Hi all,
Currently doing a course in Electrical Control Systems. We've gone through a lot of basic electrical theory. One thing we haven't really gone through though is how to select proper size wires/cables.
Was googling this topic and I came across diydata.com
Apparently, a wire with a core size of 1mm^2 can handle current up to 14A from 240V supply. Didn't think this was possible. Thought it would only take a small amount of current. How would you calculate this anyway? Is it Ohm's Law? V=IR so I=V/R (R being the resistance of the copper cable?)
Could anyone give me an example of the first time they had to select cable for a job and being a position like me, were a little unsure and what the results were?
Cheere,
Phil.
Every job is different, and there are variables to consider, but I’ll come back to that…
The simplest calculation is to divide the power rating (Watts, always given on the appliance) by the supply voltage. That tells you the designed draw of current.
If you want to size a cable correctly, you have to bear in mind the following -
The cable has to be rated at more than what the load will be demanding eg a 9kW shower will draw (9000watts/230volts) = 39 Amps. The cable has to be able to handle that, continuously.
I’d run a 10sq., which is rated at 52 Amps. People with experience of domestic installation might disagree.
And you then size your protective device so that it will protect your installation accordingly.
(statement edited out here)
So get a breaker rated between 39 and 52 Amps. Let’s say 40Amps. This will allow the current that the load needs, to pass. And will trip long before the cable is endangered.
But there’s a number of variables to consider -
a)the ambient temperature that the cables will be working in (surrounding cables in insulation means they'll be hotter than if they were clipped to a wall. This heat has obvious danger implications = ’ bad thing’ and also leads to greater resistance to the flow of current. = ‘another bad thing’)
b) the grouping factor. Cables bunched together don't cool down so efficiently.
c) the volt drop. The longer the length of run, the more the voltage drops. There’s a rule for that too.
A couple of examples and a quick formula -
Iz = In/(Ct * Cg)
Iz = continuous current-carrying capacity of the conductor - in other words, the size needed
In= Protective Device rating
Ct= Temperature correction factor
Cg= Grouping Correction factor
Ex 1
A copper cable supplying a heater that is taking 23 Amps (Watts/Volts) in an ambient temperature of 30C. It’s on it’s own, on cable tray.
Iz = 25 / (1*1)
The breaker is rated above the load being drawn, ie 25>23.
Above gives you 25 Amps carrying capacity needed. The book says that in these conditions, 2.5sq will take 30 Amps.
The (1*1) just means you don’t have to make any allowances for temperature in this job - the book quotes for conductor ratings are based on a baseline ambient of 30degreesC, and that’s what we have. And it’s on it’s own, so no correction needed.
Ex 2
Same scenario as above, but now the ambient temperature is 50degreesC
Here we have to adjust the temp. correction factor, because we’re running hotter. The obvious effect will be to increase the size of cable required. It’s still on it’s own, so don’t change the grouping factor which stays at 1.
Iz = 25 / (0.71*1) the book says if you have 50degreesC, you have to apply a correction factor of 0.71
Iz = 35.2 Amps.
You have to upgrade to a 4sq. Now, which can take 40 Amps. Because you’re in a hotter environment, 2.5sq won’t do. The job just got more expensive.
Ex 3
If you had that job to do, in 50degreesC, running on tray with say, 4 other cables you’d have to factor that in too…
Iz = 25 / (0.71*0.8) (You get the 0.8 from the regulation tables. Different situations will give you different numbers to plug in)
Iz = 44 Amps.
Now you’re up to using 6sq. which takes 51 amps. Your 4sq just got too small for the job.
The other thing to consider is Volt Drop.
You multiply...
the milliVolt drop/amp/meter for a particular cable - from the book
x
the design current of the circuit - (Watts/Volts) or Ib
x
Distance of the run, in meters
… and just hope that your answer (in Volts) is less than 4% of supply voltage. So for a domestic supply, the volt drop can’t be more than 9.2 volts. Otherwise you have to size up again.
That’s all… theory. Lads doing this stuff all the time use experience and judgement. I don’t do domestic installation so I’m just giving you the theory.
Hope that helps.
