Reference DAPPER

Power*Tools®

for Windows

™

DAPPER Reference Manual Electrical Engineering Analysis Software

for Windows

Copyright © 2006, SKM Systems Analysis, Inc. All Rights Reserved.

SKM Power*Tools for Windows

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©2006 SKM Systems Analysis, Inc. All rights reserved.

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Contents

1 DEMAND LOAD STUDY 1-1 1.1. What is the Demand Load Study? .......................................................................1-2 1.2. Engineering Methodology.....................................................................................1-3

1.2.1. Definitions of Terms and Concepts..................................................................1-3 1.2.2. Load Characteristics .........................................................................................1-6 1.2.3. Demand Load Library ......................................................................................1-6

1.3. PTW Applied Methodology..................................................................................1-6 1.3.1. Before Running the Demand Load Study ........................................................1-6 1.3.2. Motor and Non-Motor Loads ...........................................................................1-6

Non-Motor Loads...................................................................................................1-7 Demand Load Types ..........................................................................................1-7 Energy Audit Types ...........................................................................................1-7

Motor Loads ...........................................................................................................1-8 Loads Defined in a Schedule..................................................................................1-9

1.3.3. Running the Demand Load Study ....................................................................1-9 1.3.4. Demand Load Study Options .........................................................................1-10

Project Configuration ...........................................................................................1-11 1.3.5. Error Messages...............................................................................................1-11

1.4. Application Examples .........................................................................................1-12 1.4.1. Non-Coincident Demand Calculation ............................................................1-12 1.4.2. Motor Design Load Calculation .....................................................................1-13 1.4.3. Multiple Motors at a Bus................................................................................1-13 1.4.4. Motors Assigned to a Motor Control Center..................................................1-14 1.4.5. Single Phase Non-Motor Loads in a Panel Schedule .....................................1-15 1.4.6. Loads with Different Power Factors ..............................................................1-17 1.4.7. Multiple Motors on a Single Motor Component ............................................1-18 1.4.8. Motor Starting ................................................................................................1-18 1.4.9. Multiple Loops in a System............................................................................1-19 1.4.10. Example from Plant......................................................................................1-20

2 SIZING STUDY 2-1 2.1. What is the Sizing Study?.....................................................................................2-2 2.2. Engineering Methodology.....................................................................................2-3

2.2.1. Feeder Sizing....................................................................................................2-3 2.2.2. Transformer Sizing...........................................................................................2-4 2.2.3. Transformer Feeders ........................................................................................2-4

2.3. PTW Applied Methodology..................................................................................2-4 2.3.1. Before Running the Sizing Study.....................................................................2-5 2.3.2. Running the Sizing Study.................................................................................2-5 2.3.3. Sizing Study Options........................................................................................2-6


DAPPER ii Reference Manual

2.3.4. Error Messages ................................................................................................ 2-7 2.4. Application Examples........................................................................................... 2-7

2.4.1. Sizing a Simple Radial Feed............................................................................ 2-7 2.4.2. Impact of Cable Temperature Derating ........................................................... 2-9 2.4.3. Multiple Branches with Different Load Values............................................... 2-9 2.4.4. Example from Plant ....................................................................................... 2-10

3 LOAD FLOW STUDY 3-1 3.1. What is the Load Flow Study? ............................................................................ 3-2 3.2. Engineering Methodology .................................................................................... 3-2

3.2.1. The Solution Process ....................................................................................... 3-4 3.2.2. Modeling Transformers ................................................................................... 3-4 3.2.3. Utility Equivalent Impedance .......................................................................... 3-4 3.2.4. Load Characteristics ........................................................................................ 3-4 3.2.5. Voltage Drop Calculations .............................................................................. 3-5

3.3. PTW Applied Methodology ................................................................................. 3-6 3.3.1. Before Running the Load Flow Study ............................................................. 3-6 3.3.2. Running the Load Flow Study......................................................................... 3-7 3.3.3. Load Flow Study Options................................................................................ 3-7

System Modeling ................................................................................................... 3-8 Solution Method .................................................................................................... 3-8 Load Specification ................................................................................................. 3-8

Directly Connected Loads ................................................................................. 3-8 From Demand Load Study................................................................................. 3-9

Solution Criteria..................................................................................................... 3-9 3.3.4. Component Modeling ...................................................................................... 3-9

Swing Bus Generator........................................................................................... 3-10 On-Site Generation .............................................................................................. 3-10 Diversity Loads.................................................................................................... 3-11

3.3.5. Error Messages .............................................................................................. 3-12 3.4. Application Examples......................................................................................... 3-12

3.4.1. Voltage Drop and Power Losses.................................................................... 3-12 3.4.2. Modeling Transformer Losses ....................................................................... 3-14 3.4.3. Load Specifications ....................................................................................... 3-15 3.4.4. Net Branch Diversity Load ............................................................................ 3-17 3.4.5. Example from Plant ....................................................................................... 3-19

4 SHORT CIRCUIT STUDY 4-1 4.1 What is the Short Circuit Study? ................................................................... 4-2 4.2 Engineering Methodology ............................................................................... 4-2

4.2.1 Balanced Faults.......................................................................................... 4-3 Thevenin Equivalent Circuit .................................................................................. 4-3

4.2.2 Unbalanced Faults...................................................................................... 4-5 Single-Line-to-Ground Faults................................................................................ 4-5 Line-to-Line and Double-Line-to-Ground Faults .................................................. 4-7 Grounding Impedance ........................................................................................... 4-7 P ............................................................................................................................. 4-7 Per Unit Notation................................................................................................... 4-7

4.2.3 Momentary and Interrupting Fault Current................................................ 4-9 4.2.4 Asymmetrical Peak Fault Current............................................................ 4-10

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Contents DAPPER iii

4.2.5 Asymmetrical rms Fault Current ..............................................................4-11 4.2.6 Steady State Fault Current........................................................................4-11 4.2.7 Transformer Taps .....................................................................................4-12

Primary Transformer Tap Modeling ....................................................................4-12 Secondary Transformer Tap Modeling ................................................................4-13

4.2.8 Transformer Off-Nominal Voltage Modeling ..........................................4-13 4.2.9 Transformer Phase Shift ...........................................................................4-13

4.3 PTW Applied Methodology...........................................................................4-14 4.3.1 Before Running the Short Circuit Study ..................................................4-15 4.3.2 Running the Short Circuit Study ..............................................................4-15 4.3.3 Short Circuit Study Options .....................................................................4-15

Fault Type ............................................................................................................4-15 Faulted Bus...........................................................................................................4-16 Calculation Models...............................................................................................4-16

Motor Contribution ..........................................................................................4-16 Transformer Tap...............................................................................................4-16 Transformer Phase Shift ...................................................................................4-16

Report Specifications ...........................................................................................4-16 Bus Voltages ....................................................................................................4-16 Branch Currents................................................................................................4-16 Phase or Sequence............................................................................................4-17 Fault Current Calculation .................................................................................4-17 Asymmetrical Fault Current at Time................................................................4-17

4.3.4 Component Modeling...............................................................................4-18 Feeder Data ..........................................................................................................4-18 Transformer Data .................................................................................................4-18 Three-Winding Transformers...............................................................................4-18 Contribution Data.................................................................................................4-21

4.3.5 Error Messages .........................................................................................4-22 4.4 Application Examples ....................................................................................4-22

4.4.1 Fault Currents on a Radial Unloaded Feeder............................................4-23 4.4.2 Single-Line-to-Ground Fault Currents at the Secondary of a Transformer with a Grounding Reactor ........................................................................................4-25 4.4.3 Source Sequence Impedance ....................................................................4-26 4.4.4 Fault with a Generator Source with Unequal Positive-, Negative-, and Zero-Sequence Reactances................................................................................................4-28 4.4.5 Short Circuit Currents with a Motor Load at Bus 4 .................................4-30 4.4.6 Fault Duty Contribution to a Faulted Bus ................................................4-33 4.4.7 Transformer Off-Nominal Voltages and Transformer Taps.....................4-36

Case 1 ...................................................................................................................4-36 Case 2 ...................................................................................................................4-37 Case 3 ...................................................................................................................4-38 Case 4 ...................................................................................................................4-39 Case 5 ...................................................................................................................4-39

4.4.8 Modeling Transformer Connections.........................................................4-40 4.4.9 Example from Plant..................................................................................4-42



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1 Demand Load Study

The Demand Load Study calculates the total apparent power and current at each bus and within each branch in the electrical power system, excluding local generation and power lost through impedance devices. The purpose of this Study is to calculate the connected, demand and design load values for each bus, including the effects of load diversity. The calculated load currents are based on the load bus nominal system voltage. The results are used as a basis for further studies in PTW.

This chapter discusses:

• Engineering Methodology.

• PTW Applied Methodology.

• Examples.

IN

TH

IS

CH

AP

TE

R

1.1. What is the Demand Load Study? .................................................................... 1-2 1.2. Engineering Methodology ................................................................................ 1-3 1.3. PTW Applied Methodology ............................................................................. 1-6 1.4. Application Examples..................................................................................... 1-12

DAPPER 1-2 Reference Manual

1.1. What is the Demand Load Study? The Demand Load Study sums individual loads throughout a power system to size cables and transformers, provides data for the Load Flow Study, and calculates the total branch circuit loads as reported in the Load Center Summary.

The Demand Load Study requires a radial system with a single power source. If the power system contains loops, the Study detects the loops, temporarily opens them, continues calculating, and restores the original system configuration upon completing the Study. The Demand Load Study can analyze up to ten independent power systems in a single project.

Beginning at the bus farthest from the source, the Demand Load Study calculates the vector sum of all load values on the bus, and reports the connected, demand, and design load values. This process is repeated for each upstream branch in the electrical system. Demand and design load calculations are based on the principles of the National Electrical Code (NEC). A basic premise of the NEC is that multiple branch loads serviced from a single feeder may not necessarily peak concurrently. This concept is called load diversity. The Demand Load Study considers diversity when calculating feeder load values (NEC).[1]

[1] Authored by Committee, 1996 National Electrical Code. Quincy, MA: National Fire Protection Association, 1996.

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Demand Load Study DAPPER 1-3

Define System Data

Define system topology and connectionsDefine utility connections (swing bus)Define individual loads

Run Demand Load Study

Saved in Database

For each branch:Demand and design load and power factorTotal constant I demand load and power factorTotal constant Z demand load and power factorTotal constant kVA demand load and power factor

For each bus:Total connected load and power factorTotal demand load and power factorTotal design load and power factor

Datablocks

Reports

Used by Load FlowStudy

Study Setup

Demand Load LibraryDemand LoadsEnergy Audit LoadsStudy Setup

Used by Sizing Study

Used by LoadSchedules

1.2. Engineering Methodology This section explains the terms and concepts used in the Demand Load Study and this Reference Manual. The terms used when calculating the demand and design loads are defined, including sample calculations which help demonstrate the methodology. Also included are descriptions of load characteristics and the Demand Load Library.

1.2.1. Definitions of Terms and Concepts The basis of the Demand Load Study is the connected load. All of the terms in the following table are represented in the subsequent figure.

This value Is defined as Connected Load The rated size or the sum of the rated sizes of load

components at a bus.

Demand Load Value The largest or peak load value of a single load. Usually less than the connected load.

Demand Factor The ratio of the demand load value to the connected load. Usually less than one.

Design Load Value The demand load value multiplied by the long continuous load factor (LCL).

Long Continuous Load Factor The reciprocal of the cable ampacity derating factor for a continuous load (design factor).



Continuous Load A load where the maximum current is expected to continue for three or more hours.

Non-Coincident Demand The sum of the connected loads adjusted by a single or multiple level demand (diversity) factor. Also, the sum of the actual demand load values at a bus assuming that the individual demand load values do not peak at the same time. This value is referred to as the demand load throughout the Demand Load Study.

Coincident Demand The sum of the individual branch demand loads and loads attached directly to the bus.

Net Branch Diversity The difference between the coincident demand load and the non-coincident demand load reported at the bus.

Diversity Factor The coincident demand loads divided by the non-coincident demand loads. Always greater than one.

To illustrate these terms, consider a system with three 25 hp motors connected to a bus. The peak mechanical loads on the three motors are 12, 15 and 25 hp. The total connected load is 75 hp. The coincident demand load is the sum of the three peak mechanical loads, which is 52 hp:

Demand Load = 12 hp +15 hp + 25 hp = 52 hp

The demand factor is the ratio of the demand load value to the connected load:

Demand Factor 52 hp demand load75 hp connected load0.69 or 69%

=

=

The NEC defines continuous loads as loads that operate for longer than three hours at a time. To size branch and feeder circuits with continuous loads, an NEC ampacity rating (long continuous load or design factor) must be used.

For example, imagine a 30 kVA lighting load that is always on. The connected load is 30 kVA, and the demand factor is 100%. Therefore, the demand load for the lighting is 30 kVA. If the long continuous loading factor is 125%, then the design load value is 37.5 kVA, as shown:

Design Load = 30 kVA demand load 1.25 long continuous loading factor= 37.5 kVA

×

National codes may make reference to multi-level demand factors. For example, Chapter 2 of the NEC states that a receptacle load demand value is calculated based on the following rules:

• The first 10 kVA of a connected load is calculated at 100% demand factor.

• The remaining load is calculated at 50% demand factor.

If a receptacle connected load is 15 kVA, then the demand load at the bus is 12.5 kVA. The demand load is calculated as:

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Demand Load = 10 kVA 100% 5 kVA 50%12.5 kVA

× + ×

=b g b g

Next imagine two 15 kVA receptacle branch circuits fed from a single feeder circuit. The demand load value in each branch circuit is 12.5 kVA. The connected load on the feeder circuit is as follows:

Connected Load = branch load connected value

kVA +15 kVA= 30 kVA

∑= 15

The feeder circuit demand load can also be calculated. This is the non-coincident demand load, also known as the diversity load of the feeder circuit.

Demand Load = 10 kVA 100% 20 kVA 50%20 kVA

× + ×

=b g b g

The net diversity or demand load at the feeder circuit is 20 kVA. It is not the sum of the two coincident branch circuit load values of 25 kVA. The diversity factor now can be calculated as:

Diversity Factor =coincident demand load

non coincident demand12.5 kVA +12.5 kVA

20 kVA1.25 or 125%

∑

=

=

The non-coincident demand load (20 kVA) is less than the sum of the individual coincident or peak demand loads (25 kVA) at the feeder bus.

The Demand Load Study uses NEC methodology to identify each load in an electrical power system and its specific load factors. The Demand Load Study then vectorially sums the connected, demand and design load values for each upstream bus and branch. Each load may be assigned a unique power factor. To account for load diversity, the Demand Load Study re-calculates the demand and design values from the total connected load at each bus. If there are motors at the bus, the largest motor is identified and special multiplying factors are used to calculate the motor circuit design load value. For an example of multiple motors on a single motor component, see Section 1.4.7 in this Reference Manual.

The methodology depends upon the assumption that only one source services each load. When the Demand Load Study detects a loop in the power system, it determines the number of impedance components in the loop and opens the branch element that is topologically farthest from the source of supply. The Demand Load Study does not determine the branch opening point based on total equivalent impedance. If there are multiple loops, it detects them and opens them as well. With the loop(s) open, it calculates the demand and design load values at each bus and within each branch beginning at the farthest point from the source of supply. Multiple independent power systems may result from either a specific electrical design or the methodology the Demand Load Study uses to open loops in the system. The Demand Load Study Report identifies all open branches at the beginning of the Study Report.



1.2.2. Load Characteristics Loads are defined as either motor or non-motor loads, and both require a demand load value specification. For a motor load, the connected load equals the motor rated size multiplied by the number of motors. The demand load value is the connected load value multiplied by the Motor Load Factor.

Non-motor loads subdivide into Energy Audit types or Demand Load types. The demand and design load values of an Energy Audit type are the rated load size (connected load) multiplied by the Load Factor. For Demand Load types, the Demand Load Study calculates the demand and design load values using demand and design load factors in the Demand Load Library.

1.2.3. Demand Load Library Non-motor loads specified as Demand Loads must be assigned one of twenty demand load categories in the Demand Load Library. For each of these twenty demand load categories, the Library can store as many as three levels of demand factors. Usually only one or two demand levels are used. In addition, each demand load category has a specific long continuous loading factor.

1.3. PTW Applied Methodology PTW Applied Methodology discusses how the Demand Load Study applies the methodology described in the preceding section. Before beginning a Study, you need to define the system topology and load types. The different demand load types and their characteristics, options for running the Demand Load Study, demand load error statements, and procedures for correcting errors are described herein.

1.3.1. Before Running the Demand Load Study Before running the Demand Load Study, you must:

• Define the power system topology and connections.

• Define utility connections (swing bus).

• Define individual loads.

If motor components are assigned to a motor control center (MCC) and the motors are referenced in the Motor Control Center Library, then the demand and design load values are based on NEC Table 430-150, as defined in the Motor Control Center Library.

1.3.2. Motor and Non-Motor Loads Motor and non-motor loads have distinct characteristics and you should be sure to understand the differences between them to accurately model your power system.

Load Types Description Motor Load Synchronous Motor The rated size of a motor component multiplied by the number

f t ifi d i th N b f M t b f th

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Load Types Description of motors, as specified in the Number of Motors box of the Component Editor.

Induction Motor Same as above.

Non-Motor Load Demand Load type A non-motor load with a uniquely defined set of demand

factors as specified in the Demand Load Library.

Energy Audit Load type A non-motor load with a single unique load factor. Usually can be directly measured in the field.

Non-Motor Loads Non-motor loads are modeled as either a Demand Load or an Energy Audit type load.

Demand Load Types

A Demand Load type must be defined using one of the 20 load types in the Demand Load Library. Because all new non-motor loads by default are created as Energy Audit types, you will need to change the load type from Energy Audit to Demand Load type in the Component Editor under the Load Diversity subview. Select one of the twenty categories from the Component Editor, but define those categories in the Demand Load Library. Demand Loads specified without a load category are considered to have a 100% demand load factor.

The Demand Load Study calculates the connected, demand and design load values based on data from the library. The connected load value is always equal to the rated size of the load. The demand load value is generally equal to or less than the load rated size. You can add demand factors greater than one for load growth studies, although customarily the demand load factor is less than one in most practical design applications. The design load is generally equal to or greater than the demand load, and represents the design requirement that derates cables based on continuous loading.

In the Demand Load Library, the General Load category LCL factor (design factor) is set at 1.0. On the other hand, article 220 of the NEC requires that conductors that are energized for more than three hours must not carry more than 80% of the calculated demand load. For these loads, the LCL factor should be set at 1.25. Lighting type loads, for example, are continuous load types requiring a design factor equal to 1.25 in accordance with the NEC. For lighting type loads, the design load value will be 125% of the demand load.

You will receive a warning message if you do not define a demand load category in the Component Editor. If you ignore the warning, PTW makes the Demand and Design Load value equal to the Connected Load value, and posts a message telling you so after running the Study.

The Demand Load characteristic (constant power, constant impedance, constant current) is defined for each Demand Load category in the Demand Load Library.

Energy Audit Types

An Energy Audit type load is defined by a load factor and load characteristic. The term Energy Audit comes from the fact that you may have measured or audited the system loads, and may not want to use special NEC demand factors. The load characteristic for Energy Audit load types is not used in the Demand Load Study. This load characteristic is fully defined in the Load Flow Chapter of the Reference Manual.



New non-motor loads default as Energy Audit loads with a load factor of 1.0 and a constant kVA load characteristic. This default setting sets the connected, demand and design load values equal to the load rated size. The demand and design load values are always identical for Energy Audit type loads.

Motor Loads Motors are generally modeled as a constant kVA load characteristic when running, and as a constant impedance load characteristic when starting. When a motor is either running or starting, the Demand Load Study scales the individual motor rated size by multiplying the quantity of motors by the motor load factor. When a motor is modeled as starting, Demand Load Study multiplies the rated size by the number of motors, and divides the product by the subtransient reactance ( ). The default is 0.17 pu, and the default load factor is 1.0. The starting power factor is calculated based on the motor’s fault duty X/R ratio, which has a default of 10. Thus, the default motor starting condition is a locked rotor current of 5.9 multiplied by the motor’s full load amperes (FLA), and a starting power factor of 10%.

′′Xd ′′Xd

The following one-line diagram models a single 100 hp motor at Bus 2. It shows the motor parameters used to calculate demand and design load values.

BUS 1Connected 91.42 kVADemand 91.42 kVADesign 114.28 kVA C1

UTILITY 1

BUS 2Connected 91.42 kVADemand 91.42 kVADesign 114.28 kVA

MOTOR 1Size 100.0 hpPF 0.85 Lag

In this example, the motor’s mechanical to electrical energy conversion efficiency is 96% and the power factor is 85%.

The connected motor load at Bus 2 is calculated as:

Motor kVA =100 hp 746

1000 0.85 PF 0.96 efficiency91.42 kVA

wattshp

wattskW

×

× ×=

The motor design load value depends on identifying the largest motor at the bus. The Demand Load Study automatically locates the largest motor on the branch circuit and multiplies its rated size by the Largest Motor design factor. All the remaining motor rated sizes on the branch circuit are vectorially summed and then multiplied by the Remaining Motors design factor. Both the Largest Motor design factor and the Remaining Motors multiplying factor are defined in the Demand Load Study Setup dialog box, with default values of 1.25 and 1.0, respectively. The connected and demand load values are the same for motor circuits unless a load factor is specified, while the design load is generally larger than the connected load value.

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There are two instances when the motor demand load may be larger than the connected load when modeled in PTW. First, if a motor service factor (SF) is used in the Demand Load Study to determine the largest maximum system demand, then you may want to enter the SF as a load factor. Because the demand load is the motor rated size multiplied by the load factor, and because the SF might be 1.1, the demand load value could be 110% of the motor’s connected load.

Second, if a motor is placed in a motor control center and the motor’s control library data is used, then the NEC Full Load Amperes (FLA) values (Table 430-150) are used to calculate both the demand and design load values. Because the NEC FLA values tend to be conservative, the demand and design load values are usually larger than the connected load. Note that regardless of the NEC FLA values, the motor’s connected load value is always the motor’s rated size multiplied by the number of motors.

Loads Defined in a Schedule PTW provides a shortcut for assigning loads to a schedule (see the section on Load Schedules in Chapter 5, “Component Editor” for exact steps). This shortcut allows you to create loads and then assign them to the schedule, as opposed to manually creating the loads and then assigning them. However, you may not want to use the shortcut depending on your intent.

For motor loads, there is no difference between using the shortcut and manually creating them. While the motor symbols will not initially appear on the One-Line Diagram if you create them using the shortcut, you can use the Expand command and the symbols will appear connected at the bus.

For non-motor loads, the shortcut assigns them to the schedule but does not create any components for them; while the Studies will recognize that the non-motor loads exist, they will not appear either on the One-Line Diagram or in the Component Editor (which means that you cannot find them using a query). Typically, if have numerous small loads which do not need to appear on your One-Line Diagram, the shortcut provides a quick way to assign them to a schedule. However, if you have larger loads that you want to show on the One-Line Diagram and retrieve using a query, you should create them manually and then assign them to the schedule. Note that when you use the shortcut, and you do not assign a load category, PTW assumes the load’s demand and design load value equals its connected load value.

Note that motor loads, and non-motor loads created manually, are not automatically placed out of service when the schedule is placed out of service. Non-motor loads created using the shortcut, however, are placed out of service automatically when the schedule is placed out of service.

1.3.3. Running the Demand Load Study You can run the Study from any screen in PTW, and it always runs on the active project.

To run the Demand Load Study:

1. From the Run menu, choose Analysis.

2. Select the check box next to Demand Load.

3. To change Study options, choose the Setup button.



4. Choose the OK button to return to the Study dialog box, and choose the Run button.

The Demand Load Study runs, writes the results to the database, and creates a report.

1.3.4. Demand Load Study Options In the Demand Load Study dialog box under Non-Motor Loads, you can select one of three option buttons. Also enter the Motor Load Design Factors. As shown in the following figure, the default settings for the Demand Load Study are: Include All Loads, a Motor Load Design Factor for the Largest Motor of 1.25, and a Motor Load Design Factor for Remaining Motors of 1.0. The Demand Load Study defaults to the most recent setting, which is saved every time you make a change.

To choose a non-motor load in the Demand Load Study dialog box, refer to the following table:

Select this option button To do this Include All Loads Include all non-motor loads entered as Demand Loads

and/or Energy Audit loads.

Include Only Demand Loads Include non-motor loads entered as Demand Loads.

Include Only Energy Audit Loads Include non-motor loads entered as Energy Audit loads.

The Demand Load Study includes all motors that are in service. Motor Load Design Factors are used to calculate the design load value of the largest motor and all remaining motors in a branch and feeder circuit. Enter motor data using the following guidelines:

In this box Type this information Largest Motor A design load multiplying factor for the largest motor on

the branch circuit. The Demand Load Study identifies the largest motor on each branch circuit and multiplies its rated size by this design factor to determine the design load.

Remaining Motors A design load multiplying factor for all other motors on the branch circuit. The Demand Load Study vectorially

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In this box Type this information sums all of the remaining motor rated sizes on the branch circuit and multiplies them by this design factor to determine the design load.

The Demand Load Study identifies one motor in each branch as the largest motor. Both induction and synchronous motors are considered. In a case where the Number of Motors field in the Component Editor is greater than one, the Demand Load Study calculates the demand load value as the rated size multiplied by the number of motors. The design load value requires further steps. First, the Demand Load Study examines all the motors’ rated sizes on the bus to determine the largest motor on the bus. The largest motor’s rated size is then multiplied by the largest motor design factor from the Demand Load Study Setup dialog box.

Project Configuration You can use the Demand Load Study to examine any number of different project configurations. For example, during the design process you may want to examine the power system with selector switches selecting the normal or emergency system in operation. To do this, simply run the Demand Load Study on the first system configuration and save the report. Then change the system and run the Demand Load Study a second time. Prior to running the second Study, change the report name under Report Files in the Study dialog box to avoid overwriting the first report. Then compare the two reports.

The purpose of the Demand Load Study is to provide a picture of the total loads in the project. Therefore, the effects of local generation are not considered in the calculation.

Note: The Load Flow Study can use Demand Load Study information to examine the load flow in the power system with tie breakers and loops closed. The Demand Load Study provides feeder and transformer in and out of service switches to simplify opening and closing the tie breakers and selector switches.

1.3.5. Error Messages As the Demand Load Study runs, a Study Messages dialog box appears and catalogs the progress of the Study. If the Demand Load Study detects logic errors in the input data, error messages appear in the dialog box. Though not all errors are fatal, any errors should be corrected. Fatal errors must be corrected before the Demand Load Study can be completed.

For further information on using the Study Messages dialog box to resolve errors, refer to Correcting Study Errors in Chapter 9 of the User’s Guide.

The most common errors are:

• Source Bus Not Identified: No voltage source (swing bus) is defined in the system. This error is fatal and stops the Study. To correct this problem, define a utility or swing bus generator.

• Isolated Branch Detected: One or more branches is not connected to a voltage source. For example, if a single branch is placed out of service, and if downstream electrical components remain in service, then you effectively have an isolated



branch. This error is fatal. To correct the problem, check the system for components erroneously out of service.

• Incomplete System Configuration: One or more component connections is missing. This error is fatal.

• Zero Load Data: The motor load on one or more load buses is zero kVA. This error is not fatal, and the Study continues.

• Load Demand Category Not Defined, PTW will detect non-motor loads modeled as demand loads, and where the demand category is not defined. PTW assumes a 100% demand category.

1.4. Application Examples The following examples demonstrate how the methodology is applied to various system configurations. Each example changes the system, building upon the previous example, to show how the Demand Load Study meets variable conditions. Both motor load and non-motor load examples are included. In all examples except Section 1.4.6, load power factors are the same, allowing apparent power and current values to be algebraically added.

1.4.1. Non-Coincident Demand Calculation This example demonstrates how the Demand Load Study calculates non-coincident demand loads, also called demand loads. The following one-line diagram has two 1000 kVA Demand Load type loads that are connected to two branch circuits. The General Load category is selected for both loads. The total demand load at both Buses 3 and 4 is:

Demand Load = 100 kVA 100% 900 kVA550 kVA

× + ×

=b g b 50%g

g

At Bus 2, the connected load is 2000 kVA.

The methodology requires that the demand load value be calculated as:

Demand Load = 100 kVA 100% 1900 kVA 50%1050 kVA

× + ×

=b g b

The connected, demand and design loads are shown on the following one-line diagram:

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UTILITY 1

BUS 2Connected 2000.00 kVADemand 1050.00 kVADesign 1050.00 kVA C2 C3


BUS 4

Connected 1000.00 kVADemand 550.00 kVADesign 550.00 kVALOAD 1

Size 1000.0 kVAPF 0.85 Lag

LOAD 2Size 1000.0 kVAPF 0.85 Lag

In the diagram, note the coincident demand load is 550 kVA at Buses 3 and 4. The coincident peak of these values is 1100 kVA. However, the non-coincident demand load or diversity load at Bus 2 is only 1050 kVA. This is because of the methodology inherent in the NEC’s multi-level demand factors. For a review of multi-level demand factors, see the bulleted rules in Section 1.2.1 of this Reference Manual.

1.4.2. Motor Design Load Calculation This example describes how the Demand Load Study calculates the branch and feeder motor demand and design load values. The following one-line diagram is identical to that in the previous example, except that the loads at Buses 3 and 4 have been changed to induction motors. Different connected load values are used in this example to show how the Demand Load Study determines the largest motor in the branches and feeder circuits.

The design load value is calculated by multiplying the rated size of the largest motor by 125%, and adding the product to the sum of all other motor rated sizes multiplied by 100%. As shown in the following one-line diagram, Motor 2 is larger than Motor 1; Motor 2’s rated size of 75 kVA is multiplied by 125%. Motor 1 is the remaining motor; its rated size of 50 kVA is multiplied by 100%. Therefore, the sum of these two products is the design load at Bus 2, which is 143.75 kVA as shown:

Bus 2 Design Load = 75 kVA 125% 50 kVA 100%143.75 kVA

× + ×

=b g b g




UTILITY 1

BUS 2Connected 125.00 kVADemand 125.00 kVADesign 143.75 kVA C2 C3


BUS 4

Connected 75.00 kVADemand 75.00 kVADesign 93.75 kVAMOTOR 1


MOTOR 2


Since there are no loads directly connected to Buses 1 or 2, the connected, demand, and design loads are the same at Buses 1 and 2. The Demand Load Study properly calculates the individual design load values for the two branch circuits as shown on the drawing.

1.4.3. Multiple Motors at a Bus To calculate branch loads, it is critical to determine the largest motor on the branch. This example shows how the Demand Load Study determines the largest motor on the circuit when there are multiple motor loads on the branch. The following example is identical to the previous one, except that an additional 75 kVA motor is added to Bus 3.

The Demand Load Study determines that Motors 2 and 3 are both 75 kVA, and then arbitrarily selects either one as the largest motor to determine the feeder circuit size. The design load values at Buses 1 and 2 increase as a result of adding Motor 3; but, as shown, it is immaterial which motor is selected as the largest motor:

Bus 2 Design Load = Lg Mtr kVA 125% sum of remaining mtrs kVA

75 kVA 125% 75 kVA + 50 kVA

218.75 kVA

× + ×

= × + ×

=

b g bb g b g

100%

100%

g

The design load value at Buses 1 and 2 is 218.75 kVA. The data on the following one- line diagram agrees with the preceding calculations.

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UTILITY 1

BUS 2

Connected 200.00 kVADemand 200.00 kVADesign 218.75 kVA C2 C3


BUS 4

Connected 75.00 kVADemand 75.00 kVADesign 93.75 kVAMOTOR 1


MOTOR 2


MOTOR 3Size 75.0 kVAPF 0.85 Lag

1.4.4. Motors Assigned to a Motor Control Center Motors assigned to a motor control center (MCC) are handled differently than motors not placed in a motor control center. This example shows how the Demand Load Study calculates the load values given the NEC FLA values that must be used when designing motor circuits.

In the following one-line diagram, a motor control center is added to Bus 3. The two motors at Bus 3 are connected to the schedule, creating a connected load which is the sum of the two motors’ rated sizes.

Connected Load = 75 kVA + 50 kVA

The demand and design loads are based on the FLA available in the Motor Control Center Library, not on the connected motor FLA based on connected load values.



BUS 1

Connected 125.00 kVADemand 133.85 kVADesign 153.81 kVA

C1

UTILITY 1

BUS 2


C2

BUS 3

Connected 125.00 kVADemand 133.85 kVADesign 153.81 kVA MOTOR 1

Size 50.0 kVAPF 0.85

MOTOR 3

Size 75.0 kVAPF 0.85

MCC

The following portion of the Demand Load Study Report shows connected loads of 50 kVA and 75 kVA for the two motors. This is equivalent to 60.1 A and 90.2 A. The demand load current value is from the Motor Control Center Library, based on Table 430-150 of the NEC. For the 75 hp 460 V motor, the NEC FLA value is 96 A. Because the 75 hp motor is the largest motor on the bus, the design load value is 125% of the demand load value (96 A), which is 120 A. The demand and design kVA values are calculated from the NEC FLA.

LOAD SCHEDULE FOR BUS-0003 480. VOLTS LINE TO LINE ============================================================================== ITEM DESCRIPTION * CONNECTED LOAD * DEMAND LOAD * DESIGN LOAD * % KVA AMPS KVA AMPS KVA AMPS P F ============================================================================== END USE LOADS KVA TYPE MTR 50.0 60.1 54.0 65.0 54.0 65.0 85.00 LARGEST KVA MTR 75.0 90.2 79.8 96.0 99.8 120.0 85.00 ============================================================================== TOTALS 125.0 150.4 133.9 161.0 153.8 185.0 85.00

1.4.5. Single Phase Non-Motor Loads in a Panel Schedule In this example, a three-phase 208/120 V four-wire panel is located on a circuit. Seven 15 A single-phase, single-pole loads are installed in the lighting and appliance panel. Two 30 A single-phase, two-pole heater circuits are also installed in the panel. All loads are at 95% power factor, so that their amperes and apparent power can be directly added. The DLS report is shown:

LOAD SCHEDULE FOR BUS-0002 208. VOLTS LINE TO LINE SOURCE OF PWR BUS-0001 ============================================================================== ITEM DESCRIPTION * CONNECTED LOAD * DEMAND LOAD * DESIGN LOAD * % KVA AMPS KVA AMPS KVA AMPS P F ============================================================================== END USE LOADS LIGHTING 12.6 35.0 12.6 35.0 15.8 43.8 95.00 ============================================================================== TOTALS 12.6 35.0 12.6 35.0 15.8 43.8 95.00

With seven single-phase 15 A circuits, the panel is not balanced. Phase A has 45 A of lighting load, whereas Phases B and C each have 30 A of load. The Phase A power is

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5.4 kVA, and Phases B and C each have 3.6 kVA. The sum of the power and current for each phase is:

Apparent Power = 5.4 kVA + 3.6 kVA + 3.6 kVA= 12.6 kVA

The load current shown in the PTW report is an average of the three-phase currents:

Average = 45 A + 30 A + 30 A3

35.0 A=

Because the loads all have the same power factor, the arithmetical average current is the sum of the three-phase currents divided by 3.

Next, the two single-phase, two-pole water heater loads are added to the panel. Again, the load is not evenly distributed across the three phases. One heater is connected across Phases A and B, and the other is connected across Phases B and C. Combined with the seven single-pole lighting loads, the phase currents are 75 A on Phase A, 90 A on Phase B, and 60 A on Phase C. This phase current is reported at the bottom of the Schedule dialog box.

Note: In the Load Schedule Report, PTW uses the bus rated voltage to calculate the fault current, whereas in the Comprehensive Short Circuit Study PTW uses the pre-fault voltage. You probably won't notice the difference because the bus rated voltage and the pre-fault voltage are usually the same; the only time they'll be different is when you have modeled transformer tap values and turn on the "Model Transformer Taps" option in the Comprehensive Short Circuit Study.

The total heating load per heater is 6.24 kVA. Because there are two heaters modeled, the total heating load is 12.48 kVA. This load is split across the three phases as shown in the following chart:

Total Apparent Power in the Panel Schedule Panel Load (kVA)

φ A φ B φ C Total

Lighting 5.40 3.60 3.60 12.60

Heating 3.12 6.24 3.12 12.48

Total 8.52 9.84 6.72 25.08

Each heater draws 6.24 kVA split evenly across its two phases, or 3.1 kVA per phase. Phase B draws current for both heaters. The following demand load report reports the loads in kVA. Both loads are considered as continuous loads.

LOAD SCHEDULE FOR BUS-0002 208. VOLTS LINE TO LINE SOURCE OF PWR BUS-0001 ============================================================================== ITEM DESCRIPTION * CONNECTED LOAD * DEMAND LOAD * DESIGN LOAD * % KVA AMPS KVA AMPS KVA AMPS P F ============================================================================== END USE LOADS LIGHTING 12.6 35.0 12.6 35.0 15.8 43.8 95.00 HEATING 12.5 34.6 12.5 34.6 15.6 43.3 95.00 ============================================================================== TOTALS 25.1 69.6 25.1 69.6 31.4 87.1 95.00



PTW reports a three-phase current drawn by the panel by calculating:

Demand Load = 25.1 kVA3 0.208 kV

69.6 A

e jb g=

This calculation matches the results in the preceding table.

1.4.6. Loads with Different Power Factors In previous examples, all the loads at the bus were at the same power factor. In this example, the motor loads have different power factors, but the rated size and efficiency of the two motors are identical. At Motor 1, the 50 kVA load at 95% power factor is equivalent to 47.5 kW and 15.6 kvar. Motor 2 at 75% power factor is equivalent to 37.5 kW and 33.1 kvar. The sum of the real power is 85 kW and the sum of the reactive power is 48.7 kvar.

Thus the vectorial sum of the real and reactive powers is equal to the total apparent power, which is 97.95 kVA at a power factor of 88%. These are the connected and demand load values at Bus 3 as shown on the following one-line diagram. Because the power factors are different, the apparent power and associated currents in the report at the end of this example cannot be added algebraically; they must be added in vector form. Even though both motors are rated 50 kVA, the Demand Load Study accurately identifies one of two motors as the largest motor, and uses a design factor of 125% to determine the design kVA and associated design load current.

BUS 1

C1

UTILITY 1

BUS 2

C2

BUS 3

MOTOR 1Size 50.0 kVANo 2PF 0.95 Lag

MOTOR 2Size 50.0 kVANo 1PF 0.75 Lag

Connected 97.95 kVADemand 97.95 kVADesign 110.23 kVAPF 87.89 Lag



The results of the apparent power calculation appear as the demand load kVA rounded up to 98.0 kVA in the Report. The 85.62% power factor for the combined loads is based on the design load values.

LOAD SCHEDULE FOR BUS-0003 480. VOLTS LINE TO LINE ============================================================================== ITEM DESCRIPTION * CONNECTED LOAD * DEMAND LOAD * DESIGN LOAD * % KVA AMPS KVA AMPS KVA AMPS P F ============================================================================== KVA TYPE MTR 50.0 60.1 50.0 60.1 50.0 60.1 95.00 LARGEST KVA MTR 50.0 60.1 50.0 60.1 62.5 75.2 75.00 ============================================================================== TOTALS 98.0 117.8 98.0 117.8 110.2 132.6 85.62

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1.4.7. Multiple Motors on a Single Motor Component This example shows how the Demand Load Study identifies the largest motor on a bus when a single motor component is modeling multiple motors. In the following one-line diagram, two motor components are at Bus 3. The component called Motor 1 actually represents two 50 kVA motors, and the component called Motor 2 represents a single 75 kVA motor.

Before combining the demand loads, the Demand Load Study correctly recognizes the single 75 kVA motor at Motor 2 as the largest motor on the bus, even though Motor 1 has a total of 100 kVA connected load. The design load is calculated as follows:

Design Load = 75 kVA 125% 2 50 kVA 100%

93.8 kVA +100 kVA= 193.8 kVA

× + × ×

=

b g b g

The design load value for Bus 3 is 193.8 kVA, as shown in the following one-line diagram.

BUS 1Connected 175.00 kVADemand 175.00 kVADesign 193.75 kVAPF 85.00 Lag

C1

UTILITY 1

BUS 2


C2

BUS 3


MOTOR 1Size 50.0 kVANo. of Mtrs 2PF 0.85 Lag

MOTOR 2Size 75.0 kVANo. of Mtrs 1PF 0.85 Lag

The following report show the design load of the largest motor as 93.8 kVA and the total design load as 198.8 kVA.

LOAD SCHEDULE FOR BUS 3 480. VOLTS LINE TO LINE SOURCE OF PWR BUS 2 ============================================================================== ITEM DESCRIPTION * CONNECTED LOAD * DEMAND LOAD * DESIGN LOAD * % KVA AMPS KVA AMPS KVA AMPS P F ============================================================================== END USE LOADS KVA TYPE MTR 100.0 120.3 100.0 120.3 100.0 120.3 85.00 LARGEST KVA MTR 75.0 90.2 75.0 90.2 93.8 112.8 85.00 ============================================================================== TOTALS 175.0 210.5 175.0 210.5 193.8 233.0 85.00

1.4.8. Motor Starting This example shows how the Demand Load Study accounts for a single 50 kVA starting motor. Under starting conditions, the motor lock rotor current is determined from the ANSI contribution data, which is located in the ANSI Contribution subview of the Component Editor. The lock rotor current ratio is the reciprocal of the motor’s subtransient reactance. In this example, the subtransient reactance is defined as 0.25 pu,



thus the lock rotor current ratio is 4 per unit. The starting power factor is related to the motor’s X/R ratio. The motor’s X/R ratio is defined as 5, thus the starting power factor is 19.6%. The power factor result is:

Power Factor = cos arctan

cos arctan5

cos 78.7

19.61% Lag

XRd i

b ge j

=

=

=

o

The following one-line diagram shows this starting power factor. The motor load characteristic is changed from a constant kVA load to a constant impedance load during this starting simulation, although it does not effect the outcome of the Demand Load Study.

BUS 1

C1

UTILITY 1

BUS 2

C2

BUS 3

MOTOR 1

Connected 50.00 kVADemand 200. kVADesign 250.00 kVAPF 19.61

Connected 50.00 kVADemand 200.00 kVADesign 250.00 kVAPF 19.61

Connected 50.00 kVADemand 200.00 kVADesign 250.00 kVAPF 19.61

Size 50.0 kVAStarting ConditionPF 19.61

Note that the report show that 200 kVA of demand load is also at the largest motor:

LOAD SCHEDULE FOR BUS-0003 480. VOLTS LINE TO LINE SOURCE OF PWR BUS-0002 ============================================================================== ITEM DESCRIPTION * CONNECTED LOAD * DEMAND LOAD * DESIGN LOAD * % KVA AMPS KVA AMPS KVA AMPS P F ============================================================================== END USE LOADS LARGEST Z MTR 50.0 60.1 200.0 240.6 250.0 300.7 19.61============================================================================== TOTALS 50.0 60.1 200.0 240.6 250.0 300.7 19.61

1.4.9. Multiple Loops in a System The purpose of this example is to show how the Demand Load Study identifies and opens loops in the power system. In the following one-line diagram, a loop exists because of the two utility connections. Bus 6 creates a loop between Utility 1 and 2, which the Demand Load Study finds, opens, calculates the demand and design load values of, and finally closes. Whenever loops are opened, the following message appears in the Report:

*** NOTICE *** BUS BUS-0006 COMPLETES LOOP IN SYSTEM

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BRANCH BUS-0006 TO BUS-0005 IS AUTOMATICALLY TAKEN OUT OF SERVICE

Because the branch from Bus 6 to Bus 5 is temporarily out of service, the calculated connected, demand, and design loads at Bus 4 and Bus 5 are zero. Additionally, the demand load on the branch between Bus 1 and Bus 2 is 125 kVA. Utility 1 is the source of supply for all the loads in the system.

If the branch from Bus 2 to Bus 6 were open instead of the branch from Bus 5 to Bus 6, the two sources of supply would feed the loads more equally. When there are loops in the power system, pay close attention to where the branches are automatically opened. Sometimes it is best to manually open the branch to ensure proper load balance between branches.


UTILITY 1

BUS 2

Connected 125.00 kVADemand 125.00 kVADesign 143.75 kVA C2

BUS 3

Connected 75.00 kVADemand 75.00 kVADesign 93.75 kVA MOTOR 1


MOTOR 2Size 50.0 kVAPF 0.95 Lag


UTILITY 2

BUS 5

Connected 0.00 kVADemand 0.00 kVADesign 0.00 kVA C4

BUS 6


C6

C5

BUS 7


1.4.10. Example from Plant The following figure is a one-line diagram for the Plant project. The Plant project is included on the PTW diskettes.



EDIS

ON

UTIL

ITY

BUS

TX A

PRI

TX A

003-

HV S

WG

R

C1

C2

C3

C4

SYN

ASY

N B

TX B

PRI

BUS

4

005-

TXD

PRI

006-

TX3

PRI

007-

TX E

PRI

TX B

DS S

WG

1 BU

S 8

C5

C6

GEN

1G

EN 2

TX C

PRI

BUS

9HV

AC B

US

TX C

DSB

1 BU

S 14

C14

C16

C17

BUS

15BU

S 16

BUS

17

C15

018-

RA

CMP

HVA

C

TX 6

TX3

SEC

BUS

11

TX3

TER

BUS

12

C8

DS

SWG

3 BU

S 20

C9

L1M

20

GEN

3

021-

TX F

PRI

C7

DS

SWG

2 BU

S 13

L2M

13

TX F

022-

DSB

2

C12

023-

MTR

23

M23

-BM

23-A

TX E

BLDG

115

SER

V

C10

C11

026-

TX G

PRI

025-

MTR

25

TX G

DSB

3 BU

S 27

C13

L5

BUS

28 A

M25

TX D

029-

TX D

SEC

C19

TM -1

SWBD

-1

PLN

- 17

H1A

PLN

16

H2A

MCC

15

- 1A

PNL

18 R

A

C18

H3A

BU

S 19

PNL

19 H

3A

R1 CB1

R2 CB2

CB6

CB7

R8 CB8

R9

CB9

R7

R6

CB5

R5R4 CB

4

R3 CB3

R10

CB10

F1

F2 MCP

#10

LVP1

LVP2

LVP3

MCC

B1M

CCB2

PCB1

PCB2

R11

PCB3

F4

F3 MO

/L#2

5

SW1

LVP4

DEM

ONS

TRAT

ION

PRO

JEC

T FO

R PO

WER

*TO

OLS

FO

R W

IND

OW

S

CAP

1

M28

#1

& 2

C20 LV

P5

MCP

#28B

-1

MO

/L#2

8B-1

M28

#3

BUS

28 B

MCP

#28B

-2

MO

/L#2

8B-2

M28

#4

CB12

TX H

CM

P C

TR

COM

PUTE

RS

F5

LVP6

OFF

ICE

COM

PLEX

ARE

AIN

DUST

RIAL

CO

MPL

EX A

REA

3/26/2006


The following figure shows a portion of the Plant project, including Demand Load Study results.

BUS 28 A


LVP4

M28 #1 & 2

250.0 kVANo of Mtrs: 20.80 Lag PFLoad Factor 0.90

C13

DSB 3 BUS 27


TX G

L5

100.0 kVA1.00 Unity PF

C20

F4

026-TX G PRI


LVP5

BUS 28 B


C10

BLDG 115 SERV


MCP#28B-1 MCP#28B-2

MO/L#28B-1

M28 #3


MO/L#28B-2

M28 #4


C11

025-MTR 25

Connected 2500.00 kVADemand 2500.00 kVADesign 3125.00 kVASW1

F3

MO/L#25

M25


DEMAND LOAD ANALYSIS

BUILDING 115 SERVICE

The Demand Load Study reports the connected, demand and design load power and current at each bus, and the design load value power factor. Some of the data for the buses in the above one-line diagram are shown in the following report:

LOAD SCHEDULE FOR BLDG 115 SERV 4160. VOLTS LINE TO LINE SOURCE OF PWR 007-TX E PRI ============================================================================== ITEM DESCRIPTION * CONNECTED LOAD * DEMAND LOAD * DESIGN LOAD * % KVA AMPS KVA AMPS KVA AMPS P F ============================================================================== END USE LOADS BRANCH LOADS 026-TX G PRI 1081.7 150.1 981.8 136.3 1058.9 147.0 84.05 025-MTR 25 2500.0 347.0 2500.0 347.0 3125.0 433.7 80.00 029-TX D SEC 0.0 0.0 0.0 0.0 WARNING: LOAD IS ZERO ============================================================================== TOTALS 3580.3 496.9 3480.5 483.0 4125.7 572.6 81.08



LOAD SCHEDULE FOR 026-TX G PRI 4160. VOLTS LINE TO LINE SOURCE OF PWR BLDG 115 SERV ============================================================================== ITEM DESCRIPTION * CONNECTED LOAD * DEMAND LOAD * DESIGN LOAD * % KVA AMPS KVA AMPS KVA AMPS P F ============================================================================== END USE LOADS BRANCH LOADS DSB 3 BUS 27 1081.7 150.1 981.8 136.3 1058.9 147.0 84.05 ============================================================================== TOTALS 1081.7 150.1 981.8 136.3 1058.9 147.0 84.05 LOAD SCHEDULE FOR 025-MTR 25 4160. VOLTS LINE TO LINE SOURCE OF PWR BLDG 115 SERV ============================================================================== ITEM DESCRIPTION * CONNECTED LOAD * DEMAND LOAD * DESIGN LOAD * % KVA AMPS KVA AMPS KVA AMPS P F ============================================================================== END USE LOADS LARGEST KVA MTR 2500.0 347.0 2500.0 347.0 3125.0 433.7 80.00 ============================================================================== TOTALS 2500.0 347.0 2500.0 347.0 3125.0 433.7 80.00 LOAD SCHEDULE FOR DSB 3 BUS 27 480. VOLTS LINE TO LINE SOURCE OF PWR 026-TX G PRI ============================================================================== ITEM DESCRIPTION * CONNECTED LOAD * DEMAND LOAD * DESIGN LOAD * % KVA AMPS KVA AMPS KVA AMPS P F ============================================================================== END USE LOADS HEATING 100.0 120.3 100.0 120.3 125.0 150.4 100.00 BRANCH LOADS BUS 28 A 500.0 601.4 450.0 541.3 506.2 608.9 80.00 BUS 28 B 500.0 601.4 450.0 541.3 506.2 608.9 80.00 ============================================================================== TOTALS 1081.7 1301.0 981.8 1181.0 1058.9 1273.7 84.05 LOAD SCHEDULE FOR BUS 28 A 480. VOLTS LINE TO LINE SOURCE OF PWR DSB 3 BUS 27 ============================================================================== ITEM DESCRIPTION * CONNECTED LOAD * DEMAND LOAD * DESIGN LOAD * % KVA AMPS KVA AMPS KVA AMPS P F ============================================================================== END USE LOADS LARGEST KVA MTR 500.0 601.4 225.0 270.6 281.2 338.3 80.00 ============================================================================== TOTALS 500.0 601.4 450.0 541.3 506.2 608.9 80.00 LOAD SCHEDULE FOR BUS 28 B 480. VOLTS LINE TO LINE SOURCE OF PWR DSB 3 BUS 27 ============================================================================== ITEM DESCRIPTION * CONNECTED LOAD * DEMAND LOAD * DESIGN LOAD * % KVA AMPS KVA AMPS KVA AMPS P F ============================================================================== END USE LOADS KVA TYPE MTR 250.0 300.7 225.0 270.6 225.0 270.6 80.00 LARGEST KVA MTR 250.0 300.7 225.0 270.6 281.2 338.3 80.00 ============================================================================== TOTALS 500.0 601.4 450.0 541.3 506.2 608.9 80.00

The complete load summary from the Demand Load Study Report is shown below. It lists the connected, demand and design load values and the design load power factor in each load category.

TOTAL SOURCE LOAD SUMMARY ****************************************************************************** ============================================================================== LOAD DESCRIPTION UNITS CONNECTED DEMAND DESIGN POWER FACTOR TYPE LOAD LOAD LOAD % ============================================================================== GENERAL LOADS KW 59.6 59.6 59.6 KVAR 28.9 28.9 28.9 KVA 66.3 66.3 66.3 90.00 LAGGING LIGHTING KW 216.2 216.2 270.2 KVAR 85.3 85.3 106.6 KVA 232.4 232.4 290.5 93.02 LAGGING RECEPTACLES KW 80.3 44.4 44.4 KVAR 49.7 27.5 27.5

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KVA 94.4 52.2 52.2 85.00 LAGGING OFFICE EQUIPMENT KW 7.1 7.1 7.1 KVAR 4.4 4.4 4.4 KVA 8.3 8.3 8.3 85.00 LAGGING HEATING KW 540.0 540.0 675.0 KVAR 0.0 0.0 0.0 KVA 540.0 540.0 675.0 100.00 UNITY STANDBY LOADS KW 63.8 63.8 79.7 KVAR 39.5 39.5 49.4 KVA 75.0 75.0 93.8 85.00 LAGGING ENERGY AUDIT KVA KW 427.5 427.5 427.5 KVAR -1159.5 -1159.5 -1159.5 KVA 1235.8 1235.8 1235.8 -34.59 LEADING KVA TYPE MTR KW 8904.1 8831.0 8831.0 KVAR 4442.0 4405.5 4405.5 KVA 9950.6 9868.9 9868.9 89.48 LAGGING LARGEST KVA MTR KW 3200.0 3200.0 4000.0 KVAR 2400.0 2400.0 3000.0 KVA 4000.0 4000.0 5000.0 80.00 LAGGING ------------------------------------------------------------------------------ TOTAL LOADS KW 13498.5 13389.5 14394.5 KVAR 5890.4 5831.6 6462.8 KVA 14727.7 14604.3 15778.7 % PF 91.7 91.7 91.2 LAGGING LAGGING LAGGING

The demand load factors for the various demand loads are listed in the table below. Only seven of the twenty demand load categories are used in this example.

LOAD DEMAND TABLE ============================================================================== LOAD DESCRIPTION LOAD FIRST DEMAND SECOND DEMAND THIRD DEMAND DESIGN TYPE KVA % KVA % KVA % FACT ============================================================================== GENERAL LOADS K 100. 100. ALL 50. ALL 50. 1.00 LIGHTING K ALL 100. ALL 100. ALL 100. 1.25 RECEPTACLES Z 10. 100. ALL 50. ALL 50. 1.00 OFFICE EQUIPMENT Z ALL 100. ALL 100. ALL 100. 1.00 HEATING Z ALL 100. ALL 100. ALL 100. 1.25 STANDBY LOADS K ALL 100. ALL 100. ALL 100. 1.25 CAPACITOR BANK Z ALL 100. ALL 100. ALL 100. 1.35


2 Sizing Study

The Sizing Study sizes branch components such as feeders, neutrals, raceways, transformers, and equipment grounding conductors based on the results of the Demand Load Study (DLS).

The Sizing Study is the second of two Studies used to quantify the preliminary electrical design to meet national codes and standards. Combined with the DLS, the preliminary design ensures that feeders and transformers meet minimum load values. Once the preliminary design is defined, more detailed Studies (Voltage Drop, Motor Starting, Short Circuit Studies) are used to verify that electrical apparatus is sized for normal and abnormal system operation.




• Examples.

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What is the Sizing Study? ..............................................................................DAPPER 2-2 Engineering Methodology .............................................................................DAPPER 2-3 PTW Applied Methodology ..........................................................................DAPPER 2-4 Application Examples ....................................................................................DAPPER 2-7


2.1. What is the Sizing Study? The purpose of the Sizing Study is to recommend feeder and transformer sizes based on the calculated demand and design load values in a power system. Once feeder ampacities are determined, the Sizing Study recommends consistent conductor raceway and equipment grounding conductor sizes. The Sizing Study produces feeder and transformer schedules, reports information about the feeder and transformer ratings, and reports feeder percent voltage drops.

The Sizing Study follows U.S. National Electrical Code (NEC)[1] procedures; however, the critical design criteria are user-defined, and can be specified to meet local codes.

The following flow chart shows the procedure for the Sizing Study.

Define System Data

Define system topology and connectionsDefine utility connections (swing bus)Define individual loadsRun the Demand Load Study

Run Sizing Study

Saved in DatabaseFor each branch:Feeder sizeQuantity of feeders per phaseTransformer nominal kVARaceway informationTransformer full-load kVA

DXF Files

Reports

Study Setup

Cable LibraryTransformer LibraryFeeder Voltage Drop CriteriaTransformer Tap Size CriteriaStudy Setup

Used by Short Circuit Studyand Load Flow Study

Datablocks

[1] Authored by Committee, 1996 National Electric Code. Quincy, MA: National Fire Protection Association, 1996.

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Sizing Study DAPPER 2-3

2.2. Engineering Methodology This section describes the methodology and definitions used by the Sizing Study. The Sizing Study selects the conductor size based on the minimum conductor material cross-sectional area necessary to meet defined feeder current carrying capability, and associated voltage drop criteria. For more information on voltage drop, see the Load Flow Chapter in this Reference Manual. Transformers are sized based on their full load kVA rating.

2.2.1. Feeder Sizing Feeders are sized based on the design load value from the DLS. The DLS calculates the total connected demand and design load in each branch of the power system. Some loads are defined as continuous loads and, as such, the design load value is larger than its demand load value. The NEC requires branch circuits that serve continuous loads be rated so that not more than 80% of the feeder ampacity is used. Selecting a design load value of 125% ( 1

80% ) of the demand load value meets the NEC standard. Ampacity is defined as the current in amperes that a conductor may carry continuously under the conditions of use without exceeding its temperature rating.

The Sizing Study bases its calculations on two separate criteria: the minimum conductor cross-sectional area to meet feeder ampacity values and a user-defined voltage drop value. If you consider parallel feeder combinations for a specified conductor type, the Sizing Study can select multiple feeders in parallel for that conductor. Cable sizes must be specified in the Cable Library in order to be available to the Sizing Study. Derated cable ampacities are determined based on the temperature derating factor and duct bank design detail criterion.

The Sizing Study selects the cable that best meets defined ampacity values and has the smallest cross-sectional area. Once the cable is selected, the voltage drop for the cable or cable pair is calculated. If the voltage drop criterion is exceeded, the Sizing Study selects the next larger cable size and begins the comparison of cross-sectional area, rated ampacity and voltage drop.

The Sizing Study algorithm determines the feeder branch design load value in amperes, then determines the selected feeder design ampacity. The feeder design ampacity is the product of the rated ampacity, the temperature derating factor and the number of parallel cables. The design load value is the rated size of the load multiplied by specified demand factors and the long continuous load factor (or design factor).

Once the ampacity conditions are met, the Sizing Study then checks the calculated voltage drop on the cable, based on the branch design load current and power factor, cable impedance, and length. The voltage drop is calculated using the following formula:

% Voltage Drop3 I R cos I X sin

VLL=

× × + × ××

φ φb g b g100

If the voltage drop exceeds the specified level, the Sizing Study selects the next largest feeder and restarts the Sizing Study in order to select a cable or combination of parallel cables which meets the ampacity criteria of minimum cross-sectional area and acceptable voltage drop. The Sizing Study voltage drop calculation follows the methodology cited in



Chapter 3 of the IEEE Red Book.[2] This simple methodology assumes no losses in the branch.

Feeders attached to the primary or secondary of a transformer are sized differently than those associated with other branch circuits. These feeders tend to be short in length and are sized in direct relation to the transformer nominal rated size.

2.2.2. Transformer Sizing Transformer sizing is based on either the calculated branch demand or design load value, depending on which you select in the Component Editor. PTW defaults new transformers to be sized based on the demand load value. Transformers with a Transformer Key of <User Defined> or with Existing checked under Sizing Information are not sized. The sizing algorithm compares the demand or design load value to the transformer’s full load size. The transformer’s full load size is:

Full Load Size = Nominal kVA Rating Transformer Capacity Factor×

The capacity factor is stored in the Transformer Library. Typical capacity factors are:

Transformer Cooling Characteristic Capacity Factor (may be changed) Dry Type (DT) 1.0

Oil/Air Cooled (OA) 1.15

Oil/Air/Forced Air (OAFA) 1.25

Oil/Forced Air (FA) 1.33

To change a transformer’s capacity factor:

1. Open the Transformer Library.

2. Choose the Select command from the Component menu.

3. Type a new capacity factor for the selected transformer.

2.2.3. Transformer Feeders Feeders for the primary and secondary transformers require special discussion. The Sizing Study sizes the primary and secondary feeders of each transformer based on a factor (usually 125%) of the transformer’s full load rating and an allowable voltage drop criteria. This factor is defined in the Demand Load Study Setup dialog box.

2.3. PTW Applied Methodology The basis of the Sizing Study is the design load values from the DLS. The DLS is based on a methodology valid for only a radially-fed feeder circuit. If the system topology

[2] Authored by Committee. IEEE Red Book: IEEE Recommended Practice for Electric Power and Distribution for Industrial Plants, IEEE Std 141-1993. USA: Institute of Electrical and Electronics Engineers, Inc., 1994.

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contains a loop, the DLS will have temporarily opened the loop to run the demand and design load Studies. Therefore, the Sizing Study is based on the radial system used in the DLS.

Although user-defined feeders are not sized, if sufficient ampacity data is entered in the Component Editor, then feeder evaluation is accomplished.

Because the sole purpose of the DLS is to provide the vector sum of the total load at each bus and in each branch of the electrical system, it ignores local generation. Therefore, the Sizing Study, which is based on the demand and design load values resulting from the DLS, tends to oversize feeders and transformers which are upstream from local generation.

The Cable and Transformer Libraries contain feeder and transformer sizing tables that are required for the Sizing Study. You can modify the libraries to suit your individual design requirements.

2.3.1. Before Running the Sizing Study Before running the Sizing Study, you must:

• Define the power system topology and connections.



• Run the Demand Load Study (DLS). The Sizing Study sizes feeders and transformers according to the design loads calculated by the DLS. Transformers may be sized for either the design or demand load from the DLS. Only those feeders and transformers not checked as Existing are sized. Cables and transformers checked as Existing are evaluated by the Sizing Study but not sized. If they are overloaded, an Error Message is displayed in the Report. For more detailed information on Error Messages, see page 2-7.

Tip: You may run the Demand Load Study in the same step that you run the Sizing Study. All you need to do is select both Studies, and PTW will automatically run the Demand Load Study first. See “Running the Sizing Study,” following.

2.3.2. Running the Sizing Study You can run the Study from any screen in PTW, and it always runs on the active project.

To run the Sizing Study:


2. Select the check box next to Sizing. You may also select the check box next to Demand Load to run the Demand Load Study prior to the Sizing Study.



The Sizing Study runs, writes the results to the database, and creates a Report.



The Sizing Study is dependent on the system load values. Therefore, the Sizing Study bases its calculations on the results of the last Demand Load Study run.

2.3.3. Sizing Study Options When you select the Sizing Setup button, PTW displays the Feeder and Transformer Sizing Study dialog box as shown:

Following is a list of the available Study options.

You can keep the defaults, or type in new values for the Allowable Feeder Voltage Drop and Primary/Secondary Feeder percent text boxes:

In this box Type this information Allowable Feeder Voltage Drop Percent value to which you want to limit a single branch

circuit voltage drop. Used as a sizing constraint.

Primary/Secondary Feeder Percent Transformer FLA value. Used as a sizing constraint.

You can choose one of three options for the Study results:

Select this option button To do this Size and Report, Do Not Change Database

Size the existing system, generate a Report of current feeder and transformer sizes; make no changes to the database.

Size, Report and Change Database Size the existing system, generate a Report of current feeder and transformer sizes; make changes to the database based on the findings.

Report Only, Do Not Size Generate a Report of current feeder and transformer sizes; make no new sizing calculations or changes in the database.

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You can select to create a DXF file suitable for import to Computer Aided Design (CAD) software, and specify an extended print character set:

Select this check box To do this Create .DXF File Create a separate file with the dxf file extension and save

it in the projects subdirectory.

Use Extended Character Set Provide for vertical lines and boxes around the feeder and transformer schedules. You must use the PIXymbolsExtended font which is automatically loaded as a True Type font when PTW is installed.

2.3.4. Error Messages As the Study runs, the Study Messages dialog box shows the progress of the Study. If the Sizing Study detects logic errors in the input data, error messages appear in this dialog box. A single logic error may produce more than one error statement. You must correct all errors before the Sizing Study can run. The most common error is:

Feeder Size Not in Library

The Sizing Study tries to find an appropriate feeder size or parallel combination of feeders to meet either the cable ampacity or voltage drop limitation, but if none is found, the Sizing Study will broadcast a warning and leave the cable size unchanged.

Note: Because the Sizing Study may change circuit impedance values, PTW will not allow you to run the Sizing Study simultaneously with either the Load Flow Study, Comprehensive Short Circuit Study, A_FAULT Study, or IEC_FAULT Study.

2.4. Application Examples In the following sections, a few simple projects demonstrate how the Sizing Study works.

2.4.1. Sizing a Simple Radial Feed This example demonstrates how the Sizing Study works on a project with two branch circuits which are serviced by a dry type transformer. The two non-motor loads are rated as continuous constant kVA-type loads; a 125% long continuous loading factor is used to calculate the design load value. Cable C4 is ten times longer than Cable C3. The results of the Sizing Study are shown on the following one-line.



B1 4160 VDemand Load 200.00 kVA

C1Size # 6 AWG/kcmil1 Cables in Parallel30.10 AmpsAmpacity 75.0 A / CableLength 100.0 ft

B24160 VDemand Load 200.00 kVA

UTILITY

T1225.0 kVA


C2 Size # 4/0 AWG/kcmil4 Cables in Parallel260.90 AmpsAmpacity 230.0 A / CableLength 1200.0 ft B4

480 VDemand Load 200.00 kVA

C3Size # 2/0 AWG/kcmil1 Cables in Parallel128.98 AmpsAmpacity 175.0 A / CableLength 100.0 ft

C4Size # 2/0 AWG/kcmil2 Cables in Parallel131.92 AmpsAmpacity 175.0 A / CableLength 1000.0 ft



L1Design Load 125.00 kVA

L2Design Load 125.00 kVA

In the one-line diagram, note the branch circuit conductor from Bus B4 to Bus B5. The #2/0 AWG cable has an ampacity of 175 A and a load current of 129 A. The cable is loaded to 73% of its ampacity, safely under the NEC’s 80% limit. The branch from Bus B4 to Bus B6 is ten times longer than the first branch. The voltage drop is significant, and the Sizing Study selects two #2/0 AWG cables in parallel. Two are required because of the voltage drop associated with the 1000-foot cable length.

A low voltage feeder circuit is defined between Buses B3 and B4, but this is also defined as a secondary tap of the transformer. The Sizing Study sizes this feeder based on the transformer full load size. Transformer T1 was sized as 225 kVA to meet the total demand load of 200 kVA in this branch. The Sizing Study always reports when feeder cables are selected based on the transformer size. In this case, the Report listed the following message regarding the transformer secondary tap:

PRIMARY/SECONDARY TRANSFORMER FDRS SIZED AT 125. % OF TX FULL LOAD RATING *** NOTICE *** FEEDER SIZED TO 125. PERCENT OF TRANSFORMER SIZE BRANCH FROM B3 TO B4

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TX KVA: 225.0 TR FLA: 270.6 MINIMUM FEEDER AMPACITY: 338.3

From the preceding Report, the secondary feeder will be sized larger than 338.3 A because the transformer feeder sizing criterion was chosen as 125% of the transformer’s full load rated amperes. The user-defined Primary/Secondary Feeder sizing criterion is adjusted in the Feeder and Transformer Sizing Study dialog box. Thus, four parallel sets of #4/0 AWG cables are selected as the optimal solution, including the effects of the voltage drop, due to the extremely long cable length. The four #4/0 AWG cables have a combined cross-sectional area of 846.4 kcmil. Even if 750 kcmil conductors are selected, parallel circuits are required. If the transfer secondary tap were ten feet long, then two each #2/0 AWG cables, with a rated ampacity of 175 A per cable, would certainly meet the ampacity, cross sectional area, and voltage drop sizing criteria.

2.4.2. Impact of Cable Temperature Derating This example shows the impact of ambient temperature derating factors on cable sizing. Suppose that two equal-length branch circuits are affected by different ambient temperatures. Cable C3 is derated to 50oC, since it runs on the roof in direct sunlight. This reduces the conductor ampacity by 25%, in accordance with the NEC. Given equal loads on each branch, the results of this derating are as shown:

B4

C3Size 2/0 AWGNo in Parallel 1121.72 A Load Flow Current175.0 A Cable Ampacity100.0 ft

C4Size 1 AWGNo in Parallel 1122.00 A Load Flow Current130.0 A Cable Ampacity100.0 ft

B5 B6

L1100.0 kVA

L2100.0 kVA

30 Deg OperatingTemperature

50 Deg OperatingTemperature

The derating of Cable C3 is accomplished by adjusting the temperature text box in the Conductor and Raceway subview. Cable C3 must be sized differently than Cable C4, based on the 50°C derating factor of 75%. Cable C3 is sized to a single #2/0 AWG cable, which has a rating of 175 A at 30° C or 131 A at 50° C. The actual demand load value in this branch is 121.7 A. This result is compared to the identical load and cable length in the branch from Bus B4 to Bus B5. In this branch only a single #1 AWG cable, rated at 130 A, is required.

2.4.3. Multiple Branches with Different Load Values This example shows the impact of the Sizing Study when multiple branches with different load characteristics are modeled. There is no ambient compensation; thus the cable ambient temperature derating factor on Cable C4 is 1.0. The results are displayed on the following one-line diagram:



B1

C1Length 6000.0 ftAmpacity 150.0 ASize 2 No 1 LF Current 70.48 AB2

C2Length 25.0 ftAmpacity 83.0 ASize 6 No 1 LF Current 28.28 A

B3

T1

Size 225.0 kVALoad 205.00 kVA

B4

U1

C3Length 10.0 ftAmpacity 175.0 ASize 2/0 No 2 LF Current 232.86 AB5

B6

C4Length 300.0 ftAmpacity 85.0 ASize 4 No 1 LF Current 63.22 A

L155.00 kVASPECIAL LOAD

C5Length 1000.0 ftAmpacity 175.0 ASize 2/0 No 2 LF Current 175.50 A

B7


C20Length 25.0 ftAmpacity 110.0 ASize 4 No 1 LF Current 42.86 AB30

T2Size 500.0 kVALoad 305.00 kVA

B40

C30Length 10.0 ftAmpacity 200.0 ASize 3/0 No 4 LF Current 352.89 A

B50

C40

Length 300.0 ftAmpacity 175.0 ASize 2/0 No 1 LF Current 123.43 A

C50

Length 1000.0 ftAmpacity 150.0 ASize 1/0 No 3 LF Current 235.42 A

B60 B70

L10

105.00 kVASPECIAL LOAD


The above diagram shows that each branch circuit is sized according to its load and that the primary and secondary transformer feeders are sized based on the transformer full load rating.

Tip: To effectively use the Sizing Study where many transformers exist, it is best to place primary and secondary feeder taps on the transformer, as shown above.


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EDIS

ON

UTIL

ITY

BUS

TX A

PRI

TX A

003-

HV S

WG

R

C1

C2

C3

C4

SYN

ASY

N B

TX B

PRI

BUS

4

005-

TXD

PRI

006-

TX3

PRI

007-

TX E

PRI

TX B

DS S

WG

1 BU

S 8

C5

C6

GEN

1G

EN 2

TX C

PRI

BUS

9HV

AC B

US

TX C

DSB

1 BU

S 14

C14

C16

C17

BUS

15BU

S 16

BUS

17

C15

018-

RA

CMP

HVA

C

TX 6

TX3

SEC

BUS

11

TX3

TER

BUS

12

C8

DS

SWG

3 BU

S 20

C9

L1M

20

GEN

3

021-

TX F

PRI

C7

DS

SWG

2 BU

S 13

L2M

13

TX F

022-

DSB

2

C12

023-

MTR

23

M23

-BM

23-A

TX E

BLDG

115

SER

V

C10

C11

026-

TX G

PRI

025-

MTR

25

TX G

DSB

3 BU

S 27

C13

L5

BUS

28 A

M25

TX D

029-

TX D

SEC

C19

TM -1

SWBD

-1

PLN

- 17

H1A

PLN

16

H2A

MCC

15

- 1A

PNL

18 R

A

C18

H3A

BU

S 19

PNL

19 H

3A

R1 CB1

R2 CB2

CB6

CB7

R8 CB8

R9

CB9

R7

R6

CB5

R5R4 CB

4

R3 CB3

R10

CB10

F1

F2 MCP

#10

LVP1

LVP2

LVP3

MCC

B1M

CCB2

PCB1

PCB2

R11

PCB3

F4

F3 MO

/L#2

5

SW1

LVP4

DEM

ONS

TRAT

ION

PRO

JEC

T FO

R PO

WER

*TO

OLS

FO

R W

IND

OW

S

CAP

1

M28

#1

& 2

C20 LV

P5

MCP

#28B

-1

MO

/L#2

8B-1

M28

#3

BUS

28 B

MCP

#28B

-2

MO

/L#2

8B-2

M28

#4

CB12

TX H

CM

P C

TR

COM

PUTE

RS

F5

LVP6

OFF

ICE

COM

PLEX

ARE

AIN

DUST

RIAL

CO

MPL

EX A

REA



The following figure shows a portion of the Plant project.

BUS 28 A

LVP4

M28 #1 & 2

C13

DSB 3 BUS 27

TX G

L5 C20

F4

026-TX G PRI

LVP5

BUS 28 B

C10

BLDG 115 SERV

MCP#28B-1 MCP#28B-2

MO/L#28B-1

M28 #3

MO/L#28B-2

M28 #4

C11

025-MTR 25

SW1

F3

MO/L#25

M25

FEEDER AND TRANSFORMER SIZING STUDY


The Sizing Study first lists notices if feeders are identified on either the primary or secondary side of two winding transformers. If feeders are identified, they are sized in accordance to the ***Notice*** listed. Some of the data for the buses in the above one-line diagram are shown in the following Report.

FEEDER AND TRANSFORMER STUDY CRITERIA ============================================================================== VOLTAGE DROP CALCULATIONS ARE PRELIMINARY EXECUTE VOLTAGE DROP AND LOAD FLOW STUDY FOR MORE ACCURATE RESULTS *** NOTICE *** FEEDER SIZED TO 125. PERCENT OF TRANSFORMER SIZE BRANCH FROM BLDG 115 SERVICE TO 026-TX G PRI TX KVA: 1000.0 TR FLA: 138.8 MINIMUM FEEDER AMPACITY: 173.5

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*** NOTICE *** FEEDER SIZED TO 125. PERCENT OF TRANSFORMER SIZE BRANCH FROM DSB 3 BUS 27 TO BUS 28 A TX KVA: 1000.0 TR FLA: 1202.8 MINIMUM FEEDER AMPACITY: 1503.5

The Sizing Study Report also lists all the feeder sizes, as shown below, for a portion of the Plant project.

FEEDER SIZE REPORT ****************************************************************************** F E E D E R S C H E D U L E ============================================================================== FEEDER ROUTING FEEDER NO WIRE SIZE TYPE INSUL GROUND RACEWAY NO NAME VOLTAGE /PH QTY FDR MAT TYPE WIRE SIZE TYPE ============================================================================== FROM BLDG 115 SERV 4160. ********************* TO 026-TX G PRI ( 1) 3 1/0 CU XLP 3" C N TO 025-MTR 25 EX ( 1) 3 500 CU XLP 5" C N FROM DSB 3 BUS 27 480. ********************* TO BUS 28 A ( 4) 3 500 CU THWN 3" C N FROM DSB 3 BUS 27 480. ********************* TO BUS 28 B ( 4) 3 2/0 CU THWN 1 1/2" C N

The Sizing Study Report includes a feeder evaluation.

FEEDER EVALUATION ****************************************************************************** F E E D E R D E S I G N L O A D A N A L Y S I S ============================================================================== FEEDER ROUTING EXTG % QTY SIZE FEEDER DESCRIPTION DESIGN FEEDER NO NAME VD /PH FDR MAT INSUL AMBIENT LOAD CAPACITY ============================================================================== FROM 003-HV SWGR 13800. ********************* TO TX B PRI BUS 4 0.07 1 4 CU XLP 30. 82. A 110. A TO 005-TXD PRI EX 0.00 1 2/0 CU XLP 30. 0. A 225. A *** WARNING *** FEEDER LOAD IS DEFINED AS ZERO TO 006-TX3 PRI 0.12 2 1/0 CU XLP 30. 368. A 390. A TO 007-TX E PRI 0.03 2 1/0 CU XLP 30. 173. A 390. A FROM BLDG 115 SERV 4160. ********************* TO 026-TX G PRI 0.29 1 1/0 CU XLP 30. 147. A 195. A TO 025-MTR 25 EX 0.43 1 500 CU XLP 30. 434. A 585. A FROM 022-DSB 2 480. ********************* TO 023-MTR 23 0.37 4 500 CU THWN 30. 1064. A 1520. A FROM DSB 3 BUS 27 480. ********************* TO BUS 28 A 0.75 4 500 CU THWN 30. 609. A 1520. A

The Sizing Study Report also includes feeder evaluation, based on the design load value.

FEEDER EVALUATION ****************************************************************************** F E E D E R D E S I G N L O A D A N A L Y S I S ============================================================================== FEEDER ROUTING EXTG % QTY SIZE FEEDER DESCRIPTION DESIGN FEEDER NO NAME VD /PH FDR MAT INSUL AMBIENT LOAD CAPACITY ============================================================================== FROM BLDG 115 SERV 4160. ********************* TO 026-TX G PRI 0.29 1 1/0 CU XLP 30. 147. A 195. A TO 025-MTR 25 EX 0.43 1 500 CU XLP 30. 434. A 585. A FROM DSB 3 BUS 27 480.



********************* TO BUS 28 A 0.75 4 500 CU THWN 30. 609. A 1520. A FROM DSB 3 BUS 27 480. ********************* TO BUS 28 B 1.75 4 2/0 CU THWN 30. 609. A 700. A

The Sizing Study Report includes a transformer sizing section.

TRANSFORMER SIZE REPORT ****************************************************************************** T R A N S F O R M E R S C H E D U L E ============================================================================== LOCATION DESCRIPTION VOLTAGE CONN PCT. TRANSFORMER DESCRIPTION BUS NO. NAME LEVELS CODE TAP ============================================================================== FROM 026-TX G PRI 4160. D -2.5 TYPE: DT SIZE: 1000.0 KVA TO DSB 3 BUS 27 480. YG DESCRIPTION: TX G DEMAND LOAD: 981.8 KVA 5.75 %Z NOMINAL RATING 1000.0 KVA

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3 Load Flow Study

The Load Flow Study predicts the overall apparent real and reactive power distribution throughout a power system, including associated losses. Additionally, the Study calculates the voltage drop through each branch impedance component, and the associated voltages at each bus or node in the electrical system.

This chapter introduces the steady-state load flow equations used in the Load Flow Study, demonstrates PTW’s powerful algorithm used to solve these equations, and documents several case studies that validate the Study and provide you with examples of how to most effectively use the Load Flow Study.




• Examples.

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1.1. What is the Load Flow Study?.......................................................................... 3-2 1.2. Engineering Methodology ................................................................................ 3-2 1.3. PTW Applied Methodology ............................................................................. 3-6 1.4. Application Examples..................................................................................... 3-12


11/22/2006

3.1. What is the Load Flow Study? A Load Flow Study is conducted on a power system to evaluate the adequacy of continuous and emergency overload ratings of cables, transformers, and protective devices. The Study may also be used to determine anticipated low, or high, voltage levels on various sections of the power system under various loading conditions. This information can then be used to determine the associated impact of abnormal voltage on electrical apparatus. The Study can be instrumental in evaluating the impact of motor starting, and can help recommend economical sizes of local generation equipment and power factor correction equipment.

Define System Data

Define system topology and connectionsDefine utility connections (swing bus)Define individual loadsDefine feeder and transformer sizesDefine generator sizesRun the Demand Load Study (optional)

Run Load Flow Study

Saved in DatabaseFor each branch:

Voltage dropPower flow

For each bus:VoltageVoltage angleVoltage dropPower flowPower factor

Datablocks

Reports

Study Setup

Cable LibraryTransformer LibraryDemand LoadsDirectly Connected LoadsStudy Setup

3.2. Engineering Methodology The steady-state load flow solution to a power system network involves Ohm’s Law:

[I] [Y][V]= Eq. 3-1

Load Flow Study DAPPER 3-3


where

I column vector of total positive sequence currents flowing into each node (bus) in the system; Y The network admittance matrix (1/Z); V Column Vector of positive sequence voltage at each bus.

This equation is a linear algebraic equation with complex real and imaginary coefficients. The matrix may be reduced and the solution for either voltage or current reached using matrix algebra. The current flowing into any node of the system may be defined:

IP jQ *

V *ii i

i=

+LNM

OQP

b g

Eq. 3-2

where

P jQ *i i+b g complex conjugate of the apparent power flowing into the ith node; Vi* complex conjugate of the voltage of the ith node.

Combining Eq. 3-1 and Eq. 3-2 yields:

P jQ[V]*

[Y][V]−LNM

OQP

=b g

Eq. 3-3

Equation 3-3 is non-linear, and cannot be solved in closed form; thus the numerical analysis solution technique used must guess at each branch power flow, evaluate the algebraic equations, and then determine if the power into each bus equals power leaving the bus, including losses—that is, determine if Kirchoff’s Current Law is met. This iterative-type numerical analysis solution method continues until convergence is reached. (Convergence means that the power into the node equals the power out of the node.) A convergence criteria is established, based on an acceptable level of precision. For example, the algorithm might identify the largest load in the system and divide that load value by some constant, such as 20,000, to define a convergence criterion. Convergence usually occurs in less than ten guesses or numerical iterations.

The steady-state load flow Eq. 3-3 can be reduced to a set of input and output variables; knowledge about these variables aids in the solution. Three types of buses are defined when solving for the power flow, as noted below:

Bus Type Node Variable Type Independent Dependent

I Load Bus P, jQ V, α

II Generation Bus, Class A -P, ±Q V, α

Generation Bus, Class B -P, V ±jQ, α

III Swing or Slack Bus V, α ±P, ±Q

The type I bus may be either a motor or non-motor load bus, where power out of the node is defined as a positive quantity. The dependent variables are voltage magnitude and voltage angle.

The type II bus is a generation bus where real power is generated. A class A generator bus is non-regulated; the real and reactive power is fixed in magnitude. As load variations


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occur, the voltage magnitude and voltage angle vary. A class B generator type bus is a regulated bus. Automatic voltage regulation controls the bus voltage within the generator’s reactive power limits. Since the real power is fixed in magnitude, the system frequency (the resultant voltage angle) must vary.

The type III bus is called a swing or slack bus, and the voltage magnitude and angle are held constant, and the real and reactive power then must vary.

At least one bus in the electrical system must be defined as a swing bus in order to solve the steady-state load flow equation. From the preceding table, two of the four variables are always unknown, but there is a single equation (Eq. 3-3). This is why the Load Flow Study solution cannot be solved in closed form. By defining a swing bus voltage and angle, and recognizing that all type I bus voltages and angles are relative to the swing bus voltage magnitude and angles, numerical solution techniques may be used. The defining methodology assumes that the total apparent power into each node must equal the power out of the node (that is, that Kirchoff’s Current Law is valid).

3.2.1. The Solution Process PTW’s solution of Eq. 3-3 depends on the system topology, combined with knowledge of associated branch impedances and load data. PTW forms the appropriate matrixes and, through optimal ordering and standard matrix algebra techniques, solves for the dependent variables. One power flow solution technique is known as the double current injection method (PTW’s Load Flow Study uses this method). In this method, the first estimate assumes no losses and calculates the current flows in each branch, given the load values and system nominal voltages. Then the losses across the system are calculated, and the voltage drop is determined for each branch and bus. Given this new voltage at each bus, the load currents are re-calculated, and the iterative process begins. The new currents develop new losses in the branches and thus new voltage drops in each branch and bus. The iterative process continues until there is little change in the voltage at each bus between estimates, and convergence is achieved.

3.2.2. Modeling Transformers Transformer primary and secondary tap settings and transformer off-nominal voltages must be considered in the steady-state load flow solution. A negative primary tap setting raises the secondary bus voltage. Similarly, a positive secondary tap setting raises the secondary bus voltage.

3.2.3. Utility Equivalent Impedance You may want to model the utility system voltage drop, which is defined as the impedance between the swing bus and swing bus (ideal) voltage source. The utility system three-phase short circuit capacity is used to determine the positive sequence impedance for this calculation. If you want to model a tap changing under load main station transformer, and are not simulating motor starting conditions, then you would not model the utility equivalent impedance in the load flow solution.

3.2.4. Load Characteristics The load flow solution must take into account load characteristics to calculate the apparent load flow conditions in the power system. The load flow conditions are solved in



harmony with solution of the voltage conditions at each load bus, as described in the previous section.

The type of loads specified and the system losses significantly influence the results of the load flow and voltage drop calculations. Constant impedance type loads are loads that vary as the square of the applied voltage. Examples of this type of load include incandescent lighting and resistance heating elements. Constant kVA loads are loads that remain (or attempt to remain) constant within boundary limitations regardless of the applied voltage. Examples of this type of load include motor loads and some types of lighting which utilize an inductive ballast to establish constant wattage to the lamp.

It is clear that with constant kVA type loads, the actual load currents increase with decreasing voltage. With constant impedance loads, line currents decrease as the voltage is lowered. If both kVA loads and constant impedance loads are present, then the resulting voltage effects may be partially or totally canceled.

Constant current type loads hold their current constant under varying voltage conditions. Like constant impedance type loads, as the voltage drops at the bus the amount of apparent power consumed by the constant current type load decreases. Constant current type loads are affected by the fluctuations in the bus voltage angle.

3.2.5. Voltage Drop Calculations National Codes, such as the NEC in the United States, limit the total voltage drop in any one branch or the total bus voltage drop. Thus, it is critical in the design process to know the voltage drop in each branch of the power system, and the total voltage drop from the source of supply to the bus in the branch circuit. The voltage drop calculations are incorporated directly into the calculation of the steady-state load flows.

More simply, adding the receiving end voltage at each load bus and the branch voltage drop is equal to the sending end voltage, as illustrated in the following figure.


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IR

IX

IZSending End Voltage

Receiving End Voltage

Load Power Factor Angle

Load Current

Reported Voltage Drop

Actual Voltage Drop

Sending End Bus Receiving End Bus

R + jXI

∅

The sending end voltage may be expressed as:

V V I Rs r= + + jXb g Eq. 3-4

where

Vs sending bus voltage; Vr receiving bus voltage; I load current; R feeder resistance; jX feeder reactance.

PTW reports the magnitude difference between the receiving and sending end voltages as the voltage drop, expressed on a three-phase (line-to-line) basis.

VD E Es r= −

3.3. PTW Applied Methodology PTW Applied Methodology discusses how the Load Flow Study applies the methodology described in the preceding section.

3.3.1. Before Running the Load Flow Study Before running the Load Flow Study, you must:

• Define power system topology and connections.

• Define utility connection (swing bus).




• Define feeder and transformer sizes.

• Define generator sizes.

• Run the Demand Load Study (Optional). The branch load data required for Load Flow Study may either be entered, or calculated using the Demand Load Study.

Tip: You may run the Demand Load Study in the same step that you run the Load Flow Study. All you need to do is select both Studies, and PTW will automatically run the Demand Load Study first. See Section 3.3.2, “Running the Load Flow Study,” following.

3.3.2. Running the Load Flow Study You can run the Study from any screen in PTW, and it always runs on the active project.

To run the Load Flow Study:

1. From the Run menu, choose Balanced System Studies.

2. Select the check box next to Load Flow Study. You may also select the check box next to Demand Load to run the Demand Load Study prior to the Load Flow Study.



The Load Flow Study runs, writes the results to the database, and creates a report.

3.3.3. Load Flow Study Options When you run the Load Flow Study, PTW displays the Load Flow Study subview in the Studies box, as shown:

You can change any or all of the available Study Options to suit your needs. Following is a list of all of the controls available in this box, and their functions.


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System Modeling If you select the Utility Impedance check box, PTW uses the three-phase short circuit capacity to calculate an equivalent positive sequence impedance. The voltage drop at the swing bus is calculated, given the total power supplied by the swing bus generator and this positive sequence impedance. It is reported separately in the Load Flow Report. Upon opening a new Project, the PTW does not model the system equivalent impedance by default. This means the voltage at the swing bus is equal to the voltage of the swing bus generator, which is set by default to 1 pu voltage at 0°.

If you uncheck the Transformer Phase Shift checkbox, PTW reports the voltage angles relative to the swing bus voltage angle. If you check this option, PTW models the voltage angle phase shift of the transformer connections.

Solution Method PTW models either an Exact (Iterative) or Approximate Solution. Upon creating a new Project, PTW selects the Exact (Iterative) Solution method by default. It is recommended that you run the Study using the Exact (Iterative) Solution method first. This is because the solution method usually converges on most power systems. In the unlikely event that the steady-state load flow solution does not converge, you should re-run the Study using the Approximate Solution method. If it does not converge, a message in the Study Message dialog box will notify you of the problem.

When the Approximate Solution method is selected, PTW temporarily converts all loads to constant impedance type characteristics, making these system losses smaller than if constant kVA type loads were modeled. An output report is then written, and data is sent to the database. Although it is an approximate solution (since the load characteristic is approximated), this solution method may help to identify the reasons for the non-convergence.

If you have a non-convergent solution, examine the output Report’s bus voltage mismatch values and bus mismatch location, as there may be a data input problem that has caused the non-convergence. If this does not help, try adjusting the solution criteria; see “Solution Criteria,” following.

Load Specification You may select one of five options for load modeling. These options are divided into two groups: Directly Connected Loads, and loads From the Demand Load Study. The default upon opening a new Project is to model all loads as Directly Connected loads using the Connected Load values. For a more detailed discussion of the demand and design load values, see the “Engineering Methodology” section of the “Demand Load Study” chapter.

Directly Connected Loads

Directly Connected Loads can be modeled as Connected Load or 1st Level Demand or Energy Factor. When either of these two options is selected, the load flow solution calculates the load at each bus, then solves the steady state load flow equation (Eq. 3-3). Neither of these options uses results from the Demand Load Study.

When the Connected Load option button is selected, the Load Flow Study calculates the loads without considering any load or demand factors. If motor loads are identified, and if multiple motors are modeled in a single motor load object, the total motor connected load is the number of motors multiplied by the motor’s rated size. Otherwise, the load rated size is the connected load value. Motors expressed in horsepower are converted to electrical units by dividing by the efficiency.



When the 1st Level Demand or Energy Factor option button is selected, the Load Flow Study calculates the loads using the first level demand factors and energy audit load factors, as appropriate. If a non-motor load is identified with both an energy audit load factor and a demand load category, then the Study will use both the energy audit load factor and the first level demand load factor multiplied by the load’s rated size. For motor loads, the load is calculated as the number of motors multiplied by the motor rated size multiplied by the motor load factor. Load diversity resulting from identifying multiple levels of demand load factors is not taken into consideration.

From Demand Load Study

Loads selected as From Demand Load Study can be modeled as Demand Load, Design Load (All Loads in Constant kVA), or Connected Load. The Load Flow Study models the loads based on the results of the last Demand Load Study run. If you have never run a Demand Load Study, the Load Flow Study will return an error message in the Run Study dialog stating that no load has been defined.

When the Demand Load option button is selected, the Load Flow Study uses the calculated demand load values from the Demand Load Study. Upon creating a new Project, the Demand Load Study includes the results from all non-motor loads by default, unless you selected either the Include Only Demand Loads or Include Only Energy Audit Loads option button before running the Demand Load Study. Demand loads with multiple levels of demand factors, such as receptacle loads where the first 10 kVA of load have a 100% demand factor and the remaining receptacle load has a 50% demand factor result in a non-coincident demand (diversity) load that is unique. This diversity load is calculated at each branch within the load flow solution. See Example 3.4.4, “Net Branch Diversity Load” for further discussion.

Remember that when modeling loads using any of the three From Demand Load Study options, the results are based on the results from the last Demand Load Study. If you change the system topology you need to first re-run the Demand Load Study. You may run the Demand Load Study and the Load Flow Study in one step, as described in “Running the Load Flow Study” on page 3-7

Solution Criteria The two Acceleration Factor text boxes allow you to control how the Load Flow Study converges upon the solution. Generally, the Acceleration Factors do not need to be changed from their default values. However, if a non-convergent solution occurs, even after an Approximate Solution method has been run, try changing the Generation Acceleration Factor and/or the Load Acceleration Factor from their default of 1.0 to a factor between 0.1 and 1.0. This changes the guessing factor used to calculate the next iteration of the numerical solution. The smaller the factor, the smaller the step change used in the iteration solution.

The Bus Voltage Drop and Branch Voltage Drop text boxes provide a quick method to flag excessive voltage drops in the output report. In the report, PTW flags with a dollar sign ($) any bus or branch voltage drop value that exceeds the limits set in these text boxes. Upon creating a new Project, the default values are a 5% bus voltage drop, and a 3% branch voltage drop. However, you can change these percentages by typing a percent value in the appropriate text box.

3.3.4. Component Modeling The following sections describe important data required for the Load Flow Study to run.


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Swing Bus Generator As stated in Section 3.2, “Engineering Methodology,” every load flow solution requires at least one swing bus be identified. Up to ten swing buses may be modeled. At the swing bus, you must specify the voltage magnitude and voltage angle. You may model the utility equivalent impedance of the system “behind” the swing bus by selecting the Utility Impedance check box in the Load Flow Study setup dialog box. Even if you want to model a system with local generation (e.g., the utility connection is severed), you must identify one of the co-generators as the swing bus generator.

Loads may be directly connected to the co-generation bus and the utility swing bus. PTW accurately models the power flow in this situation even when directly connected loads are installed on Type II or Type III load buses.

On-Site Generation An on-site generator that operates in parallel with the utility source may be defined as either a PV- or PQ-type generator in the Component Editor. The generator may have leading or lagging reactive power. Lagging reactive kVA is plotted as a positive value and referred to as an over-excited condition of the machine. Leading reactive kVA is plotted as a negative value and referred to as an under-excited condition of the generator. Any type of balanced loading within the area bounded by the curve shown following is considered as safe by the generator manufacturer.

PF Leading

0.5 .075 1.000.25

1.0

.50

0.0

-.50

-1.0

Lagging

Leading

Kilowatts Per Unit

Rea

ctiv

e Po

wer

-- Pe

r Uni

t

Rated PF Lagging

A combined total of up to 10 Swing Bus (SB) and PV-type generators may be modeled in PTW. A total of up to 60 generators may be modeled, of which no more than 10 may be a combination of PV- and SB-type generators. When modeling SB generators, they may be modeled as generator components or as utility components.

The source connection represents the system swing bus or slack bus. It maintains a constant voltage at a constant bus angle. The kW and kvar of the swing bus vary, or “swing,” in order to satisfy the following equation:

Generation loads losses 0− − =



The swing bus generator or utility source must supply any deficiency in generation, and all losses. If there is excess generation in the system, the source bus acts to absorb the excess generation.

Usually, a system requires only a single swing bus. Multiple swing buses may be used to model large systems having multiple connections to a utility. Therefore, the exact utility voltages and angles may be specified. Additionally, multiple independent systems require that you specify a swing bus for each system. Only one swing bus may be specified at a single bus.

In modeling generation in parallel operation with a utility connection or swing bus generator, the swing bus has unlimited power capabilities to maintain its voltage magnitude and voltage angle. Therefore, regulating-type generators (PV-type) located at the swing bus or near the utility will not produce any reactive power. In this case, model co-generation as PQ-type generators.

Co-generation modeled as PV-type generators which are not located near the utility or swing bus generator will serve to regulate the voltage at the generator bus. PTW attempts to hold the bus voltage at the target level within the reactive power range. It may not be possible to obtain the target voltage. PTW will maintain the kW output of the generator and allow the voltage to float to the reactive power limit.

PTW allows for modeling more than one generator at a bus. If the generators are PV-type generators, then you need to identify how much reactive power is generated by each machine (pu Var Participation factor). Usually the reactive power is shared equally between the machines; therefore, if two PV machines are modeled on the same bus, then the participation factor is 0.5 for each machine. When multiple PV generators are modeled on individual buses separated by small cables, you may get a non-convergent solution because the reactive power calculated by PTW depends on very small bus voltage angles. If this occurs, model the multiple PV machines on a single bus, or lump the generation into a single generator model.

Diversity Loads Special attention needs to be paid when running the Load Flow Study when demand loads are considered. As discussed in the “Non-Coincident Demand Calculation” example of the “Demand Load Study” chapter, PTW considers the effects of non-coincident or diversity loads. In the example, even though each of the two branch demand loads are 550 kVA, the total demand load at Bus 2 is only 1050 kVA. The load flow solution must take this diversity load into consideration. In effect, the feeder servicing these two branches only needs to supply 1050 kVA, but each branch needs to be supplied 550 kVA. The load flow solution will report a small net negative diversity load at this upstream bus. See Example 3.4.4, “Net Branch Diversity Load”at the end of this chapter.

PTW accurately models the loads when loops are present in the power system. As was discussed in the Demand Load Study chapter, the Demand Load Study methodology must identify and calculate load diversity only in a radial system. The Demand Load Study detects loops, opens them temporarily to calculate load diversity, then closes the loops. In some cases, where extensive (nested) loops exist, the automatic opening of the loops can result in odd load flow solutions. It is recommended that if multiple loops exist in the power systems and if load diversity is not a significant issue for the Study, then the load flow solution be modeled using the Directly Connected Loads Load Specification option. Remember, this option allows for modeling the Connected Load or 1st Level Demand Load factor and Energy Factor.


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3.3.5. Error Messages If the Load Flow Study detects data errors during input data processing, error messages are provided for the errors found and the Study does not run. These errors must be corrected and the Study re-run. The most common errors are as follows:

1. Diverging Solution: If the Exact (Iterative) Solution method has been selected, and under certain extreme loading conditions, it is possible that the Load Flow Study cannot determine a steady-state load flow condition and no report will be generated. An error will be reported in the Study Messages dialog box. This can occur in power systems whenever the system voltages are not sufficient to supply constant kVA loads.

2. Load Not Defined: PTW will report a warning message if loads are defined with zero load. This is just a warning message and is provided to give you a listing of the zero load values. The Study will run, but will ignore the zero load values.

3. Load Category Not Defined: PTW will report a demand load where its demand load category is not defined. In this case, a 100% Demand Factor will be used as the default.

4. Incomplete Configuration. PTW will detect components whose nodes (connection points) are not physically connected. PTW must have all components connected.

3.4. Application Examples The following examples demonstrate how the methodology is applied to various system configurations. Each example changes the system, building upon the previous example, to show how the Load Flow Study meets variable conditions.

3.4.1. Voltage Drop and Power Losses The purpose of this example is to examine the voltage drop and power flow through a simple one-branch, single-cable circuit. The one-line is shown below. The generator is rated 100 MVA at 13.8 kV, with a subtransient impedance of 0.1+j1 pu Ω on its own base. The cable is rated at 15 kV and has an impedance of 0.1904 + j1.9044 Ω.



B113800 V13800.00 V0.00 %

C11886.52 kW1467.74 kvar5.71 kW Losses57.13 kvar Losses

100.00 A

B213800 V13573.55 V1.64 %

GENERATOR

L1

1.64 % VD

The load draws 100 A at 80% power factor and is assumed to have a constant current-type load characteristic.


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The three-phase real power loss in the cable is:

P I R 3

= 100 0.19044 35.71 kW

Loss2

2

= ×

×

=b g b g

The magnetization power required by the cable is:

Q I X 3

100 1.9044 357.1 kvar

Loss2

2

= ×

= ×

=b g b g

In a simplified form, the voltage drop through the cable is:

VD IRcos + IXsin 3

100 0.19044 100 1.9044 3

80 0.19044 60 1.9044 3

129.5 3224.3 V

L L− = ×

= + ×

= + ×

= ×

=

Θ Θb gb gb gb g b gb gb gb gb g b gb gb g

0 8 0 6. .

And on a per unit base, this is:

VD 224.3 V13800 V0.0163 pu V

pu =

=

Or an operating voltage of:

V 1 0.0163 pu

= 0.984 pupu = −

The three-phase power consumed by the load at operating voltage is then:

S 3 V I V pu operating

3 100 A 13.8 kV 0.9842351.4 kVA

Load LL L=

=

=

e jb gb g b gb gb gb g

With an 80% power factor:

S = 1881.10 kW + 1410.8 kvarLoad j

At the sending end, the power injected into cable is:

S S S

1881.10 5.7 kW 1410.8 57.1 kvar

1886.81 1467.9 kVA

Sending Load Losses= +

= + + +

= +

b g b gb g

j j

j

The datablocks posted on the one-line diagram and the hand calculations are closely matched. The difference between hand and computer solutions is due to round-off and the



use of the simplified form of the voltage drop formula, as noted in Chapter 3 of the IEEE Red Book. The Report confirms these hand calculations, as noted below.

BALANCED VOLTAGE DROP AND LOAD FLOW ANALYSIS (SWING GENERATORS) ***************************************************************************** SOURCE VOLTAGE ANGLE KW KVAR VD% (UTILITY IMPEDANCE) GEN-0001 1.000 .00 1886.52 1467.74 Gen Z Ignored BALANCED VOLTAGE DROP AND LOAD FLOW ANALYSIS ==== BUS: BUS-0001 DESIGN VOLTS: 13800 BUS VOLTS: 13800 %VD: .00 ========================= PU BUS VOLTAGE: 1.000 ANGLE: .0 DEGREES *** SWING GENERATOR: GEN-0001 1886.5 KW 1467.7 KVAR LOAD TO: BUS-0002 FEEDER AMPS: 99.9 VOLTAGE DROP: 226. %VD: 1.64 PROJECTED POWER FLOW: 1886.5 KW 1467.7 KVAR 2390.2 KVA PF: .79 LAGGING LOSSES THRU FEEDER: 5.7 KW 57.1 KVAR 57.4 KVA ==== BUS: BUS-0002 DESIGN VOLTS: 13800 BUS VOLTS: 13574 %VD: 1.64 ========================= PU BUS VOLTAGE: .984 ANGLE: -1.0 DEGREES NET BRANCH DIVERSITY LOAD: 1880.8 KW 1410.6 KVAR LOAD FROM: BUS-0001 FEEDER AMPS: 99.9 VOLTAGE DROP: 226. %VD: 1.64 PROJECTED POWER FLOW: 1880.8 KW 1410.6 KVAR 2351.0 KVA PF: .80 LAGGING LOSSES THRU FEEDER: 5.7 KW 57.1 KVAR 57.4 KVA *** T O T A L S Y S T E M L O S S E S *** 6. KW 57. KVAR

3.4.2. Modeling Transformer Losses This example demonstrates that the losses through a cable and a transformer are identical if the load and branch impedance are identical. In the example, the non-motor load is changed to a constant kVA type load, rated at 1 MVA at 80% power factor. A 13.8 kV transformer is installed in the circuit with an impedance equal to the cable on a per unit basis. The transformer has a 1:1 voltage ratio.

B113800 V13800.00 V0.00 % C1 795.54 kW

605.90 kvar1.00 kW Losses10.00 kvar Losses0.68 %41.84 A

B213800 V13705.78 V0.68 %

GENERATOR

L11.0 MVA0.80 Lag PF

T1 795.54 kW605.90 kvar1.00 kW Losses10.00 kvar Losse0.68 %41.84 A

B313800 V13705.78 V0.68 %

L21.0 MVA0.80 Lag PF

The following input data shows that on a per unit basis, the impedance of the cable equals the impedance of the transformer.


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FEEDER INPUT DATA ============================================================================== FEEDER FROM FEEDER TO QTY VOLTS LENGTH FEEDER DESCRIPTION NAME NAME /PH L-L SIZE TYPE DUCT INSUL ============================================================================== B1 B2 1 13800 1000.00 FT 1 C N +/- Impedance: 0.1904 + J 1.90 OHMS/M Length 0.1000 + J 1.0000 PU Z0 Impedance: 0.1904 + J 1.90 OHMS/M Length 0.1000 + J 1.0000 PU TRANSFORMER INPUT DATA ============================================================================== PRIMARY RECORD VOLTS * SECONDARY RECORD VOLTS FULL-LOAD NOMINAL NO NAME L-L NO NAME L-L KVA KVA ============================================================================== B1 D 13800.0 B3 YG 13800.0 0.00000 100000. Pos. Seq. Z%: 10.00 + J 100.0 0.100 + j 1.000 PU Zero Seq. Z%: 10.00 + J 100.0 0.100 + j 1.000 PU Taps Pri. 0.000 % Sec. 0.000 % Phase Shift (Pri. Leading Sec.): 30.00 Deg.

The Load Flow Report shows the power flow in both the cable and the transformer branch are identical. Losses in these two components are the same. The transformer was modeled at a 1:1 voltage ratio.

BALANCED VOLTAGE DROP AND LOAD FLOW ANALYSIS ***************************************************************************** VOLTAGE EFFECT ON LOADS MODELED ==== BUS: B1 DESIGN VOLTS: 13800 BUS VOLTS: 13800 %VD: .00 ========================= PU BUS VOLTAGE: 1.000 ANGLE: .0 DEGREES *** SWING GENERATOR: GENERATOR 1591.1 KW 1211.8 KVAR LOAD TO: B2 FEEDER AMPS: 41.8 VOLTAGE DROP: 94. %VD: .68 PROJECTED POWER FLOW: 795.5 KW 605.9 KVAR 1000.0 KVA PF: .80 LAGGING LOSSES THRU FEEDER: 1.0 KW 10.0 KVAR 10.0 KVA LOAD TO: B3 TRANSF AMPS: 41.8 VOLTAGE DROP: 94. %VD: .68 PROJECTED POWER FLOW: 795.5 KW 605.9 KVAR 1000.0 KVA PF: .80 LAGGING LOSSES THRU TRANSF: 1.0 KW 10.0 KVAR 10.0 KVA ==== BUS: B2 DESIGN VOLTS: 13800 BUS VOLTS: 13706 %VD: .68 ========================= PU BUS VOLTAGE: .993 ANGLE: -.4 DEGREES NET BRANCH DIVERSITY LOAD: 794.5 KW 595.9 KVAR LOAD FROM: B1 FEEDER AMPS: 41.8 VOLTAGE DROP: 94. %VD: .68 PROJECTED POWER FLOW: 794.5 KW 595.9 KVAR 993.2 KVA PF: .80 LAGGING LOSSES THRU FEEDER: 1.0 KW 10.0 KVAR 10.0 KVA ==== BUS: B3 DESIGN VOLTS: 13800 BUS VOLTS: 13706 %VD: .68 ========================= PU BUS VOLTAGE: .993 ANGLE: -.4 DEGREES NET BRANCH DIVERSITY LOAD: 794.5 KW 595.9 KVAR LOAD FROM: B1 TRANSF AMPS: 41.8 VOLTAGE DROP: 94. %VD: .68 PROJECTED POWER FLOW: 795.5 KW 605.9 KVAR 1000.0 KVA PF: .80 LAGGING LOSSES THRU TRANSF: 1.0 KW 10.0 KVAR 10.0 KVA

The losses through the cable and the transformer are equal, as they should be.

3.4.3. Load Specifications This example demonstrates the impact of the different load specifications on the result of the load flow solution.

You can specify the individual load type in the Load Diversity subview of the Component Editor by selecting either the Demand Load or Energy Audit option button. The Load Flow Study results are affected by which load type you select. The Load Flow Study results also depend on the Load Specification selection in the Load Flow Study setup dialog box: one of the two Directly Connected Loads, or one of the three From the Demand Load Study loads. These last three options are more global, as they allow you to model loads based on results from the Demand Load Study.

In the one-line diagram below, five different branch load types and load factors are modeled. The results of the Load Flow Study, depending on the individual load type and Load Specification selected, are shown on the one line diagram.



B113800.00 V0.00 % VD

C1

B213705.13 V0.69 % VD

GENERATOR

3115.26 kW2366.59 kvar0.80 PF

L11000.0 kVAEnergy Audit

C3 C4C2

B413705.13 V0.69 % VD

B513733.81 V0.48 % VD

B313733.81 V0.48 % VD

L31000.0 kVADemand Load

L41000.0 kVADemand Load

L21000.0 kVAEnergy Audit

General Load Special Load

C5

B613753.77 V0.33 % VD

L51000.0 kVADemand Load Special Load

42.1 A 29.4 A 42.1 A 29.4 A 20.57 A

Load Factor0.7 Energy Audit Load Factor 0.7 Energy Audit

Load Factor

1.0 Energy Audit

Each of the five loads is rated 1000 kVA; thus each load’s connected load value is 1000 kVA.

Loads L1 and L2 are Energy Audit type loads, whereas Loads L3, L4, and L5 are Demand Load types. Load L1 has an Energy Audit load factor of 1.0, while Loads L2 and L5 have an Energy Audit load factor of 0.7. Note that while Load L5 is a Demand Load type, it still has an Energy Audit load factor specified. Load L3 is selected as a General Load, while Loads L4 and L5 are selected as Special Loads; these selections were made from the Demand Load Category drop-down list box found in the Load Diversity subview of the Component Editor.

The demand load factors (which are referenced from the Demand Load Library) used in this example are:

Load Type 1st Level 2nd Level LCL Load Factor Load Factor General Loads 100 kVA 100% All 50% 1.25

Special Loads 100 kVA 70% All 100% 1.00

See the “Engineering Methodology” section of the “Demand Load Study” chapter for additional discussion on design and demand loads and long continuous loading (LCL) factors.

It is important also to note that the Load Flow Study takes into account the effects of voltage on loads. All of the loads in this example are modeled as constant kVA type loads; thus, as the voltage at the load bus decreases, the current in the branch servicing the load increases. Calculating the load current is not easily accomplished by hand methods because the solution cannot be solved in closed form.

Component Editor Data Load Flow Study Setup Specifications Directly Connected From Demand Load Study Load # Individual

Load Type Individual Load Factor

Connected Load (Amps)

1st Level Demand or Energy Factor

Demand Load (Amps)

Design Load (Amps)

Connected Load (Amps)


11/22/2006

Component Editor Data Load Flow Study Setup Specifications Directly Connected From Demand Load Study

Factor (Amps)

L1 Energy Audit

1.0 42.13 42.1 42.13 42.13 42.13

L2 Energy Audit

0.7 42.13 29.4 29.4 29.4 42.13

L3 Demand General 42.13 42.1 23.1 28.9 42.13

L4 Demand Special 42.13 29.4 40.85 40.85 42.13

L5 Demand Special with Energy Audit Load Factor of 0.7

42.13 20.57 40.85 40.85 42.13

Remember that all five loads in this example have a rated size of 1000 kVA, or 42.13 A. The calculated current in each branch is different , based on the Individual Load Type and the Individual Load Factor.

The calculated load flow branch current using the 1st Level Demand or Energy Factor for Load L1 is equal to the connected load because the load factor is 1.0, whereas the current in the branch servicing Load L2 is 0.7 of the connected load. Note that the only difference in load characteristics between Loads L4 and L5 is that, although both are modeled as demand loads, Load L4 has a default energy audit load factor of 1.0 while Load L5 has a energy audit load factor of 0.7. Therefore, the calculated load flow branch current using the 1st Level Demand or Energy Factor is 29.4 A for Load L4 but only 20.57 A for Load L5.

The calculated load flow branch current using the results From Demand Load Study should be further discussed. For example, the load flow current using the Demand Loads for both Loads L4 and L5 is equal to 40.85 A, the Energy Audit load factor of 0.7 is not used.

The data shown in the above one-line matches the load specification 1st Level Demand or Energy Factor Study setup.

3.4.4. Net Branch Diversity Load This example demonstrates how PTW calculates diversity load, and the diversity load’s impact on the load flow solution.

This example is the load flow solution for the “Non-Coicident Demand Calculation” example presented in the “Demand Load Study” chapter. Two identical demand loads of 1000 kVA (General Load Category) are connected in a system shown in the one-line following:



B1

13800.00 V0.00 % VD C1

840.03 kW630.04 kvar

B2

13799.57 V0.00 % VD

GENERATOR

840.03 kW630.04 kvar0.80 PF

L11000.0 kVADemand Load Type

C2440.00 kW330.01 kvar

B313799.38 V0.00 % VD

L21000.0 kVADemand Load Type

C3440.01 kW330.01 kvar

B4

13799.28 V0.01 % VD

From the “Non-Coicident Demand Calculation” example, note that the connected load at Bus B2 is 2000 kVA, but the demand load value is 1050 kVA. This is a diversity load value, as the sum of the two branch (coincident) demand loads is 1100 kVA (550 kVA each). It can then be concluded that only 1050 kVA of apparent power must service Bus B2, but 550 kVA must flow to each of the two loads.

Inspection of the real component of apparent power at Bus B2 is instructive. From the one-line diagram, 840 kW flows into the bus, but 440 kW flows out. A net branch diversity of -40 kW exists.

This is most clearly displayed in the Load Flow Study report, as shown following:

BALANCED VOLTAGE DROP AND LOAD FLOW ANALYSIS ***************************************************************************** VOLTAGE EFFECT ON LOADS MODELED ==== BUS: B2 DESIGN VOLTS: 13800 BUS VOLTS: 13800 %VD: 0.00 ========================= PU BUS VOLTAGE: 1.000 ANGLE: 0.0 DEGREES NET BRANCH DIVERSITY LOAD: -40.0 KW -30.0 KVAR LOAD FROM: B1 FEEDER AMPS: 43.9 VOLTAGE DROP: 0. %VD: 0.00 PROJECTED POWER FLOW: 840.0 KW 630.0 KVAR 1050.0 KVA PF: 0.80 LAGGING LOSSES THRU FEEDER: 0.0 KW 0.0 KVAR 0.0 KVA LOAD TO: B3 FEEDER AMPS: 23.0 VOLTAGE DROP: 0. %VD: 0.00 PROJECTED POWER FLOW: 440.0 KW 330.0 KVAR 550.0 KVA PF: 0.80 LAGGING LOSSES THRU FEEDER: 0.0 KW 0.0 KVAR 0.0 KVA LOAD TO: B4 FEEDER AMPS: 23.0 VOLTAGE DROP: 0. %VD: 0.00 PROJECTED POWER FLOW: 440.0 KW 330.0 KVAR 550.0 KVA PF: 0.80 LAGGING LOSSES THRU FEEDER: 0.0 KW 0.0 KVAR 0.0 KVA

At Bus B2, PTW reports the net difference in power flowing into the node (1050 kVA) and the power flowing out of the node (1100 kVA). This net difference acts to raise the system voltages, compared to the case where diversity is not considered. For example, in the above problem, change the load type by selecting the Energy Audit option button instead of Demand Load, and change the load factor to 0.55. The demand load at Bus 2 will now be 1100 kVA.


11/22/2006




EDIS

ON

UTIL

ITY

BUS

TX A

PRI

TX A

003-

HV S

WG

R

C1

C2

C3

C4

SYN

ASY

N B

TX B

PRI

BUS

4

005-

TXD

PRI

006-

TX3

PRI

007-

TX E

PRI

TX B

DS S

WG

1 BU

S 8

C5

C6

GEN

1G

EN 2

TX C

PRI

BUS

9H

VAC

BUS

TX C

DSB

1 B

US 1

4

C14

C16

C17

BUS

15BU

S 16

BUS

17

C15

018-

RA

CM

P HV

AC

TX 6

TX3

SEC

BUS

11TX

3 TE

R BU

S 12

C8

DS

SWG

3 BU

S 20

C9

L1M

20

GEN

3

021-

TX F

PRI

C7

DS

SWG

2 BU

S 13

L2M

13

TX F

022-

DSB

2

C12

023-

MTR

23

M23

-BM

23-A

TX E

BLDG

115

SER

V

C10

C11

026-

TX G

PR

I02

5-M

TR 2

5

TX G

DSB

3 B

US 2

7

C13

L5

BUS

28 A

M25

TX D

029-

TX D

SEC

C19

TM -1

SWBD

-1

PLN

- 17

H1A

PLN

16

H2A

MCC

15

- 1A

PNL

18 R

A

C18

H3A

BU

S 19

PNL

19 H

3A

R1 CB1

R2 CB2

CB6

CB7

R8

CB8

R9

CB9

R7R6

CB5

R5

R4 CB4

R3 CB3

R10

CB1

0

F1

F2 MC

P#10

LVP1

LVP2

LVP3

MCC

B1M

CCB2

PCB1

PCB2

R11

PCB3

F4

F3 MO

/L#2

5

SW1

LVP4

DEM

ONS

TRAT

ION

PRO

JEC

T FO

R PO

WER

*TO

OLS

FO

R W

INDO

WS

CAP

1

M28

#1

& 2

C20 LV

P5

MCP

#28B

-1

MO

/L#2

8B-1

M28

#3

BUS

28 B

MCP

#28B

-2

MO

/L#2

8B-2

M28

#4

CB12

TX H

CMP

CTR

COM

PUTE

RS

F5

LVP6

OFF

ICE

CO

MPL

EX A

REA

IND

USTR

IAL

COM

PLEX

ARE

A


11/22/2006

The following figure shows a portion of the Plant project, including Load Flow results.

The Load Flow and Voltage Drop Report first lists the Study setup conditions and constraints. Also included is a listing of the convergence criterion and the number of iterations required for the load flow solution to converge upon an acceptable solution.

*** SOLUTION COMMENTS *** ========================= SOLUTION PARAMETERS BRANCH VOLTAGE CRITERIA : 3.00 % BUS VOLTAGE CRITERIA : 5.00 % ACCELERATION FACTOR FOR 'PV' GENERATORS : 1.00 ACCELERATION FACTOR FOR CONSTANT KVA LOADS: 1.00 EXACT(ITERATIVE) SOLUTION : YES UTILITY IMPEDANCE : YES TRANSFORMER PHASE SHIFT : NO ALL PU VALUES ARE EXPRESSED ON A 100 MVA BASE LOAD FLOW IS BASED ON CALCULATED DEMAND LOAD RESULTS FROM THE DEMAND LOAD ANALYSIS STUDY. LOAD ANALYSIS INCLUDES ALL LOADS. <<PERCENT VOLTAGE DROPS ARE BASED ON NOMINAL DESIGN VOLTAGES>> SWING GENERATORS SOURCE NAME VOLTAGE ANGLE



================================ EDISON 1.000 0.00 PV GENERATORS SOURCE NAME VOLTAGE kW kVARMIN kVARMAX PARTICIPATION ================================================================== GEN 2 1.000 350. 0. 300. 0.25 GEN 1 1.000 600. -600. 500. 0.75 GEN 3 1.000 2400. -900. 1600. 1.00 LARGEST LOAD: 4000.00 KVA CONVERGENCE CRITERIA: 0.200 KVA LARGEST BUS MISMATCH DS SWG3 BUS 20 1517.460 KVA LARGEST BUS MISMATCH DS SWG1 BUS 8 120.509 KVA LARGEST BUS MISMATCH DS SWG1 BUS 8 24.610 KVA LARGEST BUS MISMATCH DS SWG2 BUS 13 1.128 KVA LARGEST BUS MISMATCH DS SWG1 BUS 8 1.489 KVA LARGEST BUS MISMATCH DS SWG1 BUS 8 0.199 KVA

Next the report lists the system swing bus generation and equivalent voltage drop of the swing bus generator(s) based on the the system utility fault duty contribution as you defined it.

BALANCED VOLTAGE DROP AND LOAD FLOW ANALYSIS (SWING GENERATORS) ***************************************************************************** SOURCE VOLTAGE ANGLE KW KVAR VD% (UTILITY IMPEDANCE) EDISON 1.000 0.00 10190.11 4696.85 0.55 0.00370+J 0.11105

Once the swing bus generation is known, the generation of the various PV machines can be listed:

BALANCED VOLTAGE DROP AND LOAD FLOW ANALYSIS (PV GENERATOR SCHEDULE REPORT) ***************************************************************************** ---VOLTAGE--- -KVAR LIMITS- ---ACTUAL---- PV SOURCE NAME SCHED. ACTUAL MIN MAX KW KVAR GEN 2 1.000 1.000 0. 300. 350. 98. GEN 1 1.000 1.000 -600. 500. 600. 295. GEN 3 1.000 0.996 -900. 1600. 2400. 1600.

The basic data presented in the Load Flow and Voltage Drop Report is listed below for key buses and branches associated with the Building 115 Service:

BALANCED VOLTAGE DROP AND LOAD FLOW ANALYSIS ***************************************************************************** VOLTAGE EFFECT ON LOADS MODELED VOLTAGE DROP CRITERIA: BRANCH = 3.00 % BUS = 5.00 ==== BUS: UTILITY BUS DESIGN VOLTS: 69000 BUS VOLTS: 68618 %VD: 0.55 ========================= PU BUS VOLTAGE: 0.994 ANGLE: -0.6 DEGREES *** SWING GENERATOR: EDISON 10190.1 KW 4696.9 KVAR LOAD TO: TX A PRI FEEDER AMPS: 93.8 VOLTAGE DROP: 1414. %VD: 2.05 PROJECTED POWER FLOW: 10185.5 KW 4557.0 KVAR 11158.4 KVA PF: 0.91 LAGGING LOSSES THRU FEEDER: 7.9 KW -2.9 KVAR 8.4 KVA LOAD FROM: EDISON FEEDER AMPS: 93.8 VOLTAGE DROP: 382. %VD: 0.55 PROJECTED POWER FLOW: 10185.5 KW 4557.0 KVAR 11158.4 KVA PF: 0.91 LAGGING LOSSES THRU FEEDER: 4.7 KW 139.8 KVAR 139.9 KVA ==== BUS: BLDG 115 SERV DESIGN VOLTS: 4160 BUS VOLTS: 4071 %VD: 2.13 ========================= PU BUS VOLTAGE: 0.979 ANGLE: -6.1 DEGREES LOAD FROM: 007-TX E PRI TRANSF AMPS: 502.1 VOLTAGE DROP: 46. %VD: 1.10 PROJECTED POWER FLOW: 2866.5 KW 2296.5 KVAR 3673.0 KVA PF: 0.78 LAGGING LOSSES THRU TRANSF: 26.2 KW 181.4 KVAR 183.3 KVA LOAD TO: 026-TX G PRI FEEDER AMPS: 146.3 VOLTAGE DROP: 12. %VD: 0.28 PROJECTED POWER FLOW: 834.9 KW 607.1 KVAR 1032.3 KVA PF: 0.81 LAGGING LOSSES THRU FEEDER: 2.9 KW 1.1 KVAR 3.1 KVA LOAD TO: 025-MTR 25 FEEDER AMPS: 355.8 VOLTAGE DROP: 15. %VD: 0.36 PROJECTED POWER FLOW: 2005.4 KW 1508.0 KVAR 2509.1 KVA PF: 0.80 LAGGING LOSSES THRU FEEDER: 5.4 KW 8.0 KVAR 9.6 KVA LOAD FROM: 029-TX D SEC FEEDER AMPS: VOLTAGE DROP: 0. %VD: 0.00 PROJECTED POWER FLOW: 0.0 KW 0.0 KVAR 0.0 KVA PF: 0.00 LAGGING LOSSES THRU FEEDER: 0.0 KW 0.0 KVAR 0.0 KVA ==== BUS: DSB 3 BUS 27 DESIGN VOLTS: 480 BUS VOLTS: 460 %VD: 4.09


11/22/2006

========================= PU BUS VOLTAGE: 0.959 ANGLE: -8.5 DEGREES NET BRANCH DIVERSITY LOAD: 92.0 KW 0.0 KVAR LOAD FROM: 026-TX G PRI TRANSF AMPS:1237.0 VOLTAGE DROP: 8. %VD: 1.67 PROJECTED POWER FLOW: 832.1 KW 605.9 KVAR 1029.3 KVA PF: 0.81 LAGGING LOSSES THRU TRANSF: 10.6 KW 59.9 KVAR 60.8 KVA LOAD TO: BUS 28 A FEEDER AMPS: 568.4 VOLTAGE DROP: 3. %VD: 0.70 PROJECTED POWER FLOW: 362.0 KW 272.9 KVAR 453.3 KVA PF: 0.80 LAGGING LOSSES THRU FEEDER: 2.0 KW 2.9 KVAR 3.5 KVA LOAD TO: BUS 28 B FEEDER AMPS: 574.2 VOLTAGE DROP: 8. %VD: 1.65 PROJECTED POWER FLOW: 367.5 KW 273.2 KVAR 457.9 KVA PF: 0.80 LAGGING LOSSES THRU FEEDER: 7.5 KW 3.2 KVAR 8.1 KVA ==== BUS: BUS 28 A DESIGN VOLTS: 480 BUS VOLTS: 457 %VD: 4.79 ========================= PU BUS VOLTAGE: 0.952 ANGLE: -8.6 DEGREES NET BRANCH DIVERSITY LOAD: 360.0 KW 270.0 KVAR LOAD FROM: DSB 3 BUS 27 FEEDER AMPS: 568.4 VOLTAGE DROP: 3. %VD: 0.70 PROJECTED POWER FLOW: 360.0 KW 270.0 KVAR 450.0 KVA PF: 0.80 LAGGING LOSSES THRU FEEDER: 2.0 KW 2.9 KVAR 3.5 KVA ==== BUS: BUS 28 B DESIGN VOLTS: 480 BUS VOLTS: 452 %VD: 5.74$ ========================= PU BUS VOLTAGE: 0.943 ANGLE: -8.3 DEGREES NET BRANCH DIVERSITY LOAD: 360.0 KW 270.0 KVAR LOAD FROM: DSB 3 BUS 27 FEEDER AMPS: 574.2 VOLTAGE DROP: 8. %VD: 1.65 PROJECTED POWER FLOW: 360.0 KW 270.0 KVAR 450.0 KVA PF: 0.80 LAGGING LOSSES THRU FEEDER: 7.5 KW 3.2 KVAR 8.1 KVA

A summary listing of all bus voltages and branch currents is listed at the end of the report.

BALANCED VOLTAGE DROP AND LOAD FLOW BUS DATA SUMMARY ***************************************************************************** BUS NAME BASE VOLT PU VOLT BUS NAME BASE VOLT PU VOLT UTILITY BUS 69000.00 0.9945 TX A PRI 69000.00 0.9740 003-HV SWGR 13800.00 0.9899 TX B PRI BUS 4 13800.00 0.9897 005-TXD PRI 13800.00 0.9899 006-TX3 PRI 13800.00 0.9892 007-TX E PRI 13800.00 0.9897 DS SWG1 BUS 8 4160.00 1.0000 TX C PRI BUS 9 4160.00 0.9997 HVAC BUS 4160.00 0.9977 TX3 SEC BUS 11 4160.00 0.9898 DSB 1 BUS 14 480.00 1.0029 DS SWG2 BUS 13 4160.00 0.9898 TX3 TER BUS 12 4160.00 0.9958 DS SWG3 BUS 20 4160.00 0.9957 021-TX F PRI 4160.00 0.9938 BLDG 115 SERV 4160.00 0.9787 026-TX G PRI 4160.00 0.9758 025-MTR 25 4160.00 0.9751 022-DSB 2 480.00 0.9949 BUS 15 480.00 1.0018 018-RA 480.00 0.9971 BUS 16 480.00 0.9925 BUS 17 480.00 0.9894 H3A BUS 19 480.00 0.9870 023-MTR 23 480.00 0.9914 DSB 3 BUS 27 480.00 0.9591 BUS 28 A 480.00 0.9521 029-TX D SEC 4160.00 0.9787 BUS 28 B 480.00 0.9426 CMP CTR 208.00 0.9763 BALANCED VOLTAGE DROP AND LOAD FLOW BRANCH DATA SUMMARY ***************************************************************************** FROM NAME TO NAME TYPE VD% AMPS KVA RATING% TX A PRI UTILITY BUS FDR 2.05 93.89 11158.41 UNKNOWN EDISON UTILITY BUS FDR 0.55 93.89 11158.41 UNKNOWN TX A PRI 003-HV SWGR TX2 -1.60 94.95 11152.33 44.61 003-HV SWGR TX B PRI BUS 4 FDR 0.02 30.77 728.09 27.97 003-HV SWGR 005-TXD PRI FDR 0.00 0.00 0.00 0.00 003-HV SWGR 006-TX3 PRI FDR 0.07 212.57 5029.76 54.51 003-HV SWGR 007-TX E PRI FDR 0.03 155.27 3673.94 39.81 TX B PRI BUS 4 DS SWG1 BUS 8 TX2 -1.03 30.77 727.91 38.82 007-TX E PRI BLDG 115 SERV TX2 1.10 155.27 3672.99 63.88 DS SWG1 BUS 8 TX C PRI BUS 9 FDR 0.03 108.89 784.56 55.84 DS SWG1 BUS 8 HVAC BUS FDR 0.23 69.55 501.15 63.23 DS SWG1 BUS 8 CMP CTR TX2 2.37 63.97 460.91 92.18 TX C PRI BUS 9 DSB 1 BUS 14 TX2 -0.32 106.51 784.32 78.43 DS SWG2 BUS 13 TX3 SEC BUS 11 FDR 0.01 583.10 4158.75 74.76 3WINDING TX3 SEC BUS 11 TX2 0.74 583.10 4189.93 UNKNOWN DSB 1 BUS 14 BUS 15 FDR 0.11 285.02 237.65 18.75 DSB 1 BUS 14 BUS 16 FDR 1.05 538.95 449.39 76.99 DSB 1 BUS 14 BUS 17 FDR 1.35 114.11 95.15 87.78 DS SWG3 BUS 20 TX3 TER BUS 12 FDR 0.00 109.72 787.19 16.25 3WINDING TX3 TER BUS 12 TX2 0.15 109.72 788.37 UNKNOWN BALANCED VOLTAGE DROP AND LOAD FLOW BRANCH DATA SUMMARY ***************************************************************************** FROM NAME TO NAME TYPE VD% AMPS KVA RATING% DS SWG3 BUS 20 021-TX F PRI FDR 0.20 114.39 820.72 58.66 021-TX F PRI 022-DSB 2 TX2 -0.11 114.40 819.10 81.91 026-TX G PRI BLDG 115 SERV FDR 0.28 146.39 1032.30 75.07 025-MTR 25 BLDG 115 SERV FDR 0.36 355.82 2509.12 60.82 029-TX D SEC BLDG 115 SERV FDR 0.00 0.00 0.00 0.00 026-TX G PRI DSB 3 BUS 27 TX2 1.67 146.40 1029.30 102.93



022-DSB 2 023-MTR 23 FDR 0.35 966.62 799.51 63.59 BUS 15 018-RA FDR 0.47 150.13 125.04 75.07 BUS 16 H3A BUS 19 FDR 0.54 277.53 229.00 79.30 DSB 3 BUS 27 BUS 28 A FDR 0.70 568.50 453.32 37.40 DSB 3 BUS 27 BUS 28 B FDR 1.65 574.23 457.90 82.03

Warning messages are presented at the end of the report.

NOTE: FDR RATING% = % AMPS RATING BASED ON LIBRARY FLA OR BRANCH INPUT FLA TX2 RATING% = % KVA RATING BASED ON TRANSFORMER FL KVA 31 BUSES *** T O T A L S Y S T E M L O S S E S *** 201. KW 949. KVAR ***WARNING*** STUDY CONTAINS 1 VOLTAGE CRITERIA VIOLATIONS VIOLATIONS DENOTED BY ($) AT BUS AND BRANCH %VD LOCATIONS

The voltage drop data reported is based on the criteria in the text boxes in the Load Flow Study Setup dialog box.

4 Short Circuit Study

This chapter examines the theoretical basis of the Comprehensive Short Circuit Study (referred to hereafter as “Short Circuit Study”). It includes a systematic methodology and applies the methodology to numerous practical examples. You can also purchase separately the A_FAULT and IEC_FAULT Short Circuit Study modules. The chapters provided with A_FAULT and IEC_FAULT discuss the Short Circuit Methodology as defined by the American National Standards Institute (ANSI) and the International Electrotechnical Commission (IEC), respectively.

The basis of all short circuit studies is Ohm’s Law and is referred to in this Reference Manual as the comprehensive methodology. This comprehensive methodology is exclusively examined in this chapter.




• Examples.

IN

TH

IS

CH

AP

TE

R

What is the Short Circuit Study? .................................................................DAPPER 4-2 Engineering Methodology .......................................................................... DAPPER 4-2 PTW Applied Methodology ......................................................................DAPPER 4-14 Application Examples ................................................................................DAPPER 4-22



4.1 What is the Short Circuit Study? The Short Circuit Study models the current that flows in the power system under abnormal conditions and determines the prospective fault currents in an electrical power system. These currents must be calculated in order to adequately specify electrical apparatus withstand and interrupting ratings. The Study results are also used to selectively coordinate time current characteristics of electrical protective devices.

Define System Data

Define system topology and connectionsDefine utility connection (swing bus)Define feeder and transformer sizesDefine fault contribution data

Run Short Circuit Study

Saved in Database

Three-phase fault currentsUnbalanced fault currentsMomentary and asymmetrical fault currents

Datablocks

Reports

Used by LoadSchedules

Study Setup

Cable LibraryTransformer LibrayStudy Setup

Used by Time CurrentCoordination (CAPTOR)

4.2 Engineering Methodology The systematic Short Circuit Study methodology begins by creating a system one-line diagram, thus defining all electrical characteristics of the power system. If the solution is worked by hand, the engineer must define a complete Thevenin equivalent impedance diagram and place all impedances on a common base. Selected fault points are chosen and the specific Thevenin equivalent impedances are calculated. Knowing the fault impedances and the driving point voltages, the fault currents are calculated using Ohm’s Law.

The computer-based solution methodology is slightly different. From the one-line diagram and the associated computer database of system equipment, the computer forms an admittance matrix. An admittance matrix is a square matrix of a size equal to the number of electrical buses. The matrix is well ordered and generally symmetric about the diagonal. The well ordered and sparse matrix characteristics of the admittance matrix allow for convenient, albeit computationally intensive, matrix inversion. From the

3/26/2006

Short Circuit Study DAPPER 4-3

inverted admittance (impedance) matrix, the bus fault currents are calculated using Ohm’s Law.

4.2.1 Balanced Faults The fault currents in a three-phase power system may be either equal or balanced across all three phases or unbalanced. An unbalanced fault involves one or two phases, but not all three. The three-phase symmetrical rms fault current (balanced fault) is often considered the maximum fault current at the bus. However, in certain situations an unbalanced fault may be larger. This section discusses the balanced fault current calculation procedure. First Ohm’s Law must be defined:

[E] = [Z][I]

where

E bus voltage matrix; Z bus impedance matrix; referred to as the Z Bus matrix; I bus nodal current matrix.

The impedance Z in complex notation is:

Z = R + jX

where

R resistance; jX reactance.

Thevenin Equivalent Circuit This section outlines the process used to develop a Thevenin equivalent circuit. The Short Circuit Study formulates node equations by applying Kirchoff’s Current Law. A Thevenin equivalent impedance for each fault location is then used to determine the fault current.

The following one-line diagram shows the process used to calculate the fault current at Bus 2. For a fault at Bus 2, the fault current contributions flow from the Utility and from Motors 1 and 2 into Bus 2.

B1

T1

B2

C1

B3

M1

M2

UTILITY

From the one-line diagram a Thevenin equivalent impedance diagram is drawn:



ZUtility Z Motor 1 ZMotor 2

ZCableZTransformer

Bus 1 Bus 2 Bus 3

Next, the systematic procedure calls for converting all impedances to per unit values on a common power base, such as 100 MVA. Converting impedances from units of ohms to per unit is more critical when transformers are included in the power system. This is because the transformer’s impedance in ohms depends upon whether the impedance is viewed through the primary or secondary connection. When the transformer’s impedance is in per unit, its value is the same as when viewed through either the primary or the secondary connection. From the Thevenin equivalent impedance diagram, you can create a Norton equivalent diagram by short circuiting the voltage sources and injecting a Norton equivalent current (If) into the faulted bus.

ZUtility ZMotor 1 ZMotor 2

ZCable

I

Bus 1

Bus 2

Bus 3

If

I IUtility Motor 1 Motor 2

In the above circuit diagram, the Norton equivalent current If splits at Bus 2 and flows through the three branches. The current in each branch is dependent on its impedance. The current at Bus 2 is actually the negative of the Norton equivalent current If. The sum of the three branch currents is also equal to If.

Using simple series and/or parallel impedance combinations, you can determine the single Thevenin equivalent impedance at Bus 2. If you know the Thevenin equivalent impedance, you can determine the fault current in per unit amperes.

The fault current is:

I VZf

Thevenin =

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4.2.2 Unbalanced Faults Calculating the magnitude of unbalanced faults requires using symmetrical components. Using this method, an unbalanced set of three-phase system voltages or currents is resolved into balanced sets of phases known as the positive-, negative- and zero-sequence values. Next in the unbalanced short circuit current calculation procedure, the power system is modeled as series or parallel combinations of the sequence impedance networks. Each unbalanced condition (single-line-to-ground, line-to-line, and line-to-line-to ground) requires formulation of the three sequence networks in different configurations, based on a set of boundary conditions associated with the system as faulted.

Single-Line-to-Ground Faults The following diagram shows a bolted single-line-to-ground fault in Phase A.

Phase a

Phase b

Phase c

I I I V = 0

Faultabc

The boundary conditions are:

V = 0, I = 0, I = 0a b c

where

Va line-to-ground voltage in Phase A; Ib currents in Phase B; Ic currents in Phase C.

From symmetrical components the following solution matrix is written:

I III

a

b

c

a0

a1

a2

2

2

II

1 1 1

1 a a

1 a a

L

NMMM

O

QPPP

=

L

N

MMM

O

Q

PPP

L

NMMM

O

QPPP

13



where

I = I = 0b c

and

a j

a j

= e

= e

120

2 240

Substituting the above and solving yields the single-line-to-ground fault case:

I I

I I

I I

a0 a

a1 a

a2 a

=

=

=

131313

b g

b g

b g

Therefore:

I I = Ia1 a2 a0=

Because the three symmetrical components are equal in magnitude and in phase, they may be viewed as connected in series, as shown:

Ea

Z 1

Z 0

Z 2

Ia1

Ia0

Ia2

The positive-sequence current is:

I EZ Z Za1

a

1 2 0=

+ +

where

Z1 positive-sequence impedance; Z2 negative-sequence impedance; Z0 zero-sequence impedance.

From the preceding equations we know that the positive sequence current in Phase A is one-third of the phase current.

Therefore:

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I = 3Ia a1

Substituting yields:

I Ea

a=+ +

3

1 2 0Z Z Z

Line-to-Line and Double-Line-to-Ground Faults For line-to-line and double-line-to-ground faults, the procedure is similar. The boundary conditions in conjunction with the appropriate equations establish the sequence component network connections. For line-to-line faults, the positive- and negative-sequence networks are in series; the zero-sequence network is not involved. For double-line-to-ground faults, the positive-sequence network adds to the negative- and zero-sequence networks which are in parallel.

It is entirely possible that the return current (ground current) for a line-to-line-to-ground fault could be larger than the fault current in the two faulted phase-conductors.

Grounding Impedance The zero-sequence impedance of delta-wye-grounded transformers is modeled as an infinite impedance (open connection) when viewed from the delta side, and modeled as a shunt to the reference through the transformer’s zero-sequence impedance when viewed from the wye-grounded side. The wye-grounded wye-grounded transformer, if provided with a grounding impedance, is modeled with grounding impedance on either or both sides of the transformer connection. In this configuration there is no connection to reference. Although not specifically discussed in the above formulation of the unbalanced fault networks, any grounding impedance must be modeled at three times its ohmic value in the zero-sequence network. If a grounding impedance is established on the wye-grounded side of the transformer, its impedance is added in series to the shunt path (that is, in series with the transformer’s zero-sequence impedance).

Positive-, Negative-, and Zero-Sequence Modeling Sometimes the negative-sequence data differs from the positive-sequence data. In this case, you must calculate the negative-sequence impedance network independently from the positive-sequence impedance network.

Per Unit Notation The following per unit equations are useful when calculating the short circuit currents. All impedance terms are in complex vector notation.

For utility contributions:

Z kVAkVA

MVAMVA

pu pubase

utility

base

utility= = Ω

where

kVAutility (MVAutility) the utility system three-phase short circuit capability; kVAbase (MVAbase) the system three-phase power base specified for the Study.

For motor and generator contributions:



Z X kVkV

kVAkVA

pu pu dmotor

base

2base

motor= ′′FHG

IKJFHG

IKJ Ω

where

′′Xd motor subtransient reactance on its motor rated line-to-line voltage (kVmotor) and three-phase power (kVAmotor) base; kVmotor line-to-line motor rated voltage; kVAmotor three-phase power base; kVbase line-to-line bus nominal system voltage at the point of the motor; kVAbase three-phase power base.

For feeders:

Z Zkv 1000

kVA

ZkV

MVA

pu pucable

base2

base

cable

base2

base

=×

= Ω

where

Zcable per phase (one way) feeder impedance in ohms.

For transformers:

Z Z %100

kVkV

kVAkVApu

transformer transformer

base

2base

transformer= FHG

IKJFHG

IKJFHG

IKJ

where

Ztransformer transformer percentage impedance on its self-cooled or nominal transformer three-phase power base (kVAtransformer) and rated line-to-line voltage (kVtransformer).

The three-phase fault current base is:

I VZ

pu Ashort circuitThevenin

=FHG

IKJ

Expressed in amperes:

I I pushort circuit short circuit b= I×

Where the base current is defined as:

I kVAkVbase

base

base=

×3

Sometimes the aforementioned per unit notation equations cannot be used directly. For example, if your utility fault data is not in power or current units, convert it to acceptable units. Suppose the utility equivalent impedance at the source is expressed as:

Available fault impedance = 0.0425+ 0.4250 at 4160 Vj Ω

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To express this value in fault current:

Z R

I = V3 Z

utility2 2

utility

= +

×

e jX

Substituting known quantities yields:

Zutility = +

=

0 0425 0 425

0 4272

2 2. .

. Ω

At 4.16 kV, the utility fault current is:

I V

5622 A

short circuit =

=

41603 0 4272e jb g. Ω

The utility X/R ratio is:

XR

0.4250.0425

=

=

ΩΩ

10

4.2.3 Momentary and Interrupting Fault Current The momentary fault current is defined as the short circuit peak (equivalent rms) current that flows at the first one-half cycle after the onset of the fault. Electrical apparatus must withstand the mechanical and thermal affects of this momentary current. Protective devices and switches must be capable of closing and latching into this momentary current. Momentary current can be highly asymmetrical (with the time axis).

The interrupting fault current is defined as the short circuit current that flows through a protective device at the time of its contact separation. The interrupting duty of a circuit breaker may be lower than its associated closing and latching (momentary) rating. The interrupting current tends to be more symmetrical with the time axis than the momentary fault current. Medium and high voltage circuit breakers in particular may have interrupting ratings based on contact parting times of 3 or 5 cycles after the onset of the fault.

The asymmetrical nature of the momentary fault current is a result of the instantaneous change in system X/R at the point of the fault prior to and immediately after the fault. Prior to the fault, the system generally operates at a very small X/R ratio (that is, a high power factor). However, after the fault when all the high power factor loads are ignored, the system X/R ratio can be quite large. This instantaneous change in system X/R ratio at the instant the fault occurs is exacerbated when, during the voltage sine wave, the fault occurs. If the fault occurs at a positive-increasing voltage peak, then the current wave is said to have maximum asymmetry. This asymmetric condition is known as the dc component or dc decay because the asymmetric nature of the wave shape decays exponentially over time. Also, the momentary fault current and, to a lesser degree, the interrupting fault current are dependent upon the time varying collapse in machine



voltages. This time varying collapse is known as the ac decrement and is most often modeled as a time varying machine reactance.

Current limiting fuses which operate within one-half cycle are subject to momentary currents and may open to clear the fault before the maximum prospective fault current occurs. Devices such as circuit breakers require the fault current to pass through a current zero before the arcing current is extinguished. Breakers may experience significant interrupting fault current for many cycles during operation.

To calculate the momentary and interrupting fault currents, you must evaluate both the transient and steady state conditions.

Consider a fault that occurs in a simple RL series circuit driven by an ideal sinusoidal voltage source. You can write Kirchoff’s Loop Equation as:

V sin t + = Ri + Ldidtpeak × ω Θb g

This differential equation can be solved for the instantaneous current, i:

i =V

Zsin t + -

VZ

e -peak peak-t

LRω α αΘ Θb g b− g

L

NMM

O

QPPsin

where

Z R L2 + ωb g2

α time in cycles when the fault occurs;

Θ tan LR

-1 ωFHGIKJ

ω 2πf; t time in seconds; f frequency in Hertz; Vpeak peak (crest) voltage.

The first term is clearly sinusoidal and is at an instantaneous peak if α π- =2

Θb g ±

radians when time (α) is equal to 0+. The second term is an exponentially decaying term, with a time constant of L/R. This is the dc decay. It must be subtracted from the first term to satisfy the boundary condition that at time equal to 0-, no fault current flows.

4.2.4 Asymmetrical Peak Fault Current As noted, the asymmetrical peak fault current consists of both ac and dc components, and is a function of time. As shown in the following figure, the theoretical asymmetrical peak to peak current is 2 2 multiplied by the initial symmetrical rms current. The initial symmetrical rms current is the ratio of the pre-fault no-load voltage at the bus to the Thevenin equivalent impedance. The asymmetrical nature of a fault current is best shown by the following graph.

′′Ik

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Current

Theoretical maximumPeak at 1/2 cycle

(DC decay)

Bottom envelope

Top envelopeDecaying (aperiodic) component i

Time

2

2 I"

i

k

p

2 2 I k

dc Asymmetrical valuesincluding motor contributions

Steady state value(no motor contributions)

dci

The first one-half cycle asymmetrical peak current is the sum of the de decay and ac decrement components. This can be expressed in equation form as:

i = 2 I 2 Iasymmetrical peak k k

RX

c′′ + ′′

− FHGIKJe

2π

where

iasymmetrical peak asymmetrical peak fault current; initial symmetrical rms fault current;

idc dc decaying component of fault current; Ik steady-state fault current; c time in cycles into the fault.

′′Ik

4.2.5 Asymmetrical rms Fault Current Historically ANSI standards have rated momentary current on the root mean square (rms) average of the first 1

2 cycle fault current. The rms of the asymmetrical peak current is:

I I 2 I e

= I 1 2e

asymmetrical rms k k2 R

Xc

k

RX

c

= ′′ + ′′FHG

IKJ

′′ +

−

−

b g b22

4

π

π

And at 12 cycle into the fault:

= I 1 2ek

RX′′ +

−2π

4.2.6 Steady State Fault Current The steady state fault current, usually considered as the 30-cycle fault current consists of only a symmetrical component because the dc component has already decayed to zero. Motor contributions are generally ignored.



4.2.7 Transformer Taps Transformer taps must be properly modeled in short circuit calculations. Not only does the change in the transformer tap alter the Thevenin equivalent impedance, it also changes the voltage at the fault point. Although the effect of a transformer tap on a radially fed system might be relatively easy to model in manual calculations, the procedure can become considerably more tedious when looped networks are involved.

Do not guess at the effects of transformer taps on short circuit calculations. Taps may increase or decrease fault current, even to a degree that protective device fault duties or time/current coordination is affected.

Primary Transformer Tap Modeling Visualize the circuit in the following figure when you analyze the effects of primary connection transformer taps on fault duties. As the N:1 turns ratio decreases, the Thevenin equivalent impedance at the faulted bus decreases. However, the voltage at the faulted bus increases. The rate of change of voltage increase and impedance decrease means that a fault current could be smaller or larger than the nominal or no-tap case. The tap is modeled as an ideal voltage shifter. As a result, a -5% primary tap raises the secondary voltage by 1

0 95. or 1.0526 pu V.

TapZ Transformer

Voltage Generator

Faulted BusTransformer

Xd"

The relationship between the short circuit current and primary tap setting is:

I V

TapX Z

f

dgenerator transformer

=

+FHG

IKJ × ′′

LNMM

OQPP +

11

2

where

If fault current; V fault point pre-fault no loads pre-fault voltage; Tap ideal voltage shifter in per unit;

machine reactance; Ztransformer transformer impedance.

′′Xd

The fault point no load voltage is:

V = VTapgenerator1

1+FHG

IKJ

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Secondary Transformer Tap Modeling Secondary transformer taps may also be modeled. The ideal voltage shifter is modeled as:

TapZ Transformer

Voltage Generator

Faulted Bus

Xd"

Transformer

The relationship between the short circuit current and transformer secondary tap setting is:

)()1( "2rtransformegeneratord

f ZXTapVI

++=

The fault point no-load voltage is:

V = V 1+ Tapgenerator b g

4.2.8 Transformer Off-Nominal Voltage Modeling At times you may need to model transformers that are not rated on the desired nominal system voltage. In these cases it is imperative to use proper per unit impedance modeling. Ideally, you have the opportunity to select the nominal system voltages so that the ratio of the transformer voltages equals the ratio of the primary to secondary bus nominal system voltages. However, in a multi-looped system, this might be impossible.

You can model transformer off-nominal voltages similar to transformer taps. In both cases, the modeling uses an ideal voltage shifter either on the primary, secondary, or both sides of the transformer.

In order to model off-nominal voltages, you must have the Model Transformer Taps checkbox selected in the Study Setup dialog box.

4.2.9 Transformer Phase Shift It is critical to know how transformer phase shift affects the unbalanced branch current flows. In addition to knowing the bus fault current, you also need to know the magnitude of the branch current contributions into the faulted bus. At times it is also beneficial to know the sequence fault current flows, especially the zero-sequence currents when ground fault relays are involved. The positive-sequence current through a delta-wye grounded transformer shifts the secondary side current by -30° relative to the primary side, whereas the negative-sequence current through the same transformer shifts the secondary side current by +30°. There is no phase shift in the zero-sequence network. This phase shift significantly affects the calculated branch currents in a delta-wye-grounded transformer connections. There is no phase shift across a delta-delta or wye-wye transformer.



When the transformer phase shift is modeled for a single-line-to-ground fault, the branch sequence current might be:

418 -90° 418 -90°

1388 -120°

1388 -120°

Positive-Sequence

418 -150° 418 -150°

1388 -120°

1388 -120°

Negative-Sequence

0 -0°

418 -120°

1388 -120°

1388 -120°0 -0°

0 -0°

418 -120°

Zero-Sequence

In this example, the transformer is rated 13.8 kV - 4.16 kV, and if 1388 A in the positive-sequence network flow in the secondary winding, the primary winding positive-sequence current is 418 A.

4.3 PTW Applied Methodology PTW applies the methodology described in Section 4.2. Section 4.3 describes how to run the Short Circuit Study, including explanations of the various options associated with the Study.

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4.3.1 Before Running the Short Circuit Study Before running the Short Circuit Study, you must:

• Define the system topology and connections.


• Define feeder and transformer sizes.

• Define fault contribution data.

4.3.2 Running the Short Circuit Study You can run the Study from any screen in PTW, and it always runs on the active project.

To run the Short Circuit Study:


2. Select the check box next to Short Circuit.

3. To change the Study options, choose the Setup button.

4. Choose the OK button to return to the Study dialog box, and choose the Run button

The Short Circuit Study runs, writes the results to the database, and creates a report.

4.3.3 Short Circuit Study Options The Short Circuit Study dialog box lets you select options for running the Study.

Following is a list of the available Study options.

Fault Type By default, PTW includes Three Phase Fault and Single Line to Ground fault calculations; the Line to Line Fault and Line to Line to Ground fault calculations can also be selected. Generally, the three-phase and single-line-to-ground current model the worst case fault currents in the system. You can select any combination of the four check boxes.



Faulted Bus By default, PTW faults All Buses. If you want to fault a single bus, select the One Bus option button and select the appropriate bus from the box. If All Buses is selected, limited data is written to the database; however, the Report can be extensive.

If you are running a Study on a 100 bus facility and choose to report fault duties on all buses and branches, the total report could be 300 or more pages.

Calculation Models Select the checkbox(es) for the calculations you want to include in the Study: Motor Contribution, Transformer Tap, and Transformer Phase Shift. You can select any combination of the three.

Motor Contribution

By default, PTW includes motor contributions. Unchecking the check box eliminates motor contributions from the Study results.

Transformer Tap

By default, PTW does not include tap effects in the transformer model. If this box is unchecked, all transformers appear without the effect on any taps, and the pre-fault voltage is relative to the swing bus voltage. By selecting Transformer Tap, PTW calculates the system pre-fault no load voltage profile based on the swing bus voltage and transformer tap settings. You must check this box to analyze transformer off nominal voltages properly.

Transformer Phase Shift

By default, PTW does not include Transformer Phase Shift, the transformer phase shift angle remains at 0°, and the pre-fault voltage angles in each isolated area of the power system remain at the swing bus voltage angle. To report unbalanced circuit branch flows, select the Transformer Phase Shift check box. This option calculates each transformer phase shift in degrees based on the transformer connection type; the pre-fault voltage angle includes all transformer phase shifting relative to the swing bus.

Report Specifications Make Report specification selections based on the following criteria.

Bus Voltages

By default, PTW calculates and reports voltages for the First Bus From Fault. You can also select from three other options: Second Bus From Fault, All Buses, and None. Keep in mind that if you select All Buses, the Study will take significantly longer to run.

Bus voltages are reported but not written to the database. Bus voltages are not reported when sequence branch current values are selected.

Branch Currents

You can also select from three other options: First Branch From Fault, Second Branch From Fault, and All Branches. Upon opening a new project, the default is to report branch currents one branch away from the fault point.

As shown in Figure 4-1, if you wish to display branch fault currents through 3-winding transformers, you must select the option to report 2 branches away (not just one branch away).

3/26/2006


Figure 4-1. To report branch fault currents for 3-winding transformers, select to report 2 branches away, not one.

The speed of the Study is significantly impacted by the amount of Study results written to the database and to the output report.

Branch currents are only written to the database if a fault at a single bus is selected. If a bus is faulted, then branch currents may flow from any location to the faulted bus. If a different bus is faulted, then different branch fault currents flow. Only one complete set of branch fault data is saved to the database for retrieval onto a datablock. Only initial symmetrical phase currents in the branch are reported.

Phase or Sequence

By default, PTW reports phase A, B, and C quantities. Phase current will always be written to the Report, but branch current flow will not be written to the database unless a single bus is faulted. You can also select to report the positive-, negative-, and zero-sequence quantities. Sequence currents are not reported in the database, and therefore they will not appear in datablocks. Also, the sequence currents are not reported when asymmetrical fault currents are calculated.

Fault Current Calculation

By default, PTW calculates Initial Symmetrical RMS Only (No DC offset) fault currents. You can also select to calculate the Asymmetrical Peak or Asymmetrical RMS fault currents, both with DC offset and Decay.

The Initial Symmetrical RMS Only option calculates bus voltages and branch flows as complex values. All values arise from the symmetrical. The Asymmetrical Peak option calculates the total asymmetrical peak or crest (dc offset and decay) fault current at a specified time. You must enter the time in cycles. The Asymmetrical RMS option calculates the total asymmetrical symmetrical (dc offset) rms current at a specified time. You must enter the time in cycles.

Asymmetrical Fault Current at Time

If you selected asymmetrical fault current calculations, you must enter a time in cycles. After the onset of the fault, the Study calculates the fault currents after the specified number of cycles. The default number of cycles is five. A minimum of 0.5 cycles is reported.

If you selected initial symmetrical fault current calculations, the Study ignores the Asymmetrical Fault Current at Time value.



4.3.4 Component Modeling When you run the Short Circuit Study, PTW checks for appropriate feeder sizes and lengths, and transformer sizes in the library. If the data is inappropriate or missing, error and warning messages are shown in the Study Run dialog box and included in the Report.

The following sections describe the minimum data required for the Short Circuit Study to run.

Feeder Data You must specify a cable’s positive-sequence impedance and one-way circuit length. PTW models the negative-sequence impedance as equal to the positive-sequence impedance. If a cable’s zero-sequence impedance is zero, the Short Circuit Study uses the positive-sequence value. Cable positive and zero sequence impedances may be selected from the Cable Library, or you can define them in the Component Editor.

If you make the cable User Defined, you can enter specific cable impedance in ohms per 1000 feet or ohms per 1000 meters. Cable lengths must be entered in the same units as the cable impedance data (feet or meters). If you switch the Program Options from English to Metric units, PTW converts entered cable lengths and impedances to the appropriate units. Cable impedances are unaffected by the wire circuit description characteristics.

Transformer Data You can select predefined two-winding transformers from the Transformer Library or you can define them yourself in the Component Editor. PTW defines two-winding transformers by their percentage leakage positive- and zero-sequence impedance value, cooling capacity type, and the nominal kVA rating. If a transformer's zero-sequence impedance is zero, PTW uses the positive-sequence value. Transformers' rated voltages may differ from the bus nominal voltages. PTW models those off-nominal voltages as ideal voltage shifters separate from any primary or secondary tap that is modeled. A warning message appears in the Study Messages dialog box when PTW detects a mismatch between the bus nominal voltage and the transformer rated voltage. You can also define the transformer impedance in the Component Editor using the transformer's resistance and reactance values in percent on the nominal or self-cooled kVA rating.

When you set the PTW Project for the IEC Standards, user-defined transformers can be defined in per unit on any kVA base, the Rated Short Circuit Voltage percent or on Rated Ohmic voltage percent.

Transformer negative-sequence impedance always equals the positive-sequence value in the Short Circuit Study. The primary and secondary transformer connections help determine the effect of the zero-sequence Thevenin equivalent impedance.

The wye-grounded wye-grounded zero-sequence path appears as a non-shunt primary to secondary leakage impedance. Grounding impedance may be placed on one or both of the grounded points. PTW automatically multiplies this grounding impedance by three to calculate the proper zero-sequence impedance on a per unit base. The wye-grounded wye-grounded transformer is modeled in shell form, and is defined as an infinite impedance when viewed from either connection.

Three-Winding Transformers Three-winding transformers may be modeled. Off-nominal voltage and transformer taps may be modeled in a manner similar to two-winding transformers. All three-winding

3/26/2006


transformer data must be user defined in the Component Editor. PTW models the three- winding transformer using conventional network reduction, and establishes a fictitious center point bus. Also, PTW establishes a secondary to tertiary branch. This fictitious bus and associated branch count against the total bus and branch limit in PTW.

There are two networks in the following one-line diagram. Transformer T1 is a three-winding transformer with a primary, secondary and tertiary power rating of 15 MVA, 15 MVA and 5.25 MVA, respectively.

GEN 1 GEN 2

BUS 1

C1BUS 2

T2

BUS 3 BUS 4

T1

C2 C3

BUS 5 BUS 6

BUS 7

C4BUS 8

BUS 9

T3

BUS 10

C5BUS 11

27856.53 A

22405.02 A

11150.80 A21706.12 A

20520.80 A 10860.47 A

CENTER POINT

0.8790%

21705.22 A

20520.00 A

T4

BUS 12

C6BUS 13

27856.53 A

22405.02 A

7199.51 A

6.0214%

9.9790%

11150.56 A

10860.25 A

The manufacturer’s published test data for Transformer T1 is:

Test #

Impedance Measured

into Winding

Winding Short

Circuited

Winding Open

Circuited

Short Circuit

Voltage in %

Impedance Base for

Measure (MVA)

Impedance Symbol

1 Primary Secondary Tertiary 6.9 15 ZPS

2 Primary Tertiary Secondary 5.6 5.25 ZPT

3 Secondary Tertiary Primary 3.8 5.25 ZST

It is important to note that the preceding measurements are relative to different power bases. In Test 1 when the tertiary circuit is open, short circuit current flows only in primary and secondary windings. Both of these windings have 15 MVA ratings. In the test, the voltage across the primary winding is increased until 6.9 % rated voltage causes the rated full current to flow in the secondary winding. By opening the tertiary circuit, no current flows in this winding.

In Test 2, the tertiary winding is fully loaded based on its 5.25 MVA rating, even though the primary carries only about one-third rated current on its 15.0 MVA rating. The test stopped when the 5.6 % rated voltage was applied to the primary winding and full load current was reached on the tertiary winding (corresponding to 5.25 MVA). It is critical to know on what base the short circuit voltage takes place.



The following drawing is an equivalent impedance diagram for the three-winding transformer:

Z

Z PS

ZPT

T

S

P ST

You can convert this into an equivalent wye diagram using standard network reduction techniques:

Z

Z 2

Z 1

T

S

P

3

From the network reduction diagram, we can write:

Z = Z + Z Z = Z + ZZ = Z + Z

PS 1 2

PT 1 3

ST 2 3

Network reduction yields:

Z = (Z + Z - Z )Z = (Z + Z - Z )Z = (Z + Z - Z )

11

2 PS PT ST

21

2 PS ST PT

31

2 PT ST PS

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You can solve the equations by substituting the manufacturer’s data expressed on a common 15 MVA base:

Z = 6.9 + 5.6 155.25

- 3.8 155.25

= 6.0214 %

Z = 6.9 + 3 155.25

- 5 155.25

= 0.879 %

Z = 155.25

155.25

6

= 9.978 %

11

2

21

2

31

2

j j j

j

j j j

j

j j

j

×

j

FHG

IKJ ×

LNM

OQP

×FHGIKJ ×

LNM

OQP

×FHG

IKJ + ×FHG

IKJ −

LNM

OQP

. .

. .

8 6

5 6 38 9.

The above values represent the two-winding transformer equivalent impedances that must be used in the one-line diagram on page DAPPER 4-18.

As of DAPPER Version 2.0, you cannot enter negative impedance values for the two-winding transformer component, even though these sometimes occur with network reduction of a three-winding transformer into an equivalent two-winding transformer case.

Contribution Data Fault duty contributions to the power system originate from the motor generator and utility source components. PTW provides default subtransient and X/R ratio values. You can calculate the machine kVA and voltage base using the rated size and connected bus nominal voltage. For example, if you enter a 50 hp motor with an 80% power factor and 92% efficiency, PTW calculates the rated kVA base as:

kVA =50 hp 746

1000 WkW

0.8 pf 0.92 efficiency

= 50.7 kVA

base

Whp×

× ×

This is close to the rule-of-thumb that 1 hp is equal to 1 kVA. Of course, if you have a 1000 hp synchronous motor with a unity power factor, PTW calculates the motor’s kVA base value as 746 kVA for short circuit current purposes. The fault contribution calculation remains unaffected by the motor load factor.

Fault contributions can be at any bus and there may be multiple contributions located at any bus.

Important: You may change the ANSI contribution calculated kVA base, for example to model the 50 hp motor as 50.0 kVA. However in PTW, once the machine ANSI contribution kVA base is selected (or automatically calculated by PTW if the base kVA value is 0), it will not change. Therefore, if you enter the motor load as 50 hp, run a Study, and then change the motor’s rating to 75 hp, the motor’s ANSI contribution base kVA will remain 50 kVA. You must change the base kVA to 75 kVA manually.

Induction motors are modeled as delta-connected, whereas synchronous motors and generators are modeled as wye-connected. Neutral (grounding) impedance may be modeled in the synchronous motor or generator.



4.3.5 Error Messages PTW examines the entered data for the Short Circuit Study. If PTW finds missing or incomplete information, it sends an error message to the Study Message dialog box. The Study Messages dialog box will report both fatal and warning messages. The Study will attempt to run to completion even if fatal errors are detected in order to attempt to identify all errors.

A somewhat common error is:

The calculated zero sequence impedance is negative.

It involves the entry of single-line-to-ground short circuit contribution data. PTW uses the three-phase fault data and the single-line-to-ground fault data to calculate the positive, negative- and zero-sequence impedances from the following per unit equations:

Z Z

Z 1.0I

I3 1.0

Z Z Z

Z 3I

Z Z

1 2

1f

f1 2 0

0f

1 2

3

slg

slg

=

=

=×

+ +

= − −

Φ

b gb g

Utilities often report available single-line-to-ground fault duties on an equivalent three-phase rating apparent power basis, using the equation:

kVA 3 I kVfslg= × ×

However, the actual apparent power of a single-line-to-ground fault is:

kVA = I kV3fslg

×

where

kV line-to-line voltage.

You cannot use the three-phase equivalent rating of a single-line-to-ground short circuit contribution. If you do, PTW may attempt to calculate the zero-sequence impedance as a negative value. The actual apparent power to be entered into PTW is the utility equivalent single-line-to-ground duty divided by 3. Enter the single-line-to-ground fault current X/R ratio, not the zero sequence impedance X/R ratio.

4.4 Application Examples The examples which follow illustrate how the Short Circuit Study runs on various system topologies. Unless otherwise specified, all per unit values are expressed on a 100 MVA base at the bus system nominal voltage.

3/26/2006


4.4.1 Fault Currents on a Radial Unloaded Feeder The first example validates the Short Circuit Study results against hand calculations; it is a simple three-bus radial feeder with no motor contributions. The transformer connection is delta-wye grounded and a bolted three-phase fault occurs at Bus 3. The nominal system voltage at Bus 3 is 480 V.

The following one-line diagram in Figure 4-1 shows the positive- and zero-sequence reactances of each component, and the three-phase and single-line-to-ground fault current at Bus 3 (the faulted bus) as calculated by PTW. The negative-sequence impedance is assumed to be equal to the positive-sequence impedance.

B1

C1X1 1.0000 puXo 1.0000 pu

T1X1 1.0000 puXo 1.0000 pu

B2

B3

UTILITYX1 1.0000 puXo 1.3333 pu

40,094.23 A 3 Ph51,549.68 A SLG

Figure 4-1

The short circuit capability of the Utility is given as 100 MVA at an X/R ratio of 99. This equates to an impedance of 0.010101 + j 0.999948 pu Ω on a 100 MVA base at the nominal system voltage at Bus 1 of 13.8 kV. The Utility equivalent positive-sequence impedance, X positive, is shown in Figure 4-1 as 1.0000 pu Ω, which is 0.999948 pu Ω rounded to four significant digits. Although the computer displays impedance values to four significant digits, the calculations are accomplished to 16 significant digits.

C1 is a high voltage cable between Bus 1 and Bus 2; its impedance is given as 0.0 + j1.0 pu on the 100 MVA 13.8 kV system base.

The transformer (T1) impedance is also given as 0.0 + j1.0 pu Ω, on its own base, which is the system base. The Thevenin equivalent impedance at Bus 3 is the sum of the utility, cable and transformer impedances, all expressed on the same 100 MVA base, or as:

Z = 0.010101+ 0.999948 + 0 + 1.0 + 0 + 1.0 pu = 0.010101+ 2.999948 pu

Thevenin j j jj

b g b g b g Ω

Ω

Combining the real and imaginary components of the Thevenin equivalent impedance into a single complex impedance yields:

Z = 2.999965 pu Thevenin j Ω

The short circuit current is equal to the driving point voltage divided by the Thevenin equivalent impedance. The driving point voltage is 480 V or 1 pu V on the 480 V system base. Therefore, the three-phase fault current is:

I = 12.999965

pu A

= - 0.333337 pu A

sc jj



The base current on the 480 V side of the transformer on a 100 MVA base is:

I = 100,000 kVA0.48 kV

= 120,281.3 A

base 3 ×

Thus the fault current in amperes is:

I = 120,281.3 A 0.333337 pu A= - 40,094.2 A

sc ×j

This value matches the 40,094.23 A shown in Figure 4-1.

Likewise the single-line-to-ground fault current at Bus 3 is calculated from the following equation:

I = 3VZ + Z + Z + 3Zslg

1 2 0 ground

Because there is no grounding impedance on the transformer wye-grounded connection, Zground is zero. Because there is no transformer tap modeled, the driving point voltage is 1 pu voltage.

Knowing that the transformer is connected delta-wye grounded, the zero-sequence impedance of the system network at Bus 3 is equal to only the transformer impedance of:

Z = 0 + j1.0 pu 0transformerΩ

Therefore, the single-line-to-ground fault current is:

I = 3 1 pu V0.010101+ 2.999948 + 0.010101+ 2.999948 + 0 + 1.0

pu

= 3V0.020202 + 6.999896 pu

= 3 pu V6.999925 pu

= - 0.428576 pu A

slg×

j j

j

jj

b g b g b Ω

Ω

Ω

j g

Recalling that the base current for the 480 V system is 120,281.3 A, the single-line-ground fault current at Bus 3 is:

I = I I pu

= 120,281.3 A 0.428576 pu A= 51,549.7 A

slg base sc×

×j

This single-line-to-ground calculation matches the Short Circuit Study’s result of 51,549.68 A shown in Figure 4-1.

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4.4.2 Single-Line-to-Ground Fault Currents at the Secondary of a Transformer with a Grounding Reactor The three-bus example in Section 4.4.1 is now modified by placing a grounding reactor on the transformer’s wye-grounded 480 V side to reduce the single-line-to-ground fault current to no more than 200 A. Note that in Figure 4-1 the single-line-to-ground fault current is greater than the three-phase fault current at Bus 3.

You will recall that:

I = 3VZ + Z + Zslg

1 2 0

Therefore, solving for the zero-sequence impedance yields:

Z = 3VI

- Z - Z0slg

1 2

The goal is to limit Islg to no more than 200 A or 0.00166 pu A on the 120,281.3 A current base. The positive- and negative-sequence impedances at Bus 3 are 0.010101 + j2.999948 pu Ω, respectively. The positive- and negative-sequence impedances are not affected by adding impedance in the ground path.

Ignoring resistance, the zero-sequence impedance is:

Zo = 1803.97 pu - 2( 2.999948) pu = 1797.99 pu

j jj

Ω ΩΩ

The total zero-sequence impedance is

Z = Z + 3Z0 tranformer ground0

Rearranging the equation yields:

3Z = Z - Zground 0 tranformer0

Substituting known values yields:

Z = 1797.99- 1.00

= 1796.993

pu

pu

groundj j

j3

598 99

Ω

Ω= .

But the base ohms in this part of the system is:

Z = 0.48 kV100 MVA

= 0.002304 base

2Ω



Therefore the grounding reactor is:

Z = Z Z pu

= 0.002304 598.99 pu = j1.38

generator base ×

×Ω ΩΩ

PTW allows you to enter grounding reactors to one decimal; therefore PTW rounds 1.38 Ω to 1.4 Ω. Thus, the resulting single-line-to-ground fault current at Bus 3 is slightly less than 200 A (197.71 A), as shown in Figure 4-2.

B1

C1X1 1.0000 puXo 1.0000 pu

T1X1 1.0000 puXo 1.0000 pu

B2

B3

UTILITYX1 1.0000 puXo 1.3333 pu

40094.23 A 3 Ph197.19 A SLG

1.4 Ω

Figure 4-2

The three-phase and single-line-to-ground hand calculations match PTW’s results. PTW automatically takes into account the multiplying of grounding reactance by 3.

The grounding impedance is entered in ohms: PTW automatically places it on a per unit base impedance on the nominal system voltage on the transformer’s grounded side, in this case 480 V. Likewise, if the transformer is connected wye-grounded-delta, the grounding impedance is entered in ohms; however, all per unit calculations would be based on the 13.8 kV nominal system voltage. PTW warns you in the Study Messages dialog box if a grounding impedance is entered on an unacceptable connection (a grounding reactor placed on a delta-transformer connection). Also, PTW allows two grounding reactors for the wye-grounded wye-grounded transformer connection. Each grounding impedance is converted to its respective per unit impedance value based on the bus nominal system voltage on each side of the transformer.

4.4.3 Source Sequence Impedance This example demonstrates how PTW calculates the utility system zero-sequence impedance, given a utility single-phase short circuit capability. Also, the example validates the wye-grounded wye-grounded transformer zero-sequence connection. The example system is shown in Figure 4-3.

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B1

C1

T1

B2

B3

UTILITY

Thevenin X1 2.4184 puThevenin Xo 2.1673 pu51519.40 A SLG

X1 0.4183 puXo 0.1673 pu

X1 1.000 puXo 1.000 pu

X1 1.000 puXo 1.000 pu

Figure 4-3

Given a utility three-phase fault current of 10,000 A at 13.8 kV and a single-line-to-ground fault current of 12,500 A, the equivalent Short Circuit Capacity in MVA is:

SCC = 3 10,000 A 13.8kV= 239,023 kVA

SCC = 12,500 A 13.8 kV3

99,593 kVA

3

slg

Φ × ×

FHG

IKJ

=

Whereas the single-line-to-ground fault current may be greater than the system’s three-phase fault current on an MVA base, the single-line-to-ground short circuit capability is less than the three-phase short circuit capability.

On a 100 MVA base, the three-phase short circuit capability expressed in per unit ohms is:

Z = 100,000 kVA239,023 kVA

= 0.4184 pu

utility

j Ω

Likewise, the utility zero-sequence impedance is:

Z = 3I

Z Z0slg

1 2− −

In order to determine the single-line impedance, you must first express the current as a per unit value.



I =I actual Amperes

I base Amperes pu A

I = 100,000 kVA3 13.8kv

= 4183.7 A

I = 12,500 A4183.7 A

pu A

= - 2.9878 pu A

slgslg

s

base

slg

e jb g

j

Substituting:

Z 3- 2.9878

0.4184 + 0.4184

= 0.1673 pu

0 = −j

j j

j

b gΩ

If the fault occurs at Bus 3 and the transformer connection is wye-grounded wye-grounded, then the zero-sequence impedance is the series of the transformer, cable and utility zero-sequence impedances:

Z = 0.1673+ 1.0 + 1.0 = 2.1673 pu

0Bus 3 j j jj Ω

If the driving point voltage is 1 pu, the single-line-to-ground fault current at Bus 3 is:

Z = pu VZ Z Z

3 1 pu V2.418 2.418 2.1673 per

= - 0.4284 pu A

slg1 2 0

3 1×+ +

=×

+ +j j jj

Ω

In amperes:

I I I

I = - 0.4284 120,281.3 A

= 51,524.9 A

slg pu base

slg

= ×

×j

j

This 51,524.9 A is close to the 51,519.40 A predicted by PTW in Figure 4-3. The slight difference between solutions is because the hand calculations ignored resistance.

4.4.4 Fault with a Generator Source with Unequal Positive-, Negative-, and Zero-Sequence Reactances PTW assumes that the utility system positive- and negative-sequence impedances are equal. In this example, a 100 MVA generator with a j1.0 positive-, j0.8 negative-, and j0.5 zero-sequence reactance is modeled. The generator X/R ratio is 999. The generator is modeled as a Swing Bus Generator with a driving point voltage of 1.0 pu V at an angle of 0°. The transformer grounding reactor examined in Section 4.4.2 is removed and the transformer T1 is reconnected delta-wye grounded as shown in Figure 4-4.

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B1

C1X1 1.0000 puXo 1.0000 pu

T1X1 1.0000 puXo 1.0000 pu

B2

B3

GENERATORX pos 1.0000 puX neg 0.8000 puX zero 0.5000 pu

40093.79 A 3 Ph53065.31 A SLGThevenin X1 3.0000 puThevenin X2 2.8000 puThevenin Xo 1.000 pu

Figure 4-4

The three-phase fault current at Bus 3 is:

Ik1

3.0000pu A

= - 0.3333 puI 120,281.3 AI pu A 120,281.3 A

= - 40,093.77 A

base

3

=

== ×

j

jΦ 0 3333.

Once again the difference between the hand calculated 40,093.77 A and the PTW calculated 40,093.79 A is that the hand calculation ignored resistance.

Note that the Thevenin equivalent negative-sequence impedance at Bus 3 is j2.8 pu Ω, which is slightly less than the positive-sequence value of 3.0. This is because the negative-sequence generator reactance is j0.8 pu Ω, and the cable and transformer negative-sequence impedances are equal to the positive-sequence impedance.

A portion of the single-line-to-ground Short Circuit Study report follows:

F A U L T A N A L Y S I S R E P O R T FAULT TYPE: SLG MODEL INDUCTION MOTOR CONTRIBUTION: YES MODEL TRANSFORMER TAPS: YES MODEL TRANSFORMER PHASE SHIFT: YES ============================================================================== B3 VOLTAGE BASE LL: 480.0 (VOLTS) INI. SYM. RMS FAULT CURRENT: 53065.3 / -120. ( AMPS/DEG ) THEVENIN EQUIVALENT IMPEDANCE: .002 +j 6.800 (PU) THEVENIN IMPEDANCE X/R RATIO: INFINITE SEQUENCE EQUIVALENT IMPEDANCE Z1: .001 +j 3.000 (PU) Z2: .001 +j 2.800 (PU) Z0: .000 +j 1.000 (PU) *****************************************************************************

The positive-, negative-, and zero-sequence impedance values are reported, and the total of the three impedances is 0.002 + j6.8 pu Ω. The fault point X/R ratio is 3400, or as reported in PTW, an infinite X/R ratio. Clearly, the generator negative-sequence



impedance is used to calculate the Thevenin equivalent impedance at the fault point. The Thevenin equivalent zero-sequence impedance at Bus 3 is defined as only the impedance of the wye-grounded side of the transformer, as was the case in Section 4.4.1.

4.4.5 Short Circuit Currents with a Motor Load at Bus 4 In this example a second cable is added to the original configuration in Section 4.4.1. The transformer is connected delta-wye-grounded. A synchronous motor is added to Bus 4, with the positive-, negative-, and zero-sequence subtransient reactances equal to j1.0 pu, j0.8 pu, and j0.5 pu, respectively. The short circuit is at Bus 4, and the results are shown in Figure 4-5.

B1

C1X1 1.0000 puXo 1.0000 pu

T1X1 1.0000 puXo 1.0000 pu

B2

B3

GENERATORX" X1 1.0000 puX" X2 0.8000 puX" Xo 0.5000 pu

C2X1 1.0000 puXo 1.0000 pu

B4

MOTOR

150,351.94 A 3 Ph181,177.31 A SLGThevenin X1 0.8000 puThevenin X2 0.7917 puThevenin Xo 0.4000 pu

Figure 4-5

The positive-sequence Thevenin equivalent impedance at Bus 4 is the sum of the impedance of the Generator, cables C1 and C2, and transformer T1 in parallel with the positive-sequence impedance of the Motor.

Therefore the positive-sequence impedance at Bus 4 is:

Z =1.0 + j1.0 + 1.0 + 1.0 1.01.0 + 1.0 + 1.0 + 1.0 + 1.0

= -4.05.0

pu

= j0.8 pu

1j j j jj j j j j

j

b g×

Ω

Ω

The zero-sequence impedance at Bus 4 is the sum of the transformer T1 and cable C2 impedances in parallel with the motor’s zero-sequence impedance:

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Z =1.0 + 1.0 j0.51.0 + 1.0 + 0.5

= - 1.02.5

pu

= 0.4 pu

0j jj j j

jj

j

b g×

Ω

Ω

The Thevenin equivalent positive- and zero-sequence impedances predicted by PTW match the above hand calculations.

Note that the base current in this part of the system is 120,281.3 A; therefore the calculated three-phase and single-line-to-ground fault current at Bus 4 is 150,352 A and 181,177 A, respectively.

Part of the three-phase Short Circuit Study report is shown:

***************** F A U L T A N A L Y S I S R E P O R T **************** FAULT TYPE: 3PH MODEL INDUCTION MOTOR CONTRIBUTION: YES MODEL TRANSFORMER TAPS: YES MODEL TRANSFORMER PHASE SHIFT: YES B4 VOLTAGE BASE LL: 480.0 (VOLTS) INI. SYM. RMS FAULT CURRENT: 150351.9 / -120. ( AMPS/DEG ) THEVENIN EQUIVALENT IMPEDANCE: .001 +j .800 (PU) THEVENIN IMPEDANCE X/R RATIO: INFINITE B4 ==== INI. SYM. RMS SYSTEM BUS VOLTAGES ( PU / DEG ) ======= ALL BUSES REPORTED AT TIME = .5 CYCLES ---PHASE A--- ---PHASE B--- ---PHASE C--- B1 13800.0 .7500 / 0. .7500 /-120. .7500 / 120. B2 13800.0 .5000 / 0. .5000 /-120. .5000 / 120. B3 480.0 .2500 / -30. .2500 /-150. .2500 / 90. B4 480.0 .0000 /-123. .0000 / 117. .0000 / -3. B4 ==== INI. SYM. RMS SYSTEM BRANCH FLOWS ( AMPS/DEG ) ======= ALL BRANCHES REPORTED AT TIME = .5 CYCLES VOLTS --PHASE A--- ---PHASE B--- ---PHASE C GENERATOR 13800. 1045.9/ -90. 1045.9/ 150. 1045.9/ 30. MOTOR 480. 120281.3/-120. 120281.2/ 120. 120281.3/ 0. B1 B2 13800. 1045.9/ -90. 1045.9/ 150. 1045.9/ 30. B2 B3 13800. 1045.9/ -90. 1045.9/ 150. 1045.9/ 30. B3 B4 480. 30070.6/-120. 30070.6/ 120. 30070.6/ 0.

The Study is conducted with the Study Setup defined to fault a single bus (Bus 4), and all the Bus Voltages and Branch Currents for a fault at Bus 4 are selected.

Using the datablock formats, the branch current flows can be displayed on the one-line diagram in Figure 4-6.



B1

C1

T1

B2

B3

GENERATOR

C2B4

MOTOR150,351.94 A 3 Ph

1045.94 A1045.94 A

30,070.63 A

1045.94 A

Figure 4-6

The single-line-to-ground Short Circuit Study report, for a fault at Bus 4 is shown:

FAULT TYPE: SLG MODEL INDUCTION MOTOR CONTRIBUTION: YES MODEL TRANSFORMER TAPS: YES MODEL TRANSFORMER PHASE SHIFT: YES ============================================================================== B4 VOLTAGE BASE LL: 480.0 (VOLTS) INI. SYM. RMS FAULT CURRENT: 181177.3 / -120. ( AMPS/DEG ) THEVENIN EQUIVALENT IMPEDANCE: .002 +j 1.992 (PU) THEVENIN IMPEDANCE X/R RATIO: INFINITE SEQUENCE EQUIVALENT IMPEDANCE Z1: .001 +j .800 (PU) Z2: .001 +j .792 (PU) Z0: .000 +j .400 (PU) B4 ==== INI. SYM. RMS SYSTEM BUS VOLTAGES ( PU / DEG ) ======= ALL BUSES REPORTED AT TIME = .5 CYCLES ---PHASE A--- ---PHASE B--- ---PHASE C--- B1 13800.0 .8609 / 5. .9833 /-120. .8607 / 115. B2 13800.0 .7237 / 13. .9874 /-120. .7236 / 107. B3 480.0 .3054 / -30. .9108 /-139. .9108 / 79. B4 480.0 .0000 /-123. .9135 /-139. .9135 / 79. B4 ==== INI. SYM. RMS SYSTEM BRANCH FLOWS ( AMPS/DEG ) ======= ALL BRANCHES REPORTED AT TIME = .5 CYCLES VOLTS --PHASE A--- ---PHASE B--- ---PHASE C GENERATOR 13800. 742.9/-121. 17.5/ -30. 742.9/ 61. MOTOR 480. 144438.2/-120. 501.2/-180. 505.3/ -60. B1 B2 13800. 742.9/-121. 17.5/ -30. 742.9/ 61. B2 B3 13800. 742.9/-121. 17.5/ -30. 742.9/ 61. B3 B4 480. 36739.1/-120. 500.1/ 0. 506.4/ 120.

The single-line-to-ground report shows that the voltage on Phase A of the faulted bus is 0 pu V, but the voltage on Phases B and C are 0.9 pu V. The phase current on phase A from the Utility is 3,6739 A and the motor contribution is 144,438 A. The total fault current into Bus 4 is therefore the sum of these two currents into the bus, or 181,177 A. Because the synchronous motor stator windings are connected wye-grounded configuration at Bus 4, there are currents flowing in all three phases.

You can run the same Study with a short circuit at Bus 4 again; however this time the Study Setup has been modified to ignore motor contributions.

The single-line-to-ground Short Circuit Study report ignoring motor contribution is shown:

3/26/2006


***************** F A U L T A N A L Y S I S R E P O R T **************** FAULT TYPE: SLG MODEL INDUCTION MOTOR CONTRIBUTION: NO MODEL TRANSFORMER TAPS: YES MODEL TRANSFORMER PHASE SHIFT: YES ============================================================================== B4 VOLTAGE BASE LL: 480.0 (VOLTS) INI. SYM. RMS FAULT CURRENT: 36821.3 / -120. ( AMPS/DEG ) THEVENIN EQUIVALENT IMPEDANCE: .002 +j 9.800 (PU) THEVENIN IMPEDANCE X/R RATIO: INFINITE SEQUENCE EQUIVALENT IMPEDANCE Z1: .001 +j 4.000 (PU) Z2: .001 +j 3.800 (PU) Z0: .000 +j 2.000 (PU) B4 ==== INI. SYM. RMS SYSTEM BRANCH FLOWS ( AMPS/DEG ) ======= ALL BRANCHES REPORTED AT TIME = .5 CYCLES VOLTS --PHASE A--- ---PHASE B--- ---PHASE C GENERATOR 13800. 739.4/-120. .0/ 59. 739.4/ 60. B1 B2 13800. 739.4/-120. .0/ 78. 739.4/ 60. B2 B3 13800. 739.4/-120. .0/ 78. 739.4/ 60. B3 B4 480. 36821.3/-120. .0/-120. .0/-120. *****************************************************************************

Because there is no wye-grounded motor load modeled, there is a single-line-to-ground fault current in the branch from Bus 3 to Bus 4 of 36,821 A in Phase A (the faulted phase); there are no fault currents in Phase B or Phase C. Likewise, on the delta side of the transformer, there is fault current flowing in Phase A and Phase C only. The associated sequence currents for this example are shown:

B4 ==== INI. RMS SYSTEM BRANCH FLOWS ( AMPS/DEG ) ======= ALL BRANCHES REPORTED AT TIME = .0 CYCLES VOLTS -POS SEQ- -NEG SEQ- -ZER SEQ- GENERATOR 13800. 426.9/ 90. 426.9/ 30. .0/ 45. B1 B2 13800. 426.9/ -90. 426.9/-150. .0/ 45. B2 B3 13800. 426.9/ -90. 426.9/-150. .0/ 45. B3 B4 480. 12273.8/-120. 12273.8/-120. 12273.8/-120. *****************************************************************************

On the branch from Bus 3 to Bus 4 the positive-, negative-, and zero-sequence currents are equal in magnitude and angle. The Phase A current is the sum of these three sequence values, or 36,821 A. Because the transformer is delta connected to the primary connection, there is no zero-sequence current flow on the 13.8 kV side of the system.

4.4.6 Fault Duty Contribution to a Faulted Bus This example examines four motor branches to determine their fault duty contribution to the faulted bus, Bus 2, and to demonstrate how PTW models motor impedances. The one-line diagram for this example is shown in Figure 4-7.

In each of the motor branches, the total motor load is 100 kVA. In the branch from Bus 2 to Bus 3, motor M1 is rated 100 kVA, and its associated is 0.17 pu Ω on its own base of 100 kVA and 480 V. In the second branch, motor M2 is modeled as four individual 25 kVA motors, with each of the four motors’ equal to 0.17 pu Ω on its base of 25 kVA and 480 V. Motor M3 is also modeled as four 25 kVA motors, but its ANSI contribution is entered as 0.68 pu Ω (4 × 0.17 pu Ω) on a 100 kVA base at 480 V. Finally, the fourth branch is nearly identical to the first branch; it is a single 100 kVA motor on the bus. However, note that motor M4 is rated as 0.17 pu Ω at 440 V, operating on a system at 480 V.

′′Xd

′′Xd



B1

FEEDER

C1

B3

C2

B4

M1 M2

UTILITY

C3

B5

M3

24,686.32 AC4

B6

M4

22,768.38 A

22,768.35 A

698.55 A 698.55 A 698.55 A 835.20 A

B2

Size 100.0 kVAXd" 0.1700 pu No. Mtrs 1100 Mtr kVA Base



Size 100.0 kVAXd" 0.1700 puNo. Mtrs 1100 Mtr kVA Base

Figure 4-7

It is important to note that the fault current contributions from the first three branches into the faulted bus are identical. Because each of these cases has 100 kVA of connected motor load in the branch, the motor fault duty contribution is identical. (Be aware that when you enter the data for motors M1 and M2, you need to enter only the Number of Motors and the motor Rated Size in the first subview of the Component Editor.) For both motors, the motor base kVA is the same value as the rated size. PTW correctly models the total Thevenin equivalent subtransient reactance for both cases; for M1 it models the one motor as a single motor load object and for M2 it models the four motors as a single motor load object.

Modeling motor M3 requires further explanation. The motor rated size is 25 kVA and there are four motors modeled in the single motor load object. However, when the data was entered in the ANSI Contribution subview, the motor contribution was lumped as 100 kVA. For PTW to correctly calculate the lumped impedance of these four 25 kVA motors, the motor must be multiplied by four and the product entered in the field. This is noted on the datablock below motor M3. If the motor data for M3 is entered correctly, then the results for motors M1, M2 and M3 will be identical.

′′Xd ′′Xd

In the last branch from Bus 2 to Bus 4, a single 100 kVA motor is modeled but the rated voltage of the motor is 440 V. If the motor is rated 0.17 pu Ω on its own base of 100 kVA and 440 V, then its impedance on the system base 480 V is:

′′FHG

IKJFHG

IKJX = X kV

kVkVAkVAd d

motor

base

2base

motormotor

Substituting:

′′ FHG

IKJFHG

IKJX = 0.17 0.44 kV

0.48 kV100 kVA100 kVA

= 0.143 pu

d

2

Ω

This fault duty contribution is limited by the impedance of the 50 ft of cable from Bus 2 to Bus 6. The cable is #2 AWG THHN with a copper conductor in non-metallic conduct

3/26/2006


and has a published impedance of 0.202 + j0.0467 Ω per 1000 ft. Therefore, on a 100 kVA base (machine base) the cable impedance is:

Z = 0.202 + 0.0467 1000 ft

50 ft

= 0.0101+ 0.002335

cablej

j

Ω

Ω

FHG

IKJ

On a 100 kVA, 480 V base, the base impedance is:

Z = kV baseMVA

= 0.480.100

= 2.304

base

2

base2

Ω

Therefore, the cable impedance on the system base of 100 kVA is:

Z = 0.00438 + 0.000998 pu cable j Ω

Adding the motor and cable impedances yields:

Z = 0.00438 + 0.000998 + 0.0 pu = 0.00438 + 0.14385= 0.1439 pu

branch j jj

b g b + 0143. Ω

Ω

g

The fault duty contribution to Bus 2 is:

I = VZ

= 10.1439

pu

= 6.949 pu

sc

FHG

IKJ Ω

Ω



However the Ibase is:

I = 100 kVA1.732 0.48 kV

= 120.28 A

base ×

Therefore the fault current contribution for Motor 4 is:

I = 120.28 A 6.949 pu A= 835.83 A

sc ×

PTW predicts the fault duty contribution for the motor with a base voltage that is different than the system base as 835.20 A, which is nearly identical to the hand calculation of 835.83 A.

4.4.7 Transformer Off-Nominal Voltages and Transformer Taps One of the most misunderstood issues in Short Circuit Studies is the impact of transformer off-nominal voltages and transformer taps on the per unit impedances in the system. You can easily forget that the basic premise associated with per unit calculations is that the ratio of the transformer voltages to the chosen bus nominal system voltages must be consistent. Following are five cases that demonstrate how PTW removes much of the burden of defining consistent bus nominal system voltages.

Case 1

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The one-line diagram in Figure 4-8 has three-phase transformer rated voltages that match the bus nominal system voltages, and there are no taps. The transformer reactance is rated 6% on a base of 13.8 kV and 10,000 kVA. PTW places all per unit values on a 100 MVA system base.

B1Bus System Nominal 13,800.0 VPre Fault Driving Point Voltage 1.00 pu VBus Fault Current 20,918.49 A

B2Bus System Nominal 4160.0 VPre Fault Driving Point Voltage 1.00 pu VBus Fault Current 17,348.27 A

T1Pri Rated Voltage 13,800.0 VSec Rated Voltage 4160.0 VLeakage Impedance 0.6000 puPri Tap 0.00 %Sec Tap 0.00 %

GENERATORX1 0.2000 pu

Figure 4-8

The Generator equivalent impedance is j0.2 pu and the transformer per unit impedance is j0.6 pu. The total Thevenin equivalent impedance at Bus 2 is the sum for the generator and transformer impedances, or j0.8 pu. Therefore the three-phase fault current is:

I = 10.8

pu A

= - 1.25 pu A

sc jj

However the base current at 4.16 kV is:


I = 100,000 kVA3 4.16 kV

= 13,878.6 AI = I I pu A

= 13,878.6 A 1.25 pu A= - 17,348.27 A

base

sc base

×

××

j

Case 2


You can substitute a transformer rated 13.2 kV - 3.979 kV, as shown in Figure 4-9. If the generator exciter is set to produce 13.8 kV at the transformer rated 13.2 kV bus, then the transformer on its low voltage side will produce 4.16 kV, which is greater than 3.979 kV. The transformer turns ratio is 3.317 and the ratio of nominal system voltages from buses B1 to B2 is also 3.317.



T1Pri Rated Voltage 13,200.0 VSec Rated Voltage 3979.0 VLeakage Impedance 0.6000 puPri Tap 0.00 %Sec Tap 0.00 %


Figure 4-9

The transformer is 6% on its own base; however this value must be modified by the transformer voltage to nominal system voltage ratio squared. The system base is 100 MVA. Substituting:

Z = ZkVkV

kVkV

= 0.06 pu 13.2 kV13.8 kV

100,000 kVA10,000 kVA

= 0.06 3.9794.16

100,000 kVA10,000 kVA

= 0.54896 pu

new givengiven

new

2new

given

2

2

×FHG

IKJFHG

IKJ

× FHGIKJFHG

IKJ

× FHGIKJFHG

IKJ

j

j

j Ω


The Thevenin equivalent impedance at Bus 2 is the sum of the generator and transformer impedance as shown:

Z 0.2 + j0.54896 pu = 0.74896 pu

I = 1.0 pu V0.74896 pu

= - 1.335 pu A

=

sc

jj

jj

ΩΩ

Ω

Remembering that Ibase is 13,878.6 A for a system base of 4160 V, the fault current at Bus 2 is:

I = 13,878.6 A - 1.335 pu A= - 18,531 A

sc × jj

To properly model off-nominal voltages, you must select the Transformer Taps checkbox in the Study Setup dialog box. PTW actually models transformer off-nominal voltages using two fictitious taps. The first fictitious tap is on the primary side of the transformer to model the difference between the 13.8 kV bus nominal and the 13.2 kV transformer rated voltage. The second fictitious tap is on the secondary side of the transformer to model the 3.979 kV secondary transformer rated voltage and the 4.16 kV nominal system voltage. Thus when a lower voltage transformer is placed in the system with the same transformer ratio as the bus nominal system voltage ratio, the fault current increases over the no-tap or flat-tap case.

Case 3

3/26/2006

You can also substitute a transformer rated 13.2 kV - 4.16 kV, as shown in Figure 4-10. Since the generator produces 13.8 kV without a primary tap on the transformer, the secondary voltage is 4.35 kV. To limit the secondary side voltage, the transformer is tapped at 13.8 kV. This is applied in PTW as a +4.54% primary tap. A positive primary transformer tap reduces the secondary voltage.

B1Bus System Nominal 13,800.0 VPre Fault Driving Point Voltage 1.00 puBus Fault Current 20,918.49 A

B2Bus System Nominal 4160.0 VPre Fault Driving Point Voltage 1.00 puBus Fault Current 17,348.73 A


Pri Rated Voltage 13,200.0 VSec Rated Voltage 4160.0 VLeakage Impedance 0.6000 puPri Tap 4.54 %Sec Tap 0.00 %

T1

Figure 4-10

It should be noted that the combination of the lower transformer primary voltage rating and the positive tap setting work together to maintain the system voltage at Bus 2 at 4.16 kV. While the lower voltage transformer increases the fault current (Figure 4-9), adding the positive tap value reduces the fault current. The total fault current at Bus 2 is identical to that in Figure 4-8, or 17,348 A.


Case 4 A 13.8 kV-4.16 kV transformer is placed in a system with bus nominal voltages of 13.8 - 4.16 kV, as shown in Figure 4-11. However because of heavy loading, the transformer is placed in service with a -2.5% primary tap. Under loaded conditions, the nominal system voltage with the -2.5% primary tap is 4.16 kV; however under no-load conditions, the pre-fault driving point voltage at Bus 2 is 1.0256 pu higher than the nominal system voltage.

B1Bus System Nominal 13,800.0 VPre Fault Driving Point Voltage 1.00 puBus Fault Current 20,918.49 A

B2Bus System Nominal 4160.0 VPre Fault Driving Point Voltage 1.03 puBus Fault Current 17,565.02 A

T1


Pri Rated Voltage 13,800.0 VSec Rated Voltage 4160.0 VLeakage Impedance 0.6000 puPri Tap -2.50 %Sec Tap 0.00 %

Figure 4-11

The Thevenin equivalent impedance is calculated as:

Z = 0.2 11+ Tap

+ 0.6 pu

= j0.2 11- 0.025

+ j 0.6 pu

= j(0.2 1.0519) + j0.6 pu = 0.81034 pu

I = 1.0256 pu V0.81034

pu

= - 1.26558 pu A

2

2

sc

j j

j

jj

×FHG

IKJ

× FHGIKJ

×

Ω

Ω

ΩΩ

Ω

Remembering that the base current at 4160 V is 13878.6 A, then the fault current is:

I = 13,878.6 A - 1.26558 pu A= - 17,565 A

sc × jj

It is significant to note that the pre-fault voltage at Bus 2 is 1.0256 pu above the nominal system voltage of 4160 V. Proper per unit methods should be applied if hand calculations are used.

Case 5 In this final case, the nominal system voltage at Bus 2 is set to 4.266 kV or 1.0256 higher voltage than in Case 4, as shown in Figure 4-12.





T1Pri Rated Voltage 13,800.0 VSec Rated Voltage 4160.0 VLeakage Impedance 0.6000 puPri Tap -2.50 %Sec Tap 0.00 %


Figure 4-12

The fault current at Bus 2 remains 17,565 A. The pre-fault voltage is now 1.0 pu V. The 6% transformer reactance must be placed on the system base, which is 13.8 kV. With the -2.5% primary tap, the Thevenin equivalent impedance at Bus 2 is:

Z = 0.2 + 0.06 13.455 kV13.8 kV

100,000 kVA10,000 kVA

= 0.2 + 0.5704 = 0.7704 pu

2

j j

j jj

× FHGIKJFHG

IKJ

Ω

Therefore, the fault current at Bus 2 is:

I = 1.0 pu V0.7704 pu A

= - 1.2980 pu A

sc jj

The base current at this voltage is:

I = 100,000 kVA3 4.2665 kV

= 13532.2 AI = I I pu A

= 13532.2 A 1.2980 pu A= - 17,565 A

base

sc base

×

××

j

Carefully study the differences between Cases 4 and 5. The same transformer with the same ratings and taps is analyzed in the each case. The only difference is the system voltage selected at the faulted bus. In both cases, the fault current is the same. The impedances were calculated differently because the base voltages were different. The fault current in both cases is the same.

4.4.8 Modeling Transformer Connections This example demonstrates the importance of modeling transformer primary and secondary connections properly. Although the example focuses on a two-winding transformer, similar care should be used for the three-winding transformers.

In the following one-line diagram, two transformers are connected to Bus 2.

3/26/2006



B1

C1

B2

T1

T2

B4

UTILITY

B3

100.0 MVA 3 Ph30.0 MVA SLG

4183.70 A 3 Ph3765.33 A SLG

6609.78 A 3 Ph8007.95 A SLG

3803.54 A 3 Ph3454.58 A SLG

6609.78 A 3 Ph8007.95 A SLG

Figure 4-13

If the utility and cable impedance each are 1 pu, then the total Thevenin equivalent positive-sequence impedance at Bus 3 and at Bus 4 is 3 pu.

Transformer 1 is connected delta-wye grounded, with the secondary side of Transformer 1 defined as attached to Bus 3. The Thevenin equivalent zero-sequence impedance at Bus 3 is then 1 pu. The short circuit currents at Bus 3 are reported by PTW as 6610 A (three-phase) and 8008 A (single-phase-to-ground), as noted in Figure 4-13. This is an expected solution. The Transformer 1 connections were set as defaulted with the primary as delta, and the secondary as wye-grounded as noted on the Component Editor screen shown following in Figure 4-14.

Figure 4-14


The Transformer 2 is connected between Bus 2 and Bus 4, as shown in Figure 4-13. Unlike Transformer 1, the Transformer 2 connections were changed in the Component Editor, as shown following in Figure 4-15.

Figure 4-15

PTW defines transformer connections as referred to either Node 1 or Node 2. Node 1 is defined on the Drawing as on the top of the symbol, whereas Node 2 is always defined on the Drawing as located at the bottom of the symbol. It makes no difference how you connect the transformer to the buses on the Drawing– the Node 1 connection is always referenced at the top of the symbol.

Thus, in Figure 4-13 the Node 2 (bottom) connection for Transformer 1 is Bus 3, and in the Component Editor view shown in Figure 4-14, the Node 2 connection is wye-grounded.

We know that the proper transformer connection for Transformer 2’s connection at Bus 4 is wye-grounded. Knowing that the Node 1 connection is always on the top of the symbol when viewed on the one-line diagram, or as the primary connection when viewed in the Component Editor, the connection must be set as wye-grounded. Likewise, the Node 2 (bottom) connection for Transformer 2 must be delta. Compare the connections and transformer voltages in the two Component Editor screens shown in Figure 4-14 and Figure 4-15.

It is imperative that when you model single-phase-to-ground fault currents, you understand how PTW treats the primary and secondary connections.

4.4.9 Example from Plant The following figure is a one-line diagram for the Plant project. The Plant project is included on the PTW diskettes.

3/26/2006

Short Circuit Study DAPPER 4-43 ED

ISO

N

UTIL

ITY

BUS

TX A

PRI

TX A

003-

HV S

WG

R

C1

C2

C3

C4

SYN

ASY

N B

TX B

PRI

BUS

4

005-

TXD

PRI

006-

TX3

PRI

007-

TX E

PRI

TX B

DS S

WG

1 BU

S 8

C5

C6

GEN

1G

EN 2

TX C

PRI

BUS

9H

VAC

BUS

TX C

DSB

1 B

US 1

4

C14

C16

C17

BUS

15BU

S 16

BUS

17

C15

018-

RA

CM

P HV

AC

TX 6

TX3

SEC

BUS

11TX

3 TE

R BU

S 12

C8

DS

SWG

3 BU

S 20

C9

L1M

20

GEN

3

021-

TX F

PRI

C7

DS

SWG

2 BU

S 13

L2M

13

TX F

022-

DSB

2

C12

023-

MTR

23

M23

-BM

23-A

TX E

BLDG

115

SER

V

C10

C11

026-

TX G

PR

I02

5-M

TR 2

5

TX G

DSB

3 B

US 2

7

C13

L5

BUS

28 A

M25

TX D

029-

TX D

SEC

C19

TM -1

SWBD

-1

PLN

- 17

H1A

PLN

16

H2A

MCC

15

- 1A

PNL

18 R

A

C18

H3A

BU

S 19

PNL

19 H

3A

R1 CB1

R2 CB2

CB6

CB7

R8

CB8

R9

CB9

R7R6

CB5

R5

R4 CB4

R3 CB3

R10

CB1

0

F1

F2 MC

P#10

LVP1

LVP2

LVP3

MCC

B1M

CCB2

PCB1

PCB2

R11

PCB3

F4

F3 MO

/L#2

5

SW1

LVP4

DEM

ONS

TRAT

ION

PRO

JEC

T FO

R PO

WER

*TO

OLS

FO

R W

INDO

WS

CAP

1

M28

#1

& 2

C20 LV

P5

MCP

#28B

-1

MO

/L#2

8B-1

M28

#3

BUS

28 B

MCP

#28B

-2

MO

/L#2

8B-2

M28

#4

CB12

TX H

CMP

CTR

COM

PUTE

RS

F5

LVP6

OFF

ICE

CO

MPL

EX A

REA

IND

USTR

IAL

COM

PLEX

ARE

A



The following figure shows a portion of the Plant project including Short Circuit Study results.

BUS 28 A

20786.14 A 3 PH17923.31 A SLG

LVP4

M28 #1 & 2

C13

DSB 3 BUS 27

24713.14 A 3 PH23955.28 A SLG

TX G

L5 C20

F4

026-TX G PRI

9694.91 A 3 PH9198.14 A SLG

LVP5

BUS 28 B

18508.62 A 3 PH15548.13 A SLG

C10

BLDG 115 SERV

10618.85 A 3 PH10557.23 A SLG

MCP#28B-1 MCP#28B-2

MO/L#28B-1

M28 #3

MO/L#28B-2

M28 #4

C11

025-MTR 2510060.67 A 3 PH9466.09 A SLG SW1

F3

MO/L#25

M25

SHORT CIRCUIT STUDY

FAULT ALL BRANCHES


The Short Circuit Report includes all the pre-fault voltages and associated voltage angles, given that some transformers within the system have voltage taps set. Some of the data for the buses in the above one-line diagram are shown in the following report.

************* P R E - F A U L T V O L T A G E P R O F I L E ************** BUS# NAME BASE VOLTS PU VOLTS ANGLE (D) UTILITY BUS 69000.00 1.0000 0. TX A PRI 69000.00 1.0000 0. 003-HV SWGR 13800.00 1.0256 -30. 007-TX E PRI 13800.00 1.0256 -30. BLDG 115 SERV 4160.00 1.0519 -60. 026-TX G PRI 4160.00 1.0519 -60. 025-MTR 25 4160.00 1.0519 -60. DSB 3 BUS 27 480.00 1.0789 -90. BUS 28 A 480.00 1.0789 -90. 029-TX D SEC 4160.00 1.0519 -60. BUS 28 B 480.00 1.0789 -90.

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The comprehensive Short Circuit Study (three-phase) Report for a short circuit at DSB 3 or Bus 27 is listed below.

***************** F A U L T A N A L Y S I S R E P O R T **************** FAULT TYPE: 3PH MODEL INDUCTION MOTOR CONTRIBUTION: YES MODEL TRANSFORMER TAPS: YES MODEL TRANSFORMER PHASE SHIFT: YES ============================================================================== DSB 3 BUS 27 VOLTAGE BASE LL: 480.0 (VOLTS) INI. SYM. RMS FAULT CURRENT: 24713.1 / -170. ( AMPS/DEG ) THEVENIN EQUIVALENT IMPEDANCE: 0.883 +j 5.176 (PU) THEVENIN IMPEDANCE X/R RATIO: 5.865 ASYM RMS INTERRUPTING AMPS 1/2 CYCLES 3 CYCLES 5 CYCLES 8 CYCLES 32080.4 24753.0 24713.7 24713.1 INI. SYM. RMS FAULTED BUS VOLTAGES ( PU / DEG ) AT TIME = 0.5 CYCLES ---PHASE A--- ---PHASE B--- ---PHASE C--- 0.0000 / -82.9 0.0000 / 157.1 0.0000 / 37.1 INI. SYM. RMS FAULTED CURRENT ( AMPS / DEG ) AT TIME = 0.5 CYCLES ---PHASE A--- ---PHASE B--- ---PHASE C--- 24713.1 /-170.3 24713.1 / 69.7 24713.1 / -50.3

The comprehensive Short Circuit Study (single-line-to-ground) Report for a short circuit at DSB 3 or Bus 27 is listed below.

***************** F A U L T A N A L Y S I S R E P O R T **************** FAULT TYPE: SLG MODEL INDUCTION MOTOR CONTRIBUTION: YES MODEL TRANSFORMER TAPS: YES MODEL TRANSFORMER PHASE SHIFT: YES ============================================================================== DSB 3 BUS 27 VOLTAGE BASE LL: 480.0 (VOLTS) INI. SYM. RMS FAULT CURRENT: 23955.3 / -170. ( AMPS/DEG ) THEVENIN EQUIVALENT IMPEDANCE: 2.765 +j 16.015 (PU) THEVENIN IMPEDANCE X/R RATIO: 5.791 SEQUENCE EQUIVALENT IMPEDANCE Z1: 0.883 +j 5.176 (PU) Z2: 0.883 +j 5.176 (PU) Z0: 1.000 +j 5.662 (PU) ASYM RMS INTERRUPTING AMPS 1/2 CYCLES 3 CYCLES 5 CYCLES 8 CYCLES 31011.3 23990.9 23955.7 23955.3 INI. SYM. RMS FAULTED BUS VOLTAGES ( PU / DEG ) AT TIME = 0.5 CYCLES ---PHASE A--- ---PHASE B--- ---PHASE C--- 0.0000 / -76.0 1.0940 / 148.4 1.0977 / 31.4 INI. SYM. RMS FAULTED CURRENT ( AMPS / DEG ) AT TIME = 0.5 CYCLES ---PHASE A--- ---PHASE B--- ---PHASE C--- 23955.3 /-170.2 0.0 /-170.2 0.0 /-170.2

The Short Circuit Study Report for all buses faulted is shown below. A summary three-phase and single-line-ground report is also included.

***************** F A U L T A N A L Y S I S S U M M A R Y *************** ------------------------------------------------------------------------------- ------------------------------------------------------------------------------- BUS NAME VOLTAGE AVAILABLE FAULT CURRENT L-L 3 PHASE X/R LINE/GRND X/R 003-HV SWGR 13800. 10034.8 13.6 12796.2 12.3 TX B PRI BUS 4 13800. 9830.7 7.0 12279.6 5.3 005-TXD PRI 13800. 9891.7 10.2 12452.9 8.5 006-TX3 PRI 13800. 9966.3 11.1 12526.3 8.7 007-TX E PRI 13800. 9970.8 11.5 12628.7 9.8 DS SWG1 BUS 8 4160. 6072.8 8.4 6640.8 8.6 TX C PRI BUS 9 4160. 6026.0 7.5 6559.7 7.4 HVAC BUS 4160. 5701.7 3.6 5999.3 3.0 TX3 SEC BUS 11 4160. 18318.1 9.4 18976.2 6.9 DS SWG2 BUS 13 4160. 18303.7 9.3 18946.6 6.8 TX3 TER BUS 12 4160. 20317.1 7.3 19500.2 5.1 DS SWG3 BUS 20 4160. 20307.9 7.3 19468.0 5.0



021-TX F PRI 4160. 17384.8 2.8 15443.0 2.2 BLDG 115 SERV 4160. 10618.9 8.9 10557.2 8.1 026-TX G PRI 4160. 9694.9 3.9 9198.1 3.3 025-MTR 25 4160. 10060.7 7.4 9466.1 6.3 022-DSB 2 480. 45790.9 6.4 34099.1 6.0 023-MTR 23 480. 47063.0 7.8 31374.4 5.6 DSB 3 BUS 27 480. 24713.1 5.9 23955.3 5.8 BUS 28 A 480. 20786.1 4.3 17923.3 4.0 DSB 1 BUS 14 480. 16327.5 6.3 17986.3 6.1 BUS 15 480. 15492.8 5.4 16457.9 5.2 018-RA 480. 11644.5 2.1 10984.0 1.9 BUS 16 480. 13770.7 2.6 13918.5 2.3 BUS 17 480. 6760.2 0.8 5873.2 0.7 H3A BUS 19 480. 11572.0 1.7 10988.3 1.5 029-TX D SEC 4160. 9921.5 7.3 9555.9 6.7 BUS 28 B 480. 18508.6 2.2 15548.1 1.8 UTILITY BUS 69000. 8063.3 26.9 2547.6 14.7 TX A PRI 69000. 2281.5 19.6 752.8 11.7 CMP CTR 208. 23552.7 5.3 25176.4 5.2 *********************** FAULT ANALYSIS REPORT COMPLETED ***********************

3/26/2006

Bib A-1

A Bibliography


Bibliography A

Anderson, Paul M. Analysis of Faulted Power Systems. New Jersey: IEEE Press, 1973.

Anthony, Michael A. Electric Power System Protection and Coordination. New York: McGraw-Hill, 1995.

Beeman, Donald, ed. Industrial Powers Systems Handbook. New York: McGraw-Hill, 1955.

Earley, Mark W., ed. National Electric Code Handbook. 7th ed.

Elgard, Olle I. Electric Energy Systems Theory: An Introduction. New York: McGraw-Hill, 1971.

Fitzgerald, A.E., Charles Kingsley, Jr., and Alexander Kusko. Electric Machinery. 3rd ed. New York: McGraw-Hill, 1971.

General Electric Company. “Application Engineering Information." Catalog Issue EESG II-AP-1, Dec. 1975.

---. Transformer Connections. Massachusetts: General Electric Company, 1970.

IEEE Brown Book: IEEE Std 399-1990. New York: IEEE, 1990.

IEEE Red Book: IEEE Std 141-1993. New York: IEEE, 1993.

McGraw-Edison Power Systems Division. Distribution-System Protection Manual. Pennsylvania: McGraw Edison, Bulletin # 71022.

National Electric Code, 1996 ed. Authored by Committee. Quincy, MA: National Fire Protection Association, 1996.

Skilling, Hugh Hildreth. Electrical Engineering Circuits. 2nd ed. New York: John Wiley & Sons, Inc., 1967.

Stagg, Glenn W., and Ahmed H. El-Abiad. Computer Methods in Power System Analysis. New York: McGraw-Hill, 1968.

SKM Power*Tools for Windows, Version 3.3

A-2 Reference Manual

Stevenson, William D. Elements of Power System Analysis. 3rd ed. New York: McGraw-Hill, 1975.

Stigant, S. Austin, and A.C. Franklin. The J & P Transformer Book. 10th ed. London: Newnes-Butterworths, 1973.

Wagner, C. F., and R.D. Evans. Symmetrical Components. New York: McGraw-Hill, 1961.

Westinghouse Electric Corporation. Electrical Transmission and Distribution Reference Book. East Pittsburgh, PA: Westinghouse, 1964.

3/26/2006

Index DAPPER i

Index A Admittance Matrix

definition of, 4-2 Ampacity, 2-3 ANSI, 4-1 Asymmetrical Peak Fault Current. See Faults Asymmetrical rms Fault Current. See Faults

B Balanced Faults. See Faults

C Co-Generation

and the Load Flow Study, 3-10 Coincident Demand, 1-4 Connected Load, 1-3 Constant Current. See Loads Constant Impedance. See Loads Constant kVA. See Loads Continuous Load, 1-4

D Demand Factor, 1-3 Demand Load Categories, 1-6 Demand Load Library

and the Demand Load Study, 1-6 Demand Load Study

and the Demand Load Library, 1-6 before running the Study, 1-6 error messages, 1-11 examples, 1-12

loads with different power factors, 1-17 motor design load, 1-13 motor starting, 1-18 motors assigned to a motor control center, 1-14 multiple loops in a system, 1-19 multiple motors at a bus, 1-13 multiple motors on a single motor component, 1-18

non-coincident demand, 1-12 Plant project, 1-20 single phase non-motor loads in a panel schedule, 1-

15 load categories, 1-6 loads. See Loads looped systems, 1-2, 1-5 methodology, 1-3 radial systems, 1-2 running the Study, 1-9 service factor (SF), 1-9 special considerations, 1-11 Study options, 1-10 terms and concepts, 1-3

Demand Load Type. See Loads Demand Load Value, 1-3 Design Load Value, 1-3 Diversity Factor, 1-4 Diversity Loads, 3-11 Double Line-to-Ground Fault. See Faults Dry Type Transformer, 2-4

E Energy Audit Type. See Loads Engineering Methodology. See Methodology Error Messages

Demand Load Study, 1-11 Load Flow Study, 3-12

F Fault Current. See Faults Faults

balanced, 4-3 definition of, 4-3 Thevenin Equivalent Circuit, 4-3

fault current asymmetrical peak, 4-10 asymmetrical rms, 4-11 interrupting, 4-9 momentary, 4-9 steady state, 4-11

unbalanced, 4-5 definition of, 4-3 double line-to-ground, 4-7



grounding impedance in, 4-7 line-to-line, 4-7 per unit notation use in, 4-7 positive-, negative-, and zero-sequence modeling in,

4-7 single-line-to-ground, 4-5

Feeder Sizing parallel cables, 2-3

Feeder Sizing. See Sizing Study

G Generators

PV and PQ, 3-10 Grounding Impedance. See Impedance

I IEC, 4-1 Impedance

equivalent, 3-4 grounding, 4-7

Interrupting Fault Current. See Faults

K Kirchoff's Current Law, 3-3, 3-4, 4-3

L Lead/Lag, 3-10 Line-to-Line Fault. See Faults Load Categories, 1-7 Load Flow Study

bus types, 3-3 co-generation, 3-10 definition of, 3-2 error messages, 3-12 examples, 3-12

load specifications, 3-15 modeling transformer losses, 3-14 net branch diversity load, 3-17 Plant project, 3-19 voltage drop and power losses, 3-12

load types, 3-4 methodology, 3-2 running the Study, 3-7 solution process, 3-4

special bus loads, 3-4 steady state load flow equation, 3-3 Study options, 3-7 transformer taps, 3-4 utility equivalent impedance, 3-4 voltage drop, 3-5

Loads, 1-3 constant current, 1-7, 1-8 constant impedance, 1-7, 1-8 constant kVA, 1-7, 1-8 diversity, 3-11 largest motor design factor, 1-8, 1-10 load categories, 1-6, 1-7 load characteristics, 1-6 long continuous load factor (LCL), 1-3, 1-7 motor loads, 1-6, 1-8 non-motor loads, 1-6, 1-7

Demand Load types, 1-7 Energy Audit types, 1-7

remaining motors design factor, 1-8, 1-10 Long Continuous Load Factor (LCL). See Loads Looped Systems, 1-2, 1-5, 1-19, 3-11

M Methodology

Demand Load Study, 1-3, 1-6 Load Flow Study, 3-2 Short Circuit Study, 4-2 Sizing Study, 2-3, 2-4

Momentary Fault Current. See Faults Motor Loads. See Loads Motor Service Factor (SF). See Demand Load Study

N NEC, 1-2, 1-4, 1-5, 1-7, 1-9, 2-2, 2-3, 3-5 Negative-Sequence Modeling. See Sequence Modeling Net Branch Diversity, 1-4 Non-Coincident Demand, 1-4 Non-Motor Loads. See Loads Norton Equivalent Circuit, 4-4 Norton Equivalent Diagram, 4-4

O Off-Nominal Voltage Modeling. See Transformers Ohm's Law, 3-2, 4-3 Oil/Air Cooled Transformer, 2-4 Oil/Air/Forced Air Transformer, 2-4

3/26/2006

Index DAPPER iii

Oil/Forced Air Transformer, 2-4

P Panel Schedule, 1-15 Per Unit Notation

usage in unbalanced faults, 4-7 Phase Shift. See Transformers Positive-Sequence Modeling. See Sequence Modeling Power Factor, 3-6

R Radial Systems, 1-2 Reactive Power, 3-10

S Sequence Modeling

positive-, negative-, and zero-, 4-7 Service Factor (SF). See Demand Load Study Short Circuit Study. see also Faults

before running the Study, 4-15 definition of, 4-2 error messages, 4-22 examples, 4-22

fault currents on a radial unloaded feeder, 4-23 fault duty contribution to a faulted bus, 4-33 fault with a generator source with unequal positive-,

negative-, and zero-sequence reactances, 4-28 modeling transformer connections, 4-40 Plant project, 4-42 short circuit currents with a motor load at bus 4, 4-30 single-line-to-ground fault currents at the secondary

of a transformer with a grounding reactor, 4-25 source sequence impedance, 4-26 transformer off-nominal voltages and transformer

taps, 4-36 methodology, 4-2 running the Study, 4-15 Study options, 4-15 transformer taps

primary modeling, 4-12 secondary modeling, 4-13

transformers off-nominal voltage modeling, 4-13 phase shift, 4-13

Single-Line-to-Ground Faults. See Faults Sizing Study

before running the Study, 2-5

error messages, 2-7 examples, 2-7

cable temperature derating, 2-9 multiple branches with different load values, 2-9 Plant project, 2-10 sizing a radial feed, 2-7

feeder sizing, 2-3 ampacity, 2-3

methodology, 2-3, 2-4 running the Study, 2-5 Study options, 2-6 transformer feeders, 2-4 transformer sizing, 2-4

cooling characteristics, 2-4 Slack Bus, 3-4 Steady State Fault Current. See Faults Steady State Load Flow Equation. See Load Flow Study Swing Bus, 3-4, 3-10

T Taps. See Transformer Taps Thevenin Equivalent Circuit, 4-3 Thevenin Equivalent Impedance, 4-3 Transformer Sizing. See Sizing Study Transformer Taps

modeling in Load Flow Study, 3-4 modeling in Short Circuit Study, 4-12

Transformers modeling in Load Flow Study, 3-4 off-nominal voltage modeling in Short Circuit Study, 4-

13 phase shift, 4-13

U Unbalanced Faults. See Faults Utility, 3-4

V Voltage

voltage angle, 3-4 voltage magnitude, 3-4

Voltage Angle, 3-6 Voltage Drop, 3-5, 3-6

Z


DAPPER iv Reference Manual

Zero-Sequence Modeling. See Sequence Modeling

3/26/2006

Documents

Reference DAPPER