Advanced Power SystemsAdvanced Power Systems

ECE 0909.402-02, 0909.504-02ECE 0909.402-02, 0909.504-02 Lecture 3: Electric Power FundamentalsLecture 3: Electric Power Fundamentals

5 February 2007

Dr. Peter Mark Jansson PP PEAssociate Professor – Electrical and Computer Engineering

Aims of Today’s Lecture

• Course Training Tours (Mon 3-8PM, Fri 12-5PM)

• Part One: Overview of Chapter 2 concepts• A summary of…ch. 2 concepts

• Power Factor Correction• Three Phase Systems• Power Supplies and Power Quality

• Part Two: Overview …ch. 3 concepts• Early developments• Electric industry today (NUGS, IPPs, QFs)• Polyphase synchronous generators

Aims of Today’s Lecture (con’t)

• 15 minute stretch break at 6

• Part Two: An intro to …ch. 3 concepts• Heat engines, steam cycles and efficiencies

• GTs, CCs, Baseload Plants and LDCs

• T&D

• Regulatory impacts (PUHCA, PURPA, FERC)

Power factor correction?

Why correct power factor?

1/5th of all grid losses may be due to poor power factor (>$2B/yr), the outage of 2003 was made more severe by extremely high reactive demand, all transformers are rated on kVA not watts, all these economic, efficiency and reliability benefits can be achieved at a very low cost …

Power factor correction

Adding capacitive impedance in parallel with the load enables the current to oscillate between the inductors and capacitors rather than being drawn from the utility system or the customer transformer.

Capacitors are rated by volt-amps-reactive VARs that they supply at the system voltage in which they are installed, and PF correction is a straightforward engineering design

Last Week’s pf correction example

• An industrial customer’s service entrance substation is rated at 1MVA (1,000kVA) and is at 95% capacity. The plant now experiences a power factor of 80%. A new manufacturing line is planned that will increase power demand 125kW. How many kVAR of capacitance should be added to avoid purchasing additional substation transformer capacity?• Real power (at present) = (0.8)(0.95)(1,000kVA) = 760kW• Phase angle = cos-1(0.8) = 36.87o

• Apparent power = (1,000)(0.95) = 9,500 Volt-Amps• If demand grows from 760kW to 885kW Apparent Power will

grow to Real/PF = 885/(0.8) = 1106kVA > 1MVA capacity• Reactive Power = Q = VI sin = 1106(0.6) = 664kVAR

PF correction example (con’t)• An industrial customer’s service entrance substation is rated at 1MVA (1,000kVA) and

is at 95% capacity. The plant now experiences a power factor of 80%. A new manufacturing line is planned that will increase power demand 125kW. How many kVAR of capacitance should be added to avoid purchasing additional substation transformer capacity?

• For substation to handle the growth, power factor must improve to at least PF = 885kW/1,000kVA = 0.885

• Phase angle now will be = cos-1(0.885) = 27.75o

• Reactive Power (Q) = VI sin = 1000(0.4656) = 466 kVAR

• Difference in reactive power must be supplied by the capacitor bank: 664 – 466 = 198 kVAR

• Specify a >= 200 kVAR cap bank at industrial customer’s service voltage

What is a 200 kVAR cap bank?

• Rating Capacitor Banks in terms of KVAR is quite common in power systems, however, there are times when the actual value of the capacitance is needed:• Voltage/Current in Capacitor: V=(1/C)I• Current & Power in Capacitor is all reactive:

VAR = VI = V(CV) = CV2

• C (farads) = VARs/(V2)

What is a 200 kVAR cap bank?

• C (farads) = VARs/(V2)• We need therefore to know what voltage the

capacitance will be provided at, in our case 480V

mFFVARs

kVARs

V

VARsC farads

3.2103.21086.86

102

)480(602

200

36

5

22

LM #1

• In an industrial facility (1, 240V) with a 10-kVA transformer, the real power of a motor is 4.2kW (pf=0.6). A second motor, similar to the above needs to be added.• Show transformer loads • Determine kVAR and pf required to meet

request • How much capacitance is this (farads)?

3-phase systems

• Typically, 3-phase systems are connected in one of four ways: (Supply – Load)• Wye – Wye• Wye – Delta• Delta – Wye• Delta – Delta

• What are these?

Wye (Y) connections

What is meant by Phase & Line Voltages?

• In a 3-phase Wye-connected system which is quite common (all the negative legs of generators or transformers are tied to ground) the following is true:• Voltages measured with respect to the neutral

wire are called phase voltages V (Va, Vb, Vc)

• Voltages measured between phases are called line voltages VL or VLL (Vab, Vbc, Vac )

• In power systems we typically use Line Voltage

Power in 3 Systems

• In 3-phase (3) systems to determine power - including Apparent (VA), Reactive (VAR), and Real (Watts) - we need to understand these relationships:

)_(cos3

)_(sin3

33

33

3

3

3

wattsIVP

VARIVQ

IVIV

IVS

lineline

lineline

linelinelineline

linephase

Power in 3 Systems

• Most widely used service to buildings is a 4-wire , 3, 208 V service.

• To determine line voltage of each phase alone we need to understand this relationship: VLine = 3 Vphase=

• Therefore: Vphase= = VLine/ 3 = 208V/ 3 = 120V

Power in 3 Systems

• Many buildings also use a 4-wire , 3, 480V service.

• To determine line voltage of each phase alone we use the same relationship:

• VLine = 3 Vphase=

• Therefore: Vphase= = VLine/ 3

• = 480V/ 3 = 277VSee Figure 2.15 on page 74 of our Renewable and Efficient EPS text

LM #2: 3 Power Distribution

• The Atlantic City Electric Power System distributes electricity from its substations to its customers at a 3-phase line voltage nominally 12-kV.• What is its Phase Voltage?

Delta ()

Y and current, voltage & power

• Y-connected• IL = I

• VLL = 3 V Apparent Power3 = 3 VLL IL

• Real Power3 = 3 VLL ILcos

-connected• VLL = V

• IL = 3 I Apparent Power3 = 3 VLL IL

• Real Power3 = 3 VLL ILcos

Example

• In an industrial facility (3, 208V-Y) the real power used is 80kW with single phase motors and a poor power factor is the result (0.5) leading to 5% power losses (4kW). With capacitors and 3 motors the power factor is increased to 0.9 what losses are there now?

Solution

AkAV

kVA

V

SI

kVASpfAfter

AkAV

kVA

V

SI

kVASkWSS

SIVkWP

L

L

L

L

LL

247247.02083

9.88

3

9.889.0

809.0_

444444.02083

160

3

1605.0

80805.0cos

coscos380

3

3

3

333

33

Final Solution

• Current losses are 4kW (5%)• Losses (I2R), R is constant• Losses after correction

%55.180

24.1%

24.1)444(

)247(4

2

2

Loss

kWkWPL

LM #3

• In an industrial facility (3, 208V-Y) the real power used is 250kW and a poor power factor exists (pf = 0.6).

• What is the phase current? I

• What is the line current? IL

• What phase voltage? V

• How much current is saved if pf is unity?

Power supplies

• Devices that convert ac power to dc power for electronic applications

• What devices use power supplies: tvs, pcs, copiers, portable phones, motor controls, thermostats – just about everything with an IC or digital display or any electronic control

• 6% of US electricity flows through PS

Linear and switching

• Linear PS: use transformers to drop voltage and then rectifier and filter, 50-60% efficient in conversion

• Switching PS: use rapid transistorized switches to effectively reduce available power and use a dc-dc converter to adjust dc output voltage to desired levels, 70-80% efficient in conversion

• Typical US household has ~20 devices using PS which consume between 5-8% of electricity and account for over $4B each year (500 kWh/home)

Power supplies

• Described in text (pp.77-86) and introduced in this course because they can have a significant impact on power quality as well as very sensitive to poor power quality

• Also circuits similar to the buck converter can both raise and lower dc voltages to enhance performance of photovoltaic arrays we will design later in the course

Power quality & THD

• Current and voltage waveform irregularities• Under-,over-voltage and current• Sag, swell of V and I• Surges, spikes and impulses• Electrical noise• Harmonic distortion• Outages

Power quality and interruptions

• Few cycles or momentary sags can cause major disruption for automated manufacturing equipment:• PLCs• ASDs

• Digital economy businesses can be even more disrupted by poor power quality

Sources of eroding quality

• Utility (line side):• Under-,over-voltage and current• Sag, swell of V and I• Surges, spikes and impulses• Outages

• Customer (load side):• Waveform noise (poor grounding)• Harmonic distortion

Solutions to eroding quality

• Utility (line side):• filters

• capacitors and inductors

• high energy surge arrestors

• fault current limiters

• dynamic voltage restorers

• Customer (load side):• UPS, voltage regulators, surge suppressors, filters and

various other line conditioning equipment

Overview of harmonics

• Any periodic function can be represented by a Fourier series made up of an infinite sum of sines and cosines with frequencies that are multiples of the fundamental (60Hz) frequency

• Frequencies that are multiples of the fundamental are called harmonics (ie, 5th harmonic is 300 Hz)

• Harmonic distortion does not occur from loads using the basic components of R, L & C

• Electronic loads (power supplies, electronic ballasts, adjustable speed drives, etc.) distort

• Periodic functions that are sine or cosine only have no even harmonics (called halfwave symmetry – half and full wave rectifiers will exhibit only odd harmonics)

CFLs and THD

• 60-watt Incandescent Light• V and I in phase• No THD

• 18-watt Compact Fluorescent Light• Phase shift (inductive load)• Significant wave notching• Numerous harmonics

Total harmonic distortion

1

24

23

22

I

IIITHD

Sample problem LM#4

• Calculate the THD for the following CFL:• Harmonic rms Current (A)

• 1 0.18

• 3 0.14

• 5 0.09

• 7 0.05

• 9 0.03

New homework

• HW 3 – due next Monday

• will be posted on web

• 2.9, 2.12, 2.13, 3.1, 3.2, 3.3, 3.4, 3.6, 3.7, 3.9, 3.10, 3.11

Chapter 3 concepts and content

• Part Two: An intro to …ch. 3 concepts• Early developments• T&D• Polyphase synchronous generators • Heat engines, steam cycles and efficiencies• GTs, CCs, Baseload Plants and LDCs• Electric industry today (NUGS, IPPs, QFs)• Regulatory impacts (PUHCA, PURPA, FERC)

Electricity History

• Early Development (1796 – 1838)• First Chemical Battery – Volta 1796

• Electric Arc Lamp – Davy / Moyes 1801

• Early Transformer – Schweigger 1811

• First Electromagnet – Sturgeon 1825

• First E-M Induction Generator – DC – Faraday 1831

• Thin Rod Carbon Lamp Filament – Jobart 1838

History – 2

• The First 100 Years (1840 - 1940)• Six-pole reliable commercial generator – 1840• Subdivided iron core transformer – 1850• First self-excited dynamo patented – 1855• Iron wire replaced by copper for distribution –

1877• First DC electric power system Pearl Street – 1879• First AC generation/transf./transm. Design – 1888• Rotary Converters – 1893

History – 3

• The First 100 Years (1840 - 1940) continued• 2-3 phase transformers – 1894• OCBs invented – 1897• First load dispatcher’s office (NYC) – 1903• First AC transmission 60kV – 200 miles – 1905• High voltage suspension insulator – 1906• First automatic electrical substation – 1914• First flourescent / neon tube lighting 1934

History – 4

• U.S. Shift from DC to AC• 1890: 90% DC• 1897: 62% DC• 1902: 39% DC• 1907: 28% DC• 1942: 1% DC

Transmission Lines

• Purpose: to connect power generation to loads

• Design Types: • Bulk transmission: large quantities of power to

distribution substations (Hi voltage - >= 34.5 kV)• Radial transmission: often called distribution

lines to deliver power to customer loads (Lo voltage <= 23kV)

• Rule of Thumb: Power delivered V2

Fundamentals

• Transmission / Distribution Lines• ROLE: Minimize i2R losses in system

• Transmission: 69, 138, 230, 500, 765 kV

• Distribution: 4 - 34.5 kV (12-13.8 kV common)

Types of Transmission Construction

• Overhead: most common in rural and semi-rural utility systems, least expensive, easiest to maintain, most used for bulk power transmission

• Underground: most common in center-city, urban and planned development systems, subject to higher failures and larger per unit cost length, capacitance problems over long lengths

Typical Construction – O/H

Typical Construction – O/H

• Towers / Poles – to increase phase to ground distance and limit exposure to the public

• Cross arms – separate phases from each other and from ground potential (pole is 90% ground)

• Insulators – to separate phase voltage from ground potential and maximize electrostatic creep length

• Ground/Static Wire - for lightning protection

Typical Construction – O/H

• Conductors – typically ACSR (aluminum conductor steel reinforced) one or more per phase with appropriate diameter to carry current and load at given voltage and strength to handle spans between towers and associated dead and live loads

Clearances

• Nominal line voltage - Maximum voltage, • Basic distance, Safe distance

• 50 kV - 72,5 kV 0,70 m 1,70 m• 110 kV - 123 kV 1,20 m 2,20 m• 150 kV - 170 kV 1,60 m 2,60 m• 220 kV - 245 kV 2,60 m 3,60 m• 380 kV - 420 kV 3,60 m 4,60 m• SAFE DISTANCE: divide max. voltage by 100 add 1

Underground cables

• Application: underground, underwater

• Trench Installation or Duct Bank

• Direct Buried or Cable Tray

• External sheath or ground wrap

• Riser Poles/Terminations

• Manholes for Duct Bank Installations\

Distribution Cables

Cross sections

UD Cable Cross sections

• 25kV primary: phase conductor is concentrically stranded (Cu or Al) a semi-cond. polyethylene shield, polyethylene primary insulation (white), another shield, and concentric neutral (Cu or Al) wrapped around outer shield.

• Jacketed Cable (shown below) includes an additional insulated polyethylene jacket over neutral

UD Shielded Cable

• Shielded cable: Uses a copper longitudinal corrugated tape shield in place of the concentric neutral strands

3-phase transmission cable

• 5-46 kV cable: for use in aerial, direct burial, duct bank, open tray, or underwater applications

Transmission line models

• Series resistance and inductance per unit length

• Shunt capacitance per unit length

• These values control the power-carrying capacity of the transmission line and the voltage drop at full load

Heat engines, steam cycles and efficiencies

High Temperature Sink

Low Temperature Sink

TH

TC

Heat Engine Work

QC

Low Temperature Sink

QH

Heat Engine Efficiency after Sadi Carnot

NOTE: T in oK or oR

LM #5

• If a solar pond is able to trap heat beneath is saline cover at temperature 120o C above its ambient environment, what is its maximum Carnot efficiency if the outdoor temperature is 15o C?

entropy

• A measure of the amount of energy unavailable for work in a natural process

T

QS

Sample Heat Engine Problem

• A 45% efficient heat engine operates between 2 reservoirs (750oC and 50oC) and withdraws 107 J/sec• What is rate of entropy lost in high temperature

reservoir and entropy gained at low temp reservoir ?

• Express engine’s work in Watts?• What is total entropy gain of system?

Heat Engine Solution

• A 45% efficient heat engine operates between 2 reservoirs (750oC and 50oC) and withdraws 107 J/sec• What is rate of entropy lost in high temperature reservoir?

• Sloss = Q/T = 107/1023oK = 9775 J/s-oK

• and entropy gained at low temp reservoir ?

• Sgain = Q/T = 107x55%/323oK = 17,028 J/s-oK

• Express engine’s work in Watts?

• Work = 107x45% = 4.5 x 106 J/s = 4.500kW = 4.5 MW

• What is total entropy gain of system? S = Sgain (17,028 J/s-oK) - Sloss (9775 J/s-oK) = 7253 J/s-oK

NOTE:

• The fact that there was a net increase in entropy tells us that the engine has not violated the Carnot efficiency limit (which for this device would have been what?)

Write Your Answer as LM#6

Heat Engines

• Historic devices that convert heat energy into mechanical energy• Steam Engine

• Savery 1698 (<1% efficient)

• Newcomen 1705 (~1% efficient)

• Watt 1770 (separate condenser ~2% efficient)

• Steam Turbine (Parsons 1880 ~10% efficient)

Heat Engine Efficiencies

• Modern Steam Turbines (~30% efficient)

• Gasoline Engines (max. 20% efficient)

• Diesel engine (max. 30% efficient)

• Gas Turbines (20-30% efficient)

• Heat Pumps (C.O.P. of 2-12)

• Cogeneration Systems (>70% efficient)

Sample Heat Engine Problem – LM #7

• A 65% cogeneration system operates between 2 reservoirs (750oC and 20oC) and withdraws 3 x 106 J/sec• What is rate of entropy lost in high temperature

reservoir?• Express engine’s work in Watts?

Polyphase synchronous generators

• How did we arrive at the 3 phase standard for generators?

• What does synchronous mean anyway?

• First another look back….

• BEGIN HERE

History - EM Induction Generators

• 1831 Michael Faraday’s Electromagnetic Induction Experiment

Soft iron ring

switch

batteryN

First Evolution: DC Generator

Faraday 1831

Second Evolution: AC Generator

Pixii 1832

AC Generator Output

Lenz’ Law

• When an emf is generated by a change in magnetic flux according to Faraday's Law, the polarity of the induced emf is such that it produces a current whose magnetic field opposes the change which produces it. The induced magnetic field inside any loop of wire always acts to keep the magnetic flux in the loop constant. In the examples below, if the B field is increasing, the induced field acts in opposition to it. If it is decreasing, the induced field acts in the direction of the applied field to try to keep it constant.

Lenz’ Law

synchronous

• A fixed-speed machine (generator or motor) that is synchronized with the utility grid to which it is connected…

• To generate 60Hz a two pole generator would need to rotate at 3600 rpm in order to provide synchronous output

Multi-pole machines

• Two pole machines have 1 N and 1 S pole on their rotor and their stator

• Four pole machines have 4 poles (2 N and 2 S) on both rotor and stator

p

fN

s

s

fcycles

cyclesp

revolutionN

s

s

120

min

60

)2/(

1

Synchronous machines – LM#8a

• How fast would a generator that is synchronized with the utility grid in France need to rotate to…

• Generate 50Hz if it had four (4) poles?

Finally… the 3- Wye synchronous generator

• For balanced power input and output• Input from the steam turbine• Output to the electric grid/loads

• What will be the rotation speed of this most common generator in the US? Write Your Answer as

LM#8 b

GTs, CCs, Baseload Plants

• To overcome Lenz’s Law all of these generators require motive horsepower• Gas Turbines• Steam Turbines• Hydro Turbines

Steam Electric Power Plant

Other power plant schematics

LDCs

• What is a Load Duration Curve?

• Every load hour of the year (8760 hours of system load data) arranged from the highest demand to the lowest demand

• A key design tool in determining how to match generation mix with load profiles of the utility company

US Industry structure - utilities

• Traditionally given a monopoly franchise

• In exchange, subject to regulation• State and Federal

• Most are distribution only• Many remain vertically integrated (G, T &D)• 3200 US electric utilities• Four types

US Industry structure - utilities

• Investor Owned (IOU) • 5%, generate > 2/3 of power

• Federally Owned• TVA, BPA, US Army Corps, sell power non-

profit

• Other Publicly Owned• Munis, state, 2/3 of this type, <9%

• Coops – originally set up by REA

US Industry structure – nonutilities

• Nonutility Generators (NUGs) • Prior to 1940 ~ 20% of power• By mid-1970s a small fraction• Late 1980s-1990s as regulators changed rules

• Some utilities had to sell off their assets

• Growth of NUGS in some states was significant

• By 2001 NUGs were delivering over 25%

Regulatory impacts (PUHCA, PURPA, EPAct, FERC Orders 888 & 2000)

Public Utility Holding Company Act of 19351929 16 holding companies controlled 80% of US utilities

Financial abuses in many large companiesStock Market Crash left many in bankruptcyPUHCA provided regulation and break-up of large HCs

Public Utility Regulatory Policies Act of 19781973 oil crisis led to large rise in utility retail rates

PURPA set up to encourage energy efficiency and renewable energy technologies

Regulatory impacts (PUHCA, PURPA, EPAct, FERC Orders 888 & 2000)

Energy Policy Act of 1992Created new entity EWGEPAct set up to begin opening up the grid to allow competitive generators to compete for customers hopefully to drive down costs and prices

FERC Orders 888 & 2000888 Requires IOUs to publish nondiscriminatory traiffs that can be applied to all generators/competitors

2000 Calls for the creation of regional transmission organizations RTOS to control transmission system operation

Industry today (NUGS, IPPs, QFs)

• NUG – non-utility generator

• IPP – non-PURPA-regulated NUGs

• QF – meet PURPA requirements for efficiency or renewable energy use

New homework

• HW 3 – due next Monday

• will be posted on web

• 2.9, 2.12, 2.13, 3.1, 3.2, 3.3, 3.4, 3.6, 3.7, 3.9, 3.10, 3.11