5 ratio. The fault current is 4000 A.
| PSM | 2 | 5 | 8 | 10 |
| Time | 10 | 8 | 3.2 | 2.5 |
Right Answer is:
1.92 sec
SOLUTION
The plug setting multiplier of a relay is defined as the ratio of secondary fault current to the relay current setting.
PSM = Secondary fault current/Relay current setting
Calculation:
Relay CT ratio = 400/5
Therefore, pickup value of relay = 5 A
Since the relay setting is 125%
The operating current of relay = 5 × 1.25 = 6.25 A
Given that, primary fault current, Ipf = 4000 A
Therefore, the secondary fault current is
Pulse setting Multiplier PSM is given as
PSM = Fault current/ Relay current setting × CT Ratio
= 4000/6.25 x 80 = 8 Amp
After calculating the value of PSM we have to find out the total time of operation of the relay with the help of the Time / PSM curve and for the standard 2.2-sec curve the operating time for PSM = 8 is 3.2 sec
Now in the above question, the value of TSM is given which is 0. 6 sec
Then the actual operating time of the relay is PSM × TSM = 3.2 x 0.6= 1.92 sec