Rules and Guidelines

Want to participate in the discussions? Please read our Community Rules and Guidelines. Need some help? Visit Support and Answers.

State Not Answered
Locked Locked
Replies 35 replies
Subscribers 31 subscribers
Views 4643 views
Users 0 members are here

Options

You may also be interested in

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 over 3 years ago

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

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

0 Former Community Member over 3 years ago

Thanks Everyone, if I can just summarise all of the answers:
Ct should be =((230+50)- (0.82^2-(2^2/21^2)) * (50-30))/(230+50) = 0.952
Ct is multiplied by the mV/A/M value in this instance decreasing its value and decreasing the volt drop.
This is because the load on the cable is less than maximum (under running the cable) and decreasing its value. So for this cable running at 21A with the 0.82 multiplier the CT would actually be 1.0234 at 50 degrees (increasing the mV/A/M value (but only just).
Running at 2A the 0.82 multiplier is “decreased” to 0.95 (closer to 1) because there is less self-heating in the cable, decreasing the mV/A/M value (but only just).
So all in all the specified mV/A/M is at max current and max conductor temp, ambient temp has very little effect and cable and can almost be ignored.
So a 90C Cable could still be run at max current in an 80 degree ambient albeit with a increase in voltage drop
Cancel
Vote Up 0 Vote Down

Cancel

Reply

0 Former Community Member over 3 years ago

Thanks Everyone, if I can just summarise all of the answers:
Ct should be =((230+50)- (0.82^2-(2^2/21^2)) * (50-30))/(230+50) = 0.952
Ct is multiplied by the mV/A/M value in this instance decreasing its value and decreasing the volt drop.
This is because the load on the cable is less than maximum (under running the cable) and decreasing its value. So for this cable running at 21A with the 0.82 multiplier the CT would actually be 1.0234 at 50 degrees (increasing the mV/A/M value (but only just).
Running at 2A the 0.82 multiplier is “decreased” to 0.95 (closer to 1) because there is less self-heating in the cable, decreasing the mV/A/M value (but only just).
So all in all the specified mV/A/M is at max current and max conductor temp, ambient temp has very little effect and cable and can almost be ignored.
So a 90C Cable could still be run at max current in an 80 degree ambient albeit with a increase in voltage drop
Cancel
Vote Up 0 Vote Down

Cancel

Children

No Data

Join us to get the best from IET EngX.

Joining EngX lets you personalise your experience so you stay up to date on the topics that interest you, plus you’ll be able to make connections who are looking to collaborate, exchange ideas and more.