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Cost of Electrical Energy AC Power Concepts. Housekeeping Issues. Changed way #2 is stated (replaced n 2 with T 2 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)
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Housekeeping Issues Changed 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 • 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
New Generation Costs Summary Adapted from EIA publication Electricity Market Module of the National Energy Modeling System 2010, DOE/EIA-M068(2010)
Screening Curve • Plot costs for different plants on the same graph Plot as Cost = Fixed Cost ($) + Variable Cost ($/kWh) *kWh Cost ($) Variable Costs Plant 2 Fixed Costs Plant 2 Fixed Costs Plant 1 ($) Variable Costs Plant 1 ($/kWh * kWh) Energy Produced (kWh) [or rated power (kW) x hours of operation (h)]
Using Screening Curves Combustion turbine is lowest-cost option for up to 1675 h/yr of operation Coal plant is the lowest-cost option for operation beyond 6565 h/yr The combined cycle plant is the cheapest option if it runs between 1675 and 6565 h/yr
Capacity Factor The 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)
Determining Optimum Mix Transfer crossover points onto load duration curve to identify optimum mix of power plants
Determining Optimum Mix Peaking + Reserve Intermediate Baseload
Cost of Power The CF with cost parameters allow us to determine the cost of electricity from each type of plant
Cost of Energy to User The final price is the weighted average of the costs of all generation: From example on previous slide:
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 Grid Plug 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 incentive to charge at other time peak will get higher (worse)
Smart Grid Q: 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
MICROTURBINES Generator 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 Capstone 65 kW Microturbine 80% CHP Heat Utilization 120 kW hot water 230 kW fuel 65 kW electrical 45 kW waste heat Source: http://www.capstoneturbine.com
DG/CHP: Fuel cells Example: ClearEdge Power (Hillsboro – OR!) • PEM fuel cell operating on natural gas (reformate) • No moving parts (but membrane degrades) • ~40% Efficiency, ~85% Energy Utilization
AC Power Instantaneous Power = VI What is the power transferred if V=120V, I=10A,
AC Power Note: because i is also used for current, the imaginary number basis “i” is usually replaced by “j”
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 Power Real 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 Factor A 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
Impedance In 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
Inductive Reactance • We define a quantity called the Inductance, L • From Faraday’s law
Inductive Reactance In other words, current lags voltage change • Units of inductance is the “Henry” written as H
Capacitive reactance How 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
Capacitive reactance • Therefore current leads voltage in a capacitor
You can combine impedances • In Series… • In Parallel… • Etc…