This is a systems engineering and analysis presentation from Milsoft's 2009 User Conference. It was originally presented by Bill Kersting. The Milsoft Electric Utility Solutions Users Conference is the premier event for our users and the vendors who provide interoperable solutions or services that enhance Milsoft Smart Grid Solutions. If you’d like to be on our mailing list, just email: missy.brooks@milsoft.com.
2. What is to be presented
•
•
•
•
ANSI Voltage Standards
Methods of Voltage Regulation
Example of Regulator Settings
Example of Placement of Regulators
3. 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
4. Tools for Voltage Regulation
• Shunt Capacitors
• Step-Voltage Regulators
• Substation Load Tap Changing
Transformers
5. Distribution Line Voltage Drop
R
jX
+
V
S
+
VL Load
-
I
-
VS
Im(ZI)
ZI
Real(ZI)
VL
0
jXI
RI
I
Vdrop
=
VS − Vr
Vdrop ≈ Real ( Z ⋅ I L )
=
Real ( R ⋅ I L + jX ⋅ I L )
6. Vdrop = Real(ZIL)
• 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.
7. Capacitor Voltage Rise
R
+
X
+
I L + IC
VS
VL
Load
IC
_
_
'
VS
IC
jXI C
VS
δ
ZI L
VL
θ
RI L
RI C
Im ( ZI L )
Real ( ZI L )
jXI L
IL
Vrise
=
'
VS − VS
Vrise ≈ Real ( Z ⋅ I C )
=
Real ( R ⋅ I C + jX ⋅ I C )
8. ANSI Range A Critical Voltages
Laterals
Sub
Reg
128
126
124
122
120
118
116
114
Reg
Output
First
Customer
Last
Customer
9. 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.
12. Type B Step Voltage Regulator
Preventive
Autotransformer
N1
IS
R
S
+
Reversing
Switch
L
Series
Winding
IL
VS
N2
Control
CT
Shunt
Winding
Control
PT
-
+
VL
SL
L
13. The Step Voltage Regulator Model
=
VL
1
⋅ VS
aR
I=
L
aR ⋅ I S
where: aR = 1
N2
N1
= 1 0.00625 ⋅ Tap
aR
One tap change = 0.75 V change on 120 V base
14. Three Phase Voltage Regulator Model
[ I S ]abc
[VS ]abc
=
[VS ]abc
=
[ I S ]abc
=
[VL ]abc
[ I L ]abc
aR ⋅ [VL ]abc
d R ⋅ [ I L ]abc
AR ⋅ [VS ]abc
[VL ]abc
15. Voltage Regulator Model Matrix
aR _ a
[ aR ] = 0
0
0
aR _ b
0
dR ]
AR ]
[= [ = [ aR ]−1
0
aR _ c
0
16. Compensator Circuit
MVA rating
Iline
R line + jX line
CTp − CTs
Ic
R c + jX c
kVLL hi − kVLLlow
Reg. Point
1:1
+ Vdrop +
N PT :1
Vreg
-
+
VR
-
Voltage
Relay
18. Control Circuit
Line Current
Control Current Transformer
Line D
rop
Compensator
Control Potential Transformer
V
oltage
Relay
Time
Delay
Motor
Operating
Circuit
19. 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
20. Equivalent Line Impedance
For i = a, b, c
Vregi − Vreg _ pti
Zlinei
Iregi
Ω
where: Vreg = actual line-to-neutral voltage output of regulator
Vreg _ pt = actual line-to-neutral voltage at the regulation point
Ireg = actual line current leaving the regulator
21. Compensator Impedance
Zcomp
=
Zline ⋅
CT
N pt
Volts
where: Zline = equivalent line impedance in Ohms
CT = current transformer primary rating
VLN rated
N pt = potential transformer ratio =
120
24. 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
25. 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.
28. Voltages and Currents from Power-Flow Run
eg.
V2 :=
j ⋅0
2521.87⋅ e
− j ⋅120 ⋅deg
2521.87⋅ e
2521.87⋅ ej ⋅120 ⋅deg
Ireg :=
V4 :=
2310⋅ e− j ⋅3.5 ⋅deg
− j ⋅124.6 ⋅deg
2377.5⋅ e
j ⋅116.1 ⋅deg
2284.2⋅ e
590.8⋅ e− j ⋅34.4 ⋅deg
− j ⋅150.5 ⋅deg
632.5⋅ e
651.9⋅ ej ⋅81.4 ⋅deg
29. Compensator R and X Setting
CT p = 700
Zlinei :=
N p t = 20
V2 − V4
i
i
Ireg
i
Z avg := mean( Zline)
Z set := Z avg⋅
CT p
Np t
0.1671 + 0.4037j
Zline = 0.0541 + 0.3817j
0.1426 + 0.4188j
Z avg = 0.1212 + 0.4014j
Z set = 4.2 + 14j
volts
30. 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
32. Full Load with Regs, no Caps
135
135
133
Node Voltage
131
V.a
129
127
V.b
125
V.c
123
121
119
115
117
115
1
1
2
3
Node
4
5
5
33. 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
40. 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
41. 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
52. 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.
58. 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
60. 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).
63. 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.
67. 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.
68. Sub Regulator set with R = 14.4 and X = 9.6
Voltage Output (level) = 120
69. Install Regulators at Node 5
• Set voltage level = 126
• Regulator set to infinite
• R and X = 0
72. 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.
77. Regulator at Node 5 set with V = 126
No 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.
79. 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.
80. System does not converge
4.16 Reg set with V = 122, R = 12, X = 7.2
• Use Set Regulation to compute R and X
for Reg 11 and Reg 20 with voltage output
= 122.
81. Node 5: R = 9.6, X = 4.8; Voltage Output = 122
Node 20: R = 12, X = 9.6; Voltage Output = 122
84. 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