Tab 18 - Apparent Power, Active Power, Reactive Power Distribution Systems Engineering - Course , Siemens Industry Inc., all rights reserved

Transcription

1 Tab 18 - Apparent Power, Active Power, Reactive Power Distribution Systems Engineering - Course 1 01, Siemens Industry Inc., all rights reserved

2 Some Basic Concepts Review of some important definitions: Peak Value of a sinusoidal current wave (I peak ) Occurs when the sinusoidal current wave is at its maximum amplitude RMS Value (also called effective value) of a sinusoidal current wave 1 T Ipeak IRMS Ipeak sin t dt T 0 T Average Value of a sinusoidal current wave I average 1 T I 0 peak T sin t dt T 0 I peak Course 1 Power Distribution Systems and Power Circuit Analysis 01 Siemens Industry Inc., Siemens Industry Inc., 15-

3 Voltage and current relationship COMPONENTS OF TOTAL CURRENT v(t) i(t) LOAD i R (t) i X (t) SYSTEM VOLTAGE v(t) v( t) V sin( t) V sin( t) i( t) I MAX MAX sin( t ) I RMS TOTAL CURRENT θ = Angle by which voltage v(t) leads total current i(t) RMS sin( t ) I RMS cos sin( t) COMPONENT OF CURRENT IN-PHASE WITH VOLTAGE (REAL COMPONENT) Real component of current supplies a NET energy to the LOAD i ( t) I cos sin( t) X R RMS RMS I RMS sin sin( t 90 COMPONENT OF CURRENT 90 O OUT-OF-PHASE WITH VOLTAGE (IMAGINARY COMPONENT) Imaginary component of current supplies no NET energy to the load o i ( t) I sin sin( t 90 ) o ) Course 1 Power Distribution Systems and Power Circuit Analysis 01 Siemens Industry Inc., Siemens Industry Inc., 15-3

4 Voltage and current relationship (continued) v(t) i(t) LOAD i R (t) i X (t) TOTAL CURRENT i( t) I sin( t ) RMS I RMS cos sin( t) I RMS sin sin( t 90 o ) REAL COMPONENT OF CURRENT i R ( t) I cos sin( t) RMS RMS VALUE (I R ) OF REAL COMPONENT OF CURRENT: RMS VALUE (I X ) OF IMAGINARY COMPONENT OF CURRENT: θ = Angle by which voltage v(t) leads current i(t) RMS VALUE (I RMS ) OF TOTAL CURRENT: RELATIONSHIP BETWEEN RMS VALUE OF REAL COMPONENT, IMAGINARY COMPONENT, AND TOTAL CURRENT WHEN ANGLE Θ IS POSITIVE: IMAGINARY COMPONENT OF CURRENT I i X ( t) I RMS IR I RMS cos I X I RMS sin TOTAL I RMS sin sin( t 90 I R I X I R I X o ) Course 1 Power Distribution Systems and Power Circuit Analysis 01 Siemens Industry Inc., Siemens Industry Inc., I TOTAL 15-4

5 Voltage, current, & power relationship INSTANTANEOUS AND AVERAGE POWER i(t) v(t) LOAD v( t) V sin( t) and i( t) I sin( t ) MAX MAX INSTANTANEOUS POWER, p(t), IS THE RATE AT WHICH ENERGY IS SUPPLIED TO THE LOAD p( t) v( t) i( t) V I sin( t) sin( t ) MAX MAX + p(t) means system supplies energy to load, - p(t) means load supplies energy back to system. Frequency of p(t) is twice that of system USING TRIGONOMETRIC IDENTITIES, THE EXPRESSION FOR INSTANTANEOUS POWER p(t) IS: VMAX IMAX VMAX IMAX p( t) cos 1 cos(t ) sin sin(t ) Siemens Industry Inc., 1 cos(t ) V I sin sin( ) p( t) V I cos t RMS RMS RMS RMS Note that the real component of current, I RMS cosθ, produces one component of the instantaneous power that has a non-zero average value. The imaginary component of current, I RMS sinθ, produces the second component of instantaneous power that has an average value of zero Course 1 Power Distribution Systems and Power Circuit Analysis 01 Siemens Industry Inc.,

6 Voltage, current, & power relationship (continued) EXPRESSION FOR INSTANTANEOUS POWER 1 cos(t) V I sin sin( ) p( t) V I cos t RMS RMS RMS AVERAGE POWER SUPPLIED TO THE LOAD OVER INTEGER MULTIPLES OF PERIOD T IS: RMS P AV 1 T T 0 p( t) dt V RMS I RMS cos in Watts THE INSTANTANEOUS POWER CAN ARBITRARILY BE SPLIT INTO TWO COMPONENTS CALLED p P (t) AND p Q (t) (ACTIVE AND REACTIVE POWER RESPECTIVELY) p P ACTIVE POWER, p p (t) 1 cos(t ) P 1 cos( ) ( t) V I cos t RMS RMS AVERAGE VALUE OF ACTIVE POWER = P P V RMS I RMS cos p Q REACTIVE POWER, p Q (t) ( t) V I sin sin(t ) Q sin(t ) RMS RMS PEAK VALUE OF REACTIVE POWER = Q Q V RMS I RMS sin THE APPARENT POWER IN THE CIRCUIT, S, IS THE PRODUCT OF THE RMS VALUE OF THE VOLTAGE, V RMS, AND THE RMS VALUE OF THE CURRENT, I RMS : S S V RMS I RMS Q S P Q P Course 1 Power Distribution Systems and Power Circuit Analysis 01 Siemens Industry Inc., Siemens Industry Inc., 15-6

7 Voltage, current, & power relationship (continued) EXAMPLE CALCULATION 1 A SINGLE-PHASE TWO WIRE CIRCUIT OPERATES AT 760 VOLTS RMS BETWEEN THE TWO WIRES. THE CURRENT IN THE PHASE WIRE IS MEASURED AT 5 AMPERES RMS. A METER CONNECTED TO THE CIRCUIT SHOWS THE ACTIVE (REAL) POWER SUPPLIED IS 150 KW 1. WHAT IS THE APPARENT POWER SUPPLIED BY THE CIRCUIT?. WHAT IS THE REACTIVE POWER SUPPLIED BY THE CIRCUIT? I RMS = 5 AMPS V RMS = 760 VOLTS LOAD 1. S VRMS IRMS 760 * 5 190,500 VOLT AMPERES kva GIVEN : P 150 kw. S P Q, or Q S P, Q Siemens Industry Inc., kvar IF THE LOAD WERE MODIFIED IN SOME MANNER SUCH THAT IT DRAWS ONLY 150 kw OF REAL POWER (P) AND NO REACTIVE POWER (Q = 0), WHAT WOULD THE LINE CURRENT BE IN AMPERES? THEN WITH Q 0 : I S P V RMS I RMS P 150,000 V 7,60 RMS RMS AMPERES 15-7 Course 1 Power Distribution Systems and Power Circuit Analysis 01 Siemens Industry Inc.,

8 Apparent Power (complex number) The combination of active and reactive power is referred to as apparent power, defined as follows with complex number notation: S = P + jq S = VI cos + j VI sin where S is the apparent power (VA) P is the active power (W) Q is the reactive power (VAR) θ is the angle between the voltage and the current (voltage angle current angle in this definition) P is related to energy that becomes heat, light, mechanical motion, etc. Q is related to energy that is stored in an inductor in ½ cycle, and then returned to the system in the next ½ cycle. Course 1 Power Distribution Systems and Power Circuit Analysis 01 Siemens Industry Inc., Siemens Industry Inc., 15-8

9 Course 1 Power Distribution Systems and Power Circuit Analysis 01 Siemens Industry Inc., 15-9 Siemens Industry Inc., Apparent Power Equations APPARENT POWER S IN TERMS OF VOLTAGE (V) AND CURRENT (I) AS COMPLEX NUMBERS X R X R V j V V and jv V V * X R X R I j I I and I j I I * X j R Z V = VOLTAGE PHASOR (A COMPLEX NUMBER) RMS VALUE V R = REAL PART OF VOLTAGE PHASOR V RMS VALUE V X = IMAGINARY PART OF VOLTAGE PHASOR RMS VALUE V * = COMPLEX CONJUGATE OF V I = CURRENT PHASOR (A COMPLEX NUMBER) RMS I R = REAL PART OF CURRENT PHASOR I RMS I X = IMAGINARY PART OF CURRENT PHASOR I RMS I * = COMPLEX CONJUGATE OF I - RMS * * * * * Z V Z V V Z V V I V jq P S Z I Z I I I Z I I V jq P S * * * X R R X X X R R X R X R I V I V j I V I V I j I V j V I V Q j P S * X X R R I V I V P X R R X I V I V Q APPARENT POWER S ACTIVE (REAL) POWER REACTIVE (IMAGINARY) POWER V V X V R I I X I R Z V 0 o I

10 Apparent Power Equations (continued) EXAMPLE CALCULATION V I Z I R I I X V V R V X 0 o THE OUTPUT OF A COMPUTER PROGRAM GIVES THE FOLLOWING FOR THE VOLTAGE AND CURRENT IN A SINGLE-PHASE LINE (SEE GENERALIZED SKETCH ABOVE) VOLTAGE V IS 760 VOLTS AT AN ANGLE OF 10 DEGREES LINE CURRENT I IS 5 AMPERES AT AN ANGLE OF DEGREES WHAT IS THE ACTIVE (REAL) POWER AND REACTIVE (IMAGINARY) POWER SUPPLIED TO THE LOAD? REAL AND IMAGINARY PART OF VOLTAGE PHASOR: VR 760 cos(10.0 ) Volts and VX 760 sin(10.0 ) P Q V I V I * * ,998. WATTS R R X X 1 V I V I * * ( ) 117, VARS X R R X 434 Siemens Industry Inc., o REAL AND IMAGINARY PART OF LINE CURRENT PHASOR: o IR 5.0 cos( ).06 Amps and I X 5.0 sin( ) ACTIVE (REAL) AND REACTIVE (IMAGINARY) POWER TO LOAD Note: The V, I, P, and Q values in this example are the same as in example calculation 1 o o Volts Amps Course 1 Power Distribution Systems and Power Circuit Analysis 01 Siemens Industry Inc.,

11 Apparent Power Equations (continued) EXAMPLE CALCULATION (continued) The active and reactive power, for this example, also can be calculated with the relationships below. cos P V RMS I RMS Q V RMS I RMS sin θ = Angle by which voltage v(t) leads current i(t) GIVENS: V RMS 760 Volts I RMS 5 Amperes P Q V I θ V = VOLTAGE ANGLE θ I ACTIVE (REAL) POWER: o 10.0 ( ) = CURRENT ANGLE o o * 5 cos( ) 149,999.3 Watts REACTIVE (IMAGINARY) POWER: o 760 * 5 sin( ) 117,43.8 Vars kw kvar Course 1 Power Distribution Systems and Power Circuit Analysis 01 Siemens Industry Inc., Siemens Industry Inc., 15-11

12 The Power Triangle Revisited 01, Siemens Industry Inc., all rights reserved

13 Phasor Diagram of the Power Triangle INDUCTIVE LOAD CAPACITIVE LOAD kva kvar O kw O kva POWER FACTOR IS THE COSINE OF THE ANGLE BETWEEN THE APPARENT POWER (kva) AND THE ACTIVE POWER (kw). FOR INDUCTIVE LOAD THE REACTIVE POWER (VARS) IS + IN SIGN FOR CAPACITIVE LOAD THE REACTIVE POWER (VARS) IS - IN SIGN kw kvar Course 1 Power Distribution Systems and Power Circuit Analysis 01 Siemens Industry Inc., Siemens Industry Inc., 15-13

14 Power Equations Revisited (Again!) RELATIONSHIPS BETWEEN KVA, KW, AND KVAR: INDUCTIVE LOAD kvar kva sin kva kvar kw kva cos O RELATIONSHIPS BETWEEN POWER FACTOR AND REACTIVE FACTOR: PF RF kw kw POWER FACTOR kva kvar REACTIVE FACTOR kva 1.0 PF RF kva kw kvar Course 1 Power Distribution Systems and Power Circuit Analysis 01 Siemens Industry Inc., Siemens Industry Inc., 15-14

15 Energy Billing and Load Composition 01, Siemens Industry Inc., all rights reserved

16 Customers and Loads As engineers we think of the customers as electrical loads that the power delivery system is designed to serve Generation Transmission Primary Distribution Secondary Distribution The accountants see the electrical loads as customers (sources of revenue) The customer has a meter that measures what they use, and the Utility sends them monthly bills that recover the cost of supplying the power Course 1 Power Distribution Systems and Power Circuit Analysis 01 Siemens Industry Inc., Siemens Industry Inc., 15-16

17 The kwh Meter Residential Customers An electro-mechanical type of kilowatt-hour meter to measure energy consumption Flux from current and voltage coils interact producing a torque on the disc. The rotational speed of the disk is proportional to the real power (kw). Time integration of the power thru a gear mechanism is the energy consumed, as displayed on the dials. A kwh meter only measures the active power. The amount of reactive power consumed (VARs) are not measured. Siemens Industry Inc., Course 1 Power Distribution Systems and Power Circuit Analysis 01 Siemens Industry Inc.,

18 The kva Meter Industrial and Commercial Customers Used to meter commercial and industrial loads Monitors the kw and kvar as well as the usage in time periods. Digital readouts Meter data can be downloaded to hand held computer Meter can be read remotely Course 1 Power Distribution Systems and Power Circuit Analysis 01 Siemens Industry Inc., Siemens Industry Inc., 15-18

19 What Are Electrical Loads? The electric power system elements have both a resistive and reactive component Resistive Components (R) Overhead line & Underground cable circuit conductor resistance Transformer winding resistance Reactive Components (X L, X C ) Overhead line & Underground cable circuit reactance Transformer winding leakage reactance Phase reactor reactance for limiting short circuit current Series capacitors to cancel line inductive reactance Loads are comprised of Resistive elements I * R = kw Reactive elements I * X = kvar Course 1 Power Distribution Systems and Power Circuit Analysis 01 Siemens Industry Inc., Siemens Industry Inc., 15-19

20 Load Composition Constant Power Loads Demand the same amount of kva regardless of the voltage supplied to them Constant Current Loads Demand the same current regardless of the voltage applied Constant Impedance Loads At all times, present the same impedance to the system, irrespective of voltage applied to them Course 1 Power Distribution Systems and Power Circuit Analysis 01 Siemens Industry Inc., Siemens Industry Inc., 15-0

21 Course 1 Power Distribution Systems and Power Circuit Analysis 01 Siemens Industry Inc., Load Categories Residential Single and multiple phases; seasonal Commercial Light to heavy Industrial Light to heavy with multiple shifts Agricultural Short burst of high demand Lighting Changes with time of year Siemens Industry Inc., 15-1

22 Distribution Circuits They are an aggregate of the different types of loads categories The demands fluctuate The power factor for each load type is not the same Each load has a different requirement with respect to the quality of power supply Course 1 Power Distribution Systems and Power Circuit Analysis 01 Siemens Industry Inc., Siemens Industry Inc., 15-

23 Tab 19 - Power Quality Overview Distribution Systems Engineering Course 1 01, Siemens Industry Inc., all rights reserved

24 Topics What is power quality? Voltage disturbances related to power quality Causes of PQ problems Power quality standards Course 1 Power Distribution Systems and Power Circuit Analysis 01 Siemens Industry Inc., Solutions Siemens Energy Inc., Power Technologies International 16-

25 What Is Good Power Quality? Utility perspective: The relative absence of utility-caused voltage variations at the point of common coupling (PCC). Customer perspective: Electric power which supports their operations with minimal power induced equipment disturbances and failures. Course 1 Power Distribution Systems and Power Circuit Analysis 01 Siemens Industry Inc., Siemens Energy Inc., Power Technologies International 16-3

26 Ideal Voltage Conditions Voltage magnitude is well within ANSI C84.1 Range A (normal operation) limits Sags, swells, or transient voltages are non-existent or very minor No momentary or permanent interruptions (outages) Harmonic distortion and noise are well within specified limits No observable flicker Course 1 Power Distribution Systems and Power Circuit Analysis 01 Siemens Industry Inc., Siemens Energy Inc., Power Technologies International 16-4

27 Power Quality Disturbances Long duration voltage variations steady-state overvoltages & undervoltages sustained interruptions (permanent faults) Short duration voltage variations momentary interruptions (clearing temporary faults and reclosing) sags & swells (from faults on system) Frequency variations Transients impulsive oscillatory Voltage unbalance (imbalance) Voltage fluctuations Course 1 Power Distribution Systems and Power Circuit Analysis 01 Siemens Industry Inc., Siemens Energy Inc., Power Technologies International 16-5

28 Power Quality Disturbances (continued) Waveform distortion harmonics & interharmonics noise notching Reliability issues Course 1 Power Distribution Systems and Power Circuit Analysis 01 Siemens Industry Inc., momentary outages (interruptions) permanent outages (interruptions) Siemens Energy Inc., Power Technologies International 16-6

29 Power Quality Disturbances (continued) Voltage Sags VOLTAGE SAGS ARE THE MOST COMMON POWER QUALITY DISTURBANCE EXPERIENCED BY END USERS ITIC CURVE PROVIDES GUIDANCE TO EQUIPMENT MANUFACTURERS ON VOLTAGE SUSCEPTIBILITY THAT THEY SHOULD DESIGN INTO THEIR EQUIPMENT POINTS ON ITIC CURVE ARE AN AGGREGATION OF OVER 3000 VOLTAGE EVENTS RECORDED AT 100 LARGE MANUFACTURING PLANTS IN THE USA OVER A PERIOD OF ONE YEAR Course 1 Power Distribution Systems and Power Circuit Analysis 01 Siemens Industry Inc., Siemens Energy Inc., Power Technologies International 16-7

30 Substation Bus Voltage Sag for SLG Fault on Radial Feeder FACTORS WITH SIGNIFICANT IMPACT ON SUBSTATION BUS VOLTAGE 1. AVAILABLE 3-PHASE AND SLG FAULT CURRENT ON SUB BUS (Z 0 & Z 1 ). IMPEDANCE OF FAULTED FEEDER IN OHMS PER UNIT LENGTH 3. DISTANCE FROM SUBSTATION TO FAULT (L) LINE CONFIGURATION AND IMPEDANCES FOR PLOTS SUB 13. KV BUS 477 MCM AL PHASE 4/0 CU NEUTRAL Ob 3" 3" Z 0 Z 1 UNFAULTED FEEDERS FAULTED FEEDER (SLG PHASE A) L Disk: CE Dist Course 006, #1 Sub Bus V, SLG FLT.FCW Oa DISC: CE Dist Course, 006, #1 Con Ed OH Line Z1 Z0.FCW 48" R 1 = Ohms / mile X 1 = Ohms / mile R 0 = Ohms / mile X 0 = Ohms / mile n Oc Course 1 Power Distribution Systems and Power Circuit Analysis 01 Siemens Industry Inc., Siemens Energy Inc., Power Technologies International 16-8

31 Substation Bus Voltage Sag for SLG Fault on Radial Feeder (continued) SUBSTATION LINE-TO-GROUND BUS VOLTAGES 13. kv SYSTEM FEEDER FAULTED PHASE = A, UNFAULTED PHASES = B & C Notes: Substation bus sag and swell applied to all unfaulted feeders until faulted feeder breaker opens. For selected configuration, unfaulted phase C bus voltage swell is less than unfaulted phase B bus voltage swell. In 7 kv system, voltage sag on substation bus (with same available fault current as at 13. kv), will be greater than sag at 13. kv. Course 1 Power Distribution Systems and Power Circuit Analysis 01 Siemens Industry Inc., Siemens Energy Inc., Power Technologies International 16-9

32 Power Quality Disturbances (continued) EXAMPLE OF CONDITION CREATING VOLTAGE UNBALANCE FOR SMALL THREE-PHASE PUMPING APPLICATION OPEN FUSE CUTOUT ON ONE OF THE POWER LEG TRANSFORMER FLOATING-WYE DELTA TRANSFORMER BANK FOR 4-WIRE DELTA SERVICE POWER LEG TRANSFORMERS = 10 KVA LIGHTING LEG TRANSFORMER = 15 KVA 4-WIRE DELTA SERVICE SUPPLYING A 3-PHASE PUMPING APPLICATION Course 1 Power Distribution Systems and Power Circuit Analysis 01 Siemens Industry Inc., Siemens Energy Inc., Power Technologies International 16-10

33 Then Versus Now Power quality today is not the same as power quality in 1950 With the increase in nonlinear devices, the power has become dirty with lots of voltage and current distortion It can present a localized problem that results in overheating of transformers and fuses due mainly to the presence of high harmonic currents Course 1 Power Distribution Systems and Power Circuit Analysis 01 Siemens Industry Inc., Siemens Energy Inc., Power Technologies International 16-11

34 Customers - Sensitive Equipment Industrial/commercial concerns computer drive manufacturing systems PCs for data processing and management application adjustable speed drives/motor driven assembly systems Programmable logic controllers (PLC) Computer numerical control (CNC) machines Servo Drives Robots the increasingly competitive nature of manufacturing the effect of non-linear loads on utility system, other customers and plant equipment Industrial equipment failures (Power supplies & motors) One study suggests that current surges following a voltage sag is more responsible for failures than are surge overvoltages Course 1 Power Distribution Systems and Power Circuit Analysis 01 Siemens Industry Inc., Siemens Energy Inc., Power Technologies International 16-1

35 Customers - Sensitive Equipment (continued) Residential concerns VCRs digital clocks home PCs various other sags and outage sensitive electronic devices Residential concern, open neutral in 10/40-volt service Causes steady state overvoltages /undervoltages HV O PN DISTRIBUTION TRANSFORMER H1 H X1 X X3 POLE GROUND CONNECTION 10 V 10 V OPEN NEUTRAL 1-PHASE 3-WIRE SERVICE O1 O V1 V SN 3-WIRE LOAD 10 V LOAD 10 V LOAD 40 V LOAD RESISTANCE TO GROUND AT SERVICE (ROD, H0 PIPE) Disk: CE Dist Course 006, #1 Open Neutral Secondary.FCW Course 1 Power Distribution Systems and Power Circuit Analysis 01 Siemens Industry Inc., Siemens Energy Inc., Power Technologies International 16-13

36 Origins of Power Quality Problems Utility problems weak system (High Impedance with load injecting high harmonic currents) single phasing in three-phase lines open neutral on primary or secondary system poor system design fault induced momentary outages, voltage sags and voltage swells equipment failures (transformers, splices, etc) switching surges capacitor switching transients lightning surges harmonic resonance Course 1 Power Distribution Systems and Power Circuit Analysis 01 Siemens Industry Inc., Siemens Energy Inc., Power Technologies International 16-14

37 Origins of Power Quality Problems (continued) Customer caused problems poor system design or defective wiring interaction of loads grounding problems and loops electromagnetic compatibility problems Harmonics switching of large loads (e.g. motors, arc furnaces) producing flicker Course 1 Power Distribution Systems and Power Circuit Analysis 01 Siemens Industry Inc., Siemens Energy Inc., Power Technologies International 16-15

38 Origins of Power Quality Problems (continued) Manufacturers of Utilization Equipment inadequate design of utilization equipment cost-cutting measures which make equipment more sensitive improper installation or application Course 1 Power Distribution Systems and Power Circuit Analysis 01 Siemens Industry Inc., Siemens Energy Inc., Power Technologies International 16-16

39 Who is responsible for PQ? Utilities to provide reliable service and voltage within specified limits Customers to buy less sensitive electronic devices commercial and industrial customers should understand their power environment and purchase equipment from manufacturers who are aware of PQ issues Course 1 Power Distribution Systems and Power Circuit Analysis 01 Siemens Industry Inc., Siemens Energy Inc., Power Technologies International 16-17

40 Who is responsible for PQ? (continued) Manufacturers to produce products which will operate properly on the electromagnetic environment of a typical power system Derating required of induction motors in presence of voltage unbalance DERATING FACTOR OPERATION ABOVE 5 % VOLTAGE UNBALANCE IS NOT RECOMMENDED BY NEMA STANDARDS CE 006 #1, NEMA DERATE.EP PERCENT VOLTAGE UNBALANCE Voltage unbalance at many points on radial distribution circuits will exceed 1 %. In old days, many induction motors would operate satisfactorily with voltage unbalances up to 3.0 %. Course 1 Power Distribution Systems and Power Circuit Analysis 01 Siemens Industry Inc., Siemens Energy Inc., Power Technologies International 16-18

41 Utility Power Quality Programs 1996 PQ survey of utilities done by PTI: 80% of utilities had PQ programs for commercial and industrial customers, 6% also had programs for residential customer 9% of utilities charged for PQ consulting PQ programs were usually reported as effective PQ programs monitoring (mainly, large customer loads) educational programs Course 1 Power Distribution Systems and Power Circuit Analysis 01 Siemens Industry Inc., Siemens Energy Inc., Power Technologies International 16-19

42 Power Quality Standards Course 1 Power Distribution Systems and Power Circuit Analysis 01 Siemens Industry Inc., Siemens Energy Inc., Power Technologies International 16-0

43 Power Quality Standards (continued) Course 1 Power Distribution Systems and Power Circuit Analysis 01 Siemens Industry Inc., Siemens Energy Inc., Power Technologies International 16-1

44 Power Quality Standards (continued) Course 1 Power Distribution Systems and Power Circuit Analysis 01 Siemens Industry Inc., Siemens Energy Inc., Power Technologies International 16-

45 Some Disturbances and Corresponding Standards Transients ANSI/IEEE C6 IEEE Std 1100 Harmonics IEEE Std 519 Course 1 Power Distribution Systems and Power Circuit Analysis 01 Siemens Industry Inc., Voltage sags IEEE Std P1346 CBEMA / ITIC tolerance curves Voltage Flicker IEEE Std 519 IEEE Std 141 GE Flicker Curve Siemens Energy Inc., Power Technologies International 16-3

46 Tab 0 - Voltage Unbalance, Flicker and Transients Distribution Systems Engineering Course 1 01, Siemens Industry Inc., all rights reserved

47 Voltage Unbalance Under Steady-State Conditions Voltage levels are not the same on all three phases of a feeder at the same moment in time Distribution systems are typically unbalanced The load demand at each phase is not the same The current flow in the phase conductors is not the same Line is not symmetrical The voltage drop is different for each phase The voltage on the substation bus is unbalanced This results in a voltage unbalance between the phases, and between the phases and ground Course 1 Power Distribution Systems and Power Circuit Analysis 01 Siemens Energy Inc., Siemens Energy Inc., Power Technologies International 17-

48 Voltage Unbalance Under Steady-State Conditions (continued) Unbalanced transformer bank (open-wye open-delta) causes unbalanced line currents on primary side as well as current in the primary multi-grounded neutral conductor LIGHTING LEG TRANSFORMER POWER LEG TRANSFORMER Note: Transformer bank is on a two-phase tap line supplied from a three-phase four-wire multigrounded neutral circuit. PRIMARY PHASE WIRES PRIMARY NEUTRAL Course 1 Power Distribution Systems and Power Circuit Analysis 01 Siemens Energy Inc., Siemens Energy Inc., Power Technologies International 17-3

49 Voltage Unbalance Under Steady-State Conditions (continued) Basic Definition Voltage Unbalance In percent V 100x V max AVE Where: Δ VMAX V AVE = The maximum deviation from the average phase-to-phase voltage V ab V bc V ca = The average phase-to-phase voltage 3 V ab, V bc, and V ca are the three phase-to-phase voltages Course 1 Power Distribution Systems and Power Circuit Analysis 01 Siemens Energy Inc., Siemens Energy Inc., Power Technologies International 17-4

50 Voltage Unbalance Under Steady-State Conditions (continued) Example: Suppose a customer connected to the end of a 40 V three-phase service and assume that at the PCC the phase-tophase voltages are V ab = 30 volts V bc = 3 volts V ca = 5 volts Then: V AVE = ( ) / 3 = 9 volts V 1 = 30 9 = 1 volt V = 3 9 = 3 volts V 3 = 5 9 = 4 volts V MAX = 4 volts Consequently: Voltage unbalance = (4/9) 100 = 1.75 % Course 1 Power Distribution Systems and Power Circuit Analysis 01 Siemens Energy Inc., Siemens Energy Inc., Power Technologies International 17-5

51 ANSI C84.1 Voltage Standards Classification Range A Range B Service Voltage 114 to 16 volts 110 to 17 volts Utilization Voltage 110 to 15 volts 106 to 17 volts Recommendation: Voltage unbalance at the PCC or service entrance shall not exceed 3% under no load conditions The PCC is the point of common coupling, typically at the meter or service entrance equipment The electric utilities are responsible only for satisfying the service entrance or PCC voltage requirements. Customers are responsible for maintaining proper voltage downstream of the service entrance or PCC! Course 1 Power Distribution Systems and Power Circuit Analysis 01 Siemens Energy Inc., Siemens Energy Inc., Power Technologies International 17-6

52 Voltage Unbalance Alternate Definition and Limit NEGATIVE-SEQUENCE VOLTAGE UNBALANCE (d ) % d V V 1 * 100 V = MAGNITUDE OF NEGATIVE-SEQUENCE VOLTAGE V 1 = MAGNITUDE OF POSITIVE-SEQUENCE VOLTAGE GIVEN THREE PHASE-TO-PHASE VOLTAGES V A, V B, AND V C (PHASORS WITH BOTH MAGNITUDE AND ANGLE): 1 V1 V A a VB a VC 3 O j a e j 1 V V A a VB a VC 3 RECOMMENDED LIMIT FOR d % d % Course 1 Power Distribution Systems and Power Circuit Analysis 01 Siemens Energy Inc., Siemens Energy Inc., Power Technologies International 17-7

53 Voltage Unbalance Alternate Calculation Method From Chart EXAMPLE: E a = 35.0, E b = 30.0, E c =.0 E b /E a = E c /E a = FROM CHART: d = V / V 1 = Θ = 37 Degrees FROM EXACT CALCULATIONS: V / V 1 = Θ = Degrees By knowing E a, E b and E c it is possible with chart to determine the angle θ and the unbalance factor d. Chart is applicable only when there are no zero-sequence components in the three voltages. From Westinghouse T&D Reference Book Siemens Energy Inc., Power Technologies International 17-8 Course 1 Power Distribution Systems and Power Circuit Analysis 01 Siemens Energy Inc.,

54 Voltage Flicker VOLTAGE FLICKER IS: Voltage Drop as Seen By Customers Visible With Incandescent lights and some CFL s Complaints Normally From Residential Customers Much Lower in Magnitude Than A Fault Caused Voltage Sag Usually Only a Few Volts on a 10-Volt Bas Noticeable at Low Levels and Annoying at Higher Levels VOLTAGE FLICKER DOES NOT: Normally Does Not Cause Equipment Failure or Downtime Does Not Damage Other Customers Equipment Course 1 Power Distribution Systems and Power Circuit Analysis 01 Siemens Energy Inc., Siemens Energy Inc., Power Technologies International 17-9

55 Voltage Flicker RAPID CHANGES IN THE RMS VOLTAGE CAUSE-FLUCTUATING LOADS 1. ARC FURNACES. WELDERS 3. RECIPROCATING COMPRESSORS 4. ROCK CRUSHERS 5. SAWMILLS 6. CAR SHREDDERS EFFECT OF VOLTAGE FLICKER CAN CAUSE VARIATION IN LAMP LIGHT OUTPUT THAT CAN BE EITHER PERCEPTIBLE OR ANNOYING TO THE END USER Note: 1. Flicker is worst at fluctuating load, and decreases at points up stream from the flicker source. Course 1 Power Distribution Systems and Power Circuit Analysis 01 Siemens Energy Inc., Siemens Energy Inc., Power Technologies International 17-10

56 Fluctuations and Dips EXAMPLE OF WAVEFORM PRODUCING VERY OBJECTIONABLE LAMP FLICKER ΔV = = 0.0 PER UNIT Course 1 Power Distribution Systems and Power Circuit Analysis 01 Siemens Energy Inc., Siemens Energy Inc., Power Technologies International 17-11

57 GE Flicker Curve DEFINES BASED ON TESTS BORDERLINES OF VISIBILITY AND IRRITATION WITH INCANDESCENT LIGHTS FUNCTION OF PERCENT VOLTAGE DIP AND FREQUENCY OF DIPS Notes: 1. Based upon square-wave changes to supply voltages at indicated frequencies (see previous chart).. From tests run by GE and utilities in 1930 with incandescent bulbs. 3. Used by 69 % of utilities based on 1985 IEEE survey. Siemens Energy Inc., Power Technologies International 17-1 Course 1 Power Distribution Systems and Power Circuit Analysis 01 Siemens Energy Inc.,

58 Permissible Flicker Limits (Con Edison) FROM 1958 EEI T&D COMMITTEE MEETING Course 1 Power Distribution Systems and Power Circuit Analysis 01 Siemens Energy Inc., Notes: Upper stair case curves apply to radial services and underground networks. Lower stair case curve applies to primary distribution lines. Siemens Energy Inc., Power Technologies International 17-13

59 Measuring Voltage Flicker The Flicker Meter The IEC flickermeter standard was developed in Europe and later adopted in the U.S. as IEEE The measurement methodology described in the IEC standard more accurately accounts for complex voltage fluctuations encountered in actual practice by including the effects of multiple flicker sources, frequencies and varying voltage modulation waveforms. In addition, the flickermeter approach standardizes flicker monitoring across different manufacturers IEC compliant instruments should all produce the same results for a given flicker excitation. The real advantage of the flickermeter method is inherent in its ability to accurately model the human flicker perception. This rather complex modeling is accomplished by five signal processing blocks described in the IEC standard, which represent the lamp-eye-brain response to light flicker the response of a lamp to supply voltage variations, the perception of the human eye and the memory characteristics of the human brain. Course 1 Power Distribution Systems and Power Circuit Analysis 01 Siemens Energy Inc., Siemens Energy Inc., Power Technologies International 17-14

60 Measuring Voltage Flicker The Flicker Meter The flickermeter system provides several measurement outputs. IFL, or instantaneous flicker level, represents the real time voltage modulation modified by the lamp-eye-brain response. This output can be plotted as a time interval graph and is useful for tracking down sources of voltage fluctuations. A statistical analysis block completes the human perception system by providing short and long term flicker severity indexes. Short term flicker severity P st is evaluated over a 10 minute observation period and is used to evaluate disturbances caused by flicker sources with short duty cycles. According to the IEEE 1453 standard, a P st value of 1.0 represents the system compatibility level, the level below which customer complaints are not likely to occur. P st is therefore commonly used to evaluate whether the measured voltage fluctuations are severe enough to cause flicker complaints. Long term flicker P lt is derived from 1 successive P st values, or two hours, and is more suitable for evaluating the combined effect of several randomly operating loads such as welders or motors over longer periods of time. A P lt value of 0.8 is considered the system compatibility limit according to IEEE Course 1 Power Distribution Systems and Power Circuit Analysis 01 Siemens Energy Inc., Siemens Energy Inc., Power Technologies International 17-15

61 Oscillatory Transient - Capacitor Switching LS SUBSTATION BUS FEEDER IMPEDANCE CAP BANK Vrms Vbus R F L F C CE 006 D#, Circuit Cap Sw Transient.FCW Closing capacitor switch produces high-frequency inrush current and associated high frequency component in the system voltages, that decay exponentially with time Course 1 Power Distribution Systems and Power Circuit Analysis 01 Siemens Energy Inc., Siemens Energy Inc., Power Technologies International 17-16

62 Oscillatory Transient Inductive Load Switching Course 1 Power Distribution Systems and Power Circuit Analysis 01 Siemens Energy Inc., Siemens Energy Inc., Power Technologies International 17-17

63 Impulsive Transient - Lightning Surges BOTH MEASUREMENTS MADE ON CIRCUIT OR PT SECONDARY WITH A NOMINAL VOLTAGE OF 10 VOLTS RMS OR VOLTS CREST Course 1 Power Distribution Systems and Power Circuit Analysis 01 Siemens Energy Inc., Siemens Energy Inc., Power Technologies International 17-18

64 Tab 1 - Harmonics, Notching and Noise Distribution Systems Engineering Course 1 01, Siemens Industry Inc., all rights reserved

65 Voltage Notching Course 1 Power Distribution Systems and Power Circuit Analysis 01 Siemens Energy Inc., Siemens Industry Inc., 18-

66 Electrical Noise Course 1 Power Distribution Systems and Power Circuit Analysis 01 Siemens Energy Inc., Siemens Industry Inc., 18-3

67 Harmonics Course 1 Power Distribution Systems and Power Circuit Analysis 01 Siemens Energy Inc., Siemens Industry Inc., 18-4

68 Harmonics (continued) Harmonics are multiples of the fundamental frequency TOTAL Like the waves in a pond, they will add or subtract to the fundamental Under and Over voltage Course 1 Power Distribution Systems and Power Circuit Analysis 01 Siemens Energy Inc., FUNDAMENTAL THIRD HARMONIC FIFTH HARMONIC Siemens Industry Inc., 18-5

69 Harmonics (continued) COMBINATION OF FUNDAMENTAL AND FIRST THREE ODD HARMONICS 1 Harmonic Peak Harmonic per unit Order Course 1 Power Distribution Systems and Power Circuit Analysis 01 Siemens Energy Inc., Siemens Industry Inc., 18-6

70 Harmonics (continued) COMBINATION OF FUNDAMENTAL AND FIRST THREE ODD HARMONICS - Harmonics lagged by 30 degrees from zero of fundamental Harmonic Peak 1 Harmonic Order per unit Course 1 Power Distribution Systems and Power Circuit Analysis 01 Siemens Energy Inc., Siemens Industry Inc., 18-7

71 Harmonics (continued) COMBINATION OF FUNDAMENTAL AND FIRST THREE ODD HARMONICS - Harmonics lagged by 60 degrees from zero of fundamental Harmonic Peak 1 Harmonic Order per unit Course 1 Power Distribution Systems and Power Circuit Analysis 01 Siemens Energy Inc., Siemens Industry Inc., 18-8

72 Harmonic Distortion Factor Distortion Factor in % Each harmonic component contributes to the overall distortion of the fundamental frequency wave Each harmonic component can be expressed in percent of the amplitude of the fundamental frequency wave Total harmonic distortion (THD) is the distortion factor including all relevant harmonic components (usually from nd to 50th) Sum of the Squares of the Harmonic Component Amplitudes % THD 100 Amplitude of the fundamental wave Course 1 Power Distribution Systems and Power Circuit Analysis 01 Siemens Energy Inc., Siemens Industry Inc., 18-9

73 Calculation of THD Example: Consider the voltage waveshape shown below Voltage Waveform at F9: phase A Time (ms) Course 1 Power Distribution Systems and Power Circuit Analysis 01 Siemens Energy Inc., Voltage (pu) Siemens Industry Inc., 18-10

74 Calculation of THD (continued) A harmonic spectral analysis of this wave indicates the following components: - a fundamental frequency wave (60Hz) of magnitude 1 pu - a 5th harmonic (300Hz) wave of magnitude 0.14 pu - a 7th harmonic (40Hz) wave of magnitude 0.1 pu - a 11th harmonic (660Hz) wave of magnitude 0.03 pu - a 13th harmonic (780Hz) wave of magnitude 0.03 pu Then: % THD (0.03) % Course 1 Power Distribution Systems and Power Circuit Analysis 01 Siemens Energy Inc., Siemens Industry Inc., 18-11

75 Calculation of THD (continued) THEORETICAL TOTAL WAVE CONSIDERING FUNDAMENTAL, 5 TH, 7 TH, AND 11 TH HARMONIC FROM PREVIOUS EXAMPLE FUNDAMENTAL = 1.0 PU 5 TH HARMONIC = 0.14 PU 7 TH HARMONIC = 0.1 PU 11 TH HARMONIC = 0.03 PU Course 1 Power Distribution Systems and Power Circuit Analysis 01 Siemens Energy Inc., Siemens Industry Inc., 18-1

76 Harmonic Sources Course 1 Power Distribution Systems and Power Circuit Analysis 01 Siemens Energy Inc., TVs Computers Fluorescent lights Dimmers for incandescent lights HVDC terminals UPS Systems AC adjustable speed motor drives Battery Chargers (Electric Cars) Transformers (magnetizing current) Siemens Industry Inc., 18-13

77 Fluorescent Lighting System V-I Waves OLD STYLE WITH FERRORESONANT LIGHTING BALLAST TYPICAL SCREW-IN COMPACT FLUORESCENT LAMP VOLTAGE: BLUE CURVE CURRENT: RED CURVE DISPLACEMENT POWER FACTOR 40 % Total Harmonic Distortion = 19 % VOLTAGE: BLUE CURVE CURRENT: RED CURVE DISPLACEMENT POWER FACTOR SLIGHTLY LEAD Total Harmonic Distortion = 10 % Course 1 Power Distribution Systems and Power Circuit Analysis 01 Siemens Energy Inc., Siemens Industry Inc., 18-14

78 Fluorescent Lighting System V-I Waves (continued) FLUORESCENT LAMPS WITH ELECTRONIC BALLASTS Course 1 Power Distribution Systems and Power Circuit Analysis 01 Siemens Energy Inc., VOLTAGE: BLUE CURVE CURRENT: RED CURVE DISPLACEMENT POWER FACTOR UNITY Total Harmonic Distortion = 49 % Siemens Industry Inc., 18-15

79 Harmonics in Balanced Three-Phase Four- Wire Multi-Grounded Neutral Systems Balanced Three-Phase System Current waveform in each phase is identical Fundamental component of current in each phase is displaced by 10 degrees from that in any other phase 5 th, 7 th, 11 th, 13 th, 17 th, 19 th, etc Harmonic currents are either positive- or negative-sequence No harmonic currents in neutral return path (multi-grounded neutral of primary system and earth) 3 rd, 9 th, 15 th, 1 st, etc Harmonic currents are zero-sequence At each harmonic, current in neutral return path is three-times that in the phase Do not pass from secondary to primary of delta (primary) wye-grounded (secondary) transformer Do pass from secondary to primary of wye-grounded wye-grounded transformer (flow in neutral conductor of primary system) Course 1 Power Distribution Systems and Power Circuit Analysis 01 Siemens Energy Inc., Siemens Industry Inc., 18-16

80 Harmonics in Balanced Three-Phase Four- Wire Multi-Grounded Neutral Systems ϕa I A 1.0 sin( t) 0.1sin 3t 100 % Fundamental, 10 % Third Harmonic ϕb ϕc I B I C 1.0 sin( t ) 0.1 sin3( t ) sin( t ) 0.1 sin3( t ) 3 3 I RESIDUAL 0.3 sin 3t RESIDUAL IS 3 RD HARMONIC Course 1 Power Distribution Systems and Power Circuit Analysis 01 Siemens Energy Inc., Siemens Industry Inc., 18-17

81 Harmonics in Balanced Three-Phase Four- Wire Multi-Grounded Neutral Systems ZERO-SEQUENCE HARMONIC TRANFER THROUGH DISTRIBUTION TRANSFORMERS 0 I 3RD N I 3RD NO ZERO-SEQUENCE HARMONICS IN PRIMARY PHASE CONDUCTORS N 1 OR IN PRIMARY NEUTRAL CONDUCTOR. LOADS ZERO-SEQUENCE HARMONICS CIRCULATE IN DELTA WINDING. ZERO-SEQUENCE HARMONICS IN PRIMARY PHASE CONDUCTORS AND IN PRIMARY NEUTRAL CONDUCTOR (3 TIMES THAT IN PRIMARY PHASE CONDUCTORS) PTI 010, D#1, Harmonic Transfer Thru Dist Xfrs.FCW 3 I 3RD n I 3RD n I 3RD n I 3RD n n I 3RD 3 n 3 I 3RD 1 3 I 3RD I 3RD I 3RD 3 I 3RD I 3RD I 3RD I 3RD 3 I 3RD LOAD SIDE LOAD SIDE Course 1 Power Distribution Systems and Power Circuit Analysis 01 Siemens Energy Inc., Siemens Industry Inc., 18-18

82 Harmonics Produced Problems Lighting Systems High intensity discharge lamps outside voltage threshold can shut down Transformers and Capacitor Banks Resonances Transformer derating required with high harmonics (K Factor) kvar rating of capacitor bank exceeded due to harmonic current Other problems Nuisance fuse operations Device missoperations Telephone interference (rural systems-old days) Equipment heating Eventual Component failure Course 1 Power Distribution Systems and Power Circuit Analysis 01 Siemens Energy Inc., Siemens Industry Inc., 18-19

83 IEEE 519 IEEE established a Standard for harmonic measurement and control IEEE Std. 519 It specifically set Requirements for Utilities voltage Requirements for Customers current Course 1 Power Distribution Systems and Power Circuit Analysis 01 Siemens Energy Inc., Siemens Industry Inc., 18-0

84 IEEE Std 519 Voltage Requirements For periods less than one hour, limits may be increased by 50%. Bus Voltage at PCC Individual Distortion (%) Total Distortion THD (%) Course 1 Power Distribution Systems and Power Circuit Analysis 01 Siemens Energy Inc., <69kV kV kv >138kV Siemens Industry Inc., 18-1

85 IEEE Std 519 Current Requirements AS THE STIFFNESS OF THE UTILITY SUPPLY SYSTEM INCREASES, THE LOAD IS ALLOWED TO GENERATE MORE HARMONIC CURRENTS. Course 1 Power Distribution Systems and Power Circuit Analysis 01 Siemens Energy Inc., Siemens Industry Inc., 18-

86 Tab Voltage Sags, Voltage Swells, Momentary Interruptions Distribution Systems Engineering Course 1 01, Siemens Industry Inc., all rights reserved

87 Voltage Swell This can occur, for instance, on the healthy (unfaulted) phases of a multi-grounded neutral distribution system during a single-line-to-ground fault Duration of Swell is typically between ½ cycle to 1 second LINE-TO-GROUND VOLTAGE 1.0 PER UNIT CE DE 010, Voltage Wave Sag.FCW 1.5 PER UNIT 1.0 PER UNIT Course 1 Power Distribution Systems and Power Circuit Analysis 01 Siemens Industry Inc., Siemens Industry Inc., 19-

88 Voltage Swell Unfaulted Phase-to- Ground Voltages For Single Phase-to-Ground Fault on Phase A UNFAULTED Φ-TO-GROUND VOLTAGE AT FAULT POINT Z 1 = POSITIVE-SEQUENCE IMPEDANCE AT FAULT POINT Z 0 = ZERO-SEQUENCE IMPEDANCE AT FAULT POINT Θ1 = POSITIVE-SEQUENCE IMPEDANCE ANGLE AT FAULT POINT Θ0 = ZERO-SEQUENCE IMPEDANCE ANGLE AT FAULT POINT Z1 Z0 V B a Z1 Z0 Z1 Z0 V C a Z1 Z0 a e a e j10 o j10 per per unit unit 1 j o j Course 1 Power Distribution Systems and Power Circuit Analysis 01 Siemens Industry Inc., Siemens Industry Inc., 19-3

89 Momentary Interruption Typically Caused By a Circuit Recloser or Circuit Breaker Opening and Successfully Reclosing Due to a Temporary (Transient) Fault LINE-TO-NEUTRAL VOLTAGE Course 1 Power Distribution Systems and Power Circuit Analysis 01 Siemens Industry Inc., Time in Cycles for Loss of voltage Siemens Industry Inc., 19-4

90 Momentary Interruption (continued) PERCENTAGE OF DEVICES THAT WERE ABLE TO SUCCESSFULLY RIDE THROUGH A MOMENTARY INTERRUPTION OF THE GIVEN DURATION DEVICE MOMENTARY INTERRUPTION DURATION (SECONDS / CYCLES) 0.5 / 30.0 / / 1000 Digital Clocks 70 % 60 % 0 Microwave Ovens 60 % 0 0 VCR 50 % 37.5 % 0 Computer FROM: POWER QUALITY ASSURANCE MAGAZINE, PP , 1990 Course 1 Power Distribution Systems and Power Circuit Analysis 01 Siemens Industry Inc., Siemens Industry Inc., 19-5

91 Momentary Interruptions (continued) Equipment is sensitive and can not ride through resulting in the machine shutting down System solutions faster reclosing fewer faults (tree trimming, animal & bird guards, better lightning protection) elimination of fuse saving with feeder breaker instantaneous tripping in substation installation of circuit reclosers in main feeder and on major branches closed bus tie breakers in substation (prevents momentary interruptions from loss of subtransmission sources) Customer solutions static transfer switch, UPS employ spot network systems Siemens Industry Inc., 19-6 Course 1 Power Distribution Systems and Power Circuit Analysis 01 Siemens Industry Inc.,

92 Momentary Interruptions (continued) SYSTEM MEASURES TO REDUCE NUMBER OF MOMENTARY INTERRUPTIONS: 1. ELIMINATE INSTANTANEOUS TRIPPING OF FEEDER BREAKER 5-1 IN ATTEMPTING TO PREVENT FUSE BLOWING FOR TEMPORARY FAULTS ON LATERALS. THIS MAY RESULT IN MORE PERMANENT OUTAGES FOR CUSTOMERS SERVED FROM FUSED LATERALS.. INSTALL RECLOSER IN MAIN FEEDER SO THAT TEMPORARY AND PERMANENT FAULTS ON ND HALF OF FEEDER DO NOT CAUSE A MOMENTARY INTERRUPTION TO CUSTOMERS ON FIRST HALF OF FEEDER. 3. USE SINGLE-PHASE RATHER THAN THREE-PHASE RECLOSERS IN 3-PHASE LINES SUBSTATION BUS 5-1 FUSE FEEDER FEEDER 1 LATERAL RECLOSER Disk: CE Dist Course 006, #, Momentary Reduction.FCW Siemens Industry Inc., 19-7 Course 1 Power Distribution Systems and Power Circuit Analysis 01 Siemens Industry Inc.,

93 Voltage Sag An RMS reduction in the nominal AC voltage, at the power frequency, to between 10% and 90% of nominal, for durations from 1/ cycle to 1 minute. IEEE 150: Guide on Service to Equipment Sensitive to Momentary Voltage Disturbances In three-phase systems, voltage sags may be same in each phase or different, depending on fault type or disturbance In the multi-grounded neutral system, the single line-to-ground (neutral) fault can cause voltage sags in the faulted phase, and voltage swells in the unfaulted phases Course 1 Power Distribution Systems and Power Circuit Analysis 01 Siemens Industry Inc., Siemens Industry Inc., 19-8

94 Examples of Voltage Sag Course 1 Power Distribution Systems and Power Circuit Analysis 01 Siemens Industry Inc., Even when there isn t a complete loss of voltage, customer equipment (industrial motors, ASD) are still impacted Siemens Industry Inc., 19-9

95 Causes of Voltage Sags In Distribution Systems Faults in sub transmission system Faults on adjacent distribution circuits supplied from same bus section in the distribution substation Starting of large motors Transfer of large loads EXAMPLES: 84% OF SAG EVENTS RESULT IN VOLTAGE LESS THAN 8.5% 40% OF SAG EVENTS RESULT IN VOLTAGE LESS 70% 16% OF SAG EVENTS RESULT IN VOLTAGE LESS THAN 40% Course 1 Power Distribution Systems and Power Circuit Analysis 01 Siemens Industry Inc., Siemens Industry Inc., 19-10

96 Voltage Sag on Substation Bus Single Line-to-Ground Fault on Adjacent Circuit UNFAULTED FEEDERS Z 0SUB Z 1SUB FAULTED FEEDER (SLG PHASE A) Z 0FDR V ASUB Z 1FDR Disk: CE Dist Course 006, #1 SUB SAG SLG FLT.FCW VOLTAGE ON FAULTED PHASE AT SUBSTATION BUS, V ASUB Z 1SUB = POSITIVE-SEQUENCE IMPEDANCE LOOKING INTO SUBSTATION BUS Z 0SUB Z 1FDR Z 0FDR V ASUB Z 1SUB Z Z 1FDR 0SUB Z0 Z 1FDR = ZERO-SEQUENCE IMPEDANCE LOOKING INTO SUBSTATION BUS FDR = POSITIVE-SEQUENCE IMPEDANCE OF FEEDER BETWEEN SUBSTATION & FAULT = ZERO-SEQUENCE IMPEDANCE OF FEEDER BETWEEN SUBSTATION AND FAULT Z 0FDR per unit Course 1 Power Distribution Systems and Power Circuit Analysis 01 Siemens Industry Inc., Siemens Industry Inc., 19-11

97 Voltage Sag on Substation Bus Single Line-to-Ground Fault on Adjacent Circuit (continued) EXAMPLE CALCULATION: 13. KV SYSTEM, E LN = 760 VOLTS OH LINE, SLG FAULT ON ΦA 1.5 MILES FROM SUBSTATION SYSTEM DATA: AVAILABLE 3-Φ FAULT CURRENT ON SUB BUS = I 3P = 15,000 AMP, X/R = 0 AVAILABLE SLG FAULT CURRENT ON SUB BUS = I SLG = 15,750 AMP, X/R = 0 Z 1FDR = ( j Ω/MILE)*1.5 MILE = j Ω Z 0FDR =( j Ω/MILE)*1.5 MILE = j.3763 Ω Course 1 Power Distribution Systems and Power Circuit Analysis 01 Siemens Industry Inc., Siemens Industry Inc., 19-1

98 Voltage Sag on Substation Bus Single Line-to-Ground Fault on Adjacent Circuit (continued) POSITIVE-SEQUENCE IMPEDANCE AT SUBSTATION BUS: ELN Z1SUB cos( a tan( X / R)) j sin( a tan( X / R)) I3P cos( a tan(0)) j sin( a tan(0)) j ZERO-SEQUENCE IMPEDANCE AT SUBSTATION BUS: V ASUB WHEN THE X TO R RATIO OF THE ZERO-SEQUENCE AND POSITIVE-SEQUENCE IMPEDANCE AT THE SUBSTATION BUS ARE EQUAL, THE FOLLOWING EQUATION IS USED TO FIND THE ZERO- SEQUENCE IMPEDANCE 3 3 Z 0SUB Z1SUB ( j0.5074) ISLG I P j VOLTAGE ON SUBSTATION BUS PHASE SUPPLYING THE FAULTED PHASE Z 1SUB Z Z 1FDR 0SUB Z Z 0FDR 1FDR Z 0FDR per V (0.475 j 0.788) ( j.3763) ASUB ( j ) ( j (0.475 j 0.788) ( j unit j V ASUB j per unit per unit Course 1 Power Distribution Systems and Power Circuit Analysis 01 Siemens Industry Inc., Siemens Industry Inc., 19-13

99 Voltage Sag Due to Motor Start Course 1 Power Distribution Systems and Power Circuit Analysis 01 Siemens Industry Inc., SAG FROM STARTING OF LARGE MOTORS CAN BE REDUCED WITH: 1.REACTOR STARTING.AUTO TRANSFORMER STARTING 3.WYE-DELTA MOTOR STARTING Siemens Industry Inc., 19-14

100 Voltage Sag Due to Motor Start Reduction With Wye Delta Motor Start LA LB LC TYPICAL WYE-DELTA MOTOR STARTER SEQUENCE Note: M & 1S mechanically interlocked so that M can not be closed until 1S is open 1M 1M 1M M M M 1S 1S 1S S S S 6 LEAD MOTOR T1 MOT 1 T4 Note: Motor starting current and starting torque with windings in WYE are approximately 1/3 of the values when windings are connected in DELTA across the line *** M CLOSES BEFORE S OPENS T T3 1. All Contactors Open. Motor Windings in WYE 1S CLOSED 1M CLOSED S OPEN M OPEN 3. Motor Windings in WYE, Resistors in Wye connected across motor windings 1S CLOSED 1M CLOSED S CLOSED ** M OPEN 4. Motor Windings in DELTA with resistors in series with each leg of delta 1S OPEN ** 1M CLOSED S CLOSED M OPEN 5. Motor Windings in DELTA across lines LA, LB, LC 1S OPEN 1M CLOSED S OPEN *** M CLOSED *** Siemens Industry Inc., MOT MOT 3 T5 T6 D#38, STARTER - TYPICAL Y DELTA.FCW Course 1 Power Distribution Systems and Power Circuit Analysis 01 Siemens Industry Inc.,

101 Voltage Sag Due to Motor Start Reduction With Wye Delta Motor Start 1. Motor energized with windings connected in wye at 5 seconds. Starting current and starting torque are one third of those if windings connected in delta (across the line). Motor windings reconnected from wye to delta (closed transition) at about 11.8 seconds. 3. Resistors in series with delta windings shorted out 9 cycles after windings connected in delta (windings now directly across line) 4. Motor source interrupted at about 1.4 seconds RMS CURRENT AND VOLTAGE Course 1 Power Distribution Systems and Power Circuit Analysis 01 Siemens Industry Inc., Siemens Industry Inc., 19-16

102 Voltage Sag Due to Motor Start (continued) Starting the motor inside the customer s facilities usually only impacts the customer However, If the service transformer serves more than one customer All see it Large motors can blink the primary Course 1 Power Distribution Systems and Power Circuit Analysis 01 Siemens Industry Inc., Siemens Industry Inc., 19-17

103 An Interesting Question Which is more likely to occur A complete loss of voltage (permanent outage)? Or a sag in the service voltage below equipment specifications? Course 1 Power Distribution Systems and Power Circuit Analysis 01 Siemens Industry Inc., Siemens Industry Inc., 19-18

104 How Long Do Sags Last? Duration of sag is function or time-current characteristic of clearing device, and fault current Typical Clearing Times (cycles) Expulsion Fuse Current Limiting Fuse Electronic Recloser 3 30 Oil Circuit Breaker 5 60 SF6 or Vacuum Breaker 3 60 Sometimes, there may be other sags when the system is returned to normal configuration after switching Customer experienced problems with the first sag, got every thing up and working again and then comes another sag Course 1 Power Distribution Systems and Power Circuit Analysis 01 Siemens Industry Inc., Siemens Industry Inc., 19-19

105 Estimating Number of Sags in Distribution Circuits Z LINE ( / MILE) Z XFRMR ( ) Disk: CE Dist Course 006, #1 Sag Frequency Estimator.FCW P f Z Z XFMR LINE V 1V P = NUMBER OF VOLTAGE SAGS PER YEAR WHICH DEPRESS THE SUBSTATION BUS VOLTAGE BELOW MAGNITUDE V SAG (V SAG IS IN PER UNIT) η f ρ SAG SAG = NUMBER OF FEEDERS SERVED BY THE SUBSTATION BUS = FEEDER FAULT RATE IN FAULTS PER MILE, INCLUDING FAULTS ON LATERALS FED FROM MAIN FEEDER Z XFRMR = SUBSTATION TRANSFORMER IMPEDANCE IN OHMS Z LINE = FEEDER IMPEDANCE IN OHMS PER MILE Course 1 Power Distribution Systems and Power Circuit Analysis 01 Siemens Industry Inc., Siemens Industry Inc., 19-0

106 Estimating Sags in Distribution Systems Example: Consider a distribution substation bus that serves 5 similar feeders. Consider also that: each feeder has a single-line-to-ground fault rate (including faults on laterals) equal to 1 fault/(mile year) and an impedance per phase of 1.1 ohms/mile for a single line-to-ground fault; the substation transformer which supplies the mentioned substation bus has an impedance per phase equal to 0.9 ohms; How many times per year it is expected that single-line-to-ground faults on the feeders will make the substation bus voltage sag to a value below 0.8 pu of its nominal value? Solution: Z P f Z XFMR LINE V 1V SAG SAG P = 5 1 (0.9/1.1) [0.8/(1-0.8)] 16 Sags/Year Course 1 Power Distribution Systems and Power Circuit Analysis 01 Siemens Industry Inc., Siemens Industry Inc., 19-1

107 System Changes for Reducing Voltage Sags Current limiting fuses in distribution feeders & with distribution transformers & capacitors Line (feeder) reactors at substations (Less voltage sag on bus for feeder faults) Neutral Grounding Reactors in transmission / sub-transmission system when substation transformers are connected in delta on HV side and wye-grounded on secondary side (normal configuration) Neutral Grounding Reactors in distribution system only if all distribution transformers on feeders are connected from phase-tophase (effect for single line-to-ground faults) Instantaneous breaker tripping Faster relaying on transmission circuits Connect sensitive customers to a more reliable voltage class Pulse closing technology for reclosers in overhead lines Course 1 Power Distribution Systems and Power Circuit Analysis 01 Siemens Industry Inc., Siemens Industry Inc., 19-

108 Pulse Closing with IntelliRupter Reclosers PULSE RECLOSING 1.CLOSING ANGLE CONTROLLED SUCH THAT MINOR LOOP OF CURRENT FLOWS.SOFTWARE DISTINGUISHES BETWEEN LOAD CURRENTS AND FAULT CURRENTS 3.LIMITS ARC ENERGY, VOLTAGE SAGS, THERMAL STRESSES ON CONDUCTORS 4.LIMITS MECHANICAL AND THERMAL STRESS ON SUBSTATION TRANSFORMERS FAST TRIP SLOW TRIP SLOW TRIP PERMANENT FAULT: CONVENTIONAL RECLOSER: FAST & SLOW CURVE PERMANENT FAULT: PULSE RECLOSER FAST & PULSE Course 1 Power Distribution Systems and Power Circuit Analysis 01 Siemens Industry Inc., Siemens Industry Inc., 19-3

109 System Changes for Reducing Voltage Sags (continued) NEUTRAL GROUNDING REACTOR IN SUBSTATION PHASE REACTORS ON OVERHEAD FEEDERS Course 1 Power Distribution Systems and Power Circuit Analysis 01 Siemens Industry Inc., Siemens Industry Inc., 19-4

110 System Changes for Reducing Voltage Sags (continued) CURRENT LIMITING FUSES ON LATERAL TAP FROM MAIN LINE CURRENT-LIMITING FUSES FOR CAPACITOR BANK Course 1 Power Distribution Systems and Power Circuit Analysis 01 Siemens Industry Inc., Siemens Industry Inc., 19-.5

111 Customer Changes for Reducing Voltage Sags Voltage sag problems are often best solved on the customer side! UPS, static transfer switch, ferroresonant transformer In-plant voltage regulation Load ride-through enhancements Reduced voltage motor starting Course 1 Power Distribution Systems and Power Circuit Analysis 01 Siemens Industry Inc., Siemens Industry Inc., 19-6

112 CBEMA Design Curve CBEMA stands for Computer and Business Equipment Manufacturers Association This group does not develop proprietary standards Course 1 Power Distribution Systems and Power Circuit Analysis 01 Siemens Industry Inc., ANSI/IEEE std Siemens Industry Inc., 19-7

113 New CBEMA Curve (ITIC Curve) Information Technology Industry Council Course 1 Power Distribution Systems and Power Circuit Analysis 01 Siemens Industry Inc., Siemens Industry Inc., 19-8

114 Device Tolerance Curves Course 1 Power Distribution Systems and Power Circuit Analysis 01 Siemens Industry Inc., Siemens Industry Inc., 19-9

115 Device Tolerance Curves (continued) Course 1 Power Distribution Systems and Power Circuit Analysis 01 Siemens Industry Inc., Siemens Industry Inc., 19-30

116 Digital Clock Course 1 Power Distribution Systems and Power Circuit Analysis 01 Siemens Industry Inc., Siemens Industry Inc., 19-31

117 Microwave Course 1 Power Distribution Systems and Power Circuit Analysis 01 Siemens Industry Inc., Siemens Industry Inc., 19-3

118 VCR Course 1 Power Distribution Systems and Power Circuit Analysis 01 Siemens Industry Inc., Siemens Industry Inc., 19-33

119 Tab 3 - Dealing with PQ Problems (from the utility side) Distribution Systems Engineering Course 1 01, Siemens Industry Inc., all rights reserved

120 Reducing Faults on the Utility Side Install quality equipment Load equipment conservatively (do not overload excessively) Improving lightning protection Reducing outages due to shielding failure shield wires and shielding angles (T&D) Increase overhead distribution line BIL (induced flashovers) Reducing outages due to backflashovers improve grounding (T&D) increase insulation (T&D) add lightning arresters (T&D) Improving maintenance more animal and bird protection tree trimming (or tree wire- all three phases or just top phase) equipment inspection Course 1 Power Distribution Systems and Power Circuit Analysis 01 Siemens Industry Inc., Siemens Industry Inc., 0-

121 Reducing the Impact of Faults on the Industrial/Commercial Customer Change power system to a more reliable configuration primary selective secondary selective secondary spot network Modify protective relaying to better suit the industrial customer fewer reclosing attempts on adjacent feeders eliminate instantaneous tripping for fuse saving Use substation and transmission configurations that promote PQ Course 1 Power Distribution Systems and Power Circuit Analysis 01 Siemens Industry Inc., Siemens Industry Inc., 0-3

122 Primary Selective System Using Mechanical Transfer Switches Course 1 Power Distribution Systems and Power Circuit Analysis 01 Siemens Industry Inc., Siemens Industry Inc., 0-4

123 Primary Selective System Using Medium Voltage-Class Static Switch Course 1 Power Distribution Systems and Power Circuit Analysis 01 Siemens Industry Inc., Siemens Industry Inc., 0-5

124 Secondary Selective System With Static Transfer Switch Course 1 Power Distribution Systems and Power Circuit Analysis 01 Siemens Industry Inc., Siemens Industry Inc., 0-6

125 Using Dispersed Generation as a Means of PQ Enhancement Course 1 Power Distribution Systems and Power Circuit Analysis 01 Siemens Industry Inc., Siemens Industry Inc., 0-7

126 Using Custom Power Devices Static transfer switches Static series voltage regulator Static var compensator SMES Static breaker Current limiter Course 1 Power Distribution Systems and Power Circuit Analysis 01 Siemens Industry Inc., Siemens Industry Inc., 0-8

127 Dealing with PQ Problems (from the customer side) 01, Siemens Industry Inc., all rights reserved

128 Customer PQ Improvement Power conditioners Isolation transformer Ferroresonant transformer Voltage regulator UPS (uninterruptible power supply) Motor-Generator set Surge suppressors Filters Static VAR compensators Course 1 Power Distribution Systems and Power Circuit Analysis 01 Siemens Industry Inc., Siemens Industry Inc., 0-10

129 Fast Response Voltage Regulator Course 1 Power Distribution Systems and Power Circuit Analysis 01 Siemens Industry Inc., Siemens Industry Inc., 0-11

130 Isolation Transformer Cannot remove common-mode signals or disturbances Shielding can form an LC circuit which is resonant at some high frequency Can saturate during transients Saturation currents can provoke nuisance fuse blowings What s the role of the leakage inductance? Course 1 Power Distribution Systems and Power Circuit Analysis 01 Siemens Industry Inc., Siemens Industry Inc., 0-1

131 Ferroresonant Transformer Some ability to ride through voltage sags (1:1 transformer excited high on saturation curve) Limits the short circuit current Start-up of a motor and overloads can cause the magnetic field to collapse Heavy,noisy and may cause output transients Course 1 Power Distribution Systems and Power Circuit Analysis 01 Siemens Industry Inc., Siemens Industry Inc., 0-13

132 UPS Load Load Course 1 Power Distribution Systems and Power Circuit Analysis 01 Siemens Industry Inc., Load Siemens Industry Inc., 0-14

133 Motor Generator Sets Course 1 Power Distribution Systems and Power Circuit Analysis 01 Siemens Industry Inc., Siemens Industry Inc., 0-15

134 Customer Installed UPS and Backup Generator at Factory Course 1 Power Distribution Systems and Power Circuit Analysis 01 Siemens Industry Inc., Siemens Industry Inc., 0-16

135 Customer Solutions Voltage sags that originate from utility side: UPS motor generator set Voltage sags that originate on the customer side: Course 1 Power Distribution Systems and Power Circuit Analysis 01 Siemens Industry Inc., upgrade wiring reduced voltage motor starting UPS Siemens Industry Inc., 0-17

136 Customer Solutions (continued) Incoming utility harmonics filters isolation transformers UPS motor-generator Internally generated harmonics purchase devices that generate less distortion (1 pulse vs 6 pulse rectification) increase stiffness of electric system apply above utility solutions to specific devices that are sensitive Lightning/switching surges from utility side: arresters at main transformer primary and secondary terminals arresters at service panel and at critical loads filters at service or PCC Course 1 Power Distribution Systems and Power Circuit Analysis 01 Siemens Industry Inc., Siemens Industry Inc., 0-18

137 PQ Monitoring Where to monitor Number of channels Sample rates Data Storage Communications Data analysis Course 1 Power Distribution Systems and Power Circuit Analysis 01 Siemens Industry Inc., Siemens Industry Inc., 0-19

138 Where to Monitor? Course 1 Power Distribution Systems and Power Circuit Analysis 01 Siemens Industry Inc., Siemens Industry Inc., 0-0

139 Answers for energy. Tab 4: Application of Shunt Capacitors on Distribution Systems Distribution System Engineering Course 1 01 Siemens Industry Inc., all rights reserved.

140 What is a Capacitor? IEEE Definition of Capacitor: A device, the primary purpose of which is to introduce capacitance into an electric circuit. IEEE Definition of Capacitance: The property of a system of conductors and dielectrics that permits the storage of electrically separated charges when potential differences exist between the conductors. Q C V C = Capacitance in Farads Q = Charge in Coulombs V = Potential Difference Between the Two Conductors of the System in Volts Distribution System Engineering Course for Con Edison Page 1-01 Siemens Industry, Inc. All rights reserved.

141 Shunt Capacitor Construction Distribution Circuit Application: Single bushing capacitor units (cans) can be used in the three-phase four-wire multi-grounded neutral (MGN) system, connected from phase to neutral. Two busing units can be used in any type distribution system. They can be connected: 1. Phase to neutral in the MGN system. Phase-to-phase (delta) in any system 3. Floating wye (ungrounded) in any system Internal Discharge Resistor: Must reduce the capacitor voltage to 50 volts or less in 5 minutes or less when the capacitor is disconnected from rated voltage Distribution System Engineering Course for Con Edison Page Siemens Industry, Inc. All rights reserved.

142 Common Shunt Capacitor Configurations In Distribution Systems Metal Enclosed Pad- Mounted With Switches & Overcurrent Protection Substation Open (Exposed) Rack-Type Capacitor Bank Pole Mounted With Switches and Overcurrent Protection Distribution System Engineering Course for Con Edison Page Siemens Industry, Inc. All rights reserved.

143 Example of a 3-Phase Fixed Capacitor Bank on an Overhead Distribution Feeder Surge Arresters Fuse Cutouts for Short-Circuit Protection and Connect / Disconnect Function Two Bushing Single-Phase Capacitor Units ( Cans ) \ Capacitor Bank Neutral Cutout: Allows Operation of Bank in Either Floating Wye or Grounded Wye Configuration Distribution System Engineering Course for Con Edison Page Siemens Industry, Inc. All rights reserved.

144 Example of a 3-Phase Switched Capacitor Bank on an Overhead Distribution Feeder Fused cutouts for short circuit protection Backup current-limiting fuse for energy limiting during high fault current conditions Oil switches for automatically connecting to and disconnecting bank from line Single-phase single-bushing capacitor units, two units per phase Distribution transformer for control power for oil switches Distribution System Engineering Course for Con Edison Page Siemens Industry, Inc. All rights reserved.

145 Example of a 3-Phase Fixed Capacitor Bank with a Three-Phase Capacitor Unit Advantages of three-phase unit: Lower installation cost (does not require rack for small banks) Better appearance Disadvantage of three-phase unit: Failure in one phase requires replacement of entire unit Surge arresters Fused cutouts with right-most cutout in open position Three-phase capacitor unit Distribution System Engineering Course for Con Edison Page Siemens Industry, Inc. All rights reserved.

146 Capacitor Bank in 13. kv Distribution Substation Three-Phase Vacuum Switch 1 Capacitor Units, Section 1, 3 Units per phase Capacitor Units, Section (3 Units per phase) Series Reactor To Limit High Frequency Inrush Current During Back-to-Back Switching Three-Phase Vacuum Switch Distribution System Engineering Course for Con Edison Page Siemens Industry, Inc. All rights reserved.

147 Sample Data Sheet for Single-Phase Capacitor Units Distribution System Engineering Course for Con Edison Page Siemens Industry, Inc. All rights reserved.

148 Capacitor Bank Transient Inrush Current Energizing Single Bank on Substation Bus BUS L L Vrms C t = 0 + i(t) Vrms - C t = 0 SUB CAP BANK.FCW Transient Inrush Current i(t); Closing At Voltage Peak Transient Inrush Current Frequency Page 1-10 i( t) V rms 1 sin t 1 1 L LC f radians f LC LC Hz C Example: 13.8 KV ΦΦ Station Bus, 0,000 Amps SLG Fault Current, I SLG = 0,000 3-Phase Grounded Wye Capacitor Bank, 400 KVAr, KVA C = What Is Frequency Of Inrush Current?. What Is The Peak Value Of The Inrush Current, Assuming No Damping? First, find the inductance L in the equivalent circuit X L L L f X L f L VRMS 13,800 / 3 Volts ISLG 0,000 Amps Henry H 01 Siemens Industry, Inc. All rights reserved Distribution System Engineering Course for Con Edison

149 Capacitor Bank Transient Inrush Current Energizing Single Bank on Substation Bus Capacitance of Capacitor Bank, Farads Per Phase X C 1 KV C KVA / /1000 C C X f X C C * 10 6 Farad Frequency of Inrush Current 1 1 f LC *10 * * 10 Peak Value of Inrush Current (Assuming No Damping) V rms 13,800 / 3 11,68 i p L 1057 * C * 10 SURGE IMPEDANCE Nominal Current of Capacitor Bank Hz Amps KVAC 400 INOMINAL Amps RMS 14 3 KV Peak Note: If the capacitance were increased by a factor of, the inrush frequency goes down by a factor of 1/ or by a factor of Amps Peak inrush current is 14.1 times the peak of the nominal current PEAK Distribution System Engineering Course for Con Edison Page Siemens Industry, Inc. All rights reserved.

150 Capacitor Bank Transient Inrush Current Energizing Second Bank on Substation Bus LS BUS L B Vrms C 1 C t = 0 i(t) C 1 Vrms C t = 0 = SOURCE INDUCTANCE L S SUB CAP BACK TO BACK.FCW L B = INDUCTANCE OF BUS BETWEEN CAP 1 & CAP Inrush Current i(t) Between C 1, L B, C When C Energized i( t) V L C B EQ rms 1 sin L B C EQ t C EQ C C 1 1 C C f Inrush Current Frequency 1 L B C EQ Example: 13.8 Kv ΦΦ Station Bus, 0 Feet Of Bus Between Capacitor Banks 3-phase Grounded Wye Capacitor Banks, Each Bank = 400 KVAr Assume 0.4 micro-henry Per Foot of Bus 1. What is frequency of inrush current i(t) Between C 1, L B, and C?. What is the peak value of the inrush current, assuming no damping? 6 6 L B 0.4 *10 Henry / Foot * 0 Feet 8 * 10, Henry f 1 L B C EQ Distribution System Engineering Course for Con Edison Page Siemens Industry, Inc. All rights reserved.

151 Capacitor Bank Transient Inrush Current Energizing Second Bank on Substation Bus Capacitance of Each 400 Kvar Capacitor Bank, & Equivalent Capacitance: C C EQ KVA 1000 * f C1 C C C 1 C KV 1 C * * FARADS, *10 FARADS Inductance of bus. L B, is much lower than inductance of source, L S = 1057 µh L 6 8 * 10 HENRY Frequency of Inrush Current Between Capacitor C 1 and Capacitor C f 1 L B C EQ 8 * * 10 6 B 13,761 Hz khz Peak Inrush Current to Capacitor C V rms 13,800 / 3 11,68 i p 16, 90 L B 8 * 10 C 6 EQ 16.7 * 10 Nominal Current of Each Capacitor Bank Note: There is also an inrush transient between the source inductance L S and the two capacitors in parallel when the second bank, C,is energized, similar to that which occurs when the first bank, C 1, is energized. Page 1-13 Amps Peak KVAC 400 INOMINAL Amps RMS 14 3 KV Amps Peak inrush current is times the peak of the nominal current Peak 01 Siemens Industry, Inc. All rights reserved Distribution System Engineering Course for Con Edison

152 Capacitor Bank Transient Inrush Current Two Banks on Substation Bus (From Digital Simulation) LS BUS t = 0 BUS VOLTAGE AND CURRENT IN C 1 Vrms C 1 C L S = SOURCE INDUCTANCE SUB CAP BACK TO BACK.FCW Both Capacitors Same Size C 1 =C 943 Hz (L S and C 1, period=1.06 msec) High frequency transient (16.4 khz) due to oscillation between C 1 and C thru bus inductance L B (period=.061 msec) 666 Hz transient due to inrush from source and two capacitors in parallel (L s and *C 1, period = 1.5 msec) Page 1-14 Note: If the capacitance of the capacitor is increased by a factor of, the frequency of the oscillation between L S and C is lowered by a factor of 1/ or Siemens Industry, Inc. All rights reserved Distribution System Engineering Course for Con Edison

153 Application and Operation of Capacitor Banks Distribution System Engineering Course for Con Edison Page Siemens Industry, Inc. All rights reserved.

154 Key Reasons for Applying Shunt Capacitors on Distribution Systems 1. Improves power factor. Reduce voltage drop on distribution feeders and in distribution substation transformers 3. Reduces I R and I X losses in transformers and lines 4. Aids in regulating voltage (switched capacitors only) 5. Reduce peak kvar demands on feeders and substation transformers Allows serving more load from feeder without installing new conductors (Releases Feeder Capacity) Releases capacity in distribution substation equipments and on transmission system May defer equipment upgrades/replacements Distribution System Engineering Course for Con Edison Page Siemens Industry, Inc. All rights reserved.

155 Two Ways to Operate Capacitor Banks Fixed Bank Capacitor bank is always on the system (permanently connected) supplying VARs as long as the capacitor bank fuses have not blown in any phase. The capacitor bank could be switched manually on a seasonal or other basis. Switched Bank Capacitor bank is placed in service only when the control logic determines the bank should be connected. Typically oil or vacuum switches are used to connect and disconnect the capacitor bank to distribution lines. Capacitor banks can be placed At the Transmission/Distribution substation On the utility s distribution system primary feeders (overhead or underground) On the customer s system Distribution System Engineering Course for Con Edison Page Siemens Industry, Inc. All rights reserved.

156 Control Signals for Switched Capacitor Banks 1. Voltage at bank location. Current at bank location 3. Power Factor or VAR flow at bank location 4. Time of day 5. Temperature 6. Radio Dispatch 7. SCADA / Smart Grid Logic Distribution System Engineering Course for Con Edison Page Siemens Industry, Inc. All rights reserved.

157 Control Signals for Switched Capacitor Banks Switched Capacitor Voltage Control Requires potential transformers Relatively inexpensive Difficult to apply on circuits with low voltage drop or little deviation in voltage with changing load (short urban circuits or circuits with large cables) OH line reactance (X 1 ) 0.6 Ω/mile Cable circuit reactance (X 1 ) 0. Ω/mile When capacitor vars are required for loss reduction or released capacity, may not switch on if voltage is within bandwidth Can chatter or switch in and out if circuit voltage fluctuates widely Good for emergency or contingency situations Distribution System Engineering Course for Con Edison Page Siemens Industry, Inc. All rights reserved.

158 Control Signals for Switched Capacitor Banks Switched Capacitor Current Control Simple capacitors on at heavy current load and off at light current load. Requires a current transformer (sensing devise). Relatively inexpensive. This also can switch in and out repeatedly. Can not distinguish between high and low power factor load currents. VAR Flow Control Very effective, but very expensive. Requires both potential transformer (device) and current transformer. Provides the best control for loss reduction and released capacity since it monitors the reactive volt-ampere flow (var flow) at the capacitor site Distribution System Engineering Course for Con Edison Page Siemens Industry, Inc. All rights reserved.

159 Control Signals for Switched Capacitor Banks Switched Capacitor Time Control Simple and inexpensive. Very common. Can get out of synch. Requires twice a year adjustment in the time clock. Effective for the first stage of switched VAR supply. Can be combined with radio, current or VAR controlled units to provide maximum coverage. Advantage does not chatter. Disadvantage does not respond directly to system var needs or system voltage requirements Distribution System Engineering Course for Con Edison Page Siemens Industry, Inc. All rights reserved.

160 Control Signals for Switched Capacitor Banks Switched Capacitor Radio Dispatch (SCADA) Control Controlled by the system operator More common with substation banks but is also used for distribution feeder applications Provides better control than time controlled Can respond to distribution substation and transmission system needs Can be used to optimize loss reduction More flexible to handle changes in system configuration Suitable for smart grid technologies Temperature Control Works well on systems where VARs fluctuate with the temperature fluctuations (High temperature climates with large air conditioning loads). Not recommended for systems with peaks in both summer and winter. Not directly responsive to system var or voltage needs Distribution System Engineering Course for Con Edison Page 1-01 Siemens Industry, Inc. All rights reserved.

161 Conventions for Load Current and Voltage Lagging PF Loads I = CURRENT I* = COMPLEX CONJUGATE OF CURRENT O = CURRENT ANGLE RELATIVE TO VOLTAGE ANGLE ("-" FOR INDUCTIVE LOAD) 90 o LAGGING POWER FACTOR LOADS I * -180 o S = V I* P + jq = V I* P = Real Part of V I* (+ MEANS ABSORBS WATTS) Q = Imaginary Part of V I* (+ MEANS LOAD IS INDUCTIVE) D#3, 010, Load Two Quad VI Conventions-Lagging PF.FCW -90 o O CURRENT IN QUAD. IV LOAD WATTS ARE "+" LOAD VARS ARE "+" I V I V 0 o LOAD Distribution System Engineering Course for Con Edison Page Siemens Industry, Inc. All rights reserved.

162 Conventions for Load Current and Voltage Leading PF Loads I = CURRENT I* = COMPLEX CONJUGATE OF CURRENT O = CURRENT ANGLE RELATIVE TO VOLTAGE ANGLE ("+" FOR CAPACITIVE LOAD) 90 o CURRENT IN QUAD. I LOAD WATTS ARE "+" LOAD VARS ARE "-" I V I LOAD -180 o S = V I* P + jq = V I* P = Real Part of V I* (+ MEANS ABSORBS WATTS) Q = Imaginary Part of V I* (- MEANS LOAD IS CAPACITIVE) D#3, 010, Load Two Quad VI Conventions-Leading PF.FCW -90 o O V I * LEADING POWER FACTOR LOADS 0 o Distribution System Engineering Course for Con Edison Page Siemens Industry, Inc. All rights reserved.

163 Convention for Load Active and Reactive Powers INDUCTIVE LOAD CAPACITIVE LOAD V I V I INDUCTIVE REACTIVE POWER + LOAD VARS ARE "+" INDUCTIVE REACTIVE POWER + KVAR CAPACITIVE - KVA KW 006 D#1, Load Two Quad PQ Conventions.FCW + ACTIVE POWER KVAR CAPACITIVE - KW KVA + LOAD VARS ARE "-" ACTIVE POWER Distribution System Engineering Course for Con Edison Page Siemens Industry, Inc. All rights reserved.

164 Relationship Between kva, kw, kvar Of Load REACTIVE POWER KVA KW KVAR kva = Load Apparent Power kw = Load Active Power (Real Power) KVAR + O KVA 010 D#1, KVA-KW-KVAR Relationships KW ACTIVE POWER kvar = Load Reactive Power Inductive Load is + kvar Capacitive Load is - kvar θ = Load Power Factor Angle + Angle For Inductive Load - Angle For Capacitive Load Distribution System Engineering Course for Con Edison Page Siemens Industry, Inc. All rights reserved.

165 Power Factor and Reactive Factor Concepts Power Factor (pf) = kw/kva A pf of 1 means that all power is real (Watts) (Current and voltage are in-phase ) A pf of 0 means that all power is reactive (VArs) pf cos( ) (Current lags voltage by `90 electrical degrees) Reactive Factor (rf) = kvar/kva A rf of 1 means that all power is reactive (VArs) A rf of 0 means that all power is real (Watts) θ = Load power factor angle + for inductive loads - for capacitive loads Note: Power factor can be expressed in either per unit or in percent Percent Power Factor = 100 x Per Unit Power Factor rf sin( ) Distribution System Engineering Course for Con Edison Page Siemens Industry, Inc. All rights reserved.

166 Power Factor and Reactive Factor Concepts Power Diagram Drawn For A Lagging Power Factor Load (Inductive Load) REACTIVE POWER KVAR + Note: Only the real (active) power can perform work. Page 1-8 O KVA 010 D#1, KVA-KW-KVAR Relationships KW ACTIVE POWER KVAR KW KVA 1 KVA sin KVA cos KW pf rf KVAR θ = Load power factor angle + for inductive loads - for capacitive loads 01 Siemens Industry, Inc. All rights reserved Distribution System Engineering Course for Con Edison

167 Capacitor: Steady-State Current Leads Steady- State Voltage by 90 o CURRENT or VOLTAGE Above Waveforms are for Steady-State Conditions i C Page 1-9 i (t) c ( t) I cos( t) and v ( t) V sin( t) C PEAK i (t) c v (t) c v (t) c 006 PTI, UNIT, CAP SINE WAVES.FCW C PEAK t Basic Current-Voltage Relationship for Ideal Capacitor v C ( t) 1 C t 0 i C = Capacitance in Farads V C (0) = Capacitor voltage at t = 0 C ( t) dt 01 Siemens Industry, Inc. All rights reserved. V C (0) Distribution System Engineering Course for Con Edison

168 Why the Capacitor Current Leads Voltage by 90 o? Closed at t = 0 V B R i (t) c v (t) c CURRENT or VOLTAGE V B R v (t) c i (t) c V B BATTERY Page 1-30 i C CAPACITOR t t V ( B t) e RC R v RC C ( t) VB 1 e At the instant the switch is closed at t = 0, current rushes into the capacitor to start charging it. As capacitor charges, its voltage rises, which opposes battery voltage, resulting in an exponential reduction of charging current. Peak current occurs well before peak voltage (current leads voltage). 0 Time DC CHARGE OF CAP.FCW 01 Siemens Industry, Inc. All rights reserved Distribution System Engineering Course for Con Edison

169 Inductor: Steady-State Current Lags Steady-State Voltage by 90 CURRENT or VOLTAGE Above Waveforms are for Steady-State Conditions i L i (t) L v (t) L Basic Current-Voltage Relationship for Ideal Inductor v L ( t) L dil( t) dt L = Inductance in Henry ( t) I cos( t) I cos( t ) and v ( t) V sin( t) LPEAK i (t) L v (t) L LPEAK 006 PTI UNIT,INDUCTOR SINE WAVES.FCW t L PEAK Distribution System Engineering Course for Con Edison Page Siemens Industry, Inc. All rights reserved.

170 Circuit Total Current with Parallel L and C Inductor and Capacitor Currents Assumed to be Equal in Magnitude v(t) i total i (t) L i (t) c Total current, i total, is sum of inductor and capacitor currents. Since i L and i c are 180 degrees out of phase and equal in magnitude, total current is zero in steady state, providing inductor and capacitor are lossless. Page 1-3 CURRENT or VOLTAGE v(t) i (t) L i (t) c i total 006 PTI UNIT, INDUCTOR CAPACITOR SINE WAVES.FCW t 01 Siemens Industry, Inc. All rights reserved Distribution System Engineering Course for Con Edison

171 Instantaneous Power in a Pure Inductor Notes: Frequency of power wave is twice that of voltage wave. Inductor takes energy from source during ½ of power cycle, and then returns energy to source in next ½ of power cycle. Ins tan tan eous power p( t) e( t) i( t) P AVERAGE Average power drawn from source, P AVERAGE, is zero. T 0 p( t) T Energy to Inductor Between t = 0 and t = T dt T 0 e( t) i( t) T dt Distribution System Engineering Course for Con Edison Page Siemens Industry, Inc. All rights reserved.

172 Power in a Circuit with a Power Factor of 0.60 Notes: Frequency of power wave is twice that of voltage wave. Energy taken from system during + portion of power cycle is greater than the energy returned to the system during the - portion of power cycle. Ins tan tan Note: Power is the time rate at which energy is supplied to the load. Page 1-34 CURRENT LAGS VOLTAGE BY O (0.97 RADIANS) eous power p( t) e( t) i( t) P AVERAGE P V AVERAGE RMS I RMS T 0 cos Average power supplied by source is positive. p( t) T Energy to resistor inductor combination between t = 0 and t = T dt T 0 e( t) i( t) 01 Siemens Industry, Inc. All rights reserved. T where cos 1 dt PF pu Distribution System Engineering Course for Con Edison

173 Power in a Circuit with Unity PF Notes: Voltage and current are In phase. Circuit with unity PF (resistor) always takes energy from source and never returns energy to the source Ins tan tan Note: Power is the time rate at which energy is supplied to the load. Page 1-35 eous power p( t) e( t) i( t) P AVERAGE AVERAGE Instantaneous and average power drawn from source is always positive T P V 0 p( t) RMS T I dt Energy to Resistor Between t = 0 and t = T RMS e( t) i( t) 01 Siemens Industry, Inc. All rights reserved. T 0 T dt Distribution System Engineering Course for Con Edison

174 Power Factor Correction With Capacitors The diagram below shows a capacitor bank connected to an inductive load: What size capacitor must be added to correct the power factor to a new (higher) value? S S L INDUCTIVE LOAD Q C Q L P L V P L = load active power Q L = load reactive power S L = load apparent power θ L = Load power factor angle (+ for inductive load) θ L = cos -1 (PF L ) Page 1-36 PF L = Load power factor in per unit = P L /S L P L =S L cos θ L Q L =S L sin θ L S = Resultant apparent power (kva) in the circuit after the capacitor is placed in parallel with the load. PTI 006, D#1, Capacitor Correction-1.FCW 01 Siemens Industry, Inc. All rights reserved Distribution System Engineering Course for Con Edison

175 Power Factor Correction GIVENS: Original kva load on circuit: S L Circuit s original power factor: PF L Circuits corrected power factor: PF FIND: What size capacitor, Q C, is needed θ L = load original power factor angle = cos -1 (PF L ) Q L = reactive flow before capacitor applied θ = power factor angle with capacitor θ = cos -1 (PF) Q = reactive flow after capacitor applied P L = real power before and after capacitor Page 1-37 Then: Q P tan L 1 PF PF Capacitor Size to Achieve New Power Factor (PF) = Q C Q C = Q L Q = Q L - P L tan θ Q C P L S L 1PF S L PF L 1 PF PF 1 L SL PFL PF PF O O L PTI 011 D#1, POWER FACTOR CORRECTION.FCW S L Q C P L Q L 01 Siemens Industry, Inc. All rights reserved. S Q Distribution System Engineering Course for Con Edison

176 Power Factor Correction Example A load of 1000 kva is operating at 80% power factor. What size capacitor (Q C ) must be added to raise the power factor to 98.9%? Givens: S L = 1000 kva (Loads Apparent Power) PF L = 0.80 (Loads Uncorrected Power Factor In PU) PF = (New Or Corrected Power Factor With Capacitor In PU) Solution: Q Q C Page 1-38 C S L 1000 Q C PF 1 L SL PFL 10.8 kvar 1000 * 0.80 PF PF Adding a 450 kvar capacitor to the 1000 kva load with 80% power factor will raise the power factor up to 98.3%. 01 Siemens Industry, Inc. All rights reserved Distribution System Engineering Course for Con Edison

177 Capacitors Release Capacity SUBSTATION TRANSFORMER CONDITION 1: SWITCH S1 OPEN CONDITION : SWITCH S1 CLOSED FEEDER - KVAR S1 KVA 1 L O A D Note: RELEASED CAPACITY (ΔKVA) is the amount of load at the original power factor that can be added without increasing the current in the supply feeder/substation transformer. REACTIVE POWER KVAR 1 KVAR O1 O KVA 1 KVA KW 1 KW 010 D#1, SUB XFR-FDR RELEASED CAPACITY.FCW Page 1-39 CIRCLE OF CONSTANT KVA - KVAR LOAD OF CAPACITOR RELEASED CAPACITY AT ORIGINAL POWER FACTOR (PF 1) = KVA REAL POWER KVA 1 = Feeder /Transformer Demand Before Capacitor Applied KVA = Feeder / Transformer Demand After Capacitor Applied KVAR = Reactive KVA supplied by the shunt capacitor 01 Siemens Industry, Inc. All rights reserved Distribution System Engineering Course for Con Edison

178 Capacitors Release Capacity Expression for Released Capacity REACTIVE POWER KVAR 1 KVA 1 CIRCLE OF CONSTANT KVA - KVAR LOAD OF CAPACITOR RELEASED CAPACITY IN PER UNIT (ΔKVA PU ) OF ORIGINAL LOAD (KVA 1 ) Page 1-40 KVA PU Q KVAR Q CPU O O1 KW 1 KW KVA 010 D#1, SUB XFR-FDR RELEASED CAPACITY.FCW 1 PF Per Unit of KVA 1 1 QCPU PF1 1 QC ( capacitor size in kva) KVA1 ( load size in kva) RELEASED CAPACITY AT ORIGINAL POWER FACTOR (PF 1) = KVA REAL POWER PF Original Load KVA Power Factor CPU Siemens Industry, Inc. All rights reserved Distribution System Engineering Course for Con Edison

179 Released Capacity vs Capacitor Size Lumped Load Case Released capacity is the additional load that can be added at the original power factor (PF 1 ) due to capacitor application, without exceeding the original line loading (KVA 1 ). Released Capacity: KVA PU Q CPU 1 PF Per Unit of KVA 1 1 QCPU PF1 1 1 RELEASED CAPACITY IN PER UNIT OF KVA Page 1-41 PF1 = 0.85 PF1 = 0.80 QC ( capacitor size in kva) KVA1 ( load size in kva) Released capacity is maximum when: 1 PF PF Q CPU 1 Note: 0.10 Capacitor size that yields the PF1 = 0.90 maximum released capacity is greater than the original loads kvar 0.05 PF1 = 0.95 Maximum capacity that can be released is: 0 1 PF KVAMAX PU P U of KVA PF1 CAPACITOR SIZE IN PER UNIT OF KVA 1 Q CPU 1 Per Unit 01 Siemens Industry, Inc. All rights reserved Distribution System Engineering Course for Con Edison

180 Example: Basic Concepts Power Factor Correction 1. A 13.kV circuit is supplying 6898 kva at 91 percent power factor. Calculate the kw and kvar supplied by the circuit.. Apply 1800 capacitive kvar to the circuit. 3. Calculate the new PF, and kva 4. How many kvar are required for unity PF? kw and kvar prior to application of capacitor: kw kva cos( ) kva* PF PU 6898 * kvar Page 1-4 kva sin( ) kva 1 PF PU kw, kvar, and power factor after application of capacitor: kvar kva kw 1059 kvar 677 PF kw kva 859 kva PF kw kvar D 010 CAP, EXAMPLE.FCW Siemens Industry, Inc. All rights reserved Distribution System Engineering Course for Con Edison

181 Peak Load Power Factors of Typical Loads and Systems 1. Incandescent light, resistance heating = 1.0. Induction motor = Synchronous motor = leading 0.8 (overexcitation) 4. Adjustable speed drive system = Fluorescent light (depends on ballast) = Commercial office building = Residential load = varies widely 8. Factory = Average of 0.85 ( before compensation) 9. Substation = Sum of all the load supplied looking in from Transmission System Distribution System Engineering Course for Con Edison Page Siemens Industry, Inc. All rights reserved.

182 Placement of Capacitor Banks Distribution System Engineering Course for Con Edison Page Siemens Industry, Inc. All rights reserved.

183 Lumped Load Effect of Capacitor on Circuit Voltage Drop SUBSTATION V S FEEDER I R X V L I L I C LOAD PTI Course 006, Voltage Drop Basic-1.FCW I is the total current before correction = load current I L. Once the switch is closed, I C reduces line current I by supplying all or a portion of the load VARs (reactive part of load current). Due to the reduced total current, the voltage drop between the sending point and receiving point is reduced. This is why capacitors sometimes are viewed as voltage sources or as var sources Distribution System Engineering Course for Con Edison Page Siemens Industry, Inc. All rights reserved.

184 Lumped Load - Voltage Rise Due to Capacitor Application Voltage Phasors Without Capacitor at Load (line current I = load current I L ) V S O V L IR IX 0 o I = I L Approximate Expression for Difference in Magnitude of V L and Magnitude of V S V V S V L Note: In above equation angle θ is the angle by which the load voltage leads the load current. Angle θ is + for lagging PF loads. Without capacitor, θ = θ L I R cos X sin I R cos X sin and I = I L L PTI COURSE 006, Vectors VD No Cap.FCW L L Distribution System Engineering Course for Con Edison Page Siemens Industry, Inc. All rights reserved.

185 Lumped Load - Voltage Rise Due to Capacitor Application Voltage Phasors with Capacitor at Load (I = I L + I C ) V S V S ' I O V L IR IX 0 o I C O L Approximate Expression for Difference in Magnitude of V L and Magnitude of V S V V Notes: 1. In above equation, Angle θ is + for Lagging Power Factor Loads. Approximate voltage rise caused by capacitor is product of I C and X, V RISE = I C * X 3. Voltage phasor V S is the source voltage needed to maintain the same load voltage (V L ) as shown if a capacitor were not present Page 1-47 I L S V L I R cos X sin I R cos X sin I X L L PTI COURSE 006, Vectors VD With Cap.FCW L C Voltage Rise Due to Capacitor Voltage Drop Without Capacitor 01 Siemens Industry, Inc. All rights reserved Distribution System Engineering Course for Con Edison

186 Lumped Load - Voltage Rise Due to Capacitor Application Voltage Phasors with Capacitor at Load- Leading PF on Feeder I I C O O L V L V S IR IX V ' S 0 o EFFECTS OF ADDING CAPACITOR 1. ROTATES VOLTAGE DROP TRIANGLE COUNTER CLOCKWISE. CHANGES SIZE OF VOLTAGE DROP TRIANGLE Approximate Expression for Difference in Magnitude of V L and Magnitude of V S V Notes: 1. In above equation, angle θ is + for lagging power factor loads.. Approximate voltage rise caused by capacitor is I C * X 3. Voltage phasor V S is the source voltage needed to maintain the same load voltage (V L ) as shown if a capacitor were not present. Page 1-48 I L V S V L I R cos X sin I R cos X sin I X L PTI COURSE 006, Vectors VD Overcomp with Cap.FCW L L C Voltage Drop without Capacitor 01 Siemens Industry, Inc. All rights reserved Distribution System Engineering Course for Con Edison

187 Capacitor Bank Voltage Rise and Released Capacity Calculations, Lumped Load One Mile Long Line, 350 Ampere Load At 80% Pf: 1 MILE Ob 13.8 KV P - TO - P 100 KVAR LUMPED LOAD 350 AMP 80 % PF Oa 3" 3" Oc Page 1-49 CON EDISON 13 KV ARMLESS CONSTRUCTION PHASE WIRE = 477 MCM AL R O = Ohms / mile GMR O = Feet R 1 = Ohms / mile X 1 = Ohms / mile APPROXIMATE PHASE-TO- PHASE AND PHASE-TO- NEUTRAL SPACINGS NEUTRAL WIRE = 4/0 COPPER R n = 0.78 Ohms / mile GMR n = Feet 48" 58.5" 53.57" DISC: PTI Course 006, D#1, Con Ed OH Dist Line.FCW n d bn = 86.0" 01 Siemens Industry, Inc. All rights reserved Distribution System Engineering Course for Con Edison

188 Capacitor Bank Voltage Rise and Released Capacity Calculations, Lumped Load Voltage Drop on 1 Mile of Line Without Capacitor R cos X sin * * Volts V D I NOMINAL TO NEUTRAL VOLTAGE V D % * % V D 10 VOLT BASE Volts 100 Volts Load kva, kw, and kvar (Total on All Three Phases) KVA 1 3 * KV KW 1 KVAR 1 * I AMPS PF * KVA 1 1 PF * 13.8 * kva 0.80 * kw * KVA 0.60 * kvar Nominal Current of Capacitor Bank (100 Kvar 3-Φ, 400 kvar Per Φ) I KVA C CAP KV Amps Distribution System Engineering Course for Con Edison Page Siemens Industry, Inc. All rights reserved.

189 Capacitor Bank Voltage Rise and Released Capacity Calculations, Lumped Load Voltage Rise Along Feeder Due To Capacitor Bank At End (1 Mile) VRISE ICAP * X1 LINE 50. Amps * Volts V RISE V 9.7 Volts * 100 % 7967 Volts % RISE 10VOLT BASE Volts Released Capacity By Application Of 100 kvar Capacitor Bank (400 kvar/φ) KVA PU KVA PU Q CPU PF Per Unit of KVA 1 Per 1 1 QCPU PF Unit of KVA Per KVA 1 Unit of KVA Distribution System Engineering Course for Con Edison Page Siemens Industry, Inc. All rights reserved.

190 Capacitor Bank Voltage Rise and Released Capacity Calculations, Lumped Load Maximum Capacity That Can Be Released With Application Of Capacitor KVA KVA MAX PU MAX PU 1 PF PF 1 1 Per Unit Per Unit Per Unit of of of KVA KVA KVA Capacitor Size That Will Release The Maximum Capacity, Original PF =.80 PU Q CPU 1 PF PF 1 1 Per Unitof KVA KVA 0.75 per unit of KVA 1 Q C 0.75 * KVAr KVA = 1.0 PU KW = 0.8 PU KVAR = 0.6 PU QC = 0.75 PU KVA = 0.5 PU KVA KW PTI 006, D#1, MAXIMUM RELEASED CAPACITY VECTORS.FCW KVAR 674 KVAR KVA Distribution System Engineering Course for Con Edison Page Siemens Industry, Inc. All rights reserved.

191 Lumped Load - Comments on Voltage Rise due to Capacitor Application SUBSTATION V S FEEDER I R X V L I L I C LOAD PTI Course 006, Voltage Drop Basic-1.FCW Recommendation: For large lumped loads, locate the capacitor bank as close as possible to the demand for kvar. The total voltage rise at the capacitor is dependent upon the circuit impedance (reactance) back to the source. Will be higher with overhead open-wire lines than with cables due to higher reactance of OH lines. The resulting decrease in the reactive current flow causes the voltage rise. Approximate voltage rise due to capacitor = I C X where I C is the nominal current of the capacitor at rated voltage, and X is the line reactance. Page Siemens Industry, Inc. All rights reserved Distribution System Engineering Course for Con Edison

192 Uniformly Distributed Load Effect of Capacitor on VAR Flow (reactive current flow) SUBSTATION V S Uniformly Distributed Load KVAR FLOW KVAR 500 KVAR REACTIVE FLOW WITHOUT CAPACITOR LAGGING PF KVAR FLOW Page KVAR LAGGING PF LEADING PF REACTIVE FLOW WITH CAPACITOR 500 KVAR Capacitor has no impact on var flow downstream from its point of application. LAGGING PF Note: Plot of reactive current versus distance has the same shape. PTI COURSE 006, D#3, Var Flow W & W-WO Capacitor-1.FCW 01 Siemens Industry, Inc. All rights reserved Distribution System Engineering Course for Con Edison

193 Uniformly Distributed Load Effect of Capacitor Application on Feeder Voltage Profile GENERAL PROFILE CURVES OF FEEDER VOLTAGE: SUBSTATION UNIFORMLY DISTRIBUTED LOAD CAPACITOR CASE 1 UNIFORMLY DISTRIBUTED LOAD WITH NO CAPACITOR 1.00 PU VOLTS AT SUB CASE UNIFORMLY DISTRIBUTED LOAD WITH NO CAPACITOR 1.0 PU VOLTS AT SUB CASE 3 UNIFORMLY DISTRIBUTED LOAD WITH CAPACITOR AT 66.6% OF FEEDER LENGTH 1.0 PU VOLTS AT SUB CASE 4 NO LOAD AND NO CAPACITOR 1.0 PU VOLTS AT SUB CASE 5 NO LOAD WITH CAPACITOR AT 66.6 % OF FEEDER LENGTH 1.0 PU VOLTS AT SUB Distribution System Engineering Course for Con Edison Page Siemens Industry, Inc. All rights reserved.

194 Capacitor Placement on Distribution Feeders Areas to Consider: 1. Voltage profile along the feeder and impact of capacitors on voltage profile.. Released capacity resulting from application of capacitor. 3. Impact on I R power losses in feeder at time of peak load with lumped load at end of feeder 4. Impact on I R power losses in feeder at time of peak load with load uniformly distributed along the feeder. a) Size of capacitor to minimize the power losses b) Location of capacitor to minimize the power losses 5. Impact on I Renergylosses in feeder with load uniformly distributed along the feeder a) Size of the capacitor to minimize the energy losses (maximize the reduction in energy losses) b) Location of capacitor to minimize the energy losses (maximize the reduction in energy losses) 6. Impact on loading of substation transformers Distribution System Engineering Course for Con Edison Page Siemens Industry, Inc. All rights reserved.

195 Loss Reduction-Lumped Load Distribution System Engineering Course for Con Edison At End of Feeder Page Siemens Industry, Inc. All rights reserved.

196 Lumped Load Case Power Loss Reduction Due to Capacitor Application SUBSTATION V S FEEDER I R X I C V L LOAD I L I = Line Current I L = Load Current PTI Course 006, Voltage Drop Basic-1.FCW Capacitor supplies portion or all of reactive component of load current, I L. This reduces the current in the line between the substation and load, thereby reducing the I R losses in the line (Note: I I L ) Page 1-58 I L I I C O L O V L IR V S IX PTI COURSE 006, Vectors VD With Cap.FCW V S ' 0 o 01 Siemens Industry, Inc. All rights reserved Distribution System Engineering Course for Con Edison

197 Lumped Load Case Power Loss Reduction Due to Capacitor Application V S V S ' I I C O L O V L IR IX 0 o To Minimize Line Losses: I CAP = I LOAD sin θ L I L PTI COURSE 006, Vectors VD With Cap.FCW ACTIVE CURRENT Capacitor current reduces the reactive current that flows in the feeder. Capacitor has minimal impact on the active current, and practically will not reduce losses in line from active current (real component of current). Page 1-59 REACTIVE CURRENT I LOAD I LINE I CAP O O L PTI COURSE 006, D#1, Current Phasors With Cap-Detail.FCW 01 Siemens Industry, Inc. All rights reserved Distribution System Engineering Course for Con Edison

198 Lumped Load Case Power Loss Reduction Due to Capacitor Application SUBSTATION V S FEEDER I R X V L I L I LINE ACTIVE CURRENT O I CAP PTI Course 006, Voltage Drop Basic-1.FCW I C LOAD REACTIVE CURRENT I LOAD O L Watts Loss In Feeder Without Capacitor: W I L Watts Loss In Feeder With Capacitor (Angle θ L Is Positive For Lagging PF Loads) W R WITH CAP I I I I R sin I I I I 1.0 PF R L C L C Reduction In Watts Loss From Capacitor Application (ΔW) W W W WITH CAP L L With Lumped Load, Reduction In Watts Loss (ΔW) Maximized When: I C = I L sin(θ L ) C I I sin I R I I 1.0 PF I R L C L C L C L C L L C PTI COURSE 006, D#1, Current Phasors With Cap-Detail.FCW Distribution System Engineering Course for Con Edison Page Siemens Industry, Inc. All rights reserved.

199 Lumped Load Case Power Loss Reduction Due to Capacitor Application REDUCTION IN I R LOSSES IN PU OF ORIGINAL Reduction in I R Losses in Line in Per Unit of Line Losses Without Capacitor (ΔW/W) PF L = 0.95 PF L = 0.90 SUBSTATION PF L = 0.80 PF L = 0.85 V S FEEDER PTI Course 006, Voltage Drop Basic-1.FCW CAPACITOR CURRENT IN PER UNIT OF LOAD CURRENT I L I R W W X PEAK I I PEAK IC IC 1 PF L IL IL W W I C W W V L occurs when : 1 PF C L L is : LOAD PU (CAPACITOR CURRENT = REACTIVE LOAD CURRENT) 1 PF L I L Distribution System Engineering Course for Con Edison Page Siemens Industry, Inc. All rights reserved.

200 Lumped Load Case, Power Loss Reduction Due to Capacitor Application, Example: 477 KCMIL Phase Conductor 1 MILE CON EDISON 13 KV ARMLESS CONSTRUCTION 13.8 KV P - TO - P 100 KVAR LUMPED LOAD 350 AMP 80 % PF PHASE WIRE = 477 MCM AL R O = Ohms / mile DISC: PTI Course, Unit 8, D#1, Con Ed OH Dist Line-Loss Calc.FCW Line Length = 1.0 Mile Capacitor Bank Nominal Current (100 kvar 3-Φ) I KVA C CAP KV Amps Watts Loss Per Phase In 1 Mile Line Without Capacitor W I R 350 * ,55 Watts / L Watts Loss Per Phase In 1 Mile Line When Capacitor Bank Is Applied W WITH CAP IL IC IL IC 1.0 PFL R *350 * * ,579 Watts / Distribution System Engineering Course for Con Edison Page Siemens Industry, Inc. All rights reserved.

201 Lumped Load Case, Power Loss Reduction Due to Capacitor Application, Example: 477 KCMIL Φ Conductor 1 MILE CON EDISON 13 KV ARMLESS CONSTRUCTION 13.8 KV P - TO - P 100 KVAR LUMPED LOAD 350 AMP 80 % PF PHASE WIRE = 477 MCM AL R O = Ohms / mile DISC: PTI Course, Unit 8, D#1, Con Ed OH Dist Line-Loss Calc.FCW Reduction In Watts Loss Per Mile Per Phase From Capacitor Application W IL IC 1.0 PFL IC R * 350 * ,675 Watts / Total Three-phase Reduction In Watts Losses Per Mile W PHASE 3 * W 3 * 3,675 Watts Annual Energy Loss Reduction In kwhr Assuming Constant Load KWHR = KW * 8760 HOURS = 96,63 KWHR / YEAR ANNUAL SAVINGS ASSUMING $0.05/KWHR = $4831 PER YEAR ASSUMING $6 / KVAC, 100 KVAR FIXED BANK COST IS $700 kw Distribution System Engineering Course for Con Edison Page Siemens Industry, Inc. All rights reserved.

202 Loss Reduction Uniformily Distributed Load Along Feeder Distribution System Engineering Course for Con Edison Page Siemens Industry, Inc. All rights reserved.

203 Uniformly Distributed Load Case Total Power (Watt) Loss In Feeder Source V S X dx FEEDER End L I Source I X = ( I Source 1 - X L) LOAD CURRENT MAGNITUDE r = Feeder Resistance Per Unit of Length, L = Feeder Length I SOURCE = Feeder Current at Source End dp(x) = Differential Power Loss in Section dx at Distance X Due to Load Current X X dp x I X r dx ISource r dx L L ( ) 1 Page 1-65 PTI 006 D#1, Uniformly Loaded Feeder-1.FCW 0 X Distance from source end L X 01 Siemens Industry, Inc. All rights reserved Distribution System Engineering Course for Con Edison

204 Uniformly Distributed Load Case Total Power (Watt) Loss In Feeder Total I R power loss on uniformly loaded line: P P P UNIFORM UNIFORM UNIFORM L dp( x) I SOURCE 0 0 L X ISOURCE r (1 L 0 ISOURCE r L L 3 (1 X L X ) L r ) dx dx I SOURCE r L 3 X = Distance From Sending End to Any Point on Feeder L = Feeder Length I source = Total Current at Source End Total I R power loss with load lumped at end of line: Total I R power losses with a uniformly loaded line are 1/3 of the losses with a lumped load at the end of the line. Current at source end, I SOURCE, is same in both cases. Page 1-66 P LUMPED L 0 L dp( x) I r dx I r dx 0 SOURCE SOURCE L 0 I SOURCE r L 01 Siemens Industry, Inc. All rights reserved Distribution System Engineering Course for Con Edison

205 Graphical Comparison of Total Power Losses Lumped Load and Uniformly Distributed Load Lumped load at the end of the feeder: dp(x) same at each point along feeder dp(x) dp( x) I SOURCE r dx INCREMENTAL LOSSES IN WATTS SHADED AREA REPRESENTS THE TOTAL LOSSES DUE TO THE TOTAL LOAD CURRENT 0 DISTANCE FROM SOURCE END L X Incremental Power Loss-Lumped Load.FCW Load uniformly distributed along the feeder, same current at source end dp(x) INCREMENTAL LOSSES IN WATTS dp( 0) I SOURCE SHADED AREA IS 1/3 OF AREA WITH THE LUMPED LOAD AT FEEDER END r dx X I r dx SOURCE L L dp ) 1 X X x I r dx SOURCE L L ( ) 1 0 DISTANCE FROM SOURCE END L X Distribution System Engineering Course for Con Edison Incremental Power Loss-Uniformly Dist Load.FCW Page Siemens Industry, Inc. All rights reserved.

206 Uniformly Loaded Feeder Power Loss dp(x) dp( 0) I SOURCE r dx dp X X x I SOURCE L L ( ) 1 r dx INCREMENTAL LOSSES IN WATTS SHADED AREA IS 1/3 OF AREA WITH THE LUMPED LOAD AT FEEDER END P Incremental Power Loss-Uniformly Dist Load.FCW UNIFORM Page I SOURCE 0 DISTANCE FROM SOURCE END L X POWER LOSS IN THE ENTIRE FEEDER r L POWER LOSS IN FIRST 50% OF FEEDER FROM SOURCE END P FIRST 50% OF FEEDER I SOURCE 6/7 OR 85.7% OF THE TOTAL POWER LOSS OCCURS IN THE FIRST HALF OF A UNIFORMLY LOADED FEEDER 6/7 OR 96.3% OF THE TOTAL POWER LOSS OCCURS IN THE FIRST TWO THIRDS OF THE UNIFORMLY LOADED FEEDER IN CONTRAST 50% OF THE POWER LOSS OCCURS IN THE FIRST HALF OF THE FEEDER WHEN THE LOAD IS LUMPED AT THE END OF THE FEEDER 7 r L 01 Siemens Industry, Inc. All rights reserved Distribution System Engineering Course for Con Edison

207 Uniformly Distributed Load On Feeder How to Minimize the Losses? Neagle and Samson Approach (1956) Given the capacitor size, the optimal capacitor location is the point on the circuit where the reactive power flow equals half of the capacitor VAR rating (capacitor rating is twice the reactive power). KVAR FLOW KVAR FLOW SUBSTATION V S Page KVAR 50 KVAR LAGGING PF LEADING PF Uniformly Distributed Load REACTIVE FLOW WITHOUT CAPACITOR REACTIVE FLOW WITH CAPACITOR 500 KVAR 500 KVAR LAGGING PF LAGGING PF PTI COURSE 006, D#3, Var Flow W & W-WO Capacitor-1.FCW Example: 500 kvar capacitor is applied at point in circuit where the reactive power is 50 kvar 01 Siemens Industry, Inc. All rights reserved Distribution System Engineering Course for Con Edison

208 Uniformly Distributed Load Case How to Minimize the Losses? Basis for the Neagle and Samson Approach Equation for Ratio of Power Loss Reduction from Application of Capacitor (ΔP) to the Power Loss Due to the Reactive Component of Load Current Without a Capacitor (P NO CAP ). Uniformly Distributed Load SUBSTATION I S r = resistance per unit length I Cpu L Cpu P P ΔP = Power Loss Reduction Resulting From Application Of Capacitor P = Power Loss from the Reactive Component of Load Current = (I NO CAP S r L)/3) I = Capacitor Current In Per Unit Of Peak Reactive Current (I ) CPU S L CPU = Capacitor Location From Source In Per Unit Of Feeder Length L I S = Reactive Load Current At Source (Substation) Page 1-70 NO CAP I L I L I L per unit 3 CPU CPU CPU CPU L CPU CPU Feeder Uniform with 1 Cap, Peak Power Loss Reduction.FCW 01 Siemens Industry, Inc. All rights reserved Distribution System Engineering Course for Con Edison

209 Uniformly Distributed Load Loss Reduction vs. Capacitor Location & Size The Neagle and Samson Approach P P NO CAP LOSS REDUCTION IN PU OF TOTAL LOSSES FROM REACTIVE CURRENT Page UNIFORMLY DISTRIBUTED LOAD ON FEEDER 70 % 60 % 60 % 50 % 40 % 30 % 0 % 80 % 90 % CAPACITOR SIZE = 10 % OF FEEDER VAR LOAD CAPACITOR LOCATION IN PER UNIT OF FEEDER LENGTH Given the capacitor size in PU of the reactive current at sourse end, I CPU, the location L CPU that maximizes the loss reduction is where the reactive load current is ½ of the capacitor current, or: L CPU L = Feeder Length Example: I CPU = 0.30 The capacitor is located where the reactive load current is 0.15 pu or at: L CPU I 1 CPU pu of pu of L 0.85 pu of L 01 Siemens Industry, Inc. All rights reserved. L Distribution System Engineering Course for Con Edison

210 Uniformly Distributed Load The /3 Rule To Maximize the Loss Reduction The Neagle and Samson Approach Place capacitive VARs, equal to /3 of feeder var load at the source end, at a location /3 of the way down the feeder to minimize losses (maximize the loss reduction) due to the reactive load current. P P NO CAP P P MAX NO CAP 8 9 when L CPU =/3 & I CPU = /3 The optimally sized capacitor at the optimum location will eliminate 8/9 of the losses due to the reactive component of the load current. Remember, the capacitor has practically no effect on the losses due to the active current. The /3 rule is arrived at by taking the derivatives of ΔP/P NO CAP with respect to capacitor location (L CPU ) and capacitor size I CPU, setting results to zero, and solving simultaneously Distribution System Engineering Course for Con Edison Page Siemens Industry, Inc. All rights reserved.

211 Uniformly Distributed Load Reactive Load Profiles With Optimum Capacitor Sizes KVAR FLOW KVAR FLOW KVAR FLOW SUBSTATION V S Uniformly Distributed Load REACTIVE FLOW WITHOUT CAPACITOR REACTIVE FLOW WITH CAPACITOR OF GIVEN SIZE LOCATED TO MAXIMIZE LOSS REDUCTION LAGGING PF LEADING PF LAGGING PF 500 KVAR REACTIVE FLOW WITH CAPACITOR SIZED AND LOCATED TO MAXIMIZE LOSS REDUCTION LEADING PF 500 KVAR REACTIVE LOAD AT SOURCE = 600 KVAR 50 KVAR 00 KVAR 400 KVAR 400 KVAR LAGGING PF LAGGING PF LAGGING PF PTI COURSE 006, D#3, Var Flow W-WO Capacitor-3.FCW No Capacitor on Feeder Source = 600 Capacitor of given size (500 kvar) located to maximize loss reduction (At point where reactive load = 50 kvarr) L CPU L CPU ICPU 1 pu of L pu of L pu of 1 Capacitor sized (400 kvar) & located to give the largest possible reduction in losses (At point where reactive load = 00 kvar) the /3 RULE L CPU 3 pu of L L Distribution System Engineering Course for Con Edison Page Siemens Industry, Inc. All rights reserved.

212 Uniformly Distributed Load Generalization of the /3 Rule If applying n capacitors to a uniformly distributed load circuit, to maximize the reduction in the power losses: Uniformly Distributed Load KVAR FLOW SUBSTATION V S CIRCUIT VARS AT SOURCE END REACTIVE FLOW LAGGING PF Var Flow Without Cap, Uniformly Loaded Feeder.FCW Size each capacitor to /(n+1) of the circuit VAR load at the source end. Apply them equally spaced, starting at a distance of /(n+1) of the total line length from the substation and adding the rest of the units at intervals of /(n+1) of the total line length The total VARs supplied by the capacitors are n/(n+1) of the circuit s VAR requirements at the source end Distribution System Engineering Course for Con Edison Page Siemens Industry, Inc. All rights reserved.

213 Uniformly Distributed Load Generalization of the /3 Rule Example for 3 capacitors on uniformly loaded feeder (n = 3): Each capacitor should be sized to /(n+1) or /7 of the circuit total VARs at the source end Locate them at per unit distances of /7, 4/7 and 6/7 of the line length from the substation The total VARs supplied by the three capacitors is 6/7 of the total VARs at the source end before any capacitors are applied. Example: Total VARs at source end of line = Line is uniformly loaded: Each Capacitor = (/7) of 1050 or or 300 kvar Note: These application guidelines assume that the same size conductor is used for the entire length of the circuit, and of course uniformly loaded feeder Distribution System Engineering Course for Con Edison Page Siemens Industry, Inc. All rights reserved.

214 Uniformly Distributed Load Generalization of the /3 Rule Applied to Two Capacitors Uniformly Distributed Load SUBSTATION I S I C1 I C L 1 L L PTI 006, D#1, Feeder Uniform with Caps.FCW Size and Location of Capacitors to Minimize Power Losses from Reactive Current: L 1 = (/5) L or L 1PU = 0.40 PU L = (4/5) L or L PU = 0.80 PU I C1 = (/5) I S or I C1 = 0.40 PU I C = (/5) I S or I C = 0.40 PU (I S = Reactive Current At Substation Bus) Distribution System Engineering Course for Con Edison Page Siemens Industry, Inc. All rights reserved.

215 Uniformly Distributed Load Feeder Reactive Current Profile with Two Capacitors SUBSTATION Capacitors Sized And Located To Minimize Losses From Reactive Current I S Uniformly Distributed Load I C1 L 1 L I C With Just One Capacitor On Feeder I C I S 3 P P MAX N0 CAP IS REACTIVE COMPONENT OF FEEDER CURRENT I(X) A1 I C1 + I C I C NOTE: AT OPTIMIZATION, A1 = A =A3 = A4 = A5 A L REACTIVE LOAD CURRENT WITHOUT CAPACITORS A3 PTI 006, D#1, Var Flow W-WO Two Capacitors.FCW With Two Capacitors On Feeder I C 1 I S 5 I C I S 5 P P P MAX N0 CAP NOCAP One optimally sized and located capacitor, rated 66.6 % of peak reactive load eliminates 89 % of losses due to reactive current Two optimally sized and located capacitors, rated 80% of peak reactive load eliminates 96 % of losses due to reactive current. A4 A5 DISTANCE (X) I S 4 5 r L Distribution System Engineering Course for Con Edison Page Siemens Industry, Inc. All rights reserved.

216 How to Minimize Losses at Feeders with any Load Profile The Grainger and Lee Approach Grainger and Lee (1981) Same basic idea: The optimal capacitor location is the point on the circuit where the reactive power flow equals half of the capacitor VAR rating. So, the capacitor supplies half of its VARs downstream, and half are sent upstream. Basic Steps Choose a capacitor bank size Starting from the end of the feeder, locate this capacitor bank at the point of the feeder where VAR flows on the line are equal to half of the capacitor bank VAR rating Reevaluate the VAR flow profile Move upstream until the next point where the VAR flow equals half of the second capacitor bank rating Continue placing capacitor banks until no more locations meet the criteria Distribution System Engineering Course for Con Edison Page Siemens Industry, Inc. All rights reserved.

217 Capacitors and Load Cycles Distribution System Engineering Course for Con Edison Page Siemens Industry, Inc. All rights reserved.

218 Daily or Seasonal Load Cycle The demand on the system is not constant It varies based upon the time of day and the time of the year The load cycle can be broken down into four levels Peak; High; Normal; Light The demand will also vary with circuit configuration This changing demand has a significant impact in the application and selection of capacitors Fixed capacitors can be sized and located to minimize the power losses (I R) at time of peak load on the system Fixed capacitors can be sized and located to minimize the energy losses (I R*t) throughout the load cycle Fixed capacitors can be sized and located to maximize the dollar savings in peak power losses and energy losses including the cost of the capacitors Distribution System Engineering Course for Con Edison Page Siemens Industry, Inc. All rights reserved.

219 Impact of Demand Cycles on Energy Loss Reduction Uniformly Loaded Feeder Demand cycles have a dramatic impact on the ability of a single fixed capacitor to effectively reduce feeder I R losses (power & energy) due to reactive current. A single fixed capacitor can eliminate up to 89% (8/9) of the feeder I R losses (both power and energy) due to the reactive load current if the reactive load current is constant over time and if the /3 rule is satisfied. A single fixed capacitor can eliminate up to 89% of the feeder peak I R losses due to reactive load current at time of peak load if the /3 rule is satisfied. However, the reactive load current varies with time. As a result, energy loss reduction on the order of 89% cannot be achieved with a single fixed capacitor Distribution System Engineering Course for Con Edison Page Siemens Industry, Inc. All rights reserved.

220 Impact of the Applied Voltage on Capacitor kvar Previous material has treated capacitor as a constant current source. In reality, they are constant reactance devices. When developing capacitor sizing and location guidelines, practically they can be treated as constant current sources. Current supplied by a capacitor bank is proportional to the capacitor rated current times the ratio of the applied voltage to the rated voltage. kvars supplied by capacitor bank are proportional to the capacitor rated kvar times the square of the ratio of the applied voltage to the rated voltage Distribution System Engineering Course for Con Edison Page Siemens Industry, Inc. All rights reserved.

221 Impact of Daily Demand Cycle on Capacitor Sizing Reactive Load Factor (RLF) For a uniformly loaded feeder, where should the fixed capacitor be located (L C1 ) and what should its size be (I C1 ) if the objective is to maximize the reduction in energy losses. SUBSTATION I S Uniformly Distributed Load L C1 L I C1 PTI 010, D#1, Cap Location Energy-1.FCW I S RLF I S I SMIN Page 1-83 REACTIVE LOAD CURRENT REACTIVE LOAD CURRENT AT TIME OF PEAK L AVERAGE REACTIVE LOAD CURRENT REACTIVE LOAD CURRENT AT TIME OF MINIMUM DISTANCE 01 Siemens Industry, Inc. All rights reserved Distribution System Engineering Course for Con Edison

222 Impact of Daily Demand Cycle on Capacitor Sizing Reactive Load Factor (RLF) Develop the reactive load duration curve Determine the reactive load current profile over a 4 hour period Calculate the Reactive Load Factor (RLF) which is the ratio of the average reactive load current to the peak reactive load current RLF Average Reactive Load Current Peak Reactive Load Current Distribution System Engineering Course for Con Edison Page Siemens Industry, Inc. All rights reserved.

223 Impact of Daily Demand Cycle on Capacitor Sizing Reactive Loss Factor (RLSF) Plot the square of the feeder reactive load current. Feeder reactive power losses (I R losses due to reactive load current) are proportional to the square of the feeder reactive current, with the proportionality constant being resistance. Calculate the Reactive Loss Factor (RLSF) Note: Feeder reactive power losses are the I R losses due to the reactive component of the load current. Capacitors will not eliminate losses due to the real component of current. Page 1-85 RLSF Average Loss Due to Reactive Current Peak Loss Due to Reactive Current 01 Siemens Industry, Inc. All rights reserved Distribution System Engineering Course for Con Edison

224 Capacitor Size & Location, Energy Loss Reduction ONE FIXED CAPACITOR Given capacitor size, I C1 : Where should it be located to maximize the reduction in energy losses? Uniformly Distributed Load SUBSTATION I S I C1 L C1 I S = Reactive Current At The Source End At Time Of Peak Reactive Load RLF = Reactive Load Factor = Capacitor Current I C1 Given the capacitor size, I C1, the location L C1 that maximizes the energy loss reduction is where the reactive load factor (RLF) times the reactive current at time of peak (I) is ½ of the capacitor current (the point on the feeder where the average reactive load current is ½ of the capacitor current) I I Reactive load current at time of peak at application point = I: I * C1 C RLF or I 1 * RLF LC 1 IC1 PU IC1 Distance from source: L 1 and I Page 1-86 C1PU L L * RLF C1PU I S 01 Siemens Industry, Inc. All rights reserved Distribution System Engineering Course for Con Edison

225 Capacitor Size & Location, Energy Loss Reduction ONE FIXED CAPACITOR I S RLF I S REACTIVE LOAD CURRENT PTI 010, D#1, Cap Location Energy.FCW REACTIVE LOAD CURRENT AT TIME OF PEAK AVERAGE REACTIVE LOAD CURRENT LOCATE CAPACITOR AT POINT WHERE HALF OF THE CAPACITOR CURRENT EQUALS THE AVERAGE REACTIVE LOAD CURRENT I C1 DISTANCE L Derivation of expression for distance from source to capacitor to maximize energy loss reduction when capacitor size is specified: Similar triangles is starting point. RLF I L RLF S 1 IC1 RLF IS L LC 1 L LC 1 I I IC1 L C1 LC 1 I S C1PU RLF 1 LC 1PU L L C1 L - L C1 RLF L I S L L I C1 C1 1 IC PU RLF 1 C1PU Distribution System Engineering Course for Con Edison Page Siemens Industry, Inc. All rights reserved.

226 Capacitor Size & Location, Energy Loss Reduction ONE FIXED CAPACITOR Example Calculation: Uniformly Loaded Line Peak Reactive Load at Source End = 000 kvar Reactive Load Factor (RLF) = 0.80 A 900 Kvar Capacitor is to be Added. Where Should it be Located to Maximize the Reduction in Energy Losses Due to the Reactive Current? Capacitor Current in Per Unit of Reactive Current at Source at Time of Peak IC1 KVARC1 900 IC1 PU I KVAR 000 S Location of Capacitor from Source End: S Reactive Load Current at Application Point at Time of Peak: The 900 kvar capacitor is applied at the point on the feeder where, at time of peak loading, the reactive load is * 000 or 563 kvar. Per Unit LC 1 IC1 PU 0.45 LC 1PU L * RLF *0.80 Per Unit From Source IC IC1 PU 0.45 I 1 or IPU PU of IS or IPU * RLF * RLF * 0.80 End PU of I S Distribution System Engineering Course for Con Edison Page Siemens Industry, Inc. All rights reserved.

227 Capacitor Size & Location, Energy Loss Reduction ONE FIXED CAPACITOR To Maximize the Reduction in Energy Loss on Feeder with Uniform Load Distribution: Uniformly Distributed Load SUBSTATION I S I C1 L C1 L Capacitor Location: L C1 Capacitor Size: I C1 = (/3) I S RLF Page 1-89 Feeder Uniform with 1 Cap, Energy.FCW = (/3) L ( same location as selected to maximize the reduction in power loss at time of peak load) AVERAGE REACTIVE LOAD CURRENT AT SOURCE END OF LINE Where: I = s Reactive Current At Source End at Time of Peak Load L = Section Length RLF = Reactive Load Factor 01 Siemens Industry, Inc. All rights reserved Distribution System Engineering Course for Con Edison

228 Summary, One Fixed Capacitor Uniformly Loaded Feeder of Length L Uniformly Distributed Load SUBSTATION I S L Cpu Power Loss Without Capacitor at time of peak I Cpu L Energy Loss Without Capacitor Over Time Interval T 1 ( 1 PNOCAP IS r L ) ENOCAP ( I S r L ) RLSF T 3 3 Power Loss Reduction From Capacitor at time of peak To Maximize ΔP at time of peak To Maximize ΔE During Time Interval T LCPU ICPU IC I L I RLF I I RLF S CPU CPU C S PMAX 8 IS r L EMAX I S r L RLF T 7 P P MAX N0 CAP L CPU I CPU S 8 9 = Distance To Capacitor In PU Of Circuit Length L = Capacitor Current In Per Unit Of I s At Source End r = Circuit Resistance Per Unit Length I S = Reactive Current at Source End at Time of Peak Page 1-90 CPU CPU CPU CPU CPU CPU Energy Loss Reduction From Capacitor During Time T P I r L I L I L I L E I r L I L RLF I L RLF I L T 7 E E MAX S 8 N0 CAP 9 CPU RLF RLSF CPU CPU CPU RLF = Reactive Load Factor RLSF = Reactive Loss Factor T = Time Period For Energy Calculation CPU CPU 01 Siemens Industry, Inc. All rights reserved Distribution System Engineering Course for Con Edison

229 Reactive Load Factor and Reactive Loss Factor for Hypothetical Load Curves REACTIVE CURRENT AT SOURCE I S K T I S i(t) time T PTI 006 D#, LOAD PROFILES FOR RLF AND RLSF EXAMPLE.FCW T REACTIVE CURRENT AT SOURCE I S K T I S i(t) T T time Reactive Load Factor (RLF) AVERAGE REACTIVE LOAD RLF PEAK REACTIVE LOAD Reactive Loss Factor (RLSF) 1 T T 0 i( t) dt AVERAGE LOSS FROM REACTIVE CURRENT RLSF PEAK LOSS FROM REACTIVE CURRENT I S AREA UNDER CURVE 1 T T 0 i ( t) I S dt Distribution System Engineering Course for Con Edison Page Siemens Industry, Inc. All rights reserved.

230 Reactive Load Factor and Reactive Loss Factor for Hypothetical Load Curves REACTIVE CURRENT AT SOURCE I S K T I S RLF RLSF RLF RLSF i(t) K K 1 K T T K T T 1 K T 1 K T KT T T T T T T RLF T K 1 T PTI 006 D#, LOAD PROFILES FOR RLF AND RLSF EXAMPLE.FCW T 0 1 T RLSF time PU T 1 KT TPU i( t) dt I S T 0 Reactive Load Factor Reactive Loss Factor (1 K i ( t) I S T T PU T T dt ) T PU RLF RLSF RLF RLSF K T K T K T 3 Ratio Of Reactive Load Factor Squared To Reactive Loss Factor REACTIVE CURRENT AT SOURCE I S K T I S i(t) T T T K K 1 K K T T T 1 1 time Distribution System Engineering Course for Con Edison Page Siemens Industry, Inc. All rights reserved.

231 Impact of Demand Cycles on Maximum Possible Energy Loss Reduction Triangular Demand Cycle Uniformly Distributed Load On Feeder One Fixed Capacitor Sized and Located to Maximize Energy Loss Reduction: I CPU L CPU I S = (/3) RLF = /3 Per Unit Of Feeder Length = Peak Reactive Load Current At Source End Of Feeder RLF = Reactive Load Factor RLSF = Reactive Loss Factor RLF RLSF E E MAX NOCAP 3 4 T T K K 8 9 K K T RLF RLSF T 1 1 REACTIVE CURRENT AT SOURCE K T I S K T I S RLF i(t) RLF / RLSF % OF REACTIVE CURRENT ENERGY LOSSES ELIMINATED T T time Distribution System Engineering Course for Con Edison Page Siemens Industry, Inc. All rights reserved.

232 Impact of Demand Cycles on Maximum Possible Energy Loss Reduction Square Demand Cycle Uniformly Distributed Load On Feeder One Fixed Capacitor Sized and Located to Maximize Energy Loss Reduction: I CPU L CPU RLF RLSF E E MAX NO CAP = (/3) RLF = /3 Per Unit Of Feeder Length I S = Peak Reactive Load Current At Source End Of Feeder RLF = Reactive Load Factor RLSF = Reactive Loss Factor K T RLF RLSF K T K T T K PU T 1 K T T PU (1 K T ) T PU REACTIVE CURRENT AT SOURCE K T I S K T I S RLF i(t) RLF / RLSF ΔT = 0.5 PU % OF REACTIVE CURRENT ENERGY LOSSES ELIMINATED T PTI 006 D#, LOAD PROFILES FOR RLF AND RLSF EXAMPLE.FCW T time Distribution System Engineering Course for Con Edison Page Siemens Industry, Inc. All rights reserved.

233 Impact of Demand Cycles on Maximum Possible Energy Loss Reduction Square Demand Cycle, Detail On Ratio Of Maximum Possible Energy Loss Reduction For Uniformly Loaded Feeder REACTIVE CURRENT AT SOURCE I S K T I S i(t) Optimally Sized and Located Fixed Capacitor on Uniformly Loaded Feeder: Maximum Energy Loss Reduction in Per Unit of That Without Capacitor Page 1-95 E E E 8 MAX N0 CAP 9 NO CAP T Energy RLF RLSF loss T from time PTI 006 D#, LOAD PROFILES FOR RLF AND RLSF EXAMPLE.FCW RATIO OF E MAX TO E NO CAP reactive load current without capacitor K T =0.60 K T =0.50 K T =0.40 K T =0.30 K T =0.0 ONE OPTIMALLY SIZED FIXED CAPACITOR LUMPED LOAD AT FDR END DURATION OF PEAK REACTIVE LOAD IN PU (T) 01 Siemens Industry, Inc. All rights reserved Distribution System Engineering Course for Con Edison

234 Fixed Capacitor, Example Calculation A Con Ed distribution feeder is miles in length, with 477 kcmil phase conductors. The load is uniformly distributed along the line. At the substation end, the load current at time of peak, I S, is 300 amperes, with the power factor being 0.80 per unit lagging. Shown below is the reactive load current at the substation end of the line. i(t) The system nominal voltage is 13.8 kv phase-to-phase. An 1800 kvar fixed capacitor bank is installed at the 1 mile point on the feeder (L CPU = 0.5). Find the following over a time period of 1 year. 1. The energy loss on the feeder in KWH per phase per year due to the reactive load current.. The reduction in energy loss along the feeder from application of the 1800 kvar capacitor bank at the one mile point, in KWH per phase per year 3. The size and location of the fixed capacitor that will maximize the reduction in energy loss. 4. The maximum energy loss reduction that is possible through application of a single fixed capacitor, in KWH per phase per year. 5. The ratio of the maximum energy loss reduction to the energy loss due to the reactive load current with no capacitor. r = Ω / mile for phase conductor Page 1-96 REACTIVE CURRENT AT SOURCE I S 0.6 I S EXAMPLE CALC - ENERGY.FCW 1 HR 4 HR T time 01 Siemens Industry, Inc. All rights reserved Distribution System Engineering Course for Con Edison

235 Fixed Capacitor, Example Calculation REACTIVE CURRENT AT SOURCE Feeder Length =.0 Miles Feeder Resistance = r = Ω/Mile, Reactance = x = Ω/Mile Feeder Peak Load Current at Source = % Power Factor Feeder Peak Reactive Current at Source (I S ) = 0.6 *300 = 180 Amp 1800 Kvar Capacitor Installed at 1.0 Mile Point Page 1-97 I S 0.6 I S i(t) 108 A 180 A 1 HR T PTI 006, D#3, EXAMPLE CALC - ENERGY-FIXED, PROFILE.FCW T 4 HR Reactive Load Factor And Reactive Loss Factor For Square Demand Cycle: T 1 T PU T 4 per unit time that load is at peak RLF T KT T T 80 RLSF T KT 1 KT 0.6 (1 0.6 )* 0.50 T time 1 K 0.60 (1 0.6)* RLF RLSF REACTIVE CURRENT ALONG LINE i(x) WITH CAP AT TIME OF PEAK 1 AT 60 % OF PEAK 01 Siemens Industry, Inc. All rights reserved. X Distribution System Engineering Course for Con Edison

236 Fixed Capacitor, Example Calculation Energy Loss in One Year from Reactive Component of Load Current, No Capacitor E E NOCAP 1 ( r L IS ) RLSF T *.0 * * ,476 KWH / Year Phase NO CAP / 5,476 * 10 Energy Loss Reduction from Application of 1800 Kvar Capacitor at 1 Mile Point 3 Watt Hour / Year / Phase I Page 1-98 KVA C (3PHASE ) CAP KV I CPU PU L CPU PU OF OF PEAK AMPS REACTIVE FEEDER TOTAL LENGTH E r L I I L RLF I L RLF S CPU CPU E *.0 * 180 CPU CPU 18,377 KWH / Year / Phase I LOAD CURRENT CPU L CPU T * *0.5* *0.5 *0.80 Wa tt *0.50 Hours / Year / Phase Siemens Industry, Inc. All rights reserved Distribution System Engineering Course for Con Edison

237 Fixed Capacitor, Example Calculation Size and Location for Maximum Energy Loss Reduction LCPU PER UNIT, LC ( MILES) MILES 3 3 I CPU RLF PU OF PEAK REACTIVE CURRENT AT SUBSTATION OF I C AMPS * AMPS KVA C 3PHASE 3 * 13.8 * KVAc ( Not a s tandard size) AMPS Maximum Energy Loss Reduction Possible with Optimally Sized Capacitor E MAX 8 r L IS RLF T * * * ,313 KWH / Year / Phase 1,313 *10 Watt Hours / Year / Phase Ratio of Max Possible Energy Loss Reduction to Energy Loss without Capacitor E E MAX N0 CAP 8 9 RLF RLSF Note: With the 1800 kvar capacitor at the 1 mile point, the ratio of energy loss reduction to energy loss without capacitor is 18,377/5,476 or Distribution System Engineering Course for Con Edison Page Siemens Industry, Inc. All rights reserved.

238 Fixed Capacitor, Example Calculation Reactive Load Current Profile at Peak and Off Peak with Optimally Sized Capacitor Applied (95 kvar at Mile Point) REACTIVE CURRENT AT SOURCE I S 0.6 I S Voltage Rise at 1 Mile Point with 1800 Kvar Capacitor at 1 Mile Point V I X 75.3 * Volts RISE i(t) 108 A CAP * LINE 180 A 1 HR PTI 006, D#, EXAMPLE CALC - ENERGY-FIXED, PROFILE-OPTIMAL SIZE.FCW T 4 HR % T time Voltage Rise at Mile Point with 95 Kvar Capacitor at Mile Point 74.6 VRISE ICAP * X1 LINE 96.0 * * Volts REACTIVE CURRENT ALONG LINE i(x) WITH CAP 1 AT TIME OF PEAK AT 60 % OF PEAK 4/3 % X Distribution System Engineering Course for Con Edison Page Siemens Industry, Inc. All rights reserved.

239 Fixed Capacitor Application Location & Size for Max Reduction in Power / Energy Loss Capacitor Location for Maximum Reduction in Power or Energy Loss L CPU = /3 of Feeder Length Size for Maximum Power Loss Reduction = /3 * Peak Reactive Load at Source Size for Maximum Energy Loss Reduction = (/3)* RLF* Peak Reactive Load at Source I(X) PU REACTIVE CURRENTS AT TIME OF PEAK REACTIVE CURRENT IN PU 1.0 LOAD CURRENT WITHOUT CAPACITOR 1/3 0-1/3 CAPACITOR SIZED TO MAXIMIZE REDUCTION IN ENERGY LOSS /3 RLF = 0.80 CAPACITOR SIZED TO MAXIMIZE REDUCTION IN PEAK POWER (KW) LOSS 1.0 X IN PU PTI 006, D#, 000 KVAR, 1333 KVAC, RLF-ZP8.FCW Distribution System Engineering Course for Con Edison Page Siemens Industry, Inc. All rights reserved.

240 Fixed Capacitor Application - Size Specified Location To Maximize Power & Energy Loss Reduction Capacitor Size Fixed at I cpu PU of Feeder Peak Reactive Current at Source: Location to Maximize Power Loss Reduction: Location to Maximize Energy Loss Reduction: REACTIVE CURRENT IN PU Page I(X) PU L L CPU CPU REACTIVE CURRENTS AT TIME OF PEAK CAPACITOR SIZE = 0.45 PU OF PEAK LOAD ICPU ICPU 1.0 AND I( LCPU ) PU ICPU ICPU 1.0 AND I( LCPU ) PU * RLF * RLF LOAD CURRENT WITHOUT CAPACITOR RLF = 0.80 CAPACITOR LOCATED TO MAXIMIZE REDUCTION IN ENERGY LOSS * CAPACITOR LOCATED TO MAXIMIZE REDUCTION IN PEAK POWER (KW) LOSS X IN PU PTI 006, D#, 000 KVAR, 900 KVAC, RLF=ZP8.FCW 01 Siemens Industry, Inc. All rights reserved Distribution System Engineering Course for Con Edison

241 Distribution System Engineering Course for Con Edison Impact of Switched Capacitors on System Reactive Current Page Siemens Industry, Inc. All rights reserved.

242 One Switched Capacitor Size, Location, and On Time to Maximize ΔE Uniformly Distributed Load Page REACTIVE CURRENT I S REACTIVE CURRENT AT SUBSTATION VS TIME i S X = O T ON t S T OFF T SUBSTATION t I S L Cpu L I Cpu i S t = t S I S i S (x) = [ 1- x L ] I S I Cpu X Feeder Uniform with 1 Switched Capacitor.FCW Energy Loss Without Capacitor 1 ENOCAP ( r L IS ) RLSF T 3 Reduction in Energy Loss with Switched Capacitor Connected Between T=T ON And T=T OFF E SW CAP r L I S I L RLF I L RLF I L ( T T ) CPU CPU CAP ON CAP ON I S = Reactive Load Current At Source At Time Of Peak I CPU = Capacitor Current In Per Unit Of I s T = Period Of Reactive Load Current Cycle RLF = CAP ON Reactive Load Factor When Capacitor Is Connected (Between T ON & T OFF ) RLSF = Reactive Loss Factor Of Entire Reactive Load Cycle CPU CPU CPU CPU OFF ON 01 Siemens Industry, Inc. All rights reserved Distribution System Engineering Course for Con Edison

243 One Switched Capacitor Size, Location, On Time to Maximize ΔE Uniformly Distributed Load Page REACTIVE CURRENT I S REACTIVE CURRENT AT SUBSTATION VS TIME i S X = O T ON t S T OFF T L t SUBSTATION I Cpu L Cpu L i ICPU RLFCAP ON S 3 t = t S 3 I S i S (x) = [ 1- x L ] I S CPU To Maximize ΔE SW CAP with One Switched Capacitor Between T on and T OFF : L CPU 3 E and I CPU SW CAP MAX 3 RLF S r L I CAP ON 8 7 RLF I S CAP ON I Cpu Feeder Uniform with 1 Switched Capacitor.FCW To Maximize ΔE SW CAP, The Switch On and Off Times Must Be Selected Such that the Product of the Time the Capacitor is On, (T OFF T ON ) and the Square of the Reactive Load Factor with the Capacitor Connected (RLF CAP ON ) is Maximum. Regardless, capacitor is always located at /3 point. This can be Accomplished Only When the Reactive Load Current Profile Curve is Available. T OFF T ON Function to be Maximized to Maximize the Reduction in Energy Loss, ΔE SW CAP 01 Siemens Industry, Inc. All rights reserved. X Distribution System Engineering Course for Con Edison

244 One Switched Capacitor Size, Location, and On Time (T ON ) to Maximize ΔE Example Assuming Triangular Reactive Load Cycle REACTIVE CURRENT AT SOURCE I S K T I S Page Reactive Load Factor, Entire Cycle Without Cap 1 KT RLF Reactive Loss Factor, Entire Cycle Without Cap RLSF 1 K T K T 3 I S = Maximum Reactive Load Current at Source K T I S = Minimum Reactive Load Current at Source Reactive Load Factor With Capacitor On Between T ON and T OFF 1 KT T0N RLFCAP ON 1 KT T Ratio Of Maximum Energy Loss Reduction To Energy Loss With No Capacitor ESW CAP MAX 8 RLFCAPON T RLF OFF TON 8 CAPON T 1 E 9 RLSF T 9 RLSF T NO CAP T ON PTI 006 D#1, TRIANGLE REACTIVE LOAD CURVE.FCW T OFF T time TON 1 3KT This Ratio (Energy Loss Reduction) is Maximized Whenever: T 6(1 K ) Subject To Constraint That When K T > 1/3, Capacitor Should Be On All The Time SUBSTITUTE EXPRESSIONS FOR RLF CAP ON & RLSF INTO ABOVE, TAKE DERIVATIVE WITH RESPECT TO (T ON /T), SET TO ZERO & SOLVE ON 01 Siemens Industry, Inc. All rights reserved. T Distribution System Engineering Course for Con Edison

245 Impact of Using a Single Optimally-Sized Switched Capacitor Uniformly Loaded Feeder Maximizing energy loss reduction: For K T < 1/3 TON 1 3K T 6(1 K I CPU RLF CAP 3 E E SW CAP MAX NO CAP For K T > 1/3, Capacitor is on for the Entire Load Cycle (T ON =0) to maximize energy loss reduction 1 KT ICPU RLFCAP 0N 3 3 E SW CAP MAX E NO CAP (1) With Optimally Sized Fixed Capacitor Page T T ) N K K 1 K RLF RLSF T RLF CAP-ON when T ON /T =(1-3K T ) / [6(1-K T )] 1 T T (1 K 1 K T T K T T K ) REACTIVE CURRENT AT SOURCE K T I S K T I S T ON T ON / T PTI 006 D#1, TRIANGLE REACTIVE LOAD CURVE.FCW T OFF T % OF REACTIVE CURRENT ENERGY LOSSES ELIMINATED SWITCHED CAP FIXED CAP (1) time 01 Siemens Industry, Inc. All rights reserved Distribution System Engineering Course for Con Edison

246 Maximizing Energy Loss Reduction Basics On Fixed and One Switched Capacitor Reactive Load Current Profile at Source, Lumped Load at End of Feeder REACTIVE CURRENT AT SOURCE I S K T I S time L = line length r = resistance per unit length CE 006, D#1, Lumped Load, Fixed & Switched Caps.FCW.05 L ( 1 - K T ) I S LOAD K T I S T T STEP REACTIVE LOAD CURVE.FCW One Fixed and One Switched Capacitor at End of Feeder Fixed Capacitor Current = K T x I S Switched Capacitor Current = (1-K T ) x I S Switched Capacitor on During Time of Peak Load Only Total Installed Capacitor Current = I S Energy Loss With No Capacitor E MAX ENOCAP K T IS T T IS T rl E Installation of Fixed and Switched Capacitor can Eliminate 100 % of the Energy Loss Due to the Reactive Load Current Page NOCAP 01 Siemens Industry, Inc. All rights reserved Distribution System Engineering Course for Con Edison

247 Maximizing Energy Loss Reduction Basics One Optimally Sized Fixed Capacitor Reactive Load Current Profile at Source, Lumped Load at End of Feeder REACTIVE CURRENT AT SOURCE I S K T I S STEP REACTIVE LOAD CURVE.FCW Page T T time One Optimally Sized Fixed Capacitor at End of Feeder Fixed Capacitor Current = RLF x I S E E E MAX NO CAP NOCAP RLF RLSF K T I S K T T T 1 K I S T KT T T rl K PU T 1 KT TPU (1 K One Optimally Sized Fixed Capacitor can Eliminate a Large Percentage of the Energy Losses Due to the Reactive Load Current, Dependent Upon the Reactive Load Profile. T ) T L = line length r = resistance per unit length CE 006, D#1, Lumped Load, Optimal Fixed Cap.FCW.05 PU RATIO OF E MAX TO E NO CAP L K T =0.60 K T =0.50 K T =0.40 K T =0.30 RLF K T =0.0 I S ONE OPTIMALLY SIZED FIXED CAPACITOR LUMPED LOAD AT FDR END LOAD DURATION OF PEAK REACTIVE LOAD IN PU (T) 01 Siemens Industry, Inc. All rights reserved Distribution System Engineering Course for Con Edison

248 Maximizing Energy Loss Reduction Basics Reasons for Both Fixed and Switched Capacitors 1. Voltage Rise At Light Load Periods Or On Week-ends With Fixed Capacitor May Be To High. Additional Reduction In Energy Loss With Switched Capacitors May Be Justified Economically, Considering Higher Cost Per kva Of Switched Capacitors 3. Switched Capacitors May Be Required To Regulate Voltage Within Specified Limits 4. Switched Capacitors Will Release More Capacity At Time Of Peak Load On Circuit 5. Switched Capacitors Can Be Remotely Controlled To Support VAr Requirements Of The Distribution Substation / Transmission System Remember: Voltage supplied to the customer must be within range A, or on a 10 volt basis between 114 volts and 15 volts. This requirement takes precedence and must be met irrespective of the energy or power losses on the distribution feeder Distribution System Engineering Course for Con Edison Page Siemens Industry, Inc. All rights reserved.

249 Switched Capacitors The conclusion of the preceding analysis is that many feeder with greatly varying reactive load will need fixed and switched capacitors to obtain reasonable levels of reactive loss reduction (power and energy). Fixed and switched capacitors can assist with feeder voltage regulation This is because the voltage drop due to reactive current is altered (reduced) when a capacitor is added Voltage rise at point of application is product of capacitor current and line reactance between source and capacitor Switched capacitors can regulate voltage Watch out for overcompensation Distribution System Engineering Course for Con Edison Page Siemens Industry, Inc. All rights reserved.

250 Computer Programs Calculate Optimum Capacitor Placement Programs use a variety of iterative methods to account for all sorts of complex factors: non-uniform and specific load distributions along feeder different size conductors along the feeder complex feeder topologies voltage regulators located out on the feeders Input from voltage monitors (smart meters) at select locations on feeder simultaneous influence of capacitors on voltage profile and feeder losses (both energy losses and peak power losses) economic evaluations (cost of energy, cost of generation, cost of fixed and switched capacitors, economic factors, etc.) The input and output data must be reviewed to ensure that reasonable and feasible results are obtained Page Siemens Industry, Inc. All rights reserved Distribution System Engineering Course for Con Edison

251 Summarizing Points Capacitors Are used to provide VAR support and power factor correction Release capacity to allow serving additional load Reduce both power and energy losses in feeder Reduce voltage drop from source to point of application They provide a voltage rise because they reduce the reactive current along the feeder Can be applied as fixed or switched banks Fixed banks are ON all the time and should be sized appropriately Switched banks should be inserted in stages to optimize loss reduction Switched banks can be used to regulate voltage The application of the capacitors will improve the efficiency of the system Voltage must be within specified range at all points along feeder. Loss optimization must satisfy voltage constraints at all load levels Distribution System Engineering Course for Con Edison Page Siemens Industry, Inc. All rights reserved.

252 CAPACITOR APPLICATION SUPPLEMENT 1 Application in Con Edison Area Substation: (Four 65 MVA Transformers with Peak Load of 195 MVA) CALCULATION OF: 1. Released Capacity From Addition of 60 MVAR (Three 0 MVAR Banks) on 13.8 kv Buses.. Voltage Rise on 13.8 kv Buses From Switching of a 0 MVAR Bank. 3. Loss Reduction in Substation Transformers With 60 MVAR Connected On 13.8 kv Buses Distribution System Engineering Course for Con Edison Page 1-Sup Siemens Industry, Inc. All rights reserved.

253 Released Capacity Example: Con Ed Area Substation Area Substation With Four 65 MVA Transformers - Assume: 1.Peak Load On Transformers Is 3*65 MVA or 195 MVA: (KVA 1 = 195,000). Power Factor Of Load Without Capacitors Is 80 % Lagging (PF 1 = 0.80 pu) If three 0 mva capacitor banks are added to the area substation 13.8 kv bus, what is the released capacity? Capacitor Size In Per Unit Of 195 MVA Uncompensated Load (KVA 1 ) QC ( capacitor size in kva) 60,000 QCPU KVA ( load size in kva) 195,000 Released Capacity In Per Unit Of 195 MVA Uncompensated Load (KVA 1 ) KVA PU KVA PU Installation of the Capacitor Banks (three 0 MVA Banks) Allows Serving an Additional Load with an 80 % Power Factor of * 195 MVA or 30.0 MVA Without Exceeding the Rating of the Three 65 MVA Transformers. Page 1-Sup 1- Q 1 CPU 1 1 QCPU PF1 1 per unit 1 PF PER UNIT OF KVA * PER UNIT OF KVA 1 01 Siemens Industry, Inc. All rights reserved Distribution System Engineering Course for Con Edison

254 Con Ed Area Substation, Voltage Rise on 13.8 kv Bus When One 0 MVAR Cap Bank Connected 13.8 KV Buses: 40 KA Three-phase Fault Current 0 MVAR Cap Bank NORTH SYNCHRONIZING BUS NWK. 1 NWK. NWK. 1 NWK. NWK. 1 NWK. NWK. 1 NWK. NWK. 1 NWK. NWK. 1 NWK. NWK. 1 NWK. NWK. 1 NWK. 1 SYN. N SYN. N 3 SYN. N 4 SYN. N NORMALLY OPEN BUS SECT. 1N BUS SECT. N BUS SECT. 3N BUS SECT. 4N XFR. 1 1 XFR. N XFR. XFR. N XFR. 3 3 XFR. N XFR. 4 4 XFR. N XFR. 5 IN PLACE SPARE 138 KV 13.8 KV 138 KV 13.8 KV 138 KV 13.8 KV 138 KV 13.8 KV BUS SECT. 1S BUS SECT. S BUS SECT. 3S BUS SECT. 4S NWK. 1 NWK. NWK. 1 NWK. Page 1-Sup XFR. S 1 SYN. S SYN. S 3 SYN. S 4 SYN. S NWK. 1 NWK. NWK. 1 NWK. XFR. S NWK. 1 NWK. NWK. 1 NWK. 3 XFR. S SOUTH SYNCHRONIZING BUS NWK. 1 NWK. NWK. 1 NWK. 4 XFR. S NORMALLY OPEN 01 Siemens Industry, Inc. All rights reserved Distribution System Engineering Course for Con Edison

255 Con Ed Area Substation, Voltage Rise on 13.8 kv Bus When One 0 MVAR Cap Bank Connected Thevenin Equivalent Reactance (Positive-Sequence) On 13.8 KV Bus 13,800 Phase To Neutral Voltage 3 7,967 Volts X Z Phase Short Circuit Current 40,000 40,000 Amps Nominal Current Of 0 MVAR Capacitor Bank 1,000 MVAR 0,000 I CAP KV Amperes Line-to-Neutral Voltage Rise On Substation Bus In Volts and in % Line-to-Neutral Voltage Rise On 10-Volt Base Line-to-Line Voltage Rise On Substation Bus Note: If a 10 MVAR capacitor bank were connected, the voltage rise would be one half of that when the 0 MVAR bank is connected. Page 1-Sup 1-4 VRISE N ICAP X * Volts V RISE % % 7, V RISE Volts 100 VRISE 3 VRISE N 3 * Volts 01 Siemens Industry, Inc. All rights reserved Distribution System Engineering Course for Con Edison

256 Con Ed Area Substation, Voltage Rise on 13.8 kv Bus When 0 MVAR Cap Bank Connected Alternate Formula for Calculation of Voltage Rise At Bus in % When Capacitor Bank is Connected to Bus: MVARCAP VRISE % 100 MVA SC MVAR CAP = Nominal Three-phase Rating of Capacitor Bank in MVA MVA SC = Available Three-Phase Short Circuit Capacity at Location of Capacitor in MVA KV ΦΦ KA 3Φ MVA SC 3 KV KA 3 = Phase-to-Phase Voltage In KV = Available 3-Phase Short Circuit Current On Bus in KA MVA SC 3 *13.8 * MVA V RISE %.09 % Distribution System Engineering Course for Con Edison Page 1-Sup Siemens Industry, Inc. All rights reserved.

257 Con Ed Area Substation, 13.8 kv, Loss Reduction In Sub Xfrs When 60 MVAR Cap Bank Connected Four 65 MVA Transformers Connected, 3Φ Short Circuit Current = 40 KA X/R Of Equivalent Thevenin Impedance Of 13.8 kv Bus = 5 Average Load = 3*65 MVA = % Power Factor NORTH SYNCHRONIZING BUS 60 MVAR TOTAL NWK. 1 NWK. NWK. 1 NWK. NWK. 1 NWK. NWK. 1 NWK. NWK. 1 NWK. NWK. 1 NWK. NWK. 1 NWK. NWK. 1 NWK. 1 SYN. N SYN. N 3 SYN. N 4 SYN. N NORMALLY OPEN BUS SECT. 1N BUS SECT. N BUS SECT. 3N BUS SECT. 4N XFR. 1 1 XFR. N XFR. XFR. N XFR. 3 3 XFR. N XFR KV 13.8 KV 138 KV 13.8 KV 138 KV 13.8 KV 138 KV 13.8 KV BUS SECT. 1S BUS SECT. S BUS SECT. 3S BUS SECT. 4S NWK. 1 NWK. NWK. 1 NWK. Page 1-Sup XFR. S 1 SYN. S SYN. S 3 SYN. S 4 SYN. S NWK. 1 NWK. NWK. 1 NWK. XFR. S NWK. 1 NWK. NWK. 1 NWK. 3 XFR. S SOUTH SYNCHRONIZING BUS NWK. 1 NWK. NWK. 1 NWK. 4 XFR. N 4 XFR. S XFR. 5 IN PLACE SPARE NORMALLY OPEN 01 Siemens Industry, Inc. All rights reserved Distribution System Engineering Course for Con Edison

258 Con Ed Area Substation, 13.8 kv, Loss Reduction When 60 MVAR of Cap Bank Connected Reduction in Power Loss (ΔW) in Per Unit of Power Loss Without Capacitor (W) W W IC IC 1 PFL IL I L per unit IC CAPACITOR CURRENT MVARCAP I LOAD CURRENT MVA 3 * L LOAD per unit W W * per unit Page 1-Sup 1-7 Application Of 60 MVAR Capacitor When Total Load On Substation Is 195 MVA Reduces Power Losses In The Substation Transformers By 7.46 % Calculation of Resistive Component of Substation Transformer Impedance (4 Units in Parallel). This is needed to find the reduction in power losses (ΔW) in Watts. E LN = System line-to-neutral voltage = 13,800/ 3 = 7,967 Volts Available three-phase 7,967 fault current = 40,000 Amperes RMS, X/R = 5 Z Ohms 40,000 R Z * cos(tan 1 ( X / R)) * cos 1 tan (5) Siemens Industry, Inc. All rights reserved Distribution System Engineering Course for Con Edison

259 Con Ed Area Substation, 13.8 kv, Loss Reduction When 60 MVAR of Cap Bank Connected Calculation of 195 MVA Load Current (I L ) and Reactive Current Supplied by 60 Mvar Capacitor (I C ): 1000 * MVAL 1000 * 195 IL KV 3 * * MVAR 1000 * 60 I C KV 3 * 13.8 Amperes Amperes Calculation of Reduction in Watts Loss (ΔW): W IL IC 1.0 PF 146,174 Watts / W 3 W 3 * L 3 IC R * 8158 * kw per phase kw in all three phases * Reduction in Energy Losses in KWHr Per Year (assuming constant load): 510 kwhr / year kw * # hours in year * ,84, 136 kwhr Distribution System Engineering Course for Con Edison Page 1-Sup Siemens Industry, Inc. All rights reserved.

260 Answers for energy. Tab 5: Voltage Control of Distribution Systems Distribution System Engineering Course 1 01 Siemens Industry, Inc. All rights reserved.

261 Voltage Control Objective The overall objective of voltage control is to provide each customer with a service voltage that conforms to the voltage design limitations (rms voltage, voltage unbalance, harmonics, etc.) as specified in applicable standards and regulations The engineering objective is to select and install the correct equipment that allows the system to satisfy the overall objective as the loading in the system changes and as the system topology changes On the radial distribution circuit, without corrective equipment (capacitors or voltage regulators), or without distributed generation, the voltage decreases going out the feeder from the substation. The voltage level at the first customer must not exceed the upper limit, and the voltage level at the last customer must not be below the lower limit Distribution System Engineering Course for Con Edison Page - 01 Siemens Industry, Inc. All rights reserved.

262 Voltage Regulation Definitions Service Voltage - The voltage at the point where the supplier system and the user system are connected - usually at the revenue meter. Nominal Service Voltages 1Φ 3-wire, 10/40 Volts 3Φ 4-wire, 08Y/10 Volts 3Φ 4-wire, 480Y/77 Volts or 460Y/65 Volts 3Φ 4-wire 40/10 Volt Delta Utilization Voltage - The voltage at the terminal of the utilization equipment (the electrical outlet). Base Voltage - A reference voltage; usually 10 volts Distribution System Engineering Course for Con Edison Page Siemens Industry, Inc. All rights reserved.

263 Voltage Regulation Definitions Voltage Drop - Difference between the magnitude of the sending end voltage and the magnitude of the receiving end voltage of a feeder circuit. Usually calculated using the line-toneutral voltages at the sending end and receiving end of a circuit. Voltage Regulation - The voltage drop between two points on the distribution system in percent of nominal voltage. For distribution transformers, voltage regulation is defined as the change in output voltage when the load goes from no load to full load, with the output voltage at full load being rated voltage, expressed in % of full load voltage. PCC - Point of Common Coupling Distribution System Engineering Course for Con Edison Page Siemens Industry, Inc. All rights reserved.

264 Comparison Of Line Voltage Drop (In-Phase Versus Actual From Vector Calculations) V S R 1 MILE X I VL V S Ob 3" 3" Data For Numerical Example (Line Length = 1 Mile): Load = % PF, V LOAD Φ-N = 7967 VOLTS Note: θ is the angle between the load voltage (V L ) and current I. Approximate Voltage Drop Calculation: V D-APROX ІV S І - ІV L І VD APROX I R cos X sin I R * PFPU X 1 PFPU Volts * * Volts Exact Voltage Drop Calculation - Voltage at Source End V S O V L 7967 Page -5 I ( R jx ) I (cos (0.198 j sin ) j0.583) 350 (0.8 V I ( R cos X sin ) j I X cos R sin L j 0.60) VL IR PTI 006, D#, VECTORS, EXACT VS APROX VD.FCW IX o 0 o Oa 48" R 1 = Ohms / mile X 1 = Ohms / mile PTI 006, Unit 8, D#, Con Ed OH Dist Line-Z1.FCW In-Phase Component Of Voltage Drop Oc 01 Siemens Industry, Inc. All rights reserved. n Distribution System Engineering Course for Con Edison

265 Comparison Of Line Voltage Drop (In-Phase Versus Actual From Vector Calculations) Exact Voltage Drop on 1 Mile of OH Line: V DEXACT V VL S VOLTS % Error with Approximate Calculation: V DEXACT V D APPROX % ERROR * 100 *100 V DEXACT 0.51 % Guidelines for Voltage Drop Calculations: 1. When writing software for load flow and voltage drop analysis, always perform the exact vector calculations.. When doing hand calculations for voltage drop and estimating the voltage rise caused by a shunt capacitor, only the in-phase component of the voltage drop (rise) needs to be considered Distribution System Engineering Course for Con Edison Page Siemens Industry, Inc. All rights reserved.

266 Effect of Load Power Factor on Voltage Drop OH & UG Circuits Ob 3" 3" Oa Oc 48" R 1 = Ohms / mile X 1 = Ohms / mile 750 kcmil FLAT STRAP 50 MIL JACKET 0 X 0.05" X 0.175" FLAT STRAPS R 1 = OHMS / MILE = OHMS / MILE X 1 CABLE CIRCUIT-750 PARAMETERS.FCW n Increasing Load Power Factor with Same kva Loading Reduces Voltage Drop on Circuit Distribution System Engineering Course for Con Edison Page Siemens Industry, Inc. All rights reserved.

267 ANSI Voltage Parameters on 10 Volt Base ANSI C84.1 Voltage Standard (Allowable voltage limits for systems with service voltages less than 600 volts) Classification Range A (Normal) Range B (Emerg.) Service Voltage 114 to 16 volts 110 to 17 volts Utilization Voltage 110 to 15 volts 106 to 17 volts Voltage unbalance at the PCC or service entrance shall not exceed 3% under no-load conditions. The electric utility is responsible only for satisfying the service entrance or PCC voltage requirements! Customers are responsible for maintaining proper voltage downstream of the service entrance or PCC Distribution System Engineering Course for Con Edison Page Siemens Industry, Inc. All rights reserved.

268 ANSI Voltage Parameters on 10 Volt Base TO METER AND SERVICE FROM DISTRIBUTION XFR 00 AMPERE -POLE MAIN BREAKER 40 V 15 AMPERE 1-POLE FEEDER BREAKER 10 V HOT WIRE NEUTRAL WIRE WALL TOGGLE SWITCH EACH WIRE, #14 COPPER GROUNDING WIRE CEILING FIXTURE TO FIXTURE BOX AND METAL PARTS GROUND CONNECTION TO WATER PIPE AND/OR GROUND ROD METER RANGE A 114 TO 16 VOLTS UTILIZATION RANGE A 110 TO 15 VOLTS Distribution System Engineering Course for Con Edison Page Siemens Industry, Inc. All rights reserved.

269 Primary Feeder Voltage Profile Voltage Held Constant (Regulated) at Substation Bus Uniformly Distributed Load on Feeder 1800 KVAr Capacitor At 3.4 Mile Point. Location of 1800 kvar Bank that Maximizes Energy Loss Reduction for a Reactive Load Factor Of 0.75 is 3.54 Miles from Sub Note: 300 Amp at substation corresponds to a load of 6858 kva three phase (5486 kw and 4115 kvar) Distribution System Engineering Course for Con Edison Page Siemens Industry, Inc. All rights reserved.

270 Effect of Voltage Variations on Incandescent Lamps Lamp Rating Applied 10V 15V 130V Voltage Percent Percent Percent Percent Percent Percent (volts) Life Light Life Light Life Light The life and the lumen output of the incandescent lamp is directly affected by the utilization voltage: For this reason, customers close to a substation may comment that incandescent bulbs seems to burn out very often. In future, this will be less of a concern as compact fluorescent bulbs are more widely used Distribution System Engineering Course for Con Edison Page Siemens Industry, Inc. All rights reserved.

271 Fluorescent Lamps Operate satisfactorily over range of + 10% of ballast nameplate voltage rating + 1% change in applied voltage yields + 1% change in light output Life is less affected by voltage variations than incandescent lights Distribution System Engineering Course for Con Edison Page Siemens Industry, Inc. All rights reserved.

272 Voltage Unbalance Definitions - 3Φ Circuits NEMA Motor Standards Given Phase-to-Phase Voltages (No Zero-Sequence Component in Φ-Φ Voltages) V A = = 57.0 o V C = = 6.6 o V B = 30 3 = o PTI 006, D#4, VOLTUNB-.FCW MAXIMUM VOLTAGE DEVIATION FROM AVERAGE VOLTAGE % VOLTAGE UNBALANCE x 100 AVERAGE VOLTAGE AVERAGE VOLTAGE MAXIMUM DEVIATION 7.0 % V UNBALANCE VOLTS VOLTS x % % Distribution System Engineering Course for Con Edison Page Siemens Industry, Inc. All rights reserved.

273 Voltage Unbalance Definitions - 3Φ Circuits European Voltage Unbalance Factor Given Phase-to-Phase Voltages (No-Zero Sequence Component in Φ-Φ Voltages) V A = = 57.0 o V C = = 6.6 o V B = 30 3 = o 100 * Ratio of Magnitude of V to V 1 (Ratio of Sequence Voltage Magnitudes) 100 x S ( S VA ) ( S VB ) ( S VC ) N 3 Substituting into the Above Equations: Note: This calculation method does not require the use of complex number arithmetic. V 100 x 3.98 % V Page -14 V V x PTI 006, D#4, VOLTUNB-.FCW M M N N 1 V M A VB 6 V C V S A VB V C 01 Siemens Industry, Inc. All rights reserved Distribution System Engineering Course for Con Edison

274 Voltage Unbalance Definitions - 3Φ Circuits European Voltage Unbalance Factor CHECK ON REAL NUMBER FORMULA FOR RATIO OF THE ІV І TO THE ІV 1 І V 3 3 V A o o o o VB 30.0( ) o o o VC.0( ) j10 a e j10 o a e NEGATIVE-SEQUENCE COMPONENT OF PHASE-TO-PHASE VOLTAGES 1 o o o o V a V a V ( ).0( ) 1 0 A B C V O 3 POSITIVE-SEQUENCE COMPONENT OF PHASE-TO-PHASE VOLTAGES 1 o o o o V av a V ( ).0( ) 1 0 V1 A B C 3 V V A = 35 V C = = 6.6 o V B = 30 3 = o PTI 006, D#4, VOLTUNB-.FCW O V V 1 1 = 57.0 o Φ-Φ VOLTAGES WITH V 0 o % o Distribution System Engineering Course for Con Edison Page Siemens Industry, Inc. All rights reserved.

275 Voltage Unbalance Definitions - 3Φ Circuits Magnitude of V in Percent of Nominal V A = = 57.0 o V C = = 6.6 o V B = 30 3 = o Voltage Unbalance Definition 100 x V V NOMINAL PTI 006, D#4, VOLTUNB-.FCW Negative-sequence Component (V ) Of Phase-to-Phase Voltages V 3 Page -16 o V a V a V where : a and V A B C NOMINAL Calculation Of V Assuming V A Is At Angle Zero Degrees A o V V V B o o o o o V 04 3 V x 100 x % V 40 NOMINAL C o 01 Siemens Industry, Inc. All rights reserved Distribution System Engineering Course for Con Edison

276 Impact of Voltage Unbalance on Three-Phase Induction Motors 1. Produces 10 Hz Currents in Rotor. Increases Vibration And Losses in Rotor 3. Small Voltage Unbalance Gives Much Larger Unbalance in Line Currents 1 % Negative-Sequence Voltage Gives 4 to 7 % Negative-Sequence Current Temperature Rise In % In Motor Phase With Highest Current ΔT(%) T (%) * V (%) ΔV(%) ΔT(%) For Normal Life Expectancy, Induction Motors Should be De-rated if the Voltage Unbalance Exceeds 1 % Page -17 % Unbalance Typical Motor De-rating Motor Rating (% of Nominal) (HP) See NEMA MG of NEMA MG Siemens Industry, Inc. All rights reserved Distribution System Engineering Course for Con Edison

277 Factors Causing Voltage Unbalance in Distribution Systems UNSYMMETRICAL 3- PHASE LINE CONFIG. CAPACITOR BANK WITH BLOWN FUSE IN 1 Φ UNSYMMETRICAL DIST. TRANSFORMER BANKS SINGLE- PHASE LOADS ON 3-PHASE LINES OTHER FACTORS: 1.UNBALANCED LINE-TO-LINE AND LINE-TO-NEUTRAL LOAD ON THREE-Φ DISTRIBUTION TRANSFORMER.SINGLE-PHASE LATERAL LOADS NOT BALANCED 3.UNBALANCES IN SUBSTATION BUS VOLTAGES DUE TO NON-TRANSPOSED TRANSMISSION LINES Page Siemens Industry, Inc. All rights reserved Distribution System Engineering Course for Con Edison

278 Effect of Operating Voltage on Induction Motors BELOW RATED Nominal Voltage ABOVE RATED Distribution System Engineering Course for Con Edison Page Siemens Industry, Inc. All rights reserved.

279 Other Devices that are Affected by Voltage Variations Resistance Heating Elements Lower voltage reduces the effective power output Power Factor Correction Capacitors The amount of VARs supplied is based upon the square of the applied voltage Electronic Devices Low voltage will cause the TV screen to shrink High voltage can cause microprocessor malfunctions Distribution System Engineering Course for Con Edison Page Siemens Industry, Inc. All rights reserved.

280 Other Devices that are Affected by Voltage Variations Transformer Cores If the applied voltage is greater than the nominal voltage, the transformer core operates at higher flux densities and may become saturated o Increasing voltage increases the core losses of the transformer Core total losses = hysteresis losses + eddy losses Hysteresis losses proportional to core flux to 1.6 power (Steinmetz) Eddy losses proportional to square of flux o Increasing voltage above nominal voltage also increases the harmonics in the exciting current and the likelihood that harmonic problems will develop Distribution System Engineering Course for Con Edison Page Siemens Industry, Inc. All rights reserved.

281 Distribution System Engineering Course for Con Edison Voltage Regulation Page - 01 Siemens Industry, Inc. All rights reserved.

282 Why Should the Voltage be Regulated? Regulation was introduced to improve the voltage of the distribution system (maintain service voltages within allowable range, & control voltage swings) Systems grew in size and load Increased voltage drop, especially on 5 kv class primary circuits and on long rural circuits, creates problems Page -3 Back in late 50 s, one manufacture developed a voltage regulator for application on 10/40 volt secondary circuits in urban/suburban systems (Westinghouse uno-reg ). Concept was not adopted by utility industry. The voltage level and voltage variation at a customer is a measure of service quality Consistent voltage encourages consumption Proper and consistent regulation of the distribution circuit is important Most consumer equipment operates more efficiently at nominal or slightly below nominal voltage 01 Siemens Industry, Inc. All rights reserved Distribution System Engineering Course for Con Edison

283 Voltage Regulation Techniques 115 kv Supply LTC TRANSFORMER OR BUS REGULATOR 13. kv SUBSTATION BUS BREAKER SWITCHED CAPACITOR BANKS ALSO LOCATED ON SUBSTATION MV BUSES SWITCHED CAPACITOR OUT ON FEEDER V d? FEEDER SUPPLEMENTARY REGULATOR OUT ON FEEDER DEDICATED FEEDER REGULATOR IN STATION Distribution System Engineering Course for Con Edison Page Siemens Industry, Inc. All rights reserved.

284 Voltage Regulation Equipment Load Tap Changing (LTC) transformer in substation Substation bus regulators Switched capacitor banks on medium-voltage buses in distribution substation Voltage regulators in station for distribution primary feeders Voltage regulators out on the distribution primary feeders Distribution transformers with no-load taps Switched capacitors on distribution primary feeders Factors / Equipments that have an impact on primary feeder voltage drop Conductor size and configuration (affects feeder impedance) Circuit type (OH open wire versus underground cables) Fixed capacitors on distribution feeders Feeder load current and load power factor Distributed generation sources Circuit voltage level (The same voltage drop in volts produces less percent voltage drop on higher voltage circuits [7 kv versus 13 kv] ) Page Siemens Industry, Inc. All rights reserved Distribution System Engineering Course for Con Edison

285 Substation Application Choices LTC Transformers This is a combination of a traditional transformer with no load taps (deenergized taps) and a regulating transformer in one unit Drawbacks o One phase is sampled (voltage and current) o If the regulator equipment fails, requires the transformer to be taken out of service o Frequently regulates bus supplying multiple circuits of different lengths, construction, load levels and load characteristics Independent regulators installed between the substation transformer and the circuit to be regulated Bus regulators (Either three-phase unit or three single-phase units) Circuit (feeder) regulators Distribution System Engineering Course for Con Edison Page Siemens Industry, Inc. All rights reserved.

286 Bus Regulation vs. Feeder Regulation Bus regulation Install unit to control the bus voltage in substation Same voltage applied to all feeders on regulated bus(es) Good for substations that supply just secondary networks Less substation land required Less accurate for voltage control on individual feeders Circuit (Feeder) regulation Install unit to control voltage of a single feeder Can be tailored to impedance and load characteristics of each feeder More accurate than bus regulation for feeder voltage control More substation land required Distribution System Engineering Course for Con Edison Page Siemens Industry, Inc. All rights reserved.

287 Three-phase or Single-phase Regulators? The decision involves cost and application Three-phase units Works best for circuits serving balanced three-phase loads Phase control is gang operated: therefore one phase can be outside the desired bandwidth if significant voltage unbalances are present Single-phase units Allows the ability to set the phases independent of each other Works best on circuits that serve large percentage of single-phase loads where there is unbalance between phases Higher maintenance costs If regulator control on one phase malfunctions and goes to max boost or max buck position, can create significant voltage unbalance (both negative-sequence and zero-sequence voltages) Distribution System Engineering Course for Con Edison Page Siemens Industry, Inc. All rights reserved.

288 What is in a Regulator? The main components A main or exciting winding A series winding with taps A tap/switch that changes connections to the series winding and can operate under load A reactor (preventive auto transformer) A voltage sensing circuit Requires a current transformer, potential transformer and impedance elements if line-drop compensation is to be utilized A controller to operate the tap/switch Distribution System Engineering Course for Con Edison Page Siemens Industry, Inc. All rights reserved.

289 Voltage Regulator, Single Phase, ANSI Type A Typical regulator has a range of +/- 10% voltage One core with exciting winding and coupled series winding, Separate pot. xfr Each step gives a 5/8 % change in output voltage or 0.75 Volts on 10-volt base Position of Reversing Switch determines if output voltage is raised or lowered S L With varying INPUT voltage, objective is to maintain OUTPUT voltage within a prescribed bandwidth. INPUT (SOURCE) VOLTAGE EXCITING WINDING 1 REACTOR (PREVENTIVE AUTO ) SERIES WINDING REVERSING SWITCH 1 - Boost Position - Buck Position SL POT. XFR. OUTPUT (LOAD) VOLTAGE Notes: Separate potential transformer (POT XFR) supplies voltage to controller. Current transformers for line drop compensator circuits are not shown. Rated voltage of series winding is approximately 10 % of the rated voltage of the exciting winding for a +/- 10 % regulation Distribution System Engineering Course for Con Edison CAP D 010 #1, Regulator Single Core, Type A.FCW Page Siemens Industry, Inc. All rights reserved.

290 Voltage Regulator, Single Phase, ANSI Type B Typical regulator has a range of +/- 10% voltage Single-core unit with exciting winding and coupled series winding, ANSI Type B Each step gives a 5/8 % change in output voltage or 0.75 Volts on 10-volt base Position of Reversing Switch determines if output voltage is raised or lowered With varying INPUT voltage, objective is to maintain OUTPUT voltage within a prescribed bandwidth. INPUT (SOURCE) VOLTAGE S 1 REACTOR (PREVENTIVE AUTO ) SERIES WINDING REVERSING SWITCH 1 - Boost Position - Buck Position SL EXCITING WINDING POT WINDING L OUTPUT (LOAD) VOLTAGE Notes: Does not require a separate potential transformer to monitor output voltage. POT WINDING supplies output voltage to controller. Current transformers for line drop compensator circuits are not shown. Rated voltage of series winding is approximately 10 % of the rated voltage of the exciting winding for a +/- 10 % regulation Distribution System Engineering Course for Con Edison Page -31 CAP D 010 #1, Regulator For Grd Wye, Type B.FCW 01 Siemens Industry, Inc. All rights reserved.

291 Typical Three-Phase Regulators NORMALLY APPLIED WITHIN THE SUBSTATION FOR BUS REGULATION Distribution System Engineering Course for Con Edison Page Siemens Industry, Inc. All rights reserved.

292 Three Single-Phase Regulators in Distribution Substation for Bus Regulation S (Source Side) L (Load Side) SL (Common) Four-wire MGN circuits with regulators connected in grounded wye Distribution System Engineering Course for Con Edison Page Siemens Industry, Inc. All rights reserved.

293 Example of Single-Phase Regulators Platform Mounted Voltage Regulators in a Three-Phase Distribution Feeder Distribution System Engineering Course for Con Edison Page Siemens Industry, Inc. All rights reserved.

294 Example of Single-Phase Regulators Voltage Regulators in a Three-Phase 5 kv Rural Distribution Feeder SWITCHES TO BY- PASS REGULATORS FOR MAINTENANCE SINGLE-PHASE TRANSFORMER FOR CONTROL POWER TANK GROUNDING CONDUCTORS Distribution System Engineering Course for Con Edison Page Siemens Industry, Inc. All rights reserved.

295 Basic Components of a Typical Single-Phase Regulator Position Indicator Current Transformer (CT) Tap Changer Motor Tap Changer Preventive Autotransformer (PA) Potential Transformer (PT) Main Core/Coil Assembly (SERIES WINDING AND EXCITING WINDING) Control Panel Distribution System Engineering Course for Con Edison Page Siemens Industry, Inc. All rights reserved.

296 Distribution System Engineering Course for Con Edison Typical Regulator Control Panel Page Siemens Industry, Inc. All rights reserved.

297 Electronic & Microprocessor Control for Voltage Regulators Distribution System Engineering Course for Con Edison Page Siemens Industry, Inc. All rights reserved.

298 Key Concepts Application of Regulators Conservation Voltage Regulation (Reduction) Abbreviated to CVR CVR is a generic practice of reducing power demand and energy usage by providing electric service to the consumer at the lowest possible service voltage and still meeting the ANSI C84.1 Standard requirements The concept is to flatten and lower the voltage profile of the circuit via regulation techniques Accomplished with voltage regulators, and/or voltage regulators and capacitors (both fixed and switched) Making the feeder voltage profile more flatter and lower with CVR, generally makes the regulation scheme more costly Smart Grid with Advanced Metering Infrastructure (AMI) facilitates CVR with access to real time voltages at the point of service Distribution System Engineering Course for Con Edison Page Siemens Industry, Inc. All rights reserved.

299 Key Concepts - Application of Regulators Maintaining proper voltage at first and last customer on the feeder. Voltage level is dictated by ANSI standards and public utility commission regulations Must consider voltage drop in primary feeder, distribution transformers, secondary circuits, and services up to the meter With allowance made for sum of voltage drops in distribution transformers, secondary circuits, and services, there is a maximum voltage drop allowed on the primary feeder at peak load Urban and suburban circuits with long secondary circuits on distribution transformers Rural circuits with just services from the distribution transformers (usually more voltage drop allowed on primary feeder) Line drop compensation - minimize voltage variation for each customer Voltage is held constant, not at the regulator output terminal (L), but at a point that is downstream of the regulator Page Siemens Industry, Inc. All rights reserved Distribution System Engineering Course for Con Edison

300 Feeder Voltage Profile Example with Only a Supplementary Line Regulator Set Point = 14 Volt SUBSTATION SUPPLEMENTARY REGULATOR Distribution System Engineering Course for Con Edison Page Siemens Industry, Inc. All rights reserved.

301 Operational Considerations: Station Bus Regulators Substation bus voltage regulators (either LTC transformers or separate regulators) compensate for: 1.Voltage variations in the subtransmission/transmission system that supplies the substation.voltage drop through the substation transformer. Voltage drop increases with lower PF loads, and increasing impedance 3.Voltage drop in primary feeder if the regulator is equipped with a line-drop compensator If the bus regulators have continual buck, then the station bus voltage is too high. This can be corrected by changing the de-energized (no load) taps on the substation transformer. If the bus regulators have continual boost, then the substation bus voltage is too low. This also can be corrected by changing the de-energized (no load) taps on the station transformer Distribution System Engineering Course for Con Edison Page Siemens Industry, Inc. All rights reserved.

302 Operational Considerations Time delay settings Regulator should not respond to short duration voltage sags or swells. With voltage regulators in series, time delays of downstream regulators must be coordinated. Delays in the range of 30 to 90 seconds are typical. Bandwidth or sensitivity settings Voltage regulators do not hold the voltage constant, but control the voltage within a set bandwidth, with there being a set point, normally at the middle of the bandwidth. Page -43 Bandwidth cannot be smaller than smallest possible tap change amount. Each step (5/8 %) on 10-Volt basis is 0.75 Volts Most utilities use bandwidth of 1.5 to 3 volts. 01 Siemens Industry, Inc. All rights reserved Distribution System Engineering Course for Con Edison

303 Why Do Voltage Regulators Have Time Delay? The controller of the voltage regulator has a time delay feature Without time delay, rapid changes in the load demand could require the regulator to operate unnecessarily The regulator contacts are mechanical devices subjected to an arc when performing a position change After performing a large number of position changes, preventive maintenance may be required, or failures may occur The controller s objective is to maintain the voltage regulation without performing an excessive number of position changes Distribution System Engineering Course for Con Edison Page Siemens Industry, Inc. All rights reserved.

304 30-Second Time Delay Setting The voltage level must exceed the bandwidth upper or lower limit for 30 seconds before the tap changer will respond: In the case below, no tap change occurs as voltage exceeds bandwidth for just 0 seconds VOLTAGE AT CONTROLLER Page BANDWIDTH UPPER END SET POINT BANDWIDTH LOWER END BANDWIDTH SET 1.5 VOLTS 0 SEC CONTROLLER VOLTAGE TIME IN SECONDS 01 Siemens Industry, Inc. All rights reserved Distribution System Engineering Course for Con Edison

305 Time Delay Settings Used When Cascading Regulators STATION REGULATOR SUPPLEMENTARY REGULATOR SUBSTATION 30 SECOND TIME DELAY 40 SECOND TIME DELAY When cascading regulators, be sure to allow operation room. Allow the station regulators to operate before any supplemental line regulators: The first regulator (closest to substation ) must operate first. The second regulator must be delayed long enough to allow the voltage to change but not so long as to be outside the response of the first Distribution System Engineering Course for Con Edison Page Siemens Industry, Inc. All rights reserved.

306 Selecting the Appropriate kva Ratings The key concept is that the kva rating of the regulator (KVA REG ) is related to the boost or buck voltage (KV BOOST ) of the regulator and the line current (I LOAD ): KVA REG KV BOOST I LOAD The actual kva passing through the regulator (the through kva) is equal to the output voltage of the regulator in kv multiplied by the current in amperes passing through it Distribution System Engineering Course for Con Edison Page Siemens Industry, Inc. All rights reserved.

307 Equations to Calculate Regulator kva Ratings Single-Phase Regulator (connected line-to-ground) KVA REGULATOR I LOAD KV L-G Regulator Range (PU) 1- Circuit Through KVA x Regulator Range (PU) Three-Phase Regulator on 3- Wire or 4-Wire System KVA REGULATOR I LOAD KV L-L 1.73 x Regulator Range (PU) 3 - Circuit Through KVA x Regulator Range (PU) KVA REGULATOR I LOAD KV OR L-G 3.0 x Regulator Range (PU) Open Delta - Consists of Single-Phase Regulators Each Regulator KVA ILOAD KVL -L Regulator Range(PU) Distribution System Engineering Course for Con Edison Page Siemens Industry, Inc. All rights reserved.

308 Regulator Rating (Three 1Φ Regulators) With Grounded-Wye Connection 4160 V System (3Φ), 4-Wire MGN, 000 kva 3Φ Load (78 Amps) + 10% Regulation, Three 1Φ Regulators S L 78 AMPERES Note: 000 I LOAD A INPUT MGN Calculation of Each 1Φ Regulator Rating: Range (PU) KV I / 3 x x kva Specification: Three 1Φ Regulators, Each kva, 400V, + 10% Regulation Page -49 S KVA REG 1 SL L SL SL SINGLE-PHASE REGULATOR MULTI-GROUNDED NEUTRAL S L-G L LOAD 4160 VOLT 000 KVA LOAD OUTPUT Three single-phase regulators can be connected in wye only in the 4-wire multi-grounded neutral system. The neutral of the regulators wye windings (exciting windings) must be connected to the system neutral conductor. 01 Siemens Industry, Inc. All rights reserved Distribution System Engineering Course for Con Edison

309 Regulator Rating (Two 1Φ Regulators) With Open Delta Connections 4800 V System 3Φ, 3-Wire, 000 kva 3Ø Load (40 Amps) + 10% Regulation, Two 1Φ Regulators 000 I LOAD A INPUT S SL SL L Calculation of Each 1Φ Regulator Rating: KVA S REG-1 L 40 AMPERES Range KVL-L I kva Specification: Two 1Φ Regulators, Each 115 KVA, 4800V, + 10% Regulation LOAD 4800 VOLT 000 KVA LOAD OUTPUT Note: To regulate 000 kva load in three-phase 4800 volt circuit with two singlephase regulators connected in Open Delta, the total kva rating of the two regulators is 30 kva Distribution System Engineering Course for Con Edison Page Siemens Industry, Inc. All rights reserved.

310 Regulator Rating, Three 1Φ Regulators Connected in Delta, ANSI TYPE B INPUT 4800 V System (3 Wire ), 000 kva 3Φ Load, 40 Amps Three 1Φ Regulator, Each Regulator + 10% Regulation 40 AMPERES Calculation of Each 1Φ Regulator Rating: KVA REG 1 Range (PU) KVL-L I kva Specification: Three 1Φ regulators, Each 115 kva, 4800V, + 10% regulation. Total required regulator kva is 150 % of that with open delta. Page -51 S S L SL L SL TYPE "B" BOOSTING SL L S 4800 VOLT 000 KVA LOAD LOAD OUTPUT 000 I LOAD Notes: Three single-phase regulators can not be connected in floating wye for application in a 3-wire system (ungrounded or uni-grounded system). With the range of each regulator being + 10%, 16 step, 5/8 % per step, the Φ-Φ output voltage range is + 15 %, % per step. 01 Siemens Industry, Inc. All rights reserved. A Distribution System Engineering Course for Con Edison

311 Three Single-Phase Regulators Connected in Closed Delta - Wiring Connections A A' External wiring connections are the same for ANSI Type A and ANSI Type B regulators B C INPUT S L SL S L SL S L SL B' C' OUTPUT With either type regulator (A or B), controlled voltage is the phase-to-phase on the output side (Voltage between Terminal L and Terminal SL ) TYPE A REG 1 REG REG 3 TYPE B INPUT (SOURCE) VOLTAGE S EXCITING WINDING 1 REACTOR (PREVENTIVE AUTO ) SERIES WINDING REVERSING SWITCH 1 - Boost Position - Buck Position SL POT. XFR. L OUTPUT (LOAD) VOLTAGE REG 1 (A TO C) REG (B TO A) REG 3 (C TO B) INPUT (SOURCE) VOLTAGE S 1 REACTOR (PREVENTIVE AUTO ) SERIES WINDING REVERSING SWITCH 1 - Boost Position - Buck Position SL EXCITING WINDING POT WINDING L OUTPUT (LOAD) VOLTAGE Distribution System Engineering Course for Con Edison Page -5 CAP D 010 #1, Regulator Single Core, Type A.FCW CAP D 010 #1, Regulator For Grd Wye, Type B.FCW 01 Siemens Industry, Inc. All rights reserved.

312 Regulator Rating (Three Phase Regulator) 4800 V System, 000 kva 3Φ Load, 40 Amps Three-Phase Regulator, + 10% Regulation S L 40 AMPERES 000 I LOAD A INPUT S THREE-PHASE REGULATOR +/- 10 % REGULATION OF PHASE-TO-PHASE VOLTAGE L 4800 VOLT 000 KVA LOAD OUTPUT S Calculation of 3Φ Regulator Rating: KVA REG 3 ILOAD KVL -L 1.73 x Regulator Range (PU) 3 - Circuit Through KVA x Regulator Range (PU) Page x 4.8 x 3 x kva Specification: Three-phase regulator, 00 kva, 4800V, + 10% regulation. Rating is 86.7% ( 3/) of the kva rating of the two regulators used in the open delta bank (each rated 115 kva). L 01 Siemens Industry, Inc. All rights reserved Distribution System Engineering Course for Con Edison

313 Voltage Regulator Control-Block Diagram Regulators Can Be Configured to Regulate the Voltage At: The Regulator Output Terminals A Downstream Location Remote From The Regulator Terminals LOAD CURRENT TRANSFORMER REFERENCE VOLTAGE Page -54 POTENTIAL TRANSFORMER TAP CHANGER LINE DROP COMPENSATOR TAP-CHANGING MOTOR VOLTAGE SENSOR TIME DELAY Line drop compensator (LDC) is to allow the voltage regulator to control the voltage at a point that is remote and downstream of the regulator rather than at the output terminals of the voltage regulator. 01 Siemens Industry, Inc. All rights reserved Distribution System Engineering Course for Con Edison

314 Voltage Regulator Control Regulating Voltage at a Downstream Point Voltage Applied To Regulating Relay, V REG Replicates The Voltage At The Downstream Remote Point, V E, That Is To Be Regulated. Objective is to Control Voltage V E Rather Than Regulator Output Voltage V S LOAD TAP CHANGER I L R L X L I L VS I L N CT R COM VS N VT X COM V REG CE PTI 006, REGULATOR BASICS LDC-1.FCW V E Distribution System Engineering Course for Con Edison Page Siemens Industry, Inc. All rights reserved.

315 Voltage Drop Concepts & Calculating Voltage Drop Distribution System Engineering Course for Con Edison Page Siemens Industry, Inc. All rights reserved.

316 Primary Circuit Voltage Considerations Amount of load current: peak, high, normal, light Load power factor can have a significant impact on the voltage drop along a distribution feeder Type of circuit Impedance of the circuit (OH line or cable) has a significant impact on the voltage drop What type of regulation LTC Bus Circuit Supplemental Distribution System Engineering Course for Con Edison Page Siemens Industry, Inc. All rights reserved.

317 Voltage Drop Constant Current in Line Segment R X I L V S V R LINE SIMPLE VD.FCW Approximate Voltage Drop Equation, V D (Difference between magnitude of sending end voltage V S, and magnitude of receiving end voltage V R ) V V V I R cos X sin Volts D S R L R = Line Resistance Between S and R in Ω X = Line Reactance Between S and R in Ω I L = Magnitude of Line Current in Amperes θ = Angle Between Current I L and Voltage V R (positive for lagging pf load) Percent Voltage Drop On Line (%V D ) VD % VD * 100 V Page -58 LG V L-G = NOMINAL LINE-TO-GROUND VOLTAGE IN VOLTS 01 Siemens Industry, Inc. All rights reserved Distribution System Engineering Course for Con Edison

318 FEEDER VOLTAGE PROFILE Concentrated Load at the End of the Feeder I S L I S FEEDER CURRENT VOLTS V ( x) I r cos x sin x D S L L LIGHT LOAD i( x) I S Ampere V D (x) PEAK LOAD Voltage Drop Per Unit Length Of Feeder Is Equal At Any Point Along The Feeder. Feeder Voltage Profile Is Straight Line. Voltage Drop to any Point x on Feeder, V D (x), Is: I s r L x L V cosθ x SUB CE PTI 006, VOLT PROFILE, LUMPED LOAD.FCW D Page -59 r cos x sin x Volts ( x) I S L x L = Load Current Magnitude In Amperes (Constant Along Feeder) = Resistance In Ω Per Unit Length = Reactance In Ω Per Unit Length Note: = Load Power Factor In Per Unit Voltage held constant at substation (No LDC) = Distance From Substation L 01 Siemens Industry, Inc. All rights reserved Distribution System Engineering Course for Con Edison

319 FEEDER VOLTAGE PROFILE Uniformly Distributed Load on the Feeder L I S FEEDER CURRENT VOLTS LIGHT LOAD x i ( x ) IS 1 L Ampere V D (x) PEAK LOAD SUB E PTI 006, VOLT PROFILE, UNIFORM DIST LOAD.FCW Voltage Drop Per Unit Length of Feeder is Greatest at Sub and Decreases with Increasing Distance from Substation. Voltage Drop to any Point X on Feeder, V D (x), Is: VD ( x) IS rl cos x x L sin x L I S = Load Current Magnitude In Amperes At Substation r L = Resistance In Ω Per Unit Length x L = Reactance in Ω Per Unit Length cosθ = Load Power Factor In Per Unit x = Distance From Substation Page -60 x L Note: Voltage held constant at substation (No LDC) 01 Siemens Industry, Inc. All rights reserved Distribution System Engineering Course for Con Edison

320 Feeder Voltage Profile Lumped Load & Uniformly Distributed Load Voltage Drop To Feeder End (x = L) With Same Load Current I s At Substation V D Lumped Load IS rl cos xl sin L VD ( L) IS rl cos xl sin L IS rl cos xl sin ( L) Uniformly Distributed Load L L L Distribution System Engineering Course for Con Edison Page Siemens Industry, Inc. All rights reserved.

321 FEEDER VOLTAGE PROFILE Calculation of Voltage Drop to Feeder End Feeder Data: 397 ACSR PHASE CONDUCTOR r L = 0.53 Ω/mile x L = Ω/mile L = 5 miles Substation Load Current I S : I S = 00 Ampere, PF = 0.80 PU Primary Line-to-Neutral and Secondary Nominal Voltages: Voltage Drop with Lumped Load at End: VD ( x) IS rl cos xl sin x VD (5) * *0.60* Volts Pr imary VS 10 VD Volts Secondary VP 760 (Voltage Drop to End of Feeder is 9.14 Volts on a 10-volt Base) Voltage Drop With Uniformly Distributed Load: V P =760 V V S = 10 V At Feeder End, Voltage Drop With Uniformly Distributed Load Is 1/ Of That With Lumped Load Or 4.57 Volts On 10-volt Base Distribution System Engineering Course for Con Edison Page Siemens Industry, Inc. All rights reserved.

322 Basic Application Guideline A uniformly distributed load on a feeder causing current I S at the substation will create at the feeder end half of the voltage drop of a load lumped at the end of the feeder drawing current I S. Usually, the distribution circuit is not uniformly loaded. However, it can be reduced to sections that are assumed to be uniformly loaded and then the lump sum equivalent would be used at specific distances Distribution System Engineering Course for Con Edison Page Siemens Industry, Inc. All rights reserved.

323 Line Drop Compensator Settings Determine Location On Feeder At Which Voltage Is Regulated: Compensator Can Reduce the Voltage Variation (Swing) Seen by Last Customer On Feeder Between Light Load and Heavy Load Conditions With the Resistance (R L ) and Reactance (X L ) of the Actual Circuit Between the Regulator and Regulation Point Known, a Compensator Setting is Made in Either Ohms or Volts That Allows the Regulator to Control the Voltage at the Regulation Point Rather Than at the Regulator Terminals LOAD TAP CHANGER VS I L I L N CT R COM VS N VT X COM R L X L I L V REG CE PTI 006, REGULATOR BASICS LDC-1.FCW V E Distribution System Engineering Course for Con Edison Page Siemens Industry, Inc. All rights reserved.

324 Feeder Voltage Profiles Impact of Line Drop Compensator Setting Feeder With Uniformly Distributed Load Voltage Regulated at Substation Compensator Settings = 0 Voltage Regulated at 1.6 Mile Point Compensator Finite Settings Line Drop Compensator can Reduce Voltage Swings Seen by Many Customers on the Feeder Distribution System Engineering Course for Con Edison Page Siemens Industry, Inc. All rights reserved.

325 LDC - Line Drop Compensation LDC is used to correct for voltage drop on the circuit: Best starting point for regulating voltage is the point on the feeder where the voltage drop from the source to the regulation point is equal to the voltage drop from the regulation point to the end of the feeder Requires knowing the circuit length, circuit impedances, and the voltage profile for the circuit. Need to know the existing voltage level before changing the compensation. Must factor in effect of fixed and switched capacitor banks. Watch out for the voltage regulation impact on distribution transformers with primary taps. Must know the CT and PT ratio of the regulator control circuit. Line Drop Compensator settings are made in either Ohms or Volts; Distribution System Engineering Course for Con Edison Page Siemens Industry, Inc. All rights reserved.

326 LINE DROP COMPENSATOR SETTINGS Setting With Constant Load Current to Reg. Point Power and Control Circuits LOAD TAP CHANGER I L R L X L I L I L N CT VS VS N VT R COM X COM V REG = voltage applied to regulating relay V E V S = Output Voltage Of Regulator (Load Tap Changing Xfr) In Volts V E = Voltage At Regulation Point In Volts I L = Line Current Magnitude In Amperes (Constant To Regulation Point) = Line Current Angle ( + For Lagging Power Factor Loads) R L = Line Resistance To Regulation Point In Ohms X L = Line Reactance To Regulation Point In Ohms Actual Voltage At Regulation Point: V E V S CE PTI 006, REGULATOR BASICS LDC-1.FCW cos j sin V I R cos I X sin ( RL j XL ) IL S L L L L (1) Distribution System Engineering Course for Con Edison Page Siemens Industry, Inc. All rights reserved.

327 LINE DROP COMPENSATOR SETTINGS Setting with Constant Load to Reg. Point LOAD TAP CHANGER I L R L X L I L I L N CT VS R COM X COM V E VS N VT V REG CE PTI 006, REGULATOR BASICS LDC-1.FCW Voltage (V REG ) Applied to Regulating Relay for LTC Transformer N VT N CT R COM X COM Page -68 VS IL IL VREG RCOM cos XCOM sin NVT NCT NCT = Voltage Transformer Ratio = Current Transformer Ratio = Regulator Compensator Resistance Setting In Ohms = Regulator Compensator Reactance Setting In Ohms Divide both sides of eq.(1), which gives the actual voltage at the regulation point, V E, by N VT to obtain the regulation point voltage reflected to the secondary side of the voltage transformer: V N E VT V N S VT I N L VT R L IL cos XL sin N VT () (3) 01 Siemens Industry, Inc. All rights reserved Distribution System Engineering Course for Con Edison

328 LINE DROP COMPENSATOR SETTINGS Setting with Constant Load to Reg. Point LOAD TAP CHANGER I L R L X L I L VS I L N CT R COM VS N VT X COM V REG V REG V N E VT V E CE PTI 006, REGULATOR BASICS LDC-1.FCW If the Voltage at the Regulating Relay, V REG, is to Replicate the Voltage at the Regulation Point, V E VE VREG N VT From Eq () and Eq (3), this is satisfied when: NCT N RCOM RL and X COM N N VT (4) EQ. (4) Gives the Line Drop Compensator Settings R COM and X COM in OHMS: R L = Line Resistance In Ohms To Regulation Point X L = Line Reactance In Ohms To Regulation Point CT VT X L Distribution System Engineering Course for Con Edison Page Siemens Industry, Inc. All rights reserved.

329 LINE DROP COMPENSATOR SETTINGS Setting with Constant Load to Reg. Point LDC Settings in Ω Current Transformer Parameters CT P = CT Primary Rated Current In Amperes Page -70 CT S = CT Secondary Rated Current In Amperes CTP NCT CT S Line Drop Compensator Setting in Volts (Voltage across compensator elements with rated current in CT secondary) Place (5) into (4) and multiply both sides by CT S X R COM COM CT CT S S CT N P VT CT N P VT R X L L R COM COMPENSATOR COMPENSATOR RESISTANCE REACTANCE R L = Line Resistance In Ohms To Regulation Point X L = Line Reactance In Ohms To Regulation Point Other Regulator Settings: Set Point Voltage Bandwidth Time Delay Polarity (Reversed Reactance And Reversed Resistance) N N CT VT R L and X COM N N CT VT SETTING IN X L VOLTS SETTING IN VOLTS (4) (5) 01 Siemens Industry, Inc. All rights reserved Distribution System Engineering Course for Con Edison

330 Voltage Regulating Relay for Unit Substation Applied with Load Tap Changing (Ltc) Transformer for Regulating 4160 Volt System Voltage LD Compensator Resistance Volts LD Compensator Reactance Volts LD Compensator Reverse Reactance Switch Voltage Level (set point) Bandwidth Adjust Distribution System Engineering Course for Con Edison Page Siemens Industry, Inc. All rights reserved.

331 Example - Line Drop Compensation LOAD TAP CHANGER I L R L X L I L R L = 1.5 VS I L N CT R COM VS N VT X COM V REG X L =.0 N VT = N CT = V E CE PTI 006, REGULATOR BASICS LDC-.FCW Express R COM, X COM in Ohms: Express LDC resistance and reactance settings in Volts: Page -7 R X R X COM COM COM COM CT CT C.T. RATIO R L 1.5 ( ) 37.5 P.T. RATIO C.T. RATIO X L ( ) 50 P.T. RATIO S S R X L L C.T. PRI. RATING P.T. RATIO C.T. PRI. RATING P.T. RATIO 1.5 x x NOTE: THESE CALCULATIONS ARE FOR THE SITUATION WHERE THE CURRENT IN THE LINE BETWEEN THE REGULATOR AND THE REGULATION POINT IS CONSTANT. 10 Volts Volts 01 Siemens Industry, Inc. All rights reserved Distribution System Engineering Course for Con Edison

332 LDC Settings to Hold Voltage at Any Distance x From Regulator Output Terminals Lumped Load at End of Line: Current in line section constant at I S LUMPED LOAD AT END OF LINE L r L & x L (LINE RES. & REAC. IN OHMS PER UNIT LENGTH) I S I S V D x lum ( x ) I lum S rl cos xl sin xlum CE PTI 010, LUMPED LOAD AT END.FCW X lum = distance to regulation point with lumped load. With the current constant (I S ) in the line between the regulator and the point where the voltage is to be regulated: Calculate the line-drop compensator settings from the circuit resistance and reactance from the regulator to the regulation point, where x lum = distance from the regulator to the regulation point. Page -73 CT CT R r * x X x * x COM N N VT L lum CIRCUIT RESISTANCE TO REGULATION POINT Remember: Compensator setting in volts is the voltage across the compensator element with CT rated secondary current in the compensator element. COM N N VT L lum CIRCUIT REACTANCE TO REGULATION POINT 01 Siemens Industry, Inc. All rights reserved Distribution System Engineering Course for Con Edison

333 LDC Settings to Hold Voltage at Any Distance x From Regulator Output Terminals Uniformly Distributed Load On Line (Regulator sees only current I S at source) I S r L & x L (LINE RESISTANCE & REACANCE IN OHMS PER UNIT LENGTH) Ix ( uni ) 0 I S x uni r L x uni x L x uni Ix ( uni ) L = LINE TOTAL LENGTH L I L N CT Voltage drop to regulation point at distance X uni from source end: Page -74 VS V D ( x uni ) VS N VT I S R COM X COM V REG CE PTI 006, REGULATOR BASICS LDC-3.FCW uni r L cos xl sin xuni L x X uni = distance to regulation point with uniformly distributed load on feeder of length L. V E 01 Siemens Industry, Inc. All rights reserved Distribution System Engineering Course for Con Edison

334 LDC Settings to Hold Voltage at Any Distance x From Regulator Output Terminals Uniformly Distributed Load On Line (Regulator only sees current I S at source) L r L & x L (LINE RES. & REAC. IN OHMS PER UNIT LENGTH) 0 I S xuni CE PTI 010, UNIFORMLY DISTRIBUTED LOAD.FCW With uniformly distributed loading on the feeder, the LDC settings, R COM and X COM must be made such that the voltage applied to the regulating relay, V REG, replicates that at the regulation point at a distance of x uni from the source end. EQUATIONS FOR COMPENSATOR SETTINGS IN OHMS WHERE x uni = DISTANCE TO REGULATION POINT WITH A UNIFORMLY DISTRIBUTED LOAD R COM N N CT VT r L x uni x uni L X COM N N CT VT x L x uni x uni L EQUATIONS FOR COMPENSATOR SETTINGS IN VOLTS WITH A UNIFORMLY DISTRIBUTED LOAD R COM CT S CT N P VT r L x uni xuni L Volts X COM CT S CT N P VT x L x uni xuni L Volts Distribution System Engineering Course for Con Edison Page Siemens Industry, Inc. All rights reserved.

335 LDC Settings to Hold Voltage at Midpoint of Feeder, Uniformly Loaded Line At Midpoint of Feeder (x uni = L/ in eqs on previous slide): N R 3 NCT 3 CT COM rl L X x L COM L N 8 NVT 8 VT L = Feeder Length To maintain voltage at feeder midpoint with uniform loading on line, set LDC to 3/8 Z Ohms, where Z is the Ohms to the end of the feeder. In contrast, with lumped load at end, LDC is set to 1/ Z Ohms to maintain voltage at feeder midpoint. Example below assumes L = 4 miles 3 L Distribution System Engineering Course for Con Edison Page Siemens Industry, Inc. All rights reserved.

336 LDC Settings to Hold Voltage So V D From Reg. to Reg. Point is Same as V D From to Point to Feeder End Uniformly Loaded Feeder - x = Point On Feeder Where Drop From Regulator to Point x Equals Drop From Point x to End of Feeder Voltage drop from regulator to regulation point x V D ( x) I S r L cos xl sin x L Voltage drop from regulator to end of feeder (set x = L) L V ( L) I V D D S ( x to L) V rl cos xl sin Voltage drop from regulation point x to end of feeder D ( L) V D ( x) I S x L r x I r x x L L cos L sin S L cos L sin Set equation (3) equal to equation (1) and solve quadratic equation for distance x. x 1 L L Compensator Resistance setting with uniformly distirbuted load:: R COM N N CT VT r L x x L Place value of x given by eq. (4) into eq. (5), and solve for R COM : R COM N N CT VT 1 4 r L x L (1) () (3) (4) (5) Distribution System Engineering Course for Con Edison Page Siemens Industry, Inc. All rights reserved.

337 LDC Settings to Hold Voltage So V D From Reg. to Point is Same as V D From Point to Feeder End Uniformly Loaded Feeder - x = Point On Feeder Where Drop From Regulator to Point x Equals Drop From Point x to End of Feeder x 1 L 0. 99L R COM N N CT VT 1 4 r L L X COM N N CT VT 1 4 x L L L = Feeder Length To maintain voltage at point where drop from regulator to point is the same as the drop from the point to feeder end, set LDC to 1/4 Z Ohms, where Z is the Ohms to the end of the feeder Distribution System Engineering Course for Con Edison Page Siemens Industry, Inc. All rights reserved.

338 Calculation of Line Drop Compensator Settings for Open Delta Regulator Configuration Step 1: Use the 3Φ LDC equations and determine R COM and X COM Step : Adjust the calculated results as follows: Case 1 For the Leading Regulator In the Open Delta Bank: R COM =0.866 R COM X COM X COM = X COM R COM Case For the Lagging Regulator In the Open Delta Bank: R COM = R COM X COM X COM = X COM + 0.5R R COM Distribution System Engineering Course for Con Edison Page Siemens Industry, Inc. All rights reserved.

339 Secondary Systems and Service Drops Distribution System Engineering Course for Con Edison Page Siemens Industry, Inc. All rights reserved.

340 Distribution System Engineering Course for Con Edison Secondary Layouts Page Siemens Industry, Inc. All rights reserved.

341 Single-Phase Distribution Transformer / Secondary Design Layout Considerations Size of the Distribution Transformer Number of Customers (services) per Transformer Primary Voltage Level at the Distribution Transformer Voltage Drop Through Distribution Transformer Affected by transformer impedance, load level, load power factor Voltage Drop on Secondary Lines and Services Affected by conductor size, conductor spacing (reactance), circuit length, load level, load power factor Voltage Flicker at Services During Motor Starting Affected by xfr impedance, conductor size, spacing, and circuit length Short Circuit Current Design Limits at the Service Entrance Affected by xfr impedance, conductor size, spacing, and circuit length Goal: Optimize The Design Of The Transformer/ Secondary System For Minimum Cost, Maintaining Voltage Within Allowable Limits Page Siemens Industry, Inc. All rights reserved Distribution System Engineering Course for Con Edison

342 Typical Maximum Voltage Drops of System Components Residential wiring Service drop Secondary Distribution transformer Primary feeder Total drop (sum of all) -4 volts 1- volts -3 volts -3 volts -3 volts 9-15 volts As you review the circuit voltage profile, keep in mind that, for 10 V distribution systems, the service entrance or PCC voltage should not be below 114 volts during peak load conditions. The peak load condition of the customer may not be at the same time as the peak load of the circuit. Voltage regulator bandwidth must be considered in analysis Distribution System Engineering Course for Con Edison Page Siemens Industry, Inc. All rights reserved.

343 Overall Feeder Voltage Profile (10-volt base) Substation First Customer Primary Feeder Distribution Transformer Secondary Last Customer 18 Upper ANSI limit (16V) Voltage Page First Customer This point should not exceed ANSI limits Lower ANSI limit (114V) This point should not be less than ANSI limit Last Customer When analyzing voltages at secondary, and allocating maximum voltage drops, the engineer must recognize that the regulators control the voltage within a given bandwidth. 3 Volts Primary Feeder 3 Volts Distribution Transformer 3 Volts Secondary & Service Drop 4 Volts Customer Wiring 01 Siemens Industry, Inc. All rights reserved Distribution System Engineering Course for Con Edison

344 Distribution System Engineering Course for Con Edison ANSI/IEEE C84.1 Voltage Requirements Service Voltage Ranges on 10-Volt Base Representative Equipment Rated Voltages Page Siemens Industry, Inc. All rights reserved.