Engineering Analysis and Load Imbalance

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Load Imbalance
Jennifer Taylor, P.E.
Distribution System Solutions
Perfect World
The Problem
•  Over time distribution feeders have a
tendency to increase in load imbalance:
–  Loads on single phase lines gradually increase
–  Single phase lines arbitrarily get switched to
other phases
The Problem
–  Unequal distribution of single phase loads on
three-phase lines
–  Lack of planning
The Result
• 
• 
• 
• 
Voltage shifts
Increased return currents
Increased losses
Physical ramifications
Load Imbalance Defined
•  Imbalance in current dictated by placement
of load on feeder.
•  Voltage or Current Imbalance:
A Real-Life Example
•  Rural Circuit: 20 miles long
•  Majority of backbone: 3 phase 1/0 CU and
1/0ACSR
•  Primary voltage: 12.5/7.2 kV
Phase
# of consumers
Feeder Amps
A
94
52 A
B
291
145 A
C
161
89 A
A Real-Life Example
What Happens When You Have Imbalance?
•  The imbalance in current creates a neutral
point “shift”
–  Voltage imbalance
–  Over voltage and under voltage
Voltage
•  ANSI C84.1specifies:
Voltage
imbalance not to
exceed 3%
Acceptable Voltage Range
What Causes the Shift?
•  The neutral point “shift” is a function of:
–  Per phase load
–  Per phase power factor
–  Impedance matrix
•  Not just per phase current magnitudes
Voltage Imbalance: My Assumption
•  Voltage Rise or Voltage Drop?
–  Heaviest loaded phase = Largest Voltage Drop
–  Lightest loaded phase = Largest Voltage Rise
Voltage Imbalance: In Reality
•  Voltage Rise or Voltage Drop?
–  Heaviest loaded phase = Largest Voltage Drop
–  However, the largest voltage rise doesn’t necessarily happen on the
lightest loaded phase.
–  In fact, in a majority of cases with standard construction and typical
power factors (90-100%) the phase behind the heaviest phase (c.c.
rotation) gets the voltage rise.
For Example:
For Phase “X”
VLxg = Exg - Vx
Exg = Source Voltage
Vx = Voltage Drop
VLxg = Load Voltage
In this example, phase A
was the heaviest loaded
phase. B and C were equal.
Largest voltage drop
Voltage Rise
Voltage Drop (Vd)
•  Voltage drop magnitude and angle
–  Function of the current magnitude and angle in
each phase
–  Self and mutual impedance matrices
–  Consists of a self voltage drop and 2 mutual
voltage drops
Voltage Drop (Vd)
•  For example for phase A:
–  Vda = vd1,1 + vd1,2 +vd1,3
–  Where vd1,1 = ZABC1,1.I1
vd1,2 = ZABC1,2.I2
vd1,3 = ZABC1,3.I3
ZABC
•  ZABC is the 3x3 phase impedance matrix
–  Function of resistance of conductors, GMR,
distances between positions, length of feeder
–  Consists of modified Carson’s equations (4x4
matrix including neutral)
–  Then through Kron reduction becomes 3x3
–  For a deeper explanation call 1-800-344-5647
Problems with Over/Under Voltage
•  Overvoltage/Undervoltage
–  Equipment damage
–  Motors won’t operate as efficiently
–  Overheating of induction motors
–  Tripping of sensitive loads
–  Higher no-load losses in transformers
Problems with Voltage Imbalance
•  Voltage imbalance
–  Generates high negative sequence currents
which puts back torque on motors
–  Causes adjustable speed drives to draw
significantly imbalanced currents and cause the
overload protection to trip.
–  Increases the harmonics that ASD’s produce
–  Overheating in transformers
Note on Voltage Imbalance
•  Unequal phase impedance due to
asymmetric conductor spacing can cause
voltage imbalance
–  However on distribution feeders this is quite
small; < 1% imbalance
(T.A. Short, Electric Power Distribution Handbook)
–  Most voltage imbalance is due to load imbalance.
Our Sample Circuit
% Voltage
imbalance = 7.2%
Phase angles drift
from 120o apart
105
0.0
0.1
0.3
0.7
1.1
1.6
2.0
2.2
2.7
3.1
3.6
4.3
4.8
5.1
5.4
5.6
5.9
6.4
7.1
7.6
7.8
8.3
8.8
9.1
9.5
10.0
10.5
11.0
11.6
12.2
12.8
13.5
Volts
Our Sample Circuit
Voltage Profile
130
125
120
Voltage A
115
Voltage B
Voltage C
110
Voltage Problems Reported on this Circuit
•  Low voltage complaints
•  Equipment damage
What Else Happens When You Have Imbalance?
•  Increased return current
In = Ian + Ibn + Icn
If currents are equal and
120o apart then In will be
zero. If not then In will
be non-zero.
If grounded system then
some of the return
current goes through
the earth.
Problems with Increased Return Current
•  Challenges with coordination
–  Return current may approach the minimum
ground current pickup of the protective device
Problems with Increased Return Current
•  False ground trips
–  If pickup is low enough, could get false ground
trips on substation reclosers due to imbalance
Problems with Increased Return Current
•  Stray voltage
–  Imbalance causes increased return current on
both the neutral and through the earth-return
–  This causes a potential difference between the
neutral and earth (e.g. stray voltage)
•  Worse, the closer you get to the substation
–  If there happens to be a break in the neutral also
get stray voltage
Our Sample Circuit
Total Return Current
= 81A
% Current
imbalance = 52.2%
Our Sample Circuit
Current Profile
160
140
120
Thru Amps A
80
Thru Amps B
Thru Amps C
60
Thru Amps Return
40
13.8
12.2
12.0
10.9
10.3
10.0
9.1
8.5
7.8
6.9
5.7
5.3
4.8
3.1
2.0
0
1.4
20
0.0
Amps
100
Return Current Issues Reported on this Circuit
•  Cold load pick up issues
What Else Happens When You Have Imbalance?
•  Increased losses
–  The imbalance of current will increase the I2R
losses
I2R
•  Let’s look at a simple math exercise:
•  Total Current = 600A
A
B
C
Balanced
Amps
I2
200 40,000 200 40,000 200 40,000 Total = 120,000
Imbalanced
Amps
I2
A
300 90,000 B
200 40,000 C
100 10,000 Total = 140,000
Problems with Increased Losses
•  Increased costs
•  Increased heating
Our Sample Circuit
• 
• 
• 
• 
• 
Losses before balancing = 153kW
2 tap phase changes = $500
Losses after balancing = 108kW
Reduction in losses = 45kW
30 year present worth = $37,120
That’s just one feeder!
What Else Happens When You Have Imbalance?
•  Physical Equipment Ramifications
–  Increased conductor heating
–  Uneven sagging of 3 phase conductors
–  Underutilization of equipment (kVA)
•  Unnecessary improvements
–  Premature conductor upgrades
–  Voltage regulators
Load Balance in Windmil
•  Uses Voltage Drop to
calculate losses and
establish load currents in
taps.
•  Two Methods
–  With Approximation
–  Without Approximation
Approximate Method (Fast)
•  Voltage drop runs initially to
determine load current in
taps.
•  Each taps’ I2R losses are then
added/subtracted to phases
until an optimum reduced
loss configuration is
established.
Without Approximation (Not Fast)
•  Go out for lunch
•  Runs voltage drop for each
trial of a tap change to
determine the optimum
phase for each tap to have
overall greatest loss
reduction.
Load Balance in Windmil
•  Balance all active
circuits
•  Balance downline from
a specified element
Possible Future Enhancements
•  Load Balance results automatically stored as
a project.
•  Report will be attached and saved with the
project.
•  Balance based on Amps vs. kW losses.
•  Ability to exclude specified taps from getting
phase changed.
Sample Circuit: Before and After
%
Load
Imbalance
%
Voltage
Imbalance
Line
Losses
Lowest
Primary
Voltage
Highest
Primary
Voltage
at
end
of
feeder
Before
A8er
52.20% 10.46% 7.20% 1.41% 153 kW 108 111.5V 118.1V 126.3V 121.2V Other Changes to Consider
•  In addition to changing tap phases
–  Multi-phasing
–  Backfeeding / changing open points
Before you begin
•  Verify phasing on feeders
–  Is model correct?
–  Is SCADA correct?
–  Are consumer phases correct?
Other Things to Consider
•  Downline monitoring
–  Smart devices
•  Are the recommendations feasible in the
field?
–  Crossing jumpers under/around phases
–  Bird issues with extra jumpers
–  Non-loadbreaking taps
What Load Profile to Use?
•  Many opinions on this:
–  Winter peak
–  Summer peak
–  Somewhere in the middle
–  What about seasonal loads and the impact to
imbalance?
In Conclusion
•  Benefits of load balancing
–  Improved voltage profile
•  Better for motors, demand reduction possible, easier
to obtain required voltage levels
–  Reduced return current
•  Reduce/eliminate stray voltage, avoid false ground
fault trips, ease coordination
In Conclusion
•  Benefits of load balancing continued
–  Loss savings
–  Better utilization of equipment
•  Defer improvements
The Bottom Line
•  Load balancing is a simple concept, but can
be difficult to achieve.
•  The benefits make it at least worthy of a look.
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