CDG Relay

if you have the CT ration, i think you could multiply 3.75 times CT ratio..

Take into account that relay has to successfully cater magnetic inrush at starting and harmonics created by furnace, also you have to be specific in terms of degree of inverse characterstics (curve) that has been selected.

Settng on relay (3.75) * CT ratio * (1.10/0.95 )

Will give the answer.

(It is incorrect to state both inverse and instantaneous in your question.)

WHAT IS THE CT RATIO AND THE SYSTEM VOLTAGE WITHOUT THIS NOTHING CAN BE SAID.

Shhhhhhhhhhhhhhhhhhhhhh!

CT ratio is 600/5

Assuming that the Transformer primary is 11 KV, the rated primary current is 52.5 A. Suppose you have CT ratio as 60/5A,the primary current corresponding to the relay setting of 3.75 A is 42 A ( 60 x 3.75/5). the CDG relay will narmaly set to pickup at 130 % of set current. Therefore your Relay will pick up for a primary current of 52.6 A ( 42 x 1.30) (i.e) Exactly at your Transformer's rated primary Current

I have to add a comment to my previous post that the illustration is made for transformer capacity of 1 MVA . If it has to be for 10MVA then multiply all the calculations and illustrations by 10 . Therefore the primary current will be 525 A, CT Ratio 600/5A, primary current corresponding to 3.75A become 420 A and relay pick up current becomes 546A

u r correct sir the CT ratio is 600/5 ,so the breaker should trip if the current exceeds 450. ( 600/5*3.75=450 not 420 as u told). but the breaker trips only exceeding 500A.Is it correct

how did you conduct the test?... did you prolong the injected 450+ Amps?..

we have set the relay for 3.75 A but it trips only when the current goes beyond 500A.As we have full load current of 524A we didn't took it serious.we are keeping more than 450A always but trips only when it exceed 500A

did you use secondary current injection test or the primary current injection test?. if it is secondary current injection test, then i think it would be better to change the relay, if it would be primary current injection test then, there would some error in the CT..

i have something in my mind.. it the current injected is, say, 494Amps and you held it or steady the injected current for some, will it trip?.. my thinking is that due to the long time delay at this particular current, the relay will trip at some time, considering that you are using time overcurrent relay.

however, if it will not really trip, then there is something wrong in the relay pickup..

by the way, what kind of relay is CDG?.. is it electromechanical type?

Ah.. ok the relay is an instantaneous ovecurrent type.. (device 50) so if this is the case.. if the test is done on secondary current injection, then it is best to change the relay..

if it is electromechanical type, after sometime, these relays should be calibrated reguarly these might be happening to you..

relay set current =5(CT sec amps)/6( no of taps)*600(ct ratio).is this formula correct.

If so it would trip on 500 amps on which we experiences

Dear mr. kumar, The Cdg Relay is an electro-mechnical relay. for a 5A relay you have setting possiblities, 2.5A, 3.75A, 5A (100%), 6.25A,7.5A, 8.75A and 10A. As per facory setting the relay will have starting charecteristic 130% of set value, How ever It can be made to pickup at 110%, by adjusting the spiral type controle spring. Actual pick current can be be found out by testing the relay.

As I have already told the CDG relay will narmaly pickup at 130% of set current . With 600/5A, CT ratio, the relay will pick up at 1.3 x 600 x 3.75 /5 = 585 A. However the controle spring can be adjusted to make the relay pick-up a bit early.

Please let me know how the CT ratio is calculated after knowing the primary voltage and rated current.

I have to know the CT ratio of our main incomer (11KV) to which a 10 MVA furnace transformer (rated current-524A,P voltage )and a 1000 KVA (rated current-1333A) auxiliary transformer is connected

If rated Current of 1000Kva Transformer is 1333A , then the CT ratio narmally adopted will be 1500/5A. that is a reasonable whole number such as 1500, 1600, 1800 etc.

Starting current causes the disc to rotate it would be of the order of 103% to 110% of current setting.

Current flowing for the trip contacts to operate will be 130% of setting current.

Reseting current i.e. the disc will totally reset is 90 to 95%, say 95%

Accordingly if the setting is 3.75 Relay trip would occour at (3.75 *130%), literaly a value of 4.875 A.

4.875 when multiplied with CT ratio provides an answer to be 585 Amps.

Do take it into account that for the starting inrush current you have to select a Time Multiplier Setting (TMS) otherwise inadvertant tripping will occour at this setting or in other words relay element is NOT instantaneous.

It is allowed to overload the transformer to 110% for half hour in every 24 hours and while relay setting it can be taken advantage of, there are errors in relay and measuring /testing equipment.

Dear Mr. Kumar,

You did not mention the primary voltage of the transformer; from C.T.R. as mentioned, it is noted that primary voltage is 11Kv. The relay is an elctromechanical type and the pickup value of relay is 105% to 110% of current tap setting value. For tripping of CB it requires 130% of tap setting value. The time- current characteristic curve/ scale is given in front of the relay from which one can calculate the required tap setting. Normally during switching on of transformer, the high inrush current is sustained for two cycles i.e. 40ms for 50Hz. system; so to avoid tripping during switching on operation, the relay should be insensitive for two cycles and accordingly the T.M.S. should be set. Therfore the current tap setting value is O.K. you have to calculate the maximum fault current deliverd by the transformer and accordingly instantaneous over current element should be set. After setting the above it is recommended to carry out secondary current injection test to verify the results.