NEW NCEES #509

[ndekens]

Ok this power meter thing is a bit confusing to me. Now looking at the answer in the back I understand it and why they would read the same a a PF of 1.00 heck I even guessed it was 1.00 before looking at the book. I just need some explaining done on the ins and outs of this problem as follows:

1. How did they get the +30* for W2 and -30* for W1 in the answer?

2. What is the purpose of a polarity mark in this case?

3. What the heck is a meter constant and whats it used for?

4. Since these are Watt meters, and we are talking real power, shouldn't they by default have a unity power factor and read the same if V and I were the same for both?

Thanks!

 

[ndekens]

Ok this power meter thing is a bit confusing to me. Now looking at the answer in the back I understand it and why they would read the same a a PF of 1.00 heck I even guessed it was 1.00 before looking at the book. I just need some explaining done on the ins and outs of this problem as follows:
1. How did they get the +30* for W2 and -30* for W1 in the answer?

2. What is the purpose of a polarity mark in this case?

3. What the heck is a meter constant and whats it used for?

4. Since these are Watt meters, and we are talking real power, shouldn't they by default have a unity power factor and read the same if V and I were the same for both?

Thanks!

Anyone????

 

[Flyer_PE]

1. A nice little discussion of the 2 Wattmeter method can be found here.

2. The polarity mark just gives you an indication of the polarity of the angle between the voltage and current seen by the meter. Hooking the meter up differently will result in it reading the same magnitude but the angle will change by 180 degrees.

3. Not sure what the meter constant is but I suspect it is a correction factor of some kind.

4. If I understand this question correctly, no. The power factor determines how much is actually real power and how much is reactive. For a given current and voltage magnitude, real power will be vastly different at a pf of 0.5 than it will be at a pf or 0.8.

The interesting thing about problem 509 is that almost all of the information they give you is extra information that is totally irrelevant to the actual question being asked.