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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?)

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    Andrew James Skinner: 
     

    Thanks Andy, my questioning was over the wording of the application of factor to  the mA/V/M figure  - multiplying it by 0.95 make it smaller, dividing by 0.95 makes it larger

    The mv/A/m tabulated figure is multiplied by 0.95 making it smaller and thus gaining advantage in terms of overall Ud. However, as Zoom indicates, there is no real advantage to be had so just use the tabulated value. 
    Ud is calculated using Ib, so depending on what limits you are working to use that value rather than In as it may result in a smaller csa. 

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    Andrew James Skinner: 
     

    Thanks Andy, my questioning was over the wording of the application of factor to  the mA/V/M figure  - multiplying it by 0.95 make it smaller, dividing by 0.95 makes it larger

    The mv/A/m tabulated figure is multiplied by 0.95 making it smaller and thus gaining advantage in terms of overall Ud. However, as Zoom indicates, there is no real advantage to be had so just use the tabulated value. 
    Ud is calculated using Ib, so depending on what limits you are working to use that value rather than In as it may result in a smaller csa. 

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