Module 3 Voltage Regulation. By: Dr. Hamid Jaffari

Transcription

1 Module 3 Voltage Regulation By: Dr. Hamid Jaffari

2 Primary Voltage Considerations Transmission Voltage Impacts the distribution bus voltage Amount of load current peak, high, normal, light Type of conductor Impedance of the conductor plays significant role What type of regulation LTC Bus Circuit Supplemental 2

3 Voltage Drop - DC Remember Ohm s Law? The voltage drop calculation is quite easy Only need to know the resistance of the conductor, the load current and the sending voltage V I R V L V S ( I R) 3

4 Voltage Drop DC If the voltage is constant (DC) there is only the line resistance contributing to the voltage drop Sending Voltage = V S line impedance R Receiving Voltage = V L Load V D = I*R V L V S Volt Drop Vs - V L IR 4

5 Feeder Voltage Profile The voltage profile of the circuit is not constant throughout the load cycle 5

6 Voltage Drop Equation-AC (approximate) One line representation Distribution Transformer Sending Voltage = V S line impedance R jx Receiving Voltage = V L Load Graphical Representation : VL V S I (R jx) Note VL V Reciving & V S V Source -jix -IR VSVL V Drop I (R jx) V L θ Error V S V V S V -jθ I Ie S R V0 I (R jx) IRCos jixsin I V V V V V S s S S V V V V Drop R R R R V IR COS Ø IZ IX SIN Ø (ICosθ jisinθ) x(r jx) IRCos jix Cos jir Sin IX Sin I( RCos S V R I( RCos X Sin ) ji ( X Cos R Sin ) X Sin ) negligible V V Drop Drop (Line - to-ground) (3- phase) 3 xv Drop I( RCos X Sin ) for one conductor ( Line to Ground) 6

7 Voltage Drop - AC Vector diagram for lagging & leading situations Could you identify? One line representation Distribution Transformer Sending Voltage = V S line impedance R jx -j IX Receiving Voltage = V L -IR Load V L Vector representation I θ V S Vector representation I θ V L -IR V S -jix 7

8 Voltage Drop - AC Power line has inductive (jxl ) and/or capacitive (-JXc) Reactance The voltage drop can be reduced with the application of a capacitor bank correcting for line inductance The example here still has a lagging power factor Distribution Transformer line impedance Kvar Bank -jx One line representation R jx V V L L V V S S - I ( R I R j I ( X jxl-jxc) V L XC) S - I R-j I X L j I X C Sending Voltage = V S Receiving Voltage = V L Load I Ie V S - j V0 Vector representation I θ VL VS -IR 8

9 Voltage Drop Calculation Distributed load on the line Where: L 1 is distance from source to desired point L is the total circuit distance θ is the average power factor angle or angle between Voltage & Current. Z RCos XSin is the line impedance to the point in question and θ is the angle between the voltage and the current. Three Phase % V d kva( R cos X sin ) L 10 kv l l L1 2L Single Phase % V d kva( R cos X sin ) L kvl gnd 2L 1 L 1 9

10 Facts about Power Systems Distribution systems are dynamic Circuit loads change with the load cycle Circuits have various load types Constant Impedance Constant Current Constant KVA Customers expand and contract Circuits change configuration Transmission voltages are not constant (i.e. Voltage Drop Calculator for DEED.xlsx) (i.e. VoltageDropCalculator.xls) (I.e. Motor Application Tools.xls) There are many different styles and sizes of XFMRs The distance between the loads is not the same So, how can we reduce calculations? 10

11 A Typical Radial Distribution System A Circuit supplies many loads with varying density R Main line 3Ø Lateral taps - 3Ø, 2Ø, and/or 1Ø 11

12 Case 1: Concentrated load at end of feeder The ANSI C84.1 voltage range must be maintained Distribution Transformer line impedance Receiving Voltage = V L R jx Sending Voltage = V S Concentrated Load Minimum Load Current Maximum Load Current Max Volts Min Supply Load 12

13 Basic Application Guideline A uniformly distributed load on a feeder will create half of the voltage drop of a similar load lumped at the end of the feeder. However, the distribution circuit is not uniformly loaded. However, it can be reduced to sections that are assumed uniformly loaded and then the lump sum equivalent would be used at specific distances. 13

14 Modeling Your Circuit Lumped load Method If you combine all of the load in a switchable section at the switch, a good worst case model can be developed Use peak demand for the greatest voltage drop Use light load for the least voltage drop Do not forget to add in capacitors 14

15 Lumped Load Modeling Diagram kva1 kva2 kva3 kva4 Circuit1 Station B L1 L2 L3 L4 L5 DISTRIBUTED LOAD Volts Max Min Supply Distance Tie Point 15

16 Lumped Load Modeling Steps One can assume a uniformly distributed load Assume balanced 3-phase load. Add KVA total per phase (ᶲA,ᶲB,ᶲC) Take the total load of the circuit and divide by the distance in feet. One can argue that only the three phase line needs to be used but there are many instances where a very long single phase tap can have excessive primary voltage drop. The load is attached to the model in even increments at equal distances. The smaller the distance divisions, the more intervals that require calculation 16

17 Secondary Networks and Service Drops 17

18 Secondary Layout considerations Primary voltage applied at the transformer terminals Do not assume rated voltage Distance and conductor from the transformer to the PCC or service entrance of the customer For single phase services the V d is doubled Other customers that are served on the same secondary conductor. Be aware of the load limit of the secondary cable Special equipment and motors 18

19 Secondary Layouts Try to locate the transformer at the load center This does not depend upon the number of service drops It is dependent upon the demand of each service Hs 1 Hs 3 Hs 5 Hs 7 150ft 170ft Hs 2 Hs 4 Hs 6 Hs 8 kva 19

20 Secondary Layout Problem Where would the transformer be placed if the load was as follows House #4 installs Central air Secondary is 0.16ohms per thousand feet Service cables are 100ft and 0.39ohms per thousand feet Hs 1 4.2kW Hs 3 2kW Hs 5 1.6kW Hs 7 2.9kW Hs 9 4.2kW 150ft 170ft Hs 2 3.8kW Hs 4 3.4kW Hs 6 4kW Hs 8 3.2kW Hs kW 20

21 Typical voltage drops of system components Residential wiring 2-4 volts Service drop 1-2 volts Secondary 2-3 volts Distribution transformer 2-3 volts Primary feeder 2-3 volts Total drop (sum of all) 9-15 volts As you review the circuit voltage profile, keep in mind that 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. 21

22 Voltage Typical Distribution System voltage profile Substation Primary Feeder First Customer Distribution Transformer Secondary Last Customer 128 Upper ANSI limit (126V) Volts Primary Feeder 120 This point should not exceed ANSI limits 3 Volts Distribution Transformer First Customer Lower ANSI limit (114V) This point should not be less than ANSI limit Last Customer 3 Volts Secondary & Service Drop 4 Volts Customer Wiring 22

23 Voltage Drop Example3 kva1 kva2 kva3 Circuit1 Station B L1 L2 L3 L4 For the circuit shown, what is the voltage drop and per unit value at each load center. Circuit is 13.2/7.62kV GrdY Voltage at the bus is 1.00 during peak and 1.04 at light loads Peak load demand is 6.3MVA and balanced L1, kva1 = 3.5kft, 2.1MVA L2, kva2 = 5.7kft, 2.9 MVA L3, kva3 = 2.8kft, 1.3MVA Light load demand is 40% of peak Assume Z j Line Answer: Voltage Drop Examples3.doc 23

24 Voltage Drop Example4 A single phase tap off a 12.47GrdY/7.2kV mainline is shown What is the current at each load level? What is the voltage drop? If the primary mainline is at operating 0.98 volts per unit, what is the per unit voltage level at the load? 500ft of 1/0 Al single phase w/neutral kva 250 at peak 115 at light load Circuit1 Mainline j0.139 ohm per kft Answer: Voltage Drop Examples4 (2) new.doc 24

25 What is a Capacitor? Application of Shunt Capacitors Energy storage device made of parallel conductive plates separated by an insulating medium (oil, Vacuum) Used in power system mainly to improve PF and increase voltage 25

26 Applying Shunt Capacitors Improves the power factor Reduces voltage drop Improves voltage regulation (switched capacitors only) Reduces system losses Reduces peak kw and kvar demands on system, thus: Increasing feeder capacity Releasing transmission & generation capacity 26

27 Two ways to operate the bank Fixed bank This bank is always on the system supplying VArs as long (as the fuse is in service) Switched Bank This bank is placed in service only when the VAr demand of the system exceeds the fixed supply The banks can be placed At the Transmission/Distribution substation On the Utility s distribution system On the customer s system 27

28 Methods of Control Signals Voltage Current Power Factor or VAr flow Time Temperature Radio Dispatch 28

29 Capacitor: Current Leads Voltage by 90 o 29

30 Inductor: Current Lags Voltage by 90 o 30

31 Power in a Circuit with a PF of

32 Power in a Circuit with Unity PF 32

33 Capacitors Release Capacity 33

34 Example: A 13.2kV circuit is supplying 6895kVA at 90 percent power factor. Calculate the consumed kw and kvar Apply 1800kVAr to the circuit Calculate the new pf, and kva How much kvar is required for unity pf? kva kw kvar pf

35 Placement of Capacitor banks 35

36 Tips Try to maintain the voltage rise between 2 and 5 percent. Locate as close as possible to the demand for kvar. The total voltage rise is dependent upon the circuit impedance back to the source. The resulting decrease in the reactive current flow will also contribute to the voltage increase. Pay attention to the feeder regulation and the voltage level at the application point. Keep in mind that the core losses of the distribution line transformers will increase with the increase in voltage. Use computer models for better accuracy 36

37 Voltage Profile With a Capacitor Bank V s Feeder end Substation With Capacitor Voltage Without Capacitor Sub. Cap. Bank Distance from Substation end 37

38 Voltage Rise due to Capacitor I is the total current before correction Once the switch is closed, Ic reduces I by supplying the needed VArs at the application point Due to the reduced total current, the voltage drop between the sending and receiving points is reduced This is why capacitors are viewed as a voltage source 38

39 Calculating Voltage Rise Relationship This can be seen graphically by solving the vector diagram with segregated capacitor current and load current vectors Voltage where : X I C C Rise X reactance C I C between line to neutral Note :TrytoMaintain( 2% V capacitor kvar of capacitor per phase kv and source Rise 5%) 39

40 Voltage Rise Calculation How to calculate Voltage Rise? where : The line KVACapacitorx X 10 x kv % Voltage Rise 2 X Reactance d Distance from substation in mile kv line - to-line voltage per mile Example: Calculate the Voltage Rise for a 22.86kV system with a 1800 KVAR Cap bank installed 3miles away from the substation. Note Assume XL=0.8/mile Assume X % Voltage L 0.8/ mile x d 1800 x 0.8 x3 Rise 10 x (22.86) % 40

41 Voltage calculation The effect of leading power factor or overcompensation. V R V S Voltage Drop V D I(R cos X sin ) cos (-90) 0 sin (-90) -1 V R V S I ( X ) V S IX 41

42 The modified 2/3 Rule Calculate or record the demand kva and pf between switch points on the circuit Determine the kva and pf for the four load levels Place 2/3 of the required capacitors at 2/3 of the distance down the feeder between switching points. If more capacitor banks are required for peak loads, install one or two switched banks. If one, 1/2 of the distance of the circuit If two, 1/3 and 2/3 of the distance Watch the voltage rise! 42

43 added An example of the relationship between kw and kvar demand. Assume station capacitor is on during peak. Fixed would be 300kVAr. Switched would be 450kVAr One 300 timed One 150 current Example of cap banks 43

44 Fixed and Switched Capacitors An effective program of capacitor placement for distribution feeders is to determine the size of a fixed bank by the 2/3 rule to meet the VAr demand requirements for the normal load demand located between two switch points. Note that the power factor for customer classification during normal load demand periods will not be the same for peak periods. Add a switched bank equal to the VAr demand at high load periods Then add an optimally sized switched bank which is switched on during the period of peak reactive demand Make sure that voltage rise does not exceed 5% 44

45 Example: Consider the following Find the total complex power of the system and the total power for each element of the system shown in this figure. S 1 R 1 =10 X 1 = j20 Load I 1 S 2 S 3 S 4 E 1 Source rms V 1 X 2 = -j200 X 3 = j200 R 4 =100 I 2 I 3 I 4 Source: Courtesy of Power Distribution Engineering, JJ Burke 45

46 Example 1 The net impedance is Z TOTAL R 1 jx 1 1 jx jx 3 1 R 4 10 j20 1 j200 1 j Z Total 110 j o Z The source current is I E Z TOTAL Source: Courtesy of Power Distribution Engineering, JJ Burke 46

47 Example 1 The total complex power, S, as seen by the source is E 1000 & I Complex Power : S EI * 1000 x S watts j15. 8 vars Source: Courtesy of Power Distribution Engineering, JJ Burke 47

48 Example 1 The power factor seen by the source is cos cos The voltage at the load node is V 1 E I 1 ( R 1 jx 1 ) (10 j20) j15.9 V j Source: Courtesy of Power Distribution Engineering, JJ Burke 48

49 Example 1 The power (S 2 ) consumed by X 2 ~ * ~ ~ * ~ V1 S2 V1I 2 V1 jx 2 ~ ~ * 2 VV V * ( jx ) jx 2 2 j V 1 X j j Note: The sign (-)on the power indicates that it is supplying VArs. Source: Courtesy of Power Distribution Engineering, JJ Burke 49

50 Example The power (S 3 ) consumed by X 3 is S j V1 V1 V1 j * ( jx 3) jx 3 X VArs The power (S 4 ) consumed by R 4 is 2 V1 S 4 * R watts Source: Courtesy of Power Distribution Engineering, JJ Burke 50

51 Example The power consumed by R 1 is 2 S P jq I ( R 1 ) 7.92 j0 watts The power consumed by X 1 is 2 1 S P jq I ( jx 1) 0 j15. 8 VArs Source: Courtesy of Power Distribution Engineering, JJ Burke 51

52 Example The total power of the circuit is P Q R X X X R TOTAL

53 Example 2 - remove the capacitor What is the impact if the customer s capacitor bank is taken off line (or does not install one)? S 1 R 1 =10 X 1 =j20 Load I 1 S 3 S 4 E 1 Source rms V 1 X 3 =j200 R 4 =100 I 3 I 4 Source: Courtesy of Power Distribution Engineering, JJ Burke 53

54 Example 2: The changed impedance Z total R 1 jx 1 1 jx R 4 10 j20 1 j j20 j200 1 j2 10 j20 80 j40 90 J Ztotal. Source: Courtesy of Power Distribution Engineering, JJ Burke 54

55 Example 2 continued The current is now E 1000 I Z Total The total complex power is ~ S E I * 1000 x watts 51.4 VArs Source: Courtesy of Power Distribution Engineering, JJ Burke 55

56 Example 2 continued And the voltage across the load is V 1 E I 1 ( R 1 jx 1 ) (10 j20) j10.35 V 1 92 j Source: Courtesy of Power Distribution Engineering, JJ Burke 56

57 What is the change? Current Demand 0.89 A With CAP X3 Without CAP X A Total Real Power Total Reactive Power Total supplied Power the load 87.1 W 77 W 15.8 VAr 51.4 VAr 89 VA 92.6 VA 89 V 82.7 V PF= 98.39% PF= 83.17% 57

58 Summarizing Points Capacitors Are used to provide VAr support (power factor correction) Are not a voltage source By reducing the reactive current, provide less voltage drop Can be applied as fixed or switched Fixed banks are on all the time and should be sized appropriately Switched banks should be in two stages The application of the capacitors will improve the efficiency of the system and relieve capacity 58

59 Voltage Control 59

60 Common Voltage Problems Common problem are: Low Voltage This can be corrected via line regulation, the addition of capacitor banks, transformer taps, decreasing the resistance by changing conductor or by using combinations Flickering lights More common during low voltage (peak) situations Happens due to rapid changes in the voltage seen by the light fixture High Voltage Causes increased losses in transformer cores 60

61 Voltage Control Objective Objective: to provide each customer with a voltage that conforms to the voltage design limitations of the utilization equipment Engineering objective is to maintain the ANSI defined operational voltage limits. 61

62 ANSI Voltage parameters 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 126 volts 110 to 127 volts Utilization Voltage 110 to 125 volts 106 to 127 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 62

63 Voltage Regulation Definitions Voltage Regulation The voltage drop between two sending & receiving points on the distribution system in percent of nominal voltage. PCC - Point of Common Coupling The point where utility equipment is connected to the customer equipment 63

64 Typical Feeder voltage profile 125 First Customer Upper ANSI limit (126V) Last Customer 3 Volts Primary Feeder Voltage Volts Distribution Transformer 3 Volts Secondary Drop Volts Service Drop Lower ANSI limit (114V) 110 Last Customer 64

65 Feeder voltage profile Typical profile - no voltage compensation 125 First Customer Upper ANSI limit (126V) Last Customer Voltage 120 Light Load 115 High Load Lower ANSI limit (114V) 110 Last Customer 65

66 % Voltage Fluctuation GE Flicker Curve Lights & Refrigerators, etc. Elevators, etc. Ar Furnaces, c etc. Spot Welders, etc. Per Hour Per Minute Per Second Border Line of visibility Fluctuations 66

67 Voltage Regulation Equipment Load Tap Changing (LTC) transformer Substation bus regulators Circuit regulators Supplementary line regulators Transformers with taps Equipment that has an impact on system voltage but is not a voltage source Fixed capacitors Switched capacitors Explained in the capacitor application section 67

68 Voltage Regulation Techniques 115 kv Supply 115 kv Supply Substation Bus Load tap changer 13.8 kv Side Substation Bus Bus Regulation 13.8 kv Side Breaker Breaker Breaker Breaker Breaker Breaker Breaker Breaker Circuit main line Circuit main line Circuit main line Circuit main line 68

69 Voltage Regulation Techniques 115 kv Supply 115 kv Supply 13.8 kv Side 13.8 kv Side Substation Bus Substation Bus Breaker Breaker Breaker Breaker Breaker Breaker Breaker Breaker Circuit main line Dedicated circuit regulation Circuit main line Circuit main line Circuit main line Supplemental circuit regulation 69

70 Three Phase or Single Phase? The decision involves cost and application Three Phase units Work best for circuits serving 3 phase balanced load Phase control is gang operated therefore one phase can be outside the desired bandwidth Single phase units Allows the engineers to regulate each phase independently Best application for circuits that serve large percentage of single phase load Higher maintenance costs 70

71 What Regulator is Made of? The main components A tapped autotransformer A tap/switch that can operate under load A voltage sensing circuit Requires a PT, CT & a reactor A controller to operate the tap/switch 71

72 Voltage Regulator Typical regulation has a range of +/- 10% voltage Uses taps at set new ratio, called steps, 32 of them Diagram below is only the increase steps +10% 6.25% Source (Input) Volts 0% Load (Output) Volts 72

73 Typical Regulator Control Panel 73

74 Conservation Voltage Regulation (Reduction) or CVR Definition: CVR is way to improve energy efficiency and lower electrical demand without impacting power quality. How? by lowering distribution feeder voltage while maintaining ANSI C84.1 Standard Provides peak load reduction Lowers losses 74

75 Operational Tips Utilizing Regulators Maintain proper voltage at first and last customer (114 Voltage 126) Line Drop Compensation (LDC) Maintain voltage at the desired point on a feeder Load Distribution Works best on uniform load. Non-uniform load may require supplemental equipment (Regulators??) Be aware of Capacitors 75

76 Feeder voltage profile Feeder profile with supplemental regulation Circuit1 kva1 Supplemental circuit regulation kva2 Station B First Customer Last Customer Upper ANSI limit (126V) 125 Voltage 120 Light Load 115 High Load Lower ANSI limit (114V)

77 Operational considerations Time delay settings Regulator should not respond to short duration voltage sags. Time delays of downstream regulators must be coordinated. Delays in the range of 30 to 90 seconds are typical. Bandwidth or sensitivity settings Cannot be smaller than smallest possible tap change amount. Most utilities use 1.5 to 3 volts. 77

78 Operational considerations Place Regulator to Neutral and Nonauto when switching or tying to other sources. Check to make sure the float voltage is not too high, the bandwidth is correct and that the LDC is correctly established. 78

79 How to Size 3ᶲ Regulator? 278 A 3Ø regulator 4160 V system (3 or 4 wire) 2000 kva 3Ø load 278A + 10% regulation 2000 kva 4.16kV Load KVA 3Ø Range KV REG kva 3 I LOAD kva, 3Ø, 4.16kV, + 10% Regulator 79

80 How to Size Sᶲ Regulator? 278 A 1Ø Regulators 4160 Volt 3Ø, 4-wire system 2000 kva 3Ø load, 278 amps + 10% regulator 4160 V 2000 kva Load kva 1Ø Range kv REG kva I LOAD kva, 1Ø, 2.4kV, + 10% Regulators 80

81 Line Drop Compensation (LDC) LDC is used to correct for voltage drop on the circuit Must know the CT and PT ratio of the regulator control circuit Requires knowing the distance and the impedance of the circuit (R&X value) Need to know the existing voltage levels before changing the compensation Factor in all capacitor banks Watch out for the voltage regulation impact on transformers with secondary taps Where is the best starting point? Answer: where the voltage drop from the source to the application point is equal to the voltage drop from the application point to the end of the circuit 81

82 Equations to Calculate LDC Settings In the case of 3f regulators; 3 single f regulators; or one 1f regulator on the circuit To calculate R LDC and X LDC in ohms: R LDC = R L x (CT Ratio/PT Ratio) X LDC = X L x (CT Ratio/PT Ratio) To calculate R LDC and X LDC in volts: R LDC = R L x (CT Primary Rating/PT Ratio) X LDC = X L x (CT Primary Rating/PT Ratio) 82

83 Example - Line Drop Compensation Source 100:0.2A I L R L =1.5 X L =2 E RO R LDC X LDC E R VBR 2400V 120V Express R LDC, X LDC in Ohms: R X LDC LDC C.T. RATIO RL P.T. RATIO C.T. RATIO XL 2 50 P.T. RATIO Express R LDC, X LDC in Volts: R LDC R L C.T. PRI. RATING P.T. RATIO Volts 20 X LDC X L C.T. PRI. RATING P.T. RATIO Volts 20 83

84 Why Regulators have time delay? Why controller of the regulator has a time delay feature? To avoid reaction to rapid changes in the load demand occur which can cause the voltage to exceed the settings Capacitor switching, Motor Starting To prevent arc caused by movement of regulator s contacts There is a maximum number of changes that the regulator can do before failure may occur To maintain the voltage without excessive operations of the controller 84

85 Why Regulators have Time delay settings? 30 Sec. Time Delay 40 Sec. Time Delay Cascading regulators requires proper time delay settings. Allow the station regulators to operate before any supplemental line regulators 85

86 Controller Volts 30-second time delay setting The voltage level must exceed the bandwidth for 30 seconds before the tap changer will respond In this case, no tap change occurs V V V V Seconds 86

87 Equations to Calculate Regulator Ratings Single phase application kva Regulator I load kv L-G Regulator Range Three-Phase application kva Regulator I load kv L-L Regulator Range kva Regulator I load kv L-G Regulator Range 3 Note: Open delta - consists of 2 single-phase regulators Each Regulator kva I load kv L-L Regulator Range 87

88 Summarizing points Know the ANSI C84.1 voltage standards Obtain voltage and current readings from the circuit before and after setting the voltage regulation Voltage Drop Equation Primary and secondary must be considered Be sure to check both the light and peak load conditions Distributed versus lumped load rule Understand Line Drop Compensators Watch out for season changes Regulator KVA is less than power passing through it. 88

89 References 1. J.D. Golver, M.S. Sarma, Power System Analysis and design, 4 th ed., (Thomson Crop, 2008). 2. M.S. Sarma, Electric Machines, 2 nd ed., (West Publishing Company, 1985). 3. A.E. Fitzgerald, C. Kingsley, and S. Umans, Electric Machinery, 4 th ed. (New York: McGraw-Hill, 1983). 4. P.M. Anderson, Analysis of Faulted Power systems(ames, IA: Iowa Satate university Press, 1973). 5.W.D. Stevenson, Jr., Elements of Power System Analysis, 4 th ed. (New York: McGraw-Hill, 1982). 6. J.J.Burke, Power Distribution Engineering (Marcel Dekker, INC, 1994).

90 Break Solution Time!!!!! Answer: 37.5 KVA