This discussion is locked.
You cannot post a reply to this discussion. If you have a question start a new discussion

Appendix 4, section 6.1 equation 6

Former Community Member
Former Community Member

 

85b1fc637cfb5e3825ed2b73d8b54527-original-image.png

Hi I am working through equation 6 for an ambient temperature of 50 degrees for 1mm 90degC thermoplastic cable (Table 4E2A)

11d17c7d56aee28e7d23803363b44f10-original-image.png

I get a correction factor of 0.95 for a 2A load. 

Do i divide the mV/A/m by this factor ? 

The text says multiply, but that would mean the resistance decreases with the increase in temperature (or have I got the sum wrong?)

Parents
  • Sorry -I was a bit quick with the typing when I wrote that, its not the clearest.

    Yes these are not percentage changes, but absolute multipliers /dividers, depending  what you are calculating and you need to keep your wits about you.

    The book value for  voltage drop assumes fully loaded, so 30C ambient and 70C in the copper core or r 90C for some cables. 

    Here your ambient is raised, but you will see less than the 40C rise, a lot less than a 40C rise actually (that is the thing with squared currents are really powers - I squared R and all that) The assumption is that temp rise is linearly proportional to power dissipated in the cable, so all the correction factors that are linear scaling factors for current consumption become squared for heating and power loss..

    mike

Reply
  • Sorry -I was a bit quick with the typing when I wrote that, its not the clearest.

    Yes these are not percentage changes, but absolute multipliers /dividers, depending  what you are calculating and you need to keep your wits about you.

    The book value for  voltage drop assumes fully loaded, so 30C ambient and 70C in the copper core or r 90C for some cables. 

    Here your ambient is raised, but you will see less than the 40C rise, a lot less than a 40C rise actually (that is the thing with squared currents are really powers - I squared R and all that) The assumption is that temp rise is linearly proportional to power dissipated in the cable, so all the correction factors that are linear scaling factors for current consumption become squared for heating and power loss..

    mike

Children
No Data