Power is being measured by two wattmeter method in a balanced three-phase system. The wattmeter reading are
Power is being measured by two wattmeter method in a balanced three-phase system. The wattmeter reading are
W1 = 250 kW
W2 = 50 kW
If the latter reading is obtained after reversing the connection to the current coil of W2, the power factor of the load is
Right Answer is:
0.359
SOLUTION
Reading of wattmeters
W1 = 250 kW
W2 = 50 kW
When the latter reading is obtained after the reversal of the current coil terminals of the wattmeter
W1 = 250 kW
W2 = −50 kW
Total power W = W1 + W2
W = 250 + (−50) = 200 W
The power factor of the two wattmeters is
tanØ = √3[(W1 – W2) / (W1 + W2)]
tanØ = √3[(200 – (−50)) ⁄ [250 + (−50)]
tanØ = √3[300/200]
Φ = 68.96°
Power factor = cosΦ = cos(68.96°) = 0.3590