Part 3of3 Reliability Power System Design Buenos Aires

7/29/2019 Part 3of3 Reliability Power System Design Buenos Aires

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High ReliabilityHigh Reliability

Power System DesignPower System Design

Keene M. Matsuda, P.E.Regional Electrical Manager

Senior Member IEEEIEEE/PES Distinguished Lecturer

[email protected]

Buenos Aires, ArgentinaJune 25 & 26, 2009


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Motors

Component Energy Loss, FL (%)Motors: 1 to 10 Hp 14.00 to 35.00Motors: 10 to 200 Hp 6.00 to 12.00Motors: 200 to 1500 Hp 4.00 to 7.00Motors: 1500 Hp and up 2.30 to 4.50

Variable Speed Drives 6.00 to 15.00Motor Control Centers 0.01 to 0.40MV Starters 0.02 to 0.15MV Switchgear 0.005 to 0.02

LV Switchgear 0.13 to 0.34

Reference: ANSI/IEEE Standard 141 (Red Book), Table 55


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Motors

z Motor driven systems represent about 60% of allelectrical energy used

z Energy Policy Act of 1992 set min efficiencies formotors in the U.S.

z Manufacturers have increased motor efficiencies inthe interim

z Premium-efficiency motors can therefore decrease

lossesReference: Copper Development Association


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Variable Frequency Drives

z Very common device for energy efficiency

z AC to DC to Variable output with V/Hz constant

z Not suitable in all cases

z Optimum: Must have varying load

z Or dictated by application

z Example: Chemical feed pumps, small Hp, but

precise dosing


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Variable Frequency Drives

Reference: Energy Savings in Industry, Chapter 5, UNEP-IETC


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Cables from VFDs to Motors

z VFDs convert 480 V at 60 Hz to a variable voltagewith variable frequency

z VFD holds constant the ratio of V/Hz

z Nominal is 480 V/60 Hz = 8.0 at 100% motor speed

z If you want 50% speed, reduce the voltage to 240 V

z But need to correspondingly reduce the frequency by50% or else motor wont operate

z Thus frequency is 30 Hz at 240 V, or 240 V/30 Hz = 8.0constant


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z Same for any speed in the operating range

z If you want 37% speed:

z 480 V x 0.37 = 177.6 V

z If V/Hz is held constant at 8.0,

z Then frequency is V/8.0 = 177.6 V/8.0 = 22.2 Hz


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z This shielding can easily be a steel conduit

z This works if the conduit is dedicated between theVFD and the motor

z If part of the cable run is in underground ductbank,

then the PVC conduit in the ductbank no longerprovides that shielding


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z Possible to install a steel conduit thru the ductbankto counteract

z But that would then restrict flexibility in the future tomove these VFD cables to a spare conduit whichwould then be PVC

z Too costly to install all ductbank with RGS conduit


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z Also, if the cables pass thru a manhole or pull boxalong the way, it is very difficult to keep the VFDcables sufficiently separated from the other normalcircuits

z If EMF is a problem with adjacent circuits, easy

solution is to select 600 V, 3-conductor, shieldedcables


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z However, the true nature of the EMF problem fromVFD cables is not well known or calculated

z Much depends on the type of VFD installed, 6-pulse,12-pulse, 18-pulse

z If there is an reactor on the output of the VFDz How well the reactor mitigates harmonics

z What the length of the cable run is, i.e., introducing

impedance in the circuit from the cable


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California Title 24

z Californias mandate for energy efficiency

z Three major elements: architectural design, HVAC,lighting

z Lighting: limiting watts/sq ft by room classification,

motion sensors, etc.z Title 24 revised Oct 2005 to close loopholes

z Prior: lighting indoors in air condit ioned spaces

z Now: all lighting indoors and now outdoors


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Lighting Design

z Outdoor lighting on poles more complicated

z Factors:Pole heightPole spacingFixtures per poleFixture lamps type

Fixture wattageFixture light distribution pattern

z Photometric analysis using software (Visual, AGI32,etc.)

z Calculate average fc illumination & uniformity

z Life safety illumination for egress: 1 fc average, 0.1 fcone point


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Photometric Calculations Lighting


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K-Factor Calculations Dry-Type Transformers

z K-Factor is a measure of the amount of harmonics ina power system

z K-Factor can be used to specify a dry-typetransformer such that it can handle certain levels ofharmonic content

z K-Factor rated transformers are generally built tobetter dissipate the additional heat generated fromharmonic current and voltage


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z To calculate K-Factor, must have a power systemsanalysis software program like ETAP or SKM, etc.

z Model all harmonic-producing equipment: biggestculprit is the 6-pulse VFD

z

Formula for calculating K-Factor:

K-Factor = Ih p.u.2 x h2

z Where, Ih p.u. = Current harmonic in per unit

z

Where, h = Odd harmonic (3, 5, 7, 9, 11, 13, etc.)


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System Design Summary

z A. Prepare Load Study Calculation

z

B. Size Transformer to 480 V Loadsz C. Size 480 V Motor Control Center (MCC)

z D. Select Short Circuit Rating of 480 V MCC

z E. Size 480 V Feeder from Transformer to MCC

z F. Size Transformer 12 kV Primary Disconnect

z G. Select Surge Protection at Transformer Primary

z H. Size 12 kV Feeder to Transformer (MV Cable)


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System Design: Load Study

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z View 3 of 4:

z

Connected values represent any load connected tothe power system regardless of operation or not

z Running values represent actual operating loads atmax demand

z If a pump is a standby, or backup, or spare, thispump would be turned off, or shown as zero, in therunning columns

z The Demand Factor entry of zero is what turns offany particular load


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Off

On

Off

On


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System Design: Load Flow Study

z Before a short circuit study can be performed usingpower systems analysis software, a model of the

power system must be created

z System modeling parameters include the following:

z - Utility short circuit contribution

z - Transformers

z - Motors

z - Conductor sizes and lengths

z - On-site generation, etc.


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System Design: Sample Short Circuit Study


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S t D i Sh t Ci it f 480 V MCC


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System Design: Short Circuit of 480 V MCC

z If power systems analysis software is not available,can use a conservative approximation

z The MVA method represents the worst case faultcurrent thru transformer

z Transformers naturally limit the current thrutransformer to secondary bushings

z Need transformer impedance, or assume typical is5.75%Z, plus or minus

z Assume utility supply can provide infinite short

circuit amperes to transformer primary (i.e.,substation across the street)


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S t D i 480 V F d f T f t MCC


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System Design: 480 V Feeder from Transf to MCC

z E. Size 480 V Feeder from Transformer to MCC

z First calculate IFL from transformer secondary

Transformer kVAIFL = ----------------------------

Sq Rt (3) x kV

2000 kVAIFL = ----------------------------- = 2405.7 ASq Rt (3) x 0.48 kV

z IFL x 125% = 2405.7 A x 1.25 = 3007 A

z No one makes a cable to handle 3000 A



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z Must use parallel sets of conductors

z Each conduit will have A, B, C, and GND cables, plusneutral if required for 1-phase loads

z Standard engineering practice is to use 500 kcmil(253 mm2) or 600 kcmil (304 mm2) conductors

z Why?

z Largest standard conductor that will fit easily into a

standard 103 mm2

conduitz For this example, we will use 500 kcmil (253 mm2)


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z Select grounding conductor

z Per NEC Table 250.122,

z Based on 3000 A trip rating

z Grounding conductor = 400 kcmil (203 mm2)

z Total cables = 8 sets of 3-500 kcmil (253 mm2), 1-400kcmil (203 mm2) GND

z Or, total 24-500 kcmil (253 mm2), 8-400 kcmil (203mm2) GND



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z Calculate total cross-sectional area of each set ofcables

z Per NEC Chapter 9, Table 5, for XHHW cables

z Area of 500 kcmil (253 mm2) cable = 450.6 mm2

z Area of 400 kcmil (203 mm2) cable = 373.0 mm2

z Total cross-sectional area of each parallel set =3 x 450.6 mm2 + 1 x 373.0 mm2 = 1724.8 mm2

z Select conduit to maintain FF < 40%


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NEC Chapter 9, Table 4, RMC Conduit Dimensions


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NEC Chapter 9, Table 4, RMC Conduit Dimensions



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z Per NEC Chapter 9, Table 4:

z For RMC, a conduit diameter of 103 mm has an areaof 8316 mm2

z Fill Factor = 1724.8 mm2/8316 mm2 = 20.7%

z FF < 40%, OK

z For large cables in one conduit, it is notrecommended to approach the FF = 40% due to the

excessive pulling tensions when installing the cables


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System Design: Transformer 12 kV Disconnect


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Fused Switch



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Circuit

Breaker

Transformer



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z Minimum bus rating of metal-enclosed fused loadinterrupter switches = 600 A

z Bus rating > IFL x 125%

z 600 A > 120.3 A, OK


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z Per NEC Table 450.3(A),

z For transformer typical impedance = 5.75%

z Maximum size fuse = IFL x 300%

z Maximum size fuse = 96.2 A x 3.0 = 288.7 A

z NEC allows next higher size available

z Thus, fuse = 300 A

z Although NEC dictates maximum, standardengineering practice is to select fuse at IFL x 125% =120.3 A, or round up to 150 A



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z Select OCPD relay trip setting with metal-cladvacuum breaker switchgear

z NEC governs maximum relay trip setting fortransformer protection

z NEC Table 450.3(A), Maximum Rating or Setting of

Overcurrent Protection for Transformers Over 600Volts (as a Percentage of Transformer-Rated Current)

z For transformer IFL = 96.2 A



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z Standard engineering practice is to size the CT suchthat the expected maximum current is about 2/3 of

the CT ratioz For this transformer IFL = 96.2 A

z The 2/3 point = 96.2 A/(2/3) = 144.3 A

z Select next standard available CT ratio of 150:5



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z For many years the most common type ofovercurrent relay was an induction disk type of relay

z Depending on the secondary CT current input to therelay, the disk would rotate a corresponding angle

z Todays technology uses electronic-based relays

z As such, electronic relays are more accurate insensing pick-up and contain smaller incrementalgradations of available settings than induction disk

relays



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z For example: Induction disk relays had available tapsettings in increments of 1 A or 0.5 A

z Todays electronic relays have tap settings inincrements of 0.01 A

z Thus, a more exact tap setting could be selected,

thereby making coordination with upstream anddownstream devices much easier


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System Design: Surge Protection at Transformer


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z Example, for delta circuit, most common:

z Utility Nominal Supply Voltage x 125%

z 12 kV x 1.25% = 15 kV

z Thus, surge arrester voltage rating = 15 kV, minimum

z Could select higher voltage if util ity has widelyvarying voltage supply

z Surge arrester is connected phase-to-ground

System Design: Surge Protection at Transformer


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z If wrong selection of 8.6 kV surge arrester on 12 kVcircuit, then the surge arrester would probably

explode upon energization because it will shunt toground any voltage higher than 8.6 kV

z The switchgear would be under short circuitconditions and the fuse would blow or the relaywould trip

System Design: 12 kV Feeder to Transformer


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z H. Size 12 kV Feeder to Transformer (MV Cable)

z Sizing 15 kV conductors for 12 kV circuits stil l uses

transformer IFL = 96.2 A

z IFL x 125% = 96.2 A x 1.25 = 120.3 A

z

Select conductor size based on NEC tablesz Similar to 600 V cables, depends on aboveground or

underground installation for Medium Voltage (MV)cable



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z One of the more popular 15 kV cables is rated asfollows:

z - 15 kV, 100% or 133% insulation

z - 15 kV with 133% insulation = 15 kV x 1.33 = 20 kV(optional rating for circuit voltages between 15 kV

and 20 kV)

z - MV-105 = medium voltage cable, rated for 105Cconductor temperature (previous rating was MV-90,

and had lower ampacity)



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z - EPR insulation = Ethylene Propylene Rubberinsulation (traditional insulation versus newer cross-

linked polyethylene, or XLP)z - Cu = copper conductor

z - Shielded = Copper tape wrapped around EPR

insulation (to aid in containing electric field and animmediate ground fault return path)

z - PVC jacket = overall jacket around cable

System Design: Okonite 15 kV Cable


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System Design: Okonite 15 kV Cable


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z Per NEC Table 310.73, for 15 kV, MV-105,

z 4 AWG (21.15 mm2) ampacity = 120 A


z 4 AWG (21.15 mm2) is not a common size in 15 kV

cablesz 2 AWG (33.62 mm2) is much more common and

available

z Thus, select 2 AWG (33.62 mm2) for phaseconductors



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z Select grounding conductor

z Use NEC Table 250.122

z Relay trip setting would be set to 120 A, soovercurrent rating would be 200 A per NEC table


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z Grounding conductor is 6 AWG (13.30 mm2)

z Does grounding cable for 12 kV circuit need to berated for 15 kV, same as phase cables?

z No.

z Grounding conductor is not being subject to 12 kVvoltage

z Circuit = 3-2 AWG (33.62 mm2), 15 kV, 1-6 AWG (13.30mm2) GND


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F Ok it 100% i l ti bl t di t


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z For Okonite 100% insulation, cable outer diameter =23.0 mm

z Cable cross-sectional area = Pi x d2/4

z Cable cross-sectional area = 3.14 x 23.0 mm2/4

z Cable cross-sectional area = 415.5 mm2


For Okonite 133% ins lation cable o ter diameter


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z For Okonite 133% insulation, cable outer diameter =25.3 mm

z Cable cross-sectional area = Pi x d2/4

z Cable cross-sectional area = 3.14 x 25.3 mm2/4



z For grounding conductor = 6 AWG (13 30 mm2)


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z For grounding conductor = 6 AWG (13.30 mm2)

z Use NEC Chapter 9, Table 5, XHHW Insulation



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z Per NEC Chapter 9 Table 5 for 6 AWG (13 30 mm2)


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z Per NEC Chapter 9, Table 5, for 6 AWG (13.30 mm2)


z Total cable cross-sectional area with 15 kV, 100%insulation = 3 x 415.5 mm2 + 1 x 38.06 mm2 = 1246.4mm2

z Total cable cross-sectional area with 15 kV, 133%insulation = 3 x 502.7 mm2 + 1 x 38.06 mm2 = 1508.1

mm2

z Select conduit for FF < 40%


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z Per NEC Chapter 9 Table 4:


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z For RMC, a conduit diameter of 78 mm has an area of

4840 mm2

z For 15 kV, 100% insulation:


z FF < 40%, OK


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z Per NEC Table 310.77, for 15 kV, MV-105,


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, , ,



z 4 AWG (21.15 mm2) is not a common size in 15 kVcables

z 2 AWG (33.62 mm2) is much more common andavailable

z Thus, select 2 AWG (33.62 mm2) for phaseconductors




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z Grounding conductor is stil l 6 AWG (13.30 mm2)

z Circuit = 3-2 AWG (33.62 mm2), 15 kV, 1-6 AWG (13.30mm2) GND


z Select conduit size for 12 kV circuit


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z For 15 kV cables, use Okonite data sheet

z Same as for RMC conduit


z For grounding conductor = 6 AWG (13.30 mm2)


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z Use NEC Chapter 9, Table 5, XHHW Insulation

z Same as for RMC conduit


z Per NEC Chapter 9, Table 5, for 6 AWG (13.30 mm2)


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z Total cable cross-sectional area with 15 kV, 100%insulation = 3 x 415.5 mm2 + 1 x 38.06 mm2 = 1246.4mm2

z Total cable cross-sectional area with 15 kV, 133%insulation = 3 x 502.7 mm2 + 1 x 38.06 mm2 = 1508.1

mm2

z Select conduit for FF < 40%

NEC Chapter 9, Table 4, PVC Conduit Dimensions


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z For PVC, a conduit diameter of 78 mm has an area of

4693 mm2



z FF < 40%, OK




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z For PVC, a conduit diameter of 78 mm has an area of

4693 mm2



z FF < 40%, OK


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Utility Voltage Supply Affects Reliability

z Most utility distr ibution circuits are 12 kV, 13.8 kV,etc


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etc.

z Obtaining a higher utili ty voltage circuit will increasereliability

z Dont always have a choice in utility voltage

z If available, a higher transmission voltage like 46 kV,60 kV, etc. is advantageous

Utility Voltage Supply Affects Reliability

z Higher voltage circuit means more power transfercapability


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capability

z Also means fewer direct connections to othercustomers

z Also means lesser chances for the line to fail or

impacted by other customersz Transmission circuits usually feed distribution

substations down to 12 kV


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System Optimization Siting Main Substation

z In siting the utility substation for a plant, systemoptimization helps to reduce costs


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optimization helps to reduce costs

z Most util ities are only obligated to bring service tothe nearest property line

z If you want the place the utili ty substation at the

opposite corner, you will have to pay for the extraconstruction around the plant or thru the plant

Location of Main Substation

z Electric utility circuit is usually MV


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z Voltage: 12.47 kV or 13.8 kV, 3-phase

z Capacity: 7-12 MW per circuit for bulk power

z Main substation near existing l ines

z Utility obligated to bring service to property line

z Represent large revenue stream of kWh

Reference: Rule 16, Service Extensions, per SCE,LADWP, PG&E, SMUD

Electric Utility Overhead Line



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MainSubstation

Main

Substation

Site Plan

forPlant


z You pay for extension of line around property


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z You pay for extension of line within property

z Line losses increase = square of current x resistance,or I2R

CAVEATS

z Pay for losses in longer feeder circuit as in kWh

z May be limited in choices of site plan

z Need to catch layout early in conceptual stages


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Electrical Center of Gravity

z Should optimize location of large load centerbalanced with small loads


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z Example is pump station, with 10-100 Hp pumpsz Optimized location would have pump station next to

main substation

z Minimize voltage drop and losses in feeder cables

Location of Large Load Centers

z Locate large load centers near main substation

E l P t ti ith l H t


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z Example: Pump stations with large Hp motors

z Minimize losses in feeder conductors

z Optimum: electrical center-of-gravity of all loads

z Run SKM, ETAP, etc., power systems software tooptimize system

Electric Utility Overhead Line

Location of Large Load Centers


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MainSubstation

Site Plan

forWTP or WWTP

Large HpPump Station


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Double Ended Substation

z Also known as a main-tie-main power system

z The main tie main can be both at 12 kV or 480 V to


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z The main-tie-main can be both at 12 kV or 480 V to

take advantage of two separate power sourcesz At 480 V, there are two 12 kV to 480 V transformers

feeding two separate 480 V buses with a tie breaker

between


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z For process optimization, the loads should beequally distr ibuted between the buses


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z Example, four 100 Hp pumps

z Should be Pumps 1 and 3 on Bus A, and Pumps 2and 4 on Bus B

z If all four pumps were on Bus A, and Bus A failed,

you have zero pumps available


z Normally, main breaker A and main breaker B isclosed and the tie breaker is open


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z For full redundancy, both transformers are sized to

carry the full load of both busesz Normally, they are operating at 50% load

z In the previous example, each transformer is sized at

2000 kVA, but operating at 1000 kVA when the tiebreaker is open



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12.47 kV Source 1 12.47 kV Source 2

Dual Redundant Transformers, Main-Tie-Main


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750 kVA Load 750 kVA Load

N.O.

T11500 kVA

12.47 kV-480 V

T21500 kVA

12.47 kV-480 V

Bus 1, 480 V Bus 2, 480 VN.C. N.C.

12.47 kV Source 2



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Close

T21500 kVA

12.47 kV-480 V

Bus 1, 480 V Bus 2, 480 VTrip N.C.

Lose 12.47 kV Source 1,or T1 Failure,or Prev. Maintenance

12.47 kV Source 2



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Close

T21500 kVA

12.47 kV-480 V

Bus 1, 480 V Bus 2, 480 VN.C.

All Loads Restored


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Main-Tie-Tie Main System

z For personnel safety, a dummy tie breaker is addedto create a main-tie-tie-main system

Wh ki B A f i t ll l d


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z When working on Bus A for maintenance, all loadscan be shifted to Bus B for continued operation

z Then the tie breaker is opened and Bus A is dead

z However, the line side of the tie breaker is sti llenergized

z Hence, a dummy tie is inserted to eliminate thepresence of voltage to the tie breaker


Main-Tie-Tie Main System


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N.O.

T11500 kVA

12.47 kV-480 V

T21500 kVA

12.47 kV-480 V

Bus 1, 480 V Bus 2, 480 VN.C. N.C.

N.O.


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MV vs. LV Feeders

z Recall: I2R losses increase with square of current

z Worst case is large load far away


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z Fuzzy math: increase voltage and reduce current

z Example: 1,500 kVA of load, 3-phase

z Current at 480 V = 1500/1.732/.48 = 1804 A

z Current at 4.16 kV = 1500/1.732/4.16 = 208 A

z Current at 12.47 kV = 1500/1.732/12.47 = 69 A

MV vs. LV Feeders

z Sizing feeders: 100% noncontinuous + 125% ofcontinuous


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Reference: NEC 215.2(A)(1)z Engineering practice is 125% of all loads

z Sometimes a source of over-engineering

MV vs. LV Feeders

z Example: 2-500 Hp pumps + 1-500 Hp standby

z Worst-worst: All 3-500 Hp pumps running


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z What if system shuts down or fails

z May need 4th pump as standby

MV vs. LV Feeders

z Recall: I2R losses increase with resistance

z As conductor diameter increases, resistance


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decreasesz Can increase all conductors by one size to decrease

resistance

z Thereby decreasing line losses & increase energyefficiency

z Comes at increased cost for cables/raceway

Reference: Copper Development Association

MV vs. LV Feeders

z 480 V: drop more Cu in ground w/600 V cable

z 5 kV cable: more expensive than 600 V cable


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z 15 kV cable: more expensive than 5 kV cable

z 4.16 kV switchgear: more expensive than 480 Vswitchgear or motor control centers

z 12.47 kV swgr: more expensive than 4.16 kV

z Underground ductbank is smaller with MV cables

MV vs. LV Feeders

z Previous example with 1,500 kVA load:

z At 480 V, ampacity = 1804 A x 125% = 2255 A


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z Ampacity of 600 V cable, 500 kcmil, Cu = 380 A

Reference: NEC Table 310.16

z

Need six per phase: 6 x 380 A = 2280 Az Feeder: 18-500 kcmil + Gnd in 6 conduits

MV vs. LV Feeders

z At 4.16 kV, ampacity = 208 A x 125% = 260 A

z Ampacity of 5 kV cable, 3/0 AWG, Cu = 270 A


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Reference: NEC Table 310.77, for MV-105, 1 cktconfiguration

z Feeder: 3-3/0 AWG, 5 kV cables + Gnd in 1 conduit

MV vs. LV Feeders

z At 12.47 kV, ampacity = 69 A x 125% = 87 A

z Ampacity of 15 kV cable, 6 AWG, Cu = 97 A


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z Ampacity of 15 kV cable, 2 AWG, Cu = 165 A

Reference: NEC Table 310.77, for MV-105, 1 cktconfiguration

z 2 AWG far more common; sometimes costs less

z Larger conductor has less R, hence less losses

z Feeder: 3-2 AWG, 15 kV cables + G in 1 conduit

MV vs. LV Feeders

z Use of MV-105 is superior to MV-90 cable for sameconductor size

z The 105 or 90 refers to rated temperature in C


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z The 105 or 90 refers to rated temperature in C

z MV-90 is being slowly phased out by manufacturerstoday

MV vs. LV Feeders

z Higher ampacity available from MV-105

Conductor Size MV-90 Amps MV-105 Amps


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2 AWG, 5 kV 145 A 155 A

2/0 AWG, 5 kV 220 A 235 A

4/0 AWG, 5 kV 290 A 310 A500 kcmil, 5 kV 470 A 505 A

Reference: NEC Table 310.77, 1 circuit configuration

MV vs. LV Feeders

z Multiple circuits in ductbank require derating

z Heat rejection due to I2R is severely limited


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z Worst case: middle & lower conduits; trapped

No. of Circuits Ampacity1 270 A

3 225 A6 185 A

Reference: NEC Table 310.77, for 3/0 AWG, Cu, 5 kV,

MV-105

z NEC based on Neher-McGrath (ETAP software)

Transformer Sizing

z Two basic types of transformers:

z Liquid-fil led transformers (2 types)- Pad-Mount type


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Pad Mount type- Substation type

z Dry-type transformers

Liquid-Filled: Pad-Mount Type Transformer


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Liquid-Filled: Substation Type Transformer


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Dry-Type Transformer


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Dry-Type Transformer


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Transformer Sizing

z Common mistake is to oversize transformers

z Example: Average load is 1,500 kVA, thentransformer is 1,500 or even 2,000 kVA


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z Prudent engineering: cover worst case demand

z Theres a better way and stil l use solid engineering

principles

Transformer Sizing

z Use the temperature rise rating and/or add fans forcooling

z For liquid-filled transformers in 1,500 kVA range:


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z Standard rating is 65C rise above ambient of 30C

z Alternate rating is 55/65C, which increases capacity

by 12%

Reference: ANSI/IEEE Standard 141 (Red Book), section10.4.3

Transformer Sizing

z Capacity can be further increased with fans

z OA = liquid-immersed, self-cooled

z FA f d i l d


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z FA = forced-air-cooled

Reference: ANSI/IEEE Standard 141 (Red Book), Table10-11

z In 1,500 kVA range, adding fans increases capacityby 15%

Reference: Westinghouse Electrical Transmission &Distribution Reference Book

Transformer Sizing

z Example: 1,500/1,932 kVA, OA/FA, 55/65COA, 55C = 1,500 kVAOA, 65C = 1,680 kVA (1.12 x 1,500)

FA 55C 1 725 kVA (1 15 1 500)


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FA, 55C = 1,725 kVA (1.15 x 1,500)FA, 65C = 1,932 kVA (1.15 x 1.12 x 1,500)

z Increased capacity by 29%

z Avoid larger transformer and higher losses

z Note: All we did was cool the transformer

Transformer Sizing

z Same concept for dry-type transformers

z AA = dry-type, ventilated self-cooled

z FA f d i l d


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z FA = forced-air-cooled

Reference: ANSI/IEEE Standard 141 (Red Book), Table10-11

z Adding fans increases capacity by 33.3%

Reference: ANSI Standard C57.12.51, Table 6

z Example: 1,500/2000 kVA, AA/FA

Transformer Losses

z Transformers are ubiquitous throughout water &wastewater plants

z Transformer losses = 2 components:


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z No-load losses + load losses

z No-load = constant when transformer energized

z Load = vary with the loading level

Transformer Losses

z Losses for 1,500 kVA transformer (W)

Type No-Load Full-Load Total (W)Dry-Type 4,700 19,000 23,700Liquid (sub) 3 000 19 000 22 000


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Liquid (sub) 3,000 19,000 22,000Liquid (pad) 2,880 15,700 18,580

Reference: Square D Power Dry II, Pad-Mount, &

Substation Transformers


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Transformer Losses

z Trivial difference between 98.44% (dry) and 98.78%(liquid), or 0.34%?

z Assume 10-1500 kVA transformers for 1 year at$0 14/kWh = $62 550 savings


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$0.14/kWh = $62,550 savings

Transformer Losses

z Heat Contribution for 1,500 kVA transformer atvarious loading levels (Btu/hr)

Type 100% 75% 50%Dry Type 80 860 52 510 32 240


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Dry-Type 80,860 52,510 32,240Liquid (sub) 75,065 46,700 26,445Liquid (pad) N/A N/A N/A

Reference: Square D Power Dry II & SubstationTransformers

Transformer Losses

z Energy Policy Act 2005 effective Jan 1, 2007; usesNEMA TP-1 standards as reference

z Mandates transformers meet efficiency levels,especially at low loads > larger share of total


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especially at low loads > larger share of total

z Target: higher grade of grain oriented steel

z Thinner gauge and purer material quality

z Reduces heat from eddy/stray currents

Reference: New Energy Regulations to Impact the

Commercial Transformer Market, Electricity Today,March 2007

Transformer Overloading

z Can you exceed the rating of a transformer?

z Without loss of life expectancy?

z Depends on the following conditions:


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Depends on the following conditions:

z Frequency of overload conditions

z Loading level of transformer prior/during to overload

z Duration of overload conditions

Reference: ANSI/IEEE C57.92, IEEE Guide for Loading

Mineral-Oil-Immersed Power Transformers Up to andIncluding 100 MVA


z Allowable overload forliquid-filled transformer, 1overload/day

Duration 90% 70% 50%0 5 hrs 1 80xRated 2 00xRated 2 00xRated


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0.5 hrs 1.80xRated 2.00xRated 2.00xRated1.0 hrs 1.56xRated 1.78xRated 1.88xRated2.0 hrs 1.38xRated 1.54xRated 1.62xRated

4.0 hrs 1.22xRated 1.33xRated 1.38xRated8.0 hrs 1.11xRated 1.17xRated 1.20xRated

Reference: Square D Substation Transformers


z Overloading a transformer is not strictly taboo

z Okay if you can engineer the system and control theconditions, i.e., dual redundant transformers


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z Allows purchase of smaller transformer

z Less losses, higher energy efficiency, lower energy

costs


z Spill containment issues with liquid-fil led: PCB,mineral oil, sil icone, etc.

z Mitigated by using environmentally benign fluid:

E i FR3 i b d fi i PCB


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z Envirotemp FR3 is soy-based, fire-resistant, PCB-free, can cook with it

z Meets NEC & NESC standards for less-flammable, ULlisted for transformers

Reference: Cooper Power Systems Envirotemp FR3

Fluid


z For a typical transformer: 1,500 kVA, 5/15 kV primary,480Y/277 V secondary

z Cost is about 45% to 93% higher for dry-type vs.

liquid-filled


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qu d ed

z Adding fans and temp ratings costs are incremental:capital cost only

Reference: 2000 Means Electrical Cost Data, Section16270


z Maintenance/Reliability

z Most significant and salient point

z Not advisable to have radial feed to one transformert f d ll l d


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to feed all loads

z Dual-redundant source to two transformers with

main-tie-main configuration for reliability andredundancy; transformers at 50% capacity

z Decision Point: Lower capital cost with radial system

vs. high reliability and flexibility




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N.O.

T11500 kVA

12.47 kV-480 V

T21500 kVA

12.47 kV-480 VBus 1, 480 V Bus 2, 480 V

N.C. N.C.

12.47 kV Source 2


Lose 12.47 kV Source 1,T1 F il


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Close

T21500 kVA

12.47 kV-480 VBus 1, 480 V Bus 2, 480 V

Trip N.C.

or T1 Failure,or Prev. Maintenance

12.47 kV Source 2


All Loads Restored


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Close

T21500 kVA

12.47 kV-480 VBus 1, 480 V Bus 2, 480 V

N.C.


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Emergency/Standby Engine-Generators

z Very common source of alternate power on site

z Diesel is most common choice for fuel

z Generator output at 480 V or 12 kV


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z NEC Article 700, Emergency Systems, directed at lifesafety

z Emergency: ready to accept load in 10 secondsmaximum


z NEC Article 701, Legally Required Standby Systems,directed at general power & ltg

z Standby: ready to accept load in 60 seconds

maximum


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z Both are legally required per federal, state, govt.jurisdiction

z Similar requirements, but more stringent foremergency

z

Example: equipment listed for emergency, exercisingequipment, markings, separate raceway


z NEC Article 702, Optional Standby Systems, directedat non-life safety, alternate source

z Even less stringent requirements


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Automatic Transfer Switches

z Used in conjunction with emergency/standby powersources

z Constantly sensing presence of normal power

source, utility, using UV relay


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z When normal power source fails, automatic sendssignal to start engine-generator

z When up to speed, transfers from NP to EP, in opentransition



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z Open transition: Break-Before-Make, or f inite deadtime

z Upon return of utility power, initiate time delay

z To ensure utility power is stable and not switching of


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y p gcircuits while restoring system

z

After time delay timeout, ATS transfers back to NP, inopen transition

z Plant loads will be down momentarily


z Option is Closed transition: Make-Before-Break, nodead time

z For brief time, the engine-generator is operating in

parallel with uti lity


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z Plant loads stay up

z

In closed transition, then subject to uti lity regulationsfor parallel generation


z Need to match voltage, frequency, and phase anglewith utility source

z Phase angle is most important, worst case is 180

degrees out of phase


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z Other consideration is preventing small generatorfeeding out of plant into utility distribution network

z Load would be too large for small generator

z Generator cant generate enough power andexcitation collapses

z Would trip out on low voltage and/or low frequency


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Uninterruptible Power Supply (UPS) Systems

z UPS units are very common sources of backup ACpower for a variety of uses

z They can be very large to power 100s of kWs of

crit ical loads in the power system


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z Or they can be small on the order of a few kW topower control system functions


z A true UPS is always on line

z Incoming AC is converted to DC thru a bridgerectif ier to a DC bus

z The DC bus charges a battery bank


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z Power from the DC bus is then inverted to AC for use

by loadsz If normal power fails, power to the loads is

maintained without interruption

z AC output power is being drawn from the batteries

z Battery bank is no longer being charged


z An off-line unit is technically not a UPS since there isa static switch for transferring between sources

z An off-line unit feeds the load directly from the

incoming utility AC power

A ti f th i i AC i tifi d t DC


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z A portion of the incoming AC power is rectified to DCand charges a battery bank

z If normal power fails, the static switch transfers tothe inverter AC output

z Again, the AC output power is being drawn from thebattery bank


z Some off-line units today employ very fast statictransfer switches that allege to be so fast the loadswont notice

z Need to research this carefully since some computerloads cannot handle a momentary outage


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y g

z However, a reliable power system design would

include a true on-line UPS unit so the momentaryoutage question is no longer relevant


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Switchgear Auxiliaries

z Switchgear auxiliaries are an important component inpower system reliability

z Applies to both 12 kV switchgear and 480 V

switchgear, or whatever is in the power system

z The ability to continue to operate after utility power


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z The ability to continue to operate after utility powerfails is crit ical

Switchgear Auxiliaries

z Key Components:

z Control power for tripping

z

Charging springsz Relays


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y

z PLC for automatic functions


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Switchgear Control Power for Tripping Breakers

z If there is a fault in the system, the relay must sensethe fault condition and send a trip signal to thebreaker to clear the fault

z A fault could happen at any time

z Could be minutes after the utility circuit fails


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z Could be minutes after the utility circuit fails

z Must clear the fault

Switchgear Control Power for Tripping Breakers

z The circuit breaker contactor is held closed undernormal operations

z When a fault is detected, the trip coil in the breaker

control circuit operates the charged spring to quicklyopen the contactor


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z If control power is available, the motor operated

spring immediately recharges for the next operation

z Typical demand from the charging motor is about 7 Afor about 5-10 seconds


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Switchgear Control Power

z Maintaining a secure source of power for control ofthe switchgear is essential

z If there is a fault in the system, the relay must sense

the fault condition and send a trip signal to thebreaker to clear the fault


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z Several sources of control power:

z Stored energy in a capacitor

z 120 VAC

z 125 VDC or 48 VDC


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Switchgear Control: Stored Energy (Capacitors)

z Only useful for non-crit ical systems

z Amount of stored energy is limited

z

Not commonly used


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Switchgear Control: 120 VAC

z Only operational while 120 VAC is available

z First option is obviously 120 VAC from the utility

z

If util ity fails, then could be a small UPSz Not well l iked by maintenance personnel since they

h t b ti ll h ki th bil it d


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have to be continually checking the operability andfunctionality of small UPS units all over the place

Switchgear Control: 120 VDC or 48 VDC

z Most reliable since control power is obtained directlyfrom the battery bank

z There is no conversion to AC

z Less chance of component failure


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Switchgear Relays

z Can be powered from 120 VAC

z For reliability, select 125 VDC, particularly when thereis a battery bank for switchgear control

z Relays are a critical component in order to detect thepresence of a fault on a circuit


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z Again, the fault must be cleared

PLC for Overall Substation Control

z A PLC can be just as critical to switchgear operationif there are other automatic functions carried out bythe PLC

z The PLC can also detect alarm signals and sendthem on to the central control room or dial a phonenumber for help


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number for help

z For reliabili ty, select 125 VDC as the power sourcefor the PLC

z Or the same small UPS used for switchgear control

power

Elbow Terminations for MV Cable

z Terminations for MV cable can sometimes be a pointof failure in the power system

z Most common is the use of stress cones and skirts

with bare surfaces exposed

z The concept is to prevent a flashover from the phase


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voltage to a grounded surface, or ground fault

Elbow Terminations for MV Cable

z Dirt and dust build up along the cable from thetermination can create a flashover path, especiallywith moisture

z The skirts help to break up the voltage field as it triesto bridge the gap to the grounded potential

f


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z A molded elbow has no exposed energized surfaces

z The elbow also contains the electric field withinthereby decreasing chances for corona

z The molded elbow costs a little more but provides

another level of reliabil ity in the power system


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Demand Side Management

z Managing the duty cycle on large continuous loadscan keep systems at a minimum

z Example: Clean-in-Place heater, 400 kW, 480 V

400 kW x 2 hour warm-up cycle = 800 kWh200 kW x 4 hour warm-up cycle = 800 kWhLower energy cost in dollars if off-peak


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gy p

z Program CIP via SCADA CIP to start beforemaintenance crews arrive via PLC or SCADA

z 400 kW would have increased system size

Questions .?

Keene Matsuda, P.E.

Black & Veatch


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(949) 788-4291

[email protected]

Part 3of3 Reliability Power System Design Buenos Aires

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