Electric shower

[Zinzan]

I had an electric shower, rated at 7.5 KW, put into my bathroom around 10 years ago. IT was on its own fuse box with a 30 AMP fuse. I replaced the shower 2 years ago and took advice which was to keep the shower under a 10 KW rating. My new shower is now 9.5 KW. I have had to replace the ceiling switch and had a qualified electrician do this last year. Every thing has been fine. All of a sudden the fuse has blown on two occasions in the same day. Can someone explain what maybe the problem. Is the fuse too small and if so why has it worked for a year without a problem? I noticed on the fuse packaging that a 45 AMP fuse should be used for a 9-10KW shower, but I am surprised it has worked for this long. If it is the case that the fuse is too small do I need to new fusebox as a 45 AMP fuse will not fit in the box I have? Would a circuit breaker be a better bet? Any help would be greatly appreciated.

[Bob Watkinson]

The fact of the matter is that a 10KW shower on full will draw 41.6A (10000/240) so it's little surprise your fuse is blowing. If it's only just started blowing then perhaps your local supply voltage has risen recently maybe from 220 to 240 perhaps, remember 10KW is not a constant, it all depends on the voltage and current (V x I) and I depends on the resistance of the element and the voltage (I = V/I). Also it's possible your only now using your shower on full power due to the time of year. Incidentally you should be using 10mm cable for a 10KW shower check that

6mm wasn't installed for your smaller shower as this cable is only rated at 32A.

[John]

7500 W / 240V = 31.25 A

9500 W / 240V = 39.58 A

When you are putting in a shower you need to ensure that the fuse will melt before the wire melts, and that the fuse and the wire are both "thick enough" to avoid melting under normal use.

I reckon that either the shower is drawing more power (perhaps due to colder water coming in over the winter), or the wiring is buggered, causing increased current and the fuse to blow.

If possible, I'd put in a 45A fuse and possibly re-run the job with thick wires.

I'm not an electrician BTW.

John

[Bob Watkinson]

No that can't happen. A resistive element simply works or it doesn't. It can't react to water temperature diference and adjust the output to achieve the heating effect. The OP may select the hotter setting if the water is colder though. This would add additional elements creating a higher current.

or the wiring is buggered,

agreed OP must check the cables are 10mm not 6mm or go back to a 7.5KW shower

>

[Roland Butter]

Less though, because the supply is, IIRC, now 230V. Someone will be along in a bit to put me right if I'm wrong!

[Bob Watkinson]

Well, sort of. The new figure of 230V was brought in a few years ago to bring us into line with Europe. The tolerance however is +- 10% of 230V or

216.2V - 253.0V. In reality the voltage didn't really change only the nominal voltage 'tag' did.

[Bob Watkinson]

I should of course have said 207V not 216.2V. These changes came into effect in 2003. Prior to this it was 240V +- 6%

[John]

Hmmm.

7500W / 240V = 31.25A

7500W / 230V = 32.61A

So that's more current for the same power!

John

[Bob Watkinson]

Well spotted, it doesn't work that way. Power is the current multiplied by the amps. And current is the voltage divided by the resistance. If you alter the voltage you alter the current too so the power output will also change. It follows then that A 230 V linear appliance used on a 240 V supply will take 4.3% more current and will consume almost 9% more energy. A 230 V rated

3 kW immersion heater, for example, will actually provide almost 3.27kw when fed at 240 V (nobody google that last sentence, I've plagiarised it :-)

[Palindr☻me]

So?

The manufacturer works out what resistance the heating element should have at a specified supply voltage, to give the required output power. And then fits a heater with that resistance. If he is designing for a lower voltage then yes, he will use a lower resistance element and have to allow for the higher current needed to provide the same power.

Usually though, we see it from the user, rather than designer viewpoint. The element resistance has already been fixed by the manufacturer. If the supply voltage is lower than the specified voltage, then the current and the power will be less too. If the supply voltage is higher than the specified voltage, then the current and the power will be more.

Only if the element resistance could be changed could the same power be given out for a range of input voltages. If the voltage was low, then the element resistance would have to be reduced and the load current increased to compensate.

[John]

Lets call old voltages and powers V1, P1, and new voltages and powers V2, P2. Using P = VI and V = IR for both pairs, and eliminating R, we get:

P2 = (V2/V1)**2 * P1

So if we switch a shower operating at 9500W (P1) and 240V (V1) to a

230V(V2) supply, we can find the new power (P2):

P2 = (230/240)**2 * 9500 = 8700W (2SF)

So a shower that operates at 9.5kW on 240V power will only operate at

8.7kW on 230V power.

The only question that is left to be asked is how the current is affected.

Introducing I2 and I1 as the two currents, and using P = VI for each with the previous result, we get:

I2 / I1 = V2 / V1 = 230 / 240 = 0.96

I2 = 0.96 x I1

So we can see that the new current is 4% lower than the old current.

In general, the power consumed decreases with the voltage. In general, the current decreases with the voltage.

(Which is how most of us would guess that it worked in any case).

John

[ARWadsworth]

We are on 230V +10% to -6% according to the supply note on a new intall I fitted today.

Adam

[Bob Watkinson]

I'm not clear what you mean by your double*. I'm also unclear when you refer to power if you mean the actual power consumed in the circuit or the power figure quoted by the manufacturer. The former is only variable by altering the voltage the latter is assuming a specified voltage.

[ARWadsworth]

Whilst I agree that 10mm is the the best solution, 6mm cable is rated at 32 amps on the worst case scenario when inside conduit inside insulation. If the cable is clipped direct or clipped and plastered over then it should carry 47 amps.

What is more worrying is a 30amp fuse blowing when 40 amps is been drawn. The fuse should withstand 45 amps for one hour before blowing. Either the shower is on its way out or there is some damage to the cable. The OP did not state how long the shower was running before the new fuse blew.

Adam

[Bob Watkinson]

Adam it's in his house so it's going to be method 3 isnt it in all proberbility, ie installed directly in an insulated wall. It's not going to be clipped direct and remember you need to size for the worst part in the circuit.

6mm for a 10Kw shower? It would hardly be professional to say this is ok

[ARWadsworth]

Too many probables. We have no idea how or where the cable runs. I would have said if the cable is installed directly in an insulated wall then it is installation method 15 not reference method 3 (3 may be OK but 15 would not be), but there is also a good chance the cable is embedded in plaster which case use reference method 1

I never said it was. We are refering to a 9.5kW shower not a 10kW shower. If the shower is 9.5kW @ 240 volts (another unknown) and if the cable is clipped or embedded in plaster the cable and then passes through some insulation then table 6B applies and the cable may still be sufficient.

Adam

[Bob Watkinson]

Yes sorry, typo, I was looking at the cores box instead of method box.

(3 may be OK but 15 would not

I've always considered embeded in plaster as thermal insulation. Method 15

6B? I know there's a 6B dealing with trenching, underground cable and cables alonside railway lines etc. What do you mean?

I would have stuck with 4D5A but then as you say there are too many unknowns

[Roland Butter]

I thought they'd taken the opportunity to drop the voltage slightly to reduce power consumption.

[Roland Butter]

Except that's probably not what happens. I expect they still make the element assuming 240V.

[Roland Butter]

Possibly on a new unit - but of course and old unit will still be expecting 240v.