What's your average single phase measured voltage? Is 251V at point of measurement common?

[plugwash]

If, as in practice, there were many such independent loads/thermostats, operating 'out of synch', with one another, exactly the same arithmetic would apply to each of them - so, whether considered separately or together, DNO cable energy losses would presumably be the same regardless of voltage/ That's how I see it, anyway - am I wrong?

You can't just do the sums for each load individually and then add up the results, because losses are proportional to the square of current.

 

[ericmark]

Not necessarily. Don't forget that these days power flows can reversed in the local networks due to embedded generation (e.g. solar PV). So the voltage at the end of a line can be higher than at the transformer.
The DNO likes to keep the voltage up as it minimises I2R losses. Then embedded generation comes along and reduces or reverses voltage drops.

Where I use to live, the voltage was always on the high side, and I had an old 65 watt fat tube fluorescent fitting, which was designed for 240 volt, and it worked fine most of the time, then two things happened, one could not buy 65 watt tubes, but 58 watt worked OK but shorter life, and then a load of solar panels fitted, will guess were going over voltage, and locking out, and the voltage dropped, to 230 volt and my fluorescent fitting would not fire up, so forced to fit a LED tube.

Solar must push up the volts, the national demand may increase during the day, but the local demand in a housing estate likely the reverse, this shows what I use on a typical day, below the line using from grid, above line feeding to the grid, so 1 am start charging battery, 5 am start using battery, 11:30 am the battery again full so start exporting until around 5 pm when solar not enough for my needs, until 11 pm when battery down to 10% so start using grid power, so 11:30 am to 5 pm I am exporting and likely all other homes in the estate are the same, but also likely most out at work, so no power being used, so unless the DNO transformer can auto change the tapping, then the voltage is likely to raise.

 

[JohnW2]

You can't just do the sums for each load individually and then add up the results, because losses are proportional to the square of current.

Instantaneous losses of power (I²R) are, indeed, proportional to current but the amount of time current has to flow to deliver any given amount of energy is inversely proportional to the square of current.

Hence, as I keep trying to suggest, those things appear to 'cancel' such that the losses of energy associated with the delivery of any given amount of energy (i.e. I²Rt) would appear to be the same, regardless of voltage (and current) -whether one is talking about a single load or multipleloads. Am I still wrong?

Looked at in a different way, if the loads were both 'dumb' and continuously present, then increasing voltage would increase the current, hence also the instantaneous I²R losses - and, since that situation would remain unchanged indefinitely, the energy losses would also increase. However, in that situation the 'energy delivered' would also have increased. If one then introduces thermostatic/whatever control which switches off the load when as much energy has been delivered as before the voltage increase, then that 'energy loss' would appear to get be cancelled. Still wrong?

Looked at in yet another way, I²Rt ('energy losses') in the DNO's cables whilst supplying one or more loads with a given total amount of energy will always be the same, regardless of voltage. In fact, if the resistance of the DNO's cables is Rd, the effective resistance of all the loads Rl and the total energy supplied to the loads is EN, then I think the situation (regardless of voltage) is ...

Energy loss in DNO cable = EN x (Rd / Rl)

... isn't it?

 

[SimonH2]

Looked at in a different way, if the loads were both 'dumb' and continuously present, then increasing voltage would increase the current, hence also the instantaneous I²R losses - and, since that situation would remain unchanged indefinitely, the energy losses would also increase. However, in that situation the 'energy delivered' would also have increased. If one then introduces thermostatic/whatever control which switches off the load when as much energy has been delivered as before the voltage increase, then that 'energy loss' would appear to get be cancelled. Still wrong?

OK, with non-controlled linear loads, as voltage goes up, so does power delivered, and DNO losses. Losses as percentage of energy supplied will be roughly unchanged.
If you consider controlled loads - think ovens, immersion heaters, room heaters, ... ON AVERAGE, as voltage increases, current decreases. So delivered energy remains constant, while losses reduce. It's important to note that you have to sum all the load currents BEFORE doing your I²R calculation because Σ(I²) is not the same as (ΣI)². Simple example, make I=2 twice. Σ(I²) =8, (ΣI)² =16.

 

[JohnW2]

OK, with non-controlled linear loads, as voltage goes up, so does power delivered, and DNO losses. Losses as percentage of energy supplied will be roughly unchanged.

Indeed - and, as I've said, similarly in practice when (as is usually the case) thermostatic/whatever control will prevent the 'energy delivered' increasing - so, again, the DNO energy losses as a percentage of energy supplied will remain unchanged if voltage increases..

If you consider controlled loads - think ovens, immersion heaters, room heaters, ... ON AVERAGE, as voltage increases, current decreases. ...

Yes, over a sufficient period of time, the 'average' current will decrease, because the (increased) current) will be drawn for much less of that period of time (that duration being reduced by a factor of the change in V² (or I²), won't it? However ...

So delivered energy remains constant, while losses reduce.

I don't get that. Yes, 'over a sufficient period of time' the average DNO energy losses would reduce BUT so would the average 'energy delivered' over that period decrease (by the same factor), since the energy consumed by the load will have been zero for some of that period - so I would again expect the DNO energy losses to remain the same proportion of the energy delivered. Again, I ask if I am wrong?

Consider again a hypothetical simplistic example - DNO cable resistance 1Ω, and an immersion heater with a resistance of 200Ω which (under thermostatic control) needed 1,000 Wh of energy to heat the cylinder's contents to the desired temp.

With, say, a 200 V supply, the current would be 1 A, hence a power of 200 W. Attaining the required 1,000 Wh of energy would therefore take 5 hours. During those 5 hours, the DNO energy loss will have been 5 Wh (1² x 1 x 5).

Now (hypothetically!) increase the voltage to 400 V. Current is now 2A, power 800W, so attaining the desired 1,000 of energy would take just 1.25 hours. During those 1.25 hours, the DNO energy loss will again have been 5 Wh (2² x 1 x 1.25).

Hence, with either 200 V or 200V (or, come to that, any other voltage) DNO energy loss whilst supplying 1,000 Wh to the load will be 5 Wh - which is no surprise given that, as I wrote last night, the ratio of DNO energy loss to energy supplied appears always be simply the ratio of the resistance of the DNO cables to the effective resistance of the load(s)., regardless of voltage

It's important to note that you have to sum all the load currents BEFORE doing your I²R calculation because Σ(I²) is not the same as (ΣI)². Simple example, make I=2 twice. Σ(I²) =8, (ΣI)² =16.

I don't think that I have suggested summing load currents before squaring them - everything I've said has been in terms of I² (more specifically, I²t) I think one of the problems here may be that some people (including yourself) seem to have been talking in terms of power, rather than energy (i.e. I²R, rather than I²Rt).

However, particularly given that I am now suggesting things that have never previously occurred to me, I am far from convinced/confident - which is why I keep asking whether I have 'got it wrong'!

 

[SimonH2]

I think your just looking too close - take a step back.
Let's assume all loads are controlled for a moment, and consider "lots" of end users (i.e. consuder a big chunk of network). And again consuder an impractical doubling of voltage. Yes, individual loads will draw double the current when on - but when you average hundreds or thousands of loads, then the average total current will be halved, and losses would be quartered.
Clearly the savings won't be anything like that. But remember that the DNO network is far more than just the cable passing your house. As well as cables, there's a lot of transformers - and as someone hinted at, reducing the current coukd be the difference between between making do and needing an uograde.

 

[JohnW2]

I think your just looking too close - take a step back.

Even though I've yet to see why/how, you could well be right. Until a couple of days or so ago, for many decades I had just 'blindly accepted' the accepted view that increased voltage meant decreased current and hence "less losses" in the conductors.

However, in these recent days I have more-or-less convinced myself that, whilst that is true about 'losses of power' at a point in time whilst current is flowing, it is not true of overall 'losses of energy' (which is what matters), because (all other things being equal) higher 'power losses' exist for less time (during delivery-of the same given amount of energy to load).

Let's assume all loads are controlled for a moment, and consider "lots" of end users (i.e. consuder a big chunk of network). And again consuder an impractical doubling of voltage. Yes, individual loads will draw double the current when on - but when you average hundreds or thousands of loads, then the average total current will be halved, and losses would be quartered.

Modelling, or even thinking about, "hundreds or thousands of loads", is obviously less than straightforward, since there are an infinite number of possible variations in the temporal relationships between those loads being active - so let's start with two extreme situations ...

As the first extreme. consider many loads starting simultaneous, all requiring the same amount of energy before they are 'switch off' by controls, hence all switching off' at the same time (with a given voltage supply). If one doubles the voltage then, as you say, the current will double(for as long as it flows) but, because it only flowers for one quarter of the time, the the average current will be halved, and hence the loss of power (at any point in time whilst current is flowing - i.e. I²R) will be 'quartered'. However, the amount of time for which current flows is also 'quartered', so that the loss of energy (I²Rt) will actually be unchanged by the change in voltage. ...so, again, the problem seems to be that you are thinking only of power losses, rather than the energy losses that matter.

The second extreme is perhaps more trivial, since it is merely repetition of the 'one load' scenario. Consider many identical loads occurring sequentially, with no overlaps, such that with the initial voltage, each 'next one' gets switched on immediately after the previous one one gets switched off (by thermostatic/whatever control). If one doubles the voltage, it is again true that current (whilst it is flowing) is doubled, leading to a 'quartering' of power loss (I²R) whilst current is flowing - but this time it is even more obvious that current has only flowed (hence losses occurred) for one quarter of the time - so, again, loss of energy (I²Rt) is unchanged by the change in voltage

When I have some moments, I will attempt to computer model some more complicated situations in which the many loads are 'staggered and overlapping', but I will be pretty surprised if that exercise does not confirm that the energy loss in conductors is (if 'all other things are equal') always the same when delivering a certain amount of energy to the load, regardless of voltage (and current, when it is flowing)- and, furthermore, as I've suggested, that the ratio of energy loss in conductors and energy supplied to the load is

As I've said, until just a few days ago it had never occurred to me that this would be the case - but no-one has yet persuaded me that my recent reasoning is incorrect!

 

[JohnW2]

.... When I have some moments, I will attempt to computer model some more complicated situations in which the many loads are 'staggered and overlapping',...

I was not far into trying to model Simon's "hundreds or thousands of loads" when I was reminded how unrealistic such a scenario is in terms of LV networks. As one increases the number of simultaneous/overlapping large loads, one fairly soon gets to the point at which supply voltage to consumers drops to a ridiculous extent and much, if not most, of the DNOs energy goes into heating their network cables rather than being delivered to consumers!

The same current goes through the DNO's network cables and the combined resistances of all the consumers' loads in parallel - so, at any point in time, instantaneous power loss in the network cables, as compared with instantaneous power delivered to consumers will simply be determined by the ratio of those resistances. A 3kW appliance has a resistance of about 17.6Ω and a 10.5 kW shower a resistance of about 5Ω - so it doesn't take many of those in parallel for the 'load resistance' to become similar to, or less than, the 'network resistance' (L-N 'loop impedance').

I will now re-visit the modelling with some more realistic scenarios

 

[SimonH2]

Consider something more sensible. Lets say 1000 loads, all running at say 10% duty cycle. So assuming independent loads, at any time there will be around 100 loads on - give or take a few.
Double the voltage, doubke the current for each load while it is on. But at 4 times the power, the duty cycle will drop to 2.5% - neaning that at any time there will now be just 25 loads on. So double the voltage, 2 x 1/4 factor, or half, on the current - give or take a bit as loads turn on and off independently. Half the current (give or take), 1/4 the DNO losses.
That's what I mean by stepping back a bit and taking a wider view.
As previously noted, not all loads are controlled, but I suspect they make up enough of the total load to make the savings worthwhile.

 

[ebee]

Unfortunately I have not yet had time to follow and check the details of this, just a quick glance.
Hopefully I will get time to have a look properly and digest the logic in a few days time.
I don`t think that I`m as quick on the logistics as John so I have a suspicion he has a point we might have all overlooked.
Like many I am likening it to the I2r losses the network operators use to make thinner conductors spaced further apart when increasing voltage for transmission distances.

Reminds me a bit of the old Ring Final Fig 8 Crossover Test for R1 + R2, we were taught by some great tutors that the answer was "Exactly" R1 + R2 and I happen to be one of those folk who tend to remember such things better if I understand them rather than just accept the fact,

So once I heard a few of the great and the good repeat this word "exactly" I decided to check it and found it to be slightly out so I suggested the word should be changed to "Substantially".
It is very near the correct answer but not quite.
Percentage wise "Exactly" looks quite wrong but but in real life with low measurements and then reckon measuring errors due to meter inaccuracies plus field errors then it is not wildly out in reality.
Anyway, I put it out on a forum and in the years since not one single person got back to me to say I was wrong so I assume a few of them being brain boxes they might well have checked and decided I was correct. Until someone tells me I was wrong of course.

 

[JohnW2]

Unfortunately I have not yet had time to follow and check the details of this, just a quick glance. ... Hopefully I will get time to have a look properly and digest the logic in a few days time. .... I don`t think that I`m as quick on the logistics as John so I have a suspicion he has a point we might have all overlooked.

Maybe. As I've said, I'm as 'surprised' as everyone else by what I've been suggesting in the last few days! Prior to that I have simply 'accepted', without much thought, the idea that increasing voltage, hence decreasing current, will always result in less 'losses' in the conductors.

Like many I am likening it to the I2r losses the network operators use to make thinner conductors spaced further apart when increasing voltage for transmission distances.

As I've said, there's no doubt that high voltage, hence low current, offers advantages for HV distribution networks. I am possibly coming to doubt that it offers any advantage (for a given conductor) in terms of energy loss but it certainly means that, for the same energy loss, the conductors can be of smaller CSA, hence cheaper, lighter (less difficult to support) and more 'environmentally friendly'.

To distill the essence of what I've been suggesting ...

As I keep saying, it seems that others (and myself until very recently) are thinking/talking about power losses (I²R), whereas what matters is energy loss (²IRt). With a 'simple' load, increasing voltage will increase current proportionately, hence increasing power losses at any point in time. However, in terms of loads which are 'controlled' (e.g. thermostatically) so as to draw a particular total amount of energy, the duration of the current flow will decrease by the same factor, such that total energy loss (I²R x t) remains unchanged.

In the case of a single such load, the logic and arithmetic seems trivial ('famous last words'?!), and supportive of my suggestion ... with such a load, if one doubles voltage, one also doubles current, hence increasing power losses (I²R) by a factor of 4. However, with such a 'controlled' load the duration of current flow (to deliver a certain total amount of energy) will reduce by a factor of 4 - so the energy losses (I²Rt) will remain unchanged, despite the increase in voltage. The same is obviously true of any number of identical loads which commence simultaneously, since that is no different from a single larger load.

If you agree with that last paragraph, then that, in itself, would seem to fly in the face of the traditionally-accepted view of such things. Indeed, until just a few days ago, I would have said, without question, that even with a single ('controlled') load, increasing voltage would (by decreasing current) 'reduce losses' - but that's because, like everyone else, I was thinking about power losses (at any point in time) not energy losses.

So far so good (in support of my 'suggestion'). However, what I'm discovering is that the logic and modelling gets much more complicated and confusing when one has a mixture of different (magnitudes of) load and/or only partial overlap between the times they are active. With such scenarios I am currently getting very confusing answers, and so am still 'working on it' - so 'watch this space'

I continue to keep asking myself (and all you folk out there) "where am I going wrong?" !

Kind Regards, John

 

[JohnW2]

Consider something more sensible. Lets say 1000 loads, all running at say 10% duty cycle. So assuming independent loads, at any time there will be around 100 loads on - give or take a few.

Assuming that we are talking of 'large' loads, even that is definitely 'pushing it' for any likely LV network, but OK!

Double the voltage, doubke the current for each load while it is on. But at 4 times the power, the duty cycle will drop to 2.5% - neaning that at any time there will now be just 25 loads on. So double the voltage, 2 x 1/4 factor, or half, on the current - give or take a bit as loads turn on and off independently. Half the current (give or take), 1/4 the DNO losses.

I need to think deeply about that because, at first glance, I can't see a way of arguing with your logic or arithmetic!

However, I don't (currently!) see why/how that differs from a situation in which there is just a single load, 100 times larger than each of your 1,000 ones, 'operative' just once (i.e. until the required amount of energy had been delivered to the load). In that case, doubling voltage would double DNO I²R power losses (at any point in time), but would also reduce the duration of those power losses by a factor of 4, thereby leaving energy losses unchanged.

I may have to start pulling out some of my hair soon

 

[JohnW2]

There are clearly some issues which I have yet to get my head around when there are multiple loads, differing in magnitude and/or timing, but can we at least agree on this ...

... in the case of a single ('controlled') load, the energy loss in the DNO network when supplying a given amount of energy to the load will always be the same, regardless of voltage. Furthermore, since the same current, both in magnitude and duration(s) goes through both the DNO network and the load, the energy loss in the DNO network will always be equal to the energy supplied to the load multiplied by [ R(dno) / R(load ], again regardless of voltage.

??

 

[JohnW2]

Double the voltage, doubke the current for each load while it is on. But at 4 times the power, the duty cycle will drop to 2.5% - neaning that at any time there will now be just 25 loads on. So double the voltage, 2 x 1/4 factor, or half, on the current - give or take a bit as loads turn on and off independently. Half the current (give or take), 1/4 the DNO losses.

I need to think deeply about that because, at first glance, I can't see a way of arguing with your logic or arithmetic!

I've done a bit of 'deeper thinking' and am perhaps 'onto something' - and it actually relates to something you previously wrote but which I think (at least in the context we're now discussing) you may have got 'the wrong way around' ... when you wrote....

..... It's important to note that you have to sum all the load currents BEFORE doing your I²R calculation because Σ(I²) is not the same as (ΣI)². Simple example, make I=2 twice. Σ(I²) =8, (ΣI)² =16.

As I wrote,, I think that in the context we are discussing, you may have got that the wrong way around ....

... consider a simple example of 200 V supplying a load of 100Ω resistance. That's a current of 2A, so a power of 400 W - one way of calculating which is "I²R", namely 2² x 100.

If one connects two such loads to the 200 V supply, I presume that you agree that the total power of the two will be 800 W (400 W x 2)(. However, if you added together the currents (hence getting 4 A) before squaring, you would end up calculating total power as I²R = 4² x 100, namely 1,600 W, twice as much as it actually is.

The message therefore is that one has to sum the powers (e.g. by summing I² and then multiply by R) and that one should NOT sum the currents before squaring them.

It's the same when one talks about an average current (that simply being the sum of currents divided by a constant) - again, one has to sqaure the currents and then sum those squares - i.e. determine Σ(I²) and then multiply by R to get power (and then by t to get energy).

Is that not correct? If you agree with me, then that would explain the answer that you got (and which I believe to be incorrect) in the first quote of yours above, when you squared the average current, rather than "averaging the squared currents" (which is what i believe should be done), wouldn't it ? I suspect that I probably made the same mistake with my model that was giving confusing answers, so I'm now going to 're-visit it, hopefully with a somewhat clear mind!

 

[SimonH2]

I've done a bit of 'deeper thinking' and am perhaps 'onto something' - and it actually relates to something you previously wrote but which I think (at least in the context we're now discussing) you may have got 'the wrong way around' ... when you wrote....

As I wrote,, I think that in the context we are discussing, you may have got that the wrong way around ....

... consider a simple example of 200 V supplying a load of 100Ω resistance. That's a current of 2A, so a power of 400 W - one way of calculating which is "I²R", namely 2² x 100.

OK

If one connects two such loads to the 200 V supply, I presume that you agree that the total power of the two will be 800 W (400 W x 2)(. However, if you added together the currents (hence getting 4 A) before squaring, you would end up calculating total power as I²R = 4² x 100, namely 1,600 W, twice as much as it actually is.

What is the power with 200V into a 50Ω load ? It's going to be 4A * 200V = 800W, or 4² * 50, or 16 * 50, = 800W.

The message therefore is that one has to sum the powers (e.g. by summing I² and then multiply by R) and that one should NOT sum the currents before squaring them.

Only if you miss the fact that 2off 100Ω in parallel is 50Ω


But, we were discussing losses in the network. Let's assume we have a loop impedance of 0.1Ω, and 100off such loads on the network.
At 200V, 100 of such loads will be 200A and the volt drop will be 20V, so the losses in the distribution network will be 20V * 200A = 4,000W while supplying a total load of 40kW - so network losses are around 10% of supplied energy.
If we were to double the voltage to 400V, then without any controls, each load would pull 4A and 1,600W. 100 such loads would draw 400A and 160kW. The losses in the network would now be 40V drop @ 400A, so 16kw, so still around 10% of supplied energy.

Now consider if the loads are all thermostatically controlled. Each will now change it's duty cycle to around 25% - 25% of 1,600W is 400W. At any point in time, "around" 25% of loads will be on, and so the total of 100 loads will be "around" 100A, give or take a bit - I suspect if you plot current versus frequency it'll come out looking something like a bell curve.

As an aside, there is a reason EV charge points are required by regs to have randomisation of on times. And in the USA, they also made electronic controls suppliers introduce randomisation of on times. In the USA, they suffered from having a large amount of electric heating (more common than over here), increasingly controlled by "smart" controls, with clocks set from the internet (so no spread due to clocks being out by random amounts of typically a few minutes), and switching on at default program times. The result was a massive and near instantaneous jump in electrical demand which was hard to cater for even though it was very predictable - whereas prior to all these "smart" controls, it ramped up over at least several minutes. In either case, the load would start to tail off as homes got up to temperature and the heating would start modulating on the stat.

But back to our discussion. We now have around 100A in our distribution network. That's going to give us around 10V drop. Regardless whether you do 10V * 100A, or 100A² * 0.1Ω, you get 1,000W. So for the same 40kW supplied, we now only lose 1kW in the network, only ¼ of the losses at 200V, and 2.5% of supplied energy.

As already mentioned, doubling the voltage isn't on the cards. But if you (for example) increase supply voltage by 10%, then you would reduce the current drawn by controlled loads by about 9% (1/1.1 = 0.90909). 0.90909² = 0.826, so a little over 17% reduction in network losses.

However, what we DO have is local transformers (not generally as local as they have in the USA), so a lot of that load is carried not at 200V or 400V but at 11kV. When you've transformed 200V to 11kV, your loads are now only drawing 4kW/11kV = 0.36A. 100 such loads now only draw 36.36A, and if we ignored that a chunk of the loop impedance will come from the transformer and street cable, then the losses would now only be in the region of 132W.
Which is, as we well know, why AC supply quickly killed off DC because of the ease of transforming the voltage from a high voltage for distribution with manageable losses, to low voltage for practical (and reasonably safe) use by the end user.

Does that make things clearer ?