Shower overheating no loose connection

[Ampo]

Hi all. Shower has melted on top left corner. I thought it would likely be a loose connections on the feed. On taking the cover off the 10mm coming in and connector block looks fine and is tight and I soon realised the burn Mark is from the live going from the block up to the top of the heating/pump canister. Connections are all tight. I have run it for 30 seconds and the live gets to hot to touch but the incoming 10mm is still cold so can only guess that it's an issue inside the heating unit that is then passing the heat down the live. Anyone ever come across this? I was just expending a simple loose neutral. Triton say there have been no recalls and were pretty uninterested. Sounds like a new unit needed but has left me curious if anyone has seen this before?

 

[Deuce]

No! It's a fixed resistance load, the elements resistance will not change. So if the voltage drops then the current will drop, and vice versa.

Ohms law, R (the element resistance) is the one that will not change. I = V/R

It's Ohm's law - I = V/R

R is fixed so as V gets lower, so must I.

You can think of it as the voltage pushing the current through the resistance. A lower voltage means it pushes less current through.

If you run some examples with say a resistance of 10ohms and voltages of 240 and 220, you'll see the current drop.

Edit: beaten to it!

Ok, I accept the weight of opinion and I'm off to google until I've got my head around this

On this basis then, lower voltage would mean lower amps so that the element would put out less heat?

My confusion comes from us once plugging an electric pool heater made in the UK, into a 220v system overseas and the cable quickly overheated. Why didn't the cable stay cool and the heater simply perform less well than it would at 240ish volts? It was a ~56a heater of a 63a supply.

 

[Jim_e_Jib]

Ok, I accept the weight of opinion and I'm off to google until I've got my head around this

On this basis then, lower voltage would mean lower amps so that the element would put out less heat?

My confusion comes from us once plugging an electric pool heater made in the UK, into a 220v system overseas and the cable quickly overheated. Why didn't the cable stay cool and the heater simply perform less well than it would at 240ish volts? It was a ~56a heater of a 63a supply.

Yep, that's right-lower voltage and lower current would mean lower wattage and less heat output.

I'm not sure about your pool heater - maybe the cable just wasn't able to handle such a high current or perhaps there were other factors involved - damage in transit? Was this for the aquarium...?

 

[Deuce]

Yep, that's right-lower voltage and lower current would mean lower wattage and less heat output.

I'm not sure about your pool heater - maybe the cable just wasn't able to handle such a high current or perhaps there were other factors involved - damage in transit? Was this for the aquarium...?

Not for the aquarium, but for a temporary pool so same thing..

The heaters are in regular use and are tested and still perform well. It was just that one location where both heaters struggled.

They became electrically hotter than they should around the cable entry. I can't accurately say if they heated the water more/less efficiently than normal as the issue was detected, and both disconnected too soon to make any judgement on that.

 

[telectrix]

if the element is OK, then overheating of the cable in the shower could be down to a high resistance termination, either between the cable and crimp or crimp and terminal.

 

[Ampo]

Need to dust off the old books, I think I was getting confused with the amp thing as I would be calculating say the 9.5kw of the shower where of course ohms law throws that out the window because the 9.5kw is only relevant to the expected supply voltage so less voltage less kw less amps. Have I got there

 

[telectrix]

Need to dust off the old books, I think I was getting confused with the amp thing as I would be calculating say the 9.5kw of the shower where of course ohms law throws that out the window because the 9.5kw is only relevant to the expected supply voltage so less voltage less kw less amps. Have I got there

spot on .example. a 10kW shower rated at 240V.

P = V x I, so
I = P/V = 10000/240 = 41.667
from this you can determine the resistance

R = V/I = 240/41.667 = 5.76 ohms.

now as the resistance is constant ( not allowing for temperature variations, then apply 230V to the constant resistance

I = V/R = 230/5.76 = 39.93 A

from that, P = VI = 9.184 kW

so the actual current draw and hence the power, is reduced.

can't you telli'm bored and got nowt better to do. it's pi$$ing down, so car wheel theft is not an option for this Scouser today..