Wiring for 2x 3.6kw ovens, & “Currys Team (don’t) Know How”
Discussion
Mr Pointy said:
This is all theoretical, I am not arguing the point that in reality the chances of both ovens being on and hitting this peak is extremely unlikely. If the manufacturer have said that a 16A supply to each oven is enough then I'm sure their engineers have done the calcs and it will be fine.
Initforthemoney said:
Username checks out.
It's OK, under the previous name (Mecheng84) he was an expert on comms too 
Mr Pointy said:
MECHENG84 said:
adam. said:
Has everything connected to a phone socket got an ADSL filter on it?
Does your alarm have a filter on the phone line connection? Does it 'phone home' every day around midnight?
^^ This is a good question. You've also partly answered your own question. You want a master socket on the wall as soon as BTs cable enters your property. ADSL filter into Master socket as well (or micro filter as they are sometimes called) and then router as close to Master socket as possible.Does your alarm have a filter on the phone line connection? Does it 'phone home' every day around midnight?
EVERYTHING that uses the phone lines needs to be ran through an ADSL filter, it separates the phone frequency from the internet frequency. If you plug a phone item directly into the socket without a filter it will clash with the internet frequency.
jackofall84 said:
Oh, so Neff have lied and the oven isn't capable of delivering 3600 W?
It should deliver the stated power at the stated voltage on the specs. But if the voltage is different the current and thus power output will be different, because the oven is to all intents and purposes a resistive heating device and the resistance of the element(s) is, for all practical purposes, fixed.Jambo85 said:
It should deliver the stated power at the stated voltage on the specs. But if the voltage is different the current and thus power output will be different, because the oven is to all intents and purposes a resistive heating device and the resistance of the element(s) is, for all practical purposes, fixed.
I agree with this, the element will have a fixed ohm rating and under the law V=IR if R is fixed and V drops then the only variable is I, which if V did drop I would have to drop as well to keep the balance. I can see I'm just going to be given a lot of flack here, so no point in continue arguing my case. I was simply calculating on the data provided and assuming worst case for unprovided data. In this case we are told that the power is 3600W but no voltage, Neff may well have provided this power output based on a nominal voltage of 220V and not 230V therefore increasing the current draw at max power.
jackofall84 said:
Jambo85 said:
It should deliver the stated power at the stated voltage on the specs. But if the voltage is different the current and thus power output will be different, because the oven is to all intents and purposes a resistive heating device and the resistance of the element(s) is, for all practical purposes, fixed.
I agree with this, the element will have a fixed ohm rating and under the law V=IR if R is fixed and V drops then the only variable is I, which if V did drop I would have to drop as well to keep the balance. I can see I'm just going to be given a lot of flack here, so no point in continue arguing my case. I was simply calculating on the data provided and assuming worst case for unprovided data. In this case we are told that the power is 3600W but no voltage, Neff may well have provided this power output based on a nominal voltage of 220V and not 230V therefore increasing the current draw at max power.
Mr Pointy said:
jackofall84 said:
Jambo85 said:
It should deliver the stated power at the stated voltage on the specs. But if the voltage is different the current and thus power output will be different, because the oven is to all intents and purposes a resistive heating device and the resistance of the element(s) is, for all practical purposes, fixed.
I agree with this, the element will have a fixed ohm rating and under the law V=IR if R is fixed and V drops then the only variable is I, which if V did drop I would have to drop as well to keep the balance. I can see I'm just going to be given a lot of flack here, so no point in continue arguing my case. I was simply calculating on the data provided and assuming worst case for unprovided data. In this case we are told that the power is 3600W but no voltage, Neff may well have provided this power output based on a nominal voltage of 220V and not 230V therefore increasing the current draw at max power.
jackofall84 said:
I can't really be bothered to keep going with this, but, P=IV, I did not make this formula up. If P is kept constant and V drops I must increase to balance the equation.
This is all theoretical, I am not arguing the point that in reality the chances of both ovens being on and hitting this peak is extremely unlikely. If the manufacturer have said that a 16A supply to each oven is enough then I'm sure their engineers have done the calcs and it will be fine.
" A little knowledge is a dangerous thing"This is all theoretical, I am not arguing the point that in reality the chances of both ovens being on and hitting this peak is extremely unlikely. If the manufacturer have said that a 16A supply to each oven is enough then I'm sure their engineers have done the calcs and it will be fine.
Where's Alucid when you need him?!
I run a team of 70 maintenance guys, 65 of them think they are electricians, only 5 of them really are. (much like on here)
And then there are 4 gas engineers, nobody else in the team pretends to be a gas engineer becuase they know if they mess it up they go to jail.
TLDR - Don't cut corners, get a proper sparks in, you'll sleep better knowing it was done correctly..
I run a team of 70 maintenance guys, 65 of them think they are electricians, only 5 of them really are. (much like on here)
And then there are 4 gas engineers, nobody else in the team pretends to be a gas engineer becuase they know if they mess it up they go to jail.
TLDR - Don't cut corners, get a proper sparks in, you'll sleep better knowing it was done correctly..
Jambo85 said:
Initforthemoney said:
Surely if the voltage was lower, current draw would be higher?
No? I = V / R
if you have a 200 ohm resistor and a 200 v supply.... that will draw 1 amp...
if you have a 100 ohm resistor and a 200 v supply... that will draw 2 amps...
if you have a 400 ohm resistor and a 200 v supply.... that will draw 0.5 amps....
assumption made that the resistance stays the same value.
what your thinking of is if you drop the voltage the current has to rise to maintain the same power, this is true. but that cant happen as the resistance is the same. you would have to drop the resistance to maintain the V x I ratio to give you the same power.
Initforthemoney said:
Surely if the voltage was lower, current draw would be higher?
This is a yes and no answer. From a pure P=IV calc then yes 3600W would require more current from a lower supply voltage. However seeing as this has a constant resistance in gets a little more complicated, to fully define the answer you need a 3600W @ X voltage specification from the manufacturer. Assuming its 3600W @ 240V the current draw is 15A, this would give a resistance of 16ohms. Now as the voltage decreases so does the power, due to V=IR so 230/16 = 14.375A and 220/16 = 13.75A, that would mean @230V the power output would be 3306.25W and at 220V the power output would be 3025W.ruggedscotty said:
Nope
I = V / R
if you have a 200 ohm resistor and a 200 v supply.... that will draw 1 amp...
if you have a 100 ohm resistor and a 200 v supply... that will draw 2 amps...
if you have a 400 ohm resistor and a 200 v supply.... that will draw 0.5 amps....
assumption made that the resistance stays the same value.
what your thinking of is if you drop the voltage the current has to rise to maintain the same power, this is true. but that cant happen as the resistance is the same. you would have to drop the resistance to maintain the V x I ratio to give you the same power.
You beat me to it I = V / R
if you have a 200 ohm resistor and a 200 v supply.... that will draw 1 amp...
if you have a 100 ohm resistor and a 200 v supply... that will draw 2 amps...
if you have a 400 ohm resistor and a 200 v supply.... that will draw 0.5 amps....
assumption made that the resistance stays the same value.
what your thinking of is if you drop the voltage the current has to rise to maintain the same power, this is true. but that cant happen as the resistance is the same. you would have to drop the resistance to maintain the V x I ratio to give you the same power.
For those wanting to remind themselves of the various formulae for ovens, there was a thread just last week......
https://www.pistonheads.com/gassing/topic.asp?h=0&...
https://www.pistonheads.com/gassing/topic.asp?h=0&...
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