Measurement of 3 phase power by two watt-meter method

In this slide I have explained how two watt meters can be used to measure 3 phase power. Some of the added advantage of this method is that we can calculate 3 phase reactive power and power factor of load as well.

1. ELECTRICAL ENGINEERING b y - Mohammed Waris Senan

2. Mohammed Waris Senan 2 THREE PHASE SYSTEM Measurement of 3 phase power by two wattmeter method

3. Mohammed Waris Senan 3 Wattmeter Terminals & Symbol M V C L Current Coil (CC) Potential Coil (PC)

4. Mohammed Waris Senan 4 Wattmeter Connection P = V pc I cc cos (angle between V pc & I cc ) Connection Diagram CC PC voltage supply LOAD M C V L

5. Mohammed Waris Senan 5 Measurement of 3 phase Power Connection Diagram Reading of wattmeter 1 : P 1 = V AB I A cos (angle between V AB & I A ) Reading of wattmeter 2 : P 2 = V CB I C cos (angle between V CB & I C ) For a balanced three phase STAR connected Load , Let, V phase = V, & I line = I phase = I V line = V & Load pf is cos lag   P 1 = VI cos (angle between V AB & I A ) P 2 = VI cos (angle between V CB & I C )  

6. Mohammed Waris Senan 6 Measurement of 3 phase Power           Angle between V AB & I A = 30 + A ngle between V CB & I C = 30 -   P 1 = VI cos (30 + ) P 2 = VI cos (30 - )   Total 3 phase power P 3- = P 1 + P 2 = 3VI cos  

7. Mohammed Waris Senan 7 Measurement of 3 phase Power P 1 = VI cos (30 ) P 2 = VI cos (30 )   Some Important Cases : 1. If Power factor of load is UNITY i.e. cos = 1, = 0   Reading of both wattmeter is EQUAL 2. If Power factor of load is 1/2 i.e. cos = 1/2, = 60   P 1 = VI cos (30 + 60 ) = VI cos (90 ) = 0 P 2 = VI cos (30 - 60 ) = VI cos (30 ) = VI   Reading of one wattmeter is NEGATIVE 3 . If Power factor of load is 0 and ½ i.e. 0 cos ½ , 60 90   Reading is taken by reversing the connection of Current Coil (CC)or Potential Coil (PC) Generally connection of Current Coil is reversed Reading of one wattmeter is ZERO

8. Mohammed Waris Senan 8 Measurement of Power Factor ( cos )   P 1 – P 2 = VI cos (30 + ) - VI cos (30 - ) = VI sin So, Three Phase Reactive Power , Q 3- = ( P 1 – P 2 )   Now,

9. If power factor is UNITY, cos = 1 = P 3- = V L I L cos = 400 115.4 1 = 80 kW So, Reading of watt-meters will be: P 1 = VI cos (30 + ) = I cos (30 + ) = V L I L cos (30 + ) P 1 = 400 115.4 cos (30 + 0 ) = 40 kW P 2 = VI cos (30 - ) = I cos (30 + ) = V L I L cos (30 - ) P 2 = 400 115.4 cos (30 - ) = 40 kW   Mohammed Waris Senan 9 Numerical Problems In a balanced 3-phase 400 V circuit, I line = 115.4 A. When power is measured by two watt-meter method, one of the meter reads 40 kW and another zero. What is the power factor of the load? If power factor is unity and line current is same, what would be the readings of each watt-meter? [AKU, 2019 (odd sem )] {8 marks} Question: Solution: Given data: V L = 400 V I L = 115.4 A P 1 = 40 kW P 2 = 0 power factor, cos = ?   Alternately: We know that if one of the wattmeter reads zero then pf of the load will be ½ . Total power, P 3- = P 1 + P 2 = 40 + 0 = 40 kW As, P 3- = V L I L cos 400 115.4 cos = 40000 cos = 0.5  

10. Mohammed Waris Senan 10 Numerical Problems In a balanced 3-phase star connected load, two watt-meters are connected to measure the 3-phase power as shown in figure. If impedance of each phase is 10 Ω and line-to-line voltage is 220 V ( rms ), what will be readings of two watt-meter?   Question: Solution: Reading of wattmeter 1 : P 1 = V AC I A cos (angle between V AC & I A ) Reading of wattmeter 2 : P 2 = V BC I B cos (angle between V BC & I B ) Let phase sequence be ABC

11. P 1 = 220 12.7 cos (15 ) = 2698.8 W   P 2 = 220 12.7 cos (75 ) = 723.14 W   Mohammed Waris Senan 11 Numerical Problems Solution:

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