PE exam question 49

[yaoyaodes]

A balanced 3 phase load with a power factor of 0.8 lagging is connected as show.

I am so confused with this question and solution.

 

[DLD PE]

You can do a search for any NCEES problem and find it here. Just type "NCEES #____" in the search bar. Also remember problem #49 used to be numbered #509 in the older NCEES practice exam books.

https://engineerboards.com/threads/2-watt-meter-method.30238/#post-7480437

 

[akyip]

I'll try to explain this as best as I can with what I remember Zach Stone's Electrical PE Review course taught about 2-wattmeter method.

In a 2-wattmeter method, the two wattmeters are ALWAYS connected such that:

• They share the same common negative terminal connection point.
  Here, the common negative terminal point is the B phase.
• They have different positive terminal connection points.
  Wattmeter W1's positive terminal is the C phase, and Wattmeter W2's negative terminal is the A phase.

For each wattmeter's connection going from the positive terminal to the negative terminal:

• The wattmeter that is connected in the positive or ABC sequence (AB, BC, or CA) will have its real power formula as: P = VLL * IL * cos(theta + 30).
  In this problem, Wattmeter 2 is connected from A to B, which is the positive or ABC sequence.
  So: P W2 = VLL * IL * cos(theta + 30)
• The wattmeter that is connected in the negative or CBA sequence (CB, BA, or AC) will have its real power formulas as: P = VLL * IL * cos(theta - 30)
  In this problem, Wattmeter 1 is connected from C to B, which is the negative or CBA sequence.
  So: P W1 = VLL * IL * cos(theta - 30)

The 2 wattmeters will have the same real power readings when theta = 0, which corresponds to pf = 1. The reason is that the cosine function is symmetrical about the origin theta = 0 (it is an even or evenly symmetrical function about theta = 0). So cos(-x) = cos(x).

When theta = 0:

• cos(0 - 30) = cos(0 + 30)
• cos(-30) = cos(+30) = 0.866
• Power factor: pf = cos(0) = 1

I hope that this explanation helps. I also included a few illustrations to try to reinforce my explanation.

 

[Zach Stone P.E.]

I'll try to explain this as best as I can with what I remember Zach Stone's Electrical PE Review course taught about 2-wattmeter method.

In a 2-wattmeter method, the two wattmeters are ALWAYS connected such that:

• They share the same common negative terminal connection point.
  Here, the common negative terminal point is the B phase.
• They have different positive terminal connection points.
  Wattmeter W1's positive terminal is the C phase, and Wattmeter W2's negative terminal is the A phase.

For each wattmeter's connection going from the positive terminal to the negative terminal:

• The wattmeter that is connected in the positive or ABC sequence (AB, BC, or CA) will have its real power formula as: P = VLL * IL * cos(theta + 30).
  In this problem, Wattmeter 2 is connected from A to B, which is the positive or ABC sequence.
  So: P W2 = VLL * IL * cos(theta + 30)
• The wattmeter that is connected in the negative or CBA sequence (CB, BA, or AC) will have its real power formulas as: P = VLL * IL * cos(theta - 30)
  In this problem, Wattmeter 1 is connected from C to B, which is the negative or CBA sequence.
  So: P W1 = VLL * IL * cos(theta - 30)

The 2 wattmeters will have the same real power readings when theta = 0, which corresponds to pf = 1. The reason is that the cosine function is symmetrical about the origin theta = 0 (it is an even or evenly symmetrical function about theta = 0). So cos(-x) = cos(x).

When theta = 0:

• cos(0 - 30) = cos(0 + 30)
• cos(-30) = cos(+30) = 0.866
• Power factor: pf = cos(0) = 1

I hope that this explanation helps. I also included a few illustrations to try to reinforce my explanation.

Thanks for the mention @akyip!