imbalance load

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

Our Sample Circuit

105

110

115

120

125

130

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

Voltage Profile

Voltage A

Voltage B

Voltage C

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

0

20

40

60

80

100

120

140

160

0.0

1.4

2.0

3.1

4.8

5.3

5.7

6.9

7.8

8.5

9.1

10.0

10.3

10.9

12.0

12.2

13.8

Am

ps

Current Profile

Thru Amps A

Thru Amps B

Thru Amps C

Thru Amps Return

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

Balanced ImbalancedAmps I2 Amps I2

A 200  40,000  A 300  90,000 B 200  40,000  B 200  40,000 C 200  40,000  C 100  10,000 

Total =  120,000 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

Before A8er%LoadImbalance 52.20%  10.46% %VoltageImbalance 7.20%  1.41% LineLosses 153 kW  108 LowestPrimaryVoltage 111.5V  118.1V HighestPrimaryVoltageatendoffeeder 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.