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Devices to meter kWh

Ahh, I should have said "have an inductive (rather than capacitive) component". I didn't mean to imply "purely inductive".
Yes, I assumed that but, as I said, loads have to be primary restive if they are going to actually consume any energy (i.e. 'work' :-) ).

This is the crucial difference between VA and W - if a load is primarily inductive or capacitive, VA (volts times amps) can be very high even though W (the Watts of power delivered may be close to zero!
 
Many thanks. I think I may at last have 'got it', but you seem to have made things unnecessarily complicated by connecting one end of the CT to the L side of the circuit,such that, as you say, the induced voltage in the CT either adds to or subtracts from the supply voltage. If you simply connected the voltmeter across the CT (with the CT not connected to anything else), it would simply indicate the induced voltage.
...
I'm not quite sure what you are saying here. The simple way of reading the energy (VA) being consumed is as shown, above although it's more likely to be at much lower voltage.
(my original sketch was
1747328584437.png
and totally based on the original of 20 odd years ago to explain the need for correct polarity)
 
I'm not quite sure what you are saying here. The simple way of reading the energy (VA) being consumed is as shown, above although it's more likely to be at much lower voltage.
I wasn't talking about practical implementations but, rather of the exercise of explaining why the phase of the induced voltage in the CT relative to the phase of the supply is an indication of 'flow direction' - something which could have been illustrated without having that voltage adding or subtracting from the supply voltage.
(my original sketch was ,diagram> and totally based on the original of 20 odd years ago to explain the need for correct polarity)
I'm not sure I understand that. The transformer on the right is presumably a standard 'voltage transformer' with a 10:1 turns ratio - but that means that for a 1A current going to the load, the meter would (with flow in the direction you indicate) read "34 V" - but if you connected the CT 'the wrong way around' it would read"14 V" - neither of which appear to be immediately/directly 'meaningful'?

However, that's all irrelevant to my understanding of how direction of flow can be detected (by phase relationship between supply voltage and voltage induced in the CT) - and for that enlightenment I very much thank you ;)

Kind Regards, John
 
I wasn't talking about practical implementations but, rather of the exercise of explaining why the phase of the induced voltage in the CT relative to the phase of the supply is an indication of 'flow direction' - something which could have been illustrated without having that voltage adding or subtracting from the supply voltage.
It is the simplest way to describe the operation without having to introduce some form of comparator
I'm not sure I understand that. The transformer on the right is presumably a standard 'voltage transformer' with a 10:1 turns ratio - but that means that for a 1A current going to the load, the meter would (with flow in the direction you indicate) read "34 V" - but if you connected the CT 'the wrong way around' it would read"14 V" - neither of which appear to be immediately/directly 'meaningful'?
Yes, if you look back at my initial description the reading is shown as 340V & 140V without transformer. The additive combination running at much lower voltage (typically 1 to10V min to max) is virtually the elderly detector format used to remotely measure power or feed into BMS systems.
However, that's all irrelevant to my understanding of how direction of flow can be detected (by phase relationship between supply voltage and voltage induced in the CT) - and for that enlightenment I very much thank you ;)

Kind Regards, John
It's quite easy to detect the phase relationship
No load; I=0A, induced voltage = 0V + 240V = 240V
2400Ω or 24W load; I=0.1A, induced voltage = 10V + 240V = 250V
240Ω or 240W load; I=1A, induced voltage = 100V + 240V = 340V (shown in sketch)
100Ω or 576W load; I=2.4A, induced voltage = 240V + 240V = 480V

63.16Ω or 912 W load; I=3.8A, induced voltage = 380V + 240V = 620V
51Ω or 1128W load; I=4.7A, induced voltage = 470V + 240V = 710V
50Ω or 1152W load; I=4.8A, induced voltage = 480V + 240V = 720V

It can be seen there is a voltage proportional to the load power, starting at 240V and rising.

reverse the CT and the the resulting voltage starts at 240V then drops

No load; I=0A, induced voltage = 0V + 240V = 240V
2400Ω or 24W load; I=0.1A, induced voltage = -10V + 240V = 230V
240Ω or 240W load; I=1A, induced voltage = -100V + 240V = 140V (shown in sketch)
100Ω or 576 W load; I=2.4A, induced voltage = -240V + 240V = 0V

63.16Ω or 912 W load; I=3.8A, induced voltage = -380V + 240V = -140V
51Ω or 1128W load; I=4.7A, induced voltage = -470V + 240V = -230V
50Ω or 1152W load; I=4.8A, induced voltage = -480V + 240V = -240V
Generating a graph between -+2.4A (IE staying within the red areas) against detector output voltage will produce what looks to be a straight line (I haven't tried plotting so it may be incorrect) from 0V at -2.4A through 240V at zeroA to 480V at +2.4A. It only needs to be compared with 240V to see which direction the current is flowing and my hunch also the power drawn is also linear.

Thinking further but not investigated; my hunch is referencing the zero power position of the graph to the supply voltage (rather than 240V) it may still display the correct power. Too late now to do any plotting as I have an early start.
 
It is the simplest way to describe the operation without having to introduce some form of comparator
Fair enough - but, as I said, 'implementation' (with 'a comparator' or anything else) was not what I was interested in, it being the concept that is what I was missing, and which you have kindly explained.
It's quite easy to detect the phase relationship
Sure. If you're back to 'implementation' again then, yes, there are plenty of ways to skin that particular cat.
 
weird double post deleted
 
Last edited:
This appears to be a duplicate of most of post #64, to which I replied in #65.
I have no idea how that happened as once #64 was posted the draft would have not been there.

Now deleted.
 
Very odd. I blame poltergeists, or maybe 'global warming' or (a pretty universal scapegoat at present) :)
I'm not even sure If I was still usingthe PC at that time.

Is it too early to blame Reform UK?
 

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