chemical, biological and environmental engineering cost of electrical energy ac power concepts

Chemical, Biological and Environmental Engineering

Cost of Electrical Energy

AC Power Concepts

Advanced Materials and Sustainable Energy LabCBEE

Housekeeping IssuesChanged way #2 is stated (replaced n2 with T2 for clarity)

Timeline: Midterm planned for Tuesday 2/8 (date OK?)

Class on 2/3: Intro to Nuclear Energy

Plan to issue HW4 on 2/1(due on 2/8 before midterm)

GS: Need to discuss class presentation subject

Reminder: visit to Energy center on 2/1 at class time.


Baseload, Intermediate and Peaking Supply


Baseload, Intermediate and Peaking Supply• Load demand on utilities fluctuates constantly

– During peak demand most plants are operating– During light demand many plants are idling

• Power plants are categorized as– Baseload

• Large coal, nuclear, and hydroelectric plants• Expensive to build, cheap to operate

– Intermediate• Combined-cycle plants• Cycled up during the day, cycled down during the evening

– Peaking• Simple-cycle gas turbines• Inexpensive to build, expensive to operate


Costing Power• Other fixed costs

– Regular maintenance (e.g., groundskeeping)– Administration– Insurance

• Variable Costs (primarily fuel) – Cost of fuel

• Coal ~ $2.21/MMBTU ($43.74/ton)• Gas ~ $4.74/MMBTU

– Operations and Maintenance (Repair & Spare parts, etc)

Variable Costs ($/yr) =

[ Fuel($/BTU)xHeat rate(BTU/kWh) + O&M($/kWh)] xkWh/yr


Costing Power - II

Total cost of operating power plant then is the sum:

Then, depending on how many kWh are generated in a typical year,

This levelized cost per unit of energy is useful to compare various projects

Total Annual Cost [$ / ]Levelized Cost [$ / ] $ /

Annual Output [ / ]

yrkWh kWh

kWh yr= =

Total Annual Cost [$ / ]

Total Fixed Costs [$ / ] Total Variable Costs [$ / ]

yr

yr yr

=

= +


Graphical Version of Costing Power


New Generation Costs SummaryCapital Costs ($/kW)

Fixed O&M ($/kW)

Heat Rate (BTU/kWhr)

Variable O&M(¢/kWh)

Conventional Coal 2,200 28 8,700 2.4

IGCC (Integrated Coal Gasification Combined Cycle)

2,600 40 7,500 2.0

IGCC with CCS 3,800 47 8,300 2.3

Gas Combined Cycle 1,000 12 6,300 3.2

Gas CC w/CCS 2,000 20 7,500 3.9

Gas Turbine 650 11 10,500 5.3

Nuclear 3,800 92 10,500 1.2

Wind 2,000 31 - -

Wind-Offshore 4,000 87 - -

Hydro 2,300 14 - 0.2

Geothermal (US average) 1,750 168 30,000 0.5

Solar PV 6,200 12 - -

Solar Thermal 5,100 58 - -

Adapted from EIA publicationElectricity Market Module of the National Energy Modeling System 2010, DOE/EIA-M068(2010)


Costing Power


Screening Curve• Plot costs for different plants on the same graph

Plot as Cost = Fixed Cost ($) + Variable Cost ($/kWh) *kWh

Cost

($)

Energy Produced (kWh) [or rated power (kW) x hours of operation (h)]

Fixed Costs Plant 1 ($)

Variable Costs Plant 1 ($/kWh * kWh)

Fixed Costs Plant 2Variable Costs Plant 2


Using Screening CurvesCombustion turbine is lowest-cost option for up to 1675 h/yr of operationCoal plant is the lowest-cost option for operation beyond 6565 h/yrThe combined cycle plant is the cheapest option if it runs between 1675 and

6565 h/yr


Capacity FactorThe capacity factor is defined as

CF = [produced energy per year (kWh/yr)] / [ Rated power (kW) x 8760 h/yr]

Essentially “fraction of plant on-line time at full power averaged over year”

Why would the plant not operate at full rated power for full year?1-Time down for maintenance (try to minimize this…)2-Power it produces is not cost effective (use screening curve to find out how many hours on-line)


Load-Duration Curves


Load-Duration Curve


Determining Optimum MixTransfer crossover points onto load duration curve to identify

optimum mix of power plants


LOAD-DURATION CURVES


Determining Optimum Mix

Baseload

IntermediatePeaking

+ Reserve


Baseload, Intermediate and Peaking Supply


Cost of PowerThe CF with cost parameters allow us to determine the cost of

electricity from each type of plant


Cost of Energy to UserThe final price is the weighted average of the costs of

all generation:

From example on previous slide:

Final Cost [ / ] /Output UnitCost

c kWh c kWhOutput

´/ /= =å

å

9

9

Final Cost [ / ]

27.99[10 ] 4.69[ / ] 4.94 6.23 1.14 12.87

27.99[10 ] 4.94 1.14

5.19 /

c kWh

kWh c kWh x x

kWh

c kWh

/ =

/´ + += =

+ +

/=


Costing Power• Load duration curve needs to be padded with

reserve excess capacity (reserve margin) – To deal with plant outages, sudden peaks in demand, and

other unforeseen events

• Process of selecting which plant to operate first at any given time is called dispatching

• If you have renewables, they will be dispatched first, although they are intermittent (and require extra spinning reserve)– Energy Policy Act 1992/2005


New Generation by Fuel Type(USA 1990 to 2030, GW)

Source: EIA Annual Energy Outlook 2007


Upcoming Challenges to GridPlug in Electric Vehicles: What do you think that will do to the daily load profile?

People come home at about this time

In absence of incentiveto charge at other timepeak will get higher (worse)


Smart GridQ: Why should someone using power off peak pay rates using

peak power? (“subsidizing peak power”?)

Adding information layer to power distribution grid• Allows utilities to charge differential rates depending on mix• Allows utilities to inform customers about changing rates• Allows customers to decide when to consume power• Can also be used (voluntarily) to turn off non-essential loads at

customers– Pool Pumps, Washer/Drier, A/C– “Demand Side Management”

Also• Increases ability of users to add distributed generation• Increases resilience of grid


Example of Distributed Generation: Microturbines

Very small gas turbines (“aeroderived”)

• 1kW to several 100 kW• 30 to 60 kW unit about the size of a refrigerator• Have only one moving part

– Spins at ~96000 rpm on air bearings…

• Particularly good for CHP


MICROTURBINESGenerator makes variable frequency AC that is rectified and

inverted to grid frequency ac (50 or 60 Hz)

Some microturbines are designed for power and heat


MICROTURBINES


Microturbines

Source: http://www.capstoneturbine.com

Capstone 65 kW Microturbine

230 kW fuel

80% CHP Heat Utilization

120 kW hot water

65 kW electrical

45 kW waste heat


DG/CHP: Fuel cellsExample: ClearEdge Power (Hillsboro – OR!)• PEM fuel cell operating on natural gas (reformate)• No moving parts (but membrane degrades)• ~40% Efficiency, ~85% Energy Utilization

Chemical, Biological and Environmental Engineering

Evaluating power consumption in AC


AC Power

Instantaneous Power = VI

What is the power transferred if V=120V, I=10A,


AC Power

cos sin and 2j te t j t fw w w w p= + =

Note: because i is also used for current, the imaginary number basis “i” is usually replaced by “j”


Phasor Representation


AC Power

Previous problem reduces to: “What is the instantaneous power transferred if V=120V, I=10A and f=0”

What if f=90o instead?

P=VIcosf


Real vs. Apparent PowerReal power is the capacity of the circuit for performing

work in a particular time

Apparent power is the product of the current and voltage of the circuit

If Voltage and Current out of phase, apparent power can be greater than real power


Real, Apparent, Reactive Power; Power Factor

Real Power usually written as P, given in Watts

Apparent Power is S, Given in Volt-Amperes(i,.e., based on system voltage and current)

The difference is “Reactive Power”, given as Q Basically, current flowing 90o (π/2) out of phase with voltage

Given in Volt-Amperes Reactive (or VARs)

S can be calculated as S2=P2+Q2

Power factor l=P/S (fraction of power that is useful)Can be shown that P=S cosf, therefore l=cosf


About the Power FactorA load with low power factor draws more current than

a load with a high power factor for the same amount of useful power transferred

Higher currents increase losses in the distribution system, and require larger wires and other equipment

Causes utilities to charge higher cost for low power factor – it is in your interest to correct power factor


AC Power

Further, power varies continuously

what is the power at the point V (or I) crosses 0?

Use “Root Mean Squared” (RMS) power instead


Root Mean Squared Values

2

By definition

For a sinusoid

sin(2 ) you can show that 2

, 2 2

and cos cos2

rms avg

rms

m mrms rms

m mrms rms

A A

aA a ft A

Therefore

V IV I

V IP V I


ImpedanceIn general you know that power is dissipated by

a current flowing against a voltage (V=IR)

In AC you have to consider the effect of sinusoidal voltage waveform

The generic “resistance is called “impedance”– Purely resistive (R)– Capacitive (XC)

– Inductive (XL)


Complex Impedance• In general impedance is called “Z”

– “R” is still the resistance– “X” is the “reactance”

• The relationship is

jVZ e R jX

If-= = +


Inductive Reactance• We define a quantity called the Inductance, L

• From Faraday’s law

NL

i

F=

d diN iL N L

dt dt

FÛ F = Þ =

dN V

dte

F= =

di diL V LV

dt dtÛ = Û =

0

1( ) ( ) (0)

t

i t V t dt iL

= +ò


Inductive ReactanceIn other words, current lags voltage change

• Units of inductance is the “Henry” written as H

and LL

VI X L

Xw= =

(and 2 )fw p=


Capacitive reactanceHow does capacitor work?

Current flows onto capacitor electrodes

Charge accumulation builds voltage

This causes current to lead voltage

Impedance of a capacitor is called capacitance measured in farads

dqC

dV=

( ) ( )( )

dq t dV ti t C

dt dt= =

http://blogs.courant.com/colin_mcenroe_to_wit/capacitor.jpg


Capacitive reactance• Therefore current leads voltage in a capacitor

1 and C

C

VI X

X Cw= =


In Phasor Diagrams…

( )22L CZ R X X= + -

c

tan L CX X

Rf

-=1tan L CX X

Rf - æ ö- ÷ç= ÷ç ÷÷çè ø

L CX X X= -

( )22 2L CZ R X X= + -


You can combine impedances• In Series…

• In Parallel…

• Etc…

1 2eqZ Z Z= +

1 2

1 2 1 2

1 1 1eq

eq

Z ZZ

Z Z Z Z Z= + Û =

+

2 31 2 3 1

3 3eq

Z ZZ Z Z Z Z

Z Z= + = +

+

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