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can i wire in a 9.5 kw shower unit using 6 mm sq cable if the run of cable is only 3.5 meters long & i use a 45 amp trip ?
shower cable
- Thread starter theplumber
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i think that post was a little mesleading breezer although there are many more. The simple answer to your question is yes. but use a 40A MCB not a 45A. Bear in mind that should you decide to upgrade your shower in the future you will need to upgrade the cable to 10mmsq.
Is it OK irrespective of installation method?
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Why 40A, BR?
9,500 @ 230 = 41.3A
Yes, Ban is not wrong. The max for 6mm2 (Method 1) is 46A, but if there are installation methods other than "clipped direct" along the route, the more restrictive "method" should be employed in calculating the figures.
IE if you have to install in trunking, then you should use Method 3, which restricts cable capacity to 38A, in which case (for this instance) 10mm2 should be employed. I understand that the cable run is only 3.5m, but the rules still apply.
9,500 @ 230 = 41.3A
Yes, Ban is not wrong. The max for 6mm2 (Method 1) is 46A, but if there are installation methods other than "clipped direct" along the route, the more restrictive "method" should be employed in calculating the figures.
IE if you have to install in trunking, then you should use Method 3, which restricts cable capacity to 38A, in which case (for this instance) 10mm2 should be employed. I understand that the cable run is only 3.5m, but the rules still apply.
securespark said:
Yo. And it's all very well doing the calcs using the official 230V, but when your actual voltage goes up to 250, that current goes up to 44.9A..
ninebob said:
Yup - Ohm's law. I=V/R. The resistance of the shower does not increase when the voltage increases, it stays the same, so the current goes up.
The emoticon you want is the 3rd one in the middle row.....
Ban
How would I know when this was the case for any given situation?
I've always been taught that current (in amps) = power (in watts) / voltage (v).
Are there any other exceptions aside from showers, and can you explain the theory for me a little more?
How would I know when this was the case for any given situation?
I've always been taught that current (in amps) = power (in watts) / voltage (v).
Are there any other exceptions aside from showers, and can you explain the theory for me a little more?
There are no exceptions - the shower is not an exception.
The current drawn by an appliance is indeed P/V. But when the voltage goes up, so does the current and so does the power consumed.
Start with one of the representations of basic Ohm's Law: R=V/I. We know our shower draws 41.3A at 230V, therefore its resistance is 5.57 ohms (I'm assuming for the purposes of this that the shower is a purely resistive load).
The resistance does not change when the voltage does. It is always 5.57.
One of the other ways to express Ohm's Law is I=V/R, so when the voltage goes up to 250, I=250/5.57 = 44.88 amps.
Where you went wrong with your calculation was thinking that the power consumed by the shower remains the same no matter what the voltage is - it doesn't.
P=VI, so at 250V, our shower is consuming 11.22kW, and making the water a lot hotter than normal.
P=V^2/R. 250^2/5.57 = 62500/5.57 = 11.22kW
P is proportional to V^2, so if the voltage goes up by a factor of 1.087 (250/230), the power goes up by 1.087^2, i.e. by 1.181.
9.5kW*1.181 = 11.22kW
The current drawn by an appliance is indeed P/V. But when the voltage goes up, so does the current and so does the power consumed.
Start with one of the representations of basic Ohm's Law: R=V/I. We know our shower draws 41.3A at 230V, therefore its resistance is 5.57 ohms (I'm assuming for the purposes of this that the shower is a purely resistive load).
The resistance does not change when the voltage does. It is always 5.57.
One of the other ways to express Ohm's Law is I=V/R, so when the voltage goes up to 250, I=250/5.57 = 44.88 amps.
Where you went wrong with your calculation was thinking that the power consumed by the shower remains the same no matter what the voltage is - it doesn't.
P=VI, so at 250V, our shower is consuming 11.22kW, and making the water a lot hotter than normal.
P=V^2/R. 250^2/5.57 = 62500/5.57 = 11.22kW
P is proportional to V^2, so if the voltage goes up by a factor of 1.087 (250/230), the power goes up by 1.087^2, i.e. by 1.181.
9.5kW*1.181 = 11.22kW
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If you were a non-mathematical person, you could say
Our shower draws 38A at 250V (9500/250) therefore the resistance is 6.58 Ohms.
For the benefit of people who are not maths-minded, what is wrong with the above statement?
Our shower draws 38A at 250V (9500/250) therefore the resistance is 6.58 Ohms.
For the benefit of people who are not maths-minded, what is wrong with the above statement?
Well, the actual numbers are wrong, it's 41.3A at 230V (9500/230) therefore the resistance is 5.57 ohms, but the values are irrelevant to the principle.
What is a problem with what you've written is that you show how 9500/250 = 38A, but nowhere do you explain for the non maths-minded how you get the resistance figure of 6.58 ohms.
What is a problem with what you've written is that you show how 9500/250 = 38A, but nowhere do you explain for the non maths-minded how you get the resistance figure of 6.58 ohms.
I
Igorian
theplumber said:
I noticed on the shower I just installed that although the big advert says 9.8KW, it later says this is at 240V and gives a lower figure of 9.0KW at 230V. Could this be the case with your 9.5KW shower? (ie, is it 9.5KW @ 240V?)
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But can you see my point at all?
Why must it be @230V? If this figure changes (240 250 whatever) then the whole calculation goes t*ts up.
What I'm saying is that the whole basis of your calculations rest on a given I value (whatever that may be) at a voltage of 230. If the voltage value alters, then the figures differ. And voltage is the one thing that is prone to variation!
Who says the starting point for the calcs has to be 230V?
Get my drift? (He asks, hopefully!)
Why must it be @230V? If this figure changes (240 250 whatever) then the whole calculation goes t*ts up.
What I'm saying is that the whole basis of your calculations rest on a given I value (whatever that may be) at a voltage of 230. If the voltage value alters, then the figures differ. And voltage is the one thing that is prone to variation!
Who says the starting point for the calcs has to be 230V?
Get my drift? (He asks, hopefully!)
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