Kiln cable size

Supplementary question: what size fuse/MCB is in the consumer unit protecting this cable?

At the moment a 32A Type B, but obviously I'm not going to use that.

I've been put on to a kiln installer, who I will get to look at this.

Thanks to everyone for your comments.

Cheers
Richard
 
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Supplementary question: what size fuse/MCB is in the consumer unit protecting this cable?

At the moment a 32A Type B, but obviously I'm not going to use that.

It's unlikely that you've got 10mm² cable then, more likely 6mm². Did you look at my link to determine the cable size???

6mm² will not be big enough for your load, you may need to pull some 10mm² armoured cable through that duct.
 
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At the moment a 32A Type B, but obviously I'm not going to use that.
Go back and read what I wrote. I have 10mm cable. That wasn't my question.
Since your kiln draws 47.7A, you would have to have a 50A MCB. The cable would therefore have a current-carrying-capacity of at least 50A. As has been said, 10mm² cable, if that's what you have, would be pretty marginal but would probably be just about adequate with most installation methods.

.. all of which assumes, of course, that you are not going to have any significant electrical loads in the garage other than the kiln.

Kind Regards, John
 
Hi, it may be cheaper to put the kiln in the house and move into the garage :eek:

:LOL: DS
 
. I have 10mm cable. That wasn't my question.

The big question mark, and the reason that I queried it, is that it is most unusual to install 10mm² cable and protect it with just a 32A CPD.

But you are convinced it really is 10mm².

I would be interested in the reply to BAS's question though
Why do you say that it's 10mm?
 
If wattage is voltage multiplied by amperage, surely higher voltages realise lower amperages for a given wattage?
To expand on what Detlef wrote ... you are considering the wrong thing to be the 'fixed quantity'. Wattage is voltage multiplied by current and, in turn, current is voltage divided by resistance (or impedance). If (as is the case with a kiln), the load is essentially a fixed resistance, as voltage increases, current will increase, and therefore wattage will increase.

In other words, the 'fixed quantity' is the resistance of the load, not the wattage.

Kind Regards, John
 
If wattage is voltage multiplied by amperage, surely higher voltages realise lower amperages for a given wattage?
None of those parameters is constant.

If a resistance is constant then R=V/I so if V goes down...I goes down and so does P(Wattage).
 

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