A PSERC Tutorial on Contemporary Topics in Electric Power Quality Alias G. T. Heydt Arizona State University A ‘Tutrial’ on Power Quality © 2000 Arizona State University 1 2 PROGRAM 1. Power quality indices / pitfalls / three phase phenomena and applications / ‘interharmonics’ and other non-harmonics 2. Power acceptability, when is electric power Power Quality Indices delivered ‘acceptable’, vulnerability of loads 3. Series voltage boost hardware 4. Rectifier loads 5. Power quality standards 6. Why is power quality important? The salability of power quality 3 4 Power Quality Indices Index Definition Main applications EVEN HARMONICS Total harmonic dis- ∞ II / tortion (THD) ∑ i 1 General purpose; standards i =2 PV/| || I | Power factor (PF) tot rms rms Potentially in revenue metering • THEORETICALLY IMPOSSIBLE FOR Telephone influence ∞ wI2 / I factor ∑ ii rms Audio circuit interference i =2 SIGNALS THAT ARE SYMMETRIC ABOUT ∞ cI2 / I THE TIME AXIS C message index ∑ ii rms Communications interference i =2 ∞ • PRESENCE OF EVEN HARMONICS DO NOT IT product 22 Audio circuit interference; shunt capacitor ∑ wIii = stress i 1 IMPLY DC COMPONENTS IN THE GIVEN ∞ VT product 22 ∑ wVii i =1 Voltage distortion index SIGNAL ∞ ∞ K factor hI2 22/ I ∑∑hh • MOST COMMON OCCURRENCE OF EVEN h=1 h=1 Transformer derating VV/ Crest factor peak rms Dielectric stress HARMONICS IS IN THE SUPPLY CURRENT Unbalance factor ||/||VV−+ Three phase circuit balance OF TRANSFORMERS WHOSE LOAD SIDE Incandescent lamp operation; bus voltage ∆ Flicker factor VV/| | regulation; sufficiency of short circuit ca- HAVE DC CURRENT COMPONENTS pacity 5 6 1 EVEN HARMONICS Displacement factor (True) power factor AC AC + DC COMPONENT φ LOAD DF = cos( 60) PF = V (ΣP)/|Vrms||Irms| ‘Power frequency I power factor’ ‘Total power over V total volt-amperes’ PRESENCE OF DC ON ≤ SUPPLY SIDE INDICATIVE I TPF ≤ DF OF DC ON LOAD SIDE 7 8 ΣP = (1)(1)cos30o + (0.2)(0.2)cos60o + EXAMPLE (0.05)(0.15)cos(30o) = 0.892 RMS 60 Hz 180 Hz 420 Hz (Vrms)2 = 12 + 0.22 + 0.052 V 1∠0o 0.2∠20o 0.5 ∠10o Vrms = 1.021 I 1∠-30o 0.2∠80o 0.15 ∠-20o (Irms)2 = 12 + 0.22 + 0.152 Irms = 1.031 9 S = (Vrms)(Irms) = 1.052 10 POWER FACTOR MULTIPLIERS • BILLING MULTI[PLIERS TO SEND THE CUSTOMER THE PROPER SIGNAL CONCERNING POWER FACTOR • MULTIPLIER NEAT 1.0 FOR ~86%PF LAG TPF = ΣP / S = 0.848 • MULTIPLIER INCREASES TO ~1.3 FOR DECREASING POWER FACTOR • MULTIPLIER DECREASES TO ~0.95 FOR PF NEAR UNITY o • e.g., 0.06 $/kWh AT 86% PF LAG, 0.08 $/kWh AT LOW DF = DPF = cos(30 ) = 0.866 (lag) POWER FACTOR • WHICH PF? TPF? DISPLACEMENT FACTOR? • CUSTOMERS FAVOR USE OF DF, UTILITIES FAVOR TPF • LOSSES MORE CLOSELY RELATED TO TPF THAN DF • REQUIRED kVA OF SUPPLY EQUIPMENT MORE CLOSELY RELATED TO TPF 11 • INSTRUMENTATION ISSUES 12 2 THD THD RMS 60 Hz 180 Hz 420 Hz THE THD OF A SQUARE WAVE OF AMPLITUDE ±1 IS EASILY FOUND NOTING THAT THE RMS VALUE OF SUCH A WAVE IS 1.000 AND THE V 1∠0o 0.2∠20o 0.5 ∠10o FUNDAMENTAL COMPONENT IS 4/Π (ZERO TO PEAK). THE FUNDAMENTAL COMPONENT IS I 1∠-30o 0.2∠80o 0.15 ∠-20o (0.707)(4/Π) = (0.9002). THEREFORE THE SUM OF THE SQUARES OF THE HARMONIC 2 2 2 VTHD = 0.2 + 0.05 / 1 COMPONENTS IS 12-0.90022 = (0.1896). ITHD2 = 0.22 + 0.152 / 1 THEN, VTHD = 20.62% ITHD = 25% THD2 = 0.1896/0.9002 THD = 45.89% 13 14 THD - ANOTHER EXAMPLE f |V| | I | Three Phase Considerations 60 1.00 1.00 180 0.01 0.31 Balanced THD 300 0.04 0.15 Based on positive and negative sequence THDs only 420 0.03 0.07 540 0.02 0.03 Residual THD 660 0.01 0.02 Based on zero sequence only VTHD2 = 0.012 + 0.042 + 0.032 + 0.022 + 0.012 VTHD = 5.57% ITHD = 35.33% 15 16 THD THD ADVANTAGE: EVERYONE USES IT, THE RESIDUAL THD IS GENERALLY EASY TO CALCULATE, WIDELY FAR MORE HARMFUL THAN USED IN STANDARDS AND GUIDES BALANCED THD BECAUSE THERE IS NO ‘CANCELLATION EFFECT’ OF DISADVANTAGES: DOES NOT THE THREE PHASES OUT OF PHASE ACCELERATE WITH FREQUENCY, BY 120o BALANCED AND RESIDUAL THD NOT AS WELL KNOWN, DOES NOT TRULY SHOW THE INTERFERENCE IMPACT OF THE SIGNAL 17 18 3 TOTAL DEMAND DISTORTION DISTORTION INDEX (DIN) (TDD) TOTAL DEMAND DISTORTION IS A ∞ MEASURE OF THE THD TAKING INTO 2 ACCOUNT THE CIRCUIT RATING. AS I CIRCUIT RATING VERSUS LOAD CURRENT ∑ i RISES, TDD DROPS DIN = 2 TDD = THD * (Fundamental load current / Circuit rating) I rms 19 20 TELEPHONE INFLUENCE I∗T PRODUCT FACTOR ∞ 2 2 ∑ wi Ii IT = TIF * Irms TIF = 1 Irms 21 22 PEAK VALUES V*T PRODUCT PEAK VALUES CAN BE CHARACTERIZED BY A CREST FACTOR: • DEFINED LIKE I*T PRODUCT USING CF = PEAK VALUE / RMS VALUE VOLTAGE = 1.414 FOR A PERFECT SINE WAVE • kVT = 1000 VT • BALANCED AND RESIDUAL V*T PRODUCT ABSOLUTE LARGEST VALUE CAN BE OVERESTIMATED FOR ASYNCHRONOUS • USED IN SHUNT CAPACITOR SIGNALS AS THE SIMPLE ALGEBRAIC SUM STANDARDS - TO LIMIT HARMONIC OF THE AMPLITUDES OF THE CURRENTS ASYNCHRONOUS FREQUENCIES 23 24 4 RMS VALUES RMS VALUES If the function is not periodic, take limit as T --> infinity T 1 2 = Parseval’s theorem -- for signals of FRMS f (t)dt T ∫0 different frequencies, 2 2 2 2 (Vrms) = (V1rms) +(V2rms) +(V3rms) +... 25 26 RMS VALUES RMS VALUES If signals are of the same Examples frequency, need to combine the 10cos(t) + 2cos(2t)+ sin(3t) same frequency terms using 10 2 cos(t) +10 2 sin(t) phasor arithmetic, and then apply 10cos(t) +10sin( 2t) Parseval’s theorem without regard for phase angles 440 2 cos(314t) + 50 2 sin(314t) + 80 2 sin(492t) +10cos(1570t) 27 28 RMS VALUES RMS VALUES Second example First example 1. Gather like frequency terms 10 2 1 f (t) = 10 2( 2 cos(t + 45o )) F 2 = ( )2 + ( )2 + ( )2 rms 2 2 2 2. Find RMS value of result =5.123 2 = 20 = Frms 14.14 29 2 30 5 RMS VALUES RMS VALUES Third example Fourth example This example is aperiodic -- but no change in First combine fundamental term application of Parseval’s theorem: = 2 + 2 = F1,rms 440 50 442.83 2 = 1 2 + 1 2 Frms ( ) ( ) 2 2 Then apply Parseval’s theorem = Frms 1.00 = 2 + 2 + 2 = Frms 442.83 80 10 450.11 31 32 CONSEQUENCES OF TRANSFORMER DERATING HARMONICS DEFINE PLL-R AND PEC-R AS THE FULL LOAD LOSSES AND CORE LOSSES PER- • I2R HEATING DUE TO EXCESS UNITIZED BY THE I2R LOSSES. THEN THE CURRENT DERATED TRANSFORMER MAXIMUM • TRANSFORMER MAGNETIC LOSSES CURRENT IN PER UNIT IS • INCREASED MOTOR LOSSES P • INCREASED CREST CURRENT I = LL−R • CIRCUIT BOARD HEATING derated + 1 KPEC−R 33 34 APPLICATION OF IEEE C57.110 TRANSFORMER DERATING DERATING BASIC METHOD • CALCULATE TOTAL CORE LOSSES P THIS DERATING IS CONSERVATIVE EC • CALCULATE I2R LOSSES, P IN THAT ALL CORE LOSSES ARE I2R • CALCULATE TOTAL FULL LOAD LOSSES P INCREASED BY A FACTOR OF K -- LL • PERUNITIZE P AND P BY P THIS IS AN OVERESTIMATE OF THE LL EC I2R B-H LOSSES. THE METHOD IS IN • CALCULATE THE K-FACTOR OF THE LOAD CURRENT COMMON USE AS PRESCRIBED BY • CALCULATE DERATED RATING OF LOAD IEEE STANDARD C57.110, UL 1561 CURRENT P AND UL1562. I = LL−R derated + 1 KPEC−R 35 36 6 EXAMPLE A 67.5 kVA 1Ø DISTRIBUTION IT MAY BE NECESSARY TO TRANSFORMER IS RATED 7200 / 240 V. CALCULATE I2R LOSSES USING FULL THE CORE LOSSES ARE 75 W AT RATED VOLTAGE, AND THE FULL LOAD LOSSES LOAD CURRENT AND NAMEPLATE ARE 190 W. THE WINDING RESISTANCES RATING OF RESISTANCE --OR AN ARE 0.5% TOTAL. FIND THE DERATED ESTIMATE OF THE RESISTANCE. TRANSFORMER CAPACITY TO CARRY A LOAD CURRENT OF 150% THD WHICH IS COMPOSED OF FUNDAMENTAL AND THIRD HARMONIC. 37 38 SOLUTION SOLUTION P I = LL−R derated + 1 KPEC−R Irated = (67.5 k) / 240 = 281.25 A 0.563 = Iderated = (281.25)(0.479) 1+ (6.54)(0.222) = 138 A = 0.479 pu 39 40 Power Acceptability Curves APPROXIMATION 250 200 OVERVOLTAGE CONDITIONS C 150 B 100 0.5 CYCLE PLL-R = PEC-R + 1 E 50 RATED M 0 ACCEPTABLE POWER VOLTAGE -50 A PERCENT CHANGE IN BUS VOLTAGE 8.33 ms 8.33 UNDERVOLTAGE CONDITIONS -100 0.0001 0.001 0.01 0.1 1 10 100 1000 41 TIME IN SECONDS 42 7 Power Acceptability Curves Power Acceptability Curves 250 200 BUS B OVERVOLTAGE CONDITIONS FAULT 150 I BUS A z+, z-, z0 100 T z+, z-, z0 CIRCUIT BREAKER 0.5 CYCLE 0.5 50 I +-- 10% SOURCE RATED z+, z-, z0 0 ACCEPTABLE C POWER VOLTAGE BUS C -50 PERCENT CHANGE IN BUS VOLTAGE LOAD 8.33 ms UNDERVOLTAGE CONDITIONS -100 0.0001 0.001 0.01 0.1 1 10 100 1000 43 44 TIME IN SECONDS Power Acceptability Curves Power Acceptability Curves Disturbances to loads, whether they be overvoltages or undervoltages, have an impact depending on how much excess energy is Main challenges delivered to the load (in the overvoltage case) or how much energy was not delivered to the load (in How to sell power quality as a service the undervoltage case). If the cited energy level is too great, the operation of the load will be How to sell PQ measurement services disrupted.