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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?)

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0 Former Community Member over 3 years ago

Thanks David, the cable for the application is actually a EN 50306 cable - 120degC, I just picked a 90c as an equivalent so I could work though the sum using everyday values from the tables for my understanding before adapting it. The load is electronic, pulling about 1.6A
I believe mV/A/M can be calculated (for single phase) from 2 x ohms/KM x K, where K for 70 degrees (for example) is (234.5+70)/(234.5+20)=1.197 so would be 46 for my 20 ohm/KM cable.
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0 Former Community Member over 3 years ago

Thanks David, the cable for the application is actually a EN 50306 cable - 120degC, I just picked a 90c as an equivalent so I could work though the sum using everyday values from the tables for my understanding before adapting it. The load is electronic, pulling about 1.6A
I believe mV/A/M can be calculated (for single phase) from 2 x ohms/KM x K, where K for 70 degrees (for example) is (234.5+70)/(234.5+20)=1.197 so would be 46 for my 20 ohm/KM cable.
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