Measuring Circuit (Aaron Circuit)

[papachulo]

Hello,

We have a 3 phase power supply for our Motor and we have installed measuring equipment to check our Energy demand when the motor is running. The problem is that we only have 2 transformers in-doors so we can only clip-on the measuring equipment on 2 of the 3 phases (phase 1 and phase 3).

The supplier of the measuring equipment told me that the power measurement would be done according to a Aaron Circuit Technique:

http://de.wikipedia.org/wiki/Aronschaltung

I think its called two-watt meter method in English but I am not sure.

Anyway the values do not add up. We are assuming that the real power is our total power since the motor has a power factor of 0.9. According to the Aaron circuit technique that would mean real power = phase 1 power + phase 3 power.

The possible reasons I believe the measurement are not plausible are:
i) The phase angles between phase are still there. The current and voltage must be in phase. Anyone have recommendations how we can solve this phase issue out so we have no phase angle ? This issue i reckon give us the wrong power magnitudes.

ii) I believe the measurement sensors have not been clipped on at the proper points. So instead of U32 we have U23 or something similar. This should only give us a wrong sign and correct magnitude if I am not wrong ?

any suggestion are very welcome.

 

[waross]

Start with your instrument connected single phase to one phase.
Voltage A to B. Measure current at B. For a motor this should read 1/2 of the total power. The current and voltage will not be in phase. The Watt-meter will react to the current for A phase and the A phase component of B phase.
The reading will be the motor power plus the motor losses. If the motor is not fully loaded the reading will be less than the motor rating.
When you get this working then add the other section.
The two Watt-meter method assumes equal voltages. Using one Watt-meter and multiplying by two assumes equal voltages and equal loading per phase. If the voltages are equal this is a safe assumption for most three phase motors.

Bill
--------------------
"Why not the best?"
Jimmy Carter

 

[ScottyUK]

There are plenty of Google links. Have a look at

https://www.ohiosemitronics.com/pdf/tech_papers/Two-wattmeter-method(E).pdf

as a starter.


----------------------------------

If we learn from our mistakes I'm getting a great education!

 

[electricpete]

Voltage A to B. Measure current at B. For a motor this should read 1/2 of the total power

Bill - You post a lot of great info and I have learned a lot from you.

I believe (on this rare occasion) you are incorrect on that particular point.

Try a numerical example
Assume a balanced system, A/B/C rotation.
Let the currents also be balanced but lagging 30 degrees behind the voltages-to-neutral

If we measure Vb-Va as voltage input, Ib as current input, we will see 1/3 of the total power.

If we measure Vc-Va as voltage input, Ic as current input, we will see 2/3 of the total power.

If you add the two measurements together, we get the total power. But each is not 1/2 of the total power.

If we assumed a different power factor angle, we would get a different split among the two measurements (not 1/3 and 2/3), but the total would still be correct).

Unfortunately, I am not at a computer where I can easily post those results (don't have pdf creater software). I will try to post the complete numerical example sometime later.

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(2B)+(2B)' ?

 

[waross]

Hi Pete. The current on B phase is not in phase with the A-B voltage, but, if A phase is considered as a single phase circuit the B phase current will be in phase with the A-B voltage. So the meter accurately meters the load on A-B.
However there is also a current contribution from the B-C phase. The B-C current, (unmetered phase) may be resolved into an A-B component and a C-A component. The meter on A-B will meter the A-B component of B-C and the meter on C-A will meter the C-A component of the B-C current.
Meters of Form 5S, 8S, 13S, and Form 15S all function on the two Watt-meter principal. Each of two voltage-current coil sets generates torque on the same disk.
Draw your A-B voltage vector as a reference. Now lay out the A-B phase current vector in phase with the voltage vector. Using the head to tail method, (or tail to head if you prefer) add the B-C current displaced 120 degrees. You will see that the A-B component of the B-C current is 50%.
I may have made a mistake but I suspect that it would be in my explanation rather than the application of the two Watt-meter method of measuring three phase power.

Bill
--------------------
"Why not the best?"
Jimmy Carter

 

[electricpete]

Thanks Bill. I was not contesting the 2 wattmeter method, only the suggestion that each of the two wattmeters would indicate total power.

Attached is an analysis of the power indicated on each of two wattmeters:

"Measurement B" is wattmeter reading Vb-Va as voltage input, Ib as current input
"Measurement C" is wattmeter reading Vc-Va as voltage input, Ic as current input.

The result is shown on the graph at the bottom of page 2. It shows that the sum of the "pu" measurement B and measurement C is always 1 (normalized so that 1 is the correct Ptotal).

However Measurement B and Measurment C shown in Blue and Black respectively are not always 0.5. They are 0.5 only if power factor angle is 0 (power factor 1.0). For all other power factors, the split is not 50/50, but the sum remains correct.

=====================================
(2B)+(2B)' ?

 

[electricpete]

Correction in bold:

I was not contesting the 2 wattmeter method, only the suggestion that each of the two wattmeters would indicate total power.

should have been

I was not contesting the 2 wattmeter method, only the suggestion that each of the two wattmeters would indicate excactly half of total power.



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(2B)+(2B)' ?

 

[electricpete]

The attached pdf file is similar to my previous attachment, but revised to improve readability slightly. Specifically, it uses an alternate definition for the dp (dot product) function, which leads to a much simpler expression for MeasurementB and MeasurementC. The conclusion is the same (plot at end of 2nd page)

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(2B)+(2B)' ?

 

[electricpete]

Attached is one last improvement for readability. (Got rid of Vln=1 and Im=1, which was not necessary.)

=====================================
(2B)+(2B)' ?

 

[waross]

Now I see you point Pete. I believe that you are correct.
I have used the two Watt-meter connections many times but always with a two element meter based on the two Watt-meter principle.
The meter does the summing internally and the only value available is the sum so the shift with power factor is not obvious.
I see you are using line to neutral voltages. The basic two Watt-meter circuit is best used phase to phase.
There are variations that, with the aid of CTs allow the two Watt-meter connection to meter unbalanced wye loads.
Each meter or element is connected line to neutral and uses line current. This meters only two phases. A third CT on the third phase and connected in delta with the other CTs will force a current proportional to the third phase current through the current coils of the two meters.
Alternatively the third phase conductor may be passed through each of the other CTs to meter the third phase.
This is not a Blondel solution and this solution assumes that the voltages are equal.
We used this method with form 5S meters for years in the islands to measure unbalanced wye loads. (Unbalanced loads but balanced voltages.) Then we lost one phase of our submarine cable feeding our largest customer base on an island remote from the power plant. Our original cable had three conductors in trefoil configuration. When a single conductor cable was laid near the original cable to pick up the missing phase, we had a lot more impedance on the third phase. We could no longer assume balanced voltages. We replaced the form 5S meters with form 9S meters for a true Blondel solution and accurate metering.
Interestingly, the form 6S and form 14S meters have internal current coils for three phase wye metering. Adjacent to the A phase voltage coil is an A phase current coil and a B phase current coil.
Adjacent to the C phase voltage coil is an C phase current coil and a B phase current coil. This allows the A phase and C phase current coils to meter the B phase power. This is not a Blondel solution and the voltages are assumed to be equal.
My bible in those days was:
Meter and Instrument Transformer Application Guide and Condensed Metering Course.
ALL IN ONE
Fourth edition
Westinghouse

Bill
--------------------
"Why not the best?"
Jimmy Carter

 

[electricpete]

I see you are using line to neutral voltages. The basic two Watt-meter circuit is best used phase to phase.

The phase to neutral voltages (Va, Vb, Vc) are used only as a tool to determine the behavior of "MeasurementB" and "MeasurementC", which use the phase-to-phase voltage inputs "Vb-Va" and "Vc-Va", respectively (where "-" indicates subtraction).

papachulo - I assumed the currents and voltages were balanced for simplicity in my attachment, but it is not a required assumption for the two watt-meter method to work.

I provided a proof/derivation of the method in my post 27 May 11 10:45 in the following thread
thread237-299600
It seems like a very straightforward proof to me, no messy trig. Again it does not involve any assumption that the voltages and currents are unbalanced. However, since it is based on phasors, it does involve an assumption that the voltages and currents are sinusoidal (otherwise the definition of phase and magnitude are ambiguous).

* If you think your measurement is wrong, then I think Bill and you had the right idea to double check the polarity of all the connections. Swapping either the current or voltage polarity on either wattmeter would give incorrect results. I think the proof above suggests a very straightforward way to remember the polarities: since the proof essentially just redefines our ground voltage reference to be A phase. Therefore the remaining voltages are measuremed with respect to A phase. And for load convention, we measure current flowing toward the load.

I'd like to take a step back further and ask how you came to the conclusion that your measurement is wrong?

The possible reasons I believe the measurement are not plausible are:
i) The phase angles between phase are still there. The current and voltage must be in phase. Anyone have recommendations how we can solve this phase issue out so we have no phase angle ?

I'm not sure exactly what you're suggesting, but it doesn't sound right to me. Even for power factor = 1.0, you would measure 120 degrees phase difference between Vba and Ib.
If you have phase angles available, those should provide a useful tool to analyse the measurements. Maybe you can post more of what you measured and discuss why you think it is wrong.


=====================================
(2B)+(2B)' ?

 

[electricpete]

Correction in bold:

it does not involve any assumption that the voltages and currents are unbalanced.

should have been:

it does not involve any assumption that the voltages and currents are balanced.

i.e. if connections are correct and equipment working properly, and voltages/currents are sinusoidal, then you should get correct result regardless of whether the voltages and currents are balanced or not.

=====================================
(2B)+(2B)' ?