On Sun, 25 Dec 2016 08:32:15 +0000, Chris Hogg wrote:

snip

Other than when the element loses (or reduces) it's ability to
transfer it's heat to the water, the internal resistance goes up and
the power consumption goes down and so the output energy goes down?

Yebbut... if the resistance rises due to the element getting hot, it
would take longer to dissipate the 100kW you originally proposed, and
power might fall from 100kW to 50kW, in which case it would take 2
seconds to dissipate the same amount of energy. But the efficiency
would still remain the same.

Yeahbut that was the point, it only got the 1 second, no compensation
for 'other variables'?

All the energy would end up in the water,
and bearing in mind we're only talking 3kW max for most domestic water
cylinder heaters, I can't see it being improved on in any way that
makes a difference.

No, quite, we are all pretty sure that is the case, including me. ;-)

Possibly for very high power flash heaters, if
such things exist, but not domestic stuff.

Ok.

And this would be over a sliding scale, from a single bubble to the
entire element covered.

I believe these were the sort of lines the OP was thinking re his
initial question (but not kettling specifically).

I know it's not quite the same (because it's not a closed system) but
try to accelerate a tyred vehicle past it's ability to maintain
traction and that extra power (and some of the previous power) is just
wasted.

Not the same at all.

Yes, like I said ('it's not a closed system' but the point I was
trying to make was the 'slippage' when the heat transfer process is
pushed past a particular point). Similarly, if you overcharge a
battery the surplus energy often gets dissipated in heat (or liberated
as a gas as the battery 'boils off')?

As has been said in this thread many times, all
the heat from the element goes into the water, whether you put it in
slowly or quickly.

So, extreme example ... an element that was so hot that it formed a
layer of steam round it all the time. Would all the energy still end
up in the water as heat or isn't that energy being used to convert
water to a vapour that would then be lost up the overflow pipe?

How much hotter does a kettle of water get if the thermostat gets
stuck on and the water continues boiling? If it doesn't get any
hotter, where / how is that surplus energy being dissipated?

The only factor relating to efficiency is the
external insulation on the tank. If it's poorly insulated, more heat
will be lost with slow heating than fast heating, so less efficient,
but that wasn't what the OP was asking about IIRC.

Quite.

Cheers, T i m