Max power on a new radial electrical circuit?
I have a new conservatory and the electrician has installed a radial circuit, protected by a 20A fuse with an RCD. On the circuit are three double sockets, a fused light switch, and the 3kW underfloor heating.
Will my MCB trip when my wife does the ironing?
My calculation is 3kW heating, + 2.4kW iron -> 5.4kW. 5400/240 = 22.5A.
The electrician who did the work says no, there are no domestic applicances which draw anywhere near 10A. Two other electricians say yes, it will trip.
Thanks
In response to the two answers that I have already received - thank you! I am trying to get enough facts to get the conservatory company to get another electrician in to get the work properly done, before plastering takes place! So if you are an electrician I would appreciate if you could say so - it gives me more ammo!
UK Irons: (from John Lewis - major UK department store)
http://www.johnlewis.com/230412917/Product.aspx
http://www.johnlewis.com/230232685/Product.aspx
There are a lot at 2.4, 2.6 kW.
Hopefully he does better work than his knowledge, or lack of, might indicate…
Talk soon,
Joe Beaven
P.S. To get a free report that tells you exactly how to make sure you hire the best electrician for your needs in the UK, please enter your details below:
SECURE & CONFIDENTIAL
}}}~~~~> Says:
–overstrike–Your electrician seems to be quite wrong about domestic appliances.–/overstrike–
I think that you must be in the UK.
I am a master electrician in the US.
Here we do not call circuits radial because we do not use ring circuits at all so it is assumed that they are "radial".
A 2.4 KW iron designed to be supplied at 120 volts consumes 20A, 100% of the available wattage in a 20A circuit, the larger of the two common circuit sizes for general use. (15A, 20A) We do not have irons like this in the US.
E = I R
120V = 20A x 6ohms
R = 6 ohms
H = I^2 R t
H = 400 x 6 x t
2400 joules / second
a 2.4KW iron designed for use on a 240V circuit would draw only 10A.
E=IR
240V = 10A x 24 ohms
24 ohms
H = I^2 R t
H = 100 x 24 x t
H/s = 2400
Joules per second = 2400
watts = 2400
So why would a UK iron be nearly twice as powerful as a US iron? Are you sure that your iron is that KW?
In the US we use the NEC (national electrical code) The NEC (paraphrased) states that no load may use more than 80% of a circuit unless it is a dedicated circuit. Your heating load is less than 80%. So your electrician may well be right.
References :
RE: LINKS _ Looks like you are right and your electrician is wrong!
Posted on January 11th, 2009 at 3:37 pm
Dont touch that dial Says:
Hopefully he does better work than his knowledge, or lack of, might indicate…
References :
30yrs electrical & instrumentation consulting
Posted on January 11th, 2009 at 3:44 pm
hoboson Says:
The heating should be on an isolated circuit by itself, lighting and power circuits on a separate 15 amp circuit.
References :
Electrician
Posted on January 11th, 2009 at 4:25 pm
big f Says:
i am a electrician. he is not. if you see him again ask him to show you his license. we have to carry it on every job. also he should have pulled a wiring permit with your town. your heat should be on separate circuit. and depending on how large your room is and how many plugs are installed he might need a circuit for each.
a clothes iron should only be about 8 amps
a vacuum 12 amps
coffee maker 8 amps
microwave 12 to 15 amps
toaster oven 13 amps
you get the picture.
References :
Posted on January 11th, 2009 at 4:51 pm