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SARVAJANIK OF COLLEGE OF ENGINEERING & TECHNOLOGY Subject:-Electrical measurement And Measuring instrument ( 2130903 ) Topic :-measurement of power by Two wattmeter method and Measurement of reactive power
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Branch:- Electrical (M) Code :- 09 Group No. Student Name Enrollment No J Jay Babariya130420109005 Saurabh Chaudhari 130420109011 Naren Chauhan 130420109012 Jigar Gamit130420109018 Jigar Parmar130420109021 Guided By:- Shreyanshi Desai Ami Vyas Hemin Motiwala
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Topics Measurement of power by two Wattmeter method Introduction Proof : Star connected load Delta connected load Power factor measurement Merits and Demerits Measurement of reactive power
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Measurement of power by two Wattmeter method Introduction This method uses only two wattmeters to measure the power in 3 phase circuits. Principal behind the connections: Crefully observe the way in which the current and the voltage coils of the wattmeter are connected in figure 1. The currents coils of the two wattmeters are connectesd in any two lines and the voltage coils are connected btween their own current coil terminal and the line without the current coil(from R to Y and B to Y) The wattmeter reading remain same irrespective of the type of load. The total power in a three phase circuit is equal to the algebric sum of the two wattmeter readings. total power=w 1 +w 2
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Proof : Star connected load Assuming the phase sequence to be RYB, the phase voltages are VRN, VYN and VBN. Let the phase angle between the phase voltage and phase current be Φ degree. If the load is assumed to be inductive in nature then current in each phase lags the phase voltage by Φ degrees. VRY = VRN - VYN VBY = VBN - VYN from vector diagram
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Wattmeter reading W1 = VBY IB cos (30 - Φ) Wattmeter reading W2= VRY IR cos(30 + Φ) Total power = W = W1 + W2 = VBY.IB cos (30 - Φ) +VRY IR cos (30 +Φ) But VBY = VRY = VL and IB = IR = IL W = VL IL cos (30 - Φ) + VL IL cos (30 + Φ) = VL IL [cos (30 - Φ) + cos (30+ Φ)] = VL IL [cos30.cos Φ +sin 30.sin Φ + cos 30 cos Φ-sin 30. sinΦ] = VL IL 2 cos 30 cos Φ = VL IL 2 x√ 3 / 2.cos Φ W = √3. VL IL cos Φ watts This shows that two wattmeter is sufficient to measure total power in a 3 phase star system
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Proof :Delta connected load Wattmeter reading W1, from the circuit diagram is given by, From the vector diagram, phase angle between V BY and I B is (30 - Φ) W1 = VBY. IB cos (30 - Φ)
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Wattmeter reading W2, from the circuit diagram is given by, From the vector diagram, the phase angle between VRY and IR is (30 + Φ) W2 = VRY. IR cos (30 + Φ) Therefore, total power = W = W1 +W2 W= VBY. IB cos (30 - Φ) +VRY IR cos (30 + Φ) But V RB and V YB are line voltages whereas I R and I Y are line currents. W = VLIL cos (30 - Φ) + V1I1 cos (30 + Φ) = VLIL [cos (30 - Φ) + cos (30 +Φ)] = VLIL [cos 30. cos Φ + sin 30 sin Φ + cos 30.cos Φ - sin 30. sin Φ] = VLIL.2cos 30. cos Φ √3 = VLIL 2 x --------- cos Φ 2 W = √3. VL IL cos Φ watts This shows that two wattmeter is sufficient to measure total power in a 3 phase delta system
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Power factor measurement Let the load be balanced type with a lagging power factor. W1= VL IL cos (30 – Φ) and W2 = VL IL cos (30 + Φ) W1 – W2 = VL IL cos (30 - Φ)- VL IL cos (30+Φ) = VL IL [cos (30 - Φ)-cos (30+ Φ) = VL IL [cos30.cos Φ +sin 30.sin Φ - cos 30. cosΦ + sin30.sin Φ] = VL IL [2 sin 30.sin Φ] = VL IL 2. ½ sin Φ W1 –W2 = VL ILsin Φ We know that W1+ W2 =√3 VL IL cos Φ
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Merits and Demerits Merits: 1.We can use it for the balanced as well as unbalanced loads. 2.For the star types loads,it is not necessary to connect the neutral point for connection of the wattmeter. 3.The delta load need not be opened to connect the wattmeters. 4.For the balanced loads,it is possible to measure the power factor along with the power. 5.We need to use only two wattmeters to measure the power in a 3 phase circuit. 6.It is possible to mesure the reactive volt amperes for the balanced loads. Demerits: 1.The polarities of W 1 and W 2 depend upon the power factor. so unless we identify them correctly they can lead us to wrong results. 2. This method is not applicable to three phase four wire system.
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Measurement of reactive power While obtaning the expression for the power factor in the two wattmeter method, we have seen that, W1-W2= VL IL cos Φ We also know that the total reactive volt-ampere in a 3-phase circuit is given by Q= √3 VL IL cos Φ But VL IL cos Φ= W1-W2 Q= √3(W1-W2) Thus we can obtain the reactive power by multiplying the difference between the reading of two wattmeter by √3.
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THANK YOU.
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