Electric Cooker and Induction Hob

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I am having difficulty getting correct details to fit an electric oven and induction hob. First off the manufacturer's details are a little confusing and I am uncertain of the acceptable load to carry on existing cables.

For starters the excising cooker is on its own circuit with a 32A MCB. The existing gas hob (just for the ignition) is on the kitchen ring circuit, again with a 32A MCB. The appliances have the following ratings:

Hob - 8.5kW (using all plates and with two boosted for rapid warm up)
Cooker Oven/Grill - 3.0kW/3.0kW, hence 6.0kw in total (although connected load is stated as 3.0kW)

The hob I accept is taking a large load and will therefore need to be treated carefully. However, the oven I believe will never exceed 3.0kW as the oven and hob are only ever used independently. Using this argument the MCB rating should be 12.5A I believe. If this were correct I think this would be okay on the ring circuit as this already as a rating of 32A. This will also allow the kettle and microwave to be used also, without tripping the MCB. However, my concern here is that the standard 2.5mm cable may not be able to cope with the load.

Assuming the cooker can be dealt with this way, I wished to then attach the hob to the existing cooker circuit and increase the MCB to compensate for the increased load. A MCB rated at 40A should be suitable I thought. However, again I am unsure if the cable at 6mm can take the load.

At the end of the day I wish to make use of existing cabling, as whilst I have plenty of capacity in the consumer unit for more circuits, laying new cable is out of the question. The only other possible option I can come up with is the use of 10mm cable to replace the 6mm cable, assuming I can get this through the existing trunking, and attach both appliances to the cooker circuit with the appropriate MCB, i.e. rated at the next option above 40A

Advise on the above would be greatly received.
 
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Oven will be OK if connected into a socket outlet (13Amp plug-top, taking into account that the oven is single / fan assisted).

Hob will be OK if connected to 6.0mm and upgraded MCB rating of 40Amps (6.0mm can take upto 41Amps with a max length of 38metres run in metal conduit or plastic trunking, this does include for thermal insulation)

Hob Rating 8500W / 230 Voltage = 36.9Amps

Also this is not allowing for diversity, not very often that you will using the hob under full load.
 
Many thanks il78.

I am very grateful for your advice and just to close the loop can confirm:

Oven is single and fan assisted
It is highly unlikely that each ring on the hob will be used to the max.
Whoops on my rating calculation. I am still working with old money, i.e. 240 Volts and came up with 35.4A.

Thanks again.
 
If your rating is, say 10.5kw @ 240v, at 230v it will be 9.6kw. 10.1 @240v will be 9.3kw @ 230v.
In other words the power consumption drops with the input (Well. thats what me new shower and oven instructions are telling me !!).
There must be an equation to work this out, what is it ? (Come on breezer/sparks).
Also where do I find info on "derating" (think thats what its called) of cables when passed through trunking,insulation, excessive lengths etc. etc.
TIA
 
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The equation to calculate resistance is as follows:

Power / Voltage = Resistance or P/R = I using the correct notation.

The units being measured typically as follows:

Power (P) = Kilowatts (kW)
Resistance (R) = Amperes (A)
Voltage (I) = Volts (V)

So with an output or power rating of 10500W and a voltage of 230 the resistance is 45.7A rounding up.

It should be remembered that UK voltage is now 230V as a result of alignment with the rest of the EU. It used to be 240V.

Hope this helps.
 
Just said:
The equation to calculate resistance is as follows:

Power / Voltage = Resistance or P/R = I using the correct notation.

The units being measured typically as follows:

Power (P) = Kilowatts (kW)
Resistance (R) = Amperes (A)
Voltage (I) = Volts (V)

So with an output or power rating of 10500W and a voltage of 230 the resistance is 45.7A rounding up.

It should be remembered that UK voltage is now 230V as a result of alignment with the rest of the EU. It used to be 240V.

Hope this helps.
When I went to school if you divided a number by a large number and then a small number the second would give you a bigger number as an answer,
I'd worked out that our 10.5kw shower would use more amps as a result of the change but the instructions say at 230v it's a 9.6kw shower. Or am I being really thick and missing the point??
 
Ambient Temperature, Grouping, Thermal Insulation and Re-wirable fuses are the correction factors to be accounted for (as well as volt drop and R1+R2 etc). Is that what you mean Scoby? I'll post some info if you want me to (?), but you're unlikely to use it (I rarely do in a domestic situation).
 
If its not likely that I need to use this in a domestic situation then I'll pass on your offer spark.
Ignorance is bliss as they say !!
 
Just said:
The equation to calculate resistance is as follows:

Power / Voltage = Resistance or P/R = I using the correct notation.

The units being measured typically as follows:

Power (P) = Kilowatts (kW)
Resistance (R) = Amperes (A)
Voltage (I) = Volts (V)

So with an output or power rating of 10500W and a voltage of 230 the resistance is 45.7A rounding up.

It should be remembered that UK voltage is now 230V as a result of alignment with the rest of the EU. It used to be 240V.

Hope this helps.

1) Resistance is measure in Ohms, current is measured in Amperes, Voltage is measured in Volts, power is measured in Watts
2) Resistance (R) = Voltage (V) * Current (I) (or V = IR)
3) Power (W) = Voltage (V) * Current (I) (or W = VI)


now the appliance doesn't change resistance, so if we reduce the voltage the power changes:

10500 Watts / 240 Volts = 43.75 Amps
240 Volts / 43.75 Amps = 5.49 Ohms

So the 10.5 kW shower has a resistance of 5.49 Ohms

Now

230 Volts / 5.49 Ohms = 41.89 Amps
230 Volts * 41.89 Amps = 9635 Watts

So at 230 V the shower has a power of 9.6 kW

Hope this helps
 

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