2 Tutorial Handouts 300

© Dr. P. K. Sen, PE, Fellow IEEE 1 [2012] IEEE IAS Distinguished Lecture Series Professor, Colorado School of Mines Pune, Hyderabad, Chennai, and Delhi: India Senior Consultant, NEI Electric Power Engineering, Inc. January, 2012

Workshop

Design of Industrial Power Distribution Systems:

Shortcut Methods, Quick Estimation and Application Guidelines

Pune, Hyderabad, Chennai, and Delhi: India

January, 2012

Prof. P.K. Sen, PhD, PE, Fellow IEEE Professor, Colorado School of Mines, Golden, CO 80401

Senior Consultant

NEI Electric Power Engineering, Inc. Denver, Colorado 80033

[email protected] 303.339.6750

P.O Box 1265 Arvada, CO 80001 Phone (303) 431-7895 Fax (303) 431-1836

www.neiengineering.com


Design of Industrial Power Distribution Systems: Shortcut Methods, Quick Estimation and Application Guidelines

Dr. P.K. Sen, PE, Fellow IEEE

Professor of Electrical Engineering Colorado School of Mines, Golden, CO 80401

This (multiple days) workshop has been designed for all practicing engineers (young or experienced), managers, operation and plant maintenance personnel, advanced students interested in “power and energy engineering” career and technical personnel interested in different aspects of Power Distribution Systems Design as applied to Electric Power and Energy industry. The main objective of the workshop is to introduce the “basic” tools required and utilized in designing industrial power distribution systems. The primary focus of this course is on the medium voltage (MV) and low voltage (LV) power systems with some references to the sub-transmission system. It is assumed that participants have some basic knowledge of fundamentals of electric power systems and electric machinery. Practical experience is preferable, but not required. Emphasis is given on hand calculations and estimations. Numerous real world design problems will be solved during the entire workshop. The workshop will be divided into “multiple” modules. Extensive handouts will be provided at the workshop. This introductory workshop is must for all power systems engineers, utility and no-utility alike, consulting firms, manufacturing and process plant, and designed to facilitate in educating advanced students in power and energy engineering profession.

(Tentative) Course Outline Day (Part) 1:

1) Logistics, Introduction, Background and Prerequisites, Expectations etc. 2) Scope of Electric Power Distribution Engineering and Characteristics of Power Distribution

Systems: Utility, Industrial and Commercial Users Perspective 3) Power System Fundamentals, Understanding “Load” and “Key” Design Tools:

• 3-Ph Power, Voltage-Current Calculations; • Active, Reactive Power, Apparent Power, Power Factor and Power Triangle; • Power Factor Correction and Shunt Capacitor Compensation; • Voltage Drop and Voltage Regulation • Load Characterization; • Understanding Electricity Bill; • Induction Motor Load, Torque-Speed Characteristics, Losses and Efficiency; • Selection of Plant Distribution Voltage; • Transformer Sizing; and • Motor Starting and Voltage Drop

4) Transformer Engineering, Basics and Procurement: • Equivalent Circuit and Design Fundamentals; • Performance Evaluation: Efficiency and Losses; % Impedance and Voltage Regulation; • Transformer Procurement, Specification Writing and Loss Evaluation; Testing; • Overloading, Life Assessment and Asset Management

5) Design of Industrial Power Distribution Systems and Problems: • Simplified Design Calculations, Transformer Sizing, Selection of Voltage; • Motor Starting; • One-line Diagram; Quick Cost Estimate


Day (Part) 2: (When Applicable) 1) Recap of Day 1, Questions and Answers 2) Induction Motor Performance and Procurement

• Design Fundamentals, Equivalent Circuit and Performance Evaluation; • Torque-Speed Characteristics; • Motor Starting and Voltage Drop; • Variable Frequency Drive; • Testing, Specification and Applications Guidelines.

3) 3-Phase Fault (Short-Circuit) Calculations • Per-Unit Methods of Calculations; • Sub-transient Reactance; • Source Reactance; • Shortcut Methods of Calculations for Industrial Power Systems; • Fault Current Distributions.

4) Design of an Industrial Power Distribution System and Problems • Selection of Breakers and Switchgears; • Motor Control Center - Specification and Evaluation; • System Grounding; • Reliability, Safety and Design; • Quick Cost Estimate.

5) Protection Design Philosophy 6) Emergency Power and Uninterruptible Power Supply (UPS) 7) Design Problems: Simplified Calculations, Guidelines and Techniques

Day (Part) 3: (When Applicable) 1) Recap of Days 1 and 2, Questions and Answers 2) Power Systems Protection:

• Symmetrical Components and Unsymmetrical Faults; • Instrument Transformers; • Grounding of Power Systems and Ground Fault Protection; • Utility – Industry Interface; • Design of Protection Scheme; • Power Systems Protection:

o Transformer o Induction Motor o Distribution Feeder

3) Step-by-Step Procedure in Protection Coordination and Design 4) Case Studies and Design Problems


Prof. Pankaj K. (PK) Sen, PhD, PE, Fellow IEEE Colorado School of Mines Golden, Colorado 80401

Dr. P.K. Sen, PE, Fellow IEEE has over 45 years of combined teaching, administrative, research, and consulting engineering experience. Prior to joining Colorado School of Mines, Golden, Colorado in 2000, Dr. Sen taught for 21 years at the University of Colorado, Colorado. His industrial experience

includes power plants and substation engineering design, system & feasibility studies, protection and relaying, training technical personnel at all level and solving various aspects of power systems engineering application problems. He has published over 140 technical papers on a variety of subjects related to Power Systems, Protection / and Relaying, Electric Machines, Renewable Energy and Energy Policy, Power Quality, Engineering Education and Arc Flash and Safety. Dr. Sen has supervised and mentored over 150 graduate students (including non-traditional students, and practicing engineers from the Utility Industries, Rural Electric Company’s, Consulting Engineers, and others). He is an IEEE Fellow and a Registered Professional Engineer (Electrical) in the State of Colorado. Currently Dr. Sen is a Professor of Electrical Engineering and the Site Director for the (Originally NSF funded) Industry University Cooperative Research Center (IUCRC) Power Systems Engineering Research Center (www.pserc.org) at Colorado School of Mines, Golden, Colorado. His current research interests

include application problems (safety, protection, equipment life, energy economics, asset management and policy issues, etc.) in power systems engineering, renewable energy applications and distributed generation, and engineering education. Dr. Sen is a very active member of a number of Professional Societies including IEEE PES & IAS, Rocky Mountain Electrical League (RMEL) and has been instrumental in providing seminars, short courses, conduct workshops, and provide training for technical personnel in the Rocky Mountain Region and nationwide (USA) and internationally for the past 34 years. Dr. Sen is known in the industry, locally, nationally and internationally for providing educational opportunities for practicing engineers at all level, and for both undergraduate and graduate students. He is an inspiring and prolific teacher with passion. He has authored numerous prize winning papers at the IEEE Conferences and IAS Magazine.



Presentation Outline

Part 1 Introduction, and Scope of Electric Power Distribution Systems

Engineering

Characteristics of Power Distribution Systems: Utility and

Industrial/Commercial Users Perspective Power System Fundamentals & Design Tools:

o (Review) 1-Phase and 3-Phase Power o (Review) Power, Reactive Power, Power Factor o Power Triangle o Losses and Efficiency

Selection of Voltage Power Factor Correction Percentage Impedance, Voltage Regulation and % Voltage Drop Understanding Electricity Bill Transformer

o Procurement and Specification Writing o Losses and Efficiency o Bid Evaluation o Application Guidelines o Protection Basics

Quick Cost Estimate Design Problems: Transformer Sizing, Power Factor Correction,

Voltage Drop and Voltage Regulation



Presentation Outline Part 2

Induction Motor: Characteristics, Performance

Evaluation, Specification Quick and Simplified 3-Phase Short Circuit (Hand)

Calculations for Radial System o Volt-Ampere Method o Per-Unit Method

Power System Grounding Application Guidelines -

o Motor Starting and Voltage Drop o Sizing Transformers o Capacitor Selection o Simplified Transformer Protection Considerations

Conclusions, Questions and Answers


Design of Industrial Power Distribution Systems:

Shortcut Methods, Quick Estimation and Application Guidelines

Presentation Outline Part 3

1) Recap of Day 1 and 2, Questions and Answers 2) Power Systems Protection

• Symmetrical Components and Unsymmetrical Faults; • Instrument Transformers; • Grounding of Power Systems and Ground Fault Protection; • Utility – Industry Interface; • Design of Protection Scheme; • Power Systems Protection:

o Transformer o Induction Motor o Distribution Feeder

3) Step-by-Step Procedure in Protection Coordination and Design;

4) Case Studies


Fault Contribution Rotating Machines

Synchronous Machine Direct-Axis Quantities: Sub-Transient (xd”, T d”): ~ 3 Cycles** Transient (xd’, T d’): ~ 0.6-1.0 Sec Steady-State (xd) ** Normally Used in Fault Calculations Typical Values 0.1 - 0.2 per-unit (Machine Base)

Induction Motor Fault Contributions last usually 2-3

Cycles Sub-transient Reactance varies typically

between 0.17 – 0.25 per-unit (Machine Base)

If the starting current is 6.0 times the full-load current, then the sub-transient reactance is 1/6 = 0.167 pu.


Important!! Assume “Zero Source” Reactance or “Infinite

Bus” always yields “Conservative Results” in

Fault Calculations.

3-Phase Fault Calculations (Design) Problem No. 3

Calculate the 3-Phase Fault Current at “X”

Total Fault Current @ “x” = 11,954 + 4,184 + 1,046 + 2,092 A = 19,276 A (= √√√√3 •••• 13.8 •••• 19,276 = 460.7 MVA) Assume: Finite Bus with 3-Ph Fault Level = 1,500 MVA @ 115 kV Bus Source Reactance (Xs) = 20 / 1,500 = 0.0133 pu Fault Contribution by the Source = 1.0 / (0.0133 + 0.07) pu = 12.00 pu = 12.0 x 836.8 A = 10,045 A Total Fault Current @ “X” = 10,045 + 4,184 + 1,046 + 2,092 A = 17,367 A (Compared to 19,276 A)

A418.4 13.8 3

10,000 Ib ====

••••====

1/0.1 = 10 pu = 10 x 418.4 A = 4,184 A

Assume: 1 HP ≈≈≈≈ 1 kVA

A209.2 13.8 3

5,000 Ib ====

••••====

1/0.1 = 10 pu = 10 x 209.2 A = 2,092 A

A836.8 13.8 3

20,000 Ib ====

••••====

1/0.07 = 14.3 pu = 14.3 x 836.8 A = 11,954 A

Assume: Infinite Bus Source Reactance (Xs) = 0

⇒⇒⇒⇒ 5,000 kVA

⇒⇒⇒⇒ 5,000 kVA

1/0.2 = 5 pu = 5 x 209.2 A = 1,046 A

0 A

A209.2 13.8 3

5,000 Ib ====

••••====

1/0.07 = 14.3 pu = 14.3 x 100.4 A = 1,436 A = 11,954x13.8/115 A


Important!! Assume “Zero Source” Reactance or “Infinite

Bus” always yields “Conservative Results” in Fault Calculations.

3-Phase Fault Calculations (Design) Problem No. 4

Calculate the 3-Phase Fault Current at “Y”. Total Fault Current @ “Y” = 1/0.06 = 16.7 pu

= 16.7 x 1,202 A = 20,047 A (= √√√√3 •••• 0.480 •••• 20.047 = 16.67 MVA) Assuming Fault at “X” = 460.7 MVA (or 11,954 A) Source Reactance (Xs) = 1.0 MVA (Transf. Rating) / 460.7 MVA = 0.0022 pu Fault Current at “Y” = 1.0 / (0.0022 + 0.06) = 16.085 pu = 16.085 x 1,202 A = 19,347 A ( = 16.08 MVA)


A 1,202 0.480 3

1,000 Ib ====

••••====

1/0.06 = 16.67 pu = 16.67x1,202 A = 20,047 A

1/0.06 = 16.67 pu = 16.67x41.9 A = 698 A

499 A 93 A

71 A 35 A


3-Phase Fault Calculations (Design) Practice Problem No. 7

For the given problem, calculate the fault currents at various locations (F1 to F4) using different assumptions, like using an infinite bus or neglect any other impedance, etc. Please justify your assumptions.

Fault Calcula-tions

Assumptions!!

Fault Current Values (A)

F1

(a) 17,811 (b) 17,412 (c) 16,288 (d) 16,602

F2

(e) 31,379

(f) 38,596

(g) ??

F3

(h) 5,395

(i) ??

(j) 5,910

F4 (k) 26,162 (l) 28,375


Important!! Assume “Zero

Source” Reactance or “Infinite Bus” always yields “Optimistic

Results” in Voltage Drop Calculations. Source Reactance (for minimum fault current or

highest value of Xs) must be considered, in doing the real

calculations.

Induction Motor Starting Voltage Drop Calculation

(Design) Problem No. 5

Assume: Full-load Efficiency = 0.92 and Full-load Power Factor = 0.93 (lag) Approximate Full-load Current = 363 A Assume Full Voltage (Direct-on-Line) Starting Current (I st) = 6.0 x IFL = 2,178 A @ 0.0 lagging power factor (conservative assumption) % Voltage Drop = [ % r Cos θθθθ ±±±± % x Sin θθθθ ] •••• Ipu (Loading) (+) Lagging Power Factor



(1) Assume: 1 HP ≈ 1 kVA, “Constant Current” Model and 0.0 lag power factor, % Voltage (Momentary) Drop = 28.8%

(2) Using Actual Efficiency and Power Factor in

calculating the full-load current, “Constant Current” Model and 0.0 lag power factor,

% Voltage Drop = 25.12% (3) Same as (2), but assuming a power factor (more

realistic) 0.25 lag, % Voltage Drop = 24.4%

(4) Assume, 1 HP = 1 kVA and a “Constant

Impedance” Model and 0.0 lag power factor, % Voltage (Momentary) Drop = 22.4%

All of the above simplified quick calculations (with various assumptions) produce most likely an “unacceptable” voltage drop (more than 20%) at the motor terminals. Since the motor torque is proportional to the squared of the voltage ( Tm ∞ V2 ), the starting torque will be drastically reduced (e.g., for case (4) (V = 1.0 – 0.224 = 0.776), the torque will be 0.7762 = 0.60 or 60% of the full-voltage starting torque).

So we have too much Voltage Drop!!


(1) Increase the transformer size – Doubling the transformer size to 10 MVA will reduce the voltage drop (approximately) by a factor of 2 (to 12-13%). (2) Buy an induction motor (Code Letter “D”) with starting current less than 6.0 times (say 4.0 – 4.5). This will reduce the % voltage drop proportionately (to about 14-16%). (3) Use “reduced voltage” starter (Autotransformer) or “soft start”. This will reduce the % voltage drop to a real value (much less than 10%, depending on the design). However, care must be taken to ensure that adequate motor torque is produced during starting. (4) Reduction of the % reactance of transformer may be utilized to enhance the problem. However, this will increase the fault current values, and may increase the cost of switchgear, and other associated equipment. . (5) Redesign the “process” requirements, when appropriate and possible, to reduce the motor output. (6) Using higher plant distribution voltage. Ultimately, application requirements and cost will probably

dictate the solution.


Fundamentals of Power Distribution Systems Design: Review, Simplified and Shortcut Calculations and Guidelines

Symmetrical Components Fundamentals

Any “Unbalanced” set of “3-Phase” Vectors (or Phasors) having a “Phase Sequence (or “Rotation”),” say “abc” can be replaced by (or resolved into) “three” Component set of Vectors →

(3) A set of Balanced 3-Phase Vectors (120o out-of-phase) having the same “phase sequence (or rotation)” of the original set of vectors (“abc”), called “Positive Sequence.”

(4) A set of Balanced 3-Phase Vectors (120o out-of-phase) having the same

“opposite (or negative) phase sequence (or rotation)” of the original set of vectors (“acb”), called “Negative Sequence,” and

(5) A set of 3, 1-phase Vectors (same phase and magnitude), called “Zero

Sequence.”

a

b

c

a1

b1

c1

a2

b2

c2

a0

b0

c0

a2

b2

c2

Positive Negative Zero

“Phase” Quantities “Sequence” Quantities


Mathematically, say, for a set of three “phase” (phasor) currents (I a, Ib and Ic) and the three “sequence” components of (phasor) currents (I (a,b,c)1, I(a,b,c)2 and I(a,b,c)0): I a = Ia1 + Ia2 + Ia0 Ib = Ib1 + Ib2 + Ib0 Ic = Ic1 + Ic2 + Ic0 Conversely, it can be shown by simple mathematical manipulation that the three sequence components of currents (Ia1, Ia2, and Ia0) for “phase-a” can be calculated: I a1 = 1/3 [ Ia + a Ib + a2 Ic ] Ia2 = 1/3 [ Ia + a2 Ib + a Ic ] Ia0 = 1/3 [ Ia + Ib + Ic ] where, “a” is a “unit vector” defined by, a = 1.0 ∠∠∠∠120o. Simple Numerical Examples (Graphical and Analytical Solutions): (1) Ia = 3.0 ∠0o, Ib = 0 and Ic = 0. [Incidentally, this happens when we have a

single-line-to-ground (an unsymmetrical) fault in power systems.] (2) Ia = 0, Ib = 1.732 ∠90o and Ic = -1.732 ∠90o. [Incidentally, this happens

when we have a line-to-line (another unsymmetrical) fault in power systems.]

(3) Find (both graphically and analytically) all the (nine) sequence components

of currents [positive: (Ia1 , Ib1 , Ic1), negative: (Ia2 , Ib2 , Ic2) and zero: (Ia0, Ib0, Ic0) ] from the following set of three currents in a power system.

Ia = 100 ∠90o (A), Ib = 200 ∠0o (A) and Ic = 300 ∠- 90o (A) (4) Calculate (both analytically and graphically) the “phase” currents (Ia, Ib, Ic) from the following sequence components of currents. Ia1 = 10 ∠0o (A), Ib2 = 20 ∠120o (A), and Ic0 = 10 ∠- 90o (A)


Understanding the Physical Significance of (and other Basic Facts about) the Sequence (components) Currents: (1) Vector summation of the “positive” sequence currents = 0. (Three vectors, equal in magnitude and 120o out-of-phase, and phase sequence, “abc”). (2) Statement (1) is equally true for the “negative” sequence of currents. (except the phase sequence is, “acb”). (3) Vector summation of the three “zero” sequence currents (I0, same in magnitude and phase) = 3I0 (4) Any 3-phase power system under normal operating conditions is considered

to be the “positive” sequence network conditions, even though we do not explicitly say so.

(5) Negative sequence network (or power system) conditions could be dealt with in the same fashion as the positive sequence network, except for a major difference: the voltage sources have a phase sequence opposite to the positive sequence.

(6) Synchronous generators (or alternators) connected to the normal 3-phase power network produces only positive sequence voltage (in an ideal condition).

(7) Static devices connected in a 3-phase power network (like transformers, transmission and distribution lines, capacitors, etc.) do not see any difference between positive and a negative sequence quantities. However, they behave completely differently for zero sequence quantities. (8) For a balanced 3-phase conditions (like positive and negative sequence conditions), neutral connections for Y-connections (transformer neutral, loads, generator neutrals, etc.) is immaterial. The system doesn’t see the difference between grounded (any form) and ungrounded system, since vector summation of three vectors 120o is zero. In case of currents, no current flow through the neutral conductor or ground (Kirchoff’s Current Law). (8) However, in case of zero sequence current, as an example, the current flowing through the neutral (and/or ground) equals the 3 times the zero sequence current (Kirchoff’s Current Law). This can be seen from the fundamental definition of zero sequence quantities. (9) The sequence quantities are always line-to-neutral or line-to-ground (or phase quantities).


Sequence Networks (Thevenin Equivalent): Va1 = Vf – Ia1 Z1 where, Z1 is the positive sequence impedance Va2 = – Ia2 Z2 Z2 is the negative sequence impedance Va0 = – Ia0 Z0 Z0 is the zero sequence impedance Note: (1) Only positive sequence network has any voltage source for fault

calculations. (2) For all practical conditions in power system, we assume (will be discussed

briefly in the class), Z1 = Z2 (3) Zero Sequence Impedance (Z0) is usually totally different and depends on the grounding (or neutral) connections. (4) Calculations of Z1 (= Z2) is straight forward and is the same value used for

3-phase fault calculations. (5) The sequence network used are always for “Phase-a” values and deals with

phase quantities.


Symmetrical Components Fundamentals

Sequence Connections for Typical Two-Winding Transformer Banks


Reduced Sequence Networks where Z1, Z2 and Z0 are the Equivalent Impedances of the Network to the Fault Point


System Grounding and Ground Fault Protection

39

Delta - Wye “Solidly Grounded” Transformer


40

Delta – Wye“Solidly Grounded”

P TZ S

Zero Sequence Bus

OSI



44

Wye (Grounded) –Wye (Grounded)

Zero Sequence Bus

P S

I OS

ZT


46

3-Winding TransformerWye (Grounded) - Delta - Wye (Grounded)

Zero Sequence Bus

P

T

SI OSOPI

OTIZT

ZP ZS

N


Reduced Sequence Networks where Z1 , Z2 and Z0 are the Equivalent Impedances of the Network to the Fault Point


37

Network Connections for Unsymmetrical Faults

Single-Line-to-Ground Fault

Positive

Negative

Zero

Line-to-Line-Fault

Line-to-Line-to-Ground Fault


Brain Tinker No.1

(1) A small industrial plant (receiving power at 12.47kV) has the following loads: (a) 3 x 100 HP Induction Motors (b) 2 x 50 HP Induction Motors, and (c) 300 kW of lighting, heating and other small plant loads Estimate the total plant load, typical running power factor, and size (specify) a transformer. Discuss the protection philosophy for such a plant. (2) A 3-phase transformer is rated at 5/7.5 MVA, 13.8 kV (Delta) – 4.16 kV (Grounded-Wye). Calculate (estimate) the full-load phase and line currents on both high-side and low-side of the transformer at maximum loading. What will be a typical % reactance and X/R ratio values? Also calculate the maximum available fault current on the low-side of the transformer. Discuss the protection philosophy for this transformer. (3) What is the meaning of the following ANSI/IEEE device nos.? a) Device No. 27 ___________________________

b) Device No. 51 ___________________________ c) Device No. 81 ___________________________

d) Device No. 87 ___________________________

e) Device No. 38 ___________________________


(4) Put a check (x) in front of each factor affecting transformer protection: ________ Magnetizing Inrush Current ________ Load Tap Changer ________ No-Load Tap Changer ________ Transformer Voltage Level ________ Current Transformer (CT) Ratios ________ Transformer Winding Connections (5) Draw a (typical) simple motor thermal load capability. How would you protect a 200HP induction motor? Discuss the key information you must identify in protecting the motor.


Brain Tinker No. 2 1) Draw a typical “symmetrical short-circuit current ” supplied by a

synchronous generator. Identify and briefly discuss the key points in the drawing. Also show the rms value of the current.

2) What causes the “dc offset” in a short-circuit current fed by a synchronous

generator? 3) What is the worst possible (theoretical maximum) dc offset one can expect in a

synchronous machine? 4) What is the difference between the “interrupting (short-circuit capability)”

current and “closing and latching” current in a circuit breaker?

5) Draw a typical symmetrical short circuit current supplied by an induction motor. Discuss briefly why this is distinctly different from the synchronous machine? What are typical sub-transient reactance values?

6) Identify some of the key parameters one should specify and check while procuring a medium voltage power circuit breaker.

7) How do you define the “total asymmetrical fault” current? Draw and explain. 8) What are the typical sub-transient reactance values?

a. Large Alternators – Steam or Gas Turbine (2-pole or 4-pole) _________ b. Large Alternators – Hydro (Salient Pole, slow speed) _________ c. Large (MV) Induction Motors (say, 1000 HP and above) _________ d. Smaller (LV) Induction Motors (say, 50-200 HP) _________ e. Smaller (LV) Induction Motor (say, below, 50 HP) _________


9) Answer the following questions: (a) What are the typical reactance and X/R values?

o 12.47 kV Overhead Distribution Lines ____________ o Small Distribution Transformers ____________

(Say, 500 kVA or less, 4.16 kV – 480 V) o Large Power Transformers ____________

(Say, 30/40/50 MVA, 230 kV-12.47 kV) o 115 kV Overhead Transmission Line ____________ o 500 kV Overhead Transmission Line ____________

(b) What are the typical values (@ Rated Load Condition)?

o Efficiency of a 500 MVA Generator __________ o Efficiency of a 100 MVA Transformer __________ o Efficiency of a 5,000 HP Induction Motor __________ o Efficiency of a 5 HP Motor __________ o Operating Range of Power Factor for a Large Generator __________ o Voltage of a 3,000 HP Induction Motor __________ o HP Limit of Induction Motor for 480 V System __________

10) In a large distribution power system, 3-phase fault current reported by the

utility at the incoming 69 kV point is 24.5 kA rms symmetrical. Calculate the source impedance in Ω. Also calculate the source impedance in per-unit, assume a base of 10 MVA.

11) Estimate the full-load current of a 50 HP induction motor. 12) What limits the output of a power transformer? Explain briefly. 13) What are instrument transformers? Why is it used in large scale power

systems? 14) Calculate the line and phase currents (both magnitude and phase angle) in a 2

MVA, 12.47 kV (Delta) – 480 V (Wye) transformer. Assume the load is purely resistive and the transformer is fully loaded. Use the load voltage as the reference.


More Brain Tinker

Identify “True (T),” “False (F)” or “Unclear, Unkno wn or Subjected to

Conditions and Add Brief Comments, as Appropriate.” 1) Protective relaying is utilized to protect major power equipment against the

maximum fault current (short-circuit) only. 2) Per-unit values of currents on both sides of a two-winding transformer are

same. 3) Protective relays always use both the current and voltage signals for its

operation. 4) Protective relaying costs about 30% of the major power equipment it is

protecting. 5) In order to do proper relay settings calculations in a large scale (utility

oriented) power system, it is essential that you should have both 3-phase and single-line-to-ground faults (bolted) information available at appropriate locations including the current distribution.

6) Some form of polarizing (current and/or voltage) is needed for all directional

relays. 7) In a three-phase power system (under normal operation), at any point in the

system, both positive- and zero-sequence voltages are zero.


8) To measure a 10,000A current in a normal power system application, we

need to use a CT. However, if the current is 100 A only, there is no need for a CT.

9) To design a protective relaying scheme (during the conceptual design stage),

the three-line diagram and the AC, DC schematics, are absolutely essential. 10) In modern microprocessor based relaying scheme, auxiliary relays are not

used extensively to perform a variety of control functions and logic. 11) Modern day microprocessor relays perform more than relaying like event

recording, fault location detection, other control and monitoring functions. 12) In microprocessor based relays, for each function (defined by ANSI

Numbers), you need a separate relay. 13) Knowledge of polarity and phase-sequence is essential for proper relay

applications (selection). 14) In an ungrounded “wye” connected system, if one phase goes to ground, the

voltages of the other two phases goes up to the line voltages (with respect to ground) of the system.


15) In an ungrounded wye-connected power system, the neutral is always at

ground potential. 16) In a delta-wye connected two-winding transformer, there is a phase-shift of

30o (for positive sequence) between the primary and secondary line quantities (voltage and currents), and the standard ANSI connection recommends the low-voltage quantities to “lead” the high-voltage quantities.

17) In a 3-phase power system, zero-sequence currents for all three phases are

same. 18) In a three-phase power system with a delta-wye (grounded) transformer,

zero-sequence current flows (confined) within the delta windings (in the loop).

19) For all faults involving ground, there will always be (in general) some zero-

sequence current in line. 20) Transformers of all sizes are always protected by differential relay.

2 Tutorial Handouts 300

Documents

Transcript of 2 Tutorial Handouts 300