2 Watt Meter Method

Szar said:

Greetings.  

I am missing something fundamental in regards to developing the equations for the two watt metering method.  I can develop the phasor diagrams that depict the voltages and currents and the phase angles between them, see attached.    

As I interpret the completed phasor diagram, "Ia" leads "Vac" and "Ib" lags "Vab" prior to considering power factor created from load reactance.  When considering PF, the currents should rotate counter-clockwise (assuming an inductive load) with the origin as the base point for rotation.  Also as I understand it, when the load is inductive this should always increase the angle between "Vbc" and "Ib".  For "Vab" and "Ia" the angle will initially decrease until the current eventually changes from leading to lagging.  As such, the terms I develop are cos(Ø-30°) and cos(Ø+30°), for watt meter 1 and 2 respectively.  

Where I am finding difficulty is how the equation becomes cos(30°-Ø) and cos(30°+Ø) for watt meter 1 and 2 as is commonly shown online.  

Any assistance would be greatly appreciated.  Thank you.

PS.  My apologies for the poor penmanship.

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Surf and Snow said:

While it's good to learn and know everything electric, just as an FYI, Wattmeters were removed from the 2018 test specifications. So don't stress too much, being less than 3 weeks out, I'd focus on other weak areas that are within the test specs. But that's just my $0.02

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Szar said:

Greetings.  

I am missing something fundamental in regards to developing the equations for the two watt metering method.  I can develop the phasor diagrams that depict the voltages and currents and the phase angles between them, see attached.    

As I interpret the completed phasor diagram, "Ia" leads "Vac" and "Ib" lags "Vab" prior to considering power factor created from load reactance.  When considering PF, the currents should rotate counter-clockwise (assuming an inductive load) with the origin as the base point for rotation. Place the current assuming a small phase angle less than 30 degrees.   Also as I understand it, when the load is inductive this should always increase the angle between "Vbc" and "Ib".  For "Vab" and "Ia" the angle will initially decrease until the current eventually changes from leading to lagging- Not at load, only wattmeter sees like that.  As such, the terms I develop are cos(Ø-30°) and cos(Ø+30°), for watt meter 1 and 2 respectively.  

Where I am finding difficulty is how the equation becomes cos(30°-Ø) and cos(30°+Ø) for watt meter 1 and 2 as is commonly shown online.  Cosine of any angle Theta= Cosine of negative of angle theta. Cos(Theta)= Cos(-Theta). So you can put these equations in any of the ways. Do not worry. 

Any assistance would be greatly appreciated.  Thank you.

PS.  My apologies for the poor penmanship.

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rg1 said:

You are almost right at understanding the concept. There is nothing like pf without considering load. It is always a good idea for wattmeter questions to put the currents at their places in phasors considering a small angle (less than 30 degrees either inductive or capacitive) and then form the equations. Rest is maths. Always remember pf is cosine of angle between Phase Voltage and phase current (Not the line quantities). Wattmeter reads the power as Voltage across its terminals X current through it X the pf of the angle between them. See my comments in red. Try to do it all connections for inductive as well as capacitive loads. So in all 6 connections. . You can make tables for  readings of W1, W2, for load angles of 0, 30, 45,60,90 for both capacitive as well as inductive loads with each type of connection.Take all these diagrams and equations with you to the exam as your notes so that you do not have to struggle there.

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Szar said:

rg1 & ARS,

Thanks.  In regards to the cos(Ø) vs cos(-Ø), I agree with the identify in this case.  At first I did not see how the identity applied here but after pondering for a few bit I see Cos(- (Ø-30)) yields cos(30-Ø).  Trig simplification was never my strong suit.

In regards to my other post(s) with the attached PDF involving NCEES 509, is my formulation correct?  (ie. does the NCEES have the watt equations reversed?)  

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