Question

In the balanced three phase power measurement using two wattmeter method, each wattmeter reads 22 kW, when the pf is unity. What does each wattmeter reads when the pf falls to (i) 0.866 lag and (il) 0.573 lag. The total three-phase power remains unchanged.

(i) W1 at 0.866 lag
(i) W2 at 0.866 lag
(iii) W1 at 0.573 lag
(iv) W2 at 0.573 lag

Answer 1

When the power factor falls to 0.866 lag, Wattmeter 1 reads 22 kW, and Wattmeter 2 reads 11 kW. For a power factor of 0.573 lag, Wattmeter 1 reads approximately 33.9 kW, and Wattmeter 2 reads approximately 10.1 kW.

Step-by-step explanation:

When using the two wattmeter method to measure balanced three-phase power, with each wattmeter initially reading 22 kW at unity power factor, the readings change as the power factor decreases while keeping total power the same.

At a power factor of 0.866 lag, the two wattmeters will have different readings due to the phase shift between the voltage and current. Assuming the angle φ corresponds to the power factor cos(φ), we calculate the angle as φ = cos-1(0.866), or 30 degrees. The power measured by each wattmeter can be determined by using the formula:

W1 = Total Power * (1 + cos(2 φ)) / 2

W2 = Total Power * (1 - cos(2 φ)) / 2

Since the total power is unchanged at 44 kW (22 kW from each wattmeter at unity power factor):

• 
  
• (i) W1 at 0.866 lag = 22 kW * (1 + cos(60 degrees)) / 2 = 22 kW
• 
  
• (ii) W2 at 0.866 lag = 22 kW * (1 - cos(60 degrees)) / 2 = 11 kW

When the power factor falls to 0.573 lag, the corresponding angle is φ = cos-1(0.573), or approximately 54 degrees. The wattmeter readings will again differ based on this angle:

• 
  
• (iii) W1 at 0.573 lag = 22 kW * (1 + cos(108 degrees)) / 2 ≈ 33.9 kW
• 
  
• (iv) W2 at 0.573 lag = 22 kW * (1 - cos(108 degrees)) / 2 ≈ 10.1 kW