Electric Shower tripping MCB

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

Reply to
smb
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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.

Reply to
ripper

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.

Reply to
Alistair Riddell

Reply to
ripper

??? 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.

Reply to
James Salisbury

Right enough

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.

Reply to
Alistair Riddell

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 are: High voltage, low current. Smaller cable size, cheaper. And secondly, voltage drop over long distances.

Reply to
ripper

Reply to
ripper

Reply to
ripper

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.

Reply to
Alistair Riddell

That part is true.

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.

No.

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

Reply to
Bob Eager

No, it doesn't.

LOL!

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

Reply to
Bob Eager

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.

Reply to
Bob Eager

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.

Reply to
ripper

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.
Reply to
Alistair Riddell

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?

Reply to
ripper

Reply to
ripper

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!

Reply to
Bob Eager

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?

Reply to
Bob Eager

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.

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).

Reply to
Alistair Riddell

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