aps07 lecture 3

8/8/2019 APS07 Lecture 3

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Advanced Power SystemsAdvanced Power Systems

ECE 0909.402ECE 0909.402--02, 0909.50402, 0909.504--0202

Lecture 3: Electric Power FundamentalsLecture 3: Electric Power Fundamentals5 February 2007

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


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Aims of Todays Lecture Course Training Tours (Mon 3-8PM, Fri 12-5PM)

Part One: Overview of Chapter 2 concepts

A summary ofch. 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


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Aims of Todays Lecture (cont) 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)


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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 byextremely high reactive demand, all transformers arerated on kVA not watts, all these economic, efficiency andreliability benefits can be achieved at a very low cost


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Power factor correction

Adding capacitive impedance in parallel with the load enables thecurrent to oscillate between the inductors and capacitors ratherthan being drawn from the utility system or the customer

transformer.

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


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Last Weeks pfcorrection example An industrial customers service entrance substation is

rated at 1MVA (1,000kVA) and is at 95% capacity. Theplant now experiences a power factor of 80%. A new

manufacturing line is planned that will increase powerdemand 125kW. How many kVAR of capacitance should

be added to avoid purchasing additional substationtransformer 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 Powerwill grow to Real/PF = 885/(0.8) = 1106kVA > 1MVAcapacity

Reactive Power = Q = VI sin Y= 1106(0.6) = 664kVAR


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PF correction example (cont) An industrial customers service entrance substation is rated at 1MVA (1,000kVA) and

is at 95% capacity. The plant now experiences a power factor of 80%. A newmanufacturing line is planned that will increase power demand 125kW. How manykVAR of capacitance should be added to avoid purchasing additional substationtransformer capacity?

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

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

Reactive Power (Q) = VI sinY = 1000(0.4656) = 466kVAR

Difference in reactive power must be supplied by thecapacitor bank: 664 466 = 198 kVAR

Specify a >= 200 kVAR cap bank at industrialcustomers service voltage


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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)


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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

VARsCfarads

3.2103.21086.86

102)480(602

200

3

6

5

22

!v!v

v!

vv

!!

T


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LM #1 In an industrial facility (1J, 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 andpfrequired to meetrequest

How much capacitance is this (farads)?


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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?


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Wye (Y) connections


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Power in 3J Systems In 3-phase (3J) systems to determine power -

including Apparent (VA), Reactive (VAR), and

Real (Watts) - we need to understand theserelationships:

)_(cos3

)_(sin3

3

3

33

3

3

3

wattsIVP

VARIVQ

IVIV

IVS

lineline

lineline

linelinelineline

linephase

U

U

!

!

!!!


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Power in 3J Systems Most widely used service to buildings is a

4-wire , 3J, 208 V service. To determine line voltage of each phase

alone we need to understand this

relationship: VLine = 3 Vphase=J Therefore: Vphase=J = VLine/ 3 = 208V/ 3 =

120V


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Power in 3J Systems Many buildings also use a 4-wire , 3J, 480V

service.

To determine line voltage of each phase

alone we use the same relationship:

VLine = 3 Vphase=J Therefore: Vphase=J = VLine/ 3 = 480V/ 3 = 277V

See Figure 2.15 on page 74 of our Renewable and Efficient EPS text


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LM #2: 3J Power Distribution The Atlantic City Electric Power System

distributes electricity from its substations to

its customers at a 3-phase line voltagenominally 12-kV.

What is its Phase Voltage?


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Delta ((

)


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Y and(

current, voltage & power Y-connected

IL = IJ

VLL = 3 VJ Apparent Power3J = 3 VLL IL Real Power3J = 3 VLL ILcos Y

(-connected

VLL = VJ

IL = 3 IJ Apparent Power3J = 3 VLL IL Real Power3J = 3 VLL ILcos Y


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Example In an industrial facility (3J, 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 3J motors the power

factor is increased to 0.9 what losses arethere now?


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Solution

AkA

V

kVA

V

SI

kVASpfAfter

AkAV

kVA

V

SI

kVASkWSS

SIVkWP

L

L

L

L

LL

247247.0

2083

9.88

3

9.889.0

809.0_

444444.02083

160

3

1605.0

80805.0cos

coscos380

3

3

3

333

33

!!

!

!

!!!

!!

!

!

!!@!!

!!!

J

J

J

JJJ

JJ

U

UU


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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

kWkWL


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LM #3 In an industrial facility (3J, 208V-Y) the

real power used is 250kW and a poor power

factor exists (pf= 0.6).

What is the phase current? IJ

What is the line current? IL

What phase voltage? VJ

How much current is saved ifpfis unity?


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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 electroniccontrol

6% of US electricity flows through PS


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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-dcconverter to adjust dc output voltage to desiredlevels, 70-80% efficient in conversion

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


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Power supplies Described in text (pp.77-86) and introduced

in this course because they can have a

significant impact on power quality as wellas very sensitive to poor power quality

Also circuits similar to the buck converter

can both raise and lower dc voltages toenhance performance of photovoltaic arrays

we will design later in the course


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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


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Power quality and interruptions Few cycles or momentary sags can cause

major disruption for automated

manufacturing equipment:

PLCs

ASDs

Digital economy businesses can be evenmore disrupted by poor power quality


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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


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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


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Overview of harmonics Any periodic function can be represented by a Fourier

series made up of an infinite sum of sines and cosines withfrequencies that are multiples of the fundamental (60Hz)

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

Harmonic distortion does not occur from loads using thebasic 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 noeven harmonics (called halfwave symmetry half and fullwave rectifiers will exhibit only odd harmonics)


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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


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Total harmonic distortion

1

2

4

2

3

2

2

I

IIITHD

.!


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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


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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


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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)


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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 Faraday1831

Thin Rod Carbon Lamp Filament Jobart 1838


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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


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History 3 The First 100 Years (1840 - 1940)

continued

2-3 phase transformers 1894 OCBs invented 1897

First load dispatchers 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


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History 4 U.S. Shift from DC to AC

1890: 90% DC

1897: 62% DC

1902: 39% DC

1907: 28% DC

1942: 1% DC


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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


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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 kVcommon)


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Types of Transmission

Construction Overhead: most common in rural and semi-

rural utility systems, least expensive, easiest

to maintain, most used for bulk powertransmission

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


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Typical Construction O/H


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Typical Construction O/H Towers / Poles to increase phase to ground

distance and limit exposure to the public

Cross arms separate phases from each otherand from ground potential (pole is 90%ground)

Insulators to separate phase voltage from

ground potential and maximize electrostaticcreep length

Ground/Static Wire - for lightning protection


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Typical Construction O/H Conductors typically ACSR (aluminum

conductor steel reinforced) one or more per

phase with appropriate diameter to carrycurrent and load at given voltage and

strength to handle spans between towers

and associated dead and live loads


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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


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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\


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Distribution Cables


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Cross sections


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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


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UD Shielded Cable

Shielded cable: Uses

a copper longitudinal

corrugated tape shieldin place of the

concentric neutral

strands


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3-phase transmission cable

5-46 kV cable: for use in aerial, direct

burial, duct bank, open tray, or

underwater applications


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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 thevoltage drop at full load


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Heat engines, steam cycles and

efficienciesHighTemperature Sink

Low Temperature Sink

TH

TC

Heat Engine Work

QCLow Temperature Sink

QH


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Heat Engine Efficiency after Sadi Carnot

NOTE: T in oK oroR


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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 maximumCarnot efficiency if the outdoor temperature

is 15o C?


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entropy

A measure of the amount of energy unavailable

for work in a natural process

T

Q

S!(


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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 engines 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


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NOTE:

The fact that there was a net increase in

entropy tells us that the engine has not

violated the Carnot efficiency limit (whichfor this device would have been what?)

Write Your Answer as LM#6


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Heat Engines

Historic devices that convert heat energy

into mechanical energy

Steam Engine Savery 1698 (


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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)


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Sample Heat Engine ProblemLM #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 engines work in Watts?


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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


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History - EM Induction

Generators 1831 Michael Faradays Electromagnetic

Induction Experiment

Soft iron ring

switch

batteryN


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First Evolution: DC Generator

Faraday 1831


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Second Evolution: AC Generator

Pixii 1832


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AC Generator Output


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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 inducedmagnetic 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 inthe direction of the applied field to try to keep it constant.


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Lenz Law


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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


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Multi-pole machines

Two pole machines have 1 N and 1 S pole ontheir rotor and their stator

Four pole machines have 4 poles (2 N and 2 S) onboth rotor and stator

p

fN

s

s

fcycles

cyclesp

revolutionN

s

s

120

min

60

)2/(

1

!

vv!


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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?


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Finally the 3-J 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


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GTs, CCs, Baseload Plants

To overcome Lenzs Law all of these

generators require motive horsepower

Gas Turbines

Steam Turbines

Hydro Turbines


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Steam Electric Power Plant


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Other power plant schematics


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LDCs

What is a Load Duration Curve?

Every load hour of the year (8760 hours of

system load data) arranged from the highestdemand to the lowest demand

A key design tool in determining how to

match generation mix with load profiles ofthe utility company


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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


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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,


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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%


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Regulatory impacts (PUHCA, PURPA,

EPAct, FERC Orders 888 & 2000)

Public Utility Holding Company Act of 1935

1929 16 holding companies controlled 80% of US utilities

Financial abuses in many large companies

Stock Market Crash left many in bankruptcy

PUHCA provided regulation and break-up of large HCs

Public Utility Regulatory Policies Act of 1978

1973 oil crisis led to large rise in utility retail rates

PURPA set up to encourage energy efficiency and renewableenergy technologies


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Regulatory impacts (PUHCA, PURPA,

EPAct, FERC Orders 888 & 2000)

Energy Policy Act of 1992

Created new entity EWG

EPAct set up to begin opening up the grid to allowcompetitive generators to compete for customers hopefully todrive down costs and prices

FERC Orders 888 & 2000

888 Requires IOUs to publish nondiscriminatory traiffs thatcan be applied to all generators/competitors

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


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Industry today (NUGS, IPPs,

QFs) NUG non-utility generator

IPP non-PURPA-regulated NUGs

QF meet PURPA requirements for

efficiency or renewable energy use


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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

aps07 lecture 3

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