Power*Tools
for Windows
®
™
DAPPER Reference Manual
Electrical Engineering Analysis Software
for Windows
Copyright © 2006, SKM Systems Analysis, Inc.
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SKM Power*Tools for Windows
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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
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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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Asymmetrical rms Fault Current ..............................................................4-11
4.2.5
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 ZeroSequence 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 THIS CHAPTER
•
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
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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
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Define System Data
Define system topology and connections
Define utility connections (swing bus)
Define individual loads
Study Setup
Demand Load Library
Demand Loads
Energy Audit Loads
Study Setup
Run Demand Load Study
Saved in Database
Used by Sizing Study
For each branch:
Demand and design load and power factor
Total constant I demand load and power factor
Total constant Z demand load and power factor
Total constant kVA demand load and power factor
For each bus:
Total connected load and power factor
Total demand load and power factor
Total design load and power factor
Used by Load Flow
Study
Used by Load
Schedules
Datablocks
Reports
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.
The demand load value multiplied by the long
continuous load factor (LCL).
Design Load Value
Long Continuous Load Factor
The reciprocal of the cable ampacity derating factor
for a continuous load (design factor).
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Continuous Load
Non-Coincident Demand
A load where the maximum current is expected to
continue for three or more hours.
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 noncoincident 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:
52 hp demand load
75 hp connected load
= 0.69 or 69%
Demand Factor =
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 Study
b
g b
Demand Load = 10 kVA × 100% + 5 kVA × 50%
DAPPER 1-5
g
= 12.5 kVA
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
= 15 kVA + 15 kVA
= 30 kVA
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.
b
g b
Demand Load = 10 kVA × 100% + 20 kVA × 50%
g
= 20 kVA
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 demand
12.5 kVA + 12.5 kVA
=
20 kVA
= 1.25 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.
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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
Motor Load
Synchronous Motor
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Description
The rated size of a motor component multiplied by the number
f
t
ifi d i th N b
fM t b
f th
Demand Load Study
Load Types
Induction Motor
Non-Motor Load
Demand Load type
Energy Audit Load type
DAPPER 1-7
Description
of motors, as specified in the Number of Motors box of the
Component Editor.
Same as above.
A non-motor load with a uniquely defined set of demand
factors as specified in the Demand Load Library.
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.
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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 ( X ′′d ). The default X ′′d 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%.
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.
UTILITY 1
BUS 1
Connected 91.42 kVA
Demand 91.42 kVA
Design 114.28 kVA
C1
BUS 2
Connected 91.42 kVA
Demand 91.42 kVA
Design 114.28 kVA
MOTOR 1
Size 100.0 hp
PF 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 watts hp
1000 watts kW × 0.85 PF × 0.96 efficiency
= 91.42 kVA
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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DAPPER 1-9
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.
SKM Power*Tools for Windows
DAPPER 1-10
Reference Manual
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:
3/26/2006
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
Demand Load Study
In this box
DAPPER 1-11
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
SKM Power*Tools for Windows
DAPPER 1-12
Reference Manual
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:
b
g b
Demand Load = 100 kVA × 100% + 900 kVA × 50%
g
= 550 kVA
At Bus 2, the connected load is 2000 kVA.
The methodology requires that the demand load value be calculated as:
b
g b
Demand Load = 100 kVA × 100% + 1900 kVA × 50%
g
= 1050 kVA
The connected, demand and design loads are shown on the following one-line diagram:
3/26/2006
Demand Load Study
DAPPER 1-13
UTILITY 1
BUS 1
Connected 2000.00 kVA
Demand 1050.00 kVA
Design 1050.00 kVA
BUS 2
Connected 2000.00 kVA
Demand 1050.00 kVA
Design 1050.00 kVA
C1
C3
C2
BUS 4
BUS 3
Connected 1000.00 kVA
Demand 550.00 kVA
LOAD 1
Design 550.00 kVA
Size 1000.0 kVA
PF 0.85 Lag
Connected 1000.00 kVA
Demand 550.00 kVA
Design 550.00 kVA
LOAD 2
Size 1000.0 kVA
PF 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:
b
g b
Bus 2 Design Load = 75 kVA × 125% + 50 kVA × 100%
g
= 143.75 kVA
SKM Power*Tools for Windows
DAPPER 1-14
Reference Manual
UTILITY 1
BUS 1
Connected 125.00 kVA
Demand 125.00 kVA
Design 143.75 kVA
BUS 2
Connected 125.00 kVA
Demand 125.00 kVA
Design 143.75 kVA
C3
C2
BUS 4
BUS 3
Connected 50.00 kVA
Demand 50.00 kVA
Design 62.50 kVA
C1
MOTOR 1
Connected 75.00 kVA
Demand 75.00 kVA
Design 93.75 kVA
Size 50.0 kVA
PF 0.85 Lag
MOTOR 2
Size 75.0 kVA
PF 0.85 Lag
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:
b
g b
= b75 kVA × 125%g + b75 kVA + 50 kVA g × 100%
g
Bus 2 Design Load = Lg Mtr kVA × 125% + sum of remaining mtrs kVA × 100%
= 218.75 kVA
The design load value at Buses 1 and 2 is 218.75 kVA. The data on the following oneline diagram agrees with the preceding calculations.
3/26/2006
Demand Load Study
DAPPER 1-15
UTILITY 1
BUS 1
Connected 200.00 kVA
Demand 200.00 kVA
Design 218.75 kVA
C1
BUS 2
Connected 200.00 kVA
Demand 200.00 kVA
Design 218.75 kVA
C3
C2
BUS 3
BUS 4
Connected 125.00 kVA
Demand 125.00 kVA
Design 143.75 kVA
Connected 75.00 kVA
Demand 75.00 kVA
Design 93.75 kVA
MOTOR 3
MOTOR 1
Size 75.0 kVA
PF 0.85 Lag
Size 50.0 kVA
PF 0.85 Lag
MOTOR 2
Size 75.0 kVA
PF 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.
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DAPPER 1-16
Reference Manual
UTILITY 1
BUS 1
Connected 125.00 kVA
Demand 133.85 kVA
Design 153.81 kVA
C1
BUS 2
Connected 125.00 kVA
Demand 133.85 kVA
Design 153.81 kVA
C2
MCC
BUS 3
Connected 125.00 kVA
Demand 133.85 kVA
Design 153.81 kVA
MOTOR 3
MOTOR 1
Size 75.0 kVA
PF 0.85
Size 50.0 kVA
PF 0.85
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
96.0
99.8
120.0 85.00
LARGEST KVA MTR
75.0
90.2
79.8
==============================================================================
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
12.6
35.0
12.6
35.0
15.8
43.8 95.00
LIGHTING
==============================================================================
12.6
35.0
12.6
35.0
15.8
43.8 95.00
TOTALS
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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Demand Load Study
DAPPER 1-17
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:
45 A + 30 A + 30 A
3
= 35.0 A
Average =
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:
φA
Total Apparent Power in the Panel Schedule
Panel Load (kVA)
φB
φC
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
==============================================================================
25.1
69.6
25.1
69.6
31.4
87.1 95.00
TOTALS
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DAPPER 1-18
Reference Manual
PTW reports a three-phase current drawn by the panel by calculating:
Demand Load =
25.1 kVA
e 3jb0.208 kVg
= 69.6 A
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.
UTILITY 1
BUS 1
Connected 97.95 kVA
Demand 97.95 kVA
Design 110.23 kVA
PF 87.89 Lag
C1
BUS 2
Connected 97.95 kVA
Demand 97.95 kVA
Design 110.23 kVA
PF 87.89 Lag
C2
BUS 3
Connected 97.95 kVA
Demand 97.95 kVA
Design 110.23 kVA
PF 87.89 Lag
MOTOR 1
Size 50.0 kVA
No 2
PF 0.95 Lag
MOTOR 2
Size 50.0 kVA
No 1
PF 0.75 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
==============================================================================
98.0
117.8
110.2
132.6 85.62
TOTALS
98.0
117.8
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Demand Load Study
DAPPER 1-19
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:
b
g b
g
Design Load = 75 kVA × 125% + 2 × 50 kVA × 100%
= 93.8 kVA + 100 kVA
= 193.8 kVA
The design load value for Bus 3 is 193.8 kVA, as shown in the following one-line
diagram.
UTILITY 1
BUS 1
Connected 175.00 kVA
Demand 175.00 kVA
Design 193.75 kVA
PF 85.00 Lag
C1
BUS 2
Connected 175.00 kVA
Demand 175.00 kVA
Design 193.75 kVA
PF 85.00 Lag
C2
BUS 3
Connected 175.00 kVA
Demand 175.00 kVA
Design 193.75 kVA
PF 85.00 Lag
MOTOR 1
Size 50.0 kVA
No. of Mtrs 2
PF 0.85 Lag
MOTOR 2
Size 75.0 kVA
No. of Mtrs 1
PF 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
100.0
120.3 85.00
KVA TYPE MTR
100.0
120.3
100.0
120.3
93.8
112.8 85.00
LARGEST KVA MTR
75.0
90.2
75.0
90.2
==============================================================================
193.8
233.0 85.00
TOTALS
175.0
210.5
175.0
210.5
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,
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DAPPER 1-20
Reference Manual
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:
d
i
= cosbarctan5g
= cose78.7 j
Power Factor = cos arctan XR
o
= 19.61% Lag
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.
UTILITY 1
BUS 1
Connected 50.00 kVA
Demand 200. kVA
Design 250.00 kVA
PF 19.61
C1
BUS 2
Connected 50.00 kVA
Demand 200.00 kVA
Design 250.00 kVA
PF 19.61
C2
BUS 3
Connected 50.00 kVA
Demand 200.00 kVA
Design 250.00 kVA
PF 19.61
MOTOR 1
Size 50.0 kVA
Starting Condition
PF 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
200.0
240.6
250.0
300.7 19.61
LARGEST Z MTR
50.0
60.1
==============================================================================
200.0
240.6
250.0
300.7 19.61
TOTALS
50.0
60.1
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:
***
3/26/2006
NOTICE ***
BUS
BUS-0006 COMPLETES LOOP IN SYSTEM
Demand Load Study
DAPPER 1-21
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
UTILITY 2
BUS 1
BUS 4
Connected 125.00 kVA
Demand 125.00 kVA
Design 143.75 kVA
Connected 0.00 kVA
Demand 0.00 kVA
Design 0.00 kVA
C1
BUS 2
BUS 5
Connected 125.00 kVA
Demand 125.00 kVA
Design 143.75 kVA
Connected 0.00 kVA
Demand 0.00 kVA
Design 0.00 kVA
C2
C6
BUS 3
BUS 6
Connected 75.00 kVA
Demand 75.00 kVA
Design 93.75 kVA
Connected 50.00 kVA
Demand 50.00 kVA
Design 62.50 kVA
MOTOR 1
Size 75.0 kVA
PF 0.95 Lag
C3
C4
C5
BUS 7
Connected 50.00 kVA
Demand 50.00 kVA
Design 62.50 kVA
MOTOR 2
Size 50.0 kVA
PF 0.95 Lag
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.
SKM Power*Tools for Windows
3/26/2006
MCCB2
H3A BUS 19
MCCB1
018-RA
OFFICE COMPLEX AREA
PNL 19 H3A
C18
PNL 18 RA
PLN 16 H2A
BUS 16
BUS 15
C15
C16
C14
MCC 15 - 1A
LVP2
LVP6
CB4
R4
CB5
R5
CAP 1
PLN - 17 H1A
BUS 17
C17
LVP3
CMP HVAC
MCP#10
F2
HVAC BUS
C6
GEN 2
DSB 1 BUS 14
GEN 1
CMP CTR
TX H
COMPUTERS
TX C
F5
C5
F1
DS SWG1 BUS 8
LVP1
SWBD-1
TX C PRI BUS 9
TX B
TX B PRI BUS 4
C1
CB3
R3
C19
CB12
029-TX D SEC
TX D
005-TXD PRI
C2
CB6
R6
M13
DS SWG2 BUS 13
TX3 SEC BUS 11
TX A
CB2
R2
TX A PRI
TM -1
CB1
R1
UTILITY BUS
EDISON
L2
022-DSB 2
021-TX F PRI
PCB1
C7
003-HV SWGR
M23-A
023-MTR 23
C12
TX F
C9
DS SWG3 BUS 20
PCB2
C8
TX3 TER BUS 12
INDUSTRIAL COMPLEX AREA
M23-B
TX 6
006-TX3 PRI
C3
CB7
R7
L1
M20
PCB3
R11
GEN 3
SYN A
DEMONSTRATION PROJECT FOR POWER*TOOLS FOR WINDOWS
CB8
R8
DSB 3 BUS 27
TX G
F4
C10
LVP4
C13
M28 #1 & 2
BUS 28 A
L5
026-TX G PRI
SYN B
CB9
R9
M28 #3
MO/L#28B-1
MCP#28B-1
M28 #4
BUS 28 B
M25
MO/L#28B-2
MCP#28B-2
MO/L#25
F3
SW1
025-MTR 25
BLDG 115 SERV
C11
LVP5
C20
TX E
007-TX E PRI
C4
CB10
R10
DAPPER 1-22
Reference Manual
Demand Load Study
DAPPER 1-23
The following figure shows a portion of the Plant project, including Demand Load
Study results.
DEMAND LOAD ANALYSIS
BUILDING 115 SERVICE
BLDG 115 SERV
Connected 3580.25 kVA
Demand 3480.52 kVA
Design 4125.68 kVA
C11
C10
026-TX G PRI
025-MTR 25
Connected 1081.67 kVA
Connected 2500.00 kVA
Demand 981.84 kVA
Demand 2500.00 kVA
F4
Design 1058.91 kVA
Design 3125.00 kVA
SW1
TX G
F3
MO/L#25
DSB 3 BUS 27
Connected 1081.67 kVA
Demand 981.84 kVA
C20
Design 1058.91 kVA
C13
M25
L5
100.0 kVA
LVP4
2500.0 kVA
LVP5
No of Mtrs: 1
1.00 Unity PF
0.80 Lag PF
Load Factor 1.00
BUS 28 A
BUS 28 B
Connected 500.00 kVA
Connected 500.00 kVA
Demand 450.00 kVA
Demand 450.00 kVA
Design 506.25 kVA
Design 506.25 kVA
MCP#28B-1
MCP#28B-2
MO/L#28B-1
MO/L#28B-2
M28 #1 & 2
M28 #3
M28 #4
250.0 kVA
250.0 kVA
250.0 kVA
No of Mtrs: 2
No of Mtrs: 1
No of Mtrs: 1
0.80 Lag PF
0.80 Lag PF
0.80 Lag PF
Load Factor 0.90
Load Factor 0.90
Load Factor 0.90
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
SKM Power*Tools for Windows
DAPPER 1-24
Reference Manual
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
3/26/2006
Demand Load Study
DAPPER 1-25
KVA
94.4
52.2
52.2
85.00 LAGGING
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
OFFICE EQUIPMENT
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
SKM Power*Tools for Windows
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.
This chapter discusses:
Engineering Methodology.
•
PTW Applied Methodology.
•
Examples.
IN THIS CHAPTER
•
What is the Sizing Study? ..............................................................................DAPPER 2-2
Engineering Methodology .............................................................................DAPPER 2-3
PTW Applied Methodology ..........................................................................DAPPER 2-4
Application Examples ....................................................................................DAPPER 2-7
DAPPER 2-2
Reference Manual
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 connections
Define utility connections (swing bus)
Define individual loads
Run the Demand Load Study
Study Setup
Cable Library
Transformer Library
Feeder Voltage Drop Criteria
Transformer Tap Size Criteria
Study Setup
Run Sizing Study
Saved in Database
For each branch:
Feeder size
Quantity of feeders per phase
Transformer nominal kVA
Raceway information
Transformer full-load kVA
DXF Files
Used by Short Circuit Study
and Load Flow Study
Datablocks
Reports
[1]
Authored by Committee, 1996 National Electric Code. Quincy, MA: National Fire
Protection Association, 1996.
3/26/2006
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 crosssectional 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 Drop =
b
g b
3 I × R × cosφ + I × X × sinφ
VLL
g × 100
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
SKM Power*Tools for Windows
DAPPER 2-4
Reference Manual
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.
3/26/2006
Sizing Study
DAPPER 2-5
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.
•
Define utility connections (swing bus).
•
Define individual loads.
•
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:
1.
From the Run menu, choose Analysis.
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.
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 Sizing Study runs, writes the results to the database, and creates a Report.
SKM Power*Tools for Windows
DAPPER 2-6
Reference Manual
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:
3/26/2006
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.
Sizing Study
DAPPER 2-7
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.
SKM Power*Tools for Windows
DAPPER 2-8
Reference Manual
UTILITY
B1
4160 V
Demand Load 200.00 kVA
C1
Size # 6 AWG/kcmil
1 Cables in Parallel
30.10 Amps
Ampacity 75.0 A / Cable
Length 100.0 ft
B2
4160 V
Demand Load 200.00 kVA
T1
225.0 kVA
B3
480 V
Demand Load 200.00 kVA
C2
Size # 4/0 AWG/kcmil
4 Cables in Parallel
260.90 Amps
Ampacity 230.0 A / Cable
Length 1200.0 ft
C3
Size # 2/0 AWG/kcmil
1 Cables in Parallel
128.98 Amps
Ampacity 175.0 A / Cable
Length 100.0 ft
B4
480 V
Demand Load 200.00 kVA
B5
480 V
Demand Load 100.00 kVA
C4
Size # 2/0 AWG/kcmil
2 Cables in Parallel
131.92 Amps
Ampacity 175.0 A / Cable
Length 1000.0 ft
B6
480 V
Demand Load 100.00 kVA
L1
Design Load 125.00 kVA
L2
Design 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
3/26/2006
Sizing Study
TX KVA:
225.0 TR FLA:
MINIMUM FEEDER AMPACITY:
DAPPER 2-9
270.6
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
Size 2/0 AWG
C4
No in Parallel 1
121.72 A Load Flow Current
175.0 A Cable Ampacity
100.0 ft
50 Deg Operating
Temperature
C3
B5
Size 1 AWG
No in Parallel 1
122.00 A Load Flow Current
130.0 A Cable Ampacity
100.0 ft
30 Deg Operating
Temperature
B6
100.0 kVA
100.0 kVA
L1
L2
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:
SKM Power*Tools for Windows
DAPPER 2-10
Reference Manual
U1
B1
C1
Length 6000.0 ft
Ampacity 150.0 A
Size 2
No 1
LF Current 70.48 A
B2
B3
C2
Length 25.0 ft
Ampacity 83.0 A
Size 6
No 1
LF Current 28.28 A
B30
C20
Length 25.0 ft
Ampacity 110.0 A
Size 4
No 1
LF Current 42.86 A
T1
T2
Size 500.0 kVA
Load 305.00 kVA
Size 225.0 kVA
Load 205.00 kVA
B40
B4
C3
B5
Length 10.0 ft
Ampacity 175.0 A
Size 2/0
No 2
LF Current 232.86 A
C30
Length 10.0 ft
Ampacity 200.0 A
Size 3/0
No 4
LF Current 352.89 A
B50
C40
C4
Length 300.0 ft
Ampacity 85.0 A
Size 4
No 1
LF Current 63.22 A
C5
L1
55.00 kVA
SPECIAL LOAD
Length 300.0 ft
Ampacity 175.0 A
Size 2/0
No 1
LF Current 123.43 A
Length 1000.0 ft
Ampacity 175.0 A
Size 2/0
No 2
LF Current 175.50 A
B7
B6
C50
B60
L2
155.00 kVA
SPECIAL LOAD
Length 1000.0 ft
Ampacity 150.0 A
Size 1/0
No 3
LF Current 235.42 A
B70
L10
105.00 kVA
SPECIAL LOAD
L20
205.00 kVA
SPECIAL 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.
2.4.4. 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
MCCB2
H3A BUS 19
MCCB1
018-RA
OFFICE COMPLEX AREA
PNL 19 H3A
C18
C15
PNL 18 RA
PLN 16 H2A
MCC 15 - 1A
BUS 16
C16
C14
BUS 15
LVP2
LVP6
CB4
R4
CB5
R5
CAP 1
PLN - 17 H1A
BUS 17
C17
LVP3
CMP HVAC
MCP#10
F2
HVAC BUS
C6
GEN 2
DSB 1 BUS 14
GEN 1
CMP CTR
TX H
COMPUTERS
TX C
F5
C5
F1
DS SWG1 BUS 8
LVP1
SWBD-1
TX C PRI BUS 9
TX B
TX B PRI BUS 4
C1
CB3
R3
C19
CB12
029-TX D SEC
TX D
005-TXD PRI
C2
CB6
R6
M13
DS SWG2 BUS 13
TX3 SEC BUS 11
TX A
CB2
R2
TX A PRI
TM -1
CB1
R1
UTILITY BUS
EDISON
L2
022-DSB 2
021-TX F PRI
PCB1
C7
003-HV SWGR
M23-A
023-MTR 23
C12
TX F
C9
DS SWG3 BUS 20
PCB2
C8
TX3 TER BUS 12
INDUSTRIAL COMPLEX AREA
M23-B
TX 6
006-TX3 PRI
C3
CB7
R7
L1
M20
PCB3
R11
GEN 3
SYN A
DEMONSTRATION PROJECT FOR POWER*TOOLS FOR WINDOWS
CB8
R8
DSB 3 BUS 27
TX G
F4
C10
LVP4
C13
M28 #1 & 2
BUS 28 A
L5
026-TX G PRI
SYN B
CB9
R9
M28 #3
MO/L#28B-1
MCP#28B-1
M28 #4
BUS 28 B
M25
MO/L#28B-2
MCP#28B-2
MO/L#25
F3
SW1
025-MTR 25
BLDG 115 SERV
C11
LVP5
C20
TX E
007-TX E PRI
C4
CB10
R10
Sizing Study
DAPPER 2-11
SKM Power*Tools for Windows
DAPPER 2-12
Reference Manual
The following figure shows a portion of the Plant project.
FEEDER AND TRANSFORMER SIZING STUDY
BUILDING 115 SERVICE
BLDG 115 SERV
C11
C10
026-TX G PRI
025-MTR 25
F4
SW1
F3
TX G
MO/L#25
DSB 3 BUS 27
M25
L5
C20
C13
LVP5
LVP4
BUS 28 B
MCP#28B-1
MCP#28B-2
MO/L#28B-1
MO/L#28B-2
BUS 28 A
M28 #1 & 2
M28 #3
M28 #4
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 oneline 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
***
3/26/2006
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
Sizing Study
***
NOTICE
***
DAPPER 2-13
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
*********************
TO
026-TX G PRI
TO
025-MTR 25
4160.
FROM
DSB 3 BUS 27
*********************
TO
BUS 28 A
480.
FROM
DSB 3 BUS 27
*********************
TO
BUS 28 B
( 4) 3
480.
EX
2/0
( 1)
( 1)
3
3
1/0
500
CU
CU
XLP
XLP
3" C
5" C
N
N
( 4)
3
500
CU
THWN
3" C
N
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
TO
025-MTR 25
EX 0.43
1
1
1/0
500
CU
CU
XLP
XLP
30.
30.
147. A
434. A
195. A
585. A
FROM
022-DSB 2
*********************
TO
023-MTR 23
480.
4
500
CU
THWN
30.
1064. A
1520. A
FROM
DSB 3 BUS 27
*********************
TO
BUS 28 A
480.
4
500
CU
THWN
30.
609. A
1520. A
0.37
0.75
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
TO
025-MTR 25
EX 0.43
FROM
DSB 3 BUS 27
1
1
1/0
500
CU
CU
XLP
XLP
30.
30.
147. A
434. A
195. A
585. A
480.
SKM Power*Tools for Windows
DAPPER 2-14
Reference Manual
*********************
TO
BUS 28 A
FROM
DSB 3 BUS 27
*********************
TO
BUS 28 B
0.75
4
500
CU
THWN
30.
609. A
1520. A
4
2/0
CU
THWN
30.
609. A
700. A
480.
1.75
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.
TO DSB 3 BUS 27
480.
DEMAND LOAD:
981.8 KVA
3/26/2006
D
YG
-2.5
TYPE: DT
SIZE:
DESCRIPTION: TX G
5.75 %Z NOMINAL RATING
1000.0 KVA
1000.0 KVA
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.
IN THIS CHAPTER
This chapter discusses:
•
Engineering Methodology.
•
PTW Applied Methodology.
•
Examples.
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
SKM Power*Tools for Windows
DAPPER 3-2
Reference Manual
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 connections
Define utility connections (swing bus)
Define individual loads
Define feeder and transformer sizes
Define generator sizes
Run the Demand Load Study (optional)
Study Setup
Cable Library
Transformer Library
Demand Loads
Directly Connected Loads
Study Setup
Run Load Flow Study
Saved in Database
For each branch:
Voltage drop
Power flow
For each bus:
Voltage
Voltage angle
Voltage drop
Power flow
Power factor
Datablocks
Reports
3.2. Engineering Methodology
The steady-state load flow solution to a power system network involves Ohm’s Law:
[I] = [Y][V]
11/22/2006
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;
The network admittance matrix (1/Z);
Column Vector of positive sequence voltage at each bus.
Y
V
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:
Ii =
LM bP + jQ g * OP
N V* Q
i
Eq. 3-2
i
i
where
bP + jQ g *
complex conjugate of the apparent power flowing into the ith node;
Vi*
complex conjugate of the voltage of the ith node.
i
i
Combining Eq. 3-1 and Eq. 3-2 yields:
LM bP − jQg OP = [Y][V]
N [V]* Q
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
Swing or Slack Bus
V, α
±P, ±Q
III
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
SKM Power*Tools for Windows
DAPPER 3-4
Reference Manual
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 threephase 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
11/22/2006
Load Flow Study
DAPPER 3-5
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.
SKM Power*Tools for Windows
DAPPER 3-6
Reference Manual
Sending End Bus
Receiving End Bus
I
R + jX
Reported Voltage Drop
Sending End Voltage
IZ
IX
Receiving End Voltage
∅
IR
Load Power Factor Angle
Actual Voltage Drop
Load Current
The sending end voltage may be expressed as:
b
Vs = Vr + I R + jX
g
Eq. 3-4
where
Vs
Vr
I
R
jX
sending bus voltage;
receiving bus voltage;
load current;
feeder resistance;
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 = Es − E 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:
11/22/2006
•
Define power system topology and connections.
•
Define utility connection (swing bus).
Load Flow Study
DAPPER 3-7
•
Define individual loads.
•
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.
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 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.
SKM Power*Tools for Windows
DAPPER 3-8
Reference Manual
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 nonconvergence.
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.
11/22/2006
Load Flow Study
DAPPER 3-9
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.
SKM Power*Tools for Windows
DAPPER 3-10
Reference Manual
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.
Lagging
1.0
Reactive Power-- Per Unit
Rated PF
Lagging
.50
0.0
-.50
-1.0
Leading
PF Leading
0.25
0.5
Kilowatts Per Unit
.075
1.00
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
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Load Flow Study
DAPPER 3-11
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.
SKM Power*Tools for Windows
DAPPER 3-12
Reference Manual
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.
11/22/2006
Load Flow Study
DAPPER 3-13
GENERATOR
B1
13800 V
13800.00 V
0.00 %
C1
1886.52 kW
1467.74 kvar
5.71 kW Losses
57.13 kvar Losses
1.64 % VD
100.00 A
B2
13800 V
13573.55 V
1.64 %
L1
The load draws 100 A at 80% power factor and is assumed to have a constant current-type
load characteristic.
SKM Power*Tools for Windows
DAPPER 3-14
Reference Manual
The three-phase real power loss in the cable is:
PLoss = I 2 R × 3
b g b0.19044g × 3
= 100
2
= 5.71 kW
The magnetization power required by the cable is:
Q Loss = I 2 X × 3
b g b1.9044g × 3
= 100
2
= 57.1 kvar
In a simplified form, the voltage drop through the cable is:
b
g
= b100gb0.19044gb0.8g + b100gb1.9044gb0.6g ×
= b80gb0.19044g + b60gb1.9044g × 3
= b129.5g × 3
VD L− L = IRcosΘ + IXsinΘ × 3
3
= 224.3 V
And on a per unit base, this is:
224.3 V
13800 V
= 0.0163 pu V
VD pu =
Or an operating voltage of:
Vpu = 1 − 0.0163 pu
= 0.984 pu
The three-phase power consumed by the load at operating voltage is then:
e 3jbV gbI g bV pu operatingg
= 3b100 A gb13.8 kV gb0.984g
S Load =
LL
L
= 2351.4 kVA
With an 80% power factor:
S Load = 1881.10 kW + j1410.8 kvar
At the sending end, the power injected into cable is:
SSending = SLoad + S Losses
b
g b
= b1886.81 + j1467.9gkVA
g
= 1881.10 + 5.7 kW + j1410.8 + j57.1 kvar
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
11/22/2006
Load Flow Study
DAPPER 3-15
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.
GENERATOR
B1
13800 V
13800.00 V
0.00 %
C1
795.54 kW
605.90 kvar
1.00 kW Losses
10.00 kvar Losses
0.68 %
41.84 A
B2
13800 V
13705.78 V
0.68 %
L1
1.0 MVA
0.80 Lag PF
T1
795.54 kW
605.90 kvar
1.00 kW Losses
10.00 kvar Losse
0.68 %
41.84 A
B3
13800 V
13705.78 V
0.68 %
L2
1.0 MVA
0.80 Lag PF
The following input data shows that on a per unit basis, the impedance of the cable equals
the impedance of the transformer.
SKM Power*Tools for Windows
DAPPER 3-16
Reference Manual
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
PROJECTED POWER FLOW:
LOSSES THRU FEEDER:
FEEDER AMPS: 41.8 VOLTAGE DROP:
94. %VD:
.68
795.5 KW
605.9 KVAR
1000.0 KVA PF:
.80 LAGGING
1.0 KW
10.0 KVAR
10.0 KVA
LOAD
TO: B3
PROJECTED POWER FLOW:
LOSSES THRU TRANSF:
TRANSF AMPS: 41.8 VOLTAGE DROP:
94. %VD:
.68
795.5 KW
605.9 KVAR
1000.0 KVA PF:
.80 LAGGING
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
PROJECTED POWER FLOW:
LOSSES THRU FEEDER:
FEEDER AMPS: 41.8 VOLTAGE DROP:
94. %VD:
.68
794.5 KW
595.9 KVAR
993.2 KVA PF:
.80 LAGGING
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
PROJECTED POWER FLOW:
LOSSES THRU TRANSF:
TRANSF AMPS: 41.8 VOLTAGE DROP:
94. %VD:
.68
795.5 KW
605.9 KVAR
1000.0 KVA PF:
.80 LAGGING
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.
11/22/2006
Load Flow Study
DAPPER 3-17
GENERATOR
3115.26 kW
2366.59 kvar
0.80 PF
C1
C3
C2
42.1 A
29.4 A
42.1 A
B2
13705.13 V
0.69 % VD
B3
13733.81 V
0.48 % VD
B4
13705.13 V
0.69 % VD
L1
1000.0 kVA
Energy Audit
1.0 Energy Audit
Load Factor
L2
L3
1000.0 kVA
1000.0 kVA
Energy Audit
Demand Load
0.7 Energy Audit General Load
Load Factor
B1
13800.00 V
0.00 % VD
C4
29.4 A
C5
B5
13733.81 V
0.48 % VD
B6
13753.77 V
0.33 % VD
L4
1000.0 kVA
Demand Load
Special Load
20.57 A
L5
1000.0 kVA
Demand Load
Special Load
0.7 Energy Audit
Load Factor
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
Load
1st Level
Factor
Load
2nd Level
Factor
LCL
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 #
Individual
Load Type
Individual
Load
Factor
Load Flow Study Setup Specifications
Directly Connected
From Demand Load Study
Connected
Load
(Amps)
1st Level
Demand
or Energy
Factor
Demand
Load
(Amps)
Design
Load
(Amps)
Connected
Load
(Amps)
SKM Power*Tools for Windows
DAPPER 3-18
Reference Manual
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:
11/22/2006
Load Flow Study
DAPPER 3-19
GENERATOR
840.03 kW
630.04 kvar
0.80 PF
B1
13800.00 V
0.00 % VD
C1
840.03 kW
630.04 kvar
B2
13799.57 V
0.00 % VD
C2
C3
440.00 kW
330.01 kvar
440.01 kW
330.01 kvar
B4
B3
13799.38 V
0.00 % VD
L1
13799.28 V
0.01 % VD
L2
1000.0 kVA
Demand Load Type
1000.0 kVA
Demand Load Type
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
PROJECTED POWER FLOW:
LOSSES THRU FEEDER:
FEEDER AMPS: 43.9 VOLTAGE DROP:
0. %VD: 0.00
840.0 KW
630.0 KVAR
1050.0 KVA PF: 0.80 LAGGING
0.0 KW
0.0 KVAR
0.0 KVA
LOAD
TO: B3
PROJECTED POWER FLOW:
LOSSES THRU FEEDER:
FEEDER AMPS: 23.0 VOLTAGE DROP:
0. %VD: 0.00
440.0 KW
330.0 KVAR
550.0 KVA PF: 0.80 LAGGING
0.0 KW
0.0 KVAR
0.0 KVA
LOAD
TO: B4
PROJECTED POWER FLOW:
LOSSES THRU FEEDER:
FEEDER AMPS: 23.0 VOLTAGE DROP:
0. %VD: 0.00
440.0 KW
330.0 KVAR
550.0 KVA PF: 0.80 LAGGING
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.
SKM Power*Tools for Windows
DAPPER 3-20
Reference Manual
3.4.5. Example from Plant
The following figure is a one-line diagram for the Plant project. The Plant project is
included on the PTW diskettes.
11/22/2006
COMPUTERS
MCCB2
H3A BUS 19
MCCB1
018-RA
OFFICE COMPLEX AREA
PNL 19 H3A
C18
PNL 18 RA
PLN 16 H2A
C15
BUS 16
MCC 15 - 1A
BUS 15
C16
C14
LVP6
TX C
CB4
R4
CB5
R5
CAP 1
PLN - 17 H1A
BUS 17
C17
LVP3
CMP HVAC
MCP#10
F2
HVAC BUS
C6
GEN 2
DSB 1 BUS 14
GEN 1
CMP CTR
TX H
LVP2
F5
C5
F1
DS SWG1 BUS 8
LVP1
SWBD-1
TX C PRI BUS 9
TX B
TX B PRI BUS 4
C1
CB3
R3
C19
CB12
029-TX D SEC
TX D
005-TXD PRI
C2
CB6
R6
M13
DS SWG2 BUS 13
TX3 SEC BUS 11
TX A
CB2
R2
TX A PRI
TM -1
CB1
R1
UTILITY BUS
EDISON
L2
022-DSB 2
021-TX F PRI
PCB1
C7
003-HV SWGR
M23-A
023-MTR 23
C12
TX F
C9
DS SWG3 BUS 20
PCB2
C8
TX3 TER BUS 12
INDUSTRIAL COMPLEX AREA
M23-B
TX 6
006-TX3 PRI
C3
CB7
R7
L1
M20
PCB3
R11
GEN 3
SYN A
DEMONSTRATION PROJECT FOR POWER*TOOLS FOR WINDOWS
CB8
R8
DSB 3 BUS 27
TX G
F4
C10
LVP4
C13
M28 #1 & 2
BUS 28 A
L5
026-TX G PRI
SYN B
CB9
R9
M28 #3
MO/L#28B-1
MCP#28B-1
M28 #4
BUS 28 B
M25
MO/L#28B-2
MCP#28B-2
MO/L#25
F3
SW1
025-MTR 25
BLDG 115 SERV
C11
LVP5
C20
TX E
007-TX E PRI
C4
CB10
R10
Load Flow Study
DAPPER 3-21
SKM Power*Tools for Windows
DAPPER 3-22
Reference Manual
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
11/22/2006
Load Flow Study
DAPPER 3-23
================================
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:
CONVERGENCE CRITERIA:
LARGEST BUS MISMATCH
LARGEST BUS MISMATCH
LARGEST BUS MISMATCH
LARGEST BUS MISMATCH
LARGEST BUS MISMATCH
LARGEST BUS MISMATCH
DS
DS
DS
DS
DS
DS
4000.00 KVA
0.200 KVA
SWG3 BUS 20
SWG1 BUS 8
SWG1 BUS 8
SWG2 BUS 13
SWG1 BUS 8
SWG1 BUS 8
1517.460
120.509
24.610
1.128
1.489
0.199
KVA
KVA
KVA
KVA
KVA
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
SKM Power*Tools for Windows
DAPPER 3-24
Reference Manual
========================= 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
PROJECTED POWER FLOW:
LOSSES THRU FEEDER:
FEEDER AMPS: 568.4 VOLTAGE DROP:
3. %VD: 0.70
362.0 KW
272.9 KVAR
453.3 KVA PF: 0.80 LAGGING
2.0 KW
2.9 KVAR
3.5 KVA
LOAD
TO: BUS 28 B
PROJECTED POWER FLOW:
LOSSES THRU FEEDER:
FEEDER AMPS: 574.2 VOLTAGE DROP:
8. %VD: 1.65
367.5 KW
273.2 KVAR
457.9 KVA PF: 0.80 LAGGING
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
11/22/2006
Load Flow Study
022-DSB 2
BUS 15
BUS 16
DSB 3 BUS 27
DSB 3 BUS 27
023-MTR 23
018-RA
H3A BUS 19
BUS 28 A
BUS 28 B
FDR
FDR
FDR
FDR
FDR
0.35
0.47
0.54
0.70
1.65
966.62
150.13
277.53
568.50
574.23
799.51
125.04
229.00
453.32
457.90
DAPPER 3-25
63.59
75.07
79.30
37.40
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.
SKM Power*Tools for Windows
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.
This chapter discusses:
Engineering Methodology.
•
PTW Applied Methodology.
•
Examples.
IN THIS CHAPTER
•
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
SKM Power*Tools for Windows
DAPPER 4-2
Reference Manual
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 connections
Define utility connection (swing bus)
Define feeder and transformer sizes
Define fault contribution data
Study Setup
Cable Library
Transformer Libray
Study Setup
Run Short Circuit Study
Saved in Database
Three-phase fault currents
Unbalanced fault currents
Momentary and asymmetrical fault currents
Used by Load
Schedules
Used by Time Current
Coordination (CAPTOR)
Datablocks
Reports
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
Z
I
bus voltage matrix;
bus impedance matrix; referred to as the Z Bus matrix;
bus nodal current matrix.
The impedance Z in complex notation is:
Z = R + jX
where
R
jX
resistance;
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.
UTILITY
B1
T1
B2
C1
M1
B3
M2
From the one-line diagram a Thevenin equivalent impedance diagram is drawn:
SKM Power*Tools for Windows
DAPPER 4-4
Reference Manual
Bus 1
Bus 2
Z Transformer
Z Utility
Bus 3
Z Cable
Z Motor 1
Z Motor 2
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.
Bus 2
Bus 1
Bus 3
Z Cable
If
Z Utility
I
Z Motor 1
Utility
Z Motor 2
I Motor 1
I
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:
If =
3/26/2006
V
Z Thevenin
Short Circuit Study
DAPPER 4-5
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
Ic
I
b
Ia
Fault
V=0
The boundary conditions are:
Va = 0, I b = 0, I c = 0
where
line-to-ground voltage in Phase A;
currents in Phase B;
currents in Phase C.
Va
Ib
Ic
From symmetrical components the following solution matrix is written:
LMI
MMI
NI
a0
a1
a2
OP 1 LM1
PP = 3 MM1
Q N1
OPLI O
a P MMI PP
P
a Q MNI PQ
1 1
a
a
a
2
b
2
c
SKM Power*Tools for Windows
DAPPER 4-6
Reference Manual
where
I b = Ic = 0
and
a = e j120
a 2 = e j240
Substituting the above and solving yields the single-line-to-ground fault case:
b g
b g
b g
1
Ia
3
1
I a1 = I a
3
1
I a2 = I a
3
I a0 =
Therefore:
I a1 = I a2 = I a0
Because the three symmetrical components are equal in magnitude and in phase, they may
be viewed as connected in series, as shown:
Z1
Ia1
Z2
Ia2
Z0
Ia0
Ea
The positive-sequence current is:
I a1 =
Ea
Z1 + Z 2 + Z 0
where
Z1
Z2
Z0
positive-sequence impedance;
negative-sequence impedance;
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:
3/26/2006
Short Circuit Study
DAPPER 4-7
I a = 3I a1
Substituting yields:
Ia =
3E a
Z1 + Z 2 + Z 0
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 negativesequence networks are in series; the zero-sequence network is not involved. For doubleline-to-ground faults, the positive-sequence network adds to the negative- and zerosequence 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 wyegrounded 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 pu =
kVA base
MVA base
=
pu Ω
kVA utility MVA utility
where
kVAutility (MVAutility)
kVAbase (MVAbase)
the utility system three-phase short circuit capability;
the system three-phase power base specified for the Study.
For motor and generator contributions:
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F kV IJ FG kVA IJ pu Ω
= X ′′ G
H kV K H kVA K
2
Z pu
motor
base
base
motor
d
where
X ′′d
motor subtransient reactance on its motor rated line-to-line voltage (kVmotor)
kVmotor
kVAmotor
kVbase
kVAbase
and three-phase power (kVAmotor) base;
line-to-line motor rated voltage;
three-phase power base;
line-to-line bus nominal system voltage at the point of the motor;
three-phase power base.
For feeders:
Z pu =
Z cable
kv base × 1000
kVA base
2
=
Z cable
pu Ω
2
kVbase
MVA base
where
Zcable
per phase (one way) feeder impedance in ohms.
For transformers:
Z pu =
FG Z
H
transformer %
100
IJ FG kV
K H kV
transformer
base
IJ FG kVA
K H kVA
2
base
transformer
IJ
K
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:
Ishort circuit =
FG
HZ
V
Thevenin
IJ pu A
K
Expressed in amperes:
Ishort circuit = Ishort circuit pu × I b
Where the base current is defined as:
I base =
kVA base
3 × kVbase
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 + j0.4250 Ω at 4160 V
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To express this value in fault current:
Z utility =
I=
eR
2
+ X2
j
V
3 × Z utility
Substituting known quantities yields:
Z utility = 0.04252 + 0.4252
= 0.4272 Ω
At 4.16 kV, the utility fault current is:
Ishort circuit =
4160 V
e 3jb0.4272 Ωg
= 5622 A
The utility X/R ratio is:
X 0.425 Ω
=
R 0.0425 Ω
= 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
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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:
b
g
Vpeak × sin ω t + Θ = Ri +
Ldi
dt
This differential equation can be solved for the instantaneous current, i:
i=
Vpeak
Z
b
LM V
g MZ
N
sin ω t + α - Θ −
peak
-t
e
L
R
b
sin α - Θ
OP
gP
Q
where
Z
α
Θ
ω
t
f
Vpeak
b g
R 2 + ωL
2
time in cycles when the fault occurs;
L
tan -1 ω
R
2πf;
time in seconds;
frequency in Hertz;
peak (crest) voltage.
FG IJ
H K
b
g
The first term is clearly sinusoidal and is at an instantaneous peak if α - Θ = ±
π
2
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 I ′′k 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.
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Current
Theoretical maximum
Peak at 1/2 cycle
Top envelope
idc
Asymmetrical values
including motor contributions
2
ip
2 I"k
Decaying (aperiodic) component
(DC decay)
i dc
2
2I k
Time
Steady state value
(no motor contributions)
Bottom envelope
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 asymmetrical peak = 2 I ′′k + 2 I ′′k e
−2π
FG R IJ c
H XK
where
asymmetrical peak fault current;
initial symmetrical rms fault current;
dc decaying component of fault current;
steady-state fault current;
time in cycles into the fault.
iasymmetrical peak
I ′′k
idc
Ik
c
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 asymmetrical rms =
F
bI′′ g + GH b
2
= I ′′k 1 + 2e
And at
1
2
2 I ′′k e
k
−4π
−2π
I
JK
2
R
c
X
R
c
X
cycle into the fault:
= I ′′k 1 + 2e
−2π
R
X
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.
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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.
Tap
X d"
Z Transformer
Faulted Bus
Transformer
Voltage Generator
The relationship between the short circuit current and primary tap setting is:
If =
LMF 1 I
MNGH 1 + Tap JK
V
2
OP
PQ
× X ′′d generator + Z transformer
where
If
V
Tap
X ′′d
Ztransformer
fault current;
fault point pre-fault no loads pre-fault voltage;
ideal voltage shifter in per unit;
machine reactance;
transformer impedance.
The fault point no load voltage is:
V = Vgenerator
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FG 1 IJ
H 1+ Tap K
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DAPPER 4-13
Secondary Transformer Tap Modeling
Secondary transformer taps may also be modeled. The ideal voltage shifter is modeled as:
Tap
X d"
Z Transformer
Faulted Bus
Transformer
Voltage Generator
The relationship between the short circuit current and transformer secondary tap setting is:
If =
V
(1 + Tap ) ( X
2
"
d generator
+ Z transformer )
The fault point no-load voltage is:
b
V = Vgenerator 1+ Tap
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.
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When the transformer phase shift is modeled for a single-line-to-ground fault, the branch
sequence current might be:
1388 -120°
418 -90°
418 -90°
1388 -120°
Positive-Sequence
1388 -120°
418 -150°
418 -150°
1388 -120°
Negative-Sequence
1388 -120°
0 -0°
418 -120°
0 -0°
1388 -120°
418 -120°
0 -0°
Zero-Sequence
In this example, the transformer is rated 13.8 kV - 4.16 kV, and if 1388 A in the positivesequence 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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DAPPER 4-15
4.3.1 Before Running the Short Circuit Study
Before running the Short Circuit Study, you must:
•
Define the system topology and connections.
•
Define utility connections (swing bus).
•
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:
1.
From the Run menu, choose Analysis.
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.
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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).
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DAPPER 4-17
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 zerosequence 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.
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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
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DAPPER 4-19
transformer data must be user defined in the Component Editor. PTW models the threewinding 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 threewinding transformer with a primary, secondary and tertiary power rating of 15 MVA, 15
MVA and 5.25 MVA, respectively.
GEN 2
GEN 1
BUS 1
BUS 7
27856.53 A
27856.53 A
C4
C1
BUS 2
BUS 8
22405.02 A
22405.02 A
T2
T1
BUS 3
21706.12 A
BUS 5
7199.51 A
CENTER POINT
11150.80 A
C2
6.0214%
BUS 9
BUS 4
C3
T3
T4
0.8790%
BUS 10
21705.22 A
11150.56 A
C5
BUS 6
C6
BUS 11
20520.80 A
9.9790%
BUS 12
BUS 13
10860.47 A
20520.00 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)
1
Primary
Secondary
Tertiary
6.9
2
Primary
Tertiary
Secondary
5.6
5.25
ZPT
3
Secondary
Tertiary
Primary
3.8
5.25
ZST
15
Impedance
Symbol
ZPS
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.
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The following drawing is an equivalent impedance diagram for the three-winding
transformer:
T
Z PT
ZST
P
Z PS
S
You can convert this into an equivalent wye diagram using standard network reduction
techniques:
Z3
T
Z1
P
S
Z2
From the network reduction diagram, we can write:
Z PS = Z1 + Z 2
Z PT = Z1 + Z3
ZST = Z 2 + Z3
Network reduction yields:
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Z1 =
1
2 (Z PS
+ Z PT - ZST )
Z2 =
1
2 (Z PS
+ ZST - Z PT )
Z3 =
1
2 (Z PT
+ ZST - Z PS )
Short Circuit Study
DAPPER 4-21
You can solve the equations by substituting the manufacturer’s data expressed on a
common 15 MVA base:
LM j6.9 + FG j5.6 × 15 IJ - j3.8 × 15 OP
5.25 Q
N H 5.25K
= j6.0214 %
LM j6.9 + FG j3.8 × 15 IJ - j5.6 × 15 OP
=
5.25 Q
N H 5.25K
Z1 =
1
Z2
1
2
2
= j0.879 %
Z3 =
1
2
LMFG j5.6 × 15 IJ + FG j38. × 15 IJ − j6.9OP
NH 5.25K H 5.25K Q
= j9.978 %
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 twowinding 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:
50 hp × 746 W hp
1000 W
× 0.8 pf × 0.92 efficiency
kW
= 50.7 kVA
kVA base =
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.
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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:
Z1 = Z 2
1.0
Z1 =
I f3Φ
b g
3 × 1.0
Z1 + Z 2 + Z 0
I fslg =
b
Z0 =
3
− Z1 − Z 2
I fslg
g
Utilities often report available single-line-to-ground fault duties on an equivalent threephase rating apparent power basis, using the equation:
kVA = 3 × I fslg × kV
However, the actual apparent power of a single-line-to-ground fault is:
kVA = I fslg ×
kV
3
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
Short Circuit Study
DAPPER 4-23
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.
UTILITY
X1 1.0000 pu
Xo 1.3333 pu
B1
X1 1.0000 pu
Xo 1.0000 pu
C1
B2
X1 1.0000 pu
Xo 1.0000 pu
T1
B3
40,094.23 A 3 Ph
51,549.68 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:
b
g b
g b
g
Z Thevenin = 0.010101+ j0.999948 + 0 + j1.0 + 0 + j1.0 pu Ω
= 0.010101+ j2.999948 pu Ω
Combining the real and imaginary components of the Thevenin equivalent impedance into
a single complex impedance yields:
Z Thevenin = j2.999965 pu 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:
1
pu A
j2.999965
= - j0.333337 pu A
Isc =
SKM Power*Tools for Windows
DAPPER 4-24
Reference Manual
The base current on the 480 V side of the transformer on a 100 MVA base is:
100,000 kVA
3 × 0.48 kV
= 120,281.3 A
I base =
Thus the fault current in amperes is:
Isc = 120,281.3 A × 0.333337 pu A
= - j40,094.2 A
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:
Islg =
3V
Z1 + Z 2 + Z 0 + 3Z 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 0transformer = 0 + j1.0 pu Ω
Therefore, the single-line-to-ground fault current is:
Islg =
b
3 × 1 pu V
pu Ω
0.010101+ j2.999948 + 0.010101+ j2.999948 + 0 + j1.0
g b
g b
g
3V
0.020202 + j6.999896 pu Ω
3 pu V
=
j6.999925 pu Ω
= - j0.428576 pu A
=
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:
Islg = I base × Isc pu
= 120,281.3 A × 0.428576 pu A
= j51,549.7 A
This single-line-to-ground calculation matches the Short Circuit Study’s result of
51,549.68 A shown in Figure 4-1.
3/26/2006
Short Circuit Study
DAPPER 4-25
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:
Islg =
3V
Z1 + Z 2 + Z 0
Therefore, solving for the zero-sequence impedance yields:
Z0 =
3V
- Z1 - Z 2
Islg
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 = j1803.97 pu Ω - 2( j2.999948) pu Ω
= j1797.99 pu Ω
The total zero-sequence impedance is
Z 0 = Z tranformer0 + 3Z ground
Rearranging the equation yields:
3Z ground = Z 0 - Z tranformer0
Substituting known values yields:
j1797.99 - j1.00
3
j1796.99
=
pu Ω
3
= 598.99 pu Ω
Z ground =
But the base ohms in this part of the system is:
Z base =
0.48 kV 2
= 0.002304 Ω
100 MVA
SKM Power*Tools for Windows
DAPPER 4-26
Reference Manual
Therefore the grounding reactor is:
Z generator = Z base × Z pu
= 0.002304 Ω × 598.99 pu Ω
= j1.38 Ω
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.
X1 1.0000 pu
Xo 1.3333 pu
UTILITY
B1
X1 1.0000 pu
Xo 1.0000 pu
C1
B2
X1 1.0000 pu
Xo 1.0000 pu
T1
1.4 Ω
B3
40094.23 A 3 Ph
197.19 A SLG
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.
3/26/2006
Short Circuit Study
DAPPER 4-27
UTILITY
X1 0.4183 pu
Xo 0.1673 pu
B1
C1
X1 1.000 pu
Xo 1.000 pu
X1 1.000 pu
Xo 1.000 pu
B2
T1
B3
Thevenin X1 2.4184 pu
Thevenin Xo 2.1673 pu
51519.40 A SLG
Figure 4-3
Given a utility three-phase fault current of 10,000 A at 13.8 kV and a single-line-toground fault current of 12,500 A, the equivalent Short Circuit Capacity in MVA is:
SCC3Φ = 3 × 10,000 A × 13.8kV
= 239,023 kVA
SCCslg = 12,500 A
FG 13.8 kV IJ
H 3 K
= 99,593 kVA
Whereas the single-line-to-ground fault current may be greater than the system’s threephase 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:
100,000 kVA
239,023 kVA
= j0.4184 pu Ω
Z utility =
Likewise, the utility zero-sequence impedance is:
Z0 =
3
Islg
− Z1 − Z 2
In order to determine the single-line impedance, you must first express the current as a per
unit value.
SKM Power*Tools for Windows
DAPPER 4-28
Reference Manual
Islg =
I base =
Islg actual Amperes
Is base Amperes
100,000 kVA
pu A
e 3jb13.8kvg
= 4183.7 A
12,500 A
=
pu A
4183.7 A
= - j2.9878 pu A
Islg
Substituting:
b
3
− j0.4184 + j0.4184
- j2.9878
= j0.1673 pu Ω
Z0 =
g
If the fault occurs at Bus 3 and the transformer connection is wye-grounded wyegrounded, then the zero-sequence impedance is the series of the transformer, cable and
utility zero-sequence impedances:
Z 0 Bus 3 = j0.1673+ j1.0 + j1.0
= j2.1673 pu Ω
If the driving point voltage is 1 pu, the single-line-to-ground fault current at Bus 3 is:
3 × 1 pu V
Z1 + Z 2 + Z 0
3 × 1 pu V
=
j2.418 + j2.418 + j2.1673 per Ω
= - j0.4284 pu A
Zslg =
In amperes:
Islg = I pu × I base
Islg = - j0.4284 × 120,281.3 A
= j51,524.9 A
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.
3/26/2006
Short Circuit Study
B1
X1 1.0000 pu
Xo 1.0000 pu
DAPPER 4-29
GENERATOR
X pos 1.0000 pu
X neg 0.8000 pu
X zero 0.5000 pu
C1
B2
X1 1.0000 pu
Xo 1.0000 pu
T1
B3
40093.79 A 3 Ph
53065.31 A SLG
Thevenin X1 3.0000 pu
Thevenin X2 2.8000 pu
Thevenin Xo 1.000 pu
Figure 4-4
The three-phase fault current at Bus 3 is:
1
pu A
3.0000
= - j0.3333 pu
= 120,281.3 A
= 0.3333 pu A × 120,281.3 A
= - j40,093.77 A
Ik =
I base
I3Φ
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
FAULT
MODEL
MODEL
MODEL
R E P O R T
TYPE: SLG
INDUCTION MOTOR CONTRIBUTION: YES
TRANSFORMER TAPS: YES
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
SKM Power*Tools for Windows
DAPPER 4-30
Reference Manual
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
X1 1.0000 pu
Xo 1.0000 pu
C1
X1 1.0000 pu
Xo 1.0000 pu
T1
GENERATOR
X" X1 1.0000 pu
X" X2 0.8000 pu
X" Xo 0.5000 pu
B2
B3
X1 1.0000 pu
Xo 1.0000 pu
C2
B4
MOTOR
150,351.94 A 3 Ph
181,177.31 A SLG
Thevenin X1 0.8000 pu
Thevenin X2 0.7917 pu
Thevenin 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:
Z1 =
b j1.0 + j1.0 + j1.0 + j1.0g × j1.0
j1.0 + j1.0 + j1.0 + j1.0 + j1.0
-4.0
=
pu Ω
j5.0
= j0.8 pu 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:
3/26/2006
Short Circuit Study
Z0 =
DAPPER 4-31
b j1.0 + j1.0g × j0.5
j1.0 + j1.0 + j0.5
- j1.0
=
pu Ω
j2.5
= j0.4 pu 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
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
B4
B1
B2
B3
B4
B4
GENERATOR
MOTOR
B1
B2
B3
==== INI. SYM. RMS SYSTEM BUS
ALL BUSES REPORTED
---PHASE A--13800.0
.7500 /
0.
13800.0
.5000 /
0.
480.0
.2500 / -30.
480.0
.0000 /-123.
VOLTAGES ( PU / DEG ) =======
AT TIME =
.5 CYCLES
---PHASE B-----PHASE C--.7500 /-120.
.7500 / 120.
.5000 /-120.
.5000 / 120.
.2500 /-150.
.2500 / 90.
.0000 / 117.
.0000 / -3.
==== INI. SYM. RMS SYSTEM BRANCH FLOWS ( AMPS/DEG ) =======
ALL BRANCHES REPORTED
AT TIME =
.5 CYCLES
VOLTS --PHASE A-----PHASE B-----PHASE
13800.
1045.9/ -90.
1045.9/ 150.
1045.9/
480. 120281.3/-120. 120281.2/ 120. 120281.3/
B2
13800.
1045.9/ -90.
1045.9/ 150.
1045.9/
B3
13800.
1045.9/ -90.
1045.9/ 150.
1045.9/
B4
480. 30070.6/-120. 30070.6/ 120. 30070.6/
C
30.
0.
30.
30.
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.
SKM Power*Tools for Windows
DAPPER 4-32
Reference Manual
GENERATOR
1045.94 A
B1
1045.94 A
C1
B2
1045.94 A
T1
B3
30,070.63 A
C2
B4
150,351.94 A 3 Ph
MOTOR
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
ALL BUSES REPORTED
---PHASE A--13800.0
.8609 /
5.
13800.0
.7237 / 13.
480.0
.3054 / -30.
480.0
.0000 /-123.
B1
B2
B3
B4
B4
GENERATOR
MOTOR
B1
B2
B3
VOLTAGES ( PU / DEG ) =======
AT TIME =
.5 CYCLES
---PHASE B-----PHASE C--.9833 /-120.
.8607 / 115.
.9874 /-120.
.7236 / 107.
.9108 /-139.
.9108 / 79.
.9135 /-139.
.9135 / 79.
==== INI. SYM. RMS SYSTEM BRANCH FLOWS ( AMPS/DEG ) =======
ALL BRANCHES REPORTED
AT TIME =
.5 CYCLES
VOLTS --PHASE A-----PHASE B-----PHASE C
13800.
742.9/-121.
17.5/ -30.
742.9/ 61.
480. 144438.2/-120.
501.2/-180.
505.3/ -60.
B2
13800.
742.9/-121.
17.5/ -30.
742.9/ 61.
B3
13800.
742.9/-121.
17.5/ -30.
742.9/ 61.
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
Short Circuit Study
***************** F A U L T
A N A L Y S I S
DAPPER 4-33
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
13800.
739.4/-120.
.0/ 59.
739.4/ 60.
B2
13800.
739.4/-120.
.0/ 78.
739.4/ 60.
B3
13800.
739.4/-120.
.0/ 78.
739.4/ 60.
B4
480. 36821.3/-120.
.0/-120.
.0/-120.
GENERATOR
B1
B2
B3
*****************************************************************************
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
GENERATOR
B1
B2
B3
==== INI.
RMS
SYSTEM BRANCH FLOWS ( AMPS/DEG ) =======
ALL BRANCHES REPORTED
AT TIME =
.0 CYCLES
VOLTS
-POS SEQ-NEG SEQ-ZER SEQ13800.
426.9/ 90.
426.9/ 30.
.0/ 45.
B2
13800.
426.9/ -90.
426.9/-150.
.0/ 45.
B3
13800.
426.9/ -90.
426.9/-150.
.0/ 45.
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 oneline 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 X ′′d 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’ X ′′d 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.
SKM Power*Tools for Windows
DAPPER 4-34
Reference Manual
UTILITY
22,768.38 A
B1
FEEDER
22,768.35 A
B2
C1
24,686.32 A
C2
698.55 A
B3
C3
698.55 A
B4
M1
M2
Size 100.0 kVA
Xd" 0.1700 pu
No. Mtrs 1
100 Mtr kVA Base
Size 25.0 kVA
Xd" 0.1700 pu
No. Mtrs 4
25 Mtr kVA Base
C4
698.55 A
835.20 A
B5
B6
M3
M4
Size 25.0 kVA
Xd" 0.6800 pu
No. Mtrs 4
100 Mtr kVA Base
Size 100.0 kVA
Xd" 0.1700 pu
No. Mtrs 1
100 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 X ′′d must be multiplied by four and the product entered in the X ′′d 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.
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:
FG kV IJ FG kVA IJ
H kV K H kVA K
2
X ′′d = X d motor
motor
base
base
motor
Substituting:
FG 0.44 kV IJ FG 100 kVA IJ
H 0.48 kV K H 100 kVA K
2
X ′′d = 0.17
= 0.143 pu 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
Short Circuit Study
DAPPER 4-35
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 cable =
FG 0.202 + j0.0467 Ω IJ 50 ft
H 1000 ft K
= 0.0101+ j0.002335 Ω
On a 100 kVA, 480 V base, the base impedance is:
Z base =
kV base2
MVA base
0.482
0.100
= 2.304 Ω
=
Therefore, the cable impedance on the system base of 100 kVA is:
Z cable = 0.00438 + j0.000998 pu Ω
Adding the motor and cable impedances yields:
b
g b
g
Z branch = 0.00438 + j0.000998 + 0.0 + j0143
.
pu Ω
= 0.00438 + j0.14385
= 0.1439 pu Ω
The fault duty contribution to Bus 2 is:
Isc =
=
V
Z
FG 1 IJ pu Ω
H 0.1439 K
= 6.949 pu Ω
SKM Power*Tools for Windows
DAPPER 4-36
Reference Manual
However the Ibase is:
100 kVA
1.732 × 0.48 kV
= 120.28 A
I base =
Therefore the fault current contribution for Motor 4 is:
Isc = 120.28 A × 6.949 pu A
= 835.83 A
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
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.
GENERATOR
X1 0.2000 pu
Pri Rated Voltage 13,800.0 V
Sec Rated Voltage 4160.0 V
Leakage Impedance 0.6000 pu
Pri Tap 0.00 %
Sec Tap 0.00 %
Bus System Nominal 13,800.0 V
B1 Pre Fault Driving Point Voltage 1.00 pu V
Bus Fault Current 20,918.49 A
T1
Bus System Nominal 4160.0 V
B2 Pre Fault Driving Point Voltage 1.00 pu V
Bus Fault Current 17,348.27 A
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:
1
pu A
j0.8
= - j1.25 pu A
Isc =
However the base current at 4.16 kV is:
3/26/2006
Short Circuit Study
I base =
DAPPER 4-37
100,000 kVA
3 × 4.16 kV
= 13,878.6 A
Isc = I base × I pu A
= 13,878.6 A × 1.25 pu A
= - j17,348.27 A
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.
GENERATOR
X1 0.2000 pu
Pri Rated Voltage 13,200.0 V
Sec Rated Voltage 3979.0 V
Leakage Impedance 0.6000 pu
Pri Tap 0.00 %
Sec Tap 0.00 %
B1
Bus System Nominal 13,800.0 V
Pre Fault Driving Point Voltage 1.00 pu V
Bus Fault Current 20,918.49 A
B2
Bus System Nominal 4160.0 V
Pre Fault Driving Point Voltage 1.00 pu V
Bus Fault Current 18,531.12 A
T1
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:
FG kV IJ F kV I
H kV K GH kV JK
F 13.2 kV IJ FG 100,000 kVA IJ
= j0.06 pu × G
H 13.8 kV K H 10,000 kVA K
F 3.979 IJ FG 100,000 kVA IJ
= j0.06 × G
H 4.16 K H 10,000 kVA K
2
Z new = Z given ×
given
new
new
given
2
2
= j0.54896 pu Ω
SKM Power*Tools for Windows
DAPPER 4-38
Reference Manual
The Thevenin equivalent impedance at Bus 2 is the sum of the generator and transformer
impedance as shown:
Z = j0.2 + j0.54896 pu Ω
= j0.74896 pu Ω
1.0 pu V
j0.74896 pu Ω
= - j1.335 pu A
Isc =
Remembering that Ibase is 13,878.6 A for a system base of 4160 V, the fault current at Bus
2 is:
Isc = 13,878.6 A × - j1.335 pu A
= - j18,531 A
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
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.
GENERATOR
X1 0.2000 pu
Bus System Nominal 13,800.0 V
B1 Pre Fault Driving Point Voltage 1.00 pu
Bus Fault Current 20,918.49 A
Pri Rated Voltage 13,200.0 V
T1
Sec Rated Voltage 4160.0 V
Leakage Impedance 0.6000 pu
Pri Tap 4.54 %
Sec Tap 0.00 %
Bus System Nominal 4160.0 V
B2 Pre Fault Driving Point Voltage 1.00 pu
Bus Fault Current 17,348.73 A
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.
3/26/2006
Short Circuit Study
DAPPER 4-39
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 prefault driving point voltage at Bus 2 is 1.0256 pu higher than the nominal system voltage.
GENERATOR
X1 0.2000 pu
Bus System Nominal 13,800.0 V
B1 Pre Fault Driving Point Voltage 1.00 pu
Bus Fault Current 20,918.49 A
Pri Rated Voltage 13,800.0 V T1
Sec Rated Voltage 4160.0 V
Leakage Impedance 0.6000 pu
Pri Tap -2.50 %
Sec Tap 0.00 %
Bus System Nominal 4160.0 V
B2 Pre Fault Driving Point Voltage 1.03 pu
Bus Fault Current 17,565.02 A
Figure 4-11
The Thevenin equivalent impedance is calculated as:
F 1 IJ + j0.6 pu Ω
Z = j0.2 × G
H 1+ Tap K
F 1 IJ + j 0.6 pu Ω
= j0.2 × G
H 1- 0.025K
2
2
= j(0.2 × 1.0519) + j0.6 pu Ω
= j0.81034 pu Ω
1.0256 pu V
pu Ω
j0.81034
= - j1.26558 pu A
Isc =
Remembering that the base current at 4160 V is 13878.6 A, then the fault current is:
Isc = 13,878.6 A × - j1.26558 pu A
= - j17,565 A
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.
SKM Power*Tools for Windows
DAPPER 4-40
Reference Manual
GENERATOR
X1 0.2000 pu
B1
Pri Rated Voltage 13,800.0 V
Sec Rated Voltage 4160.0 V
Leakage Impedance 0.6000 pu
Pri Tap -2.50 %
Sec Tap 0.00 %
Bus System Nominal 13,800.0 V
Pre Fault Driving Point Voltage 1.00 pu V
Bus Fault Current 20,918.49 A
T1
B2
Bus System Nominal 4266.5 V
Pre Fault Driving Point Voltage 1.00 pu V
Bus Fault Current 17,565.02 A
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:
FG 13.455 kV IJ FG 100,000 kVA IJ
H 13.8 kV K H 10,000 kVA K
2
Z = j0.2 + j0.06 ×
= j0.2 + j0.5704
= j0.7704 pu Ω
Therefore, the fault current at Bus 2 is:
1.0 pu V
j0.7704 pu A
= - j1.2980 pu A
Isc =
The base current at this voltage is:
I base =
100,000 kVA
3 × 4.2665 kV
= 13532.2 A
Isc = I base × I pu A
= 13532.2 A × 1.2980 pu A
= - j17,565 A
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
Short Circuit Study
DAPPER 4-41
UTILITY
100.0 MVA 3 Ph
30.0 MVA SLG
B1
B4
4183.70 A 3 Ph
3765.33 A SLG
C1
6609.78 A 3 Ph
8007.95 A SLG
T2
B2
T1
3803.54 A 3 Ph
3454.58 A SLG
B3
6609.78 A 3 Ph
8007.95 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 (threephase) 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
SKM Power*Tools for Windows
DAPPER 4-42
Reference Manual
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 wyegrounded.
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
COMPUTERS
H3A BUS 19
018-RA
OFFICE COMPLEX AREA
PNL 19 H3A
MCCB2
MCCB1
PNL 18 RA
C18
C15
PLN 16 H2A
BUS 16
BUS 15
MCC 15 - 1A
C16
C14
LVP6
TX C
CB4
R4
CB5
R5
CAP 1
PLN - 17 H1A
BUS 17
C17
LVP3
CMP HVAC
MCP#10
F2
HVAC BUS
C6
GEN 2
DSB 1 BUS 14
GEN 1
CMP CTR
TX H
LVP2
F5
C5
F1
DS SWG1 BUS 8
LVP1
SWBD-1
TX C PRI BUS 9
TX B
TX B PRI BUS 4
C1
CB3
R3
C19
CB12
029-TX D SEC
TX D
005-TXD PRI
C2
CB6
R6
M13
DS SWG2 BUS 13
TX3 SEC BUS 11
TX A
CB2
R2
TX A PRI
TM -1
CB1
R1
UTILITY BUS
EDISON
L2
022-DSB 2
021-TX F PRI
PCB1
C7
003-HV SWGR
M23-A
023-MTR 23
C12
TX F
C9
DS SWG3 BUS 20
PCB2
C8
TX3 TER BUS 12
INDUSTRIAL COMPLEX AREA
M23-B
TX 6
006-TX3 PRI
C3
CB7
R7
L1
M20
PCB3
R11
GEN 3
SYN A
DEMONSTRATION PROJECT FOR POWER*TOOLS FOR WINDOWS
CB8
R8
DSB 3 BUS 27
TX G
F4
C10
LVP4
C13
M28 #1 & 2
BUS 28 A
L5
026-TX G PRI
SYN B
CB9
R9
M28 #3
MO/L#28B-1
MCP#28B-1
M28 #4
BUS 28 B
M25
MO/L#28B-2
MCP#28B-2
MO/L#25
F3
SW1
025-MTR 25
BLDG 115 SERV
C11
LVP5
C20
TX E
007-TX E PRI
C4
CB10
R10
Short Circuit Study
DAPPER 4-43
SKM Power*Tools for Windows
DAPPER 4-44
Reference Manual
The following figure shows a portion of the Plant project including Short Circuit Study
results.
SHORT CIRCUIT STUDY
FAULT ALL BRANCHES
BUILDING 115 SERVICE
BLDG 115 SERV
10618.85 A 3 PH
10557.23 A SLG
C11
C10
026-TX G PRI
025-MTR 25
10060.67 A 3 PH
9694.91 A 3 PH
9198.14 A SLG
9466.09 A SLG
F4
SW1
F3
TX G
MO/L#25
DSB 3 BUS 27
M25
24713.14 A 3 PH
23955.28 A SLG
L5
C20
C13
LVP5
LVP4
BUS 28 B
18508.62 A 3 PH
15548.13 A SLG
MCP#28B-2
MCP#28B-1
BUS 28 A
20786.14 A 3 PH
MO/L#28B-1
MO/L#28B-2
17923.31 A SLG
M28 #1 & 2
M28 #3
M28 #4
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
BUS#
3/26/2006
V O L T A G E
PU VOLTS
P R O F I L E **************
NAME
BASE VOLTS
ANGLE (D)
UTILITY BUS
TX A PRI
003-HV SWGR
69000.00
69000.00
13800.00
1.0000
1.0000
1.0256
0.
0.
-30.
007-TX E PRI
13800.00
1.0256
-30.
BLDG 115 SERV
026-TX G PRI
025-MTR 25
4160.00
4160.00
4160.00
1.0519
1.0519
1.0519
-60.
-60.
-60.
DSB 3 BUS 27
BUS 28 A
029-TX D SEC
BUS 28 B
480.00
480.00
4160.00
480.00
1.0789
1.0789
1.0519
1.0789
-90.
-90.
-60.
-90.
Short Circuit Study
DAPPER 4-45
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
1/2 CYCLES
32080.4
INTERRUPTING AMPS
3 CYCLES
5 CYCLES
24753.0
24713.7
8 CYCLES
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
1/2 CYCLES
31011.3
INTERRUPTING AMPS
3 CYCLES
5 CYCLES
23990.9
23955.7
8 CYCLES
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 threephase 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
TX B PRI BUS 4
005-TXD PRI
006-TX3 PRI
007-TX E PRI
13800.
13800.
13800.
13800.
13800.
10034.8
9830.7
9891.7
9966.3
9970.8
13.6
7.0
10.2
11.1
11.5
12796.2
12279.6
12452.9
12526.3
12628.7
12.3
5.3
8.5
8.7
9.8
DS SWG1 BUS 8
TX C PRI BUS 9
HVAC BUS
TX3 SEC BUS 11
DS SWG2 BUS 13
4160.
4160.
4160.
4160.
4160.
6072.8
6026.0
5701.7
18318.1
18303.7
8.4
7.5
3.6
9.4
9.3
6640.8
6559.7
5999.3
18976.2
18946.6
8.6
7.4
3.0
6.9
6.8
TX3 TER BUS 12
DS SWG3 BUS 20
4160.
4160.
20317.1
20307.9
7.3
7.3
19500.2
19468.0
5.1
5.0
SKM Power*Tools for Windows
DAPPER 4-46
Reference Manual
021-TX F PRI
BLDG 115 SERV
026-TX G PRI
4160.
4160.
4160.
17384.8
10618.9
9694.9
2.8
8.9
3.9
15443.0
10557.2
9198.1
2.2
8.1
3.3
025-MTR 25
022-DSB 2
023-MTR 23
DSB 3 BUS 27
BUS 28 A
4160.
480.
480.
480.
480.
10060.7
45790.9
47063.0
24713.1
20786.1
7.4
6.4
7.8
5.9
4.3
9466.1
34099.1
31374.4
23955.3
17923.3
6.3
6.0
5.6
5.8
4.0
DSB 1 BUS 14
BUS 15
018-RA
BUS 16
BUS 17
480.
480.
480.
480.
480.
16327.5
15492.8
11644.5
13770.7
6760.2
6.3
5.4
2.1
2.6
0.8
17986.3
16457.9
10984.0
13918.5
5873.2
6.1
5.2
1.9
2.3
0.7
H3A BUS 19
029-TX D SEC
BUS 28 B
UTILITY BUS
TX A PRI
480.
4160.
480.
69000.
69000.
11572.0
9921.5
18508.6
8063.3
2281.5
1.7
7.3
2.2
26.9
19.6
10988.3
9555.9
15548.1
2547.6
752.8
1.5
6.7
1.8
14.7
11.7
208.
23552.7
5.3
25176.4
5.2
CMP CTR
*********************** FAULT ANALYSIS REPORT COMPLETED ***********************
3/26/2006
Bib
A Bibliography
SKM Power*Tools for Windows
A-1
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: McGrawHill, 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, 115
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
SKM Power*Tools for Windows
DAPPER ii
Reference Manual
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
3/26/2006
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
Index
Oil/Forced Air Transformer, 2-4
DAPPER iii
Radial Systems, 1-2
Reactive Power, 3-10
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
S
T
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
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, 413
phase shift, 4-13
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
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
SKM Power*Tools for Windows
DAPPER iv
Reference Manual
Zero-Sequence Modeling. See Sequence Modeling
3/26/2006