volt drop rule of thumb

[Sparky Ninja]

The 1.2 is the multiplying factor used in volt drop.
If you measure resistance R1 + R2 at 20C or use tables measured at 20C you have to multiply the figure by 1.2 to give you the resistance at operating temperature 70C.
This R1+R2 *1.2 is then used in the equation V=IR.
.....I..... being the design current for the circuit or preferably the measured load current.

V is the volt drop along the cable.

The verification of voltage drop will require a measurement of the 'circuits impedance'. Therefore it will be R1&Rn, and not R1&R2.

 

[johnboy6083]

ive left my electricians design guide in the van, and i know that there is a formula for working out resistance of a conductor from mV/A/m figures. Does anyone have the calc to hand please?

 

[La Poste]

The verification of voltage drop will require a measurement of the 'circuits impedance'. Therefore it will be R1&Rn, and not R1&R2.

Spot on.
Thanks for that.

 

[JUD]

ive left my electricians design guide in the van, and i know that there is a formula for working out resistance of a conductor from mV/A/m figures. Does anyone have the calc to hand please?

The volt drop figures are the values of resistance of the line and neutral conductors per meter at 70°C.

The formula to work out the volt drop figures is (R"1 + R"n) x Ct

R"1 = resistance per meter of line conductor at 20°C.
R"n = resistance per meter of neutral conductor at 20°C.
Ct = correction factor for operating temperature of cable.

Example

Resistance per meter of 2.5mm² is 7.41mΩ/m so the formula would be (7.41 + 7.41) x 1.2 = 17.78 (rounded up to 18).

This is the value of the mV/A/m for 2.5mm² from the volt drop tables.

 

[johnboy6083]

Does this wrk with cables larger than 16mm2? will this be afected by reactance?ive transposed tha above to get miliohm/m = mV/A/m X 0.833
2

is this correct? ive onl;y got my BGB to test this with, and i dont think that it lists the resistance of cables in it.

Can somebody post me a few values for resistance of single core cables?

Thanks,

John

 

[johnboy6083]

ive just found my 16th edition OSG, so used the resitance values there. my above calc works oncables rated at 70degrees c. does anybody have the derating factors for other temps? thanks,


john

 

[gap30]

i am a bit stuck on volt drop myself, could someone break it down a bit for me please

i is the design current ie 300w

r is the resistance say 10m x 2.5 7.41 x2 x100 x1.2x10/1000 = 0.178

v = ixr 300w x 0.178 = 53.4 volts?

 

[JUD]

i am a bit stuck on volt drop myself, could someone break it down a bit for me please

i is the design current ie 300w

r is the resistance say 10m x 2.5 7.41 x2 x100 x1.2x10/1000 = 0.178

v = ixr 300w x 0.178 = 53.4 volts?

Design current (Ib) isn't measured in watts, it's measured in amps i.e 300W ÷ 230V = 1.3A (I = P ÷ V)

Volt drop = (mV/A/m) x Ib x L so for 10m of 2.5mm² carrying 1.3A it would be (18 ÷ 1000) x 1.3 x 10 = 0.23V

 

[gap30]

Design current (Ib) isn't measured in watts, it's measured in amps i.e 300W ÷ 230V = 1.3A (I = P ÷ V)

Volt drop = (mV/A/m) x Ib x L so for 10m of 2.5mm² carrying 1.3A it would be (18 ÷ 1000) x 1.3 x 10 = 0.23V

how do you get the 18 figure?

 

[JUD]

how do you get the 18 figure?

From the Volt Drop table in BS 7671 for 2.5mm²

 

[gap30]

From the Volt Drop table in BS 7671 for 2.5mm²

ok just got that so 20m x 2.5 and 300w load

300/230 = 1.3a

v = ixr

18/1000 x 1.3 x 20 = 0.46v

so in the exam they are going to provide you with the volt drop table figure i assume?

 

[JUD]

ive just found my 16th edition OSG, so used the resitance values there. my above calc works oncables rated at 70degrees c. does anybody have the derating factors for other temps? thanks,


john

Here's a tip. Remember this...


The rating factors for cable and ambient temperatures are worked out like this:

For a rise or fall in temperature you multiply or divide by 1 + (0.004 x temperature difference).

That's where the 1.2 comes from.

You start with a resistance at 20°C and then multiply it by (1 + (0.004 x 50)) to get to 70°C (0.004 x 50 = 0.2). 50 is the temperature difference between 20°C and 70°C.

And the other way, if you start with a resistance at 70°C, this time you divide by (1 + (0.004 x 50)) to get to 20°C.

 

[johnboy6083]

im remebering the theory now, from AC theory in college. Ro=1x fish shaped symbolxt.

cheers mate.

great help.

its one thing learning the formulae, but another about knowing when to put it in practice.

 

[JUD]

ok just got that so 20m x 2.5 and 300w load

300/230 = 1.3a

v = ixr

18/1000 x 1.3 x 20 = 0.46v

so in the exam they are going to provide you with the volt drop table figure i assume?

Yes. All info should be provided with these types of questions.