Understanding Reactive Power Measurement with Wattmeter Configuration

AI Thread Summary
The discussion clarifies how a wattmeter configuration measures reactive power by responding to the current that is in phase with the voltage. When the power factor is 1.00, the current vector remains at 90 degrees to the voltage vector, but as the power factor decreases, the current vector rotates, affecting the wattmeter's response. The equations presented illustrate the relationship between voltage, current, and power factor in calculating reactive power (VAr) in a three-phase system. A key takeaway is that multiplying the wattmeter reading by √3 is necessary to obtain the three-phase reactive power. This understanding is essential for practical applications in the electrical power industry.
mnada
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Does anybody understand how this wattmeter configuration is measuring reactive power?

One%20Wattmeter.jpg


Thanks
 
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The wattmeter responds to current that is in phase with the voltage.
In your picture, The reference voltage is Y-B. (picture this vector as horizontal, the base of the delta
triangle). The current is from phase R. Its vector would be vertical or at 90 deg to the voltage vector.
If the power factor is 1.00, then the vector representing current would remain at 90 deg in relation to the voltage vector. As the power factor decreases however, the current vector rotates right or left (inductive or capacitive) and comes closer to being in phase with the voltage vector. As it does, the meter begins to respond. It will measure and display VAr. This is common and has practical application in the electrical power industry.
 
Thank you very much, your explanation is crystal clear, and helped me so.
Again, thank you.
 
That varmeter regarding to special connection is a wattmeter inherently, and we can write:

P = K1. VYB . IR . COS ø (output of measurement device)

VYB = √3.VR< 90˚ (In three phase balance system)

P = K1. √3.VR . IR . COS (ø - п/2) = K2. VR . IR . Sin ø

If K2 = 3 we can write:

P = 3. VR . IR . Sin ø = Q3PH

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Another excellent Explanation. Thanks
 
you need to multiply \sqrt{3} to the wattmeter reading to get 3-phase reactive power...
 
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