is there a different b\w ground fault CB and earth leakage
Answer Posted / s.sundar rajan.
Yes, there is diff. Ground fault means it is a dead short
of the line into the ground terminals and whereas the earth
leakage means it is not short, it is due to leakage b/w
connectors or line to neutral or poor insulation of the
conductor which has leakage current will flow to earth.
| Is This Answer Correct ? | 8 Yes | 0 No |
Post New Answer View All Answers
1. Why is the speed of shunt motor practically constant? 2. Why dc series motor use to start with heavy loads? 3. advantages & disadvantages of separately excited generator over self excited generator?
How to calculate total harmonic distortion level(THD) in VFD? and how can we reduce harmonics?
How many types of interlocking use in screw type air compressor,make-Kaser(BSD MODEL)?
limitation of conductor height on ground when we consider EMC
I want d.g.set our company
Why arc quenching is used in Diode ?
why on axial hitting up of air blast on hydrogen produced in oil circuit breaker,temperature rises of the core of the arc?
How to calculate ant test the burden of Current transformer
what do you mean by type zero control sytem ?Give an practical example of type zero contol sytem?
How can I calculate per unit (KWH) fuel (Gas/ Diesel both) consumption 1MW unit?
what is the responsibility of C license holder in industry?
different aspects of voltage stability
How to decide whether A / B / C / D curve type MCB should be used?
We have drive feeding an output contactor for a 5000 hp motor @ 5kV application. The motor FLA is 629amps x 1.25% service factor = 786.25 amps The cable that we current have in place between the drive and the output contactor is 350kcmil- MV-105 (parallel cables). Per the NEC table 310.60 (C) (69). This show the cable ampacity is at 615 amps per cable for a rating of 1230 amps. We have derated the cable for the cable tray and the multiple cables application and our calculations show that after the correction factor factor we are at 922amps. 1230 amps x .75% correction factor factor = 922.5 amps. The customer is driving us to then use another safety factor of .25% off of the already derated cable. Can you please provide a longhand calculation for your solution on this application?
What is lighting kitts.