2 Watt Meter Method

[Szar]

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]

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.

gView attachment 10968

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

 

[Surf and Snow]

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]

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]

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]

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.

View attachment 10968

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]

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 & 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]

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?)  

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]

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.