Really stupid voltage drop question

[john2512_00]

Hey all, reading through the 17th trying to get my head around some areas of it. I did physics at school and some at uni so electrical calcs usualy dont bother me the slightest.

BUT, If were calculating voltage drop over a reasonable length radial, for example twin and earth or SWA feeding a garage, say its a 20 meter run, do you calculate voltage drop using 20m, or 40m to take account for the path down to garage and then back up to source again?

thanks in advance

J

 

[balbecdaze]

Good question, keep up the thinking! Now I'm not a "time served" electrician or anything but to my way of thinking you are interested in the voltage at the load and so its 20m you would use. Its an AC circuit (obv I know but that's the way I thought about this)

 

[oldAL]

were you paying attention for any of the 4 years?

 

[Richard Burns]

Agree with balbec you are considering the voltage drop that will occur at the end of the circuit.
So there is the resistance of the cable from source to load over which to consider how much voltage is dropped.
Consider the cable as a resistor in series with, and before, the load.
If you took this back to the source (through the load) then by definition the voltage drop will be 230V since the source end connection is to neutral at 0V

 

[telectrix]

the mV/A/m figures in BS7671 take into account the L and N, so in your case, you use 20m as the length to calculate the VD.

 

[ackbarthestar]

the mV/A/m figures in BS7671 take into account the L and N, so in your case, you use 20m as the length to calculate the VD.

Yep, i agree.

The circuit is made up of basically three series resistors. R1, load, Rn. To determine the total current flow in the circuit you will need to take into consideration all three resistors. so the volt drop generated at any point Is a combination of both phase and neutral.
If you use tables 4xx.B from app 4 of BS7671 that includes both conductors. If you choose to calculate it by using the resistance of each conductor times the current flow then you need to multiply by 2

 

[lauriehurman]

Come on chaps - you're going to give us University types a bad name. The one thing you should have learned in 3 years is how to think clearly

You have a conductor, it has a resistance. You have a load, with it's own resistance and then you have another conductor, again with a resistance. That's the circuit.
The voltage drop refered to in the regulations is the voltage available to the load, i.e. 230 minus the drop in each of the two conductors.
Think about it; if there was no voltage drop in the neural conductor what would drive the current back to the CU?

The table in the "On-site-guide" is for a CABLE, i.e. two conductors, out and back.

Laurie

 

[HandySparks]

There will be voltage dropped over both the line and neutral conductors. The voltage at the load will be the supply voltage less the drop in both conductors.

However, the tables in Appendix 4 of the regs take into account the drop in both conductors. Read para 6 on page 312 of BS7671:2011. So in your calc, the VD (in mV) is the tabulated drop (in mV/A/m) multiplied by the circuit length (in m, 20 in your case), multiplied by the design current (in A).

Edit: Clearly I'm too slow this morning and there are too many of us sitting in front of our computers and not out grafting. My excuse is that I'm catching up on the paperwork and ordering stuff.

 

[telectrix]

my excuse is that the van's broke. stupid immobiliser is locked out. hope to sort it for monday.

 

[jaresquire]

Sort it for Friday, slacker!
I'm at home doing, er, admin.

 

[Guest55]

i'm at home counting the £52,000 that the training providers say i must be earning each year.
;-D