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2 comments
Anonymous Poster

Two Wattmeter Method

08/09/2010 9:09 AM

Hey! Would anyone help me? What is the necessary formula derivation for 2 wattmeter method of 3-phase power measurement?

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Power-User

Join Date: Nov 2008
Location: India
Posts: 333
Good Answers: 6
#1

Re: Two Wattmeter Method

08/09/2010 2:11 PM

dear ,

Watts = (W1 - W2 )* MF ( i.e. scale factor) of meters * CT ratio

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Guru

Join Date: Feb 2009
Location: Houston, USA
Posts: 971
Good Answers: 244
#2
In reply to #1

Re: Two Wattmeter Method

08/09/2010 7:05 PM

Using two wattmeters, the power can be measured for both balanced and unbalanced loads of the 3-phase 3-wire system. However, for the 3-phase 4-wire system (with having neutral), the two wattmeter method doesn't work for unbalanced loads. So, the two wattmeter method is supposed to be used for measuring the power of the 3-phase 3-wire system. The total power of the two wattmeter method is:

P = W1 + W2

Where, P = Total power, W1 = the power measured by wattmeter 1 and W2 = the power measured by wattmeter 2

It may happen that the values of both W1 and W2 are positive, or either W1 or W2 is negative.

For a balanced system, the power factor also can be calculated by using the 2 wattmeter readings with this formula:

tan θ = √3 (W2-W1)/(W2+W1)

For more detail for the formula derivation, see these links:

http://www.voltech.com/support/articles/51/Three%20Phase%20Measurements%20(104-022).pdf

http://books.google.com/books?id=GXJ2tfECwV0C&pg=SA7-PA32&lpg=SA7-PA32

http://books.google.com/books?id=0QcUUlyUQjwC&pg=RA1-PA257

Similarly, the two element energy meter is used for measuring the energy used (KWH) for 3-phase 3-wire system.

See this thread http://cr4.globalspec.com/thread/56190 for detail.

- MS

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