Quick Electrics Question

lucernarian

Hey there,
Would a 32A MCB/RCD be enough for an 8.5KW shower? That rating is for 240 Volts and when the mains is on load the voltage falls to only 210V or less. Would the amperage also fall because of this or would it increase to make up the full 8.5 kW?

It's connected with 6mm TCE buried in plaster. It was fitted today but we're not sure on the electrical safety. (it was a replacement shower).

Any opinions would be really appreciated as the fella fitting it wasn't sure to go with a 32A MCB or a 40A one.

Thanks in advance!

Btw the current setup is an old cartridge fuse @ 35A and there are scorch/melt marks around it for at least 5 years now:(

Hoagy

The maximum size of MCB/RCD is governed by the size of the circuit cable, in the case of 6sqmm 32amps is about it.
The voltage is dropping because the cable is too small for the load.
The current ( and the power) will also drop.
You need 10sqmm for a 8.5KW shower.
Because the shower is an intermittent load there isn't a safety issue but the water won't be as hot as it could be otherwise.

lucernarian

The cable is only 6 metres long though? And it was worse with the old 7 kW shower for some reason. The voltages were measured from an ordinary socket as the lights dim and remain dimmed whenever even a hoover is switched on. Would the thin cable affect the voltage for the whole house?

We usually have the shower on full power so would that not mean it's a continuous load?

The cable size could be an issue in any case but it will be very messy to change it. I'm more worried about the lack of an RCD and how we will go about fitting one. Any ideas on how to fit an RCD in a 30 year old siemens fuseboard?

Thanks for the reply:)

Hoagy

If the lights are dimming with a hoover running it sounds like a problem with the incoming supply, maybe it's an old overhead ESB cable? You probably need to get that checked out.

The shower is not a continuous load because it isn't in use for very long.

If your 6sq cable is only 6mtrs long the volt drop won't be too bad provided you're not losing volts on the supply (ESB) side.

You don't need to fit an MCB/RCD for the shower into the fuseboard, it can go in a small enclosure beside the board.

If there is no RCD for the sockets either it would be better to replace the consumer unit entirely and probably cheaper too.

FX Meister

You should have the shower on it's own RCBO. That's an RCD and MCB combined into one.

rogue-entity

An 8.5kW shower operating at our nominal line voltage (230V) will draw 36-38Amps at its peak. For safey reasons you should never connect a shower to a standard fuse, you should always connect it to a 40Amp double-pole RCD.

6sq is probably too thin for an 8.5kW shower, I would have thought it two thin for a 7kW shower too and would suggest that you run a lenght of 10sq from the fuseboard to the shower isolator switch. Make sure your isolator switch is rated to carry 40A and higher. We have a shower that uses a standard cooker switch which we needed to replace as the old one melted.

The more currant you intend to draw, the thicker the cable needs to be because electric cables will heat up as the load increases, and if the cable isnt thick enough, it will begin melt and eventually trip the RCD which will fail to prevent a fire.

Also, as an electric cable heats up, its resistance rises which exacerbates the problem and eventually causes an overload, tripping the protection device.

You say that when you plug in a hoover, the voltage drops.. you could have a poor supply to start with, but you may be better off getting your house rewired in parts as well as getting the fuseboard replaced. That said, it would do no harm to have the ESB check your supply and see if it is the reason for the voltage problems.

Lex Luthor

I = P/v

lucernarian

Thanks for the advice everyone.

I thought about the 10mm cable to the isolator option but I wasn't sure if it was okay to leave the 6mm tail to the shower.

Btw I just saw printed on the bottom of the shower that it uses 7.8 kW at 230V so the usage is 34A. It wasn't printed in the manual though.

Would 34A be safe enough for 6mm? I suppose if I consider the voltage drop, the shower is using even less electricity than that.

mathias

As posted I ( current ) = P ( power ) / V ( voltage )

If the voltage drops , the current goes up , not down , power remains constant regardless , or should on a mains supply.
For example , at 240 then current is 7500 / 240 = 31.5
at 210 the current is 7500 / 210 = 35.71 .
Even worse at 8500 but you get the idea !!

So you need a 40 A RCD for safety , ratings are for instantaneous not constant and this cannot be stressed enough ! You dont want a fire , and get the supply wiring looked into , it definitely shouldnt be that unstable.

charlie@w.d

6 square T+E is fine for an 8.5kw shower, 32 amp MCB/RCD is not. u need a 40amp MCB/RCD.
volt drop wouldnt be an issue.
how old is the house wiring?

FX Meister

mathias wrote:

So you need a 40 A RCD for safety , ratings are for instantaneous not constant and this cannot be stressed enough ! You dont want a fire , and get the supply wiring looked into , it definitely shouldnt be that unstable.

There's no point putting a 40A RCD on that circuit if the cable is only rated to take 32A. FACT

Hoagy

mathias wrote:

If the voltage drops , the current goes up , not down , power remains constant regardless , or should on a mains supply.
For example , at 240 then current is 7500 / 240 = 31.5
at 210 the current is 7500 / 210 = 35.71 .
Even worse at 8500 but you get the idea !!

That's not what they taught us in Kevin St.
The relevant equation for power is P=Isquared R
A heating element rated at 7500 watts at 240volts has a resistance of approx 7.68 ohms
If you put 210 volts into 7.68 ohms you will get a current of 27.3 amps with an equivalent power output of 5740 watts.

charlie@w.d

http://www.tlc-direct.co.uk/Technical/Charts/VoltageDrop.html
put ure specs in there

Copper

If the voltage drops , the current goes up , not down , power remains constant regardless , or should on a mains supply.
For example , at 240 then current is 7500 / 240 = 31.5
at 210 the current is 7500 / 210 = 35.71 .

I think that if the voltage across a load is reduced, the current and the power dissipated are both reduced. Otherwise if I connected my multimeter (1.5V battery) across the shower around 6000 amps would flow through it! The power rating of an appliance is the power dissipated by it at a specified voltage.

6 square fused at 32A and protected by an RCD is perfectly safe. A 32A supply for a 7.8kW load is borderline, and probably should be increased to 10 square / 40A. If this is not possible though it would be ok to run the shower on the existing supply.

mathias

That's not what they taught us in Kevin St.
The relevant equation for power is P=Isquared R

Same thing , its all ohms law , V = I x R , and P = I^2 x R
But I squared X R = (I x I ) x R and if V =I x R then I x I X R = I x V .
So theres no difference , P =VxI is more direct for this case.

There's no point putting a 40A RCD on that circuit if the cable is only rated to take 32A. FACT

Your either missing the point or overlooking it , the load looks likes its going to draw more than 32A , in which case 32A rated cable is going to overheat and burn , the cable can in no way regulate the current !! If the load is going to be more than 32A the rating of the cable and the breaker needs to be increased !!

I think that if the voltage across a load is reduced, the current and the power dissipated are both reduced. Otherwise if I connected my multimeter (1.5V battery) across the shower around 6000 amps would flow through it! The power rating of an appliance is the power dissipated by it at a specified voltage.

Each mains supply point has a maximum power or V x I product , if you max out the current the voltage will drop but the power cannot exceed its rating , the more current you draw , the more the voltage will drop , to the point where a dead short = 0 voltage but max current !

By the way , Multimeter test voltage and current is a constant current source , and limited to a very small amount regardless of the load being measured !!

Hoagy

mathias wrote:

Each mains supply point has a maximum power or V x I product , if you max out the current the voltage will drop but the power cannot exceed its rating , the more current you draw , the more the voltage will drop , to the point where a dead short = 0 voltage but max current !

Shakes head in disbelief - zero volts equals max current!

Since you are so fond of Ohm's law consider V=I X R. You're suggesting that if V drops then something magical happens to R to make I increase?

I feel sorry for the OP hoping to get sensible advice from this thread.

mathias

Ok then hoagy , lets check ,

Say you got two resistors in parallel , same value , it doesnt really matter , say theres 10 v across them , an you are measuring with a standard multimeter ,

You should see 10 v across each resistor right ? Also half the current will flow along each branch ,

Now take a piece of wire and short around one of the resistors , , what happens to the values ?

No voltage now across the shorted resistor .... but where is all the current going ?

That should be enough for you to grasp it ?

Nothing magical happens to R , its power we are talking about , for a given wattage , V and I will change depending on conditions in the load , thats what Im talking about , it shouldnt be too hard for you if you were in Kevin St !!
And Ohms law is what you use to calculate this type of problem isnt it ? or did you learn some other way ! Perhaps one I should be fonder of ?

Sparky

Its a rule that if volts drop, current increases.

The shower must be protected either by a 40amp RCD together with same MCB or an RCBO.

6mmsq is rated at 34 amps, while 10mmsq is 46amps
Volt drop in 6sq is 7.3 mV per amp per meter, where as it is 4.4 mV per amp per meter. So on such a borderline cable, you must consider this.

If your using the shower on full load like you said, then I would use 10sq as it works out that you would be pulling ~38amps if juice was at 220v. At 230v its ~36 amps thats at 8500watts.