In order to calculate the actual relay operating time, the following things must be known :
Time/P.S.M. curve
Current setting
Time setting
Fault current
Current transformer ratio
The procedure for calculating the actual relay operating time is as follows :
Convert the fault current into the relay coil current by using the current transformer ratio.
Express the relay current as a multiple of current setting i.e. calculate the P.S.M.
From the Time/P.S.M. curve of the relay, read off the time of operation for the calculated P.S.M.
Determine the actual time of operation by multiplying the above time of the relay by time setting multiplier in use.
Example for Relay Operating Time
Determine the time of operation of a 5-ampere, 3-second overcurrent 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 A.
Solution:
Rated secondary current of C.T. = 5 A
Pickup current = 5 × 1.25 = 6.25 A
Fault current in relay coil = 4000* 5/ 400 × = 50 A
Plug-setting multiplier (P.S.M.) = 50 /6.25 = 8
Corresponding to the plug-setting multiplier of 8 ,the time of operation is 3.5 seconds.
Actual relay operating time = 3.5 × Time-setting = 3.5 × 0·6 = 2.1 seconds
Important Terms for Relay Operating Time
Pick Up current
It is the minimum current in the relay coil at which the relay starts to operate. So long as the current in the relay is less than the pick-up value, the relay does not operate and the breaker controlled by it remains in the closed position.
However, when the relay coil current is equal to or greater than the pickup value, the relayØ operates to energies the trip coil which opens the circuit breaker.
Current Setting
It is often desirable to adjust the pick-up current to any required value. This is known as current setting and is usually achieved by the use of tapping on the relay operating coil. The taps are brought out to a plug bridge. The plug bridge permits to alter the number of turns on the relay coil.
This changes the torque on the disc and hence the time of operation of the relay. The values assigned to each tap are expressed in terms of percentage full-load rating of C.T. with which the relay is associated and represents the value above which the disc commences to rotate and finally closes the trip circuit.
Pick-up current = Rated secondary current of C.T. × Current setting
For example, suppose that an overcurrent relay having current setting of 125% is connected to a supply circuit through a current transformer of 400/5.
The rated secondary current of C.T. is 5 amperes. Therefore, the pick-up value will be 25% more than 5 A i.e. 5 × 1·25 = 6·25 A. It means that with above current setting, the relay will actually operate for a relay coil current equal to or greater than 6·25 A.
The current plug settings usually range from 50% to 200% in steps of 25% for overcurrent relays and 10% to 70% in steps of 10% for earth leakage relays. The desired current setting is obtained by inserting a plug between the jaws of a bridge type socket at the tap value required.
Plug Setting Multiplier (PSM)
It is the ratio of fault current in relay coil to the pick-up current i.e.
P.S.M. = Fault current in relay coil / Pick – up current =Fault current in relay coil /Rated secondary current of Current setting.
For example, suppose that a relay is connected to a 400/5 current transformer and set at 150%. With a primary fault current of 2400 A, the plug-setting multiplier can be calculated as under :
Pick-up value = Rated secondary current of CT × Current setting = 5 × 1.5 = 7.5 A
Fault current in relay coil = 2400 *5/ 400 × = 30 A
P.S.M. = 30/7.5 = 4