electric system losses - coursewebs

© 2010 Siemens Energy, Inc. All rights reserved

Electric System LossesJuly 2010 Siemens


Tab 1 –

Course Outline

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3Siemens Energy Inc., Siemens Power Technologies International

Exclusive Copyrighted Property

Course notes provided to course participants in any form, electronic or otherwise, are the exclusive copyrighted property of Siemens Energy, Inc., Siemens Power Technologies International. Course participants may only use the course notes for completion of the course and for each participants own future reference. Course participants may not make copies or share the course notes in any way.Siemens

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4Siemens Energy Inc., Siemens Power Technologies International

Course Outline

Basic Concepts

Data Requirements

Loss Calculations Basics and Terminology

Equations and Definitions

Methodologies for Loss Calculations

Siemens


Tab 2 -

Basic Concepts

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Siemens Power Transmission & Distribution, Inc., Power Technologies International 2

Why electric losses?

Electric losses occur as energy is transformed into waste heat in electrical conductors and equipment

Energy (kWh) and power (kW) losses result from the normal operation of the power system

Electric losses should be not higher than the absolute minimum dictated by the economics of the power system operation

Losses are measured or calculated for a given time period, usually a year

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Why the Need to Know Electric Losses?

Electric loss is a waste of money and resources

Electric losses can be reduced –

not eliminated

Loss reduction may requires money investments

Money saved in loss reduction ≥

Money invested

In loss studies the following is determined,

Total electric system losses

How the losses are distributed in the system. Therefore, elements or areas of the system having unusually high losses can be targeted for loss reduction measures

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Why the Need to Perform Loss Calculations?

Electric losses occur in every

system element

System is composed of a very great number

of elements

Loss measurement in every element is impossible –

at

least for now

The calculation of losses, using standard engineering equations, yield fairly accurate results

SCADA and Data Management System records are very valuable sources of information for loss studies

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Load and Loss Equations for Loss Calculation

ii LossiLoad PLP

220 iLoss LCCP

i

2210 iiLoss LCLCCP

i

2210 iiiLoad LCLCCLP

i

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Transmission System Load and Losses

1500

1700

1900

2100

2300

2500

2700

2900

3100

3300

3500

100 600 1100 1600 2100

Hours

Load

(MW

)

Transmission System Load


Difference are Transmission Losses

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Energy Losses

Energy losses

Do we really lose energy?

Energy is never lost, it only changes its form. This is a basic law of Physics –

Principle of Conservation of Energy

Energy is lost when is converted from one form of energy into another form of energy:

Mechanical to electrical (generators, etc.)

Electrical to mechanical (motors, etc.)

Electric energy is also lost when it is transported, for example, when electricity flows through conductors

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Energy Losses (continued)

Electric energy losses result from the transmission and distribution of energy from generators or tie-line sources to the ultimate customers

Starting or input point

Ending or output point

Process, event

Energy input

Energy output

Loss

EnergyLoss EnergyInput EnergyOutput

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Electric System Losses

T ransm ission S ystem

D istribu tion S ystem

G eneration S ystem

S ales

H igh V oltage C ustom ers (assum es tha tnon-techn ica l lo sses m ay be neg lec ted)

O w n-use (no t considered to be losses

T echn ica l L osses N on-T echn ica l L osses

T echn ica l L osses

Losses occur at each delivery, service, or voltage level, from the transmission system down to the 120/240-volt servicedrops and customer meters

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Electric System Losses (continued)

Losses on network elements

i o LE E E

System losses

Network elementEnergy input

Energy output

Loss

i o SLE E E Siemens

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Energy and Power Losses

Energy losses are composed of:

Power losses (expressed in Watt, kW, or MW)

Energy losses (expressed in kWh or MWh)

Power loss is energy lost per unit of time. It expresses how quickly the energy is being lost

Electric Power loss is also referred to as “demand loss”

Energy loss is energy lost in a period of time, for example, a year

(kWh) (kW) (hour)Energy Power Time

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Again, Why Energy Losses Need to be Determined ?

There is an increasing interest in the study of electric system losses in general, and distribution losses in particular, due to:

Increase of cost of electric energy and power

Increase of capital cost

Difficulty of siting

new generation

Increased pressure from regulatory bodies

to improve loss management techniques and reduce losses in general

Electric utilities are assessing the effects of electric system losses on the operation and planning of power systems

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Economic Requirements

Overall Annual Electric Utility Cost for Typical U.S. Utility

Aprox. %

O&M $ Capital& Labor 10%

Distr. Powr. Purch. Sales 5%

Losses 10%

Functional 5%

70%

Equipment Costs

$ Capital

Labor

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Due to energy losses there

are:

&

Basic Loss Concept

Energy losses have two main components in their financial impact:

Lost energy has a direct cost

To produce and transport the energy that will be lost, at certain power losses, additional capacities have to be built-in the power systems, at a cost

Energy costs

Capacity costsSiemens

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Incentives to Reduce Losses

The cost of supplying energy is impacted by,

Cost and availability of capital

Availability of site locations for new generation in relation to

the electrical grid. Additional land and right of ways need to be secured for substations and electric lines

Location of the load centers in relation to the generation

Pressure for the regulating bodies to keep the price of electricity low

Pressure by the investors for a higher rate of return

Distribution companies keep checking their systems trying to find ways to reduce losses.

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Who Pays, Who is Responsible?

Electric losses have a financial impact. Question is, who does pay for the financial impact of losses?

Answer: Consumers pay

Who is responsible for the losses?, where do they happen?

Answer: Power companies (system operators and planners) are the ones responsible, as the losses occur in their networks

Consumer and network users pay for the network improvements and upgrades that may result in a reduction of losses

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Minimum Cost Objective…

Losses can and should be reduced to decrease their economic and financial impact

On one side, relatively inexpensive network upgrade solutions, may result in higher losses

On the other side, the best network solutions, from the point of

point of reducing losses, may be relatively expensive

The good news is that, there is always room for efficiency improvements based on good and creative management

Consumer and network users are interested in an overall optimization and in network upgrade alternatives that minimize the overall cost

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…Minimum Cost Objective

It is a balancing act to find the optimal solution

Only relevant costs should be taken into account and over the right time horizon

Quality of input data is very important

Inefficient EfficientNetwork efficiency with regard to losses

Cost of losses plus cost of network solution

Optimum

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Again, Why Energy Losses Need to be Determined?…

Also, electric energy losses are determined for rate making purposes

Energy losses must be accounted for in rate design and allocated

to the parties responsible for creating the losses as energy travels over the delivery system. For example,

The transmission customer is responsible only for those losses that its load causes on the transmission system

The primary distribution customer is responsible for the losses resulting from its load on both the primary and the transmission systems

The secondary distribution customer is responsible for losses that its load creates in all three systems (transmission, distribution primary and distribution secondary systems

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…Again, Why Energy Losses Need to be Determined?

For example, if a residential customer requires one kWh of energy, the generation system would have to provide more that one kWh, let’s say 1.071 kWh, to supply this customer and to cover the losses. Therefore, the residential customer should pay for the production and transportation of 1.071 kWh

Information above is provided by power and energy multipliers

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Loss Analysis

Loss analysis start with information available

Energy sales and inputs

Adjust and take the difference

Results are total energy losses

Determine demand and energy losses by subsystem

Allocate demand and energy losses by subsystem

Information available is not always good or sufficient. Approximations or assumptions are sometimes are required to perform the loss analysis

As technological advances are implemented, the need for assumptions or approximations are less necessary

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Get the System Configuration

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Identify the Service Territory

Utility A

Utility B

Utility C

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Identify the Control Areas

Control A

Control B

Control C

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System Power Flows

ImportsExports

Wheeling

Loop Flow

Wheeling

Loop Flow

Losses

ToDistribution

Primary

ToDistributionSecondary

Losses

Industrial

Other

Losses

Residential

Commercial

Industrial

Transmission

System

Distribution

Primary

System

Generation

Generation

Distribution

Secondary

Other

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Locate the Meters and Sources of Data

Generator step up transformers

Control area, utility boundaries

Distribution substation transformers

Distribution circuits

Most customer load

Company use, street lights, traffic lights

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Classification of Losses

Technical losses

Energy units that are physically lost due to the physical nature

of the transmission and distribution networks

Non-technical losses

Some energy is actually delivered and consumed but, for some reason, is not recorded as sales; this energy is lost in the sense that it is not charged for by neither the suppliers nor the distribution businessesSiemens

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Technical Losses

Technical losses can be calculated. They occur in lines and transformers in the transmission and distribution systems

Transmission and sub-transmission systems (usually operating at voltages 69 kV or higher

Transmission to Distribution transformers

Primary Distribution system (usually operating at voltages 34 kV

or lower. Upper end of voltage range varies, depending on the utility or transmission operator)

Secondary Distribution system (usually operating at voltages such as 240/120, 480/277, etc.). Losses in customer meters are included

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Non-Technical Losses…

Non-technical losses include factors such as:

Un-accounted loads (substation light and power, government use, street lights, traffic lights, etc.)

Metering errors (equipment calibration, etc.)

Meter reading errors

Energy diversion (theft)

Non-technical losses are small (usually less than 1%)

Non-technical losses can be estimated

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…Non-Technical Losses

Meter errors

E.g. their accuracy could be -3.5% to +2.5%

Recently manufactured meters are more accurate

Measurement errors in the settlement system

Depends on electricity market arrangements and technology applied

Errors in accounting for the un-metered supply

Depend on how much consumption is un-metered (e.g. street lights, traffic lights and cameras, telecom masts, distribution utility’s consumption etc.) and how good are inventory data

Energy Diversion or Theft

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System Losses From Actual Studies in U.S.

Electric System Losses-%

COMPANY

1 2 3 4

Maximum Demand - MW 3,384 987 6,419 4,000

Total Energy- Consumption - GWH 21,456 4,918 31,680 22,800

Transmission Losses - % 29.2 31.6 48.8 35

Transmission to Distribution Losses - % 16.5 15.6 8.7 -

Primary Distribution Losses - % 20.3 25.4 20.4 28.7

Secondary Distribution Losses - % 33.1 27.3 22 36.3

Non-Technical Losses - % 0.9 0.1 0.1 0

TOTAL Losses - % 100.0 100.0 100.0 100.0

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In Some Cases Losses are Very High…

Energy Losses by Origin (37% of demand)

3%

8%

26%

Subtransmission losses

Distribution lossestechnical

Distribution losses nontechnical

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Tab 3 -

Typical Data Requirement and Loss Calculations

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Typical Data Requirements and Calculations…

1.

Active electric services

2.

Energy use by rate class or service level

3.

Average energy use by customer

4.

Energy balance

5.

Power pool interchange summary

6.

Hourly loads at every service level for entire study period

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…Typical Data Requirements and Calculations…

7.

Load/loss factor for entire system and for each sub-system

8.

If not provided, load and loss factors can be calculated or estimated

9.

Transmission tie line summary

10

.

Generation output summary

11.

Transmission power flow cases

12.

Data for Corona loss calculation

13.

Transmission transformer inventory (load & no-load data)

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14.

Generator Step-Up (GSU) transformer inventory

Load losses of GSU transformers may or may not be included in the loss calculation depending on where the plant meter is located. If the plant meter is on the generator side, the loss of GSU transformers should be included as part of the transmission losses.

15.

Calculation of Transmission System Losses

16.

Distribution substation transformer inventory

17

.

Calculation of substation transformer losses (Load and No-load)

18.

Distribution secondary transformer inventory

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19. Calculation of Secondary Transformer losses (load and no-load losses)

20. Distribution primary circuit characteristics and data

21. Calculation of primary circuit losses

22. Secondary and service drops characteristics and data

23. Calculation of secondary and service drops losses

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…Typical Data Requirements and Calculations

Customer meter inventory

Calculation of customer meter losses

Unmetered load and energy use estimation

Energy diversion estimation

Calculation of Total System Losses

Siemens


Tab 4 -

Loss Calculation Basics and Terminology

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Sub-systems for Loss Calculations…

Loss calculations are performed for each of eight sub-

systems:

1.

Generator step-up transformers (GSU’s)

2.

Transmission lines (usually, 69kV and above)

3.

Transmission transformers (69kV and above)

4.

Distribution primary transformers (with transmission level primary voltage and distribution level secondary voltage)

5.

Distribution primary lines (at voltages such as 34.5, 25.0. 13.8, 13.2, 12.7, 4.16 kV, etc.)

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…Sub-systems for Loss Calculations

Distribution secondary transformers (with primary distribution level primary voltage and secondary distribution level secondary

voltage)

Distribution secondary lines (440, 240, 120 volts, etc.)

Customer meters

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General Procedure

Calculate demand or power (kW) losses by sub-system

Calculate energy loss (kWh) for each subsystem from the demand loss (kW)

Adjust energy losses to match total system loss as determined from sales data

Calculate demand loss multipliers

Calculate energy loss multipliers

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Load Data Collection

Metered customers sales

Generator output –

kW (time series –

hour by hour)

Interconnection points in and out –

kW (time series-

hour by hour)

Sales to Large customers –

kW (time series –

hour by hour)

Customer load research –

where metered data is not available, (time series –

hour by hour estimation of load using statistical methods)

Transformer meters (hourly kW demand for each transformer)

Distribution line loadings per phase or total (amps, kW, Kvar, kWh)

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Definitions and Terms Used in Loss Studies

Load factor

Loss factor

Loss multipliers

Coincident factor

Diversity factor

Power factor

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Load Factor and Loss Factor…

Load factor

is the ratio of total energy (kWh) supplied during the period of study to the peak demand (kW) during the same period multiplied by the number of hours in the period (for example, 8,760 hours for a year). Load factor is calculated for each sub-system and the entire system

Load Factor is equal to the average demand (kW) divided by the peak demand (kW) in the same period

Loss factor

is the ratio of the total losses (kWh) during the period of study to the peak loss (kW) during the same period multiplied by

the number of hours in the period (for example, 8,760 for a year). Loss factor is calculated for each sub-system and the entire system

Loss Factor is equal to the average loss (kW) divided by the peak loss (kW) in the same period

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…Load Factor and Loss Factor…

unitperLoss

losskWhFactorLoss ,760,8*max

unitperP

loadkWhFactorLoad ,760,8max*

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…Load Factor and Loss Factor

The load and loss factors can be calculated for every sub-

system

The kWh load and the maximum demand, Pmax, are determined from metered data, or from load research data

The maximum demand loss, LossMAX

, can be determined from power flow solutions, or equipment manufacturer’s data, for non-coincident peak demand conditions, using metered or load research data

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Calculation of Energy Losses from Loss Factors…

Load factor for every sub-system is calculated from the total kWh load and the peak load for a given period of time, usually a year

Loss factor for every sub-system can be calculated from the total kWh loss and the peak loss for a given period of time, usually a year. If these values are not available, the loss factor can be estimated from the load factor

The basic data for the calculation of load and loss factors is the hourly load of the sub-system

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…Calculation of Energy Losses from Loss Factors

Load factors can always be calculated, even if the hourly loads are not available. This is because the total kWh load and the peak kW load are always known. These values are very basic pieces of data

If the hourly loads are not known, the loss factor can be estimated from the load factor. The estimation is sufficiently accurate, usually with an error or less than 1%. This type of calculation works well in radial systems but it should not be used in looped systems

Once the loss factor is known, the energy losses for the sub-system can be calculated from the non-coincident peak demand loss of the sub-system

For the energy loss calculation of the transmission system other method is used

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Load and Loss Factors Examples

Load and Loss Factors –

typically, these factors do not

change significantly from one year to the next

SUBSYSTEM LOAD AND LOSS FACTORS

Subsystem for 2004 Load Factor Loss Factor

Secondary 0.462 0.229

Primary 0.499 0.263

Control Area (EEI) 0.503 0.265

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Hourly Loads from Research Data

0

100

200

300

400

500

600

0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24

Time (Hours)

Load

(MW

)

Load research data integrated over some time period such as one hour.Siemens

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Load Factor

0

100

200

300

400

500

600

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24

Time (Hours)

Syst

em L

oad

(MW

)

Average Load

Hourly Load

Load Factor = 0.815

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Calculation of Load Factor from Hourly Loads

i

T

iLoad L

TF

1

*1

FLoad

= Load Factor

T = Period of Calculation (i.e. 8,760 hours)

i = Unit of time (i.e

one hour)

Li

= Average load during time “i”, in per unit of the

maximum load during the period

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Calculation of Loss Factor from Hourly Loads

2

1*1

i

T

iLoss L

TF

FLoss

= Loss Factor

T = Period of Calculation (i.e. 8,760 hours)

i = Unit of time (i.e

one hour)

Li

= Average load during time “i”, in per unit of the

maximum load during the period

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Per Unit Load/Loss Relationship

0.00

0.20

0.40

0.60

0.80

1.00

1.20

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24

Time (Hours)

Per

Unit

Per Unit Load

Per Unit Load Squared

Load Factor = 0.815Loss Factor = 0.679Siemens

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Empirical Calculation of Loss Factor

If the determination of loss factors is inconvenient or the data for such calculation is not available, empirical relationships can be used

Calculation of load factors is always possible because it uses very basic data that is always available. Therefore, if the loss factor cannot be calculated accurately, it can always be estimated

Once the loss factor is known, the kWh losses can be calculated from the kW losses for every sub-system

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Gustafson’s Loss Equation

The exponent is applicable to USA systems only. For other countries, a new exponent may need to be determined from actual load data

L L

L

L

Loss Load

Loss

Load

1 912.

Where: Is the Loss Factor.

Is the Load Factor.Siemens

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Buller

and Woodrow’s Loss Equation

Where:

x = A constant between 0.7 and 0.85

2)1( LoadLoadLoss xLLxL

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Loss Terminology…

Demand

Losses –

are power losses (kW).

Demand losses can always be calculated using suitable equations and data

The demand loss may be coincident

or non-coincident

The loss is coincident when it occurs a the time of the system peak.

Non-coincident loss occurs at non-peak times

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…Loss Terminology

Coincident Factor

–

Ratio of the coincident demand (kW)

to the non-coincident demand (kW)

Technical Losses

–

Losses that can be technically

calculated using equations or measured by equipment

Non-Technical Losses

–

Unaccounted losses that cannot

be determined using equations or a rational method. Usually called “energy diversion”. Electric energy theft is considered a non-technical loss

Total Losses = Technical + Non-Technical

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Coincidence Factor

The Coincident Factor

is the ratio of the maximum

demand of a set of users to the summation of the set’s individual maximum demand

CF = D1tp+2tp+….+Ntp

/(D1t1

+D2t2

…+DNtn

)

Coincident Factor is the reciprocal of the Diversity Factor

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Diversity Factor

The Diversity Factor

is the ratio of the sum of the non-

coincident maximum demands of two or more loads to their coincident maximum demand for the same period

DF = (D1t1

+D2t2

…+DNtn

)/ D1tp+2tp+….+Ntp

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Power Factor

Power Factor

is the ratio of real power (kW) to apparent

power (kVA) for any given load and time. Generally, it is expressed in percent or per unit

PF = cos

(kW/kVA)

Kva

KwSiemens


Tab 5 -

Equations and Definitions

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Important Equations

Real power balance

Reactive power balance

Power and VAr

equations

Basic equations

Per unit system

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Real Power Balance

F f MW MW MWgen ld ls

F

gen

ld

ls

( ) Frequency (Hz)

MW Input Power Generated (MW)

MW System Load (MW)

MW System Loss (MW)Siemens

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Reactive Power Balance

V =V = Voltage (kV)

= Reactive Power Generated (MVAR)

= Reactive Power Load (MVAR)

= Reactive Power Losses (MVAR)

f MVAR MVAR MVARgen ld ls

MVAR

MVAR

MVAR

gen

ld

ls

( )

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Power and VAR Equations

ES ER

XSR

RSS Sin

XEEP **

)*(* RSRSR

S ECosEXEQ

PS

QS

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Power Equations

S = E x I*

Where:

S = Apparent power (complex number or phasor)

I* = Conjugate of Current (complex number)

E = Voltage (complex number)

S = E x I x Cos φ

–

j E x I x Sin φ

S = P + jQ

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Basic Equations

I E RP I R

P ER

I

MVA

KVLL

/2

2

1

3

3

31000

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Voltage Drop Diagram

Vs R + jX VL

LOAD

VL = Vs - IZ = VS – I(R+jX)

I

θ

VL

-j IX-IR

VS

I

Assuming that VS

= 1 pu

Note that VL

< 1 pu

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Per Unit System

Per Unit Value = Physical ValueBase Value

Transmission Base Value = 100 MVA

Distribution Base Value = 1,000 kVA

Typical Base Power-Selected

Typical Base Voltage is Nominal Voltage at System Location

Typical Base Impedance –

Requires Calculation

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Per Unit Base Change

Z ZSSPUnew PUold

basenew

baseold

3

3

Typical use is to change transformer impedance, R and X.Siemens

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Per Unit Impedance

S3φ

= 3 x S1φ

= 3 x VLN

x IL

= √3 x VLL

x IL

BaseLLbaseL V

SI

33

base

BaseLL

BaseLN

BaseLNBase S

VIVZ

3

2Siemens

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The instantaneous power is

Using well known trigonometric identities and with a little work ...

p(t) = VI[cosθ(1

-

cos(2t)) -

sinθ

sin(2t)]

Where, rms values are used defined as follows:

tsinItsinV

titvtp

maxmax

maxV2

1V maxI

21

I

Instantaneous Power in AC Circuits

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Instantaneous Voltage, Current & Power in AC Circuits

Positive p(t): the source supplies energy

Negative p(t): the source takes energy

Frequency of p is 2 times that of V or I.

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Active and Reactive Power

P -

the active powerpP

(t) = |V| |I| cosθ

(1 -

cos (2t))which is always positive

the peak value of pP

(t) is P:

P = |V| |I| cos θ

[watts, W]

Which is also the average value

Q -

the

reactive powerpQ

(t) = |V| |I| sinθ

sin (2t)

Which is positive and negative during equal times and has a zero average value

the peak value of pQ

(t) is Q:Q = |V| |I| sin θ, [volt-amperes

reactive, or VAR]

Average value is zero

The instantaneous power

equation,

p(t)=VI[cosθ(1-cos(2t))-sinθsin(2t)]

Equation above can be arbitrarily split into two components, called active and reactive power components as shown below:

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Instantaneous P & Q in AC Circuits

The instantaneous power as a function of the active and reactive power is:

p(t) = P (1 -

cos (2t)) -

Q sin (2t) [W]

The useful information in this expression is contained in P and Q. Therefore, the trigonometric terms are neglected and the so called “apparent”

power may be interpreted as

a phasor whose real and imaginary components are P and Q:

S = P + jQ

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Power from V & I in AC Circuits

The total apparent

power from a voltage V and current l under steady state linear AC conditions is defined as the following phasor:

S = VI*

Where P and Q are the rectangular components:P = |V| |I| cos θQ = |V| |I| sin θ

Where|V| = rms voltage magnitude in Volts|I| = rms current magnitude in Ampsθ

= phase angle between V and I

Assumption:

Power going into circuit element is positive (loads)

Power going out of load is negative (generators)

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Apparent Power in AC Circuits

The steady state apparent power in AC circuits can be expressed as

a phasor with a complex value

S = P + jQ

Where,

S = apparent power in VA, kVA or MVA

P = active power in Watts, kW or MW

Q = reactive power in VAr, kVAr, or MVAr

P is related to energy that becomes heat, light, mechanical motion, etc

Q is related to energy that is temporarily stored in an inductance or capacitance

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Power Factor

Power Factor is the ratio of the active power P to the apparent power S. It represents the fraction of VI doing actual work:

Power factor is the cosine of the angle between the voltage and current functions. It is also the angle between the apparent and

active powers.

A load that has a unity (1.00) power factor has a power factor angle of 0 and has a Q of 0 (entirely resistive circuit)

A load that that has a power factor less than 1.00 (partially inductive or capacitive) draws more current than an entirely resistive load with the same total resistance in order to establish the electromagnetic fields in the load inductance or capacitance

cosSP

.F.P

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Power Triangle

The active and reactive powers can be imagined as the sides of a

triangle whose hypotenuse is equal to Voltage times the “conjugate”

of Current, and the power angle, α,

is the sum of the phase angles of these functions.

Assuming that voltage is at zero angle, in capacitive circuits, the current leads voltage (its angle is positive) and the power angle α

is negative and Q is negative. Therefore, power engineers usually think of a capacitor as a generator of reactive power.

α

Q=VI sin αS=VI*

P=VI Cos αSiemens

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Some Basic Concepts

Let’s review some important definitions:

Peak Value of a sinusoidal current wave (Ipeak

)•

Occurs when the sinusoidal current wave is at its maximum amplitude

RMS Value (also called effective value) of a sinusoidal current wave

Average Value of a sinusoidal current wave

02sin10

T

peakaverage dttT

IT

I

22sin1

022 peakT

peakRMSI

dttT

IT

I Siemens

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Voltage and Current Relationship

)90sin(sin2)sin(cos2)sin(2)sin()( oRMSRMSRMSMAX tItItItIti

)sin(2)sin()( tVtVtv RMSMAX

)sin(cos2)( tIti RMSR

v(t)

i(t) LOADiR(t) iX(t)

• COMPONENTS OF TOTAL CURRENT

• TOTAL CURRENT

)90sin(sin2)( oRMSX tIti

• SYSTEM VOLTAGE v(t)

• COMPONENT OF CURRENT IN-PHASE WITH VOLTAGE (REAL COMPONENT)Real component of current supplies a NET energy to the LOAD

• COMPONENT OF CURRENT 90O

OUT-OF-PHASE WITH VOLTAGE (IMAGINARY COMPONENT)Imaginary component of current supplies no NET energy to the load

θ

= Angle by which voltage v(t) leads current i(t)Siemens

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Voltage and Current Relationship (continued)

)90sin(sin2)sin(cos2)sin(2)( oRMSRMSRMS tItItIti

)sin(cos2)( tIti RMSR

v(t)

i(t) LOADiR(t) iX(t)

• TOTAL CURRENT

)90sin(sin2)( oRMSX tIti

• REAL COMPONENT OF CURRENT • IMAGINARY COMPONENT OF CURRENT

• RMS VALUE (IR

) OF REAL COMPONENT OF CURRENT:

• RMS VALUE (IX

) OF IMAGINARY COMPONENT OF CURRENT:

cosRMSR II

sinRMSX II

• RMS VALUE(IRMS

) OF TOTAL CURRENT: 22XRRMSTOTAL IIII

IR

IX

ITOTAL

RELATIONSHIP BETWEEN RMS VALUE OF REAL COMPONENT, IMAGINARY COMPONENT, AND TOTAL CURRENT WHEN ANGLEΘ

IS POSITIVE:

θ

= Angle by which voltage v(t) leads current i(t)Siemens

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Voltage, Current, & Power Relationship

)sin()()sin()( tItiandtVtv MAXMAX

)sin()sin()()()( ttIVtitvtp MAXMAX

)2sin(sin22

)2cos(1cos22

)( tIVtIVtp MAXMAXMAXMAX

• INSTANTANEOUS AND AVERAGE POWER

• USING TRIGONOMETRIC IDENTITIES, THE EXPRESSION FOR INSTANTANEOUS POWER p(t) IS:

)2sin(sin)2cos(1cos)( tIVtIVtp RMSRMSRMSRMS

• INSTANTANEOUS POWER, p(t), IS THE RATE AT WHICH ENERGY IS SUPPLIED TO THE LOAD

“+”

p(t) means system supplies energy to load, “-”

p(t) means load supplies energy back to system. Frequency of p(t) is twice that of system

Note that the real component of current, Irms

cosθ, produces one component of the instantaneous power thathas a non-zero average value. The imaginary component of current, IRMS

sinθ, produces the second componentof instantaneous power that has an average value of zero.

v(t)

i(t)

LOAD

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Voltage, Current, & Power Relationship (continued)

)2sin(sin)2cos(1cos)( tIVtIVtp RMSRMSRMSRMS

)2sin()2sin(sin)( tQtIVtp RMSRMSQ )2cos(1)2cos(1cos)( tPtIVtp RMSRMSP

• EXPRESSION FOR INSTANTANEOUS POWER

• THE INSTANTANEOUS POWER CAN ARBITRARILY BE SPLIT INTO TWO COMPONENTS CALLED pP

(t) AND pQ

(t) (ACTIVE AND REACTIVE POWER RESPECTIVELY)

ACTIVE POWER , pp

(t) REACTIVE POWER , pQ

(t)

WattsinIVdttpT

P RMSRMS

T

AV cos)(1

0

• AVERAGE POWER SUPPLIED TO THE LOAD OVER INTEGER MULTIPLES OF PERIOD T IS:

• AVERAGE VALUE OF ACTIVE POWER = P • PEAK VALUE OF REACTIVE POWER = QcosRMSRMS IVP sinRMSRMS IVQ

• THE APPARENT POWER

IN THE CIRCUIT, S, IS THE PRODUCT OF THE RMS VALUE OF THE VOLTAGE,VRMS

,

AND THE RMS VALUE OF THE CURRENT, IRMS

:

RMSRMS IVS

22 QPS P

QS

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Voltage, Current, & Power Relationship (continued)

•

EXAMPLE CALCULATION 1

A SINGLE-PHASE TWO WIRE CIRCUIT OPERATES AT 7620 VOLTS RMS BETWEEN THE TWO WIRES.THE CURRENT IN THE PHASE WIRE IS MEASURED AT 25 AMPERES RMS. A METER CONNECTEDTO THE CIRCUIT SHOWS THE ACTIVE (REAL) POWER SUPPLIED IS 150 KW

1.

WHAT IS THE APPARENT POWER SUPPLIED BY THE CIRCUIT?2.

WHAT IS THE REACTIVE POWER SUPPLIED BY THE CIRCUIT?

LOADVRMS = 7620 VOLTS

IRMS = 25 AMPS

kVAAMPERESVOLTIVS RMSRMS 5.190500,19025*7620.1

kWPGIVEN 150:

kVArQPSQorQPS 43.1171505.190,,.2 222222

IF THE LOAD WERE MODIFIED IN SOME MANNER SUCH THAT IT DRAWS ONLY

150 kW OF REAL POWER (P) AND NO REACTIVE POWER (Q = 0), WHAT WOULD THE LINE CURRENT BE IN AMPERES?

RMSRMS IVPSQWITHTHEN :0

AMPERESV

PIRMS

RMS 69.19620,7000,150

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Apparent Power (complex number)

The combination of active and reactive power is referred to as apparent power, defined as follows with complex number notation:

S

= P

+ jQ

S = VI

cos

+ j VI

sin

where

S is the apparent power (VA)

P is the active power (W)

Q is the reactive power (VAR)

θ

is the angle between the voltage and the current (voltage

angle –

current angle in this definition)

P is related to energy that becomes heat, light, mechanical motion,

etc.

Q is related to energy that is stored in an inductor in ½

cycle, and then returned to the system in the next ½

cycle.

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Lumped Load Case –

Loss Reduction Due to

Capacitor Application

Voltage Drop Basic-1.FCW

R X

LOAD

IL

IC

I VLVS

FEEDERSUBSTATION

Vectors VD With Cap.FCWIL

IX

IR

VL

VS

O

IC

I

VS'

0o

OL

CAPACITOR SUPPLIES PORTION OR ALL OF REACTIVE COMPONENT OF LOAD CURRENT, IL

.

THIS REDUCES THE CURRENT IN THE LINE BETWEEN THE SUBSTATION AND LOAD, THEREBYREDUCING THE I2

R LOSSES IN THE LINE (Note: | I | ≤

|

IL

|)

I = LINE CURRENTIL

= LOAD CURRENT

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Capacitor Application (continued)

Vectors VD With Cap.FCWIL

IX

IR

VL

VS

O

IC

I

VS'

0o

OL

Current Phasors With Cap-Detail.FCW

ILOAD

O

ICAP

I

OL

LINE

ACTIVE CURRENT

REA

CTIV

ECU

RREN

T

CAPACITOR CURRENT REDUCES THE REACTIVE CURRENT THAT FLOWSIN THE FEEDER.

CAPACITOR HAS MINIMAL IMPACTON THE ACTIVE CURRENT, AND PRACTICALLY WILL NOT REDUCE LOSSES IN LINE FROM ACTIVE CURRENT.

TO MINIMIZE LINELOSSES:ICAP

= ILOAD

sin θL

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R X

LOAD

IL

IC

I VLVS

FEEDERSUBSTATION

WATTS LOSS IN FEEDER WITHOUT CAPACITOR

WATTS LOSS IN FEEDER WITH CAPACITOR (ANGLE ΘL

IS POSITIVE FOR LAGING PF LOADS)

REDUCTION IN WATTS LOSS FROM CAPACITOR APPLICATION (ΔW)

RIW L2

RPFIIIIRIIIIW LCLCLLCLCLCAPWITH

22222 0.12sin2

RIPFIIRIIIWWW CLCLCLCLCAPWITH222 0.12sin2

WITH LUMPED LOAD, REDUCTION IN WATTS LOSS MAXIMIZED WHEN: IC

= IL

SIN(θL

)

Current Phasors With Cap-Detail.FCW

ILOAD

O

ICAP

I

OL

LINE

ACTIVE CURRENT

REAC

TIVE

CURR

ENT

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0

0.05

0.10

0.15

0.20

0.25

0.30

0.35

0.40

0 0.2 0.4 0.6 0.8 1.0

PFL = 0.95

PFL = 0.90

PFL = 0.85

PFL = 0.80

CAPACITOR CURRENT IN PER UNIT OF LOAD CURRENT IL

RE

DU

CTI

ON

IN

I2 R

LO

SS

ES I

N P

U O

F O

RIG

INA

L





R X

LOAD

IL

IC

I VLVS

FEEDERSUBSTATIONREDUCTION IN I2R LOSSES

IN LINE IN PER UNIT OF LINELOSSES WITHOUT CAPACITOR(ΔW/W)

21

:

LPF

ISWWPEAK

PUPFII

WHENOCCURSWWPEAK

LL

C 21

:

2212

L

CL

L

C

II

PFII

WW

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Lumped Load Case, Power

Loss Reduction Due To

Capacitor Application, Example, 477 KCMIL PHASE

CON EDISON 13 KV ARMLESSCONSTRUCTION

PHASE WIRE = 477 MCM ALRO = 0.198 Ohms / mile

DISC: PTI Course, Unit 8, D#1, Con Ed OH Dist Line-Loss Calc.FCW

1 MILE

LUMPEDLOAD350 AMP80 % PF

1200KVAR

13.8 KVP - TO - P

LINE LENGTH = 1.0 MILE

AMPSKV

KVAI CCAP 2.50

8.1331200

3

/255,24198.0*35022 WATTSRIW L

/579,20198.0*8.00.12.50*350*22.50350

0.12

222

222

WATTS

RPFIIIIW LCLCLCAPWITH

CAPACITOR BANK NOMINAL CURRENT (1200 KVA 3-Φ)

WATTS LOSS PER PHASE IN 1 MILE LINE WITHOUT CAPACITOR

WATTS LOSS PER PHASE IN 1 MILE LINE WHEN CAPACITOR BANK IS APPLIED

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kWWATTSWW PHASE 03.11675,3*3*33

/675,3198.02.508.00.12.50*350*2

0.12

22

22

WATTS

RIPFIIW CLCL

Lumped Load Case, Power

Loss Reduction Due To Capacitor Application, Example, 477 KCMIL Φ

(continued)

CON EDISON 13 KV ARMLESSCONSTRUCTION

PHASE WIRE = 477 MCM ALRO = 0.198 Ohms / mile

DISC: PTI Course, Unit 8, D#1, Con Ed OH Dist Line-Loss Calc.FCW

1 MILE

LUMPEDLOAD350 AMP80 % PF

1200KVAR

13.8 KVP - TO - P

REDUCTION IN WATTS LOSS PER MILE PER PHASE FROM CAPACITOR APPLICATION

TOTAL THREE-PHASE REDUCTION IN WATTS LOSSES PER MILE

ANNUAL ENERGY LOSS REDUCTION IN KWHR ASSUMING CONSTANT LOAD

KWHr

= 11.03 KW * 8760 HOURS = 96,623 KWHR / YEAR

ANNUAL SAVINGS ASSUMING $0.05/KWHR = $4831 PER YEAR

ASSUMING $6 / KVAC, 1200 KVAR FIXED BANK COST IS $7200.00

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Uniformly Distributed Load Case Total Power (Watt) Loss In Feeder

r = FEEDER RESISTANCE PER UNIT OF LENGTHdP(x) = DIFFERENTIAL POWER LOSS IN SECTION dx

AT DISTANCE X DUE TO LOAD CURRENT

dxrLX

LXIdxrIxdP SourceX

2

222 21)(

Uniformly Loaded Feeder-1.FCW

VS

Source

FEEDERdx End

L

LOAD CURRENTMAGNITUDE

ISource

Distance from source end X

ISource 1 - IX =

X

L0

X

XL)(

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Graphical Comparison of Total Power Losses

Lumped Load and Uniformly Distributed Load

Lumped load at the end of the feeder

Same load uniformly distributed along the feeder

SHADED AREA REPRESENTS THE TOTAL LOSSESDUE TO THE TOTAL LOAD CURRENT

INCREMENTALLOSSES IN

WATTS

dP(X)

DISTANCE FROM SOURCE END0 L XIncremental Power Loss-Lumped Load.FCW

INCREMENTALLOSSES IN

WATTS

dP(X)

DISTANCE FROM SOURCE END0 L XIncremental Power Loss-Uniformly Dist Load.FCW

SHADED AREA IS1/3 OF AREA WITH THE

LUMPED LOAD AT FEEDER END

dxrIxdP SOURCE2)(

dxrLX

LXIxdP SOURCE

2

22 21)(

dxrIdP SOURCE2)0( Siemens

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Assuming a Uniformly Distributed Load

If the load is evenly distributed along the circuit as shown below

Total losses can be determined by calculating the losses for each section and summing the results. Not hard but one needs to know the current in each section

kW

kW

kW

kW

kW

kW

L

I1 I2 I3 I4 I5 I6Siemens

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Assuming a Uniformly Distributed Load (continued)

Without knowing the current in each section, or how many sections required, an approximation of the kW losses can be made with the following diagram.

Determine the length of the main line three phase.

Use the resistance (R) per unit length of the main line

Use the total sending current. Then,

Losses in kW = (I)²(RL/3)

kW

1/3 L

R=?

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If There is an Express Section

The losses can be approximated by

Losses = I12(R)(L1 ) + I2

2(R)(L2 /3)

kW

kW

kW

kW

kW

L2

I1

L1

kW

1/3L2

I2Siemens

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Assuming a Uniformly Distributed Load

But you must use caution when making these calculations.

The circuit conductor may not be the same type for the whole circuit length.

Most distribution circuit loads are not uniform. In fact they can be very lumpy

It can only get you an approximate result

This method can prove to be a reasonable method for screening circuits for a more accurate review of the losses. Siemens

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Another Approximation Tool

Without the advantage of computers and load readings, approximations can be made

This is just one way, there are other ways

Divide the circuit into areas based upon maximum load or distance values

•

For example one megawatt of maximum connected load or a distance of three thousand feet

•

Lump the load at the end. Do this for each section

•

Scale connected load based upon the readings at the station.

•

Determine the average “r”

value

Use the uniformly distributed load technique to estimate losses per section

•

Sum the section losses

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Another Approximation Tool (continued)

Here is a simple diagram of what I just did.

This will be a little more accurate than the previous methods but it is still an approximation!

kW1

kW2

kW3

kW4

kW5

I1

L1

L2/3

I2 I3 I4 I5

L3/3

L4/3L5/3

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Equations

Section 1losses

I12

x R1 x L1

Section 2 losses

I22

x R2 x (L2

/3)

Section 3 losses

I32

x R3 x (L3

/3)

Section 4 losses

I42

x R4 x (L4

/3)

Section 5 losses

I52

x R5 x (L5

/3)

The line sections are not equal in length but the units are in kilo-feet (kft)

The resistance is the weighted average of the types in the section based upon

contribution to the length. Units are ohms per unit length

The power factor is assumed constant.

Siemens


Tab 6 -

Loss Calculation Methodologies

Siemens

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Calculation of Electric System Losses…

The calculations are based on the power and energy system balance equations:

For power,

( ) ( ) ( )input output lossesEnergy kWh Energy kWh Energy kWh

)()()( kWPowerkWPowerkWPower lossesoutputinput

Similarly, for Energy:Siemens

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…Calculation of Electric System Losses…

First, define the area of study

Second, determine for a period of time, usually a year, the total energy flowing into the study area from all sources

Third, determine for the same period, the total energy delivered

to all customers through the known customer metering points

Find the difference between the two numbers found above. The difference is, normally, a combination of:

System losses –

which is what the loss study needs to find

Unmetered loads –

which should be known and can be estimated. Losses caused by unmetered loads need to be estimated

Data from inaccurate meters –

which can be minimized by re-calibration of meters

Reading errors –

which can be minimized by optical reading and computerized recording

Energy Diversion (theft) –

which can be minimized with adequate supervision

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Load and no-load losses are calculated The calculations are performed for:

GSU transformers –

if plant meter is on low voltage side of GSU transformer

Transmission system

Transmission to Distribution transformers

Primary Distribution System

Secondary Distribution System

Losses in the Secondary Distribution system include:

Secondary lines

Service Drops

Distribution transformers

Customer Meters

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Due to the vast number of primary and secondary circuits, it is virtually impossible to calculate the losses in each and every one of them. One method is to calculate in detail the losses in a selected number of circuits, and apply the results to the whole population of circuits using regression curve techniques

The number of distribution transformers is, typically, very large. In many companies, transformer data management systems monitor the load of most distribution transformers. Peak loads or hourly

loads can be recorded. The load of transformers that are not monitored

can be estimated or loading assumptions can be made.

The number of service drops maybe very large. Therefore, the calculation of losses in service drops can be performed for a selected group of service drops with typical designs. Then, the results can be applied to all service drops in the system using regression techniques

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…Calculation of Electric System Losses

The higher the number of primary circuits or service drops that are calculated in detail, the more accurate the results will be

The number of circuits to be considered for the detailed calculations is a tradeoff between the budget and time allocated for the loss study and the desired accuracy of the loss results Siemens

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General Methodology…

Demand and energy losses are used in the design of electric rates

In the United States, power companies are required to report their annual energy losses on page 401a of FERC Form 1 (Annual Report of Major Electric Utilities)

Total electric losses can be determined from metered data by taking all inputs to the electric system, including inter-tie flow “in”

and the system internal generation,

and subtracting the sales and inter-tie flow “out”, plus any un-metered use. This calculation requires data that is readily available, usually

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…General Methodology

Input (kWh) = Inter-tie flows in + internal generation + purchases (internal and external)

Output (kWh) = Sales + Inter-tie flows out + un-metered use

Metered Energy Losses (kWh) = Input -

Output

For inputs, good hourly accounting records usually exist

For outputs, not always accounting records exist

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Outputs

Customer sales are usually the largest component of output

Output also includes meter company use and non-metered use (street lights, traffic signals, consumption of some buildings, station use etc.).

Consumption of non-metered output is normally estimated

Energy diversion is typically very small in United States and many countries. It can be estimated

Small customer classes have their meters usually read on a monthly bases. Data is recorded on a monthly schedule

Meter reading timing issues usually arise due to the fact that readings are taken throughout the month and not at the end of every month

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General Methodology

Non-coincident power demand losses are calculated for each sub-system

Energy losses are determined for each sub-system from the corresponding non-coincident power demand losses

With the exception of the transmission system (for which a different approach is used), for the calculation of energy losses of each sub-system, the “loss factor”

method is used

Energy losses for all sub-systems are added up to determine the total system losses

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July

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Metered vs. Calculated Energy Losses

Losses in every sub-system can be determined from metered data, to the extent this is possible. Losses can also be calculated

Calculated Energy Losses for each sub-system are compared to the metered losses for that sub-system. Any difference between calculated should be evaluated

Demand and Energy Loss Multipliers are calculated. Multipliers show how much power or energy is required at the supply level to deliver a unit of power or energy at the customer service level

Loss multipliers are used in electric rate design

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Demand and Energy Loss Multipliers…

Multipliers show how much power or energy is required at the supply level to deliver one kW or one kWh at every customer service level

Each customer level has a different multiplier for demand and energy

For example, an energy multiplier of 1.06 at the residential level means that if a residential customer required one kWh of energy, the generation system would have to provide 1.06 kWh to cover the one kWh and the energy losses

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… Demand and Energy Loss Multipliers…

Similarly, a demand multiplier of 1.08 at the residential level means that if a residential customer placed a demand of one kW, the generation system would have to provide 1.08 kW to cover the one kW load and the demand losses

Transmission customers are only responsible for their share of losses that result from their service on the transmission system

Substation customers are responsible of their share of losses on the substation system and the transmission system

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… Demand and Energy Loss Multipliers

Primary service customers are responsible for losses resulting from their load on the primary system, substation system and the transmission system

Secondary customers are responsible for losses that their load creates on all four systems

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Loss Multipliers

Loss Multipliers are sometimes called Loss Factors. Referring to loss multipliers as loss factors may lead to confusion

Demand or energy multipliers specify how much generation is required to serve one unit of energy (kW or kWh) at the respective service level

Secondary

Primary

Transmission

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Example of Multipliers

Loss multipliers by service level

Energy/demand

Total-secondary-primary-transmission

Secondary

Primary

Transmission

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Energy Loss Multipliers

Energy Multipliers

Input:Native

Generation +

Purchases

Transmission Lines

&Transformers

Transmission To

Primary Substations

DistributionPrimarySystem

DistributionSecondary

System

DistributionSecondary

Sales

DistributionPrimary

Sales

11,401,5211,835,946

SubstationSales

531,563

TransmissionSales

431,794

Losses:230,950

Losses:128,278

Losses:161,331

Losses +

Diversion:266,234

Total Losses:786,793 kWH

14,987,617 14,324,873

System Input

13,665,032 11,667,756

Sector Loss Multiplier 1.0157 1.0090 1.01191.0234

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Sector Loss Multiplier Calculation

Multiplier = OutputSalesSectorOutputLossSectorSalesSector

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Calculation of Energy Loss Multipliers

Transmission Lines & Transformers Cumulative

Transmission to Primary Substations Cumulative

Distribution Primary System Cumulative

Distribution Secondary System Totals

System Input - kWh 14,987,617 14,324,873 13,665,032 11,667,755Losses - kWh 230,950 230,950 128,278 359,228 161,331 520,559 266,234 786,793Sales - kWh 431,794 431,794 531,563 963,357 1,835,946 2,799,303 11,401,521 14,200,824System Output - kW 14,324,873 13,665,032 11,667,755 0

Energy Loss Multiplier

Sector 1.0157 1.0090 1.0119 1.0234Cumulative 1.0157 1.0246 1.0360 1.0554

DataCalculated

Energy Loss Multipliers

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Type of Customers…

Power companies connect their customers to the electric system depending on the customer demand requirements

•

Large demand customers are connected directly to the transmission system and are considered industrial class customers or requirement sales customers

•

Intermediate demand customers are connected to the distribution primary system and are normally commercial and small industrial in nature

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…Type of Customers

•

Smaller commercial and residential customers are connected to the distribution secondary system

•

Requirement Sales Customers purchase energy for re-sale. Requirement sales customers require energy on an ongoing basis and their energy needs are included in the system resource planning of the supplier or utility company. Examples are, other

utilities, municipalities, cities, cooperatives, etc.Siemens

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Where are the electric losses?...

Losses occur in transmission lines, transformers and power equipment at all voltage levels

Equipment includes, customer meters, potential transformers, current transformers, grounding transformers, etc

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Other equipment with losses

Reactors

Capacitors

DC Terminals

DC Lines

AC-DC-AC Back to Back

Synchronous Condensers

Line Regulators

Filters

FACTS DEVICES

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Transformers

Generation

22 kV to 230 kV

13.8 kV to 115 kV

Transmission

230 kV to 115 kV

500 kV to 230 kV

Distribution substation

230 kV to 13.8 kV

69 kV to 13.8 kV

Distribution secondary

13.8 kV to 120/240 volts

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Ideal Transformer Model

Load

E 2

I2I1

E 1

P1

= E1

I1

P2

= E2

I2b

•a

•c

d

Magnetic Core

N1N2

Sou

rce

Mutual Flux

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Transformer Model

Load

2 le

ak

1 le

ak

1 le

ak

E 2

I2I1

E 1

M

Sou

rce

Mutual Flux

Magnetic Core

2 le

akSiemens

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Transformer Impedance

E1 LoadSource

I

Rp Xp E3 Rs Xs

Im

Ih&e

REDDY

+

RHYSTERESIS

LMAGNETIZING

MAGNETIZING IMPEDANCE (CORE LOSS OR NO LOAD LOSS)

LEAKAGE (WINDING LOSS OR LOAD LOSS)

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Excitation Losses

W W WWhereWWW

iron h e

iron

h

e

: = Iron Loss or Excitation Loss

= Hysteresis Loss = Eddy Current LossSiemens

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Transformer Voltage a Function of Flux

E f n AB

EfnAB

s

s

22

1028

2

max

max

Where: = Winding rms Induced Voltage

= Frequency = Turns in Winding = Magnetic Circuit Area

= Maximum Flux DensitySource: Electical T & D Reference Book

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July

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Transformer No-Load Loss

Note: Losses –

Watts/pound

W K B f

W K B f t

K K

t

h hx

e e

h e

max

max2 2 2

2

Where:, , x = Steel Quality Factors

= Thickness of Steel LaminationsSource: Electical T & D Reference Book

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July

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Iron Losses a Function of V

Iron loss at 100 percent voltage

90,290 watts source (from manufacturer’s test report)

Iron loss at 110 percent voltage

125,200 watts

Calculated voltage function

(125,200/90,290) = (1.1/1.0)x

X = 3.4

Use transformer specific, or

Recommend use square function

Siemens

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Transformer Losses

0

100,000

200,000

300,000

400,000

500,000

600,000

700,000

0 20 40 60 80 100 120 140 160 180 200

Percent Loading

Loss

es W

atts

Transformer No-Load & Load Losses230/115 kV, 150, 200, 250, 280 MVA

Minimum

Load Siemens

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Approximate Iron Loss

Generator step-up

.00156 pu

500/230 kV

.00046 pu

230/115 kV

.00100 pu

115/13.8 kV

.00138 pu

34.5/13.8 kV

.00148 pu

120/240 volts

.00228 pu

The above values are not a statistical collection, check the manufacturer for actual values.Siemens

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July

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Load and No-Load Losses…

Load losses depend on the load level

Also known as “copper losses”

Depend on the square of the current flowing in the circuit

Occur in all system components (lines, transformers, etc.)

No-Load losses exist even if the circuit is not supplying any electrical load

In transformers, no-load losses are also known as “iron”

or “core”

losses

No-load losses depend on the voltage of the circuit

No-load losses may be assumed constant (as long voltages don’t vary much)

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July

201

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…Load and No-Load Losses…

Load losses-

they are variable in nature because depend on the load

They are often referred to as copper losses, occur mainly in lines and cables, but also in the copper parts of transformers

Usually, between 2/3 and 3/4 of the technical (or physical) losses on distribution networks are variable

Due to the proportionality between losses and the square of the current, the level of losses on a network will be affected by the utilization of its capacity

Losses will fall if the cross sectional area of lines and cables

for a given load is increased

There is a trade-off between cost of losses and cost of capital. Optimal average utilization rate on a distribution network that considers the cost of losses in its design could be as low as 30 per cent; however,

those low utilization rates are not, probably, economical

Siemens

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July

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…Load and No-Load Losses

No-load losses. They are, approximately, fixed in nature

Occur mainly in the transformer cores (thus, also called iron losses) and do not vary according to current (note that there are other fixed losses, like customer meter losses)

Fixed losses do not vary according to current. They take the form of heat and noise on transformers and occur as long as a transformer is energized

Fixed losses vary with voltage, but are relatively constant if the voltage is relatively constant

Typically, between 1/4 and 1/3 of technical losses on distribution networks are fixed

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Transformer Auxiliary Loss

Transformers rating OA/FA/FA/FOA

OA –

Oil to Air natural convection

FA –

Oil to Forced Air (provided by first group of fans)

FA –

Oil to Forced Air (provided by second group of fans)

FOA –

Forced oil to Air natural convection (oil pumps are used)

Auxiliary load at the substation is part of the substation service power. It may or may not be metered.Siemens

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July

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GSU Losses…

The losses in the step-up transformers at power plants need to be calculated only if the meters are located on the generator side of the transformers (low voltage side of the GSU transformer). Transformer losses that have high side metering are considered part of the plant efficiency and are not part in the electric rate design procedure

Load and no-load losses are calculated using manufacturer’s data or typical data. No-load losses are determined in the same manner as described above for other classes of transformers

Hourly plant outputs are used to determine the hourly transformer loads. From the hourly loads and the transformer resistance or rated load loss data, the hourly transformer load losses can be determined (kW). The hourly transformer losses are summed to determine the energy losses

Siemens

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July

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…GSU Losses…

Generating type, mode of operation and planned and forced outages of the generators supplying the GSU transformers are used in the

calculation of GSU losses

Generator types and modes of operation are, typically,

Peaking, load following few hours

Intermediate, load following shoulders, full load peak

Base, full load when in service

Load shapes of GSU transformer are a function of the above

Planned and forced outages

Results in zero or null loads making loss factor inaccurate

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…GSU Losses

0

100

200

300

400

500

600

0 5 10 15 20

Hours

Load

MW

Base loaded

Intermediate loaded

Peaking

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Transformer Losses –

No Load Losses

In general, losses in transformers have two components:

No-load losses component, in Watts, sometimes referred to as “Iron Losses”. No-load losses are losses in the iron core of the transformer and are caused by the excitation and magnetizing currents

No-load losses are in the form of heat energy and noise and are present whenever the transformer is energized

No-load losses vary with the square of the applied voltage. Assuming that the operating voltage remains relatively constant throughout the year, the iron losses are relatively constant

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Transformer Losses –

Load Losses

Load Losses component, in Watts, sometimes referred to as “copper losses”. Load losses vary with the square of the electric current flowing through the transformer

To calculate the total transformer losses, data for the entire transformer inventory of all units in operation during the year of study must be obtained

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Transformer Data

Transformers are classified by size, year of manufacturer and manufacturer. For every transformer, the manufacturer provides the test report data which contains, at least, the following:

•

Size, kVA

•

Rated voltage, primary and secondary

•

Impedance, in Ohms or per unit

•

No-load losses, in Watts or per unit, at rated voltage

•

Load losses, in Watts or per unit, at rated load conditionsSiemens

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Transformer Losses…

For every transformer that is energized, with load or without load, the non-coincident demand, in kW, can be determined using metered load data. If load data is not available, the non-coincident demand can be estimated from the transformer size

Information for transformers with no manufacturer’s data usually can be estimated using data from similarly sized transformers. In these cases, knowing the manufacturer and year of manufacture is useful

Once the non-coincident demand (kW) is determined or estimated for every transformer, the corresponding load losses (kW) can be calculated

Siemens

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…Transformer Losses

Energy load losses (kWh) can be determined from the load and loss factors calculated from load research or metered data for the particular sub-system where the transformer is located (transmission, distribution, etc.)

Demand no-load losses, in Watts, for every transformer is provided by the manufacturer

Energy no-load losses for every transformer are obtained by multiplying the demand no-load losses by the number of hours the transformer is energized

Total load and no-load demand and energy losses for all transformers in respective sub-systems are calculated by summing the corresponding losses

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GSU Transformers

GSU transformers

Only included if meter is on the generator side of the transformer

Many companies include the losses for the transformer in the overall plant efficiency

Be sure to check how the company accounts for these losses•

No-load, iron, or excitation losses

•

Load or copper losses

Siemens

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Procedure for Calculation of Transformer Load Loss

For GSU transformers

Determine load loss from manufacturer

Determine hours of operation

Calculate demand loss

Calculate energy loss a function of type/output characteristic

•

Base

•

Intermediate

•

Peaking

Sum losses of all GSU transformers

Siemens

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Procedure for Calculation of Transformer No-Load Loss…

Determine no-load loss from manufacturer

Determine if transformer is off when generator is out of service

Determine hours of operation

Determine voltage magnitude schedule

Adjust loss a function of square of voltage

Calculate demand loss

Calculate energy loss

Siemens

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… Procedure for Calculation of Transformer No-Load Loss

For Transmission transformers

Inventory by voltage and size

Obtain no-load loss from manufacturer or estimate

Apply voltage correction factor

Determine demand loss

Determine energy loss

Siemens

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July

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Losses in Transmission Lines

Corona•

Voltage

•

Rain

Current squared•

Load

•

Generation

•

Schedules with neighbors

•

Loop flow

•

WheelingSiemens

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Lines in General

Transmission lines

Distribution lines•

Three phase•

Two phase•

Single phase to neutral

Secondary and service drops

Overhead

Underground

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July

201

0


Electrical Model of Transmission Lines

Bus i Bus j

Rij j Xij+I i

G li j Bli+ jBch2

jBch

2Glj jBlj+

Ij

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Dc Resistance of One Wire

ra

r

dc

dc

12*

DC Resistance of a Conductora = Conductor Radius

= Resistivity at a Specific TemperatureSiemens

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19 Strand Conductor

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Resistivity as a Function of Temperature

In general, the resistivity, ρ, of a conductor is given by:

ρ

= ρt1

*[1+α*(t-t1

)]

Where,

α

= Resistance temperature coefficient for the conductort1

= Temperature reference for which the reference resistivity, ρt1

, is given t = Actual temperature of the conductorρt1

= Resistivity at reference temperature, t1ρ

= Resistivity at an actual temperature, t

Siemens

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Conductor Characteristics

Name –

dahlia

Size –

556.5 kcmil

AAC

Stranding –

19

Strand diameter –

0.1711 inch

Conductor diameter –

0.856 inch

Dc resistance @ 20°

c –

0.1640 Ohms /mile

AC resistance @ 25 °

c –

0.1691 Ohms /mile

AC resistance @ 50 °

c –

0.1855 Ohms /mile

Ampacity

@ 75 °

c –

703 amps

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Resistance Increases

Resistance increases due to stranding

Stranding

Skin effect

Length of catenary

Temperature

Temperature of conductor is a function of,

Current (load)

Temperature of surrounding air

Speed of the wind –

temperature decreases when the wind speed increases

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Resistance and Reactance

0.0000

0.1000

0.2000

0.3000

0.4000

0.5000

0.6000

0.7000

0.8000

0.9000

1.0000

0.0000 0.2000 0.4000 0.6000 0.8000 1.0000 1.2000 1.4000 1.6000 1.8000 2.0000

Conductor Area (Square Inches)

Ohm

s/M

ile

115 kV Reactance

115 kV ResistanceSiemens

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ACSR Conductor Resistance

Conductor Name

Size kcmil

R@25°C Ohms

R@50°C Ohms

R@75°C Ohms

Increase Per

Degree

Quail 2/0 0.6810 0.8430 0.9320 0.74%Penguin 4/0 0.4290 0.5730 0.6280 0.93%Junco 266.8 0.3413 0.3748 0.4083 0.39%Oriole 336.5 0.2708 0.2974 0.3240 0.39%Lark 397.5 0.2294 0.2518 0.2743 0.39%Hen 477 0.1913 0.2100 0.2288 0.39%Eagle 556.5 0.1642 0.1802 0.1963 0.39%Scoter 636 0.1439 0.1579 0.1719 0.39%Drake 795 0.1166 0.1278 0.1390 0.38%Ortolan 1033.5 0.0921 0.1013 0.1102 0.39%Bittern 1272 0.0759 0.0832 0.0903 0.38%Nuthatch 1510.5 0.0649 0.0709 0.0769 0.37%Bluebird 2156 0.0477 0.0516 0.0555 0.33%

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AAC Conductor Resistance

Conductor Name

Size kcmil

R@25°C Ohms

R@50°C Ohms

R@75°C Ohms

Increase Per

Degree

Aster 2/0 0.7000 0.7692 0.8384 0.40%Oxlip 4/0 0.4407 0.4842 0.5277 0.39%Laurel 266.8 0.3499 0.3843 0.4188 0.39%Tulip 336.5 0.2779 0.3052 0.3325 0.39%Canna 397.5 0.2355 0.2586 0.2817 0.39%Syringa 477 0.1968 0.2159 0.2352 0.39%Misteltoe 556.5 0.1691 0.1855 0.2020 0.39%Orchid 636 0.1485 0.1311 0.1771 0.39%Lilac 795 0.1197 0.1311 0.1425 0.38%Larkspur 1033.5 0.0934 0.1020 0.1107 0.37%Narcussus 1272 0.0772 0.0841 0.0912 0.36%Galdious 1510.5 0.0663 0.0720 0.0778 0.35%Sagebrush 2250 0.0482 0.0519 0.0556 0.31%

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Wind and Temperature

40.0

50.0

60.0

70.0

80.0

90.0

100.0

110.0

120.0

0 200 400 600 800 1000 1200 1400

Current Amps

Tem

pera

ture

Cen

tigra

de

Wind 2 Ft/Sec

Wind 10Ft/Sec

Wind 6 Ft/Sec

Drake 795 kcmil ACSR @40°C AmbientSiemens

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Resistance a Function of Current

0.1200

0.1250

0.1300

0.1350

0.1400

0.1450

0.1500

0.1550

0.1600

0 200 400 600 800 1000 1200 1400

Current Amps

Res

ista

nce

Ohm

s Wind 2 Ft/Sec

Wind 6 Ft/Sec

Wind 10 Ft/Sec

Drake 795 kcmil ACSR @40°C AmbientSiemens

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Resistance and Temperature

0.1600

0.1650

0.1700

0.1750

0.1800

0.1850

0.1900

0.1950

0.2000

0.2050

20 30 40 50 60 70 80

Temperature C

Res

ista

nce

(ohm

s/m

ile)

Dahlia 556.6 kcmilSiemens

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July

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Probability Density Function

Line Rating Distribution(1 Year of Data, Drake Conductor)

0

10

20

30

40

50

60

70

80

90

100

0 500 1000 1500 2000 2500 3000

Line Rating (amps)

920 ampStatic Rating(based on "worstcase" weatherassumptions)

Actual Ratingsover 1 year

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July

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Conductor Resistance Adjustment

Resistance of conductors is usually provided at 25°

c

Transmission lines

Distribution lines

•

Run power flow using resistances at 25 °

c

•

Adjust resistance as a function of current

•

Rerun power flow with adjusted resistances

Resistance of Transformers is usually provided at 75°

c

•

Run power flow using resistances at 75 °

c

•

Adjust resistance as a function of current

•

Rerun power flow with adjusted resistances

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July

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Resistance Adjustment

Needed information for resistance adjustment

Conductor resistance at specific temperature

Maximum conductor temperature

Select wind speed, ambient temp.

Calculate temperature/resistance for wind/ambient/current

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July

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Transmission System Losses…

Siemens

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July

201

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…Transmission System Losses…

The transmission system is comprised of transmission lines and transmission transformers

Losses in the transmission system can be calculated with the help of a power flow program

Power flow cases are developed representing a number of system conditions of load, generation dispatch and tie flows. These different system conditions should be representative of system conditions that occurred during the year being studied. The more cases are considered, the more accurate the transmission loss calculation is going to be. Typically, between 20 and 30 system conditions are represented in power flow cases.

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July

201

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Transmission Procedure

The idea is to develop power flows to capture

Hourly load during the year

Generation schedules as they vary by day or season

Imports, exports, wheeling, and loop flow

Power flow modeling normally includes transmission transformers (500/345 kV, 230/115 kV)

Make sure the resistance is modeled in the transmission transformers

No-load transformer losses can be included in the power flow. Normally they are not

Integrate the curve as a function of the load duration curve

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July

201

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Transmission Transformers

Transmission transformers

•


•

Load or copper losses (included in the transmission lines procedure)

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Conditions representing summer peak, summer shoulder load, summer light load, winter peak, minimum load, etc. are analyzed,

with different generation schedules, wheeling levels, import and

exports. These cases should be modeled with the greatest care using sound engineering judgment. The transmission losses for each of these system conditions can be determined by solving the power flow cases

Pairs of points representing the system load and the transmission loss for every system condition modeled will be obtained. A regression curve can be fitted to these pairs of points and a regression equation can be derived to calculate the transmission

loss for any value of system load between peak and minimum system load

The loss equation so derived for the transmission system is a function of the system load and the transfer of energy to and from neighboring utilities

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0.0

10.0

20.0

30.0

40.0

50.0

60.0

1000 1500 2000 2500 3000 3500

System Demand - MW

Loss

es -

MW

Power Flow Results

Regression Analysis Results

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July

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Transmission System Losses

0.00

0.20

0.40

0.60

0.80

1.00

1.20

1.40

1.60

1.80

40 50 60 70 80 90 100

Percent Transmission Load

Perc

ent L

osse

s

Actual System Losses

Theoretical System Losses

Hypothetical System Losses

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Estimation of Transmission Losses

Transmission Losses

Curve Point Area Calculated Estimated 1.000632 6.317448No. Generation Tie Flow Losses Losses m b1 3384 110 53.7 53.5 1.592 3266 126 50.3 49.7 1.543 3166 104 48.5 46.6 1.534 3065 170 44.8 43.7 1.465 2970 188 42.4 41.2 1.436 2870 314 38.2 38.7 1.337 2773 408 35.5 36.4 1.288 2669 413 33.4 34.1 1.259 2571 485 30.1 32.0 1.1710 2470 460 29.1 30.0 1.1811 2368 211 30.5 28.2 1.2912 2270 661 23.3 26.5 1.0313 2169 232 27.1 24.8 1.2514 2072 188 26.3 23.4 1.2715 1970 579 20.3 21.9 1.0316 1870 636 18.8 20.6 1.0117 1770 693 17.4 19.3 0.9818 1670 129 21.4 18.1 1.2819 1570 493 16.7 17.0 1.0620 1470 568 15.3 16.0 1.0421 1369 474 14.9 15.0 1.0922 1270 628 13.5 14.1 1.0623 1170 529 13.7 13.2 1.1724 1070 465 13.3 12.4 1.24

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July

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…Transmission System Losses

From the hourly system loads, available from metered data, the hourly transmission losses can be calculated from the regression equation. Energy losses can be calculated from the regression equation by integrating the hour-by-hour demand losses over the 8,760 hours in a year

A direct integration of the loss vs. system demand curve is not possible because there are multiple load data points with the same value between the minimum and maximum loads. Therefore, a load duration curve has to be used

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July

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Load Duration Curve

0

500

1000

1500

2000

2500

3000

3500

0 1000 2000 3000 4000 5000 6000 7000 8000

Time (Hours)

Dem

and

(MW

)

Maximum Demand = 3384 MW

Minimum Demand = 1070 MW

Average Demand 1701 MW

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1500

1700

1900

2100

2300

2500

2700

2900

3100

3300

3500

100 600 1100 1600 2100

Hours

Load

(MW

)

Transmission System Load


Difference are Transmission Losses

Siemens

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Transmission Losses

Most power flow cases don’t have the no-load or iron losses of the transmission transformers represented. Therefore, no-load losses are not included in the calculation of transmission losses and have to be calculated separately

In many power flow cases, the resistance of the transmission transformers is not properly modeled. If this is the case, the resistance must be added or, if not available, typical values should be usedSiemens

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July

201

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Corona Losses

Corona loss is an electric discharge to the air surrounding an energized conductor. Corona loss is negligible for voltages 69-kV and below. It is also small during fair weather conditions. The amount of corona discharge is principally a function of:

•

Voltage level •

Diameter of the conductor•

Conductor bundling•

Length of the circuit•

Conductor spacing•

Elevation•

Presence of shield wire•

Adverse weather conditions (rain increases the corona loss substantially)

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July

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Procedure for Calculation of Corona Losses

Determine hours of rain

Prepare Inventory of transmission lines 11kV and above (corona losses are not significant for lower voltage transmission lines)•

Voltage

•

Conductor size

Determine the corona loss by voltage•

Normal weather

•

Adverse weather

•

Voltage

•

Conductor sizes

Sum losses

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July

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Corona Losses

Corona loss is a function of these important factors:

•

Voltage

•

Conductor diameter

•

Conductor bundling

•

Conductor smoothness

•

Spacing and clearance

•

Weather conditions

•

Length of line

•

Altitude

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July

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Corona Loss –

No Rain

0.000

0.500

1.000

1.500

2.000

2.500

3.000

3.500

4.000

4.500

0 100 200 300 400 500 600 700 800

Voltage kV

Loss

es k

W/M

ile

Drake at 230 kV

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Corona Loss –

Rain 0.25 In/Hr

0.000

50.000

100.000

150.000

200.000

250.000

300.000

350.000

400.000

0 100 200 300 400 500 600 700 800

Voltage kV

Loss

es k

W/M

ile Drake at 230 kV

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July

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Corona Loss/Rain Intensity

0.000

2.000

4.000

6.000

8.000

10.000

12.000

14.000

0 0.2 0.4 0.6 0.8 1 1.2

Rain in/hr

Loss

es k

W/M

ile

Drake at 230 kVSiemens

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July

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Load Duration Curve

0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1

Per Unit Cumulative Time (One Year)

Per U

nit L

oad

Load Factor 0.582Loss Factor 0.365

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July

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Transmission Transformer No-Load Losses

No-load demand losses (kW) in transmission transformers are calculated using the data from the transformer inventory and actual transformer test results provided by the manufacturer

If manufacturer’s data is not available, typical data should be used

No-load energy (kWh) losses are obtained by multiplying the no-load demand loss for the transformer by the number of hours the transformer is in operation in the year

Siemens

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July

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Transmission Losses Summary

Transmission Lines Transformers Totals

Load Losses Corona Losses

Load Losses

No-Load Losses

Load Losses No-Load Losses

Non-Coincident Peak (kW)

58,722 153 2,489 4,699 61,211 4,852

Coincident Peak (kW)

57,756 153 2,489 4,699 60,245 4,852

Energy Losses (kWh)

179,706,861 2,896,493 5,543,466 34,461,595 185,250,327 37,358,088Siemens

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July

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Distribution Primary Transformer Losses…

Distribution primary transformers have distribution low-

side voltages such as 25.0 kV, 13.8 kV or 4.16 kV. The high-side voltage is a transmission voltage, such as 345 kV, 230 kV, 161 kV, 138 kV or 115 kV, or a sub-

transmission voltage such as 69 kV or even 34.5 kV in some systems

No-load demand loss (kW) is calculated using the transformer inventory data and by applying actual transformer test results, if available

No-load energy losses (kWh) are obtained by multiplying the individual transformer no-load losses by the number of hours in the year

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July

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…Distribution Primary Transformer Losses…

Typical data is used if manufacturer’s data is not available

Calculation of load losses in each transformer requires the knowledge of the non-coincident peak load of each unit. The coincident peak is calculated by multiplying the non-coincident peak by the coincident factor

Coincident factors for every subsystem are usually obtained from load research dataSiemens

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…Distribution Primary Transformer Losses

Transformer test results provide the resistance of the transformer at 75°C

Copper (or load) losses (kW) are calculated from load and resistance data

Energy load losses (kWh) are calculated using the load factor/loss factor methodSiemens

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Distribution Substation Transformers

Distribution substation transformers



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Procedures

Prepare transformer inventory by voltage and size

Obtain the no-load loss from manufacturer or estimate if the data is not available

Apply voltage correction factor, if needed



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Losses in Distribution Primary Lines…

A typical system has many distribution feeders or primary lines operating at different primary voltages

Feeders typically have different cable sizes, using different materials (copper or aluminum)

Circuit maps showing the location of distribution secondary transformers, feeder conductor sizes, distances between nodes, location of capacitors, etc., are used to allocate load and model the primary and lateral runs of the distribution primary circuit

Operating temperature, load, power factor, phase balance influence the losses significantly

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… Losses in Distribution Primary Lines…

There may be many distribution primary lines and, in many cases, it may be impractical to calculate the losses for each and every distribution primary circuit

Therefore, a sample of primary distribution lines, at all distribution primary voltages levels, is carefully selected for detailed analysis. The more the number of primary lines studied, the more accurate the results will be

Typically, 10 to 15 circuits are selected for detailed study

Power flow programs, specifically designed to work with distribution lines may be used to model the operation of each of the selected primary distribution circuits

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… Losses in Distribution Primary Lines

Distribution line amperes, kW or kVA

load are used in determining demand losses (kW)

The selected circuits are modeled and demand (kW) losses are calculated for each selected circuit using the non-coincident peak demand for the circuit. Pairs of results of circuit demand vs. circuit losses are obtained

A regression equation is fitted to the pairs of demand vs. losses. The regression equation allows the calculation of the demand losses (kW) for every primary distribution circuit in the system for the known non-coincident peak demand of the circuit

The energy losses are calculated using the load factor/loss factor method

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Procedures

Distribution primary lines

Model lines (statistical sample)•

Include different voltages, 13.8 kV, 4.8 kV

•

Include different densities, lengths, load levels

•

Include overhead and underground systems

Include distribution secondary transformers

Calculate demand losses (power flow)

Calculate energy losses (loss factor)

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Distribution Primary Lines

Distribution primary lines

Three phase

Two phase

Resistance a function of load/weather

Single phase (ground or neutral)

•

Normal voltages range from 2.4 to 34.5 kV

•

Most popular range 11.5 to 13.8 kV

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Capacitor Placement

Fixed capacitors

Switch capacitors

Load on simple system:

3,130 kW

1,704 kVAr

Demand losses w/o caps: 100.53 kW

Demand losses with caps: 87.00 kW

Loss savings with caps: 13.53 kW

Percent savings: 13.5 %

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Power Factor Effects

26.0 Percent Increase

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Unbalanced Effects

8.2 Percent Increase

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Procedure…

Select representative sample of primary lines including,

Rural, city, and urban lines

Representative voltages

From all available lines, choose a statistical sample to match the budget and time available

Model distribution lines

Three phase, 2 phase, and single phase

Calculate demand losses

Calculate energy losses

Allocate the losses to all lines

Obtain load research information by customer class for loads

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…Procedure

Model with power flow or spreadsheet

Calculate demand losses for service drops

Use load/loss factor for energy

Estimate the subsystem total quantity (demand and energy) Siemens

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Calculated Distribution Circuit Losses

0

50

100

150

200

250

300

0 2000 4000 6000 8000 10000 12000

Circuit Load (KW)

Circ

uit L

oss

(KW

)

Exponential Regression Analysis Fit to Data

Calculated Data

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Losses in Distribution Secondary Transformers…

Primary voltages are usually in the 15-kV class

Secondary voltages use consumer voltages (120 V, 240 V, etc)

As all transformers, distribution secondary transformers have two loss components, copper and iron losses

The inventory data of all existing distribution secondary transformers is used to calculate the load and no-load losses

Data in the transformer inventory usually include size, year of manufacturer, manufacturer, impedance, and test report loss information

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… Losses in Distribution Secondary Transformers

If known, the non-coincident peak demand for each transformer is used to calculate the load losses (kW). If not known, the non-coincident peak demand has to be estimated. Coincident peak demand is calculated from coincident or diversity factors derived from load research data

Energy losses for each transformer are calculated using the load factor/loss factor method

No-load losses are calculated separately, using manufacturer’s data or typical data. Iron losses are assumed constant over the 8,760 hours of the year

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Secondary Service Transformers

Distribution voltages

Primary 34.5 to 2.4 kV

SECONDARY 380 volt

Secondary 240/120, 480/277, 208/120 v



Statistical information•

Number•

Size•

Loss characteristics

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Procedure No-Load Loss

Secondary service transformers

Inventory by voltage and size

Obtain no-load loss from manufacturer or estimate

Apply voltage correction factor



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12.47-kV and 13.2-kV Circuit Loading Frequency Distribution (Example)

0

20

40

60

80

100

120

0 2 4 6 8 10 12 14 16

Circuit Load MVA

Freq

uenc

y

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Losses in Distribution Secondary Lines and Service Drops…

The most uncertain and difficult calculation is the calculation of losses for distribution secondary lines and the service drops to the customer. The sheer number of secondary and service drop circuit configurations prohibits a detailed calculation of all circuits

Usually, power companies have distribution standard for the secondary services that describe the standard conductor configuration to be used for each kind of customer. While each customer’s electric service is slightly different from the standard, the standards serve as a guide

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… Losses in Distribution Secondary Lines and Service Drops…

Typically, a sample of typical secondary circuits is used to calculate the demand losses, some of them residential, some of them commercial

Hourly research data for each customer class is used to create a secondary system load set

Key factors influencing the losses are cable size, power factor and load balance on the two-wire and neutral system

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… Losses in Distribution Secondary Lines and Service Drops…

Even considering a statistical sample, the uniqueness of the feeders in this subsystem would require calculations for many different combinations which is time consuming and costly

Typically, due to budget and time constraints, calculations are performed for a number of configurations, let’s say, ten configurations, with certain cable sizes

The customer loads modeled in the calculation come from load research data and non-coincident demand losses (kW) can be calculated for every configuration

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… Losses in Distribution Secondary Lines and Service Drops

Energy loss (kWh) is calculated for every configuration using the load factor/loss factor method

Demand and energy losses calculated for every circuit configurations are used to project the losses for the entire system at this service level

Diversity or coincident factors, calculated from load research data, are used to account for circuit peaks that do not occur at the time of the system peak

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Secondary Service Drops

Residential 120/240 v or three phase 380 v

Secondary lines to meter

Three wire or two wire

Utility standard wire sizes

Overhead and underground

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Typical Secondary Service Drop 120/240 V

Neutral

+120 V

-120 V

240 V Meter

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Typical Secondary Service Drop 220 V

A Phase

B PhaseC Phase

Neutral

Meter220 V

Phase to Phase Voltage 380

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Secondary Service Lines

120/240 Volts

Have short runs from transformer

May run in several directions

Service drops are attached

Three phase mostly have radial systems with two phase

and single phase laterals

Small industrial, commercial, and residential are supplied

from same transformer

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Procedure

Develop characteristic systems

Select representative sample•

Rural, city, urban

•

Statistical sample, match budget

Obtain load research information by customer class for loads

Determine power factor

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Procedure (continued)

Model with power flow or spread sheet

Calculate demand losses for secondary service lines

Use load/loss factor for energy

Allocate losses to the subsystem total (demand and energy) Siemens

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Meter Losses…

Most meter losses are no-load losses

Small amount

energy is lost in each individual

mechanical or electronic meter at the customer’s delivery point

Meter losses are small in magnitude on an individual basis, but these losses add up in a large system

Friction losses in the rotating disk of mechanical meters and excitation losses in the mechanical and electronic meters cause the meter losses

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…Meter Losses

Utilities usually have data regarding the losses for the typical meters used. Otherwise, this information should be available from manufacturers

Meter losses for each kind of meter are multiplied by the total number of meters of each kind to obtain the total meter losses

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Revenue Meters

Customer sales meters (revenue)

Excitation and dead zone losses

Check manufacturer for losses

Find the watt loss for each meter and multiply by 8760 hours per

year (or number of hours of the study period)

Source is the company records for inventory

Include substation and the company use metersSiemens

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Procedure

Determine meter inventory

Demand=number*1 watt (for example)

Energy= demand*8,760 hours

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Example of Customer Meter Loss Calculation

Meter Type Quantity Loss/Meter Watts Demand Loss Watts

Energy Loss kWh

Single-Phase Mechanical 488,597 0.80 390,878 3,433,469

Three-Phase Mechanical14,215 1.00 14,215 124,865

Three-Phase Electronic 27,390 0.25 6,848 60,148

Sum530,202 411,940 3,618,482

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Unaccounted Losses

Un-metered company use

Un-metered substation use

Service without meters (street lights & traffic signals)•

Sometimes estimated (no meters)

Free service (charity, churches)

Accounting practices

Energy diversion (meter tampering, meter bypass)

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Questions?

Any questions?

Siemens

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