Hi all.

My brother is having a problem with his electric shower. You know the
sort, the instantaneous ones which heat the mains cold water as it passes
through. It keeps tripping the MCB. There is nothing else on the
circuit, which is also RCD protected. It hasn't always happened, but it
is now occuring very frequently. The MCB is rated at 40amps, and the
shower is a 9.5 kilowatt unit (at 240volts).

Any ideas on what could cause the MCB to trip before the RCD, given that
the shower shouldn't be pulling more than 40amps?

I guess the simple answer is that somehow the shower *is* pulling more
than 40amps, or the MCB is naff and trips too easily.

Cheers
Simon

Your MCB is an overcurrent protection device. It is tripping because it is
undersised.
For a 9.5 kW shower fit a 45 Amp MCB.

"smb" wrote in message
...
Hi all.

My brother is having a problem with his electric shower. You know the
sort, the instantaneous ones which heat the mains cold water as it passes
through. It keeps tripping the MCB. There is nothing else on the
circuit, which is also RCD protected. It hasn't always happened, but it
is now occuring very frequently. The MCB is rated at 40amps, and the
shower is a 9.5 kilowatt unit (at 240volts).

Any ideas on what could cause the MCB to trip before the RCD, given that
the shower shouldn't be pulling more than 40amps?

I guess the simple answer is that somehow the shower *is* pulling more
than 40amps, or the MCB is naff and trips too easily.

Cheers
Simon

On Sat, 17 Jan 2004, smb wrote:

Any ideas on what could cause the MCB to trip before the RCD, given that
the shower shouldn't be pulling more than 40amps?

In theory a 9.5kW shower will draw 39.6 Amps at 240 volts, however the
supply voltage is allowed to go as high as 253 volts (230volts +- 10%
IIRC) in which case it will draw significantly more. The MCB may also be
on the sensitive side.

You should fit a 45 Amp MCB after checking that the cable to the shower,
incoming supply and consumer unit are all properly sized.

--
Alistair Riddell - BOFH
Microsoft - because god hates us

You are a little wrong here, the supply voltage is allowed to fluctuate
between +10% and -6%.
However, should your voltage be higher than the 230 Volts, the less current
your load will draw.
This is a simple application of ohms law. The higher the voltage the less
current the load will draw.
That aside, as I said earlier - your MCB is too small. 45 Amps is the
correct size for this shower.

"Alistair Riddell" wrote in message
g.uk...
On Sat, 17 Jan 2004, smb wrote:

Any ideas on what could cause the MCB to trip before the RCD, given that
the shower shouldn't be pulling more than 40amps?

In theory a 9.5kW shower will draw 39.6 Amps at 240 volts, however the
supply voltage is allowed to go as high as 253 volts (230volts +- 10%
IIRC) in which case it will draw significantly more. The MCB may also be
on the sensitive side.

You should fit a 45 Amp MCB after checking that the cable to the shower,
incoming supply and consumer unit are all properly sized.

--
Alistair Riddell - BOFH
Microsoft - because god hates us

"ripper" wrote in message
...
You are a little wrong here, the supply voltage is allowed to fluctuate
between +10% and -6%.
However, should your voltage be higher than the 230 Volts, the less
current
your load will draw.
This is a simple application of ohms law. The higher the voltage the less
current the load will draw.
That aside, as I said earlier - your MCB is too small. 45 Amps is the
correct size for this shower.

???
Since when did a shower have negative resistance!!!!!!

Volts=AMPS*RESISTANCE

Power=Volts*amps

If the resistance remains approximatly constant such as in the case of a
shower if the voltage goes up the current goes up P=(V*V)/R

In the very very special case of a swich mode powersupply the output power
and the input power is aprox constant regardless of supply voltage so ONLY
in this special case the volts go up and the amps go down.

On Sat, 17 Jan 2004, ripper wrote:

You are a little wrong here, the supply voltage is allowed to fluctuate
between +10% and -6%.

Right enough

However, should your voltage be higher than the 230 Volts, the less current
your load will draw.
This is a simple application of ohms law. The higher the voltage the less
current the load will draw.

I'm afraid you have it back to front - ohm's law is V=IxR (or I=V/R)
therefore for a given resisitive load the current will increase
proprotionally with the voltage.

--
Alistair Riddell - BOFH
Microsoft - because god hates us

You are wrong, P=V*I according to ohms law.
Therefore, I= P/V
Power remains constant @ 9.5 kW
Therfore
Amps = 43.2 @ 220 Volts
Amps = 41.3 @ 230 Volts
Amps = 39.6 @ 240 Volts
Amps = 38 @ 250 Volts
And just for the heck of it, Amps = 86.4 @ 110 Volts.

Do you see the reducing current with increasing voltage?
Why do you think power is transmitted at such high voltages?
Well the two main reasons a
High voltage, low current. Smaller cable size, cheaper. And secondly,
voltage drop over long distances.


"James Salisbury" wrote in message
...

"ripper" wrote in message
...
You are a little wrong here, the supply voltage is allowed to fluctuate
between +10% and -6%.
However, should your voltage be higher than the 230 Volts, the less
current
your load will draw.
This is a simple application of ohms law. The higher the voltage the
less
current the load will draw.
That aside, as I said earlier - your MCB is too small. 45 Amps is the
correct size for this shower.

???
Since when did a shower have negative resistance!!!!!!

Volts=AMPS*RESISTANCE

Power=Volts*amps

If the resistance remains approximatly constant such as in the case of a
shower if the voltage goes up the current goes up P=(V*V)/R

In the very very special case of a swich mode powersupply the output power
and the input power is aprox constant regardless of supply voltage so ONLY
in this special case the volts go up and the amps go down.

Alistair
You are trying to find current from two known quantities, voltage and power.
Power remains constant, voltage varies.
P=V*I
P=9500 Watts
Voltage = 230
Therefore I = P/V which is 9500/230 = 41.3 Amps the shower will take at 9.5
kW.
Using the same power with a voltage of 240 Volts,
I = P/V which is 9500/240 = 39.6 Amps.
Now at 250 Volts, I = 9500/250 = 38 Amps.
And if you use 110 Volts, the amps now increase to 86.4 Amps

"Alistair Riddell" wrote in message
g.uk...
On Sat, 17 Jan 2004, ripper wrote:

You are a little wrong here, the supply voltage is allowed to fluctuate
between +10% and -6%.

Right enough

However, should your voltage be higher than the 230 Volts, the less
current
your load will draw.
This is a simple application of ohms law. The higher the voltage the
less
current the load will draw.

I'm afraid you have it back to front - ohm's law is V=IxR (or I=V/R)
therefore for a given resisitive load the current will increase
proprotionally with the voltage.

--
Alistair Riddell - BOFH
Microsoft - because god hates us

And just to put you all at ease, if you go here you will find a simple ohms
law calculator.
http://measurementsconverter.co.uk/ohmslaw.html

"ripper" wrote in message
...
Alistair
You are trying to find current from two known quantities, voltage and
power.
Power remains constant, voltage varies.
P=V*I
P=9500 Watts
Voltage = 230
Therefore I = P/V which is 9500/230 = 41.3 Amps the shower will take at
9.5
kW.
Using the same power with a voltage of 240 Volts,
I = P/V which is 9500/240 = 39.6 Amps.
Now at 250 Volts, I = 9500/250 = 38 Amps.
And if you use 110 Volts, the amps now increase to 86.4 Amps

"Alistair Riddell" wrote in message
g.uk...
On Sat, 17 Jan 2004, ripper wrote:

You are a little wrong here, the supply voltage is allowed to
fluctuate
between +10% and -6%.

Right enough

However, should your voltage be higher than the 230 Volts, the less
current
your load will draw.
This is a simple application of ohms law. The higher the voltage the
less
current the load will draw.

I'm afraid you have it back to front - ohm's law is V=IxR (or I=V/R)
therefore for a given resisitive load the current will increase
proprotionally with the voltage.

--
Alistair Riddell - BOFH
Microsoft - because god hates us

On Sat, 17 Jan 2004, ripper wrote:

You are wrong, P=V*I according to ohms law.
Therefore, I= P/V
Power remains constant @ 9.5 kW

That's where you are wrong, power does not remain constant.
There are various means for modulating power to electric heating elements
but electric showers do not generally incorporate anything more
sophisticated than the ability to switch either or both of a pair of
heating elements on.

Many heating appliances these days have rating plates showing the current
consumption / power output for different input voltage e.g. 230v and 240v.
In all cases you will find the power output (and current) higher for a
higher input voltage.

--
Alistair Riddell - BOFH
Microsoft - because god hates us

On Sat, 17 Jan 2004 19:16:09 UTC, "ripper"
wrote:

You are wrong, P=V*I according to ohms law.
Therefore, I= P/V

That part is true.

Power remains constant @ 9.5 kW

No, it doesn't. This is your fundamental error. The resistance of the
shower element is approximately constant during operation. Increased
voltage means increased current through the element, thus increased
power. The power rating on the element is purely nominal.

Do you see the reducing current with increasing voltage?

No.

Why do you think power is transmitted at such high voltages?
Well the two main reasons a
High voltage, low current. Smaller cable size, cheaper. And secondly,
voltage drop over long distances.

No argument with that....but it's a different topic.

--
Bob Eager
rde at tavi.co.uk
PC Server 325*4; PS/2s 9585, 8595, 9595*2, 8580*3,
P70...

On Sat, 17 Jan 2004 19:28:23 UTC, "ripper"
wrote:

You are trying to find current from two known quantities, voltage and power.
Power remains constant, voltage varies.

No, it doesn't.

And if you use 110 Volts, the amps now increase to 86.4 Amps

LOL!

So if I apply a 1.5 volts from a Duracell AA cell, I can get about 6300
amps? Lovely!

--
Bob Eager
rde at tavi.co.uk
PC Server 325*4; PS/2s 9585, 8595, 9595*2, 8580*3,
P70...

On Sat, 17 Jan 2004 19:31:34 UTC, "ripper"
wrote:

And just to put you all at ease, if you go here you will find a simple ohms
law calculator.

Nothing wrong with Ohm's Law...I learned it many yaers ago.

BUT...your application is wrong because of your erroneous assumption
that power remains constant no matter what the applied voltage.
--
Bob Eager
rde at tavi.co.uk
PC Server 325*4; PS/2s 9585, 8595, 9595*2, 8580*3,
P70...

I am only using the 9.5 kW as an example. If you want to be fussy, the
resistance will also change with temperature.
So go find yourself an ohms law calculator, type in 9500 Watts, 240 Volts
and tell us what the current is.
Then try it with 9500 Watts, 230 Volts and tell us what the current is.
I = P/V Yes? Well it did when I was at school.

"Bob Eager" wrote in message
...
On Sat, 17 Jan 2004 19:16:09 UTC, "ripper"
wrote:

You are wrong, P=V*I according to ohms law.
Therefore, I= P/V

That part is true.

Power remains constant @ 9.5 kW

No, it doesn't. This is your fundamental error. The resistance of the
shower element is approximately constant during operation. Increased
voltage means increased current through the element, thus increased
power. The power rating on the element is purely nominal.

Do you see the reducing current with increasing voltage?

No.

Why do you think power is transmitted at such high voltages?
Well the two main reasons a
High voltage, low current. Smaller cable size, cheaper. And secondly,
voltage drop over long distances.

No argument with that....but it's a different topic.

--
Bob Eager
rde at tavi.co.uk
PC Server 325*4; PS/2s 9585, 8595, 9595*2, 8580*3,
P70...

On Sat, 17 Jan 2004, ripper wrote:

And just to put you all at ease, if you go here you will find a simple ohms
law calculator.
http://measurementsconverter.co.uk/ohmslaw.html

If you put 6.06 ohms into your calculator, this being a guess at the
resistance of the heating element of the shower and the only constant [1]
value in the equation, and then input different voltage values, then you
will see the current and power increasing with higher voltage.


[1] I know it's not completely constant but near enough.

--
Alistair Riddell - BOFH
Microsoft - because god hates us

Bob see my earlier post, I was not using the 9.5 kW as an assumption, merely
an example on using ohms law to calculate current at various voltages for a
fixed power rating.
Perhaps it might be worth pointing out that the resistance of the heating
element will also change with temperature.
So at what temperature do we calculate the current draw for a 9.5 kW shower?

"Bob Eager" wrote in message
...
On Sat, 17 Jan 2004 19:31:34 UTC, "ripper"
wrote:

And just to put you all at ease, if you go here you will find a simple
ohms
law calculator.

Nothing wrong with Ohm's Law...I learned it many yaers ago.

BUT...your application is wrong because of your erroneous assumption
that power remains constant no matter what the applied voltage.
--
Bob Eager
rde at tavi.co.uk
PC Server 325*4; PS/2s 9585, 8595, 9595*2, 8580*3,
P70...

Yes that is what happens when you put a constant resistance into the
calculator, and the wattage of the shower goes up.
But the resistance does not remain constant, it varies with temperature.

"Alistair Riddell" wrote in message
g.uk...
On Sat, 17 Jan 2004, ripper wrote:

And just to put you all at ease, if you go here you will find a simple
ohms
law calculator.
http://measurementsconverter.co.uk/ohmslaw.html

If you put 6.06 ohms into your calculator, this being a guess at the
resistance of the heating element of the shower and the only constant [1]
value in the equation, and then input different voltage values, then you
will see the current and power increasing with higher voltage.


[1] I know it's not completely constant but near enough.

--
Alistair Riddell - BOFH
Microsoft - because god hates us

On Sat, 17 Jan 2004 19:45:09 UTC, "ripper"
wrote:

I am only using the 9.5 kW as an example. If you want to be fussy, the
resistance will also change with temperature.
So go find yourself an ohms law calculator, type in 9500 Watts, 240 Volts
and tell us what the current is.
Then try it with 9500 Watts, 230 Volts and tell us what the current is.
I = P/V Yes? Well it did when I was at school.

And me too. But THE POWER ISN'T CONSTANT. I know the resistance will
change with temperature. But not a lot.

Your ideas are so ridiculous, you must be NICEIC registered!

--
Bob Eager
rde at tavi.co.uk
PC Server 325*4; PS/2s 9585, 8595, 9595*2, 8580*3,
P70...

On Sat, 17 Jan 2004 19:53:19 UTC, "ripper"
wrote:

Yes that is what happens when you put a constant resistance into the
calculator, and the wattage of the shower goes up.
But the resistance does not remain constant, it varies with temperature.

Indeed. It goes up a bit. Which reduces the current slightly. But it's
not linear, and not enough to give the change you propose.

I despair. Somebody else, please?
--
Bob Eager
rde at tavi.co.uk
PC Server 325*4; PS/2s 9585, 8595, 9595*2, 8580*3,
P70...

On Sat, 17 Jan 2004, ripper wrote:

Bob see my earlier post, I was not using the 9.5 kW as an assumption, merely
an example on using ohms law to calculate current at various voltages for a
fixed power rating.

That is where you are going wrong, the 9.5kW is NOT fixed, it is the power
output quoted by the manufacturer for a given supply voltage.

The equations you are quoting are correct, but the constant value is the
resistance of the element, NOT the power output.

Perhaps it might be worth pointing out that the resistance of the heating
element will also change with temperature.
So at what temperature do we calculate the current draw for a 9.5 kW shower?

It will, but not by much. A shower heater element is not like a light bulb
whose resistance increases hugely when hot. An incandescent light bulb
filament operates at thousands of degrees C, whereas a water heater
element cannot go over 100 degrees C or you would be generating steam
rather than hot water. (hopefully a safety cutout would intervene long
before that point was reached).

--
Alistair Riddell - BOFH
Microsoft - because god hates us

On 17/01/2004 ripper a wrote :
You are a little wrong here, the supply voltage is allowed to fluctuate
between +10% and -6%.
However, should your voltage be higher than the 230 Volts, the less current
your load will draw.
This is a simple application of ohms law. The higher the voltage the less
current the load will draw.
That aside, as I said earlier - your MCB is too small. 45 Amps is the
correct size for this shower.

WRONG!

The current drawn will INCREASE in proportion as the voltage rises. The
9.5Kw will be its rating at either 230v or 240v depending upon the
appliances age. As the voltage increases the current drawn will also
increase, as will the Kw.

The current (I) and Kw drawn will only be dependent upon the
resistance. The resistance (R) will stay approximately the same.

I= V/R

--

Regards,
Harry (M1BYT) (Lap)
http://www.ukradioamateur.org

From a Triton specification sheet for what they advertise as a 9.5kW shower:

SPECIFICATIONS
Nominal power rating at 240V
9.5kW – (40A MCB rating)

Nominal power rating at 230V
8.7kW – (40A MCB rating)

--
Toby.

'One day son, all this will be finished'

Back to the subject, at lower voltages (less than 240V) the stated power
rating of showers is reduced.
For example a 9.8 kW shower @ 240 Volts is only rated at 9kW @ 230 Volts.
A 9 kW @ 240V shower is 8.2 kW @ 230 Volts
A 7.5 kW shower @ 240V is 6.85 kW @ 230 Volts
My example of constant power was only to demonstrate using ohms law to
calculate current and various voltages.

"ripper" wrote in message
...
Yes that is what happens when you put a constant resistance into the
calculator, and the wattage of the shower goes up.
But the resistance does not remain constant, it varies with temperature.

"Alistair Riddell" wrote in message
g.uk...
On Sat, 17 Jan 2004, ripper wrote:

And just to put you all at ease, if you go here you will find a simple
ohms
law calculator.
http://measurementsconverter.co.uk/ohmslaw.html

If you put 6.06 ohms into your calculator, this being a guess at the
resistance of the heating element of the shower and the only constant
[1]
value in the equation, and then input different voltage values, then you
will see the current and power increasing with higher voltage.


[1] I know it's not completely constant but near enough.

--
Alistair Riddell - BOFH
Microsoft - because god hates us

ripper wrote on Saturday (17/01/2004) :
I am only using the 9.5 kW as an example. If you want to be fussy, the
resistance will also change with temperature.
So go find yourself an ohms law calculator, type in 9500 Watts, 240 Volts
and tell us what the current is.
Then try it with 9500 Watts, 230 Volts and tell us what the current is.
I = P/V Yes? Well it did when I was at school.

You are working on the false assumption that the 9.5Kw will remain
constant irrespective of a change in voltage. It does not remain
constant, the only constant is the resistance, but even this is not
absolutely constant. As temperature rises, the resistance will also
rise very slightly.

--

Regards,
Harry (M1BYT) (Lap)
http://www.ukradioamateur.org

"Bob Eager" wrote in message
...
On Sat, 17 Jan 2004 19:53:19 UTC, "ripper"
wrote:

Yes that is what happens when you put a constant resistance into the
calculator, and the wattage of the shower goes up.
But the resistance does not remain constant, it varies with temperature.

Indeed. It goes up a bit. Which reduces the current slightly. But it's
not linear, and not enough to give the change you propose.

I despair. Somebody else, please?

I was going to join in and mention motors
running at under their rated voltage drawing
more current but didn't as you may never
speak to me again ~)

You are exactly right Chris.

"Chris Oates" none wrote in message
...

"Bob Eager" wrote in message
...
On Sat, 17 Jan 2004 19:53:19 UTC, "ripper"
wrote:

Yes that is what happens when you put a constant resistance into the
calculator, and the wattage of the shower goes up.
But the resistance does not remain constant, it varies with
temperature.

Indeed. It goes up a bit. Which reduces the current slightly. But it's
not linear, and not enough to give the change you propose.

I despair. Somebody else, please?

I was going to join in and mention motors
running at under their rated voltage drawing
more current but didn't as you may never
speak to me again ~)

On 17/01/2004 ripper opined:-
Using the same power with a voltage of 240 Volts,
I = P/V which is 9500/240 = 39.6 Amps.
Now at 250 Volts, I = 9500/250 = 38 Amps.
And if you use 110 Volts, the amps now increase to 86.4 Amps

No it does not!

For an unvaring value of resistance, half the voltage and the current
and the wattage drawn would also be reduced by half.

If the current and the wattage stayed the same, then their would be no
need to produce different appliances for different voltages and
different markets would there? You could quite happily plug the bulbs
from your car into the mains, after all a 50w bulb is a 50w bulb if the
voltage rating doesn't matter :-)

Simple proof....

What happens to your lights during a 'brown out'?

During a 'brown out' your supply voltage will decrease, your lights
draw less current and less than their rated wattage. You are trying to
argue that the wattage would remain constant and therefore your lights
would not dim.

--

Regards,
Harry (M1BYT) (Lap)
http://www.ukradioamateur.org

On Sat, 17 Jan 2004, ripper wrote:

My example of constant power was only to demonstrate using ohms law to
calculate current and various voltages.

Maybe it was, but the thread was about the current drawn by an electric
shower, which is definitely not a constant power device.

IMO Your posts will not have helped the OP reach an understanding of why
his MCB may be tripping!

--
Alistair Riddell - BOFH
Microsoft - because god hates us

On 17/01/2004 ripper a wrote :
And just to put you all at ease, if you go here you will find a simple ohms
law calculator.
http://measurementsconverter.co.uk/ohmslaw.html

Yes, but you do need to know how and when to apply it.

--

Regards,
Harry (M1BYT) (Lap)
http://www.ukradioamateur.org

On 17/01/2004 ripper a wrote :
Yes that is what happens when you put a constant resistance into the
calculator, and the wattage of the shower goes up.
But the resistance does not remain constant, it varies with temperature.

Yes it does, but the amount by which it varies is tiny by comparison to
the other values which vary.

--

Regards,
Harry (M1BYT) (Lap)
http://www.ukradioamateur.org

On 17/01/2004 Chris Oates a wrote :
I was going to join in and mention motors
running at under their rated voltage drawing
more current but didn't as you may never
speak to me again ~)

No, please don't, in fact go get your coat :-)

--

Regards,
Harry (M1BYT) (Lap)
http://www.ukradioamateur.org

On Sat, 17 Jan 2004 20:10:06 UTC, "Chris Oates" none wrote:

I was going to join in and mention motors
running at under their rated voltage drawing
more current but didn't as you may never
speak to me again ~)

I was deliberately keeping well clear of that one!

--
Bob Eager
rde at tavi.co.uk
PC Server 325*4; PS/2s 9585, 8595, 9595*2, 8580*3,
P70...

"Toby" wrote in message
...
From a Triton specification sheet for what they advertise as a 9.5kW
shower:

SPECIFICATIONS
Nominal power rating at 240V
9.5kW - (40A MCB rating)

Nominal power rating at 230V
8.7kW - (40A MCB rating)

saw that - the accompanying .pdf
is slightly different it says 40/45A

ripper wrote on Saturday (17/01/2004) :
Do you see the reducing current with increasing voltage?
Why do you think power is transmitted at such high voltages?

Because as V increases, I decreases for an identical load.

Well the two main reasons a
High voltage, low current. Smaller cable size, cheaper. And secondly,
voltage drop over long distances.

But that has nothing at all to do with the present discussion.

--

Regards,
Harry (M1BYT) (Lap)
http://www.ukradioamateur.org

"smb" wrote in message
...
Hi all.

My brother is having a problem with his electric shower. You know the
sort, the instantaneous ones which heat the mains cold water as it passes
through. It keeps tripping the MCB. There is nothing else on the
circuit, which is also RCD protected. It hasn't always happened, but it
is now occuring very frequently. The MCB is rated at 40amps, and the
shower is a 9.5 kilowatt unit (at 240volts).

Any ideas on what could cause the MCB to trip before the RCD, given that
the shower shouldn't be pulling more than 40amps?

the thermal component of the MCB rather than the overcurrent
(the bit that's like a fuse) - the 40A rating is really to low for
9.5kw

I guess the simple answer is that somehow the shower *is* pulling more
than 40amps, or the MCB is naff and trips too easily.

"ripper" wrote in news:7HgOb.25323$qx2.2799204
@stones.force9.net:

You are exactly right Chris.

I bet that makes you feel better

mike r

Harry Bloomfield wrote in
.uk:


Simple proof....

What happens to your lights during a 'brown out'?

During a 'brown out' your supply voltage will decrease, your lights
draw less current and less than their rated wattage. You are trying to
argue that the wattage would remain constant and therefore your lights
would not dim.



No, that's not true.

As the filament cools it's resistance diminishes, allowing more current to
pass, so increasing the power again.

Stands to reason

mike r

In uk.d-i-y, Chris Oates none wrote:

My brother is having a problem with his electric shower. You know the
sort, the instantaneous ones which heat the mains cold water as it passes
through. It keeps tripping the MCB. There is nothing else on the
circuit, which is also RCD protected. It hasn't always happened, but it
is now occuring very frequently. The MCB is rated at 40amps, and the
shower is a 9.5 kilowatt unit (at 240volts).

Any ideas on what could cause the MCB to trip before the RCD, given that
the shower shouldn't be pulling more than 40amps?

the thermal component of the MCB rather than the overcurrent
(the bit that's like a fuse) - the 40A rating is really to low for
9.5kw

I guess the simple answer is that somehow the shower *is* pulling more
than 40amps, or the MCB is naff and trips too easily.

Piggin 'eck, this thread has generated - to use a painfully appropriate
saying - lots of heat and no light: a ridiculous misapplication of a
misnamed law, and crap advice. Sorry to be blunter than usual, but there
it is.

To deal with the practical aspects first: there is something *wrong* with
at least one of: the shower; the cable feeding it; the MCB; the incoming
supply - and that's more or less in order of likelihood (with change in the
supply being massively less likely than the other three). The original
poster's brother could do a lot worse than spend 50 quid on getting a
competent local electrician in: the mixture of lots of incoming amperage,
wet skin, and well-earthed metalwork is *not* to be treated lightly. Yes,
the RCD - *if* it's working - should protect your brother and any other
members of the household from the worst consequences. How lucky are you
feeling? And replacing the 40A MCB with a higher-rated one, even the
slightly higher rating of 45A, could be

D A N G E R O U S

leading in a quite plausible case to a

F I R E.

Again I ask: how lucky are you feeling?

Now, a little more reasoning. To all intents and purposes, the supply
voltage across the UK *is* 240V. The FAQ explains that for product design
purposes, we've 'harmonised' across Europe on a wider range of voltages
for which kit should behave, but the *actual* supply voltages in the different
European countries is unchanged, for now and for a number of years to come.
So let's stop dicking about with "well if the voltage was X, the current
would be Y, but if it was Xprime, the current would be Yprime", 'cos in
just about all reasonable foreseeable circumstances, the voltage will be
within a gnat's whisker of 240V. (And another reason for dropping it is
that one of the incessant responders got it all arse about face, as Bob
tried to explain: more on this below).

So, as it's sensible to say "V = 240", what's I for a shower rated at
9.5kW? Well, P = IV, so I = P/V (/ meaning "divided by") = P/240 =
9500/240 = 39.6A near as damn it. Follow? So a "40A" MCB is *NOT*,
*NOT*,
N O T
too low (or even 'to low' [sic]) for a 9.5kW shower. Thus we conclude
that either the whole circuit - shower+supply cables - is pulling *more*
than the 9.5kW it ought to, or the MCB has lost its marbles, and "become
oversensitive". But "becoming oversensitive" is not a particularly
well-known failure mode of MCBs. It could be that the 40A MCB is sandwiched
between a couple of other MCBs also running close to, or a little over,
their rated currents, so the mutual warming among these is making the 40A
one trip: but that's not likely.

What else could be causing a fault? Well, maybe the heater element has
somehow developed some internal shorts, so its resistance is now lower,
making it pull more current. Possible - happens in electric blankets -
but not likely in a leccy shower, where there's not much chance of the
heating-element wire touching itself inappropriately. (Pauses for the
giggles from the back of the class; looks severely over glasses; continues,
nary a glimmer of a smile passing across his face). More likely is some
other internal fault in the shower, allowing substantial extra current to
pass from L to N, which after a while makes the MCB trip. (The marking
of 40A on the MCB is a 'nominal' one; it'll pass 42A or 43A for hours and
hours - it won't suddenly cut in at 40.001A! You can look up the detailed
"overload versus time to break circuit" curves in any MCB supplier's
technical data sheet.) Worse, the fault could be a Live to Earth fault,
pulling 5A or more (that's about what you'd need for the MCB to trip
not at once but after a little while). Surely the RCD would cut in if
it was a L to E fault, wouldn't it? Why yes, of course it would... *if*
it's still working. Is it?

What else? Well, maybe the cable was put in a while ago, and was sized for
the kind of shower more common 10-15 years back - a 7.2kW model, say; with
a 30A MCB. Then someone put in a new, more powerful shower - the current
9.6kW one; and read the instruction about "needs a 40A MCB", and put one in.
After all, the cable looked pretty beefy, right? What would happen then?
Contrary to Hollywood special-effects pictures, the somewhat overloaded
cable will *not* instantly turn into a molten plasma, vapourising all
within a 10-mile radius. No, it just runs warm; a bit too warm. Hot, in
fact; hot enough - maybe somewhere out of sight, with a bit of thermal
insulation - to make the PVC insulation start to soften. Somewhere where
this is happening, the cable is bent; and the softening insulation allows
the live and neutral conductors to come closer together. If they do that
suddenly - bang, the trip will short. If they do it gradually, though,
you could get just the behaviour you're seeing: L and N nearly touching
when cold (shower off), bugger-all insulation between them now, just a
small air gap. Then you turn the shower on; the conductors heat up
gradually, expand, and now they touch. Off pops the MCB.

And *that* is one of the possible reasons for the MCB going off.

And *that* is the situation in which some posters are suggesting 'whack
a (slightly) bigger MCB in'. *Not* all that sound a suggestion, that.
The cabling fault - if such it be - need not relate to an undersized
cable, as the detailed picture drawn above had it: similar effects
can occur from connections which have worked loose, so they're now
getting too hot, maybe sparking a bit, again softening/weakening/charring
insulation so it doesn't do its job anymore. No, I don't *know* if a
worsening cable fault is causing the MCB to blow more often than before;
but it *could* be; and you'd be a fool to just cover up the symptom
by whacking in a bigger MCB. If you don't have the knowledge to
work out what's going on, for pity's sake pay for an hour's time
of someone who does.

Oh, the "mains supply" possibility? Well, it's remotely possible that
the incoming supply voltage has been increased recently: maybe someone
further away from the substation has complained of too low a voltage,
and the supply company's shifted up to the next tap on the transformer;
or some local heavy user's no longer heavy-using (any cannabis-in-the-loft
growers been busted round your way recently? or a little engineering
company's gone bust and no longer fires up its heavyish loads at the
times the shower's in use?) A higher incoming voltage will (as we
work out in remedial-level detail below) cause increased current to
flow. But any change in incoming voltage won't be dramatic - otherwise
you'd be seeing lightbulbs glowing extra-bright and burning out noticeably
more often; so as an explanation for the MCB behaviour it's the least
plausible.

Finally, the electrickle theory bit. The two relevant equations are
"V = IR" - that's the one we most usually call Ohm's Law - and "P = IV".
Each one tells you how *three* quantites are interrelated, so to ask 'how
does one quantity vary as the other one varies' is unanswerable, *unless*
you can *justify* the assumption that the third one will stay *constant*.
Geddit? The guy rabbiting on about the greater current being drawn as
voltage fell in order (by some unexplained, and in this case totally
non-present) to magically maintain a constant power was talking utter tosh.
As Bob tried to explain, the most nearly constant quantity here - as a
matter of physical truth, given the way a shower element is built - is
the resistance of the element when it's up to its operating temperature.
So, knowing the resistance is fixed, we can do some Really Complicated
Algebra. Ready?

V = IR

We can actually stop the analysis right here. Given that R is constant,
this says that as V goes up, so does I, 'linearly' - i.e. double V, and
you double I; knock V down by 5% and I comes down by 5%. End of story.

If we want to go doing Deep Mathematics, we can go on to work out how the
power developed across the fixed resistance will change as the voltage
changes. Coo, that's going to be advanced mathematical reasoning, since
we haven't been given a single formula which has both P and R in it.
But we're going to *derive* it, i.e. work it out. You see, we know

P = IV

And we're going to monkey with V. Can we assume I will stay constant?
Can we hell - we just reasoned a couple of paras ago that I was linearly
dependent on V: that's what "V = IR" means, remember? Now, given that
V = IR, we can write V/R = I (/ meaning Division), by dividing both sides
of the Ohm's Law equation by R. (It's mathematically and physically valid
to do so, since we know R isn't 0: no superconductors in sight ;-).
So, armed with the manipulated flavour of Ohm's Law saying "I = V/R", we
can replace the I in "P = IV" with "V/R", to get - wait for it -

P = (V/R)V

or as we more usually write it,

P = V*V*R or P = V-squared times R or P = V^2*R

Thus we see that, *across a constant resistance*, the power developed
is proportional to the *square* of the voltage. So, if the voltage
sags by 10%, so it's only 0.9 of the previous value, the power across
constant-resistance elements will be 0.9*0.9 = 0.81 times its previous
value, i.e. nearly 20% down. Fascinating? Maybe. Relevant? Not to the
original poster's brother, really; to the unteachable who kept on hoping
the constant-power fairy was in the room? yes, but will they ever realise
its relevance?

OK? Are we settled on what to do?
- For the original poster's brother: get someone genuinely competent in
to find the fault. There is a fault. Find it before it has more
serious consequences than it has till now.
- For the wannabe electrickle theorists:
- get your theory straight in your head;
- align it with physical reality;
- consider the consequence of ****ty advice when it concerns heavy-duty
volts, amps, and wet skin.

Stefek, shirtier than his normal self...

wrote:

snip

Well reasoned * well said = well done

--
Toby.

'One day son, all this will be finished'

In uk.d-i-y, Toby wrote:

Well reasoned * well said = well done

blushthanks, Toby/blush Another possibility is that there's more
than the shower running off the single MCB! Of course, there *shouldn't*
be; just like the immersion in this place *shouldn't* have been put
onto the upstairs ring back at the CU when someone needed a couple of
new ways in the only-one-spare-way-left CU. If there was another load
sharing the shower feed - some recent hack-it-about alterations - then
that too would account for the behaviour reported. And the MCB would
be *right* to trip to protect the cable...

Stefek