Devices to meter kWh

[JohnW2]

An Introduction to AC Power — OpenEnergyMonitor 0.0.1 documentation

"Determining the direction of power flow".

Thanks. Well ....

... If the house is generating power, the direction of the current is reversed compared to our previous example. Now when the voltage is positive, (above the X axis) the current is negative (below the X axis), and when the voltage is negative, the current is positive. The power always negative - all of the power curve is below the X axis.

... certainly sounds straightforward enough but I need to do some thinking to get my head around what 'voltage' they are talking about - since, if one connects an AC source to an essentially resistive (i.e. 'power consuming') load, then the current through that load will be essentially 'in phase' with the voltage, won't it?

 

[JohnW2]

You can't do it with a CT alone. But you *can* do it with a CT in combination with a voltage sensor. You multiply the instantaneous voltage by the instantaneous current and integrate.

Yes, that's what the article linked to by Why Not Indeed - but, as I have just written, I'm still confused. Can you help me understand what voltage it is which is out of phase with the current it is resulting in (and why) ?

 

[Harry Bloomfield]

Thanks. Well ....

... certainly sounds straightforward enough but I need to do some thinking to get my head around what 'voltage' they are talking about - since, if one connects an AC source to an essentially resistive (i.e. 'power consuming') load, then the current through that load will be essentially 'in phase' with the voltage, won't it?

and certainly news to me..

 

[plugwash]

Lets forget the grid a minute and think about a traditional car (with an internal combustion engine).

Lets say we have measuring devices that measure the battery voltage, and the current flowing from/to of the battery.

We get into the car, turn on some accessories and engage the starter. This results in current flowing "out" of the positive terminal of the battery and "in" to the negative terminal of the battery. The battery is being discharged, that is energy is being transffered from the battery to the rest of the car.

Then the engine starts, the starter is disengaged and the alternator starts to generate electricity. Current is now flowing "in" to the positive terminal of the battery and "out" of the negative terminal of the battery. The battery is being charged, that is energy is being transffered from the rest of the car to the battery.

The key observation, that the directions of energy transfer depend on the relative directions of voltage and current. This still holds in an AC system, the volage and current reverse every half-cycle, but their relative direction is still what defines the direction of energy transfer. On the "positive" half-cycle current flows from source to load in the "live" conductor and from load to source in the "neutral" conductor. On the "negative" half-cycle current flows from load to source in the "live" conductor and from source to load in the "neutral" conductor.

 

[JohnW2]

The key observation, that the directions of energy transfer depend on the relative directions of voltage and current. This still holds in an AC system, the volage and current reverse every half-cycle, but their relative direction is still what defines the direction of energy transfer. On the "positive" half-cycle current flows from source to load in the "live" conductor and from load to source in the "neutral" conductor. On the "negative" half-cycle current flows from load to source in the "live" conductor and from source to load in the "neutral" conductor.

Thanks. What you say is essentially obvious, but I'm afraid that I still don't seem to have "got it"

My first reaction is to say that the fact that during the (voltage) positive half cycle current goes:

Source -> L conductor -> Load -> N conductor -> Source​

.... and during the negative half cycle goes:

Source -> N conductor -> Load -> L conductor -> Source​

... arises because in the first case the L conductor voltage is positive with respect to the N conductor and in the second because the N conductor voltage is positive with respect to the L conductor - i.e. as I would expect, voltage and current are 'in-phase' - and that would presumably remain the case if Source and Load were to be swapped?

What am I missing?

Kind Regards, John

 

[Munroast]

the voltage does not change but the current flow does. imagine 2 sine waves in sync, then one of those sine waves is flipped upside down, then they would be 180° out

 

[JohnW2]

the voltage does not change but the current flow does.

Eh? The potential difference between L and N ('the voltage') surely changes from positive to negative every half cycle.? You surely don't think that this (AC) L-N potential difference ('voltage') is always positive or always negative, do you?

imagine 2 sine waves in sync, then one of those sine waves is flipped upside down, then they would be 180° out

Yes, that's the definition of "180° out of phase' - but so what?

 

[Munroast]

Eh? The potential difference between L and N ('the voltage') surely changes from positive to negative every half cycle.? You surely don't think that this (AC) L-N potential difference ('voltage') is always positive or always negative, do you?

Yes, that's the definition of "180° out of phase' - but so what?

the "Sine Wave" of the voltage does not change when the flow direction of the current changes - it remains the same. (yes its still doing its AC stuff) but it is still the same voltage sine wave!

The "So What" bit is when the sine wave of the current goes out of sync with the sine wave of the voltage then the flow has changed. Not only is the CT clamp measuring current, it is also seeing if both sine waves are in sync

 

[JohnW2]

the "Sine Wave" of the voltage does not change when the flow direction of the current changes - it remains the same. (yes its still doing its AC stuff) but it is still the same voltage sine wave!

I don't really know what you are on about. Your comment came amidst exchanges between plugwash and myself about the fact that, in any AC circuit, the direction of current flow changes every half cycle, regardless of which end of the circuit is the source and which is the load. The polarity of the voltage resulting in that current (L-N potential difference) also changes every half cycle.

The "So What" bit is when the sine wave of the current goes out of sync with the sine wave of the voltage then the flow has changed.

Yes, but can you not see that my problem is that I don't understand how any such 'out-of-synchness' arises

Not only is the CT clamp measuring current, it is also seeing if both sine waves are in sync

As has been said, the CT itself cannot know anything about any voltage - but the circuitry to which the CT is connected could.

 

[boringoldcodger]

AIUI when we have an inductive load the current sine wave lags behind (peaks after) the voltage sine wave, but in a capacitive load the current sine wave leads (peaks before) the voltage sine wave.
However most loads are inductive, any coil of wire is an inductor so transformers (e.g. taking 240V to 12V), electric fires, electric motors etc. all give rise to an inductive load. So I'm just wondering whether the meters which claim to measure power direction look at the 2 sine waves and assume power is flowing in the direction which gives a lagging current sine wave.

Have I just fallen off the edge of my knowledge?

 

[JohnW2]

AIUI when we have an inductive load the current sine wave lags behind (peaks after) the voltage sine wave, but in a capacitive load the current sine wave leads (peaks before) the voltage sine wave.

Very true.

However most loads are inductive ....

No. A purely inductive (or capacitive) load consumes no energy - so whenever energy is transferred to a load, it must have a major resistive component, with voltage and current not that far off being 'in phase'.

, Have I just fallen off the edge of my knowledge?

Only to the extent of one false premise - and I'm in no position to criticise, since there clearly must be something fairly fundamental which I am still 'missing'

Kind Regards, John

 

[SUNRAY]

As has already been stated a CT on it's own cannot determine current direction as ACchanges 50 times a second and requires some sort of reference.

Don't for one moment think this explanation is an accurate description of a power meter. I have 'copied' it straight from my note book where I explained to an electrician why the wiring between the power meter and the CT and the direction of curent flow were critical despite his remonstrations that as it was AC it didn't matter.

A simple circuit of a voltmeter supplied by a current transformer from the 240V circuit (ratio 1A to 100V, the brown winding being the wire being monitored and the orange being the clip-on CT) and and an adjustable resistive load:

The red dots indicate phase relationship across the windings, for the sake of explanation I'll consider it to be the positive during the positive half cycle and therefore add to the original 240V:

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.




Let's reverse the CT

or apply power from the other end

Hopefully it can be seen the CT's induced voltage is now out of phase with the supply voltage in both situations and is therefore subtracted from the original 240V.

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

It can be seen the voltage is no longer proportional to the load power and worse; as the voltage is AC we have duplicated measured values.

Getting back to the original notes we found big discrepancies between phase power on 3ph motors, if it wasn't for that fact he would have quite happily left site thinking it was a job well done.

 

[Harry Bloomfield]

on it's own cannot determine current direction as ACchanges 50 times a second

100 times per second..

 

[JohnW2]

A simple circuit of a voltmeter supplied by a current transformer from the 240V circuit (ratio 1A to 100V, the brown winding being the wire being monitored and the orange being the clip-on CT) and and an adjustable resistive load:
[with current going one way]The red dots indicate phase relationship across the windings, for the sake of explanation I'll consider it to be the positive during the positive half cycle and therefore add to the original 240V:
[with current going other way] Hopefully it can be seen the CT's induced voltage is now out of phase with the supply voltage in both situations and is therefore subtracted from the original 240V.

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.

At the very start of this discussion, when people started talking about the phase relationship between current and 'voltage' being an indication of which direction 'the current was flowing', I asked "what voltage?" (the only obvious one being the supply voltage - which made no sense in terms of what was being said about detecting the 'direction of flow'), and no-one has really addressed that question - until now!

However, I think you have now made me realise that the 'voltage' in question is actually the induced voltage in the CT - which I think will, indeed, be in-phase with the current (hence in-phase with supply voltage) with one direction of current flow and 180° out-of-phase with current (hence 180° out-of-phase with supply voltage) with the other direction of current flow.

I need to think a little more to be sure I agree with what I've just written but it certainly seems that one can, indeed, determine the direction of current flow by looking at the phase relationship between supply voltage and voltage induced in the CT.

Kind Regards, John

 

[boringoldcodger]

No. A purely inductive (or capacitive) load consumes no energy

Ahh, I should have said "have an inductive (rather than capacitive) component". I didn't mean to imply "purely inductive".