2.16 Numerical on Over-current relay

1. Determine the time of operation of a 5 ampere, 3 second over-current relay having a current setting of 125 % and a time setting multiplier of 0.6 connected to supply circuit through a 400 / 5 current transformer when the circuit carries a fault current of 4000 amp. Use the curve shown in figure 2.8.

Solution: - Rated secondary current of CT = 5 amp

From figure 2.8, corresponding to the plug setting multiplier of 8, the time of operation is 3.5 seconds.

Now, for TSM = 0.6 -

Therefore,

Actual relay operating time = Time of operation x Time setting multiplier

= 3.5 x 0.6 = 2.1 seconds

2. An over-current relay having a current setting of 125 % is connected to a supply circuit through a current transformer of ratio 400 / 5. Find the pick-up value.

Solution: -

Pick-up current = Rated secondary current in CT x Relay current setting

= 5 x 1.25 = 6.25 A

3. The pick-up value of a relay is 7.5 and fault current in relay coil is 30 A. Find its plug setting multiplier.

Solution: -

Plug setting multiplier (PSM) = Fault current in relay coil / Pick-up current = 30 / 7.5 = 4

4. Determine the operating time of the relay for the current rating of a relay is 5 ampere having a current setting of 150 % and a time setting multiplier of 0.4 connected to supply circuit through a 400 / 5 current transformer when the circuit carries a fault current of 6000 amp. At TMS = 1, operating time at various PSM are:

PSM: |
2 |
4 |
5 |
8 |
10 |
20 |

Operating time in seconds |
10 |
5 |
4 |
3.2 |
2.8 |
2.4 |

Solution: - Rated secondary current of CT = 5 amp

Corresponding to PSM of 10, from a given table operating time is 2.8 seconds at TSM = 1.

Now, for TSM = 0.4 -

Therefore,

Actual relay operating time = Operating time from table x Time setting multiplier

= 2.8 x 0.4 = 1.12 seconds

5. A 10 A IDMT relay has a current setting of 150 % and has a time multiplier setting of 0.5. The relay is connected in a circuit through a current transformer having ratio 400 : 5 A. Calculate the time of operation of the relay if the circuit carries a fault current of 6000 A. The relay characteristic for TSM = 1 is given as:

PSM: |
2 |
3.6 |
5 |
10 |
15 |
20 |

Operating time in seconds |
10 |
6 |
3.9 |
2.8 |
2.2 |
2.1 |

Solution: - Rated relay current = 10 A

Corresponding to PSM of 5, from the given table, operating time is 3.9 seconds at TSM = 1.

Now, for TSM = 0.5 -

Therefore, Actual relay operating time = Operating time x Time setting multiplier

= 3.9 x 0.5 = 1.95 seconds

6. An IDMT over-current relay has a current setting of 150 % and has a time multiplier setting of 0.5. The relay is connected in a circuit through a current transformer having ratio 500 : 5 A. Calculate the time of operation of the relay if the circuit carries a fault current of 6000 A. The relay characteristic for TSM = 1 is given as:

PSM: |
2 |
4 |
5 |
8 |
10 |
20 |

Operating time in seconds |
10 |
5 |
4 |
3 |
2.8 |
2.4 |

Solution: - Rated secondary current of C T = 5 A

Corresponding to PSM of 8, from a given table operating time is 3 seconds at TSM = 1.

Now, for TSM = 0.5 -

Therefore, Actual relay operating time = Operating time x Time setting multiplier

= 3 x 0.5 = 1.5 seconds

7. Consider a system:

Given that: Fault current = 2000 A, CT ratios for each relay = 200 / 1, R_{B} is set at 100 % and R_{A }is set at 125 %.

For discrimination the time grading margin between the relays are 0.5 sec. Determine the time of operation of the two relays and TSM of relay R_{A}. Assuming relay R_{B} has TSM = 0.2. At TSM = 1 operating time at various PSM are:

PSM: |
2 |
4 |
5 |
8 |
10 |
20 |

Operating time in seconds |
10 |
5 |
4 |
3.2 |
2.8 |
2.4 |

Solution: -

Part-I: – When Fault is near B, the PSM is –

Corresponding to PSM of 10, from a given table operating time is 2.8 seconds at TSM of 1.

Now, for TMS = 0.2 -

Therefore, Actual relay operating time = Operating time x Time setting multiplier

= 2.8 x 0.2 = 0.56 seconds

Part-II: – When Fault is near B, the PSM for relay R_{A} is –

Corresponding to PSM of 8, from a given table operating time is 3.2 seconds at TSM = 1.

Now operating time for relay R_{A} = operating time for relay R_{B} + time grading margin

= 0.56 + 0.5 = 1.06 sec

Therefore, Actual relay R_{A} operating time = Operating time x Time setting multiplier

1.06 = 3.2 x TMS

=> TMS = 0.33125

8. Design the time current grading scheme for the following system:

Relay Point |
CT ratio |
Fault current |

A |
400 / 5 |
6000 A |

B |
200 / 5 |
5000 A |

C |
200 / 5 |
4000 A |

At TSM = 1, the time current characteristic of the relay is as given below:

PSM: |
2 |
3 |
5 |
10 |
15 |
20 |

Operating time in seconds |
10 |
6 |
3.9 |
3 |
2.2 |
2.1 |

Solution: -

Part-1: - Assume current setting of 100 % and TMS is 0.1

For relay R_{C} -

Corresponding to PSM of 20, from a given table operating time is 2.1 seconds at TSM = 1.

Now, for TMS = 0.1 -

Therefore, Actual relay operating time = Operating time x Time setting multiplier

= 2.1 x 0.1 = 0.21 seconds

Now for the same fault current i.e. 4000 A the relay at B must be set to operate as a back-up protection, so assume 0.5 second (time delay step) longer than the relay at C i.e. 0.21 + 0.5 = 0.71 sec

The current setting at B must be increased as compared to that at C to allow for any increase in load caused by load at C, setting the relay current setting at B as 125 %. Thus for 4000 A –

Corresponding to PSM of 16, from a given table operating time is 2.2 seconds (i.e. near about same as PSM of 15) at TSM = 1.

Now operating time for relay R_{B} = operating time for relay R_{C} + time grading margin

= 0.21 + 0.5 = 0.71 sec

Therefore, Actual relay R_{B} operating time = Operating time x Time setting multiplier

0.71 = 2.2 x TMS

=> TMS = 0.323

Part-2: - Assume current setting of 125 % and TMS calculated is 0.323

When Fault is near B, the PSM is –

Corresponding to PSM of 20, from a given table operating time is 2.1 seconds at TSM = 1.

Now, for TMS = 0.323 -

Therefore, Actual relay operating time = Operating time x Time setting multiplier

= 2.1 x 0.323 = 0.6783 seconds

Now for the same fault current i.e. 5000 A the relay at A must be set to operate as a back-up protection, so assume 0.5 second (time delay step) longer than the relay at B i.e. 0.6783 + 0.5 = 1.1783 sec

The current setting at A automatically gets increased without increasing the setting of the relay by virtue of its CT ratio. So keeping the relay current setting at A as 125 % i.e. same as that of the relay at B. Thus PSM of relay A, when fault takes place at relay B i.e. for 5000 A –

Corresponding to PSM of 10, from a given table operating time is 3.0 seconds at TSM = 1.

Now operating time for relay R_{A} = operating time for relay R_{B} + time grading margin

= 0.6783 + 0.5 = 1.1783 sec

Therefore, Actual relay R_{A} operating time = Operating time x Time setting multiplier

1.1783 = 3.0 x TMS

=> TMS = 0.393

Part-3: - Assume current setting of 125 % and TMS calculated is 0.393

When Fault is near A, the PSM is -

Corresponding to PSM of 12, from a given table operating time is 2.6 seconds at TSM = 1.

Now, for TMS = 0.393 -

Therefore, Actual relay operating time = Operating time x Time setting multiplier

= 2.6 x 0.393 = 1.0218 seconds

9. Determine the operating time of the relay for the current rating of a relay is 5 ampere having a current setting of 200 % and a time setting multiplier of 0.3 connected to supply circuit through a 400 / 5 current transformer when the circuit carries a fault current of 4000 amp. At TMS = 1, operating time at various PSM are:

PSM: |
2 |
4 |
5 |
8 |
10 |
20 |

Operating time in seconds |
10 |
5 |
4 |
3 |
2.8 |
2.4 |

Solution: - Rated secondary current of CT = 5 amp

Corresponding to PSM of 5, from a given table operating time is 4 seconds at TSM = 1.

Now, for TSM = 0.3 -

Therefore, Actual relay operating time = Operating time x Time setting multiplier

= 4 x 0.3 = 1.2 seconds

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