volt drop rule of thumb

[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.