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Power & Energy Measurements Chapter 3 Part B 3-Phase Power System Line current Line voltage Vs1=√2Vs(rms)sin(wt) Z1 Phase voltage Phase current Vs2=√2Vs(rms)sin(wt-120º) Vs3=√2Vs(rms)sin(wt-240º) 3-phase Generator Reference: http://en.wikipedia.org/wiki/Three-phase_electric_power Z2 Z3 3-phase Load 2 3-Phase Load Phase difference between the phase voltage 3 3-Phase Voltage and Current Relationship For star (Y) connected Load VL = 3 VP IL = IP For Mesh () connected Load VL = VP IL = 3 IP Example 3B.1: A balanced star-connected load of (8+j6)Ω per phase is connected to a balanced 3-phase 400V supply. Find the phase voltage and the line current? Solution 3B.1 The impedance per each phase is: Z p 82 62 10 Vp VL / 3 400 / 3 231V I p Vp / Z p 231/10 23.1A I L I p 23.1A 4 Power Measurement in 3-Phase circuits Following methods are used for measuring power in 3-load: 1. Single Wattmeter - Balanced case (equal impedances). 2. Three Wattmeters - Unbalanced case 3. Two Wattmeters - Balanced or Unbalanced case One Wattmeter Method: In this method, the total power is three times the power consumed by one of the load elements. PT = 3 × P1 Wattmeter M VA IA L C V VAN ZA N 5 Three Wattmeters Method: In this case, the sum of the readings of the three wattmeters will give the total power consumed by the three-phase load. PT = P1 + P2 + P3 IA M L C V VA ZA VAN N VB VC IB IC M M ZC L C V VBN C V VCN ZB L 6 Two Wattmeters Method: In this case, the sum of the readings of the two wattmeters will give the total power consumed by the three-phase load. PT = PW1 + PW2 IA M L VA W1 ZA VAB IB VC VB ZB ZC VCB IC M L W2 7 3-Phase Diagram The voltage VAB = VAN – VBN and VCB = VCN – VBN are indicated in the phase diagram. VAB VAN IA 30o IC VCB 30 o VBC VCN VBN The phase angle between VAB and VAN is 300. If current lags behind their respective phase voltage by , then the phase angle between iA and vAB is (300+). IB VCA 8 Two Wattmeters Method The reading of first wattmeter W1 and second wattmeter W2 : PW 1 VAB I A cos AB A Real{v AB iA* } PW 2 VCB I C cos CB C Real{vCB iC* } Derivation !!! Thus, the power measured (PW1+PW2) is the sum of real power consumed in the three phases. PW 1 PW 2 Real{vAN i*A} Real{vBN iB* } Real{vCN iC* } PZ A PZB PZC (Proven!!!) 9 Reading of wattmeter W1: P1 = VAB IA cos AB-A = VL IL cos ( 30º) VAB Reading of wattmeter W2: P2 = VCB IC cos CB-C = VL IL cos ( – 30º) VAN IA 30o IC VCB 30 o VBC VCN VBN For balanced system, the magnitudes: VAB = VCB = VCA = VL and IA = IB = IC = IL IB VCA 10 The sum of the two reading of two wattmeters': P1 + P2 = VL IL cos ( 30 ) VL IL cos ( 30 ) = 3 VL IL . cos cos(A+B) = cosAcosB - sinAsinB The difference between the readings of two wattmeters': P2 - P1 = VL IL cos ( 30 ) VL IL cos ( 30 ) = VL IL . sin power factor OR 11 Example 3B.2 On a 3-phase balanced-delta connected load supplied at 240V a.c., the two Wattmeters readings are: –1710 and 3210 watts. Find the power factor (P.F) and the current. Solution 3B.2: cosφ = 0.1733 P1 + P2 = 3 VL IL . cos OR P2 – P1 = VL IL . sin IL = 20.82A 12 Exercise 3.2 (a) A balanced star-connected load of (3+j4)Ω per phase is connected to a balanced 3-phase 200V supply. Find the phase voltage and the line current. (b) In a 2 wattmeter method of measurement of 3-phase power, if the total power is 240kW and the power factor is 0.8, calculate the reading of each wattmeter. 13 Answer 14 Variation in Wattmeter Readings The readings of the two wattmeters are: P1 = VL IL cos ( 30 ) and P2 = VL IL cos ( 30 ) Wattmeter, W1 readings: Both meters indicate equal and positive readings: P1 = P2 = VL IL cos 30 P1 = VL IL cos (30+60) = 0 0º 0 < < 60 60º 60 < < 90 P. F = cos 1 1 > P.F> 0.5 0.5 0.5 >P.F> 0 P1 +Ve +Ve 0 -Ve P2 +Ve P1 = P2 +Ve P1 P2 +Ve +Ve Both meters have positive readings but not equal: P1 P2 What happen if = 90º? Readings for both meters: P2 is +ve but P1 is -ve 15 Wattmeter Errors Power Loss in Pressure Coil and Fixed coil Moving coil Inductance: the moving coil current will tend to lag the supply voltage. Moving coil Capacitance: the moving coil current will tend to lead the supply voltage. (will not be discussed) Eddy Current: alternating magnetic field of the fixed coil induces eddy current in the solid metal parts near the fixed coil. These eddy currents set up their own magnetic field and thus effect the magnitude and phase of the magnetic field causing the deflection. (will not be discussed) Stray magnetic fields: electrodynamic wattmeter has to be shielded for protection. 16 Error due to Moving coil Inductance Let RM , LM and RS are the moving coil resistance, inductance and the series connected resistance respectively. Total resistance of pressure coil circuit is RP = RM + RS. Hence, the impedance of the moving coil: Wattmeter ZM IF And the current: k1iL iM vL ikM1i=L | Z M | ZM XM S O U R C E L IL M C V RM + j wLM L O A D VL RS IM RP 17 iM is lagging behind the vL by an angle of tan 1 wLM Rp Thus, the angle between the load current iL and iM will be ( - β). The deflecting torque is given as: Td k1iL iM vL k1iL | Z M | VL = VM IM ( ) vL Td k1iL cos kd iL vL' cos RP IL Moving coil phase diagram where vL' 2VL sin(wt ) and iL 2I L sin(wt ) cos RP | ZM | 18 The average value of deflecting torque: 1 T ' Td ( average ) kd cos iL vL dt kd I LVL cos cos( ) T 0 This gives the power metered by the wattmeter due to inductance of the moving coil. If the inductance LM of moving coil circuit were zero, then at all frequencies, the phase angle between voltage vL and current iM, namely would be zero and average wattmeter deflection torque would be equal to: Td ( average ) kd I LVL cos which gives the true power without error. 19 With the former as metered reading and the latter as the true power that should have been metered: kd I LVL cos True Power cos Metered Power kd I LVL cos cos( ) cos cos( ) Applying correction, the true power that should have been metered is: True Power The term cos Metered Power cos cos( ) cos cos cos( ) is called as the correction factor and the error is cos Error Metered Power True Power Metered Power 1 cos cos( ) % Error cos cos( ) Metered Power True Power 100% 1 100% True Power cos 20 Example 3B.3 The inductive reactance of the moving coil circuit of a dynamometer wattmeter is 0.4% of its resistance at normal frequency and the capacitance is negligible. Calculate the percentage error and correction factor due to reactance for loads at 0.7071 and 0.5 P.F. lagging 21 Solution 3B.3 22 Example 3B.4 A dynamometer type wattmeter is rated at 10A and 100V with a full-scale reading of 1000W. The inductance of the moving coil circuit is 5mH and its total series resistance is 3000Ω. If the voltage drop across the fixed coil of the wattmeter is negligible, what is the error in the wattmeter at the full rated voltamperes at zero power factor? 23 Solution 3B.4 Watmeter reading = VLILcosβcos(- β) = (100)(10)cos0.03ºcos(90º- 0.03º) = 0.5236W 24