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Topic Title: Volt Drop r x z ? Topic Summary: Created On: 21 May 2008 10:02 PM Status: Read Only 
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21 May 2008 10:02 PM


Hi all.
Was just looking through the regs book and i have a question about the volt drop choices when cable's get to 25mm and above. What is the difference between r x and z? been a long time since my college days. Cheers all. Edited: 21 May 2008 at 10:04 PM by jimncc1701a 



21 May 2008 10:49 PM


From memory
r = resistive loads x = inductive loads z = mixed loads or use if not sure if either of above. Dr2366 



21 May 2008 11:28 PM


R= resistive load
x= inductive reactance z= impedance Do you all remember A.C. circuits, as cable size increases other effects need to be taken into account. 



22 May 2008 08:35 AM


Impedance (Z) = R + jX where R = Resistance and X = Reactance either positive for Inductance; or negative for Capacitance. j is the symbol for "j notation" this method is known as rectangular notation. These components are vectors.
Another common method for such calculations is by polar notation. If you search for these terms on Wiki, you'll probably get more information. Regards 



22 May 2008 08:42 AM


Do you all remember A.C. circuits, as cable size increases other effects need to be taken into account. If the manufacturers' tables are referred to for current ratings, then these will include the allowance for skin effect at 50Hz, for other frequencies e.g. VSD's which operate at higher frequencies, the VSD supplier will dictate the larger cable sizes due to this effect. Again, Wiki will probably give more information. Regards 



22 May 2008 01:48 PM


The only two things that generally affect voltage drop is temperature and power factor, when designing a circuit these are the two generally unknowns, if this is the case the equations would be;
<=16mm² (mV/A/m) x Ib x L / 1000 >=25mm² (mV/A/m)z x Ib x L / 1000 The only problem with most of the equations in BS7671 is it leads to oversized cables which when you move above 25mm it can get quite expensive, the figures in BS7671 are at a temperature factor of 1 (30 deg), this is probably ok for cables of less than 16mm but bigger cables operate at much lower temperatures therefore voltage drops are less than stated, say for instance you had a cable operating in an ambient temperature of 20 deg then you could use a reduction factor in the voltage drop of say 0.985.  As a quick example, if you knew say the load power factor then the equation would change somewhat, if you wanted a more accurate representation of the voltage drop; <=16mm² (mV/A/m) x Ib x L x cos(theta) / 1000 >=25mm² (mV/A/m)(r) x cos(theta) + (mV/A/m)(x) x sin(theta) x Ib x L / 1000 Where sin(theta) = (sqrt) 1  cos²(theta)  As I said there are other factors which affect cable selection and voltage drop such as temperature, installation methods such as grouping, parallel cables etc; but if you are just designing a single circuit like 20m of 35mm² armoured down a workshop the equation m/V/A/m(z) x Ib x L / 1000 is generous enough rather than getting bogged down in complicated calculations. regards  "Take nothing but a picture, leave nothing but footprints!"  "Oh! The drama of it all."  "You can throw all the philosophy you like at the problem, but at the end of the day it's just basic electrical theory!"  



22 May 2008 02:22 PM


Out of interest, you may also use the tabulated voltdrop values of R, X and Z for short circuit calcs on larger cables  they are effectively resistance reactance and impedance values for conductors  just remeber to divide through by the correct factor if you need to use them for the values of one core
Regards OMS  Let the wind blow you, across a big floor. 


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Volt Drop r x z ?

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