Heat loss in cable

[joner1]

Hi Guys, first of all sorry if this is posted in wrong section.
I am doing a building services design course at southbank university and have an assignment on cable sizing for a submain, I need to calculate the heatloss in this cable.

As i remember heat loss is calculated using P=I*R,
Do i have to calculate the resistance of the Line AND Neutral conductor for this calculation or is it just the resistance of the Line conductor?

Many thanks in advance

 

[telectrix]

line and neutral IMO. if you think about it, the circuit consists of both L and N. heat loss is dissipated in both conductors.

 

[malcolmsanford]

Think you may have things a little mixed up here son, that formula is ohms law for power (watt) and should be I2 (square)R

Are you referring to VD and the use of both conductors, or the impedance (R1 + R2)

 

[telectrix]

what's a star between friends?

 

[Chr!s]

Try P= V.D X Ib

Regards Chris

 

[joner1]

Thanks for the replies everyone, Malcolm, thanks for clearing that up(I realise i put multiply instead of ²), i seem to remeber years ago being taught that I²R losses were heat losses, but i've never had to calculate heat loss in a cable before so didn't know if it applied.
Telectrix, I thought that it was L+N too but with all the other replies i'm confused now

 

[Chr!s]

Well either I2R or P =V.D X IB will be correct, use the mV/a/m tables in BS7671

Regards Chris

 

[joner1]

Thanks Chris, Will try them both and see how the answers compare.

 

[i=p/u]

so say 6.9v x 4A = 27.6 what??? degreess???? hmmm

is this important for an electrician and in what way??? curious geroge

 

[Chr!s]

so say 6.9v x 4A = 27.6 what??? degreess???? hmmm

is this important for an electrician and in what way??? curious geroge

Watts cost money

 

[joner1]

I=p/u the answer will be in watts, not really sure how important this is but it's part of my assignment and so i need to answer it,

 

[Chr!s]

Say your losses are 1500 watts for 12 hours 360 days a year

so 12 x 360 = 4320, lecy costs 10p per kw/hour

so 4320x10x1500/100000 = £648

May of be worth installing that bigger cable after all.

 

[joner1]

OK have got a different answer from each formula-

Using I²R i have 65.2² * 0.0414 = 176W
Using V.d * Ib i have 2.73 * 65.2 = 118W

 

[Chr!s]

OK have got a different answer from each formula-

Using I²R i have 65.2² * 0.0414 = 176W
Using V.d * Ib i have 2.73 * 65.2 = 118W

0.0414? what cable and size are you using?

 

[i=p/u]

i thought we encouaged heat loss...... from cables...