Two ovens and induction hob wiring?

[JohnW2]

Don't forget that what matters is 'average current for appreciable periods of time'. If absolutely everything were 'turned on full' from cold simultaneously (which is virtually never going to happen) then yes, the full 80A would initially be drawn, but only for a very short period of time. Even if you continued using every single bit of the cooking appliances, the thermostats would rapidly start turning things on and off, and you would soon reach the situation in which the average current over any appreciable period of time would be a small fraction of 80A.

I don't have anything to hand that shows what the situation would be like if you were boiling some foodstuff in a litre or two of water in a saucepan on a hob but, on a much larger (volume and time) scale, I can show you what happens when an immersion heater heats up 140 litres of water and (by means of thermostat) keeps it as the desired temp.

In the below, the power when the element is 'on' is, as expected, 3000W (about 13.0A) but averaged over the 7 hours displayed, the average power (to keep the water at the desired temp) is 414W (about 1.8A), about 13.8% of the 'peak load/current' (ignore the small twiddly bits at the bottom of the graph - due to something else use very small amounts of power from the same circuit) ....

If you mentally 'scale that down' to the much smaller situation with much smaller volumes kept hot (usually 'boiling') for much shorter periods of time, you should get some idea of how low the 'average current' is likely to be in relation to the peak current.

Kind Regards, John

 

[plugwash]

Don't forget that what matters is 'average current for appreciable periods of time'.

What actually matters is average power dissipated in the cable over an appreciable period of time. This depends on RMS current over that same appreciable period of time, not on mean current.

While RMS will be lower than peak it can still be substantially higher than mean.

In the below, the power when the element is 'on' is, as expected, 3000W (about 13.0A) but averaged over the 7 hours displayed, the average power (to keep the water at the desired temp) is 414W (about 1.8A), about 13.8% of the 'peak load/current'

However the RMS current in the cable will be about 37.1% of full load or about 4.83A.

 

[JohnW2]

What actually matters is average power dissipated in the cable over an appreciable period of time.

Indeed.

This depends on RMS current over that same appreciable period of time,...

Indeed.

... not on mean current.

It depends upon the mean RMS current over the period of time.

I think you are probably seriously confusing the issue. Everything we deal with in relation to AC current in electrical installations (cable CCCs, trip thresholds of MCBs and RCDs, In of MCBs, 'ratings of items' etc,. etc.) is in terms of RMS current. We never talk about the mean current of the waveform, or the peak current of the waveform - since those measures are not in any way useful

The fact that one can describe the voltage of a sinusoidal waveform in other ways (e.g. peak and mean) is totally irrelevant - since, as above, we never think or talk in terms of those sort of things in relation to electrical installations.

However the RMS current in the cable will be about 37.1% of full load or about 4.83A.

I haven't bothered to try to work out where you got those figures from but, as above, all the currents I talked about WERE RMS currents (and the 'average' current about was the average of the RMS current (which is proportional to the average power dissipated) over a period of time..

Kind Regards, John

 

[JohnW2]

I don't have anything to hand that shows what the situation would be like if you were boiling some foodstuff in a litre or two of water in a saucepan on a hob but, on a much larger (volume and time) scale, I can show you what happens when an immersion heater heats up 140 litres of water and (by means of thermostat) keeps it as the desired temp. .... If you mentally 'scale that down' to the much smaller situation with much smaller volumes kept hot (usually 'boiling') for much shorter periods of time, you should get some idea of how low the 'average current' is likely to be in relation to the peak current.

OK. To save people doing 'mental scaling down', I've done it for you ...

... on the basis of the immersion heater situation, I have 'scaled down' the time periods involved from 7 hours for 140 litres of water to 20 minutes of cooking, assuming (obviously only very roughly correct) that the cooking appliances exhibit roughly the same (pro-rata) pattern of thermostatically ons/offs.

I have assumed that 80A worth of cooking loads are turned on (from cold) simultaneously (exceedingly unlikely in practice) and, assuming that same patter of thermostatic ons/offs, have determined the average current draw over 1-minute and 10-minute periods. The data shows that, as one would expect, the average current drawn during the first couple of minutes is 80A but rapidly declines thereafter, and the average is down to about 25 A by the fourth minute. As for averages over 10-minute periods (much more of relevance to cable heating etc.), the average over the first 10 minutes is about 27 A, and that falls to about 4A after a further 4 mins, and to even low currents thereafter.

I don't pretend that these figures are going to be anything like accurate, but they give some insight into the concept of diversity for cooking appliances.

Edit: it's just occurred to me that although my choice of 10-minute periods over which to average was essentially arbitrary, if one uses that period, and also assumes the on/off pattern I have described, this approach leads to results pretty similar to what one would get with standard diversity calculations (i.e. the 10-minute average is, at most, a bit under 30A when the 'maximum load' is 80A).

Kind Regards, John

 

[ban-all-sheds]

assuming (obviously only very roughly correct) that the cooking appliances exhibit roughly the same (pro-rata) pattern of thermostatically ons/offs.

Not sure how valid that is - your stored hot water vessel is not uninsulated and open at the top. Nor is the water in it kept boiling.

 

[JohnW2]

Not sure how valid that is - your stored hot water vessel is not uninsulated and open at the top.

Of course .. as I said ...

I don't pretend that these figures are going to be anything like accurate, but they give some insight into the concept of diversity for cooking appliances.

... but, as I said, it gives a bit of insight into the concept of diversity.

... Nor is the water in it kept boiling.

I'm not sure that is particularly important - keeping water boiling (and providing the necessary energy for the phase change) is not conceptually all that different from maintaining water at any other temperature. The lack of insulation (and possibly 'open top') are far more important. However, it's probably more applicable to an oven, which has a reasonable amount of insulation and no 'open top'. Again, however, I'm merely trying to demonstrate the concept, primarily 'qualitatively'. Having said that, as I observed in my edit, it's interesting that what I did produced results which are fairly consistent with standard diversity calculations.

 

[plugwash]

(and the 'average' current about was the average of the RMS current (which is proportional to the average power dissipated) over a period of time..

Power dissipated in the cable is not proportional to current, it is proportional to the SQUARE of current.

Lets say our cable has a resistance of 0.1 ohm (total for live and neutral) and we draw 10 amps through it continuously for one second. The power dissipated in the cable will be 10 watts.

On the other hand lets say we draw 20 amps through it 50% of the time and nothing the remainder of the time. When the load is on the power dissipated in the cable will be 40 watts. So the average power dissipated in the cable will be 20 watts.

That is why we use RMS in the first place and it is equally valid whether we are talking about short-term averaging or long term averaging. Taking a short-term RMS followed by a long term mean makes no sense when you are talking about power dissipated in the cable (it *does* make sense when talking about power delivered to the load, because in that case on the short-term current is proportional to voltage, while on the long term it is not).

 

[JohnW2]

Power dissipated in the cable is not proportional to current, it is proportional to the SQUARE of current.

Yes, needless to say, I know that - and it was just a 'typo' on my part - my mind and typing fingers were was concentrating on my main point that it is the square of RMS (not mean) current which is proportional to power dissipated - so I forgot to type the "square of".

Lets say our cable has a resistance of 0.1 ohm (total for live and neutral) and we draw 10 amps through it continuously for one second. The power dissipated in the cable will be 10 watts. On the other hand lets say we draw 20 amps through it 50% of the time and nothing the remainder of the time. When the load is on the power dissipated in the cable will be 40 watts. So the average power dissipated in the cable will be 20 watts.

Yes, you're right - apologies for misunderstanding you (and getting things a bit wrong!). I now see what you meant - you appeared to be talking about the difference between the RMS and mean of the waveform - but I now realise that you actually meant that I should be looking at "the RMS of (the RMS of the waveform)", rather than "the RMS of (the RMS of the waveform)" - and, as I've just said, you're totally right about that.

I think I've got it right now, and it doesn't change things much. Given that I was working with (since it's what I have!) 1-minute data, that does not alter the first of the graphs I presented. For the 10-bit moving averages (again, working with 1-minute data), when I change the (incorrect) "mean of RMS" to the (correct) "RMS of RMS", it obviously does increase the 'average' (RMS) over the first 10 minutes a bit, but only to about 40 A, and it's down to under 32 A for the 10-minute period from 1 to 11 mins, after which it rapidly decreases. Indeed, as I recently wrote, the 10-minute 'averaging periods' were chosen pretty arbitrarily, and if one moves to 15-minute averages (which I would think would be very reasonable), the 'average' (RMS) over the first (highest current) 15-minute period is then about 32 A (see graphs below)

I'm going to try to undertake some measurements on something a bit closer to a 'cooking appliance', and over a longer time period, and will present some figures in due course!

Kind Regards, John

 

[JohnW2]

I'm going to try to undertake some measurements on something a bit closer to a 'cooking appliance', and over a longer time period, and will present some figures in due course!

'Due course' as come ...

.... OK. Although it was all the data I had to hand at the time, my previous attempts to illustrate ‘diversity’ with data relating to heating a large (insulated) cylinder of water with an immersion heater was not very analogous to cooking appliances, and my attempts to ‘scale it down’ to much shorter time periods went all wrong!

I have no electric cookers, ovens or hobs, so I have undertaken experiments using a 1,700 W deep fat fryer, which I switched on from cold and ran for an hour or so at ‘chip frying temperature’. Whilst this is probably not an ideal model of a hob, it’s probably not a bad model of the likely behaviour of an oven. What I found is illustrated in the graphs below.

Note that, following plugwash’s comments, the 'average power' over 15-minute periods has now been calculated correctly - i.e. as the RMS (rather than mean) of the data during that period.

[ note also that the few brief periods of ‘low' (<1,700 W) consumption result from the fact that, although data is collected once every ~12 seconds, the stored data I have toi work with are ‘1-minute averages'. Hence, if the element was only ‘on’ for part of a 1-munute period, the indicated coinsumption will be a corresponding fraction of 1,700 W ]

As can be seen, the average (RMS) power consumed by the 1,700 W appliance fairly quickly settles down to around 600 W, meaning that the heating of cables etc. will be approximately the same as with a load drawing ~600 W continuously.

Scaling that up to larger cooking appliances (or collections of cooking appliances), assuming that they 'behave' (in terms of on/off thermostatic cycling) similarly to the fryer, if one had 11 of my fryers (total of 18,700 W, about 81.3 A at 230 V) turned on from cold simultaneously, they ought to settle down to a total average load of about 6,600 W (about 28.7 A at 230V).

Hence, if my fryer is ‘typical’ of cooking appliances, a collection of cooking appliances whose maximum loads represented a total of 81.3 A (at 230V (18.7 kW) should result in an average effect on (i.e. heating of) cables which is the same as would be achieved by a load constantly drawing about 28.7 A at the same voltage (i.e. roughly a 6.6 kW load).

This is very similar to (in fact, a bit more 'conservative' than) the conclusion one would come to by use of the standard domestic cooking appliance diversity calculation, which would say that the after-diversity current for a total cooking appliance load of 81.3A would be about 31.4 A.

Hence, both my empirical observations and the standard diversity calculation indicate that a total cooking appliance load of 81.3 A (about 18.7 kW) should be OK on a '32 A circuit'.

If (as would nearly always be the case in domestic practice), there were some 'staggering (i.e. not absolutely everything switched on from cold simultaneously), then the average power consumption would be likely to be a bit less than these figures suggest.

Edit: for anyone reading this, please note that I may well fairly shortly be posting some revisions to the above!

Kind Regards, John

 

[ChrisZx]

A lot of effort gone into this! It's interesting, thanks. Think you've all convinced me 32A would be sufficient!