Transcript Final Review

ECE 333
Final Review, Spring 2015
Notes
• This review does not cover all of the course material and IS NOT a
reflection of exactly what will be covered the exam
• Final Exam is Friday, May 8th, 7-10PM
• A-J in ECEB 3017
• K-Z in ECEB 1013
• Final is comprehensive, will cover everything with an emphasis on
more recent material. You are allowed 3 handwritten notesheets
• Similar format to midterms, only slightly longer – MC, T/F, 3 longer problems,
some short answer
ECE 333 Syllabus
• Introduction, fundamentals of electric power
• Electric Power Grid, Conventional Generation
• Wind Power Systems
• Wind/Grid Integration, Introduction to Power Flow
• The Solar Resource
• Photovoltaic Materials and Systems
• Smart Grid Integration Issues
• Distributed Generation Technologies (e.g., fuel cells)
• Economics of Distributed Resources
• Energy Storage including Electric/Hybrid Cars
3
Notation - Energy
• Energy: Integration of power over time; energy is
•
what people really want from a power system
Energy Units
–
–
–
–
Joule = 1 Watt-second (J)
kWh = Kilowatthour (3.6 x 106 J)
Btu
= 1055 J; 1 MBtu=0.292 MWh; 1MWh=3.4MBtu
One gallon of gas has about 0.125 MBtu (36.5 kWh); one
gallon ethanol as about 0.084 Mbtu (2/3 that of gas)
• U.S. electric energy consumption is about 3600
billion kWh (about 13,333 kWh per person)
4
Electric Systems in Energy Context
• Class focuses on renewable electric systems, but we
•
first need to put them in the context of the total
energy delivery system
Electricity is used primarily as a means for energy
transportation
• Use other sources of energy to create it, and it is usually
converted into another form of energy when used
• Concerns about need to reduce CO2 emissions and
fossil fuel depletion are becoming main drivers for
change in world energy infrastructure
5
Energy Economics
• Electric generating technologies involve a tradeoff
between fixed costs (costs to build them) and
operating costs
• Nuclear and solar high fixed costs, but low operating costs
•
•
(though cost of solar has decreased substantially recently)
Natural gas/oil have low fixed costs but can have higher
operating costs (dependent upon fuel prices)
Coal, wind, hydro are in between
• Also the units capacity factor is important to
determining ultimate cost of electricity
6
Power System Structure
•
•
•
•
•
All power systems have three major components:
Load, Generation, and Transmission/Distribution.
Load: Consumes electric power
Generation: Creates electric power.
Transmission/Distribution: Transmits electric
power from generation to load.
A key constraint is since electricity can’t be
effectively stored, at any moment in time the net
generation must equal the net load plus losses
7
Large-Scale Power Grid Overview
8
GENERATION
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•
•
•
Large plants predominate, with sizes up to about 1500
MW with wind a rapidly growing source.
Coal is still the most common source but with a value
falling from 56% a few years ago to 39% now. Natural
gas has rapidly grown due to low costs, now making
27% of total. Nuclear (20%), hydro (6%), wind (4.3%),
wood (1.0%), solar (0.4%, high growth)
New construction is mostly natural gas and wind with
economics highly dependent upon the gas price
Generated at about 20 kV for large plants, around 600
V for many wind turbines; solar PV is dc.
9
Carnot Efficiency of Heat Engines
• Heat engines use differences in temperature to
convert part of the heat from a high temperature
source, QH, into work, W, with output heat QC
–
Examples are fossil fuel generators, nuclear generators,
concentrated solar generators and geothermal generator
Thermal Efficiency =  =
Net work output W

Total heat input QH
QH  W  QC
Carnot Maximum Efficiency = 1 -
TC
TH
10
Coal Plant Example
• Assume capital costs of $4 billion for a 1600 MW coal
plant with a FCR of 10% and operation time of 8000
hours per year. Assume a heat rate of 10
Mbtu/MWh, fuel costs of 1.5 $/Mbtu, and variable
O&M of $4.3/MWh. What is annualized cost per
kWh?
Fixed Cost($/kW) = $4 billion/1.6 million kW=2500 $/kW
Annualized capital cost = $250/kW-yr
Annualized operating cost = (1.5*10+4.3)*8000/1000
= $154.4/kW-yr
Cost = $(250 + 154.4)/kW-yr/(8000h/yr) = $0.051/kWh
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Complex Power
V=VRMS  V
I=I RMS  I
Asterisk denotes complex conjugate
VI*  VRMS I RMS  V   I 
Power triangle
S
Q
(θV-θI)
P
VI*  VRMS I RMS cos V   I   jVRMS I RMS sin V   I 
Q
S
P
Reactive
Apparent
Real
S
=
P+jQ
Power
Power
power
12
Power Factor Correction Example
• Assume we have a 100 kVA load with pf = 0.8 lagging,
and would like to correct the pf to 0.95 lagging
We have:
1


cos
(0.8)  36.9
S  80  j 60 kVA
We want:
desired  cos 1 (0.95)  18.2
Qdes.
tan(18.2) 
P
S
Qdes.=?
18.2
Qdes.  tan(18.2) * 40  26.3 kVAr
P=80
This requires a capacitance of:
P
Q=60
P
Qdes=26.3
Q=-33.7
Qcap  60  26.3  33.7 kVAr
13
Balanced 3 Phase () Systems
• A balanced 3 phase () system has
• three voltage sources with equal magnitude, but with an
angle shift of 120
• equal loads on each phase
• equal impedance on the lines connecting the generators
to the loads
• Bulk power systems are almost exclusively 3
• Single phase is used primarily only in low voltage,
low power settings, such as residential and some
commercial
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Advantages of 3 Power
• Can transmit more power for same amount of wire
•
•
•
(twice as much as single phase)
Torque produced by 3 machines is constrant
Three phase machines use less material for same
power rating
Three phase machines start more easily than single
phase machines
15
Per Phase Analysis Procedure
To do per phase analysis
1. Convert all  load/sources to equivalent Y’s
2. Solve phase “a” independent of the other phases
3. Total system power S = 3 Va Ia*
4. If desired, phase “b” and “c” values can be
determined by inspection (i.e., ±120° degree phase
shifts)
5. If necessary, go back to original circuit to determine
line-line values or internal  values.
16
Transformers Overview
• Power systems are characterized by many different
•
•
•
voltage levels, ranging from 765 kV down to
240/120 volts.
Transformers are used to transfer power between
different voltage levels.
The ability to inexpensively change voltage levels is
a key advantage of ac systems over dc systems.
In 333 we just introduce the ideal transformer,
with more details covered in 330 and 476.
17
Ideal Transformer
• We’ll develop the voltage/current relationships
for an ideal transformer
–
–
–
no real power losses
magnetic core has infinite permeability
no leakage flux
• We’ll define the “primary” side of the transformer
as the side that usually takes power, and the
secondary as the side that usually delivers power.
–
primary is usually the side with the higher voltage, but
may be the low voltage side on a generator step-up
transformer.
18
Key Problems with Harmonics
• A key problem with the third harmonic is neutral
current since the fundamental 120 degree phase
shift becomes 360 degrees for the third harmonic so
the third harmonic values do not cancel (also true
for other triplen harmonics)
–
Delta-grounded wye transformers prevent triplen
harmonic currents from flowing into the power grid
• Harmonics cause transformer overheating since core
•
losses are proportional to frequency
Harmonic resonance, particularly with shunt
capacitors (can be around 5th or 7th harmonic values)
19
Power Flow
• A common power system analysis tool is the power flow
–
It shows how real and reactive power flows through a
network, from generators to loads
• Solves sets of non-linear equations enforcing
"conservation of power" at each bus in the system (a
consequence of KCL)
–
–
Loads are usually assumed to be constant power
Used to determine if any transmission lines or transformers
are overloaded and system voltages
• Educational version PowerWorld tool available at
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http://www.powerworld.com/gloversarmaoverbye
20
Generator Costs
• There are many fixed and variable costs associated
•
•
•
•
with power system operation.
The major variable cost is associated with generation.
Cost to generate a MWh can vary widely.
For some types of units (such as hydro and nuclear) it
is difficult to quantify.
Many markets have moved from cost-based to pricebased generator costs
21
Types of Wind Turbines
• “Windmill”- used to grind grain into flour or pump
•
•
•
water
Many different names - “wind-driven generator”,
“wind generator”, “wind turbine”, “wind-turbine
generator (WTG)”, “wind energy conversion system
(WECS)”
Can have be horizontal axis wind turbines (HAWT)
or vertical axis wind turbines (VAWT)
Groups of wind turbines are located in what is
called either a “wind farm” or a “wind park”
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Power in the Wind
Combining previous equations,
1
Power through area A    Av  v 2
2
1
PW   Av3 (6.4)
Power in the wind
2
PW (Watts) = power in the wind
ρ (kg/m3)= air density (1.225kg/m3 at 15˚C and 1 atm)
A (m2)= the cross-sectional area that wind passes through
v (m/s)= windspeed normal to A (1 m/s = 2.237 mph)
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Power in the Wind (for reference solar is
about 600 w/m^2 in summer)
• Power increases with
•
•
•
the cube of wind speed
Doubling the wind
speed increases the
power by eight
Energy in 1 hour of 20
mph winds is the same
as energy in 8 hours of
10 mph winds
Nonlinear, so we cannot
use average wind speed
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Power in the Wind
•
•
•
•
1
PW   Av3
2
Power in the wind is also proportional to A
For a conventional HAWT, A = (π/4)D2, so wind
power is proportional to the blade diameter
squared
Cost is somewhat proportional to blade diameter
This explains why larger wind turbines are more
cost effective (plus, as we shall see, because they
are higher, the winds are stronger)
25
Tip-Speed Ratio (TSR)
• Efficiency is a function of how fast the rotor turns
• Tip-Speed Ratio (TSR) is the speed of the outer tip
of the blade divided by wind speed
Rotor tip speed rpm   D
Tip-Speed-Ratio (TSR) 
=
(7.30)
Wind speed
60v
• D = rotor diameter (m)
• v = upwind undisturbed wind speed (m/s)
• rpm = rotor speed, (revolutions/min)
• One meter per second = 2.24 miles per hour
26
Electric Machines
• Electric machines can usually function as either a
•
motor or as a generator
Three main types of electric machines
–
–
DC machines: Advantage is they can directly operate at
variable speed. For grid application the disadvantage is they
produce a dc output. Used for small wind turbines.
AC synchronous machines

–
Operate at fixed speed. Used extensively for traditional power
generation. The fixed speed had been a disadvantage for wind.
AC induction machines

Very rugged and allow some speed variation but usually not a lot for
efficient operation.
27
Slip
• The difference in speed between the stator and the
rotor
NS  NR
NR

 1
NS
NS
• s = rotor slip – positive for a motor, negative for a
•
•
•
•
generator
NS = no-load synchronous speed (rpm)
f = frequency (Hz)
120 f
NS 
p = number of poles
p
NR = rotor speed (rpm)
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Rayleigh Statistics – Average Power in the
Wind
• To figure out average power in the wind, we need
to know the average value of the cube of velocity:
1
1
3
Pavg    Av    A  v 3 
avg
2
avg 2
• With Rayleigh assumptions, we can write the (v3)avg
in terms of vavg and the expression for average
power in the wind is just
3
6 1
Pavg    A  vavg 
 2
• This is an important and useful result
29
Real Data vs. Rayleigh Statistics
This is why it is important to gather as much real
wind data as possible
30
Wind Farms
• Normally, it makes sense to install a large number
•
of wind turbines in a wind farm or a wind park
Benefits
–
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Able to get the most use out of a good wind site
Reduced development costs
Simplified connections to the transmission system
Centralized access for operations and maintenance
• How many turbines should be installed at a site?
31
Bus Admittance Matrix or Ybus
• First step in solving the power flow is to create what is
known as the bus admittance matrix, often call the Ybus.
• The Ybus gives the relationships between all the bus
current injections, I, and all the bus voltages, V,
I = Ybus V
• The Ybus is developed by applying KCL at each bus in the
system to relate the bus current injections, the bus
voltages, and the branch impedances and admittances
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Power Flow Slack Bus
• We can not arbitrarily specify S at all buses because
total generation must equal total load + total losses
• We also need an angle reference bus.
• To solve these problems we define one bus as the
"slack" bus. This bus has a fixed voltage magnitude
and angle, and a varying real/reactive power
injection.
• A “slack bus” does not exist in the real power
system.
33
Real Power Balance Equations
n
Si  Pi  jQi  Vi  Yik*Vk* 
k 1

n
 Vi Vk
k 1
n
j ik
V
V
e
(Gik  jBik )
 i k
k 1
(cos ik  j sin  ik )(Gik  jBik )
Resolving into the real and imaginary parts
Pi 
Qi 
n
 Vi Vk (Gik cos ik  Bik sinik )  PGi  PDi
k 1
n
 Vi Vk (Gik sin ik  Bik cosik )  QGi  QDi
k 1
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Power Flow Variables
Assume the slack bus is the first bus (with a fixed
voltage angle/magnitude). We then need to determine
the voltage angle/magnitude at the other buses.
 2 




 n 
x  

V
 2




 Vn 
 P2 (x)  PG 2  PD 2 




 Pn (x)  PGn  PDn 
f ( x)  
Q2 (x)  QG 2  QD 2 






 Qn (x)  QGn  QDn 
N-R Power Flow Solution
The power flow is solved using the same procedure
discussed last time:
Set v  0; make an initial guess of x, x( v )
While f (x( v ) )   Do
( v 1)
( v ) 1
x
 x  J (x ) f (x )
v
 v 1
End While
(v)
(v)
Newton-Raphson Comments
• When close to the solution the error decreases
•
•
•
•
quite quickly -- method has quadratic convergence
f(x(v)) is known as the mismatch, which we would
like to drive to zero
Stopping criteria is when f(x(v))  < 
Results are dependent upon the initial guess. What
if we had guessed x(0) = 0, or x (0) = -1?
A solution’s region of attraction (ROA) is the set of
initial guesses that converge to the particular
solution. The ROA is often hard to determine
37
The Solar Resource
• Before we can talk about solar power, we need to
•
•
•
•
•
talk about the sun
Need to know how much sunlight is available
Can predict where the sun is at any time
Insolation : incident solar radiation
Want to determine the average daily insolation at a
site
Want to be able to chose effective locations and
panel tilts of solar panels
38
Solar Position at Any Time of Day
• Described in terms of altitude angle β and azimuth
•
•
angle of the sun ϕS
β and ϕS depend on latitude, day number, and time
of day
Azimuth angle (ϕS ) convention
–
–
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positive in the morning when sun is in the east
negative in the evening when sun is in the west
reference in the Northern Hemisphere (for us) is true
south
• Hours are referenced to solar noon
39
PV System Overview
•Solar cell is a diode
•Photopower coverted to DC
Shadows
•Shadows & defects convert
generating areas to loads
•DC is converted to AC by an
inverter
•Loads are unpredictable
•Storage helps match
generation to load
40
General PV Cell Equivalent Circuit
• The general equivalent circuit model considers both
parallel and series resistance
Equivalent Circuit
Vd
I  ( I SC  I d ) 
RP
I  ( I SC  I 0 (e
+
V
Load
Vd
(maximize)
+
RP
Id
ISC
q / kT V  RS I 
I
(minimize)
Photocurrent
source
V  Vd  I  RS
Rs
-
V  RS I
-1)) 
RP
41
Shading Mitigation
• Solution - bypass diodes to bypass the shaded cell
• Instead of an I(RP+RS) voltage drop across the
•
resistors, only a fixed voltage drop due to the
diode, about 0.6 V (or even smaller with special
diodes)
No effect during normal operation
42
Annualized Investment
•
Then, the equal annual payments are given by
d (1  d )
A  P
(1  d )n  1
n
A  P  CRF(i ,n)
•
•
•
Capital Recovery
Factor (CRF)
(5.20)
The capital recovery factor, CRF(i,n), is the inverse
of the present value function PVF
CRF measures the speed with which the initial
investment is repaid
Capital recovery function in Microsoft Excel:
PMT(rate,nper,pv)
43
Internal Rate of Return
• Until now, we have always specified the interest
•
•
rate or discount rate
Now we’ll “solve for” the rate at which it makes
sense to do the project
This is called the internal rate of return, also called
the “break-even interest rate”
–
Higher is better because a higher IRR means that even if
the interest rate gets higher, the project still makes sense
to do
• Note there is no closed form solution - use a table
(or Excel, etc.) to look it up
44
Amortizing PV Costs
• Simple payback is the easiest analysis, which assumes
there is no cost for money and no inflation. Annual
cost is just total cost divided by lifetime
This can give a quick ballpark figure
–
• Example: Assume 5 kW system with a capacity factor
of 18%, an installed cost of $ 5/W (after tax credits),
and a lifetime of 20 years with no maintenance costs.
What is $/kWh?

$5/W×5000 W

20 yrs
$1250/yr
=
=$0.158/ kWh
5 kW×0.18×8760
7884 kWh/yr
45
Amortizing PV Costs
• More detailed analysis uses the capital recovery factor
•
using an assumed discount rate
Redo the previous example using a discount rate of 5%
per year
d (1  d )n
0.05  1.0520
A  P
 25, 000
n
(1  d )  1
1.0520  1
0.1327
A  25, 000
 2006.6
These values vary
1.6533
linearly with the
$2006.6/yr
assumed PV
=$0.2545/ kWh
installed cost
7884 kWh/yr
46
Batteries and PV Systems
• Batteries in PV systems provide storage, help meet
•
surge current requirements, and provide a constant
output voltage.
Lots of interest in battery research, primarily driven by
the potential of pluggable hybrid electric vehicles
–
$2.4 billion awarded in August 2009
• There are many different types of batteries, and which
one is best is very much dependent on the situation
–
Cost, weight, number and depth of discharges, efficiency,
temperature performance, discharge rate, recharging rates
47
DC Motors
• Main types are based on how the stator is
powered:
–
–
–
–
Separately excited (separate windings)
Permanent magnet
Shunt connection (field winding is in parallel with the
armature); these motors have near constant speed
regardless of load
Series connection (field winding is in series with
armature); these motors have high torque at low speed,
but can have high speed with low torque
48
Example: Energy to Pump Water from
Shallow Well
• How many kWh/day are required to pump 250
gallons/day with a 66 ft head + 230 ft pressure head
assuming 35% efficiency?
With efficiency given, the below equation works for any rate, so
calculate power required if the 250 gallons is pumped in one hour
0.1885  H ( ft )  Q( gal / min)
Pelec (W ) 
Efficiency
250gal / day
Q( gal / min) 
 4.167 gal / min
60 min/ day
0.1885  296  4.167
Pelec (W ) 
 664 W (for one hour)
0.35
Cost is about $0.05 or
Answer  0.664 kWh
$0.02 per 100 gallons
49
Types of Hydro
• Impoundment (or Storage) is the most common,
usually with a dam used to store river water in
reservoir
–
–
–
–
Multiple dams can be chained together
Management of water usage can get quite complex, with lots
of side issues such as flood control and maintaining flow
rates for fish
Water released from one
reservoir becomes available
to the one downstream
reservoirs (and plants)
Constraint on total water
available over a year (or longer)
Image: http://energy.gov/eere/water/types-hydropower-plants
50
Types of Hydro
• Pumped Storage
–
–
–
–
–
Uses the potential energy of water to "store" electricity
Part of the time it works as a conventional impoundment
hydro plant with water in a high reservoir flowing through
the turbine to a lower reservoir (or lake/river)
Part of the time it functions as a large load as water is
pumped from the lower reservoir back to the higher reservoir
Works as a generator when the price of electricity is high
(e.g., during the day) and as a load when price of electricity is
low (e.g., during the night)
Round trip efficiency can be up to 80%
51
Homer Lake Hydro Example
• 80 acres, 30 ft head, say we get
4488 gal/minute out, and capacity
factor is 100%
• What is power/energy impact for
100 ft of 10” vs. 12” pipe?
10.472 44881.852
10” H L [ft]=

×100=7.61 ft
1.852
4.871
150
10
Hazen-Williams
(30-7.61)
Loss Equation
P=4488 
=18.96 kW
5.3
Efficiency η is 50%: P = 18.96kW  0.5 = 9.48 kW
Capacity factor is 100%: 83.04 MWh  50$/MWh = $4152/yr
image: http://www.ccfpd.org/about/Map_HL_final_vert_042010.pdf
52
Tidal Power
• Tidal power can be extracted either by building dams,
•
and is hence equivalent to traditional hydro, or by
placing stream (not steam!) turbines in the tidal flow
World’s largest tidal power station
had been in La Rance, France, built
from 1960-66, with a capacity of
240 MW
–
It uses a 330 meter long dam, built in
front of a 22 square km basin. Tidal
differences average about 8 meters.
• Largest is now 254 MW Sihwa Lake Tidal Power, South
Korea. Uses a 30 square km basin
Photo source: http://www.rise.org.au/info/Tech/tidal/image019.jpg
53
Geothermal Power
• Geothermal power creates electricity by leveraging
temperature differences between surface and below
the surface
–
Should not be confused with ground source heat pumps,
which are also sometimes called geothermal systems
• In most areas of world heat differences are not
•
•
sufficient to be practical (temperature increases only
by 30º C on average per km)
Some areas have rocks close to the surface
Liquid is used to transfer the heat energy
54
Biomass -- Wood
• Wood has long been used as a electric fuel source,
primarily wood waste
–
–
In 2014 the US got about 1% of its electricity from wood,
more than two times the total from solar PV and CSP
Significant growth in wood generation is not expected
• Largest wood-fired
plant in the US is the
100 MW Nacogdoches
Generating Facility in
Sacul, Texas
http://www.power-eng.com/content/dam/pe/print-articles/2012/oct/f6-Photo-4-1210pe.jpg
55
Combined Heat and Power (CHP)
• Some generation technologies produce usable
waste heat, and some industries have heat or
steam demands
–
Examples: refineries and chemical plants
• Higher temperature waste heat is more versatile
• Using waste heat can displace the need to buy
•
•
electricity or an expensive fuel such as propane
One challenge is to appropriately time the use of
the electric and thermal energy
Sometimes called cogeneration
56
Distributed Generation (DG)
• Provide small increments in generation which can
•
•
•
•
•
•
•
track load growth closely to reduce costs of unused
capacity
Can ease bottlenecks in distribution networks
Improve voltages
Improve power factor
Reduce losses
Provide power during outages
Reduce emissions
But they lose economics of scale!
57
The Smart Grid
• As defined by DOE a Smart Grid has the following
seven characteristics
–
–
–
–
–
–
–
Self-healing from power disturbance events
Enabling active participation by consumers in demand
response
Operating resiliently against physical/cyber attack
Providing power quality for 21st century needs
Accommodating all generation and storage options
Enabling new products, services, and markets
Optimizing assets and operating efficiently
58
Good Luck!