jps1
Electrical
- Mar 1, 2011
- 10
From the attach JPEG file with single transformer my understanding as it relates to calculating the relay 50/51 elements are as follows:
1) Determine max transformer FLA, NATURAL COOLING (55deg.C)1mva/ (1.73x3.3kv)=175A.
2) Select phase OC relay(51) pickup in the range 150-300% of transformer OA rating, ie, (1.5xIfl)<(51 pickup)<(3xIfl)--(1.5x175)<(51 pickup)<(3x175)gives; (263A)<(51 pickup)<(525A)---Select pickup setting = 280A, 160% FLA.
3) Select HV CT ratio above Trxf. FLA: 400/5A
4) 51 Pickup in sec. current: 280/(400/5)= 3.5A
5) Assume a max trxf. sec. SC of 30ka.
6) Max. primary infeed to a sec. fault = 30000/(3300x420)=3.8ka
7) Select device 50 setting >1.6x3.8ka=6.1ka
8) Element 50 setting sec. = 6.1/(400/5)=76A delay =0
Finally to my question:
Assuming the the above calculation is correct!!!
1) With two transformers in parallel as in the other JPEG image would line one above be x2?
2) Would we use the max secondary fault current for one transformer or do we add the fault current for both trxf. secondaries.
3) With a configuration of a transformer and motors in parallel as shown in one of the JPEG image; would we add both the trxf. max sec. fault current and the motor bus fault current to determine the infeed max current? In addition, one motor is used as a spare. It only comes on when one of the other motor is being switch. Can we use a factor of say, 25% FLA for this motor to add to the bus total current draw?
I know the question is a bit long but I would love to clarify the points raised.
Regards
JPS
1) Determine max transformer FLA, NATURAL COOLING (55deg.C)1mva/ (1.73x3.3kv)=175A.
2) Select phase OC relay(51) pickup in the range 150-300% of transformer OA rating, ie, (1.5xIfl)<(51 pickup)<(3xIfl)--(1.5x175)<(51 pickup)<(3x175)gives; (263A)<(51 pickup)<(525A)---Select pickup setting = 280A, 160% FLA.
3) Select HV CT ratio above Trxf. FLA: 400/5A
4) 51 Pickup in sec. current: 280/(400/5)= 3.5A
5) Assume a max trxf. sec. SC of 30ka.
6) Max. primary infeed to a sec. fault = 30000/(3300x420)=3.8ka
7) Select device 50 setting >1.6x3.8ka=6.1ka
8) Element 50 setting sec. = 6.1/(400/5)=76A delay =0
Finally to my question:
Assuming the the above calculation is correct!!!
1) With two transformers in parallel as in the other JPEG image would line one above be x2?
2) Would we use the max secondary fault current for one transformer or do we add the fault current for both trxf. secondaries.
3) With a configuration of a transformer and motors in parallel as shown in one of the JPEG image; would we add both the trxf. max sec. fault current and the motor bus fault current to determine the infeed max current? In addition, one motor is used as a spare. It only comes on when one of the other motor is being switch. Can we use a factor of say, 25% FLA for this motor to add to the bus total current draw?
I know the question is a bit long but I would love to clarify the points raised.
Regards
JPS
