For techies - Potential Flaw in Zs/ADS calculations?

[JohnW2]

2 points, one, I think with tn-s the situation would be better, as only the L has a load on it, the E is in the no load situation.

You may be right - I'll try when I have some moments. The L does, of course, carry boththe 'load' and the 'fault' currents.

The other point about why the error is so small, i think it's because (or maybe the reason why) the permitted voltage drop is small. If it helps, consider a situation where the transformer voltage is 460v and the supply voltage is 230v due to other loads.

I'm not so sure about that. If you look at the last simulation I posted (message 71), although I was using a transformer voltage of 250V, what I presented went all the way to a supply voltage (without fault) of just 82V (due to high loads), and even then the 'error' (in PFC) was still under 10% - and that was a much more dramatic VD than from 460V to 230V. However, when I have a moment I'll post a simulation of what you describe.

Kind Regards, John

 

[John D v2.0]

Interesting, good point!
So given the fault current want affected much, that means at the installation, a reduced normal voltage through load doesn't affect the fault voltage much. That's not what I expected either! Which means the voltage during a fault must be lower than the 82 volts.

 

[JohnW2]

The other point about why the error is so small, i think it's because (or maybe the reason why) the permitted voltage drop is small. If it helps, consider a situation where the transformer voltage is 460v and the supply voltage is 230v due to other loads.

This is very interesting - there is seemingly a limiting value to the error. No matter how heavily one loads the LV network, the 'error' (underestimation of true PFC by 'conventional' Zs-based calculation) never seems to exceed 13%. In the below, with a transformer voltage of 460V, the error is still under 13% (actually 12.99%) even if I load the network so heavily that the supply voltage at the premises (without fault) falls to about 20 μV! Since there clearly is a limiting value, it must be possible to determine that value algebraically - so I haven't given up that exercise, not the least because it might help me to understand the reason for that 'limit'!

Interesting, good point! So given the fault current want affected much, that means at the installation, a reduced normal voltage through load doesn't affect the fault voltage much. That's not what I expected either! Which means the voltage during a fault must be lower than the 82 volts.

Indeed so. If you look at that simulation in msg #71, you'll see that the supply voltage (without fault) of "82V" (actually 81.97V) falls to 71.79V during the fault, corresponding to a fault current of 62.1A).

Kind Regards, John

 

[John D v2.0]

There must be a mistake, surely if the supply voltage at the premises (without fault) falls to about 20 μV then if r1+r2 is around 1 ohm then the fault current must be less than 20 micro amp? (Edit: corrected)
However that's made me realise why the fault current doesn't drop as much - as the load on the network increases, the fault current has a proportionally smaller effect on the network.
Try a higher shared ze 0.8ohm and a lower r1+r2 say 0.2ohm for a b45 cooker circuit, and no local ze.

 

[JohnW2]

There must be a mistake, surely if the supply voltage at the premises (without fault) falls to about 20 μV then if r1+r2 is around 1 ohm then the fault current must be less than 1 micro amp?

20μV across 1Ω is actually 20μA, not "less that 1 microamp" In fact, in my simulation (with R1+R2 = 1.2395Ω), the fault current was 17.17μA - but so what, and why do you regard it as a 'mistake'?? Such ridiculously small supply voltages are obviously of no practical interest - the point is that it seems clear from my simulations that as the supply voltage gets lower and lower, the error in PFC estimation approaches 13%, but never seems to exceed that value.

Try a higher shared ze 0.8ohm and a lower r1+r2 say 0.2ohm for a b45 cooker circuit, and no local ze.

OK, watch this space. However, I think I could make a reasonable guess of what it is going to show

Kind Regards, John

 

[John D v2.0]

Sorry I meant less than 20 micro amps. I'll edit my post for clarity for others.
And also I was forgetting there are two sources of error, one being the voltage used is not the transformer voltage and the other being the total load taken by the other consumers is reduced by the fault.
So maybe the ceiling is based on the difference between the two errors. And as the first grows, the second will also grow but at a different rate.
I actually tried to calculate actualpfc = k•regspfc where k only depends on ZeShared, Zeprivate+r1+r2 and Zotherloads, but I ended up in algebra hell.

 

[JohnW2]

Try a higher shared ze 0.8ohm and a lower r1+r2 say 0.2ohm for a b45 cooker circuit, and no local ze.

Hmmmm - my guess was certainly wrong, and now I'm getting more confused/uncertain. I changed to a shared Ze of 0.8Ω (and a zero 'local Ze'). However, I had to reduce the transformer voltage to 300V and use an (R1+R2) of 0.409Ω to get it to be 'right' for a B45 at 230V (225A PFC). As you can see below, there was again a limiting value to the error, but this time the 'limit' was a bit over 66%, very different from the 13% I had before. Indeed, all the errors (underestimates of PFC) were 'high' - at a 230V supply current the error is nearly 35A (about 15%). It therefore looks as if, if I can ever get my head around the algebra, the error (or, at least, it's limiting value) may well to prove to be a function of the ratio of Ze to (R1+R2), or something like that. However, we are now seeing errors in the conventional method of estimating PFC which are somewhat more 'serious', hence perhaps with more practical implications.

Sorry I meant less than 20 micro amps. I'll edit my post for clarity for others.

Fair enough.

I actually tried to calculate actualpfc = k•regspfc where k only depends on ZeShared, Zeprivate+r1+r2 and Zotherloads, but I ended up in algebra hell.

Indeed, I have been living in 'algebra hell' for a few days, and it reminds me of periods of my education in the distant past, when I would stay up all night fighting with scribbles on very large sheets of paper. At least this present exercise only involves very simple/basic (albeit very tedious) algebra, without all the complications of calculus which usually pervaded those sleepness nights in the past!!

Kind Regards, John

 

[John D v2.0]

OK I've successfully worked out a formula for
Ir/If (REGS_PFC/ACTUAL_FC) in terms of

1. the effective shared Ze
2. the effective resistance of the other load Ro, and
3. the non shared Ze+R1+R2 (Rf)

So we would get 1 if they are the same, but if the regs calculated value is (say) 13% lower, we would get 0.87.
I graphed the ratio with each of the three terms varying and I found the resistance of the other loads on the network always causes the same percentage error.

Here shows the bigger the Ze the bigger the error (obviously, since with no Ze we'd be measuring true FC)

Here shows the bigger the R1+r2, the more accurate things are

And here shows the resistance/current of the other loads has a small effect

EDITED the graphs due to a spreadsheet error

 

[JohnW2]

And here shows the resistance/current of the other loads has no effect on the *percentage* error of the PFC

Thanks. As you say, your first two graphs are 'as expected'. However, this third one is clearly at variance with my simulations - which showed a marked effect of the 'other loads' (as reflected by the supply voltage on my x-axis) on the percentage error in fault current (which is what I plotted). What were the other 'conditions' (in terms of Ze, R1+R2 etc.) for this third plot?

Kind Regards, John

 

[John D v2.0]

Here is the formula I used - the definitons here:

The actual forumla here:

 

[John D v2.0]

Sorry, somehow all my "B" in the spreadsheet got turned to "C", I'll edit the original post to have the correct graphs

Edit: I have edited my above post. See below for an intutiton.

 

[John D v2.0]

ok I've got it, for a low error given a certain Ze, you need either the other load or the fault loop to have a relatively high resistance.

The intuition works because if there's almost no other load, there's almost no error on any huge fault current. Also if there is almost no fault current (high R1+R2) then there's going to be almost no error regardless how loaded the network is. However if you make Ze tiny, doesn't matter if there's a heavy load or a huge fault current, there won't be an error.

So the key to low error is making the *shared* Ze low compared with the resistance of the other loads. IE not too many people on the same cable unless you make it thick!

If Ze, Ro and Rf are equal, there's always a 25% error.

Do we need a new thread to discuss the 0.95/0.94 thing?

 

[JohnW2]

The intuition works because if there's almost no other load, there's almost no error on any huge fault current.

True ... but that's not going to happen in practice.

Also if there is almost no fault current (high R1+R2) then there's going to be almost no error regardless how loaded the network is.

Again true, but again that is not going to happen in practice, since 'almost no fault current' means that there is no hope of the requirements for ADS being satisfied!

However if you make Ze tiny, doesn't matter if there's a heavy load or a huge fault current, there won't be an error.

Yet again true, but yet again pretty irrelevant in practice, since the designer has no control over Ze, and it's only going to approach 'tiny' if one has a substation in one's garden!

So the key to low error is making the *shared* Ze low compared with the resistance of the other loads. IE not too many people on the same cable unless you make it thick!

As above, it's all very well saying that it's the "key", but the designer and consumer have no control over Ze. The regs' estimate of PFC is a 'blanket' one, and which will be applied as much to someone at the end of a long LV network, with a high Ze and low supply voltage, as to someone sitting on top of the substation. From what we've discovered, it looks as if, in the forrner case, in some situations the under-estimation of PFC by the regs' method could be very substantial - perhaps to the extent of causing designers to specify a larger cable CSA than was actually necessary.

Kind Regards, John

 

[John D v2.0]

Hey you can't talk about "in practice" cos this is a brain game! If you're being serious, the extra cost of somehow calculating an accurate FC would far outweigh any cable saving I was just trying to help myself intuitively think of it that way to convince myself it was right.

Anyway, for my final flourish, I've proved that our lighting circuit is the most dangerous in the house - you were right that 0.95 doesn't make sense and it should be 0.94 for the voltage to use:

As you can see, when Ze is below 0.5 ohms, our lighting circuit that (just) passed the Zs test will not be guaranteed to trip the MCB in time!

That assumes Vtrans=253, our origin voltage is 0.94x, and the MCB needs 5x to trip quickly.

 

[JohnW2]

As you can see, when Ze is below 0.5 ohms, our lighting circuit that (just) passed the Zs test will not be guaranteed to trip the MCB in time!

Hmmm. Are you sure that your model is correct, since that's something which I have not yet ever seen with any of my simulations. I've looked at a fairly wide range of situations and the ('per regs') Zs-based method has always under-estimated actual PFC, never overestimated it - which is why I thought I had at least reassured myself that errors in the estimates were always going to be in the 'safe' direction. When I have a little time, I'll see if I can reproduce your equation.

Kind Regards, John