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big willie

I have been asked to connect up a mobile home on a large transportation site. The run of cable is approximately 250meters. In the regs book for 25mm armoured cable there are three values x y and z which one should I use
 
I have been asked to connect up a mobile home on a large transportation site. The run of cable is approximately 250meters. In the regs book for 25mm armoured cable there are three values x y and z which one should I use


I have replied to this question, on the end of another thread.......


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Originally Posted by big willie
I have been asked to connect a mobile home on a large transport site. The supply cable run is approx 250 meters. When I try to work out the voltage drop for the swa cable from size 25mm upwards there are three values in the regs book x,y,and z which shoul I use

Thanks anybody




"Upto 35mm, I continue to use the resistive (r) component for calculating the VD. By moving to the impedance (z) component you will end up with a pessimistically large conductor size.

Going above 35mm and bringing reactance & impedance in to play makes it all a bit more complicated as power factors (if known) and temperature corrections come into play.

Have a read from page 312 (new green book) onwards (appendix 4)."
 
Last edited by a moderator:
Like Lenny, in normal circumstances for <35mm I only usually apply the resistive component. However, if it is known that the load is heavy, continuous and complex e.g. non-sinusoidal, as in a switched mode converter then the reactive components should be considered.
 
Another volt drop calculation that i could do with some help with is the 3 phase VD

I fully understand the single phase vd calculations but just wanted to get a confirmation when using 3 phase.

e.g

Say a large load of 30Kw is 200m from the point of distribution and you need to find out the volt dropped.

Do you use 30000/230 - 130 Amps or

do you use 30000/415 - 72 Amps????


Now lets say using a 50mm conductor the differences are below.

230 ---- 0.84 x 130 X 200 (m) /1000 = 21.84 volts dropped
415 ---- 0.84 x 72 x 200 (m) /1000 = 12.0 volts dropped

At 415 v this would be approx 3% vd!!

Have a got this correct or would some body like to correct me?

Just looking for some help................

many thanks
 

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voltage drop calculations
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