# 2 Watt Meter Method



## Szar (Mar 25, 2018)

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.


----------



## ARS (Mar 25, 2018)

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.
> 
> ...


cos (-Ø) = cos (Ø) is one relationship that I can think of. Hope it helps.


----------



## Surf and Snow (Mar 26, 2018)

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


----------



## Szar (Mar 26, 2018)

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


Sample Question NCEES  509 is still watt meters. 

The equations they show are similar but also opposite then mine.   So I believe I am doing something wrong with the phasors.


----------



## Szar (Mar 26, 2018)

This is the phasor diagram I develop for the NCEES 509 watt meter problem.  From my understanding, the equations for power should be as stated in the diagram.

The NCEES Solution has the equations with cos(Ø-30) for W1 and cos(Ø+30) for W2, opposite of what I develop.

View attachment Figures-Model.pdf


----------



## rg1 (Mar 26, 2018)

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.
> 
> ...


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.


----------



## Szar (Mar 26, 2018)

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.


rg1 &amp; 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?)


----------



## rg1 (Mar 26, 2018)

Szar said:


> rg1 &amp; 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?)


I do not have the NCEES question but,  from your explanation I understand that you are understanding the concept, which is more important. You can get any of the combinations of connections in the exam so prepare yourself that way, making phasors is best way to understand and solve these questions.


----------



## ARS (Mar 27, 2018)

If the load is inductive with Ia lags Va by less than 30 degree, say there, then wattmeter 

W1 = Ia . Vac cos (30 - theta) 

And if Ia lags Va by more than 30 degree then

W1 = Ia . Vac cos(theta - 30) 

You can do the same for capacitive load to get the equation.

Hope this helps.


----------

