Calculating Voltage Drops and Symmetrical RMS Fault Currents

Engineering Encyclopedia Saudi Aramco DeskTop Standards

Calculating Voltage Drops And Symmetrical R MS Fault Currents

Note: The source of the technical material in this volume is the Professional Engineering Development Program (PEDP) of Engineering Services. Warning: The material contained in this document was developed for Saudi Aramco and is intended for the exclusive use of Saudi Aramco’s employees. Any material contained in this document which is not already 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, Engineering Services, Saudi Aramco.

Chapter : Electrical File Reference: EEX10204

For additional information on this subject, contact W.A. Roussel on 874-1320

Engineering Encyclopedia

Electrical Calculating Voltage Drops and Symmetrical RMS Fault Current

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 Saudi Aramco DeskTop Standards



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 Saudi Aramco DeskTop Standards



Exact Calculations ...............................................................38 Approximate Calculations (Neglect R).................................39 Per-Unit Model Selection Factors...................................................39 One-Line Diagram................................................................39 Base Values.........................................................................41 Per-Unit Calculations ...........................................................42 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 Saudi Aramco DeskTop Standards



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 Per-Unit Model Selection Factors...................................................69 One-Line Diagram................................................................69 Base Values.........................................................................70 Per-Unit Calculations ...........................................................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 AN ELECTRICAL 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 LINE VOLTAGE DROPS ...........................................................90 Work Aid 2A: IEEE Standard 141-1986 (Red Book) ......................90 Saudi Aramco DeskTop Standards



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 LARGE MOTOR STARTING VOLTAGE DROPS .........................92 Work Aid 3A: IEEE Standard 141-1986 (Red Book) ......................92 Work Aid 3B: IEEE Standard 399-1990 (Brown Book)...................92 Work Aid 3C: Applicable Calculation Procedures ..........................93 WORK AID 4: RESOURCES USED TO CALCULATE SYMMETRICAL RMS FAULT CURRENTS (BALANCED FAULT CONDITIONS)................................94 Work Aid 4A: IEEE Standard 141-1986 (Red Book) ......................94 Work Aid 4B: Applicable Calculation Procedures...........................94 GLOSSARY ..............................................................................................98

Saudi Aramco DeskTop Standards



MODELING AN ELECTRICAL POWER SYSTEM IN PER-UNIT Modeling Methods In order to calculate short-circuits, line voltage drops and motor starting currents, the different impedance values for circuit elements (transformers, cables, motors, generators) must be determined from nameplates, handbooks, and catalogs. These impedance values can be modeled in ohms, percent, or per-unit on a chosen base value. In most calculations the per-unit modeling of circuit elements is used because it is easier and more convenient to solve the problems when the system contains several voltage levels. Impedances that are modeled in per-unit can be combined in parallel or series 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 almost always the case, the ohmic value will change as the square of the ratio of the voltage levels. In other words, the ohmic values change from one side of a transformer to the other. For example, if a power system has three voltage levels, which is not uncommon, each electrical circuit element (cable, transformer, motor, etc.) will have three ohmic values (see Figure 1). For this reason, power system studies are typically not performed using ohmic models. Percent Model

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


1



Figure 1. Ohmic Model Impedances


2



Per-Unit Model

When impedances of circuit elements are expressed in per-unit, the problems mentioned for both the ohmic and percent models are eliminated. Impedances expressed 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 kV secondary sides of the transformer, and ZT2 =5.5% on both sides of the 4.16 kV primary and .48 kV secondary sides of the transformer. By definition, the per-unit quantity (ohms, voltage, current, etc.) equals the ratio of the actual quantity to the base quantity (p.u. quantity = actual quantity/base quantity). The following steps are use to model an electrical system in per-unit: Note: Work Aid 1 has 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-unit modeling of any power system regardless of the type of power study being performed.

One-Line Diagram (Step 1) An accurate one-line diagram (system road map) should be prepared that shows all sources (utility tie, generators, and motors) and significant circuit elements (transformers, cables, busway etc.). Note: The study to be performed will only be as accurate as the one-line diagram. Figure 2 is an example one-line diagram of a power system that will be used throughout the remainder of this Module.


3



Figure 2. Example One-Line Diagram


4



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 to make the mathematical calculations somewhat less tedious. For purposes of this Module, select kVAb = 100000 kVA or MVAb = 100 MVA (see Work Aid 1). Note: Perunit 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 the base 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 shown in 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


5



Base Current

Base current (Ib) is calculated through use of the apparent power relationships developed in the previous Module. 3 x kVb x Ib

•

kVAb =

•

Ib = kVAb/( 3 x kVb)

Example B: Using Work Aid 1, what are the base currents for the power system shown 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, and power relationships developed in the previous Module. These relationships are the following: •

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 in figure 2? Answer C: Bus 50 Bus 100 Bus 250 Bus 300

shown

Zb = (34.5)2/100 = 11.902 Zb = (13.8)2/100 = 1.9044 Zb = (0.48)2/100 = 0.0023 Zb = (4.16)2/100 = 0.173


6



Table of Base Values

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

Bus

kVA b

kV b

Ib

Zb

Numbers

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 elements of the power system that is being studied. The following circuit elements will be modeled: •

Utility or source


7



•

Transformers

•

Conductors (cables, busway, aerial lines)

•

Generators

•

Motors (over 1000V)

•

Motors (under 1000V)


8



Utility or Source

The utility system is modeled as a voltage (EU) behind an impedance (ZU) as shown in Figure 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)


9



Example D: Figure 2 modeled the utility at 450 MVA, X/R = 11.4. What are the other three 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


10



Grounding Resistors

Grounding resistors (RR) are modeled as shown in Figure 5. Data for a grounding resistor 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 following data 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)


11



Conductors

Single-Conductors are modeled as an impedance (ZC) as shown in Figure 7. The following 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 of aluminum 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 except the configuration is 1-3/c. Busway is modeled as an impedance (ZB), also as shown in Figure 7. The following data 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)

12



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 other conductors, the reactance (XOHL) consists of two components (Xa and Xd) . Xa is found in a cable handbook, like most other cable data. However, Xd is a function of the 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 and spacing (See Work Aid 1)

Figure 8. Aerial Line Model (ZOHL)


13



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 Figure 10). However, the impedance data that are required to model a motor depends on the system and motor voltages.

Figure 10. Motor Model (ZM)


14



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 are also 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), low voltage 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 or assume 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 assume 1.2X”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 assume 3.0X”d = 50% on motor rating. Note: For purposes of this Module use 3.0X”d = 50%.


15



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 hp rating 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 was identified 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 is provided as kVA, MVA, current, or ohms. Work Aid 1 lists the procedures and formulas to calculate the per-unit impedance of the utility depending on the given data.


16



Example E: Using Work Aid 1, what is the per-unit impedance of the utility shown in Figure 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 the transformer’s kVA (kVAb(old)) and voltage (kVb(old)) ratings. Calculating the per-unit impedance 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 in Work Aid 1. Example F: Using Work Aid 1, what are the per-unit impedances of the 3 transformers 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


17



ZT2 = (0.0550)(100/0.75)(0.48/0.48)2 = 7.333 pu X/R ratio = 6 (Handout 1, page 344) 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 X/R ratio = 8 (Handout 1, page 344) 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 the nameplate base kVA and voltage ratings. Calculating the reactance values also requires 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 X/R ratio = 20 (Handout 1, page 344) 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 ohmic value by the base ohms (Zb) at the given voltage. Work Aid 1 lists the procedures for calculating conductor per-unit impedance values.


18



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 a given 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 line reactance (Xa) that is based on the type of material is also provided in ohms per unit of length (1 mile). Aerial line reactance (Xd) must be calculated based on the configuration and spacing of the overhead lines. Work Aid 1 lists the procedures for calculating the per-unit impedance [R + j(Xa +Xd)] of aerial lines.


19



Example I:

Using Work Aid 1, what is the per-unit impedance of overhead aerial line C1?

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/parallel network of resistances and reactances). Complex Diagram

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


20



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

See Figure 12.

Figure 12. Complex Impedance Diagram


21



R-Only Diagram

The R-only resistance diagram requires redrawing and labeling the complex impedance diagram with the reactances omitted (X = 0). Example K: Using Work Aid 1 and referring to Figure 12, draw and label the diagram.

R-only

Answer K: See Figure 13.

Figure 13. R-Only Diagram


22



X-Only Diagram

The X-only reactance diagram requires redrawing and labeling the complex impedance diagram with the resistances omitted (R = 0). Example L: Using Work Aid 1 and referring to Figure 12, draw and label the X-only diagram. Answer L:

See Figure 14.

Figure 14. X-Only Diagram


23



CALCULATING LINE VOLTAGE DROPS Introduction Designers of power systems must have practical knowledge of voltage drop calculations, not only to meet required codes and standards, but to ensure that the required voltage of a particular piece of equipment, for example a motor, is kept within manufacturer’s specified tolerances to prevent damage to the equipment. Note: The effects of voltage drops on different types of electrical equipment was discussed in EEX 102.02. NEC Article 210-9(a), FPN No. 4, states that “conductors for branch circuits as defined in Article 100, sized to prevent a voltage drop exceeding 3 percent at the farthest outlet of power, heating, and lighting loads, or combinations of such loads, and where the maximum total voltage drop on both feeders and branch circuits to the farthest outlet does 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 in Article 100, sized to prevent a voltage drop exceeding 3 percent at the farthest outlet of power, heating, and lighting loads, or combinations of such loads, and where the maximum total voltage drop on both feeder and branch circuits to the farthest outlet does not exceed 5 percent, will provide reasonable efficiency of operation”. Saudi Aramco Standard SAES-P-100 limits the voltage drops of the electrical power system 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 voltage drop (VD), and the receiving end or load voltage (VL) are shown in Figure 15.


24



Figure 15. Line Voltage Drop Phasor Diagram Exact Calculations

Due to the phasor relationships shown in Figure 15, exact methods of calculating line voltage drops require extensive knowledge of complex phasor algebra as discussed in Module EEX 102.03. However, for line lengths less than 50 km and system voltages below 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 Aramco installations, the approximate method formula, as presented below, is adequate for most 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 power factor loads.


25



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) =

3 VD

Ohmic Model Selection Factors Calculating line voltage drops is typically the only power system study that is performed using the ohmic method. The rationale for using the ohmic method is that the calculations are performed incrementally one voltage level at a time. Note: Work Aid 2B has been developed to teach ohmic method procedures for calculating line voltage drops. One-Line Diagram

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


26



Figure 16. Typical One-Line Diagram


27



Load Current (I L)

For the purposes of line voltage drop calculations, the load current is first calculated based on load voltage equal to the sending end voltage (VS). In reality, the actual load current should be based on the actual load voltage (VL = VS - VD). However, this requires the exact method calculations, which once again, is not typically necessary for line 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 for lagging power factor loads, sin should also be a negative number. However, because the line drops are considered to be scalar (non-phasor) quantities, by convention sin is considered to be a positive number. Line Impedance (Z = R + jX)

Line impedance is modeled (calculated) simply as R + jX ohms. Because the approximate method formula is based on scalar quantities, the j operator (1900) is ignored, 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 a one-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 line voltage 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) =


3 VD (line-to-neutral)

28



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 method instead of the ohmic method. Because the circuit elements are already in per-unit form from use in the other studies, they may be used to perform the line voltage drop calculations as well. Note: Work Aid 2C has been developed to teach per-unit method procedures for calculating line voltage drops. One-Line Diagram

The one-line diagram used for the per unit model is no different from the ohmic method model 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, as mentioned in the previous Information Sheet, most studies are performed on a 100 MVA base (MVAb = 100 MVA or kVAb = 100000 kVA). Per-Unit Calculations

Load Current - The load current (IL) must be calculated in amperes and converted to per-unit amperes by dividing by the base current (Ib). Impedance - The line impedance must be calculated in ohms (Z)and converted to perunit ohms by dividing by the base ohms (Zb).


29



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 base voltage (kVb). No additional calculation, such as 3 x VD, is required when using perunit procedures, because the line-to-line or line-to-neutral relationships are already factored 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 drop by 100(%VD = 100VD pu). Again, the result should be compared to the NEC and SAES-P-100 limits. Example M: Using Work Aid 2B and referring to the one-line diagram shown in Figure 16, calculate the line voltage drop assuming the circuit conductor is spaced 2.5 feet on center. Answer M: 1. One-Line Diagram (Figure 16) 2. Calculate IL:

IL = kW/( 3 = 104.6 A

x

kV

x

p.f.)

=

2000/( 3

x

13.8

x

0.80)

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)


30



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


31



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 in this 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) 2 2 1/2 - [(7967 - ((104.6)(1.193)(0.8) - (104.6)(1.332)(0.6)) ]

= 7967 + 111.462 + 74.873 - 7966.983 = 186.352 V This example shows the exact method calculation was unnecessary.


32



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%


33



CALCULATING LARGE MOTOR STARTING VOLTAGE DROPS Introduction Alternating current (AC) motor characteristically draw much higher current during starting than they draw under normal running conditions (see Figure 17). This sudden draw in current results in excessive voltage drops on the entire power distribution system, resulting in unacceptable conditions for other electrical equipment on the power system.

Figure 17. Motor Starting Currents


34



Motor Starting Effects Critical Voltage Levels (Minimum)

To develop sufficient starting torque, alternating current motors require a full voltage starting current (ILRA) that is between four and ten times their normal full-load running current (IFLA). Such large increases in current, when suddenly drawn, can prevent acceleration of the motor to normal speed and can cause other plant equipment such as 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 is common 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 Standard SAES-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 Module the maximum voltage drop for medium voltage motors is also 15%. Figure 18 lists the critical voltage levels of different plant equipment. Saudi Aramco’s 15% voltage drop limits under motor starting conditions should prevent problems on most of the other equipment in their facilities.

Voltage Drop Location or Problem

Minimum Allowable 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%

35



Solid-State Control Devices

Noticeable Light Flicker

90%

3% Change

Source: IEEE Brown Book Figure 18. Critical Voltage Levels


36



Effects on Plant Equipment

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

Figure 19. Induction Motor Speed Torque Characteristics Other Plant Motors are also adversely affected by a system voltage drop throughout the plant. The other motors slow down and their horsepower load may exceed breakdown torque at the reduced voltage. Motor contactors may drop out and the low voltage at the motor terminals cause higher currents to be drawn, causing even higher system voltage drop (“domino effect”).


37



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 may permit 10% drop without damaging the computer system. Electronic Control equipment, similar to computers, have wide ranges of voltage drop tolerance. It is best to check with the equipment vendor to determine if sudden drops in voltage are harmful to the equipment. Light Flicker as a result of voltage drops is a perception issue. If it occurs often enough, it becomes objectionable. This is not typically a problem since most large motors are not frequently stopped and started. In fact, the duty cycles of large motors usually require cooling down periods before restarting. Calculation Methods Similar to line voltage drop studies, large motor starting voltage drop studies may be performed using exact and approximate methods. Note: Work Aid 3 has been developed 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 the system 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 any load is the initial voltage (kVI) times the ratio of the load impedance (ZM) to the entire system impedance [(RM + RS2) + (XM + XS)2]1/2.


38



Approximate Calculations (Neglect R)

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

kVS = (XMkVI)/(XM + XS)

•

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

Per-Unit Model Selection Factors Large motor starting voltage drops are usually accomplished through use of the perunit calculation procedures because more than one voltage level is present in the system. Note: Work Aid 1 was developed to teach per-unit modeling of an electrical power system. The following factors should be considered for calculating large motor starting voltage drops: •

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 starting point. Figure 20 is a typical one-line diagram used for calculating large motor starting voltage drops.


39



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


40



Base Values

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

Bus

kVA b

kVb

Ib

Zb

100, 140

100000 kVA

13.8 kV

4183.7 A

1.9044

145

100000 kVA

4.16 kV

13878.6 A

0.173

Numbers

Figure 21. Motor Voltage Drop Study Base Values


41



Per-Unit Calculations

The per-unit calculation procedures for the utility, transformers, motors, and cables are identical to the procedures taught in Work Aid 1. Example P: Using Work Aid 1 and referring to Figure 20, model the system in perunit. 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


42



Impedance Diagram

If the exact method calculations are used, the impedance diagram model should include both resistance (R) and reactance (X). If approximate method calculations are used, 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-line diagram shown in Figure 20. Answer Q: See Figure 22.

Figure 22. Example Q Impedance Diagram


43



Voltage Drop

The voltage drop at the terminals of the starting motor is calculated by performing a voltage divider calculation. Since the per-unit voltage at starting is assumed to be 1.0 pu, 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 is neglected. Percent Voltage Drop

Although the actual voltage drop in volts is important, the drop in percent is compared against codes and standards. For Saudi Aramco installations, as mentioned earlier, the voltage drop is limited to 15% under starting conditions. The percent voltage drop is 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 operating problems with other equipment in the plant. The procedure for calculating the voltage drop at other buses also uses the voltage divider rule as described above, and in Work Aid 3. For example, the voltage drop at the other buses shown in Figure 22 are the following: •

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

•

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


44



Example R: Using Work Aid 3 and referring to Figure 22, calculate the voltage drop at the motor terminals (Bus 145) using both the exact and approximate calculation 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 M

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

kVS

= (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


45



CALCULATING SYMMETRICAL RMS FAULT CURRENTS (BALANCED FAULT CONDITIONS) Purposes of Fault Calculations No matter how well an electrical power system is designed, it will occasionally experience short circuits resulting in abnormally high current flows. Fault calculations are performed to determine device interrupting and withstand ratings, and to determine relay pickup and sensitivity settings. Protective Device Interrupting Ratings

The maximum calculated short circuit current magnitudes are used to select adequate protective device (fuses and breakers) interrupting ratings to minimize the fault damage and plant down time. Equipment Component Withstand Ratings

Other electrical equipment (e.g. buses and cable) must be capable of withstanding the high 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 the pickup settings of relays and low voltage breakers. For example, the instantaneous trip setting of an upstream relay should be set » 10% above the maximum available short 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 relay settings for sensitivity. For example, the maximum setting for ground fault relays in medium voltage system is typically selected based on a minimum fault current of 10% of the available maximum fault current.


46



Effects of Short Circuits Many things happen on an electrical power system when a short circuit occurs--all of them 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) that varies as a function of the current squared (I2) and the fault duration (t). The temperature rise (t) is an integrated effect as shown below:

z

t2

(σ cυA ) 2

•

T=

•

where:

-1

i (t) dt = I2t

t1

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 amperes t = time in seconds

Figure 23. Balanced Loads


47



Mechanical Stress

All components carrying electrical current are also subjected to mechanical stress related to the square of the current and the fault duration [i2(t)]. The primary difference from thermal stress is that the mechanical stress is an instantaneous effect as shown below: •

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 directly proportional to the magnitude of the fault current. The worst case drop is of course, at the point of the fault, where the drop is 100% (V=0), but all parts of the power system are subjected to some level of voltage drop.


48



Characteristics of Fault Current Sources When calculating the fault current, it is important to include all sources (Figure 25) of short circuit current. The fault current sources are generators (local), motors (induction and synchronous), and utility (“large” generators).

Figure 25. Fault Current Sources


49



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 because the field excitation is maintained and the prime mover continues to drive the generator at normal speed. Stated another way, the generator “sees” the fault as an instantaneous load demand. The only impedance that limits the fault current magnitude (Figure 26) flowing from a generator is the impedance of the generator and any system impedance between the generator terminals and the point of the fault.

Figure 26. Local Generator Fault Source


50



Motors

Synchronous Motors are constructed similar to generators, which is that they have a separate field excited by dc current and current flows in the stator winding. Normally, a synchronous motor converts electrical energy (input kW) to mechanical energy (output hp). Under fault conditions, the opposite occurs, the synchronous motor acts like a generator and delivers fault current with the load inertia acting as a prime mover. As a result 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 any impedance between the motor terminals and the point of the fault.

Figure 27. Synchronous Motor Fault Source


51



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

Figure 28. Induction Motor Fault Source


52



Utility System

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

Figure 29. Utility (Generator) Fault Source


53



Total Fault Current

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

Figure 30. Total Fault Current


54



Machine Reactance Modeling The impedance of generators and motors consist primarily of reactance and, unlike cables and transformers, is not a simple value, but is more complex and also varies with 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, the waveshape is as shown in Figure 31. The magnitude starts out relatively high and decays to a symmetrical or steady state value after a definite time lapse depending on the types of sources and the system X/R ratio.

Figure 31. Short Circuit Current Oscillogram


55



Subtransient Reactance (X” d)

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

0 < t< 3 to 6 cycles

A synchronous motor has the same kind of reactance as a generator, but it is of a different value. Induction motors have no field coils, but the rotor bars act like the amortisseur winding in a generator; therefore, induction motors are said to have subtransient reactance only. Transient Reactance (X’ d)

Transient reactance (X’d) determines the current following the period when subsransient reactance is the controlling value. Transient reactance is effective up to one-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 a steady state condition is reached. It is not effective until several seconds after the short circuit occurs; consequently, it is not generally used in short-circuit calculations in industrial 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 to symmetrical currents.


56



Symmetrical Fault Currents

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

Figure 32. Symmetrical Fault Current


57



Asymmetrical Fault Currents

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

Figure 33. Asymmetrical Fault Currents


58



System Power Factor and X/R Ratios

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


59



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 lagging approximately 180 (see Figure 35) to instantaneously (t=0) lagging by approximately 850 (see Figure 36) is for the current waveshape to offset itself from the zero axis. The greater the ratio of reactance to resistance (X/R ratio), the greater the offset and the slower the rate of decay.


60



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)


61



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

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


62



Sample Waveshapes

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

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


63



Low X/R Ratio - Figure 40 shows a typical short circuit oscillogram in a low voltage circuit 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 current waveshapes 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


64



AC and DC Components

Asymmetrical fault currents are analyzed in terms of two components, an AC or symmetrical component, and a DC or exponential component. Figure 41, once again shows 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 B q

= E sin ( q - f)/(Rs2 + Xs2)1/2 = E/( Rs2 + Xs2)1/2 = E/Zs = tan-1 Xs/Rs

•

DC Component - Aexp[(-Rs/Xs) t]

•

AC Component - B sin(w t + f - q )


65



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

Figure 42. Time Domain Model of Voltage and Current


66



Types of Faults/Magnitudes Three-Phase Faults

Three-phase faults (Figure 43) usually are the least often occurring, but are also often the worst case in industrial power systems. Note: This Module is restricted to calculating 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-phase fault values. Accordingly, fault calculations involving line-to-line faults (Figure 44) are seldom required because it is not the maximum value.

Figure 44. Line-to-Line Faults


67



Line-to-Ground Faults

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

Figure 45. Line-to-Ground Faults Arcing Faults

Many low voltage power systems experience low level arcing faults, and in many cases, they are the most damaging faults, especially at the 480 volt level. Because of the very high arc impedance, arcing fault current magnitudes tend to be too low for the protective devices to detect and clear. Arcing fault current magnitudes are typically estimated 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 device ranges rather than in the fault (instantaneous or short-time) ranges of the protective devices.


68



Per-Unit Model Selection Factors Short circuit studies are usually accomplished using per-unit calculation procedures because more than one voltage level is present in the system. Note: Work Aid 1 was developed 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 starting point. Figure 46 is a typical one-line diagram used for calculating short circuit currents.

Figure 46. One-Line Diagram for Short Circuit Calculations


69



Base Values

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

Bus

kVA b

kVb

Ib

Zb

100, 140

100000 kVA

13.8 kV

4183.7 A

1.9044

145

100000 kVA

4.16 kV

13878.6 A

0.173

Numbers

Figure 47. Short Circuit Study Base Values Per-Unit Calculations

The per-unit calculation procedures for the utility, transformers, motors, and cables are identical to the procedures explained in Work Aid 1. Example S: Using Work Aid 1 and referring to Figure 46, model the system in perunit. 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


70



Impedance Diagrams

If the exact method calculations are used, the impedance diagram model should include both resistance (R) and reactance (X). If approximate method calculations are used, 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 line diagram 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 reflect the momentary duty and interrupting duty of medium voltage breakers (Figure 49).


71



Figure 49. Medium Voltage Breaker Duties Momentary Duty Network - Motors and generators are modeled for the momentary duty network (first cycle) as follows: Note: See Figure 50. •

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

•

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

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

Induction motors less than 50 hp - neglect

•

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

•

Synchronous motors and induction motors > 250 hp X = 1.5 X”d


72



Type of Rotating Machine

Interrupting Duty

Momentary Duty

(Per Unit)

(Per Unit)

All turbo-generators, all hydrogenerators with amortisseur windings, and all condensers 1.0X”d

1.0X”d

0.75X”d

0.75X”d

1.5X”d

1.0X”d

1.5X”d

1.0X”d

3.0X”d

1.2X”d

Neglect

Neglect

Hydro-generators without amortisseur windings

All synchronous motors

Induction Motors

Above 1000 hp at 1800 r/min or less Above 250 hp at 3600 r/min

All others, 50 hp and above

All smaller than 50 hp (a)


73



Induction Motors (Note 2) All others, 50 hp and above All smaller than 50 hp

3.0X”d (Note 5)

1.2X”d (Note 3)

Neglect

1.67X”d (Note 4)

(b) Notes:1.

Resistance values should also use same multipliers.

2.

For comprehensive multivoltage calculations.

3.

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

4.

Or X = .28 p.u. based on motor rating.

5.

Or X = .50 p.u. based on motor rating. * Source IEEE Red Book

Figure 50. Rotating Machine Reactance Multipliers


74



Example U: Using Work Aid 4 and referring to Figures 46 and 48, model the interrupting 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/parallel network 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 by using the momentary duty reactances of any machines in the system, followed by the same 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 Thevenin reactance. Note: See Work Aid 4 for the procedures.


75



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

1. Thevenin Circuit: See Figure 52.

Figure 52. Example V Thevenin Circuit


76



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)


77



WORK AID 1:

RESOURCES USED TO MODEL AN ELECTRICAL POWER SYSTEM 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 is typically 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, cable handbooks, equipment nameplates, national or internal standards, and code books. 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)]


78



(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 (page 344) 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 found in IEEE Red Book Figure N1.2 (page 344) and in IEEE Red Book Figure N1.3 (page 344) for induction motors.


79



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’s handbook. For purpose of this Module, use Figures 55 and 56 for cable values. (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 and Figure 58 for ACSR aerial lines impedance data. (2) R2 @ t10C = R[(T + t20C)/(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)


80



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 network of resistance and reactances. a. Draw and label the X-only reactance diagram. b. Draw and label the R-only resistance diagram.


81



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

In Magnetic Duct and Steel Interlock Armor

AW

In Nonmagnetic Duct and Aluminum Armor

600 V & 5 kV

5 kV Shielded &

600 V & 5 kV

5 kV Shielded &

Non Shielded

15 kV

Nonshielded

15 kV

(/1000 ft.)

(/1000 ft.)

(/1000 ft.)

(/1000 ft.)

G kcm il

R

X

Z

R


X

Z

R

X

Z

R

X

Z

82



8

0.811

0.0754 0.814

0.811

0.0860

0.816

0.811

0.0603 0.813

0.811

0.0688 0.814

*8

0.786

0.0754 0 790

0.786

0.0860

0.791

0.786

0.0603 0.788

0.786

0.0688 0.789

6

0.510

0.0685 0.515

0.510

0.0796

0.591

0.510

0.0548 0.513

0.510

0.0636 0.514

*6

0.496

0.0685 0.501

0.496

0.0796

0.502

0.496

0.0548 0.499

0.496

0.0636 0.500

4

0.312

0.0632 0.327

0.312

0.0742

0.329

0.312

0.0506 0.325

0.312

0.0594 0.326

*4

0.312

0.0632 0.318

0.312

0.0742

0.321

0.312

0.0506 0.316

0.312

0.0594 0.318

2

0.202

0.0585 0.210

0.202

0.0685

0.214

0.202

0.0467 0.207

0.202

0.0547 0.209

1

0.160

0.0570 0.170

0.160

0.0675

0.174

0.160

0.4560 0.166

0.160

0.0540 0.169

1/0

0.128

0.0540 0.139

0.128

0.0635

0.143

0.127

0.0432 0.134

0.128

0.0507 0.138

2/0

0.102

0.0533 0.115

0.103

0.0630

0.121

0.101

0.0426 0.110

0.102

0.0504 0.114

3.0

0.085

0.0519

0.0814 0.0605

0.101

0.0766 0.0415

0.0958 4/0

0.0871 0.0650 0.0583

0.064

0.0497

0.0895 0.0484 0.0939

0.0929

0.0810

0.0640 0.0466 0.0792 0.0633 0.0396 0.0748

250

0.0547 0.0456 0.0712 0.0557 0.0570

300 350 400

0.0552 0.0495

0.0797

0.0742

0.0541 0.0396 0.0670

0.0473 0.0564 0.0464 0.0493

0.0736

0.0677

0.0451 0.0394 0.0599

0.0460 0.0451 0.0644 0.0375 0.0450 0.0586 0.0348 0.0438 0.0559

0.0386 0.0562 0.0378 0.0491 450 500

0.0681

0.0617

0.0368 0.0393 0.0536

0.0362 0.0548 0.0356 0.0490

0.0657

0.0606

0.0342 0.0392 0.0520

600 750

0.0312 0.0430 0.0531 0.0284 0.0421 0.0508 0.0246 0.0412 0.0479

0.0328 0.0538 0.0322 0.0480

0.0630

0.0578

0.0304 0.0384

0.0203 0.0396 0.0445

0.0490 0.0300 0.0526

0.0294 DeskTop 0.0466 Standards 0.0505 Saudi Aramco 0.0551

0.0264 0.0516

0.0276 0.0373 0.0464

83



* Solid Conductor

Source:

IEEE Gray

Book

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


84



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

In Magnetic Duct and Steel Interlock Armor

AW

In Nonmagnetic Duct and Aluminum Armor

600 V & 5 kV

5 kV Shielded &

600 V & 5 kV

5 kV Shielded &

Non Shielded

15 kV

Nonshielded

15 kV

(/1000 ft.)

(/1000 ft.)

(/1000 ft.)

(/1000 ft.)

G kcm il

R

X

Z

R


X

Z

R

X

Z

R

X

Z

85



8

0.811

0.0577 0.813

0.811

0.0658

0.814

0.811

0.0503 0.812

0.811

0.0574 0.813

*8

0.786

0.0577 0 788

0.786

0.0658

0.789

0.786

0.0503 0.787

0.786

0.0574 0.788

6

0.510

0.0525 0.513

0.510

0.0610

0.500

0.496

0.0457 0.512

0.510

0.0531 0.513

*6

0.496

0.0525 0.499

0.496

0.0610

0.500

0.496

0.0547 0.498

0.496

0.0531 0.499

4

0.312

0.0483 0.325

0.312

0.0568

0.326

0.312

0.0422 0.324

0.312

0.0495 0.325

*4

0.312

0.0483 0.316

0.312

0.0508

0.317

0.312

0.0422 0.315

0.312

0.0495 0.316

2

0.202

0.0448 0.207

0.202

0.0524

0.209

0.202

0.0390 0.206

0.202

0.0457 0.207

1

0.160

0.0436 0.166

0.160

0.0516

0.168

0.160

0.3800 0.164

0.160

0.0450 0.166

1/0

0.128

0.0414 0.135

0.128

0.0486

0.137

0.127

0.0360 0.132

0.128

0.0423 0.135

2/0

0.102

0.0407 0.110

0.103

0.0482

0.114

0.101

0.0355 0.107

0.102

0.0420 0.110

3.0

0.085

0.0397

0.0814 0.0463

0.0898

0.0936

0.0766 0.0346 0.0841

4/0

0.0640 0.0389 0.0749 0.064

0.0381

0.0745

0.0650 0.0446 0.0788

0.0633 0.0332 0.0715

250 300 350 400

0.0547 0.0380 0.0666 0.0552 0.0379 0.0670 0.0464 0.0377 0.0598 0.0376 0.0373

450 500

0.0539 0.0356 0.0371 0.0514

0.0557 0.0436 0.0707 0.0473 0.0431 0.0640 0.0386 0.0427 0.0576 0.0362 0.0415 0.0551

0.0541 0.0330 0.0634 0.0451 0.0329 0.0328

0.0492 0.0375 0.0530 0.0375 0.0375 0.0530 0.0348 0.0366 0.0505

0.0368 0.0328 0.0492 0.0342 0.0327 0.0475

600 750

0.0805 0.0403 0.090

0.0312 0.0359 0.0476 0.0284 0.0351 0.0453 0.0246 0.0344 0.0422

0.0322 0.0361 0.0484

0.0328 0.0404 0.0520

0.0294 DeskTop 0.0349 Standards0.0300 0.0394 Saudi Aramco 0.0456

0.0495

0.0304 0.0320

0.0203 0.0332 0.0389

0.0441 0.0276 0.0311 0.0416

86



* Solid Conductor

Source:

IEEE Gray

Book

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


87



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

Size

Approx.

cmil

Ampacity

Resistance (R)

AWG*

amps**

60 Hz

AC

DC 0 25 C

or

Reactance (X a)

0 50 C

ohms/mile

0 25 C

0 50 C

ohms/mile

1 ft spacing ohms/mile

4(1)

170

1.374

1.053

1.374

1.503

.609

4(3)

180

1.388

1.518

1.388

1.518

.599

2(1)

220

.864

.945

.864

.945

.581

2(3)

240

.873

.955

.873

.955

.571

2(7)

230

.881

.964

.882

.964

.574

1/0(7)

310

.555

.606

.555

.606

.546

2/0(7)

360

.440

.481

.440

.481

.532

3/0(7)

420

.349

.381

.350

.382

.518

4/0(7)

480

.276

.302

.278

.303

.503

*

number of strands as indicated by (x)

0 0 ** for conductor at 75 C, air at 25 C, wind 1.4 mi/hr

Figure 57. Copper Aerial Line Impedance Data


88



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

Resistance (R) Size

Approx.

cmil

Ampacity

amps**

60 Hz

AC

DC 0 25 C

or AWG*

Reactance (X a)

0 50 C

ohms/mile

0 25 C

0 50 C

ohms/mile

1 ft spacing ohms/mile

6

100

3.560

3.920

3.560

3.980

.673

4

140

2.240

2.470

2.240

2.570

.659

2

180

1.410

1.550

1.410

1.690

.665

1

200

1.120

1.230

1.120

1.380

.665

1/0

230

.885

.974

.888

1.120

.656

2/0

270

.702

.773

.706

.895

.641

3/0

300

.556

.612

.560

.723

.621

4/0

340

.441

.485

.445

.592

.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


89



WORK AID 2:

RESOURCES USED TO CALCULATE LINE VOLTAGE DROPS

Work Aid 2A: IEEE Standard 141-1986 (Red Book) For the content of IEEE Standard 141-1986, refer to Handout 1. 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 cable handbook. 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 =

3 VD (3 system)

VD = 2VD (1 system) Step 9. Calculate the load voltage (VL). VL = VS - VD Step 10. Calculate VD as a percentage (VD%). VD% = 100[(VS - VL)/VS]


90



Work Aid 2C: Applicable Calculation Procedures (Per-Unit Method) Step 1. Obtain the specified one-line diagram. 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 6. Draw and label the impedance diagram. 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.


91



WORK AID 3:

RESOURCES USED TO CALCULATE STARTING VOLTAGE DROPS

LARGE

MOTOR

Work Aid 3A: IEEE Standard 141-1986 (Red Book) 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 in the system where voltage is assumed to be constant. For example, the utility or source bus.

XS

= Total reactance of the circuit between the motor and the point in the system where voltage is assumed to be constant. For example, 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 IFLA VM

= Motor line-to-line voltage in volts


92



Work Aid 3C: Applicable Calculation Procedures Step 1. Obtain the specified one-line diagram. Step 2. Select kVAb, kVb. Step 3. Calculate Ib, Zb. Ib = (kVAb)/[( 3 )(kVb)]

Zb = (kVb)2/(MVAb) = Rb = Xb

Step 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 the new 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 7. Draw and label the impedance diagram. 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 - kVS kVD (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)


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WORK AID 4:

RESOURCES USED TO CALCULATE SYMMETRICAL RMS FAULT CURRENTS (BALANCED FAULT CONDITIONS)

Work Aid 4A: IEEE Standard 141-1986 (Red Book) For the content of IEEE Standard 141-1986, refer to Handout 1. 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 is typically 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 hand books, equipment nameplates, national or internal standards, and code books.


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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’s handbook. For purpose of this Module, use Figures 56 and 57 for cable values. (2) X = [(jX) /1000 ft](number of feet) (3) XC = X/Xb = j(X/Xb)


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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 and Figure 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


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Step 5. Draw and label the reactance diagram. •

This step consists of graphically modeling and labeling the series/parallel network 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 + X2th)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 are short 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


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GLOSSARY American National Standards Institute (ANSI)

An organization whose members approve various standards for use in American industries.

apparent power (S)

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

asymmetrical current (IASY) always

Envelopes of the peaks of the current waves are not symmetrical about the zero axis. Most short-circuit currents are nearly asymmetrical during the first few cycles after the fault occurs.

asymmetrical factor (AF)

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

base apparent power (kVAb)(MVAb)

An arbitrary kilovolt-ampere value, (e.g., 100000 kVA or 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’s rated voltage at that level.

full load amperage (IFLA)

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

horsepower (hp)

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

impedance (Z)

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

impedance triangle

A right-angle triangle developed geometrically from the equations associated with resistance, inductive reactance, and capacitive reactance.

induction motors

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


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Institute of Electrical and Electronics Engineers (IEEE)

A worldwide society of electrical and electronics engineers.

interrupting duty

The short-circuit rating which medium and high voltage breakers can safely interrupt. Sometimes called the contact parting duty. For interrupting duty calculations, machines are modeled using their 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 to calculate line voltage drops.

locked rotor amperage times (ILRA)

Current drawn by a motor during starting (usually four to six

low voltage (LV)

Voltage levels less than 1000 volts. Usually called utilization level voltages.

lump sum motor

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

medium voltage (MV)

Voltage levels greater than or equal to 1000 volts and less than 100000 volts, usually called distribution level voltages.

momentary duty

The short-circuit rating assigned to medium and high voltage circuit 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 using their subtransient reactance.

motor impedance

Impedance of a large motor being started. Ratio of motor full load amperage to motor locked rotor amperage.

National Electric years Code (NEC)

An electrical safety code developed and approved every three

overcurrent device

An electrical device inserted in a circuit to protect it protective against damage from an overload or short-circuit. This protection is achieved by automatic interruption.


full load amperage).

by the National Fire ProtectionAssociation (NFPA 70).

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percent voltage drop (VD%)

The ratio of voltage drop in a circuit to voltage delivered by the 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 between voltage and current.

power triangle

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

reactance (X)

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

reactive power (Q)

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

real power (P)

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

receiving end end voltage (VL)

Voltage at the terminal of the load. The difference of sending

resistance (R)

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

root mean square (rms)

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

sending end voltage (Vs)

Voltage at the source of power, used in line voltage drop calculations.

single-phase voltage (VD)

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

short-circuit current the (Isc)

Current (usually very large) flowing in an electrical system as

short-circuit current source (ISCC)

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


voltage and receiving end voltage is the line voltage drop.

result of a three-phase, phase-to phase, double-phase-to-ground, or single phase-to-ground fault.

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induction motors, and the utility (really a generator). Sometimes called short circuit current available (ISCA). short-circuit study

Calculations of short-circuit currents resulting from fault conditions. These calculations can be accomplished manually (hand-calculations), but are usually done with the aid of a personal computer.

symmetrical current (ISYM)

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

synchronous motor

A motor having a field excited by direct current and a stator winding in which alternating current flows.

system reactance (Xs)

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

system X/R ratio

The ratio of reactance to resistance in the system from the fault point looking back toward the source.

subtransient reactance (X”d)

The apparent reactance of the stator winding at the instant short-circuit occurs. X”d determines the current flow during the first few cycles.

synchronous reactance (Xd)

The apparent reactance of the stator winding under steady-state 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 determines the current flowing three to eight cycles after a fault occurs. voltage drop(line)(VD)

Voltage drop calculated at the terminals of a load under normal conditions. Usually expressed as a percent.

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


101

Calculating Voltage Drops and Symmetrical RMS Fault Currents

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