Voltage Regulation

Bill Kersting

What is to be presented

• ANSI Voltage Standards• Methods of Voltage Regulation• Example of Regulator Settings• Example of Placement of Regulators

The ANSI Voltage Standards

• Range A– Nominal Utilization Voltage = 115 volts– Maximum Utilization Voltage = 126 volts– Minimum Service Voltage = 114 volts– Minimum Utilization Voltage = 110 volts

• Range B– Nominal Utilization Voltage = 115 volts– Maximum Utilization Voltage = 127 volts– Minimum Service Voltage = 110 volts– Minimum Utilization Voltage = 107 volts

Tools for Voltage Regulation

• Shunt Capacitors• Step-Voltage Regulators• Substation Load Tap Changing

Transformers

Distribution Line Voltage DropR jX

Load+

-VS

+VL

-

I

IRI

VS

VL jXI

ZI

0

Im(ZI)

Real(ZI)

( ) ( )Real Realdrop S r

drop L L L

V V V

V Z I R I jX I

= −

≈ ⋅ = ⋅ + ⋅

• Impedance (Z) and current (I) must be computed as accurately as possible.

• Impedance best computed using Carson’s Equations

• Current is a function of “load.”• If Z and I are not computed accurately, all bets

are off on the calculated system voltages.

Vdrop = Real(ZIL)

Capacitor Voltage RiseR X

Load+

_SV

+

_LV

LILRI

LjXI( )Real LZI

( )Im LZI

LV

LZI

SV

δ

θ

CI CRI

CjXI'SV

CIL CI I+

( ) ( )

'

Real Real

rise S S

rise C C C

V V V

V Z I R I jX I

= −

≈ ⋅ = ⋅ + ⋅

ANSI Range A Critical Voltages

LastCustomer

FirstCustomer

RegOutput

126124122120118

116114

Sub Reg

128

Laterals

Voltage Drop Assumptions

• 1 Volt drop on the service drop• 2 Volt drop on the secondary• 3 Volt drop through the transformer

• Minimum Voltage at the Transformer Primary Terminals will be 120 volts.

Voltage Profiles

LastCustomer

FirstCustomer

RegOutput

126124

122120118

116114

PointReg. Last

Xfm

Min Load

Max Load

Sub Reg

128

Laterals

Step Voltage Regulator

Type B Step Voltage Regulator

R

L

SeriesWinding

ReversingSwitch

PTControl

ControlCT

+

-

SL

-

+

ShuntWinding

S

L2N

PreventiveAutotransformer 1N

SV

LV

SI

LI

The Step Voltage Regulator Model

2

1

1

where: 1

L SR

L R S

R

V Va

I a INaN

= ⋅

= ⋅

=

1 0.00625 TapRa = ⋅

One tap change = 0.75 V change on 120 V base

Three Phase Voltage Regulator Model

[ ]L abcI[ ]S abcI

[ ]S abcV [ ]L abcV[ ] [ ][ ] [ ][ ] [ ]

S R L abcabc

S R L abcabc

L R Sabc abc

V a V

I d I

V A V

= ⋅

= ⋅

= ⋅

Voltage Regulator Model Matrix

[ ]

[ ] [ ] [ ]

_

_

_

1

0 0

0 0

0 0

R a

R R b

R c

R R R

a

a a

a

d A a −

=

= =

Compensator Circuit

+

-

+

-

line lineR Xj+

Reg. Point

dropV

regVPTN :1

1:1

p sCT CT−

RV VoltageRelay

cIc cR + jX

lineI

+ -

hi lowkVLL kVLL−

ratingMVA

Control Panel

Control Circuit

Line DropCompensator Relay

Voltage TimeDelay

MotorOperatingCircuit

Control Current TransformerLine Current

Control Potential Transformer

Regulator Control Settings

• Voltage Level – voltage to hold at the regulation point

• R and X setting (volts) – Equivalent impedance from the regulator to the regulation point

• Time Delay – time after a tap change required before the tap is changed

• Bandwidth – allowed deviation from the set voltage level

Equivalent Line Impedance

For , ,_

where: = actual line-to-neutral voltage output of regulator _ = actual line-to-neutral voltage at the regulation point

i ii

i

i a b cVreg Vreg ptZline

IregVregVreg ptIreg

=−

= Ω

= actual line current leaving the regulator

Compensator Impedance

Volts

where: equivalent line impedance in Ohms CT = current transformer primary rating

N potential transformer ratio = 120

pt

ratedpt

CTZcomp ZlineN

Zline

VLN

= ⋅

=

=

Bandwidth

Bandwidth = 2 V

123

122121Vo

ltage

Level

T∆

Modified IEEE 13 Node Test Feeder

1

13

2

3

5

6 789

14

1012 114

Modifications

• Line 4-12 changed to phases B-C• Transformer 6-7 changed to Ungrounded Wye – Delta• Load at Node 7 converted to Delta-PQ• Load at Node 8 converted to Delta-PQ• Load at Node 14 changed to phase B with constant Z

load• Load added at Node 5 phase c: 300 + j145.3 kVA• Interchange phase a and c distributed loads on line 3-4

Step 1

• Select regulation point to be Node 4.• Turn off regulator in Analysis Manager.• Run power-flow with source set to 126

volts (IEEE 13 Node Test Feeder Start.wm).• Display Voltage Profile.• Compute compensator impedance.

Step 1 Voltage Profile

1 2 3 4 5110

112.5

115

117.5

120

122.5

125

127.5

130

132.5

135

Node

Volt

age

135

110

V.a

V.b

V.c

51 Node

Voltages and Currents from Power-Flow Run

eg.

V2

2521.87 ej 0⋅⋅

2521.87 e j− 120⋅ deg⋅⋅

2521.87 ej 120⋅ deg⋅⋅

:= V4

2310 e j− 3.5⋅ deg⋅⋅

2377.5 e j− 124.6⋅ deg⋅⋅

2284.2 ej 116.1⋅ deg⋅⋅

:=

Ireg

590.8 e j− 34.4⋅ deg⋅⋅

632.5 e j− 150.5⋅ deg⋅⋅

651.9 ej 81.4⋅ deg⋅⋅

:=

Compensator R and X Setting

CTp 700= Npt 20=

Zlinei

V2iV4i

−

Iregi

:= Zline

0.1671 0.4037j+

0.0541 0.3817j+

0.1426 0.4188j+

=

Zavg mean Zline( ):= Zavg 0.1212 0.4014j+=

Zset ZavgCTp

Npt⋅:= Zset 4.2 14j+= volts

IEEE 13 with Regulator Set

• Set source voltage to 120 V.• Set regulator control.

– R and X = 4.2 + j14– Set voltage output (level) to 121.

• Analysis Manager– Set regulator to step.

• Run voltage drop.– Show results– Show profile

IEEE 13 with Regulator Set

Full Load with Regs, no Caps

1 2 3 4 5115

117

119

121

123

125

127

129

131

133

135

Node

Volt

age

135

115

V.a

V.b

V.c

51 Node

Use WindMil “Set Regulation”

• Select Voltage Drop.– Analysis Manager

• Set regulators to infinite. • Set source to 126 volts.

• Select Set Regulation.– Analysis Manager

• Select substation regulator.• Select Node 4 as load center.• Most desirable voltage = 121• Tolerance 2%• Unbalanced study

Set Voltage Regulation

WindMil R and X settings

WindMil with R 4.8, X = 14.4 and no feeder caps

• Set source voltages to 120.• Set voltage level (output voltage) to 121 V.• Run Voltage Drop.

WindMil R and X setting with no feeder capacitors

WindMil with regs, no caps

1 2 3 4 5115

117

119

121

123

125

127

129

131

133

135

Node

Volt

age Va

Vb

Vc

Node

WindMil Voltage Profile

Observations

• Regulator taps– Phase a: 12– Phase b: 13– Phase c: 15

• Concern that Phase c is near maximum tap• Concern about high voltage at Node 2• Need to add shunt capacitors

Shunt Capacitors

• Source reactive power– Phase A: 834 kVAr– Phase B: 805 kVAr– Phase C: 1040 kVAr

• Install shunt capacitors– Node 3: 100 kVAr per phases a,b,c– Node 4: 300 kVAr per phases a,b,c– Node 4: Switched 300 kVAr per phases a,b,c

WindMil R and X settings with capacitors

Full Load with Regs and Caps

Observations

• Regulator taps– Phase a: 6– Phase b: 6– Phase c: 8

• Concern for voltage unbalance at Node 4

Node 4 Voltage Unbalance

V4

119.8

124

121.4

:= Vavg mean V4( ):= Vavg 121.7333=

Devi V4iVavg−:= Dev

1.9333

2.2667

0.3333

=

Vunbalancemax Dev( )

Vavg100⋅:= Vunbalance 1.862= %

Minimum load of 50%

• Analysis manager– Set load growth to -50%

• Run voltage drop– Observe power factor at source– Switch 900 kVAr at Node 4

50% load reduction with all capacitors

50% load with all Capacitors

50% load with 900 kVAr at Node 4 switched off

1 2 3 4 5115

117

119

121

123

125

127

129

131

133

135

Node

Volt

age Va

Vb

Vc

Node

10% Growth with Original Capacitors

10% Load Growth

• Analysis Manager– Set load growth to 10%.

• Run voltage drop– Voltage profile– Check kVAr supplied by sub.– Install new shunt capacitors if necessary.

10% load growth with original capacitors

50% load reduction, switch off 900 kVAr at Node 4

10% Load Growth with original caps

10% Growth with 300 kVAr added at Node 10

10% Load Growth100 kVAr per phase added at Node 10

IEEE 34 Node Test Feeder

• Will be used to:– Determine location of downstream step

voltage regulators– Voltage level– R and X settings

• My method• WindMil method

Modified IEEE 34 Node Test Feederhttp://ewh.ieee.org/soc/pes/dsacom/testfeeders.html

9

301

8

21

22

23

2414

33

11 12 25

19 20

2615

27

28

29

16 1718

31

Sub

32

5

10

13

2 3 4 6 7

To Start

• System is very unbalanced.• System is very long (35 miles).• Voltage level is 24.9 kV.• Set substation output voltage to 126 volts.• Run power flow for the IEEE 34 node system

with no regulators or shunt capacitors (IEEE 34 Node Bare Bones).

IEEE 34 with no regulators and no capacitors

IEEE 34 with no regulators or capacitors

Install substation regulators

• Install 3 Step Voltage Regulators connected in grounded Y in the substation to start the regulation process.

• Potential transformer ratio = 14,400/120• Current transformer ratio = 100/0.1• Voltage level = 126 volts• Bandwidth = 2 volts• R and X = 0• Run power flow.

Modified IEEE 34 Node Test Feederhttp://ewh.ieee.org/soc/pes/dsacom/testfeeders.html

30

32

21

22

23

2414

33

13

10 11 12 25

19 20

2615

27

28

29

16 1718

31

Sub

1

2 3 4 5 6 7

8

9

IEEE 34 with Y connected sub regulators,Voltage Output (level) = 126, R and X = 0

Voltage Profiles with Substation Regulators with Voltage Level = 126

Observations and next step• Node 5 is the first node downstream where the voltage

drops below 120.• Select Node 5 as the regulation point for the substation

regulator.• Set regulators to infinite.• Run Set Regulator to compute R and X settings.• Set R and X on the sub regulator control.• Set voltage level on regulator to 120 volts.• Run power flow with regulators set as step.

Sub Regulator set with R = 14.4 and X = 9.6Voltage Output (level) = 120

Install Regulators at Node 5

• Set voltage level = 126• Regulator set to infinite• R and X = 0

IEEE 34 Node Test Feederhttp://ewh.ieee.org/soc/pes/dsacom/testfeeders.html

30

32

21

22

23

2414

33

13

10 11 12 25

19 20

2615

27

28

29

16 1718

31

Sub

1

2 3 4 5 6 7

8

9

Substation Regulators SetRegulators Installed at Node 5 Voltage Level = 126

Observations

• All voltages at Node 5 are between 119 and 121 volts.• The first downstream node where all of the voltages drop

below 120 V is Node 11.• Set regulators to Infinite.• Run Regulation Set to compute R and X from Node 5 to

Node 11.• Set regulators to step.• Run power flow.

Sub and Node 5 (R=16.8, X = 7.2) Regulators Set

Observations

• Install a regulator at Node 11.• Set voltage level to 126.• Set regulators to step.• Run Voltage Drop.

Regulator Installed at Node 5

30

32

21

22

23

2414

33

13

10 12 25

19 20

2615

27

28

29

16 1718

31

Sub

1

2 3 4 5 6 7

8

9

11

Reg at Node 11 Set to 126 Volts, R and X = 0

Regulator at Node 5 set with V = 126No R and X

• With the regulator set at 126 volts, all of the downstream voltages in the main feeder are greater/equal to 120 volts.

• No need to set R and X for this regulator• The only problems occur on the 4.16 kV

line from 19 to 20.

Profile including the 4.16 kV line

Install regulator at secondary terminals of the transformer

• Potential Transformer Ratio = 2400/120• Primary CT Rating = 100 amps• Calculate R and X.• Set regulators to Infinite.• Load center is Node 20. • Run Set Regulation.

System does not converge4.16 Reg set with V = 122, R = 12, X = 7.2

• Use Set Regulation to compute R and Xfor Reg 11 and Reg 20 with voltage output = 122.

Node 5: R = 9.6, X = 4.8; Voltage Output = 122 Node 20: R = 12, X = 9.6; Voltage Output = 122

Set source voltage to 120

• Run with no capacitors.• Add capacitors.

Source set to 120, no capacitors

Correct feeder power factor to near 1

• Display P and Q on 4.16 kV line.– Install a three phase capacitor bank to supply most of

the 4.16 kV kVAr load.– 75 kVAr/phase at Node 20

• Need to add 200 kVAr/phase– Node 16, 100 kVAr/phase– Node 822: 100 kVAr/phase A– Node 848: 100 kVAr/phase

Final with capacitors

Final Regulator Tap Positions

• Sub Regulator: 9, 8, 7• Node 812: 10, 6, 7• Node 830: 5, 6, 7• Node 888: 12, 12, 12

Final kVAr supplied by source

• Source power factor:– Phase a: 43 (PF = 99.8 %)– Phase b: 71 (PF = 99.5 %)– Phase c: 16 (PF = 99.9 %)

To be continued by you

• Minimum load– Which capacitors to switch

• Load growth– Where and how big new capacitor banks