![If power factor is UNITY, cos𝜙 = 1 ⇒ 𝜙 = 0°
P3-𝜙 = 3VLILcos𝜙 = 3 × 400 × 115.4 × 1 = 80 kW
So, Reading of watt-meters will be:
P1 = 3 VI cos (30° + 𝜙) = 3
𝑉𝐿
3
I cos (30° + 𝜙) = VLIL cos (30° + 𝜙)
P1 = 400 × 115.4 × cos (30° + 0°) = 40 kW
P2 = 3 VI cos (30° - 𝜙) = 3
𝑉𝐿
3
I cos (30° + 𝜙) = VLIL cos (30° - 𝜙)
P2 = 400 × 115.4 × cos (30° - 0°) = 40 kW
Mohammed Waris Senan 9
Numerical Problems
In a balanced 3-phase 400 V circuit, Iline = 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:
VL = 400 V
IL = 115.4 A
P1 = 40 kW
P2 = 0
power factor, cos𝜙 = ?
Alternately:
We know that if one of the wattmeter reads zero then
pf of the load will be ½ .
Total power, P3-𝜙 = P1 + P2 = 40 + 0 = 40 kW
As, P3-𝜙 = 3VLILcos𝜙
⇒ 3 × 400 × 115.4 × cos𝜙 = 40000
∴ cos𝜙 = 0.5](https://image.slidesharecdn.com/3phasepowermeasurement-200423083026/85/Measurement-of-3-phase-power-by-two-watt-meter-method-9-320.jpg)
Measurement of 3 phase power by two watt-meter method
- 1. ELECTRICAL ENGINEERING by - 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 VC L Current Coil (CC) Potential Coil (PC)
- 4. Mohammed Waris Senan 4 Wattmeter Connection P = Vpc Icc cos (angle between Vpc & Icc ) Connection Diagram CC PC voltage supply L O A D LI LV M C V L
- 5. Mohammed Waris Senan 5 Measurement of 3 phase Power Connection Diagram Reading of wattmeter 1 : P1 = VAB IA cos (angle between VAB & IA) Reading of wattmeter 2 : P2 = VCB IC cos (angle between VCB & IC) For a balanced three phase STAR connected Load, Let, Vphase = V, & Iline = Iphase = I ∴ Vline = 𝟑 V & Load pf is cos𝝓 lag P1 = 𝟑 VI cos (angle between VAB & IA) P2 = 𝟑 VI cos (angle between VCB & IC)
- 6. Mohammed Waris Senan 6 Measurement of 3 phase Power 𝝓 𝝓 𝝓 BNV ANV CNV CBV ABV AI CI BI BNV 𝟑𝟎° 𝟑𝟎° Angle between VAB & IA = 30° + 𝝓 Angle between VCB & IC = 30° - 𝝓 P1 = 𝟑 VI cos (30° + 𝝓) P2 = 𝟑 VI cos (30° - 𝝓) Total 3 phase power P3-𝜙 = P1 + P2 = 3VI cos𝝓
- 7. Mohammed Waris Senan 7 Measurement of 3 phase Power P1 = 𝟑 VI cos (30°) P2 = 𝟑 VI cos (30°) Some Important Cases: 1. If Power factor of load is UNITY i.e. cos𝝓 = 1, 𝝓 = 0° Reading of both wattmeter is EQUAL 2. If Power factor of load is 1/2 i.e. cos𝝓 = 1/2, 𝝓 = 60° P1 = 𝟑 VI cos (30° + 60°) = 𝟑 VI cos (90°) = 0 P2 = 𝟑 VI cos (30° - 60°) = 𝟑 VI cos (30°) = 𝟑 𝟐 VI Reading of one wattmeter is NEGATIVE 3. If Power factor of load is 0 and ½ i.e. 0 ≤ cos𝝓 < ½ , 60° < 𝝓 ≤ 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 𝝓) P1 – P2 = 3 VI cos (30° + 𝜙) - 3 VI cos (30° - 𝜙) = 3 VI sin𝜙 So, Three Phase Reactive Power, Q3-𝝓 = 3 (P1 – P2 ) 3VIcos VIsin3 PP PP 21 21 3 tan 21 211 PP PP 3tan Now,
- 9. If power factor is UNITY, cos𝜙 = 1 ⇒ 𝜙 = 0° P3-𝜙 = 3VLILcos𝜙 = 3 × 400 × 115.4 × 1 = 80 kW So, Reading of watt-meters will be: P1 = 3 VI cos (30° + 𝜙) = 3 𝑉𝐿 3 I cos (30° + 𝜙) = VLIL cos (30° + 𝜙) P1 = 400 × 115.4 × cos (30° + 0°) = 40 kW P2 = 3 VI cos (30° - 𝜙) = 3 𝑉𝐿 3 I cos (30° + 𝜙) = VLIL cos (30° - 𝜙) P2 = 400 × 115.4 × cos (30° - 0°) = 40 kW Mohammed Waris Senan 9 Numerical Problems In a balanced 3-phase 400 V circuit, Iline = 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: VL = 400 V IL = 115.4 A P1 = 40 kW P2 = 0 power factor, cos𝜙 = ? Alternately: We know that if one of the wattmeter reads zero then pf of the load will be ½ . Total power, P3-𝜙 = P1 + P2 = 40 + 0 = 40 kW As, P3-𝜙 = 3VLILcos𝜙 ⇒ 3 × 400 × 115.4 × cos𝜙 = 40000 ∴ cos𝜙 = 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∠45° Ω and line-to-line voltage is 220 V (rms), what will be readings of two watt-meter? Question: Solution: Reading of wattmeter 1 : P1 = VAC IA cos (angle between VAC & IA) Reading of wattmeter 2 : P2 = VBC IB cos (angle between VBC & IB) V01270 3 V V,.V L ANAN Soceas referenTake A457.12 4510 0127 Z V I, AN A Now Let phase sequence be ABC
- 11. P1 = 220×12.7×cos (15°) = 2698.8 W P2 = 220×12.7×cos (75°) = 723.14 W Mohammed Waris Senan 11 Numerical Problems Solution: A1657.12)120-45(7.12I A457.12I B A V30220V V90220V AC BC 1530-45I&V AAC betweenAngle 7590-165I&V BBC betweenAngle
- 12. Mohammed Waris Senan 12 30V3VAC ANV CNV BNV CNV BCV 9𝟎° ACV 𝟑𝟎° 90V3VBC
- 13. Mohammed Waris Senan 13 120VVor240VV 120VV 0VV cnCN BN AN 150VVor210V3V 90V3V 30V3V cnCA BC AB ANV CNV BNV CAV ANV 𝟑𝟎° 𝟑𝟎° CNV BNV ABV ANV 𝟑𝟎° CNV 𝟑𝟎° 30V3VAC