Could someone please explain how the value of sin is calculated when working out voltage drop for cables above 16mm? Do we use the values provided in BS 7671 appendix 4 such as resistance reactance and impedance of a conductor?
Thank you
Could someone please explain how the value of sin is calculated when working out voltage drop for cables above 16mm? Do we use the values provided in BS 7671 appendix 4 such as resistance reactance and impedance of a conductor?
Thank you
To answer my own question a - there are some limiting Ze values here: Earth fault loop impedance revision of ENA Engineering Recommendation P23 e.g. 90% of 100A supplies will have a supply impedance ≤ (0.25+j0.23)Ω
- Andy.
the resultant of that of course is sqrt (.25^2 + .23^2) or about 0.34 ohms, and at a touch off 45 degrees. (it would have been be 45 degrees if the two were exactly equal.)
Mike
which if I remember my O-level trig correctly means we can equally write it as:
0.3397 = cos(42.614º)*0.25 + sin(42.614º)*0.23
(perhaps slightly perverse in the immediate context, but might compare with the appendix 4 approach)
- Andy.
which if I remember my O-level trig correctly means we can equally write it as:
0.3397 = cos(42.614º)*0.25 + sin(42.614º)*0.23
(perhaps slightly perverse in the immediate context, but might compare with the appendix 4 approach)
- Andy.
yes, the appendix 4 seems to go out of its way to avoid mentioning complex phase, which is for me at least, a much easier way to visualise it.
M.
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