calculating voltage drops and symmetrical rms fault currents

Note: The source of the technical material in this volume is the ProfessionalEngineering Development Program (PEDP) of Engineering Services.

Warning: The material contained in this document was developed for SaudiAramco and is intended for the exclusive use of Saudi Aramco’semployees. Any material contained in this document which is notalready in the public domain may not be copied, reproduced, sold, given,or disclosed to third parties, or otherwise used in whole, or in part,without the written permission of the Vice President, EngineeringServices, Saudi Aramco.

Chapter : Electrical For additional information on this subject, contactFile Reference: EEX10204 W.A. Roussel on 874-1320

Engineering EncyclopediaSaudi Aramco DeskTop Standards

Calculating Voltage Drops AndSymmetrical R MS Fault Currents

Engineering Encyclopedia Electrical

Calculating Voltage Drops andSymmetrical RMS Fault Current

Saudi Aramco DeskTop Standards

TABLE OF CONTENTS PAGES

MODELING AN ELECTRICAL POWER SYSTEM IN PER-UNIT ..............1

Modeling Methods ............................................................................1

Ohmic Model..........................................................................1

Percent Model........................................................................1

Per-Unit Model .......................................................................3

One-Line Diagram (Step 1) ..............................................................3

Base Values (Step 2)........................................................................5

Base Kilovolt-Amperes (kVAb) ..............................................5

Base Kilovolts (kVb)...............................................................5

Base Current..........................................................................6

Base Ohms (Zb) ....................................................................6

Table of Base Values.............................................................7

Data Collection and Modeling (Step 3).............................................7

Utility or Source......................................................................9

Grounding Resistors ............................................................11

Transformers........................................................................11

Conductors...........................................................................12

Aerial Lines ..........................................................................13

Generators ...........................................................................14

Motors ..................................................................................14

Per-Unit Values (Step 4).................................................................16

Utility or Source....................................................................16

Transformers........................................................................17

Motors and Generators........................................................18

Conductors...........................................................................18

Aerial Lines ..........................................................................19

Impedance Diagrams (Step 5)........................................................20




Complex Diagram ................................................................20

R-Only Diagram ...................................................................22

X-Only Diagram ...................................................................23

CALCULATING LINE VOLTAGE DROPS................................................24

Introduction.....................................................................................24

Calculation Methods .......................................................................24

Exact Calculations ...............................................................25

Approximate Calculations ....................................................25

Ohmic Model Selection Factors......................................................26

One-Line Diagram................................................................26

Load Current (IL)..................................................................28

Theta () Calculations............................................................28

Line Impedance (Z = R + jX)................................................28

Voltage Drop Line-to-Neutral ...............................................28

Voltage Drop Line-to-Line(VD).............................................28

Load Voltage (VL) ................................................................29

Percent Voltage Drop (%VD)...............................................29

Per-Unit Model Selection Factors...................................................29

One-Line Diagram................................................................29

Base Values.........................................................................29

Per-Unit Calculations ...........................................................29

Voltage Drop (VD)................................................................30

Percent Voltage Drop (%VD)...............................................30

CALCULATING LARGE MOTOR STARTING VOLTAGE DROPS .........34

Introduction.....................................................................................34

Motor Starting Effects .....................................................................35

Critical Voltage Levels (Minimum) .......................................35

Effects on Plant Equipment..................................................37

Calculation Methods .......................................................................38




Exact Calculations ...............................................................38

Approximate Calculations (Neglect R).................................39


One-Line Diagram................................................................39

Base Values.........................................................................41


Impedance Diagram.............................................................43

Voltage Drop ........................................................................44

Percent Voltage Drop...........................................................44

Voltage Drop (Other Buses) ................................................44

CALCULATING SYMMETRICAL RMS FAULT CURRENTS(BALANCED FAULT CONDITIONS) ........................................................46

Purposes of Fault Calculations.......................................................46

Protective Device Interrupting Ratings ................................46

Equipment Component Withstand Ratings..........................46

Protective Relay Pickup Settings.........................................46

Maximum Relay Settings for Sensitivity...............................46

Effects of Short Circuits..................................................................47

Arcing and Burning ..............................................................47

Current Flow ........................................................................47

Thermal Stress.....................................................................47

Mechanical Stress ...............................................................48

Voltage Drops ......................................................................48

Characteristics of Fault Current Sources .......................................49

Generators (Local)...............................................................50

Motors ..................................................................................51

Utility System .......................................................................53

Total Fault Current ...............................................................54

Machine Reactance Modeling ........................................................55

Typical Short Circuit Oscillogram.........................................55




Subtransient Reactance (X”d) .............................................56

Transient Reactance (X’d) ...................................................56

Synchronous Reactance......................................................56

Symmetrical Versus Asymmetrical Fault Currents.........................56

Symmetrical Fault Currents .................................................57

Asymmetrical Fault Currents ...............................................58

System Power Factor and X/R Ratios .................................59

Sample Waveshapes...........................................................63

AC and DC Components .....................................................65

Types of Faults/Magnitudes ...........................................................67

Three-Phase Faults .............................................................67

Line-to-Line Faults ...............................................................67

Line-to-Ground Faults ..........................................................68

Arcing Faults ........................................................................68

Overloads.............................................................................68


One-Line Diagram................................................................69

Base Values.........................................................................70


Impedance Diagrams...........................................................71

Modification of Per-Unit Values............................................71

Thevenin Equivalent Network ..............................................75

Fault Current Calculations ...................................................75

WORK AID 1: RESOURCES USED TO MODEL ANELECTRICAL POWER SYSTEM IN PER UNIT...............78

Work Aid 1A: IEEE Standard 141-1986 (Red Book) ......................78

Work Aid 1B: Applicable Modeling Procedures..............................78

WORK AID 2: RESOURCES USED TO CALCULATE LINEVOLTAGE DROPS ...........................................................90





Work Aid 2B: Applicable Calculation Procedures(Ohmic Method) .......................................................90

Work Aid 2C: Applicable Calculation Procedures(Per-Unit Method) ....................................................91

Work Aid 2D: SAES-P-100.............................................................91

WORK AID 3: RESOURCES USED TO CALCULATE LARGEMOTOR STARTING VOLTAGE DROPS .........................92


Work Aid 3B: IEEE Standard 399-1990 (Brown Book)...................92

Work Aid 3C: Applicable Calculation Procedures ..........................93

WORK AID 4: RESOURCES USED TO CALCULATESYMMETRICAL RMS FAULT CURRENTS(BALANCED FAULT CONDITIONS)................................94


Work Aid 4B: Applicable Calculation Procedures...........................94

GLOSSARY ..............................................................................................98



Saudi Aramco DeskTop Standards 1

MODELING AN ELECTRICAL POWER SYSTEM IN PER-UNIT

Modeling Methods

In order to calculate short-circuits, line voltage drops and motor starting currents, thedifferent impedance values for circuit elements (transformers, cables, motors,generators) must be determined from nameplates, handbooks, and catalogs. Theseimpedance values can be modeled in ohms, percent, or per-unit on a chosen basevalue. In most calculations the per-unit modeling of circuit elements is used because itis easier and more convenient to solve the problems when the system contains severalvoltage levels. Impedances that are modeled in per-unit can be combined in parallel orseries regardless of the number of voltage levels.

Ohmic Model

Modeling an electrical circuit element (for example, a cable) in ohms is quite simple.However, if the power system contains more than one voltage level, which is almostalways the case, the ohmic value will change as the square of the ratio of the voltagelevels. In other words, the ohmic values change from one side of a transformer to theother. For example, if a power system has three voltage levels, which is notuncommon, each electrical circuit element (cable, transformer, motor, etc.) will havethree ohmic values (see Figure 1). For this reason, power system studies are typicallynot performed using ohmic models.

Percent Model

The percent method of modeling an electrical system differs from the per-unit methodby a factor of 100 (percent = 100 x p.u.). It is also not typically used for power systemstudies calculations because it leads to simple math errors. For example, 50% x 100%= 50% and not 5000% as the math indicates (50 X 100).




Figure 1. Ohmic Model Impedances




Per-Unit Model

When impedances of circuit elements are expressed in per-unit, the problemsmentioned for both the ohmic and percent models are eliminated. Impedancesexpressed in per-unit on a defined base are the same on both sides of a transformer.For example, referring to Figure 1, ZT1 = 7% on both the 13.8 kV primary and 4.16 kVsecondary sides of the transformer, and ZT2 =5.5% on both sides of the 4.16 kVprimary and .48 kV secondary sides of the transformer. By definition, the per-unitquantity (ohms, voltage, current, etc.) equals the ratio of the actual quantity to the basequantity (p.u. quantity = actual quantity/base quantity).

The following steps are use to model an electrical system in per-unit: Note: Work Aid 1has been developed to teach per unit procedures.

• STEP 1 - One-line Diagram

• STEP 2 - Select/Calculate Base Values

• STEP 3 - Data Collection and Modeling

• STEP 4 - Per-Unit Impedance Calculations

• STEP 5 - Impedance Diagram Note: These five steps are the same for per-unitmodeling of any power system regardless of the type of power study beingperformed.

One-Line Diagram (Step 1)

An accurate one-line diagram (system road map) should be prepared that shows allsources (utility tie, generators, and motors) and significant circuit elements(transformers, cables, busway etc.). Note: The study to be performed will only be asaccurate as the one-line diagram. Figure 2 is an example one-line diagram of a powersystem that will be used throughout the remainder of this Module.




Figure 2. Example One-Line Diagram




Base Values (Step 2)

In the per-unit system, there are four base quantities: base kilovolt-amperes (kVAb),base kilovolts (kVb), base amperes(Ib), and base ohms or base impedance (Zb).

Base Kilovolt-Amperes (kVA b)

The selection of kVAb is arbitrary. However, a convenient value is usually selected tomake the mathematical calculations somewhat less tedious. For purposes of thisModule, select kVAb = 100000 kVA or MVAb = 100 MVA (see Work Aid 1). Note: Per-unit formulas are often listed in terms of MVAb versus kVAb where 1MVA = 1000 kVA.

Base Kilovolts (kV b)

kVb is selected to match one of the system transformer’s rated voltages and then thebase voltages at other levels are established (calculated) by transformer turns ratios,which also equal the transformer voltage ratios.

Example A: Using Work Aid 1, what are the base voltages for the transformers shownin Figure 2?

Answer A: Bus 100 Let kVb = 13.8 kV

Bus 50 kVb = 13.8(34.5/13.8) = 34.5 kV

Bus 250 kVb = 13.8(0.48/13.8) = 0.48 kV

Bus 300 kVb = 13.8(4.16/13.8) = 4.16 kV




Base Current

Base current (Ib) is calculated through use of the apparent power relationshipsdeveloped in the previous Module.

• kVAb = 3 x kVb x Ib

• Ib = kVAb/( 3 x kVb)

Example B: Using Work Aid 1, what are the base currents for the power systemshown in Figure 2?

Answer B: Bus 50 Ib = 100000/( 3 x 34.5) = 1673.5 A

Bus 100 Ib = 100000/( 3 x 13.8) = 4183.7 A

Bus 250 Ib = 100000/( 3 x 0.48) = 120281.3 A

Bus 300 Ib = 100000/( 3 x 4.16) = 13878.6 A

Base Ohms (Z b)

Base ohms (Zb) is also calculated through use of voltage, current, impedance, andpower relationships developed in the previous Module. These relationships are thefollowing:

• Zb = (kVb)2/MVAb = [(kVb)2 x 1000]/kVAb

• Zb = Rb = Xb

Example C: Using Work Aid 1, what are the base ohms for the power system shownin figure 2?

Answer C: Bus 50 Zb = (34.5)2/100 = 11.902

Bus 100 Zb = (13.8)2/100 = 1.9044

Bus 250 Zb = (0.48)2/100 = 0.0023

Bus 300 Zb = (4.16)2/100 = 0.173




Table of Base Values

For easier data retrieval, the base values that are selected and calculated for aparticular power system study are usually listed in tabular form. For the one-linediagram shown in Figure 2, the base values are listed in Figure 3.

Bus

Numbers

kVAb kVb Ib Zb

50 100000 kVA 34.5 kV 1673.5 A 11.902

100, 150, 200 100000 kVA 13.8 kV 4183.7 A 1.9044

300 100000 kVA 4.16 kV 13878.6 A 0.173

250 100000 kVA 0.480 kV 120281.3 A 0.0023

Figure 3. Table of Base Values

Data Collection and Modeling (Step 3)

This step involves collecting data and modeling each of the electrical circuit elementsof the power system that is being studied. The following circuit elements will bemodeled:

• Utility or source




• Transformers

• Conductors (cables, busway, aerial lines)

• Generators

• Motors (over 1000V)

• Motors (under 1000V)




Utility or Source

The utility system is modeled as a voltage (EU) behind an impedance (ZU) as shown inFigure 4. The utility can provide the data in any of the following forms:

• kVASCA and X/R ratio

• MVASCA and X/R ratio

• Amperes (ISCA) and angle (-)

• R and X in ohms ()

• R and X in per-unit (p.u.) on a given base

Figure 4. Utility or Source Model (Zu)




Example D: Figure 2 modeled the utility at 450 MVA, X/R = 11.4. What are the otherthree possible models?

Answer D: (1) |ISCA| = 450000/( 3 x 34.5) = 7531 A

q = tan-1 (11.4) = -85°

ISCA = 7531 _-85° A

(2) VLN = 34500/ 3 = 19918.6 V

|Z| = 19918.6/7531 = 2.645 W

Z = 2.645 _ 85° W

= 0.231 + j2.635 W

(3) |Zpu| = 100/450 = .2222 pu

q = 85°

Zpu * = .2222 _ 85° pu

= .0194 + j.2212 pu

* Note: 100 MVA, 34.5 kV base




Grounding Resistors

Grounding resistors (RR) are modeled as shown in Figure 5. Data for a groundingresistor are limited to the following:

• Current ratings (e.g., 200 A, 400 A)

• Resistance (e.g., 40 , 6 )

• Time (e.g., 10 sec)

Figure 5. Grounding Resistor Model (RR)

Transformers

Transformers are modeled as an impedance (ZT) as shown in Figure 6. The followingdata is needed to model transformers:

• kVA or MVA self-cooled (OA) rating

• Nameplate impedance (e.g., 5.75%, 7.5%)

• Voltage ratings (primary and secondary)

• X/R ratio (if available) (see Handout 1, page 344)

• Connections ( delta or wye) (L-G calculations)

• Method of grounding (solid or resistance) (L-G calculations)

Figure 6. Transformer Model (ZT)




Conductors

Single-Conductors are modeled as an impedance (ZC) as shown in Figure 7. Thefollowing data are needed to model a conductor:

• Size (kcmil or AWG) and length (per 1000 ft.)

• Conductors per phase

• Conductor material (CU or AL) Note: Saudi Aramco does not permit the use ofaluminum conductors except ACSR aerial lines.

• Conductor configuration (3-1/c)

• Shielding

• Conduit type (magnetic, non-magnetic) or direct burial cable

• Ohmic values of resistance and reactance per unit of length (R + jX)

Figure 7. Conductor or Busway Model (ZC or ZB)

Multi-Conductor cables are modeled exactly the same as single-conductors exceptthe configuration is 1-3/c.

Busway is modeled as an impedance (ZB), also as shown in Figure 7. The followingdata are needed to model busway:

• Type (plug-in, feeder, or current limiting) and length (per 1000 ft)

• Ampacity

• Conductor material (AL or CU)

• Ohmic values of resistance and reactance per unit of length (R + jX)




Aerial Lines

Aerial (overhead) lines are modeled as an impedance (ZOHL) as shown in Figure 8.Although the resistance (ROHL) of an aerial line is modeled exactly the same as otherconductors, the reactance (XOHL) consists of two components (Xa and Xd) . Xa isfound in a cable handbook, like most other cable data. However, Xd is a function ofthe overhead line configuration and spacing and must be calculated (see Work Aid 1).The following data are needed to model an aerial line.

• Size (kcmil, AWG) and length (per mile)

• Conductor material (CU, AL, ACSR)

• Conductors per phase

• Conductor configuration and spacing

• Resistance (ROHL) per unit of length @ t 0C (per mile)

• Operating temperature (t 0C)

• Reactance (Xa) per unit of length (per mile)

• Reactance (Xd) per unit of length based on the conductor configuration andspacing (See Work Aid 1)

Figure 8. Aerial Line Model (ZOHL)




Generators

Generators, similar to the utility, are modeled as a voltage (EG) behind an impedance(ZG) as shown in Figure 9. The following data are required to model a generator:

• kVA or MVA ratings or KW and power factor

• Subtransient reactance (X”d)

• Assume 15% if unknown for 4 pole generators and 9% for 2 pole generators

• Speed (rpm)

• Voltage rating

• X/R ratio (if available) (See Handout 1, page 344)

Figure 9. Generator Model (EG)

Motors

Motor impedance models are modeled the same as utility and generators (see Figure10). However, the impedance data that are required to model a motor depends on thesystem and motor voltages.

Figure 10. Motor Model (ZM)




Less Than 1000 Volts - For system studies less than 1000 volts (low voltage),motors are modeled as follows:

• Lump sum (for motors applied at 480 V and below, each less than 50 hp).

• Individual motors (all other motors). Note: Induction motors 50 to 250 hp arealso often treated as “lump sum” motors.

• Assume X”d = 25%, R”d = 2.5%, X/R = 10

Over 1000 Volts - For system studies over 1000 volts (multivoltage studies), lowvoltage motors are modeled using a 1st cycle or interrupting network.

Multivoltage system studies’ motors (1st cycle network) are modeled as follows:

• For LV induction motors <50 hp neglect reactance; or assume X = 1.67X”d orassume 1.67X”d = 28% on motor rating. Assume an X/R ratio of 10. Note:For purposes of this Module use 1.67X”d = 28%.

• For LV induction motors greater than 50 hp assume X = 1.2X”d or assume1.2X”d = 20% on motor rating. Use an X/R ratio of the largest motor. Note:For purposes of this Module use 1.2X”d = 20%.

Multivoltage system studies’ motors (interrupting network) are modeled as follows:

• For LV induction motors < 50 hp neglect reactance.

• For LV induction motors greater than 50 hp assume X=3.0X”d or assume3.0X”d = 50% on motor rating. Note: For purposes of this Module use 3.0X”d= 50%.




Large motors over 1000 volts (medium voltage) are modeled as follows:

• Horsepower (hp)


• FLA and LRA; X”d = FLA/LRA

• Assume X”d = 17% if the actual data are not available for induction motors,and 20% for synchronous motors.

• Speed (RPM)

• Synchronous or induction

• Power factor (if synchronous), kVA = hp rating if p.f. = 80% kVA = 0.8 x hprating if p.f. = 100%

• X/R ratio (if available) (Handout 1, page 344)

• For synchronous motors use generator/synchronous motors’ X/R ratio table(See Handout 1, page 344).

Per-Unit Values (Step 4)

This step involves calculating the per-unit values of each circuit element that wasidentified in step 3. In practice, steps 3 and 4 are usually accomplished concurrently.Note: Work Aid 1 has been developed to teach the calculation procedures.

Utility or Source

The per-unit impedance calculation of the Utility (ZU) depends on whether the data isprovided as kVA, MVA, current, or ohms. Work Aid 1 lists the procedures andformulas to calculate the per-unit impedance of the utility depending on the given data.




Example E: Using Work Aid 1, what is the per-unit impedance of the utility shown inFigure 2?

Answer E: Zu = MVAb/MVASCA pu

Ru = Zpu cos[tan-1(X/R)]

Xu = (Rpu)(X/R ratio) pu or

Xu = Zpu sin [tan-1 (X/R)] pu

Zu = 100/450 = 0.2222 pu

Ru = 0.2222 cos [tan-1 (11.4)] = 0.0194 pu

Xu = (0.0194)(11.4) = j0.2212 pu

Transformers

The nameplate impedance of a transformer is provided in per-unit (Zpu(old)) on thetransformer’s kVA (kVAb(old)) and voltage (kVb(old)) ratings. Calculating the per-unitimpedance values of the transformer to the new kVA (kVAb(new)) and voltage(kVb(new)) base ratings requires use of the change-of-base formula as presented inWork Aid 1.

Example F: Using Work Aid 1, what are the per-unit impedances of the 3transformers shown in Figure 2?

Answer F: Zpu(new) = Zpu(old)(kVAb(new)/kVAb(old))(kVb(old)/kVb(new))2 = ZT

RT = ZT cos[tan-1(X/R)] pu

XT = (Rpu) (X/R ratio) pu or

XT = ZT sin[(tan-1(X/R)] pu

ZT1 = (0.070)(100000/5000)(34.5/34.5)2 = 1.40 pu or

ZT1 = (0.070)(100/5)(13.8/13.8) 2 = 1.40 pu

X/R ratio = 14 (Handout 1, page 344)

RT1 =1.4 cos[tan-1(14)] = 0.0997 pu

XT1 = (0.0997)(14) = j1.3958 pu




ZT2 = (0.0550)(100/0.75)(0.48/0.48)2 = 7.333 pu


RT2=7.333 cos[tan-1(6)] = 1.2056 pu

XT2 = (6)(1.2056) = j7.234 pu

ZT3 = (0.05)(100/2)(4.16/4.16)2 = 2.500 pu


RT3 = 2.500 cos[tan-1(8)] = 0.3101 pu

XT3 = (8)(0.3101) = j2.481 pu

Motors and Generators

Motor and generator reactance data are also provided in per-unit (X”pu(old)) on thenameplate base kVA and voltage ratings. Calculating the reactance values alsorequires use of the change-of-base formula as presented in Work Aid 1.

Example G: Using Work Aid 1, what is the per-unit impedance of the motor (M1)shown in Figure 2?

Answer G: X”pu(new) = (X”pu(old))(kVAb(new)/kVAb(old))(kVb(old)/kVb(new))2 =

XM

RM = (XM)/(X/R ratio)

XM1 = (0.15)(100/1.0)(4.16/4.16)2 = j15.00 pu


RM1 = (15.0)/(20) = 0.75 pu

Conductors

Conductor impedance data are provided in ohms (R + jX) per unit of length (1000 ft).Calculating the per-unit impedance of the conductors requires dividing the given ohmicvalue by the base ohms (Zb) at the given voltage. Work Aid 1 lists the procedures forcalculating conductor per-unit impedance values.




Example H: Using Work Aid 1, what is the per-unit impedance of conductor C2?

Answer H: Z = R + jX per 1000 ft (Figure 55 or 56)

Z = (R + jX per 1000 ft)(no of ft)

Zc = (R + jX )/(Zb) pu

Zc2 = (0.2020 + j.0524)(500/1000) = 0.1010 + j.0262

Zc2 = Z/Zb = (0.1010 + j.0262)/(1.9044) = 0.053 + j.0138 pu

Aerial Lines

Aerial line resistance data are provided in ohms (R1) per unit of length (1 mile) at agiven temperature (t1

0C). If the operating temperature (t2 0C) is different, the

resistance must be changed to a value (R2) for the operating temperature. Aerial linereactance (Xa) that is based on the type of material is also provided in ohms per unit oflength (1 mile). Aerial line reactance (Xd) must be calculated based on theconfiguration and spacing of the overhead lines. Work Aid 1 lists the procedures forcalculating the per-unit impedance [R + j(Xa +Xd)] of aerial lines.




Example I: Using Work Aid 1, what is the per-unit impedance of overhead aerial lineC1?

Answer I: R = (R per mile)(number of miles)

Xa = (X per mile)(number of miles)

deq = (dab x dbc x dca)1/3

Xd = (0.2794 log10 deq) per mile

X = [(Xa + Xd) per mile](number of miles)

ZC = (R + jX )/(Zb) pu

Rc1 = 1.12 /mi @ 500C (Figure 58)

Xac1 = 0.656 /mi (Figure 58)

deq = (3.5 x 3.5 x 7.0)1/3 = 4.41 ft (see Figure 11)

Xdc1 = 0.2794 log10 (4.41) = .180 /mile

Zc1 = 1.12 + j(0.656 + 0.180) = 1.12 + j.836 /mile

Zc1 = (1.12 + j.836)(1.25)/(1.9044) pu = 0.7351 + j.5487 pu

Figure 11. Aerial Line Configuration

Impedance Diagrams (Step 5)

The final step in per unit modeling is to draw the impedance diagrams (series/parallelnetwork of resistances and reactances).

Complex Diagram

The complex diagram consists of both resistance and reactance values as shown inFigure 12.




Example J: Using Work Aid 1, draw and label the complex impedance diagram for theone-line diagram shown in Figure 2.

Answer J: See Figure 12.

Figure 12. Complex Impedance Diagram




R-Only Diagram

The R-only resistance diagram requires redrawing and labeling the compleximpedance diagram with the reactances omitted (X = 0).

Example K: Using Work Aid 1 and referring to Figure 12, draw and label the R-onlydiagram.

Answer K: See Figure 13.

Figure 13. R-Only Diagram




X-Only Diagram

The X-only reactance diagram requires redrawing and labeling the complex impedancediagram with the resistances omitted (R = 0).

Example L: Using Work Aid 1 and referring to Figure 12, draw and label the X-onlydiagram.

Answer L: See Figure 14.

Figure 14. X-Only Diagram




CALCULATING LINE VOLTAGE DROPS

Introduction

Designers of power systems must have practical knowledge of voltage dropcalculations, not only to meet required codes and standards, but to ensure that therequired voltage of a particular piece of equipment, for example a motor, is kept withinmanufacturer’s specified tolerances to prevent damage to the equipment. Note: Theeffects of voltage drops on different types of electrical equipment was discussed inEEX 102.02.

NEC Article 210-9(a), FPN No. 4, states that “conductors for branch circuits as definedin Article 100, sized to prevent a voltage drop exceeding 3 percent at the farthest outletof power, heating, and lighting loads, or combinations of such loads, and where themaximum total voltage drop on both feeders and branch circuits to the farthest outletdoes not exceed 5 percent, will provide reasonable efficiency of operation”.

NEC Article 215-2(b), FPN No. 2, states that “conductors for feeders as defined inArticle 100, sized to prevent a voltage drop exceeding 3 percent at the farthest outlet ofpower, heating, and lighting loads, or combinations of such loads, and where themaximum total voltage drop on both feeder and branch circuits to the farthest outletdoes not exceed 5 percent, will provide reasonable efficiency of operation”.

Saudi Aramco Standard SAES-P-100 limits the voltage drops of the electrical powersystem depending on the system voltage as follows:

• Systems rated 600 V and below

• Systems rated above 600V

• Direct current systems

Note: For the contents of SAES-P-100, refer to Work Aid 2D (Handout 2).

Calculation Methods

The phasor relationships between the sending end (VS) or bus voltage, the voltagedrop (VD), and the receiving end or load voltage (VL) are shown in Figure 15.




Figure 15. Line Voltage Drop Phasor Diagram

Exact Calculations

Due to the phasor relationships shown in Figure 15, exact methods of calculating linevoltage drops require extensive knowledge of complex phasor algebra as discussed inModule EEX 102.03. However, for line lengths less than 50 km and system voltagesbelow 40 kV, the exact method formula, as presented below, is not typically used.

• Exact Method Formula

VD = VS + IR cos + IX sin - [VS2-(IX cos - IR sin )2]1/2

Approximate Calculations

In most industrial electrical distribution systems, such as found on Saudi Aramcoinstallations, the approximate method formula, as presented below, is adequate formost line voltage drop calculations.

• Approximate Method Formula:

VD = IR cos + IX sin = I(R cos + X sin ) = IZ

Note: By convention, sin is considered to be positive for lagging powerfactor loads.




where: VD = line-to-line neutral voltage drop, volts (V),(one-way)

I= line current, amperes (A)

R = circuit line resistance, ohms ()

X = circuit line reactance, ohms ()

Z = circuit line impedance, ohms ()

Z = R cos + X sin

q = load power factor angle, degrees

cos q = load power factor, decimals

sin q = load reactive factor, decimals

VS = sending end or bus voltage, line-to neutral, volts (V)

VL = receiving end or load voltage, line-to-neutral, volts (V)

VR = VL

VD(line-to-line, 1 system) = 2VD

VD(line-to-line, 3 system) = 3VD

Ohmic Model Selection Factors

Calculating line voltage drops is typically the only power system study that isperformed using the ohmic method. The rationale for using the ohmic method is thatthe calculations are performed incrementally one voltage level at a time. Note: WorkAid 2B has been developed to teach ohmic method procedures for calculating linevoltage drops.

One-Line Diagram

As in any other type of power systems study, a one-line diagram must be obtained ordrawn that accurately describes the line under study. Figure 16 shows a typical one-line diagram for a 1.5 mile ACSR aerial line.




Figure 16. Typical One-Line Diagram




Load Current (I L)

For the purposes of line voltage drop calculations, the load current is first calculatedbased on load voltage equal to the sending end voltage (VS). In reality, the actual loadcurrent should be based on the actual load voltage (VL = VS - VD). However, thisrequires the exact method calculations, which once again, is not typically necessary forline lengths less than 50 km and voltages less than 40 kV.

Theta () Calculations

Both cos and sin must be known or calculated. Since is considered negative forlagging power factor loads, sin should also be a negative number. However, becausethe line drops are considered to be scalar (non-phasor) quantities, by convention sin isconsidered to be a positive number.

Line Impedance (Z = R + jX)

Line impedance is modeled (calculated) simply as R + jX ohms. Because theapproximate method formula is based on scalar quantities, the j operator (1900) isignored, and the reactance (X) is treated as a “real” number versus an “imaginary”number.

Voltage Drop Line-to-Neutral

The voltage drop line-to-neutral (VD) calculated using the approximation formula is aone-way drop.

Voltage Drop Line-to-Line(V D)

The voltage drop line-to-line is the quantity needed to determine whether or not the linevoltage drop (expressed as a percentage) is within the NEC or SAES-P-100 limits.This drop depends on the system being single-phase or three-phase as follows:

• VD (single-phase system) = 2VD (line-to-neutral)

• VD (three-phase system) = 3VD (line-to-neutral)




Load Voltage (V L)

The load voltage calculated is simply the sending end voltage minus the voltage drop(VL = VS - VD).

Percent Voltage Drop (%V D)

The percent voltage drop [%VD = 100 (VD/VS)] calculated is compared to the code(NEC) and standard (SAES-P-100) limits.

Per-Unit Model Selection Factors

If the line voltage drop calculations are just part of a complete power system study(short-circuit, motor-starting, etc.), they are performed using the per-unit methodinstead of the ohmic method. Because the circuit elements are already in per-unit formfrom use in the other studies, they may be used to perform the line voltage dropcalculations as well. Note: Work Aid 2C has been developed to teach per-unit methodprocedures for calculating line voltage drops.

One-Line Diagram

The one-line diagram used for the per unit model is no different from the ohmic methodmodel and again, it must be an accurate representation of the line under study.

Base Values

Any set of base values may be used for the per-unit procedures. However, asmentioned in the previous Information Sheet, most studies are performed on a 100MVA base (MVAb = 100 MVA or kVAb = 100000 kVA).

Per-Unit Calculations

Load Current - The load current (IL) must be calculated in amperes and converted toper-unit amperes by dividing by the base current (Ib).

Impedance - The line impedance must be calculated in ohms (Z)and converted to per-unit ohms by dividing by the base ohms (Zb).




Voltage Drop (V D)

Voltage drop in per-unit uses the same approximation formula as the ohmic method(VD = IZ). To convert per-unit volts to actual volts requires multiplication by the basevoltage (kVb). No additional calculation, such as 3 x VD, is required when using per-unit procedures, because the line-to-line or line-to-neutral relationships are alreadyfactored in during the selection of the base values.

Percent Voltage Drop (%V D)

The percent voltage drop is simply calculated by multiplying the per-unit voltage dropby 100(%VD = 100VD pu). Again, the result should be compared to the NEC andSAES-P-100 limits.

Example M: Using Work Aid 2B and referring to the one-line diagram shown in Figure16, calculate the line voltage drop assuming the circuit conductor isspaced 2.5 feet on center.

Answer M: 1. One-Line Diagram (Figure 16)

2. Calculate IL:

IL = kW/( 3 x kV x p.f.) = 2000/( 3 x 13.8 x 0.80)= 104.6 A

3. Calculate : = cos-1 (p.f.) = cos-1(0.80) = 36.90

4. Calculate sin : sin = sin 36.90 = 0.60

5. Determine line impedance from a cable handbook.

R = 0.888 /mi (Figure 58, Work Aid 1)

Xa = 0.656 /mi (Figure 58, Work Aid 1)

Xd = 0.2794 log10 deq = 0.2794 log10 (2.5 x 2.5 x 5.0)1/3

= 0.139 /mi

X = Xa + Xd = (0.656 + .139) = 0.795 /mi

6. Calculate the line impedance:

Z = (R + jX /mi)(number of miles)




Z = (0.888 + j.795)(1.5) = 1.332 + j1.193




7. Calculate VD:

VD = I (R cos + X sin )

= 104.6 [(1.332)(0.8) + (1.193)(0.6)] = 186.3 volts

8. Calculate VD line-to line:

VD (line-to line) = 3 (VD)

= ( 3)(186.3) = 0.323 kV

9. Calculate the load voltage (VL).

VL = VS - VD = 13.8 - 0.323 = 13.477 kV

10. Calculate VD as a percentage:

VD% = 100[(VS - VL)/VS]

= 100[(13.8 - 13.477)/13.8] = 2.3%

Example N: Repeat Example N using the exact method formula described earlier inthis Information Sheet.

Answer N: 1. VS = VL-L/( 3) = 13800/( 3) = 7967 volts

2. All other quantities remain the same as in Example M.

3. VD = VS + IR cos + IX sin - [(Vs2 - (IX cos - IR sin )2]1/2

= 7967 + (104.6)(1.332)(0.8) + (104.6)(1.193)(0.6)

- [(79672 - ((104.6)(1.193)(0.8) - (104.6)(1.332)(0.6))2]1/2

= 7967 + 111.462 + 74.873 - 7966.983 = 186.352 V

This example shows the exact method calculation was unnecessary.




Example O: Using Work Aid 2C, repeat Example M using per-unit procedures(methods).

Answer O: 1. Base Values: 100000 kVA, 13.8 kV, 4183.7 A, 1.9044

2. IL(pu) = (104.6)/(4183.7) = 0.025 pu

3. Z(pu) = (1.332 + j1.193)/(1.9044) = 0.6994 + j.6264 pu

4. VD(pu)= I(R cos + X sin )

= 0.025[(0.6994)(0.8) + (0.6264)(0.6)] = 0.0234 pu

5. VD = VD(pu) x kV(b) = (0.0234)(13.8) = 0.323 kV

6. VD% = 100 VD(pu) = (100)(0.0234) = 2.34%




CALCULATING LARGE MOTOR STARTING VOLTAGE DROPS

Introduction

Alternating current (AC) motor characteristically draw much higher current duringstarting than they draw under normal running conditions (see Figure 17). This suddendraw in current results in excessive voltage drops on the entire power distributionsystem, resulting in unacceptable conditions for other electrical equipment on thepower system.

Figure 17. Motor Starting Currents




Motor Starting Effects

Critical Voltage Levels (Minimum)

To develop sufficient starting torque, alternating current motors require a full voltagestarting current (ILRA) that is between four and ten times their normal full-load runningcurrent (IFLA). Such large increases in current, when suddenly drawn, can preventacceleration of the motor to normal speed and can cause other plant equipment suchas relays and contactors to misoperate. Although usually not a serious problem,voltage drops due to large motors starting can also cause annoying light flicker. It iscommon practice to restrict voltage drops to 30 percent or less during motor starting;some plant managers restrict the drops to 10 percent or less. Saudi Aramco StandardSAES-P-100 limits voltage drop during starting to 15% for systems rated 600 V or less.Although not explicitly stated in SAES-P-100, it is assumed for purposes of this Modulethe maximum voltage drop for medium voltage motors is also 15%. Figure 18 lists thecritical voltage levels of different plant equipment. Saudi Aramco’s 15% voltage droplimits under motor starting conditions should prevent problems on most of the otherequipment in their facilities.

Voltage Drop

Location or Problem

MinimumAllowable

Voltage (% Rated)

At Terminals of Starting Motor 80%

All Terminals of Other Motors that Must Reaccelerate 71%

AC Contactor Pick-up (By Standard) 85%

Contactor Hold-In (Average of Those in Use) 60-70%




Solid-State Control Devices 90%

Noticeable Light Flicker 3% Change

Source: IEEE Brown Book

Figure 18. Critical Voltage Levels




Effects on Plant Equipment

Starting Motor When an induction or synchronous motor starts, it draws current sixto eight times normal full load running current, with a corresponding severe drop at themotor terminals. This drop in voltage greatly reduces the available motor torque sincetorque (T) is directly proportional to current squared (I2). For example, NEMA Bmotors have a starting torque of 150% of rated torque (see Figure 19). If the voltagedrops to 80%, starting torque is reduced to 100%(T = (.802)(150)). This drop in starting torque may stall the motor and/or overheat themotor windings.

Figure 19. Induction Motor Speed Torque Characteristics

Other Plant Motors are also adversely affected by a system voltage drop throughoutthe plant. The other motors slow down and their horsepower load may exceedbreakdown torque at the reduced voltage. Motor contactors may drop out and the lowvoltage at the motor terminals cause higher currents to be drawn, causing even highersystem voltage drop (“domino effect”).




Computer systems have very wide ranges of voltage drop tolerance. For example,even a 5% drop may be unacceptable to some systems, whereas another system maypermit 10% drop without damaging the computer system.

Electronic Control equipment, similar to computers, have wide ranges of voltagedrop tolerance. It is best to check with the equipment vendor to determine if suddendrops in voltage are harmful to the equipment.

Light Flicker as a result of voltage drops is a perception issue. If it occurs oftenenough, it becomes objectionable. This is not typically a problem since most largemotors are not frequently stopped and started. In fact, the duty cycles of large motorsusually require cooling down periods before restarting.

Calculation Methods

Similar to line voltage drop studies, large motor starting voltage drop studies may beperformed using exact and approximate methods. Note: Work Aid 3 has beendeveloped to teach procedures for calculating large motor starting voltage drops.

Exact Calculations

The exact method calculations involves modeling the power system as an impedance,both R and X, between the terminals of the starting motor and another point in thesystem where the voltage is assumed to be constant. In most plant electrical systems,this other point, is the utility bus. The exact method formula is:

• kVs =(ZM x kVI)/[(RM + RS)2 + (XM + XS)2)]1/2

The voltage drop formula is simply a voltage divider formula, where the voltage at anyload is the initial voltage (kVI) times the ratio of the load impedance (ZM) to the entiresystem impedance [(RM + RS

2) + (XM + XS)2]1/2.




Approximate Calculations (Neglect R)

The approximate formula as shown below is identical to the exact formula, except thatresistance of the motor and the system is neglected. This is very seldom, if ever, aproblem in medium voltage systems because the X/R ratio ranges from 6 to 15 inmedium voltage systems. Whenever the X/R ratio equals or exceeds 4, the percenterror in the calculated result will be 3% or less. The following are the approximatecalculation formulas:

• kVS = (XMkVI)/(XM + XS)

• kVs = XM/(XM + XS) if kVI = 1.0 (most cases)


Large motor starting voltage drops are usually accomplished through use of the per-unit calculation procedures because more than one voltage level is present in thesystem. Note: Work Aid 1 was developed to teach per-unit modeling of an electricalpower system.

The following factors should be considered for calculating large motor starting voltagedrops:

• One-line diagram

• Base values

• Per-unit calculations

• Impedance (reactance only) diagram

• Voltage drops

One-Line Diagram

As with all types of power system studies, an accurate one-line diagram is the startingpoint. Figure 20 is a typical one-line diagram used for calculating large motor startingvoltage drops.




Figure 20. One-Line Diagram for Motor Voltage Drop Calculations




Base Values

Any set of base values may be used, however most power system studies areperformed on a 100 MVA base as shown in Figure 21.

Bus

Numbers

kVAb kVb Ib Zb

100, 140

145

100000 kVA

100000 kVA

13.8 kV

4.16 kV

4183.7 A

13878.6 A

1.9044

0.173

Figure 21. Motor Voltage Drop Study Base Values





The per-unit calculation procedures for the utility, transformers, motors, and cables areidentical to the procedures taught in Work Aid 1.

Example P: Using Work Aid 1 and referring to Figure 20, model the system in per-unit.

Answer P:

Utility: Zpu = MVAb/MVASCA = 100/186 = 0.5376 pu

Xpu = Z sin = (0.5376) sin[tan-1(17)]

= j.5376 pu

Rpu = Xpu/(X/R ratio) = 0.5376/17

= 0.0316 pu

Cable: Z = (0.0650 + j.0583)(1640/1000)

= 0.1066 + j.0956

Zpu = Z/Zb = (0.1066 + j.0956)/1.9044

= 0.056 + j.0502 pu

Transformer: Zpu = (0.055)(100/5) = 1.100 pu

Xpu = (1.100)sin[tan-1(12)] = j1.0962 pu

Rpu = (1.0962/12) = 0.0914 pu

Motor: ZM() = (4160)/( 3 x 1390) = 1.7279

ZM(pu) = (1.7279/.173) = 9.9878 pu

XM(pu) = 9.9878 sin [tan-1 (22)] = j9.245 pu

RM(pu) = (9.245/22) = 0.4202 pu or

XM(pu) = (.20)(100/2) = j 10.00 pu

RM(pu) = (10/22) = .4545 pu




Impedance Diagram

If the exact method calculations are used, the impedance diagram model shouldinclude both resistance (R) and reactance (X). If approximate method calculations areused, the system should be modeled using an X-only diagram.

Example Q: Using Work Aid 3, draw and label the impedance diagram for the one-linediagram shown in Figure 20.

Answer Q: See Figure 22.

Figure 22. Example Q Impedance Diagram




Voltage Drop

The voltage drop at the terminals of the starting motor is calculated by performing avoltage divider calculation. Since the per-unit voltage at starting is assumed to be 1.0pu, the voltage at the motor terminals in per-unit is simply the ratio of motor impedance(ZM) to the total impedance (ZM + ZS) or the ratio of XM to (XM + XS) if resistance isneglected.

Percent Voltage Drop

Although the actual voltage drop in volts is important, the drop in percent is comparedagainst codes and standards. For Saudi Aramco installations, as mentioned earlier,the voltage drop is limited to 15% under starting conditions. The percent voltage dropis calculated by multiplying the per-unit voltage drop by 100 (VD% = 100 x VD p.u.).

Voltage Drop (Other Buses)

If the voltage drop is too severe at other buses in the system, it may cause operatingproblems with other equipment in the plant. The procedure for calculating the voltagedrop at other buses also uses the voltage divider rule as described above, and in WorkAid 3. For example, the voltage drop at the other buses shown in Figure 22 are thefollowing:

• Bus 140 - (XT + XM)/(XU + XC + XT + XM)

• Bus 100 - (XC + XT + XM)/(XU + XC + XT + XM)




Example R: Using Work Aid 3 and referring to Figure 22, calculate the voltage drop atthe motor terminals (Bus 145) using both the exact and approximatecalculation methods.

Answer R:

• Exact Calculation

ZM = 9.9878 pu

kVI = (4.16/4.16) = 1.0 pu

RS = RU + RC + RT = (.0316 + .056 + .0914) = 0.179 pu

XS = XU + XC + XT = (.5376 + .0502 + 1.0962) = 1.6840 pu

M = cos-1(p.f.) = cos-1 (.15) = 81.370, sin M = .987

RM = ZMcos M = (9.9878)(.15) = 1.4982 pu

XM = ZMsin M = (9.9878)(.987) = 9.8580 pu

kVS = (ZM x kVI)/[(RM + RS)2 + (XM + XS)2]1/2

= [(9.9878)(1.0)]/[(1.4982 + .179)2 + (9.8580 + 1.6840)2]1/2

= (9.9878)/(11.6623) = 0.8564 pu

kVS = kVS(pu) x kVb = (0.8564)(4.16) = 3.563 kV = 3563 Volts

kVD = 1 - 0.8564 = 0.1436 pu

kVD% = (100)(0.1436) = 14.36%

kVD = kVD(pu) x kVb = (0.1436)(4.16) = 0.597 kV = 597 Volts

kVM% of rated voltage = (100)(3563/4000) = 89%

• Approximate Calculation (Neglecting R)

kVS = [(XM)(kVI)]/(XM + XS)

= [(9.858)(1.0)]/(9.858 + 1.6840) = 0.8542 pu

kVD = (1 - 0.8542)(4.16) = (.1458)(4.16) = 0.607 kV = 607 Volts




CALCULATING SYMMETRICAL RMS FAULT CURRENTS (BALANCED FAULTCONDITIONS)

Purposes of Fault Calculations

No matter how well an electrical power system is designed, it will occasionallyexperience short circuits resulting in abnormally high current flows. Fault calculationsare performed to determine device interrupting and withstand ratings, and to determinerelay pickup and sensitivity settings.

Protective Device Interrupting Ratings

The maximum calculated short circuit current magnitudes are used to select adequateprotective device (fuses and breakers) interrupting ratings to minimize the faultdamage and plant down time.

Equipment Component Withstand Ratings

Other electrical equipment (e.g. buses and cable) must be capable of withstanding thehigh thermal and mechanical stresses (I2t) associated with the short circuit current.

Protective Relay Pickup Settings

The maximum value of the available short circuit current is used to determine thepickup settings of relays and low voltage breakers. For example, the instantaneoustrip setting of an upstream relay should be set » 10% above the maximum availableshort circuit current “seen” by the downstream protective device.

Maximum Relay Settings for Sensitivity

Minimum values available of short circuit current are required to set maximum relaysettings for sensitivity. For example, the maximum setting for ground fault relays inmedium voltage system is typically selected based on a minimum fault current of 10%of the available maximum fault current.




Effects of Short Circuits

Many things happen on an electrical power system when a short circuit occurs--all ofthem bad!

Arcing and Burning

At the point of the fault, arcing and burning occurs resulting in hot, molten copper,burning insulation, etc. In many instances, the end result is a fire.

Current Flow

Large currents flow under short circuit conditions form various sources (utility, motors,generators). These currents are independent of load current.

Thermal Stress

All components carrying short circuit currents are subject to thermal stress (heat) thatvaries as a function of the current squared (I2) and the fault duration (t). Thetemperature rise (t) is an integrated effect as shown below:

• T = (σ υ c A ) i (t) dt = I t2 - 1 2

t

t

1

2z• where:

T = temperature rise in 0C= electrical conductivity in ohm-1/meter

C = specific heat in joules/0C x m3

A = conductor cross-sectional area in m2 (Figure 23)i(t) = rms instantaneous current in amperest = time in seconds

Figure 23. Balanced Loads




Mechanical Stress

All components carrying electrical current are also subjected to mechanical stressrelated to the square of the current and the fault duration [i2(t)]. The primary differencefrom thermal stress is that the mechanical stress is an instantaneous effect as shownbelow:

• F/L = 0 i2(t)/2d

• where:

F/L = force per unit of length in Newtons/meter

0 = magnetic permeability of air = 4 x 10-7 henries/meter

i(t) = instantaneous current in amperes (A)

d = conductor spacing in meters (m) (Figure 24)

Figure 24. Parallel Conductors (Simplest Case)

Voltage Drops

The system voltage drops throughout the entire power system and is directlyproportional to the magnitude of the fault current. The worst case drop is of course, atthe point of the fault, where the drop is 100% (V=0), but all parts of the power systemare subjected to some level of voltage drop.




Characteristics of Fault Current Sources

When calculating the fault current, it is important to include all sources (Figure 25) ofshort circuit current. The fault current sources are generators (local), motors (inductionand synchronous), and utility (“large” generators).

Figure 25. Fault Current Sources




Generators (Local)

Generators are driven by some type of prime mover (turbine, diesel engine, etc.).When a fault occurs on the circuit, the generator continues to produce voltage becausethe field excitation is maintained and the prime mover continues to drive the generatorat normal speed. Stated another way, the generator “sees” the fault as aninstantaneous load demand. The only impedance that limits the fault currentmagnitude (Figure 26) flowing from a generator is the impedance of the generator andany system impedance between the generator terminals and the point of the fault.

Figure 26. Local Generator Fault Source




Motors

Synchronous Motors are constructed similar to generators, which is that they have aseparate field excited by dc current and current flows in the stator winding. Normally, asynchronous motor converts electrical energy (input kW) to mechanical energy (outputhp). Under fault conditions, the opposite occurs, the synchronous motor acts like agenerator and delivers fault current with the load inertia acting as a prime mover. As aresult the synchronous motor delivers fault current for many cycles (Figure 27) and,like a generator, the current flowing is limited by the motor impedance and anyimpedance between the motor terminals and the point of the fault.

Figure 27. Synchronous Motor Fault Source




Induction Motors act much like synchronous motors under fault conditions. Theinertia of the load acts like a prime mover and the motor delivers fault current, but onlyfor a few cycles (Figure 28). The duration is very short because the field of theinduction motor is from the stator winding rather than a dc winding, as is the case witha synchronous motor. The short circuit current contribution must be considered, butonly for momentary ratings of medium voltage switchgear, and interrupting andwithstand ratings of low voltage equipment. The fault current magnitude delivered bythe induction motor, like generators and synchronous motors, is limited by the motor’sown impedance and any impedance between the motor terminals and the point of thefault.

Figure 28. Induction Motor Fault Source




Utility System

The utility system, strictly speaking, is not a source of fault current. The actual sourceis the utility company’s generators, which are typically the largest source of short circuitcurrent in an industrial power system. Like the other short circuit current sources, theonly impedance limiting the fault current from the utility is the self-impedance of thegenerator(s), and the impedance between the generators’ terminals and the point ofthe fault. For example, the impedance of step-up and step-down transformers,transmission and distribution lines, etc. Unlike the other sources, the utility deliversfault current without any noticeable decay for an indefinite period (Figure 29).

Figure 29. Utility (Generator) Fault Source




Total Fault Current

The total short circuit current flowing (Figure 30) into the fault point is the sum of theindividual sources plus the dc component. Note: The DC component will bediscussed later in this Module.

Figure 30. Total Fault Current




Machine Reactance Modeling

The impedance of generators and motors consist primarily of reactance and, unlikecables and transformers, is not a simple value, but is more complex and also varieswith time. Machines are modeled by three reactance values as follows:


• Transient reactance (X’d)

• Synchronous reactance (Xd)

Typical Short Circuit Oscillogram

For example, if a short circuit occurs across the terminals of a generator, thewaveshape is as shown in Figure 31. The magnitude starts out relatively high anddecays to a symmetrical or steady state value after a definite time lapse depending onthe types of sources and the system X/R ratio.

Figure 31. Short Circuit Current Oscillogram




Subtransient Reactance (X” d)

Subtransient reactance (X”d) is the apparent reactance of the stator winding at theinstant short circuit occurs, and it determines the current flow during the first few cyclesafter short circuit.

• 0 < t< 3 to 6 cycles

A synchronous motor has the same kind of reactance as a generator, but it is of adifferent value. Induction motors have no field coils, but the rotor bars act like theamortisseur winding in a generator; therefore, induction motors are said to havesubtransient reactance only.

Transient Reactance (X’ d)

Transient reactance (X’d) determines the current following the period whensubsransient reactance is the controlling value. Transient reactance is effective up toone-half second or longer, depending upon the design of the machine.

• 3 to 6 cycles < t < 30 to 200 cycles

Synchronous Reactance

Synchronous reactance (Xd) is the reactance that determines the current flow when asteady state condition is reached. It is not effective until several seconds after theshort circuit occurs; consequently, it is not generally used in short-circuit calculations inindustrial power systems.

• t > 30 to 200 cycles

Symmetrical Versus Asymmetrical Fault Currents

Symmetrical and asymmetrical describe the ac current waveshape about the zero axis.Virtually all short circuit currents begin as asymmetrical currents and decay tosymmetrical currents.




Symmetrical Fault Currents

If the envelopes of the peaks of the fault current waveshape are symmetrical about thezero axis, they are called symmetrical currents as shown in Figure 32.

Figure 32. Symmetrical Fault Current




Asymmetrical Fault Currents

If the envelopes of the peaks of the fault current waveshape are not symmetrical aboutthe zero axis, they are called a symmetrical currents as shown in Figure 33.

Figure 33. Asymmetrical Fault Currents




System Power Factor and X/R Ratios

Most fault currents are asymmetrical because the system power factor before a faultoccurs is very high (implying low X/R ratios) and then changes instantaneously to avery low system power factor after a fault occurs (implying high X/R ratios).

Circuit Model Figure 34 shows a typical system circuit model.

Figure 34. Circuit Model

Before a fault occurs the following conditions apply:

• Load impedance dominates

• ZS very small

• ZL very large and mostly R(RL>XL)

• High power factors (70% - 95%)

• Very low X/R ratios (<1)

• Current lags voltage by 18 - 45 degrees (Figure 35)

Figure 35. E vs. ILOAD Relationship (Before Fault Occurs)




After a fault occurs the following conditions apply:

• System impedance dominates

• ZL equals 0

• ZS dominant factor and mostly X(XS>>RS)

• Low power factors (5% - 45%)

• Very high X/R ratios (5-20 or larger)

• Current lags voltage by 79 - 87 degrees (Figure 36)

Figure 36. E vs. ISC Relationship (After Fault Occurs)

The only possible means a current/voltage relationship can change from laggingapproximately 180 (see Figure 35) to instantaneously (t=0) lagging by approximately850 (see Figure 36) is for the current waveshape to offset itself from the zero axis. Thegreater the ratio of reactance to resistance (X/R ratio), the greater the offset and theslower the rate of decay.




Maximum System Voltage If a fault occurs in a zero resistance circuit at Es =Emax, the fault current will be totally symmetrical as shown in Figure 37.

Figure 37. Zero Power Factor Circuit (Es = Emax)




Zero System Voltage If a fault occurs in a zero resistance circuit at Es = 0, the faultcurrent will be totally asymmetrical (offset) as shown in Figure 38.

Figure 38. Zero Power Factor Circuit (Es = 0)




Sample Waveshapes

High X/R Ratio - Figure 39a shows a typical short circuit current oscillogram in amedium voltage circuit where X/R = 15 (power factor of 7%). Figure 39b shows thevoltage current relationships under the same conditions.

Figure 39. Short Circuit Current Waveshapes at X/R = 15




Low X/R Ratio - Figure 40 shows a typical short circuit oscillogram in a low voltagecircuit where X/R = 2 (power factor of 45%).

Figure 40. Short Circuit Current Waveshape at X/R = 2

The following conclusions can be made concerning the decay of the currentwaveshapes under fault conditions.

• Rate of decay is function of system X/R ratio

• Low X/R ratios imply fast decay

• High X/R ratios imply slow decay




AC and DC Components

Asymmetrical fault currents are analyzed in terms of two components, an AC orsymmetrical component, and a DC or exponential component. Figure 41, once againshows the typical system circuit model.

Figure 41. Circuit Model

• where:

e(t) = E sin ( wt + f ) = R i(t) + L di(t)/dt

E = crest voltage of sinusoidal source

w = 2pf = angular frequency of source

f = angle between E equals zero and t = 0

= angle of source voltage when fault occurs

• solution set:

i(t) = Aexp[(-R/X) w t] + B sin ( wt + f - q)

• where:

A = E sin ( q - f)/(Rs2 + Xs

2)1/2

B = E/( Rs2 + Xs

2)1/2 = E/Zs

q = tan-1 Xs/Rs

• DC Component - Aexp[(-Rs/Xs) t]

• AC Component - B sin(w t + f - q )




Time Domain Model Figure 42 describes the time domain model of the systemvoltage (Figure 42a) and current (Figure 42b) for the circuit model shown in Figure 41.

Figure 42. Time Domain Model of Voltage and Current




Types of Faults/Magnitudes

Three-Phase Faults

Three-phase faults (Figure 43) usually are the least often occurring, but are also oftenthe worst case in industrial power systems. Note: This Module is restricted tocalculating three-phase fault currents.

Figure 43. Three-Phase Faults

Line-to-Line Faults

Line-to-line fault current magnitudes are approximately 87% ( 3 /2) of the three-phasefault values. Accordingly, fault calculations involving line-to-line faults (Figure 44) areseldom required because it is not the maximum value.

Figure 44. Line-to-Line Faults




Line-to-Ground Faults

Line-to-ground faults (Figure 45) are the types of faults that most often occur: theirmagnitudes range from a few percentage points to 125% of the three-phase values. Inindustrial systems, line-to-ground faults exceeding three-phase fault values are theexception rather than the rule. Symmetrical components modeling techniques arerequired to analyze line-to-ground faults. These techniques are considered beyond thescope of this Module.

Figure 45. Line-to-Ground Faults

Arcing Faults

Many low voltage power systems experience low level arcing faults, and in manycases, they are the most damaging faults, especially at the 480 volt level. Because ofthe very high arc impedance, arcing fault current magnitudes tend to be too low for theprotective devices to detect and clear. Arcing fault current magnitudes are typicallyestimated as follows:

• 0.89 at 480 V and 0.12 at 208 V for three-phase arcing.

• 0.74 at 480 V and 0.02 at 208 V for line-to-line single-phase arcing.

• 0.38 at 277 V and 0.01 at 120 V for line-to-neutral single-phase arcing.

Overloads

Overloads are not technically regarded as faults. Very low level faults (<250% of FLA)are often called overloads because they are detected in the overload protective deviceranges rather than in the fault (instantaneous or short-time) ranges of the protectivedevices.





Short circuit studies are usually accomplished using per-unit calculation proceduresbecause more than one voltage level is present in the system. Note: Work Aid 1 wasdeveloped to teach per-unit modeling of an electrical power system.

One-Line Diagram

As with all types of power system studies, an accurate one-line diagram is the startingpoint. Figure 46 is a typical one-line diagram used for calculating short circuit currents.

Figure 46. One-Line Diagram for Short Circuit Calculations




Base Values

Any set of base values may be used; however, most power system studies areperformed on a 100 MVA base as shown in Figure 47.

Bus

Numbers

kVAb kVb Ib Zb

100, 140

145

100000 kVA

100000 kVA

13.8 kV

4.16 kV

4183.7 A

13878.6 A

1.9044

0.173

Figure 47. Short Circuit Study Base Values


The per-unit calculation procedures for the utility, transformers, motors, and cables areidentical to the procedures explained in Work Aid 1.

Example S: Using Work Aid 1 and referring to Figure 46, model the system in per-unit. Ignore resistance.

Answer S:

Utility: Xpu = MVAb/MVASCA = 100/186 = j.5376 pu

Cable: X = (j.0583)(1640/1000) = j.0956 ohms

Xpu = X/Xb = (j.0956)/1.9044 = j.0502 pu

Transformer: Xpu = (.055)(100/5) = j1.100 pu

Motor XM(pu)(old) = 278/1390 = j.20 pu

XM(pu) = 0.20(100/2) = j10.00 pu




Impedance Diagrams

If the exact method calculations are used, the impedance diagram model shouldinclude both resistance (R) and reactance (X). If approximate method calculations areused, the system should be modeled using an X-only diagram.

Example T: Using Work Aid 4, draw and label the impedance diagram for the one linediagram shown in Figure 48.

Answer T: See Figure 48.

Figure 48. Example T Reactance Diagram

Modification of Per-Unit Values

For multivoltage calculations, the reactances of machines must be modified to reflectthe momentary duty and interrupting duty of medium voltage breakers (Figure 49).




Figure 49. Medium Voltage Breaker Duties

Momentary Duty Network - Motors and generators are modeled for the momentaryduty network (first cycle) as follows: Note: See Figure 50.

• Induction motors less than 50 hpX = 1.67 X”d or X = 28% on motor rating or neglect

• Induction motors greater than 50 hpX = 1.2 X”d or X = 20% on motor rating

Interrupting Duty Network - Motors and generators are modeled for the interruptingnetwork as follows: Note: See Figure 50.

• Induction motors less than 50 hp - neglect

• Induction motors greater than 50 hpX = 3.0 X”d or X = 50% on motor rating

• Synchronous motors and induction motors > 250 hpX = 1.5 X”d




Type of

Rotating Machine

InterruptingDuty

(Per Unit)

MomentaryDuty

(Per Unit)

All turbo-generators, all hydro-generators with amortisseurwindings, and all condensers

Hydro-generators without

amortisseur windings

All synchronous motors

Induction Motors

Above 1000 hp at 1800 r/min orless

Above 250 hp at 3600 r/min

All others, 50 hp and above

All smaller than 50 hp

1.0X”d

0.75X”d

1.5X”d

1.5X”d

3.0X”d

Neglect

1.0X”d

0.75X”d

1.0X”d

1.0X”d

1.2X”d

Neglect

(a)




Induction Motors (Note 2)

All others, 50 hp and above

All smaller than 50 hp

3.0X”d (Note 5)

Neglect

(b)

1.2X”d (Note 3)

1.67X”d (Note 4)

Notes:1. Resistance values should also use same multipliers.

2. For comprehensive multivoltage calculations.

3. Or X = .20 p.u. based on motor rating.



* Source IEEE Red Book

Figure 50. Rotating Machine Reactance Multipliers




Example U: Using Work Aid 4 and referring to Figures 46 and 48, model theinterrupting duty reactance of the 2000 hp motor.

Answer U: Xpu = 1.5 X”d =(1.5)(j10) = j15.00 pu or

Xpu = (1.5)(j.20)(100/2) = j15.00 pu

Thevenin Equivalent Network

The Thevenin equivalent network (Figure 51) involves combining the series/parallelnetwork of impedances into a single Thevenin equivalent impedance (reactance).Note: See Work Aid 4 procedures.

Figure 51. Thevenin Circuit

Fault Current Calculations

The momentary fault current duty of medium voltage breakers must be calculated byusing the momentary duty reactances of any machines in the system, followed by thesame exact calculation using the interrupting duty reactances as modified in Figure 50.The actual fault current is simply the quotient of the Thevenin voltage and the Theveninreactance. Note: See Work Aid 4 for the procedures.




Example V: Using Work Aid 4 and referring to Figure 48, calculate the short circuitcurrent (momentary and interrupting) for a fault at Bus 145.

Answer V: 1. Thevenin Circuit: See Figure 52.

Figure 52. Example V Thevenin Circuit




2. Thevenin Reactance (Momentary Duty)

Let X1 = XU + XC + XT = j.5376 + j.0502 + j1.0962

= j1.684 pu

Xth = (X1XM)/(X1 + XM) = (j1.684)(j10)/(j1.684 + j10)

= j1.4413 pu

3. Fault Current (Momentary)

IF = 1.0/Xth = 1.0/1.4413 = 0.6938 pu

IF = Ipu x Ib = (.6938)(13878.6)

= 9.6 kA (symmetrical)

IF(mom) = IF x 1.6 = (9.6)(1.6)

= 15.4 kA (momentary)

4. Thevenin Reactance (Interrupting Duty)

Xth = (X1XM)/(X1 + XM) = j1.684)(j15)/(j1.684 + j15)

= j1.5140 pu

5. Fault Current (Interrupting)

IF = 1.0/Xth = 1.0/1.514

= .6605 pu

IF = Ipu x Ib = (.6605)(13878.6)

= 9.2 kA (symmetrical)




WORK AID 1: RESOURCES USED TO MODEL AN ELECTRICAL POWERSYSTEM IN PER UNIT

Work Aid 1A: IEEE Standard 141-1986 (Red Book)

For the content of IEEE Standard 141-1986, refer to Handout 1.

Work Aid 1B: Applicable Modeling Procedures

Step 1. Obtain the system one-line diagram.

Step 2. Select/calculate the base values.

a. Select kVAb or MVAb. Note: kVAb = 100,000 kVA or MVAb = 100 MVA istypically selected for power systems studies.

b. Select kVb to match one of the systems transformers rated voltage.

c. Calculate the other base voltages based on the transformer turns ratio.

d. Calculate Ib = kVAb /( 3 x kVb).

e. Calculate Zb = (kVb)2/MVAb or [(kVb)2 x 1000]/kVAb.

f. Construct a table of base values.

Step 3. Model and collect circuit element data from the one-line diagram, cablehandbooks, equipment nameplates, national or internal standards, and codebooks.

Step 4. Calculate the per-unit impedance values (R + jX) of each circuit element.

a. Utility or Sources:

(1) Zpu = kVAb/kVASCA or

= MVAb/MVASCA or

= Ib/ISCA or

= ZW/Zb

(2) Rpu= Zpu cos[tan-1(X/R)]




(3) Xpu= Zpu sin[tan-1(X/R)] or

= (Rpu)(X/R ratio)

b. Transformers:

(1) ZT = Zpu(old)(kVAb(new) /kVAb(old))(kVb(old) /kVb(new))2

= Zpu(old)(kVAb(new) /kVAb(old))

Note: For purposes of this Module kVb(old) = kVb(new)

(2) XT = ZT sin[tan-1(X/R)]

Note: Typical X/R ratios of transformers can be found in Figure N1.1 (page344) of the IEEE Red Book (Handout 1).

(3) RT = ZT cos[tan-1(X/R)] or

= XT /(X/R ratio)

c. Motors or Generators:

(1)XG or XM = Xpu(old)(kVAb(new)/kVAb(old))

Note: Assumes kVb(old) = kVb(new)

(2) RG or RM = X/(X/R ratio)

Note: Typical X/R ratios of generators and synchronous motors can be foundin IEEE Red Book Figure N1.2 (page 344) and in IEEE Red Book Figure N1.3(page 344) for induction motors.




d. Cables or Busway:

(1) Z = (R + jX) /1000 ft (Figure 55 or 56)

Note: The best source for cable or busway impedance data is a vendor’shandbook. For purpose of this Module, use Figures 55 and 56 for cablevalues.

(2) Z = [(R + jX) /1000 ft](number of feet)

(3) ZC = Z/Zb = (R/Zb) + j(X/Xb)

e. Aerial Lines:

(1) R1 @ t10C (Figure 57 or 58)

Note: For purposes of this Module, use Figure 57 for aerial copper lines andFigure 58 for ACSR aerial lines impedance data.

(2) R2 @ t10C = R[(T + t2

0C)/(T + t10C)] = R

Where: T = 234.5 for annealed copper - 100% conductivity

= 241.0 for hard drawn copper - 97.3% conductivity

= 228.0 for hard drawn aluminum - 61% conductivity

R1, R2 = resistance of conductor at temperatures t1, t2

(3) X = Xa + Xd

Where: Xa = reactance of conductor per mile (Figure 57 or 58)

Xd = 0.2794 log10 deq

deq= (dab x dbc x dca)1/3 (See Figure 54)

(4) Z = [(R + jX)/mile](number of miles)

(5) ZOHL = Z/Zb = (R/Zb) + j(X/Zb)




Figure 54. Aerial Lines Configuration and Spacing

Step 5. Draw and label the impedance diagram.

This step consists of graphically modeling and labeling the series/parallel networkof resistance and reactances.

a. Draw and label the X-only reactance diagram.

b. Draw and label the R-only resistance diagram.




Figure 55 lists resistance, reactance, and impedance data for copper cables configuredas three single conductors (3 -1/c) in conduit.

In Magnetic Duct and Steel Interlock Armor In Nonmagnetic Duct and Aluminum Armor

AWG

kcmil

600 V & 5 kV

Non Shielded

(/1000 ft.)

R X Z

5 kV Shielded &

15 kV

(/1000 ft.)

R X Z

600 V & 5 kV

Nonshielded

(/1000 ft.)

R X Z

5 kV Shielded &

15 kV

(/1000 ft.)

R X Z




8

*8

6

*6

4

*4

2

1

1/0

2/0

3.0

4/0

250

300

350

400

450

500

600

750

0.811 0.0754 0.814

0.786 0.0754 0 790

0.510 0.0685 0.515

0.496 0.0685 0.501

0.312 0.0632 0.327

0.312 0.0632 0.318

0.202 0.0585 0.210

0.160 0.0570 0.170

0.128 0.0540 0.139

0.102 0.0533 0.115

0.085 0.05190.0958

0.064 0.04970.0810

0.0552 0.04950.0742

0.0464 0.04930.0677

0.0378 0.04910.0617

0.0356 0.04900.0606

0.0322 0.04800.0578

0.0294 0.04660.0551

0.811 0.0860 0.816

0.786 0.0860 0.791

0.510 0.0796 0.591

0.496 0.0796 0.502

0.312 0.0742 0.329

0.312 0.0742 0.321

0.202 0.0685 0.214

0.160 0.0675 0.174

0.128 0.0635 0.143

0.103 0.0630 0.121

0.0814 0.0605 0.101

0.0650 0.05830.0929

0.0557 0.05700.0797

0.0473 0.05640.0736

0.0386 0.05620.0681

0.0362 0.05480.0657

0.0328 0.05380.0630

0.0300 0.05260.0505

0.0264 0.0516

0.811 0.0603 0.813

0.786 0.0603 0.788

0.510 0.0548 0.513

0.496 0.0548 0.499

0.312 0.0506 0.325

0.312 0.0506 0.316

0.202 0.0467 0.207

0.160 0.4560 0.166

0.127 0.0432 0.134

0.101 0.0426 0.110

0.0766 0.04150.0871

0.0633 0.03960.0748

0.0541 0.03960.0670

0.0451 0.03940.0599

0.0368 0.03930.0536

0.0342 0.03920.0520

0.0304 0.03840.0490

0.0276 0.03730.0464

0.811 0.0688 0.814

0.786 0.0688 0.789

0.510 0.0636 0.514

0.496 0.0636 0.500

0.312 0.0594 0.326

0.312 0.0594 0.318

0.202 0.0547 0.209

0.160 0.0540 0.169

0.128 0.0507 0.138

0.102 0.0504 0.114

0.0895 0.0484 0.0939

0.0640 0.0466 0.0792

0.0547 0.0456 0.0712

0.0460 0.0451 0.0644

0.0375 0.0450 0.0586

0.0348 0.0438 0.0559

0.0312 0.0430 0.0531

0.0284 0.0421 0.0508

0.0246 0.0412 0.0479

0.0203 0.0396 0.0445




* Solid Conductor Source: IEEE GrayBook

Figure 55. Copper Cable Impedance DataThree Single Conductors (3 - 1/c)




Figure 56 lists resistance, reactance, and impedance data for copper cables configuredas a single three-conductor cable (1 - 3/c) in conduit.

In Magnetic Duct and Steel Interlock Armor In Nonmagnetic Duct and Aluminum Armor

AWG

kcmil

600 V & 5 kV

Non Shielded

(/1000 ft.)

R X Z

5 kV Shielded &

15 kV

(/1000 ft.)

R X Z

600 V & 5 kV

Nonshielded

(/1000 ft.)

R X Z

5 kV Shielded &

15 kV

(/1000 ft.)

R X Z




8

*8

6

*6

4

*4

2

1

1/0

2/0

3.0

4/0

250

300

350

400

450

500

600

750

0.811 0.0577 0.813

0.786 0.0577 0 788

0.510 0.0525 0.513

0.496 0.0525 0.499

0.312 0.0483 0.325

0.312 0.0483 0.316

0.202 0.0448 0.207

0.160 0.0436 0.166

0.128 0.0414 0.135

0.102 0.0407 0.110

0.085 0.03970.0898

0.064 0.03810.0745

0.0552 0.03790.0670

0.0464 0.03770.0598

0.0376 0.03730.0539

0.0356 0.03710.0514

0.0322 0.03610.0484

0.0294 0.03490.0456

0.811 0.0658 0.814

0.786 0.0658 0.789

0.510 0.0610 0.500

0.496 0.0610 0.500

0.312 0.0568 0.326

0.312 0.0508 0.317

0.202 0.0524 0.209

0.160 0.0516 0.168

0.128 0.0486 0.137

0.103 0.0482 0.114

0.0814 0.04630.0936

0.0650 0.04460.0788

0.0557 0.04360.0707

0.0473 0.04310.0640

0.0386 0.04270.0576

0.0362 0.04150.0551

0.0328 0.04040.0520

0.0300 0.03940.0495

0.811 0.0503 0.812

0.786 0.0503 0.787

0.496 0.0457 0.512

0.496 0.0547 0.498

0.312 0.0422 0.324

0.312 0.0422 0.315

0.202 0.0390 0.206

0.160 0.3800 0.164

0.127 0.0360 0.132

0.101 0.0355 0.107

0.0766 0.03460.0841

0.0633 0.03320.0715

0.0541 0.03300.0634

0.0451 0.03290.0328

0.0368 0.03280.0492

0.0342 0.03270.0475

0.0304 0.03200.0441

0.0276 0.03110.0416

0.811 0.0574 0.813

0.786 0.0574 0.788

0.510 0.0531 0.513

0.496 0.0531 0.499

0.312 0.0495 0.325

0.312 0.0495 0.316

0.202 0.0457 0.207

0.160 0.0450 0.166

0.128 0.0423 0.135

0.102 0.0420 0.110

0.0805 0.0403 0.090

0.0640 0.0389 0.0749

0.0547 0.0380 0.0666

0.0492 0.0375 0.0530

0.0375 0.0375 0.0530

0.0348 0.0366 0.0505

0.0312 0.0359 0.0476

0.0284 0.0351 0.0453

0.0246 0.0344 0.0422

0.0203 0.0332 0.0389




* Solid Conductor Source: IEEE GrayBook

Figure 56. Copper Cable Impedance DataSingle Three-Conductor Cable (1 - 3/c)




Figure 57 lists resistance and reactance (impedance) data for copper aerial lines.

Size Approx. Resistance (R) Reactance (X a)

cmil

or

AWG*

Ampacity

amps**

DC

250

C 500

C

ohms/mile

AC

250

C 500

C

ohms/mile

60 Hz

1 ft spacing

ohms/mile

4(1)

4(3)

2(1)

2(3)

2(7)

1/0(7)

2/0(7)

3/0(7)

4/0(7)

170

180

220

240

230

310

360

420

480

1.374 1.053

1.388 1.518

.864 .945

.873 .955

.881 .964

.555 .606

.440 .481

.349 .381

.276 .302

1.374 1.503

1.388 1.518

.864 .945

.873 .955

.882 .964

.555 .606

.440 .481

.350 .382

.278 .303

.609

.599

.581

.571

.574

.546

.532

.518

.503

* number of strands as indicated by (x)

** for conductor at 750C, air at 25

0C, wind 1.4 mi/hr

Figure 57. Copper Aerial Line Impedance Data




Figure 58 lists resistance and reactance (impedance) data for aluminum conductor -steel reinforced (ACSR) aerial lines.

Size Approx.

Resistance (R)

Reactance (X a)

cmil

or

AWG*

Ampacity

amps**

DC

250

C 500

C

ohms/mile

AC

250

C 500

C

ohms/mile

60 Hz

1 ft spacing

ohms/mile

6

4

2

1

1/0

2/0

3/0

4/0

100

140

180

200

230

270

300

340

3.560 3.920

2.240 2.470

1.410 1.550

1.120 1.230

.885 .974

.702 .773

.556 .612

.441 .485

3.560 3.980

2.240 2.570

1.410 1.690

1.120 1.380

.888 1.120

.706 .895

.560 .723

.445 .592

.673

.659

.665

.665

.656

.641

.621

.581

* Conductor at 750C, air at 250C, wind at 1.4 mi/hr

** Current approx. 75% capacity which will produce 500C conductor temperature (250C rise) with 250C ambient, wind at 1.4 mi/hr

Figure 58. ACSR Aerial Line Impedance Data




WORK AID 2: RESOURCES USED TO CALCULATE LINE VOLTAGE DROPS



Work Aid 2B: Applicable Calculation Procedures (Ohmic Method)

Step 1. Obtain the specified one-line diagram.

Step 2. Calculate line current (IL).

IL = (kVA)/[( 3)(kV)] = (kW)/[( 3)(kV)(p.f.)]

Step 3. Calculate the load power factor angle. = cos-1 p.f.

Step 4. Calculate the load reactive factor (sin ). sin = sin[cos-1 p.f.]

Step 5. Determine line impedance (Z) per 1000 feet or per mile from a cablehandbook. Note: Use the cable impedance tables from Work Aid 1.

Step 6. Calculate the line impedance.

Z = ((R + jX) per 1000 ft)(5280 ft per mi)(number of miles)

or = ((R + jX) per mi)(number of miles)

Step 7. Calculate VD line-to-neutral.

VD = I(R cos + X sin )

Step 8. Calculate VD line-to-line.

VD = 3VD (3 system)

VD = 2VD (1 system)

Step 9. Calculate the load voltage (VL).

VL = VS - VDStep 10. Calculate VD as a percentage (VD%).

VD% = 100[(VS - VL)/VS]




Work Aid 2C: Applicable Calculation Procedures (Per-Unit Method)


Step 2. Select kVAb, kVb.

Step 3. Calculate Ib, Zb.

Ib = (kVAb)/[( 3)(kVb)], Zb = (kVb)2/(MVAb) = Rb = Xb

Step 4. Calculate line current IL in per-unit.

IL = (kVA)/[( 3)(kV)] = (kW)/[( 3)(kV)(p.f.)]

IL pu = IL/Ib

Step 5. Calculate line impedance in per-unit.

Z = ((R + jX) per 1000 ft)(5280 ft per mile)(number of miles) or

Z = ((R + jX) per mile)(number of miles)

Zpu = Z/Zb


Step 7. Calculate voltage drop in per unit.

VD = I(R cos + X sin )

where cos = pf and sin = sin[cos-1(pf)]

Step 8. Calculate VD in volts.

VD (volts) = (VDpu)(kVb)

Step 9. Calculate VD in percent.

VD% = 100 x VD(pu)

Work Aid 2D: SAES-P-100

For the content of SAES-P-100, refer to Handout 2.




WORK AID 3: RESOURCES USED TO CALCULATE LARGE MOTORSTARTING VOLTAGE DROPS


For the contents of IEEE Standard 141-1986, refer to Handout 1.

Work Aid 3B: IEEE Standard 399-1990 (Brown Book)

1. Formula for exact method calculations:

kVS = (ZM x kVI)/[(RM + RS)2 + (XM + XS)2]1/2

where: kVS = Voltage at motor when it starts

kVI = Initial voltage at starting

ZM = Impedance of the motor being started at rated voltage

= kVM /( 3 x kILRA)

RM = ZM cos and XM = ZM sin

cos = Power factor of the motor during starting

= .20 (less than 1000 hp)

= .15 (greater than or equal to 1000 hp)

RS = Total resistance of the circuit between the motor and the point inthe system where voltage is assumed to be constant. Forexample, the utility or source bus.

XS = Total reactance of the circuit between the motor and the point inthe system where voltage is assumed to be constant. Forexample, the utility or source bus.

2. Formula for the approximate method (R=0) calculations:

kVS = (XMkV)/(XM + XS)

where: XM(pu) = IFLA/ILRA or XM = VM /( 3 x ILRA)

IFLA = Motor full load amperes

ILRA = Motor locked rotor amperes or starting current in amperes

= 4 to 10 times IFLAVM = Motor line-to-line voltage in volts




Work Aid 3C: Applicable Calculation Procedures


Step 2. Select kVAb, kVb.

Step 3. Calculate Ib, Zb.

Ib = (kVAb)/[( 3)(kVb)] Zb = (kVb)2/(MVAb) = Rb = XbStep 4. Calculate the utility or source impedance in per-unit (see step 11).

Zpu(source) = (kVAb)/(kVASCA) = (Ib)/(ISCA)

Step 5. Convert electrical element impedances from the old (nameplate) base to thenew base.

Zpu(new) = Zpu(old)(KVAb(new)/kVAb(old))(kVb(old)/kVb(new))2 -

(transformers)

Xpu(new) = Xpu(old)(KVAb(new)/kVAb(old))(kVb(old)/kVb(new))2 - (motors)

Step 6. Determine cable impedance (Z) in ohms per 1000 ft from a cable handbook,and calculate cable impedance in per-unit.

Z(cable) = (Z per 1000 ft)(number of feet) = Z

Zpu(cable) = (Z)/(Zb)


Step 8. Calculate voltage at the motor during starting.

kVS pu = (ZM x kVI)/[(RM + RS)2 + (XM + XS)2]1/2

Step 9. Calculate voltage drop (kVD).

kVD pu = kVI - kVSkVD (volts) = kVD(pu) x (kVb)

Step 10. Calculate voltage drop in percent.

kVD% = 100 x kVD(pu)Step 11. If neglecting resistance (assuming R=0), replace Z in steps 4,5,6, and 8 by X

and calculate the voltage drop at motor starting as follows:

kVS = (XM)(kVI)/(XM + XS)




WORK AID 4: RESOURCES USED TO CALCULATE SYMMETRICAL RMSFAULT CURRENTS (BALANCED FAULT CONDITIONS)



Work Aid 4B: Applicable Calculation Procedures

Step 1. Obtain the system one-line diagram.

Step 2. Select/calculate the base values.

a. Select kVAb or MVAb. Note: kVAb = 100000 kVA or MVAb = 100 MVA istypically selected for power systems studies.

b. Select kVb to match one of the systems transformers rated voltage.

c. Calculate the other base voltages based on the transformer turns ratio.

d. Calculate Ib.

• Ib = kVAb /( 3 x kVb)

e. Calculate Zb.

• Xb = (kVb)2/MVAb or [(kVb)2 x 1000]/kVAb

Note: For purposes of this Module, ignore resistance.

f. Construct a table of base values.

Step 3. Model and collect circuit element data from the one-line diagram, cable handbooks, equipment nameplates, national or internal standards, and code books.




Step 4. Calculate the per-unit reactance values (jX) of each circuit element.

a. Utility or Sources:

Xs kVAb/kVASCA or

MVAb/MVASCA or

Ib/ISCA or

Z/Zb

b. Transformers:

XT Zpu(old)(kVAb(new) /kVAb(old))(kVb(old) /kVb(new))2

Zpu(old)(kVAb(new) /kVAb(old))

Note: For purposes of this Module kVb(old) = kVb(new)

c. Motors or Generators:

XG or XM = Xpu(old)(kVAb(new)/kVAb(old))

Note: Assumes kVb(old) = kVb(new)

d. Cables or Busway:

(1) X = (R + jX) /1000 ft (Figure 55 or 56)

Note: The best source for cable or busway reactance data is a vendor’shandbook. For purpose of this Module, use Figures 56 and 57 for cablevalues.

(2) X = [(jX) /1000 ft](number of feet)

(3) XC = X/Xb = j(X/Xb)




e. Aerial Lines:

(1) XW = Xa + Xd

• where: Xa = reactance of conductor per mile (Figure 57 or 58)

Xd = 0.2794 log10 deq

deq = (dab x dbc x dca)1/3 (See Figure 59)

Note: For purposes of this Module, use Figure 57 for aerial copper lines andFigure 58 for ACSR aerial lines impedance data.

(2) X = [(jX)/mile](number of miles)

(3) XOHL = X/Xb = j(X/Xb)

Figure 59. Conductor Configuration




Step 5. Draw and label the reactance diagram.

• This step consists of graphically modeling and labeling the series/parallelnetwork of reactances.

Step 6. Modify the machine reactances per Tables 24 or 25 of the IEEE Red Book(Handout 1).

Step 7. Combine all reactances into a single Thevenin-equivalent reactance.

• Ith = IF = Eth/Zth = 1.0/(R2th + X2

th)1/2 = IF pu

• Ith = Eth/Xth = 1.0/Xth (assumes R=0)

• Imom = IF x IB x 1.6 = Imom (amperes)

• Iint = IF x IB = Iint (symmetrical amperes)

Note: Assumes 1.0 p.u. voltage at the faulted bus and all other source voltages areshort circuited.

• Reactances in series: Xeq = X1 + X2 + ... + Xn

• Reactances in parallel: 1/Xeq = 1/X1 + 1/X2 + ... + 1/Xn

• Two reactances in parallel: Xeq = (X1) (X2)/(X1 + X2)

Step 8. Calculate Ipu at each bus.

• Ipu = Epu /Xpu where Epu = 1.0

Step 9. Calculate I(amps) at each bus.

• I(amps) = Ipu x Ib




GLOSSARY

American National An organization whose members approve various standardsStandards Institute for use in American industries.(ANSI)

apparent power (S) The product of voltage and current. The units of S are volt-ampere (va) and kilovolt-ampere (kVA).

asymmetrical current Envelopes of the peaks of the current waves are not symmetrical(IASY) about the zero axis. Most short-circuit currents are nearlyalways

asymmetrical during the first few cycles after the fault occurs.

asymmetrical factor The ratio of asymmetrical current to symmetrical current after a(AF) fault occurs; a function of the system X/R ratio.

base apparent power An arbitrary kilovolt-ampere value, (e.g., 100000 kVA or(kVAb)(MVAb) 100 MVA).

base current (Ib) The ratio of base kVA to the product of 1.732 times base kV.

base impedance(Zb) The ratio of base kV squared to base MVA.

base voltage (kVb) A voltage selected at a specific level to match the transformer’srated voltage at that level.

full load amperage Current drawn by a motor under full load conditions(IFLA) (e.g., rated horsepower and rated voltage).

horsepower (hp) The mechanical power rating of the machine. One hp equals 746watts.

impedance (Z) The ratio of voltage to current. With sinusoidal voltage andcurrent, impedance is a complex number and will have both amagnitude and an angle. The units of impedance are ohms ().

impedance triangle A right-angle triangle developed geometrically from theequations associated with resistance, inductive reactance, andcapacitive reactance.

induction motors A motor in which the field is produced by induction from thestator rather than from a direct current winding.




Institute of Electrical A worldwide society of electrical and electronics engineers.and ElectronicsEngineers (IEEE)

interrupting duty The short-circuit rating which medium and high voltage breakerscan safely interrupt. Sometimes called the contact parting duty.For interrupting duty calculations, machines are modeled usingtheir transient reactance.

lagging power factor Where the current lags the voltage, as in an inductive circuit.

leading power factor Where the current leads the voltage, as in a capacitive circuit.

line impedance Impedance (resistance and reactance) of an overhead line used tocalculate line voltage drops.

locked rotor amperage Current drawn by a motor during starting (usually four to sixtimes(ILRA) full load amperage).

low voltage (LV) Voltage levels less than 1000 volts. Usually called utilizationlevel voltages.

lump sum motor The size given to a motor. This size equals the sum of all motorson a bus less than 50 hp.

medium voltage Voltage levels greater than or equal to 1000 volts and less than(MV) 100000 volts, usually called distribution level voltages.

momentary duty The short-circuit rating assigned to medium and high voltagecircuit breakers during the first cycle after a fault occurs.Sometimes called the close-and-latch capability of a breaker.For momentary duty calculations, machines are modeled usingtheir subtransient reactance.

motor impedance Impedance of a large motor being started. Ratio of motor fullload amperage to motor locked rotor amperage.

National Electric An electrical safety code developed and approved every threeyearsCode (NEC) by the National Fire ProtectionAssociation (NFPA 70).

overcurrent An electrical device inserted in a circuit to protect it protectivedevice against damage from an overload or short-circuit. This

protection is achieved by automatic interruption.




percent voltage drop The ratio of voltage drop in a circuit to voltage delivered by the(VD%) circuit, multiplied by 100 to convert to a percent.

per-unit quantity (p.u.) Ratio of actual quantity to base quantity.

power factor (p.f.) The term cosine theta (cos ) where theta () is the angle betweenvoltage and current.

power triangle A right angle triangle developed geometrically from theequations associated with real power (P), apparent power (S),and reactive power (Q).

reactance (X) The imaginary portion of the complex impedance (Z). The unitsof reactance are ohms ().

reactive power (Q) The product of voltage, current, and sin theta (q). The units of Qare vars (VAR) and kilovars (kVAR).

real power (P) The product of voltage, current, and cosine theta (q). The unitsof P are watts (W) and kilowatts (kW).

receiving end Voltage at the terminal of the load. The difference of sendingendvoltage (VL) voltage and receiving end voltage is the line voltage drop.

resistance (R) The real portion of the complex impedance (Z). The units ofresistance are ohms ().

root mean square In a sinusoidally varying waveshape, the rms value is equal to(rms) .707 (1/ 2 ) times the maximum (peak) value of the

waveshape.

sending end voltage Voltage at the source of power, used in line voltage drop(Vs) calculations.

single-phase voltage The product of phase current and conductor impedance.(VD) Sometimes called line-to-neutral voltage drop (VD).

short-circuit current Current (usually very large) flowing in an electrical system asthe(Isc) result of a three-phase, phase-to phase, double-phase-to-ground,

or single phase-to-ground fault.

short-circuit current A source of current which provides short-circuit current to a faultsource (ISCC) point. The four sources are generators, synchronous motors,




induction motors, and the utility (really a generator). Sometimescalled short circuit current available (ISCA).

short-circuit study Calculations of short-circuit currents resulting from faultconditions. These calculations can be accomplished manually(hand-calculations), but are usually done with the aid of apersonal computer.

symmetrical current Envelopes of the peaks of the current waves are symmetrical(ISYM) about the zero axis.

synchronous motor A motor having a field excited by direct current and a statorwinding in which alternating current flows.

system reactance Total reactance of a circuit between a motor and the point in the(Xs) system where voltage is assumed to be constant.

system X/R ratio The ratio of reactance to resistance in the system from the faultpoint looking back toward the source.

subtransient reactance The apparent reactance of the stator winding at the instant(X”d) short-circuit occurs. X”d determines the current flow during the

first few cycles.

synchronous reactance The apparent reactance of the stator winding under steady-state(Xd) conditions (> 30-200 cycles).

three-phase voltage The product of phase current and conductor impedance times

1.732 ( 3). Sometimes called line-to-line voltage drop.

transient reactance (X’d) The apparent reactance of the stator winding which determinesthe

current flowing three to eight cycles after a fault occurs.

voltage drop(line)(VD) Voltage drop calculated at the terminals of a load under normalconditions. Usually expressed as a percent.

voltage drop(motor)(VD) The voltage drop calculated at the terminals of a motor understarting conditions. Usually expressed as a percent.

calculating voltage drops and symmetrical rms fault currents

Documents

modeling step

b values step

diagram step

diagrams step

base values

technical material

modeling methods

tr lines