10mm Cable Capacity.

[megawatt]

For a resistive load if the voltage is higher then the current is higher and the wattage disapated at the element is higher.
If you have a rated wattage at 230 volts that means you can calculate the current at 230 volts.
If the actual voltage is 240 then the current will then be higher

Bernard ... I'm sorry but there's no other way of saying this ... The quote above is quite simply the biggest load of nonsense I have ever read by anyone posting on this forum.

Please produce formulaic evidence to support what you have said ... You can use the OP's P, V and I if you wish.

MW

 

[ricicle]

Bernard's statement is absolutely correct.Re-read it again and if you want to prove me wrong I'll quite happily go through it with you.

 

[Softus]

...the biggest load of nonsense I have ever read by anyone posting on this forum.

Er, even it was, it was immediately usurped.

 

[DESL]

For a resistive load if the voltage is higher then the current is higher and the wattage disapated at the element is higher.
If you have a rated wattage at 230 volts that means you can calculate the current at 230 volts.
If the actual voltage is 240 then the current will then be higher

Bernard ... I'm sorry but there's no other way of saying this ... The quote above is quite simply the biggest load of nonsense I have ever read by anyone posting on this forum.

Please produce formulaic evidence to support what you have said ... You can use the OP's P, V and I if you wish.

MW

MW I think Bernard is quoting 'Ohms' Law

V = I x R..............eq 1

Assuming resistance is constant then

V is proportional to I

ergo - increase in V means increase in I

For resistive loads

P (power) = V x I............eq 2

Re arranging this formula gives

I = P/R

For a shower that is rated at 10.5kW at 240 V

then I = 10500/240 A hence I = 43.75 A (from eq 1)

If a shower with this rating is then run at 230 V the current will be

I = (230/240) x 43.75 Hence I = 41.9 A

The Power for this case will be 9.7 kW (from eq 2)

Oh me - well I'm just a mere electrical engineer

 

[Albert]

I think that someone is quiet confused now. If you read the original post you have only one parameter and it is the load in Watts. The best you can do with it is to give your advice so it will be on the safe side. As the voltage is not stable anyhow you cannot base your calculation on theoretical figures.
What we are interested in is the current so we can choose the adequate cable size, the difference between 240V designed unit to 230V is about 2A (in this case).
This is not so significant as the influence of thermal insulation where the de-rating factor length for of 500mm will half the carrying capacity of the cable...

 

[megawatt]

DESL Wrote:

For a shower that is rated at 10.5kW at 240 V
then I = 10500/240 A hence I = 43.75A

If a shower with this rating is then run at 230 V the current will be

I = (230/240) x 43.75 Hence I = 41.9A

BG Wrote:

If you have a rated wattage at 230 volts that means you can calculate the current at 230 volts.
If the actual voltage is 240 then the current will then be higher

DOH!

 

[DESL]

If you read the original post you have only one parameter and it is the load in Watts.

But Albert this is a variable which is dependent on the supply voltage.


The important factor is the stated rating of load i.e. the Watts at what voltage.

OK so I'm being pedantic but if one is doing design calculations the correct parameters/information should be used. Erring on the safe side when making a subsequent judgement is always advisable

 

[ricicle]

The important factor is the stated rating of load i.e. the Watts at what voltage.

Exactly.

 

[DESL]

Out of interest I was looking in an installation manual for a well known make of shower and they had a table of currents for different supply voltages.
Guess what they'd used the Wattage as the constant which showed an increase in current for a reduction of supply voltage.

"Och Jim you can't change the Law of Physics" But one shower manufacturer has!!!

 

[Albert]

I agree, and I did not say that the nice example you posted is wrong. I am just concerned that this is a DIY forum, and the advice have to be practical and simple so people can follow...

 

[megawatt]

DESL ... If you wish to maintain a constant power and you reduce voltage you must increase the current since P = V x I

In reality though a reduction of supply voltage simply produces less power and draws less current.

This is the reason I questioned Bernard's explanation ... Not because he doesn't know what he's talking about but because what he said didn't make much sense.

MW

 

[ricicle]

DESL ... If you wish to maintain a constant power and you reduce voltage you must increase the current since P = V x I

You cannot directly just increase current.To maintain a constant power with a reduced voltage you must also reduce the resistance (if that were possible) and then the current would increase.

 

[Softus]

If you read the original post you have only one parameter and it is the load in Watts.

But Albert this is a variable which is dependent on the supply voltage.

Utterly correct.

The important factor is the stated rating of load i.e. the Watts at what voltage.

Also utterly correct.

OK so I'm being pedantic but if one is doing design calculations the correct parameters/information should be used. Erring on the safe side when making a subsequent judgement is always advisable

A hat trick.

 

[DESL]

I agree, and I did not say that the nice example you posted is wrong. I am just concerned that this is a DIY forum, and the advice have to be practical and simple so people can follow...

Agree wholeheartedly but to carry out any DIY electrics some basic understanding is required. I know it's not rocket science but ....... I make good living out of people who thought they would dabble with electrics.

 

[plugwash]

In reality though a reduction of supply voltage simply produces less power and draws less current.

Indeed the real confusion comes from the fact that manufacturers specifiy power rather than resistance for resistive loads.