Voltage droop

[Kris uk]

Hi guys and girls I'm new to the forum and it's probably on here somewhere but could someone please tell me a simple way of calculating volt drop as I'm not having much joy with the Maths side of things and I've got my 2391-10 in two weeks as easy as possible plse

 

[Spudnik]

Welcome to electriciansforums.net. Stick around I'm sure you'll like the forum.

 

[telectrix]

there's only 1 way. multiply the mV/m by the length then the current then divide by 1000

 

[Kris uk]

Thanks for the fast reply

 

[Kiers1970]

Happens to even the best of us mate.

 

[Sintra]

Hi & welcome.

 

[Guest123]

WElcome along.:hippy:

 

[telectrix]

the calculation gets more complex if all the load is not at the end of the circuit, say, a lighting circuit with the loads spaced along it. e.g. 10 lights, 100watts each. spaced along the run. VD at 1st light will be calculated at 1000watts for the length from source, 2nd VD will be 900watts fot it's length , etc. till the last will be 100watts at the full length of the cct.

 

[tony mc]

There was a post re max length of ring final cct but I cant seem to find it.

The calc was Max VD x 4 x 1000 / Ib x mv/A/m can someone shed some light on this one?

 

[telectrix]

post deleted... brain fart.

 

[telectrix]

The calc was Max VD x 4 x 1000 / Ib x mv/A/m can someone shed some light on this one?

that would give an approx. length of a 2.5mm RFC at 30A of 90m, which seems about right. not seen the formula before now, though.

 

[JUD]

there's only 1 way. multiply the mV/m by the length then the current then divide by 1000

It's not the only way.... ((R1+Rn) x 1.2) x Ib is another way (V = I x R)

The calc was Max VD x 4 x 1000 / Ib x mv/A/m can someone shed some light on this one?

that would give an approx. length of a 2.5mm RFC at 30A of 90m, which seems about right. not seen the formula before now, though.

It's in the Electricians Installation Design Guide. For a radial it is (MaxVD x 1000) ÷ (Ib x mV/A/m)

 

[MarkieSparkie]

Hi, welcome to the forum.

 

[i=p/u]

the mv/A/m values for say 2.5mm is 18. so can use this as your R1+R2 per meter of 2.5mm

so an immersion heater circuit would be 11.5 x 1000= 11500 / 13.04 x 18 =11500 / 234.72 = 48m length

It's not the only way.... ((R1+Rn) x 1.2) x Ib is another way (V = I x R)



It's in the Electricians Installation Design Guide. For a radial it is (MaxVD x 1000) ÷ (Ib x mV/A/m)

 

[Kris uk]

Can any one tell me how to calculate the milli volts per meter from theese measurements load current 45amps .circuit length 80m .live conductors 10mm.m ohm/m at 20 degrees c 1.83

 

[telectrix]

deleted due to mis reading post

 

[JUD]

Can any one tell me how to calculate the milli volts per meter from theese measurements load current 45amps .circuit length 80m .live conductors 10mm.m ohm/m at 20 degrees c 1.83

VD = (((R"1+R"n) x Cr) x L x Ib) ÷ 1000
VD = (((1.83+1.83) x 1.2) x 80 x 45) ÷ 1000 = 15.81V

Cr = Correction factor for 70°C
R"1 = mΩ/m of line conductor
R"n = mΩ/m of neutral conductor

 

[Kris uk]

Thanks dude that I can understand

 

[Des 56]

tell me a simple way of calculating volt drop as I'm not having much joy with the Maths side of things and I've got my 2391-10 in two weeks as easy as possible plse

You asked for easy:28:
Measure the voltage at the supply and deduct the voltage at the extremity,simples :blush5:​

 

[telectrix]

bloody 'ell ,des. you just made me choke on my beer.

 

[Kris uk]

Sorry to be a pain jud but where did y get the correction factor is it always 1.2

 

[JUD]

Sorry to be a pain jud but where did y get the correction factor is it always 1.2

The values of resistance in the regs are at 20°C, so to correct these values for 70°C then yes you would use 1.2

Resistance increases with temperature by 0.4% per 1°C from its 20°C value so an increase in temperature of 50°C (the difference between 20 and 70°C) would increase the resistance by 20% (0.4% x 50°C) hence you multiply the 20°C resistance value by 1.2.

If you want to correct measured values then you would need to know the temperature at the time you measured them.

Let's say the ambient temperature was 25°C when you took an R1+Rn measurement. This time the temperature difference is 45°C (0.4% x 45°C = 18% so multiply by 1.18).

Hope that makes sense.

 

[DurhamSparky]

Hi, Welcome to the forums