What shower power?

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I need to replace my leaking and slightly melted electric shower. It's a Newlec NL9216 Calypso Plus. The label on the top gives a supply rating of 9.5kW and a loading rating of 8.7kW. The supply cable is 6mm, only 5m long and not through insulation. Protected by a B32 Wylex breaker. I think a loose connection on the neutral to the elements caused the overheating.

Can I replace with a 9.5kW shower, or should it be a 8.5kW max?

Thanks for replies.
 
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I believe 6mm cable can only handle a 7.5 kw shower but wait til someone else with the knowledge replies.
 
I believe 6mm cable can only handle a 7.5 kw shower but wait til someone else with the knowledge replies.
If the cable did not have to be 'de-rated' (e.g. because it was running through insulation, or in conduit), 6mm² cable would be OK for a shower up to about 11 kW.

However, the 32A breaker the OP has is not OK for a shower bigger than about 7.5 kW - it would need to be upgraded to 40A or 45A (cable permitting) for a 9.5 kW shower.

Kind Regards, John
 
Well thanks for the replies, but those aren't the answers I want.

I've only been in this place for a couple of weeks, and the shower's been used maybe 10 times and I don't suspect the previous owner of doing anything dodgy with it when he moved out.

So the 9.5kW shower has been working fine with 6mm cable and the 32A breaker, presumably for several years, presumably without tripping.

Ideally I would like to replace with a 9.5 unit, but it would be a pain if the new one started tripping the breaker.

Realistically, I'd probably get a 8.6kW unit, but I wanted some clarification first before buying.

Thanks for your thoughts.
 
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So the 9.5kW shower has been working fine with 6mm cable and the 32A breaker, presumably for several years, presumably without tripping. ... Ideally I would like to replace with a 9.5 unit, but it would be a pain if the new one started tripping the breaker.
A 9.5 kW shower will take at most just under 40A. As I said, unless the 6mm cable is buried in insulation or suchlike, it's fine for that current. Also, a 32A MCB will stand 40A for a fair while (probably long enough for a shower) before tripping - so, although it's 'bad' to have a 32A MCB, it's not very surprising that it never trips.

Probably all that needs to be done in order for you to 'put things right' (for a 9.5 kW shower) is to have the 32A MCB upgraded to a 40A or (cable permitting) 45A one.

Kind Regards, John
 
Thanks for your advice John. I think I will go for the 9.5 and hope the current set up will suffice.
Can I ask what the label on the old unit means where it says "Supply 240V 9.5kW, Loading 230V 8.7kW. ?
Many thanks.
 
Thanks for your advice John. I think I will go for the 9.5 and hope the current set up will suffice.
It's your decision, but I have to tell you that it's not good practice to 'deliberately overload' the 32A breaker. It might get hot, hence theoretically might catch on fire, and certainly might have a reduced lifespan. If (albeit very unlikely) there were a fire, your insurers may well not be too impressed by the overloaded breaker!
Can I ask what the label on the old unit means where it says "Supply 240V 9.5kW, Loading 230V 8.7kW. ?
I presume that it simply means what it says - that the shower will produce 9.5kW if the supply voltage is 240V (about 39.6A current) and 8.7 kW if the supply voltage is 230V (about 37.8A current).

Kind Regards, John
 
Oh alright then, I'll also order a B40 breaker.

Thanks again.
 
I presume that it simply means what it says - that the shower will produce 9.5kW if the supply voltage is 240V (about 39.6A current) and 8.7 kW if the supply voltage is 230V (about 37.8A current).
Yes, its a heating element, which is going to be basically ohmic.
- So for a fixed resistance, I, element, as the voltage,V, goes up (more push) so does the current, I, (more flow) and the power, P, (quality used). V=IR, I=V/R. P=IV , P = V^2 / R

Daniel
 
I presume that it simply means what it says - that the shower will produce 9.5kW if the supply voltage is 240V (about 39.6A current) and 8.7 kW if the supply voltage is 230V (about 37.8A current).
Yes, its a heating element, which is going to be basically ohmic. - So for a fixed resistance, I, element, as the voltage,V, goes up (more push) so does the current, I, (more flow) and the power, P, (quality used). V=IR, I=V/R. P=IV , P = V^2 / R
Quite so - but I assumed the OP just wanted an answer, not the mathematical explanation :)

[I take it that by 'ohmic' you mean 'resistive' - any sort of impedance is measured in ohms!]

Kind Regards, John
 
I take it that by 'ohmic' you mean 'resistive' - any sort of impedance is measured in ohms!
Indeed, but you had already given him that.

Ohmic, as in, follows Ohms law some things do not.


Daniel
 
I feel sorry for the scientist who discovered super conductivity.

Ohm less so he has no where to live......
 

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