ColinMcDonnell finds an apparent anomaly in BS7671 cable sizing calculations and asks whether they are correct? His question is answered by the NICEIC (Theme – Regulations & Legislation):
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Question: A colleague asked me to check a cable calculation based on the attached sheet (click on the link below to see it), which I believe is from an Annex to BS7671. We came across what seem to be a number of anomalies and I would greatly appreciate any guidance that you can offer.
1) Section 7.1 states that the design mV/A/metre(z) is equal to the vector sum of the resistive and reactive mV/A/metre figures. Should this not mean that the mV/A/metre(z) is the square root of the sum of ((resistive mV/A/metre)xCos(Phi) squared) and ((reactive mV/A/metre)xCos(Phi) squared)?
2) The equation in 7.2 does not use a >square root of the sum of the squares< format - merely an arithmetic sum of the (resistive x Cosine) and (reactive x sine) values. This seems intuitively wrong. If I apply the formula using a >square root of the sum of the squares< format, I get a reduction in calculated voltage drop, but if I apply it as written in Section 7.2 I don't see that reduction. Is there an error in the formula as presented in the attached sheet or am I missing something?
3) 2) Equation 10, Section 7.1 is stated as being applicable for ambient temperatures of 30oC and above. Is this because the value of 30oC is contained in the formula? Could you remove the 30oC value and use the formula down to 0oC?
Answer: Firstly, this is a precise calculation for the correction for operating temperatures, and also correction for load power factor. These calculations are concise and if correctly equated would bring your voltage down to an absolute minimum and, therefore, may concisely size your cable for its loading.
However, if you use the current carrying tables for cables, and if load, and if power factor and temperature are not factors, then the 'z' values should be used for mV. This sum will take into account all circumstances and the volt drop will be greater because of this factor.
The calculated value will always come back as less because more accurate information has been used to evaluate the equation.
In short, there are no errors in the calculations provided. However, due care must be taken when calculating using this formula - as outlined above.
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