# Power Cycles and Energy Storage I talk about the puzzle of designing future power systems It's a"puzzle," because so many factors must be balanced to:

## Presentation on theme: "Power Cycles and Energy Storage I talk about the puzzle of designing future power systems It's a"puzzle," because so many factors must be balanced to:"— Presentation transcript:

Power Cycles and Energy Storage I talk about the puzzle of designing future power systems It's a"puzzle," because so many factors must be balanced to: Provide enough affordable power Sustain a high quality biosphere Earlier lectures have shown that renewables are not silver bullets If only because the demand huge natural resource commitments Suggesting that future power systems MUST combine different technologies Today you will see how the natural cycles of power consumption and production ALSO drive the use of different technologies And how energy storage will become critical

An Introduction to Sustainable Energy Systems: www.virlab.virginia.edu/Energy_class/Energy_class.htm It all starts with this: The cycles in our consumption of electrical power (and its real time cost): Note the almost sinusoidal cycles: Peaking at 6pm (cook dinner, cool down/heat up house, do laundry... ) With amplitude of the oscillation equaling ~ 40% of peak value www.eia.gov/todayinenergy/detail.cfm?id=12711 YELLOW midnight lines added

An Introduction to Sustainable Energy Systems: www.virlab.virginia.edu/Energy_class/Energy_class.htm Simplifying a little bit, over one day: We have a constant component at 60% of the peak And an oscillating component moving from there up to 100% at about 6 pm With oscillating component's average at 80% (from its symmetry) Giving us a time averaged power = 80% of the peak power Peak Power 60% Midnight Noon Midnight 80%

Constant "base" electrical load: Plus a variable "dispatchable" component ( so called because a dispatcher controls it? ) An Introduction to Sustainable Energy Systems: www.virlab.virginia.edu/Energy_class/Energy_class.htm Or dividing these two up: 60% of Peak Power Midnight Noon Midnight 0% 40% of Peak Power 0% Midnight Noon Midnight

Scenario #1) Use only conventional power plants to meet this load: This is TODAY's scenario, now in almost universal use Power companies want to minimize their (and ultimately your) costs Costs come in two principal flavors: FIXED COSTS and VARIABLE COSTS A fixed cost is just that, something you're going to have to pay no matter what For power plants, the biggest begin as construction (capital) costs Which are then translated into ongoing mortgage / bond payments Which have to be paid, whatever power you are now producing A variable cost is a cost that varies with the plant's power output Which includes the labor cost of the operators But is more often dominated by fuel costs Your power company owns different types of power plants:

An Introduction to Sustainable Energy Systems: www.virlab.virginia.edu/Energy_class/Energy_class.htm Which types of plants would it use for "dispatchable" power? Dispatchable power plants will be turned off (or run at low power) most of the time Which suggests (at least, after a little thought): 1) You don't want plants that take lot of energy to heat up Because it will just be wasted when the plant is shutdown: 2) You don't want plants that take a huge amount of money to build High mortgage payments don't go away when plant is turned down/off! Those payments are the power plant's biggest fixed cost 3) But you CAN use plants w/ expensive fuel, because they'll be off most of the time Fuel is the power plant's biggest variable cost

An Introduction to Sustainable Energy Systems: www.virlab.virginia.edu/Energy_class/Energy_class.htm Putting this all together: For dispatchable (variable) power you want to target plants with: 1) Lower start-up/shut-down energy losses 2) Lower fixed capital/build cost, and hence lower mortgage payments 3) Higher variable fuel/operating costs From first point, avoid plants that heat and boil things to drive turbines Coal plants, nuclear plants, solar thermal plants, geothermal plants AND Combined cycle natural gas plants (where the 2 nd cycle is steam turbines) From second point (on lower fixed capital/build costs) avoid avoid NULCEAR From third point (on tolerance of fuel costs) use NATURAL GAS TURBINE PLANTS Where combustion gas expansion alone directly powers those turbines

All three points actually suggest gas turbine power plants: Not because the natural gas itself is cheaper – It's more expensive! But because simple, cheap, low thermal mass, turbine generators will burn it! And, secondarily, because it is a bit cleaner: Because it doesn't expend chemical energy on fuel vaporization, it liberates fewer carbons per watt generated This choice of short-term, but expensive, power leads to the cost spiking below: Natural gas (\$) Natural gas (\$)

Scenario #2) Eliminate dispatchable plants in favor of storage Why pay for dispatchable power plants that sit idle almost all of the day! INSTEAD: Choose the CHEAPEST CLEANEST TYPE(S) OF POWER PLANTS Let them run ALL DAY AT FULL POWER (maximizing their bang / capital buck) Size them so their power = 80% of peak power = average daily power And STORE EXCESS POWER produced early in the day for use later in the day Old Scenario #1:New Scenario #2: Peak Power 60% Midnight Noon Midnight 80% Base load power plants Dispatchable load power plants Peak Power Midnight Noon Midnight 80% Most effective power plants, always at full power Stored energy

But how MUCH energy do we have to store in the mornings? From earlier discussion, average power per day = 80% of peak power So daily energy = E daily = 0.8 P peak x 1 day Assuming shape of energy stored & moved curve below is indeed sinusoidal: From the diagram, amplitude of the sine = (0.2 x peak power) Then, stored early day energy = integral (from 0 to day/2) of 0.2 P peak Sin = 0.2 P peak x day / Pi = 0.064 P peak x day = (0.064)(E daily /0.8) = 8% of E daily Peak Power Midnight Noon Midnight 80% Most effective power plants, always at full power Stored energy

1) www.eia.gov/tools/faqs/faq.cfm?id=65&t=2 U.S. energy per day value: From the Power Plant Land and Water Requirements lecture: Total US energy production (2012) = 4,047,765 GW-h (1) per year Which translates into US energy production = 11,089 GW-h per day So we'd need to store 8% of that early in the morning (for use later in that day): Scenario #2: US energy storage required per day = 887 GW-h So with that target figure in mind, what are the alternatives? Peak Power Midnight Noon Midnight 80% Most effective power plants, always at full power Stored energy 887 GW-h

An Introduction to Sustainable Energy Systems: www.virlab.virginia.edu/Energy_class/Energy_class.htm Scenario 2a: Energy storage via Pumped Storage Hydro (PSH) The best current way to store power! Which, in my Hydropower / Windpower lecture I represented thusly: Small dam and reservoir Lake/river"Power House" located far below With the largest example (according to the co-owner) being:

The Bath County Virginia pumped hydro storage plant: From Dominion Power's website: 1 "The World's Biggest Battery" – but no numbers! Supplies "over ½ million customers with (peak) power" At rate "exceeding Hoover Dam" One reservoir 1262 feet (385 m) above other reservoir Max rate of flow back down = 13.5 million gallons / minute (852 m 3 / sec) Area of upper reservoir = 265 acres (1.07 x 10 6 m 2 ) Its "water level fluctuates 105 feet (32 meters) during operation" OK, now we have finally got enough information to calculate energy stored! 1 km 1) Link to webpage: www.dom.com/about/stations/hydro/bath-county-pumped-storage-station.jspwww.dom.com/about/stations/hydro/bath-county-pumped-storage-station.jsp

An Introduction to Sustainable Energy Systems: www.virlab.virginia.edu/Energy_class/Energy_class.htm Stored water's gravitational potential energy: Measured relative to its final energy down in the lower reservoir:  E = M g h =  water x (volume of water moved) x g x (height difference)  water = 1000 kg / m 3 g = 9.8 m / sec 2 height difference = 385 meters volume of water moved = (1.07 x 10 6 m 2 ) (32 meters)  E = (1000 kg / m 3 ) (1.07 x 10 6 m 2 ) (32 meters) (9.8 m / sec 2 ) (385 m) = 1.29 x 10 14 kg m / s 2 = 1.29 x 10 14 Joules = 1.29 x 10 14 / (3,600,000,000) MW-h = 35.8 MW-h

An Introduction to Sustainable Energy Systems: www.virlab.virginia.edu/Energy_class/Energy_class.htm But we now need to bring in efficiencies: Having to do with the HUGE electrical motor/generators That first powered the turbines to move the water UP the hill And then (as generators) are powered by the turbines as it falls DOWN Also, conceivably, having to do with turbulence-induced heating of the water itself So what is the overall power recovery, after accounting for these losses? Wikipedia:"the electric Power Research Institute (EPRI) reports that PSH accounts for more than 99% of bulk storage capacity worldwide" Wikipedia:"the electric Power Research Institute (EPRI) reports that PSH accounts for more than 99% of bulk storage capacity worldwide" "reported efficiency varies in practice between between 70% and 80%, 1,2,3,4 with some claiming up to 87%. 5" Whoa! Aside from my intrinsic caution about this source, that's a huge range!

An Introduction to Sustainable Energy Systems: www.virlab.virginia.edu/Energy_class/Energy_class.htm What do other sources have to say? Trying to avoid Wikpedia's over reliance on vendor data and/or student theses: Efficiency:Source: 70% Energy and the Environment – Fay & Golomb (p. 100) 75%Sustainable Energy without the Hot Air (p. 191 re. Dinorwig site) 70-75%The Economist ( www.economist.com/node/21548495?frsc=dg|a ) 70-85%European Commission ( etis.ec.europa.eu/publications/jrc-setis-reports/ pumped-hydro-energy-storage-potential-transformation-single-dams ) 80.472%NREL (http://www.nrel.gov/docs/fy14osti/60806.pdf) OK, back to where I started, but now a lot happier about using Wikipedia numbers

An Introduction to Sustainable Energy Systems: www.virlab.virginia.edu/Energy_class/Energy_class.htm Using the NREL study number of ~ 80%: So to pump up enough water at the Bath plant to recover 35.8 MW-h Bath probably had to expend (35.8 MW-h) / 0.8 = 44.75 MW-h From above, across the entire US, each day we'd need to store: 887 GW-h The Bath PSH stored 35.8 MW-h (using 44.75 MW-h to do so) So we'd require 887 GW-h / 35.8 MW-h = 24,781 Bath PSH equivalents Bath upper reservoir = 265 acres = 1.07 km 2 Assume lower reservoir is reduced to same size => 2.14 km 2 total Nationally we'd need 53,031 km 2 of equivalently separated reservoir area = (230 km) 2 = 0.54% of U.S. land area = 68 New York City areas (Not impossible, but not terribly plausible either!)

Scenario 2b: Energy storage via Capacitors or Super-capacitors As mentioned in an earlier lecture, capacitors side step Maxwell's 1 st equation: Trying to push more charge into an object doesn’t really work: Unless you fold the two plates over on top of one another: No longer need charge balance on each plate Excess +’s on top plate don’t like one another But repulsion balanced by attraction to -’s below Becomes more effective the closer the plates are: Capacitance = Area x (dielectric constant in gap) / (gap thickness) = A  / d A "Super-capacitor" = Capacitor with extremely small plate and gap thicknesses An Introduction to Sustainable Energy Systems: www.virlab.virginia.edu/Energy_class/Energy_class.htm +- + -

A Hands-on Introduction to Nanoscience: www.virlab.virginia.edu/Nanoscience_class/Nanoscience_class.htm First need figure out how much energy a capacitor can store Easiest to understand via a gas storage analogy: Move a fixed amount of air into tanks of different size: Will pressure changes be the same? Of course not: MUCH more room for added air to spread out in bigger tank! Pressure induced by addition of quantity of air is inversely proportional to volume: That's what the ideal gas law states: PV = nRT or: Pressure  Quantity in tank / Volume or using Q for quantity: P(Q)  Q / Vol ∆P 1 ?

A Hands-on Introduction to Nanoscience: www.virlab.virginia.edu/Nanoscience_class/Nanoscience_class.htm Work done to reach a given pressure? It's NOT just proportional to amount of air added: Because as air was added, we had to fight increased pressure to add more! So increment of work to add increment of air increases with pressure: W (P)  P(Q) or from preceding page W (Q)  Q / Volume = Work to add MORE air when have Q in tank increases as value of Q So integrated energy expended in adding total quantity of air Q to tank is ∆ E stored  Q 2 / 2 Volume Recapping: P ~ Q / Volume and E ~ ½ Q 2 / Volume

A Hands-on Introduction to Nanoscience: www.virlab.virginia.edu/Nanoscience_class/Nanoscience_class.htm Translating back into the electrical world: VOLTAGE corresponds PRESSURECAPACITANCE corresponds to tank VOLUME Capacitance is measure of a capacitor’s ability to store charge For capacitors that are (at least on a local scale) more or less flat: C = A (  /d) Putting this all together, expect: FOR GAS TANK:FOR CAPACITOR: P ~ Q gas / Volume =>Voltage = Q charge / Capacitance E ~ Q gas 2 / 2 Volume=>E = Q charge 2 / 2 Capacitance (using Q for quantity of air stored / quantity of charge stored) Which can be re-expressed as Energy stored = ½ Capacitance x Voltage 2

An Introduction to Sustainable Energy Systems: www.virlab.virginia.edu/Energy_class/Energy_class.htm So now to make a supercapacitor: Supercapacitor's trick lays in making all those layers incredibly small But the two metal "plates" can't touch! So you somehow introduce an insulating spacer, giving total structure (from side): Or, for even better energy storage density Stacked design: 4 gaps => 4X capacitance

An Introduction to Sustainable Energy Systems: www.virlab.virginia.edu/Energy_class/Energy_class.htm Which can be manufactured more easily by: 1) Not trying to ADD the insulating spacer layer(s) But instead GROWING it from the metal plates. How? Can heat metal in oxygen to form its oxide: 2) Add second un-oxidized metal sheet: 3) Then, for a particularly easy assembly, roll the whole thing up: The above actually describes a conventional present day tantalum foil capacitor www.electronics- tutorials.ws/capacitor/cap_2.html

What voltage should this super-capacitor be designed for? Insulator has to be thick enough so that voltage across it will not arc through it Critical number is the insulator's "dielectric breakdown field" = Maximum electric field strength it can withstand without arcing What about just converting power system's 110 VAC to 110 VDC and using it? Is this large voltage the best choice? An argument in its favor of large voltages: Insulator will breakdown if exceed above dielectric breakdown field But electric field =  Voltage / thickness, so if increase voltage by X Must also increase insulator thickness by X => 1/X such capacitors will fit inside a given volume But, from above, energy per capacitor goes up as voltage squared (X 2 ) So energy stored in volume of capacitors increases linearly with their design voltage

What materials to choose? Many are hoping that plates can be made of all-carbon graphene sheets Which are only one atom thick And through which electricity flows extraordinarily easily => Plates would heat up less than normal metal Let's be similarly aggressive and assume insulator is all-carbon diamond Which has a dielectric breakdown of = 2000 MV / m ~ 50 - 100X larger than most "normal" insulators So 100 Volts / thickness of our diamond insulators must be < 2000 M-Volts / m Which means that we require: Thickness insulator > 100 nanometers (Robust thickness would also make it less vulnerable to flaws or dust inclusions!)

An Introduction to Sustainable Energy Systems: www.virlab.virginia.edu/Energy_class/Energy_class.htm Putting this together to figure out capacitance per volume: With graphene, the metallic plate layers would be < 1 nanometer thick But diamond insulator = 100 nm, so capacitor's thickness is almost all insulator: T capacitor ~ T insulator ~ 100 nm = 10 -7 meters A 1 meter cube could thus contain ~ a stack of 10 7 capacitors of size 1m x 1m And for each of those capacitors, its capacitance = A  / d Diamond's dielectric constant  = 5.7 x  o  o is the "permittivity of free space" = 8.85 x 10 -12 Farads/m So TOTAL capacitance per 1 meter cube is: C meter_cube = 10 7 x (1 m) 2 (5.7)(8.85 x 10 -12 Farads/m)/(10 -7 m) = 5044 Farads (Farad = Coulomb / Volt = Joule / Volt 2 )

Which, when charged to 100 Volts DC, gives: Energy = ½ Capacitance x Voltage 2 = ½ (5044 Farads)(100 Volts) 2 = 25.22 mega-Joules = 25.22 mega Watt-sec = 7.0 kW-hours = 25.22 mega Watt-sec = 7.0 kW-hours Far above we found that, ideally, in US we wanted to store/shift 887 GW-h daily So need this volume of super capacitors: 887 GW-h/7 kW-h => 1.26 x 10 8 m 3 = 1 m x (11.2 km) x (11.2 km) Huge, but not impossible An Introduction to Sustainable Energy Systems: www.virlab.virginia.edu/Energy_class/Energy_class.htm Peak Power Midnight Noon Midnight 80% Most effective power plants, always at full power Stored energy 887 GW-h

Sanity check / potential problems with super-capacitors: Volume ended up being smaller than I'd anticipated => Too much heat/volume? Electrical "resistance" heats things up as electrons flows in and out But choice of graphene is ~ ideal as it has extraordinary low resistance Little heat produced would also be diluted in much thicker insulator layers A bigger worry is my hypothetical diamond insulator Placing thin pieces of diamond, layer by layer, would take forever (10 7 !) So would be essential to figure out way of GROWING diamond on graphene Or of growing another extraordinarily high breakdown "dielectric" Because breakdowns of normal insulators are 50-100X lower If use more conventional insulator, thickness/per capacitor increases by 50-100X And energy stored per stack of capacitors drops by 50-100X per volume

Scenario 2c: Energy storage via super-batteries From lecture on Batteries and Fuel Cells, top candidate might be flow batteries: Which, recall, have this strange and complex configuration: With two different electrolytes circulated in (via pumps) from external tanks To a central cell with an "ion selective membrane" Plus simple metal electrodes to either side An Introduction to Sustainable Energy Systems: www.virlab.virginia.edu/Energy_class/Energy_class.htm

Central structure for the vanadium ion version: CENTER: H + consumed on left is replaced by H + selectively crossing membrane from right That is, on left (cathode) side:And on the right (anode) side: That is, on left (cathode) side:And on the right (anode) side: VO 2 +1 + 2 H + + e - => VO +2 + H 2 OV +2 => V +3 + e - These electrodes are just simple, dumb, inert, slabs of metal RIGHT SIDE: V +2 ion is pumped in e- is released to electrode Converting ion to V +3 LEFT SIDE: VO 2 +1 ion is pumped in It reacts with H + ion plus electron from electrode Becoming VO 2 +2 ion and releasing water

Figure: Electrochemical Energy Storage for Green Grid, Yang et al., Chemical Reviews 111, 3577–3613 (2011) 1) http://en.wikipedia.org/wiki/Molten_salt_battery Molten sodium beta alumina batteries were a second candidate With overall structure: - Central (anode) reservoir of molten sodium (green) - Membrane capable of passing Na + ions (gray) Typically: Al 2 O 3 "beta alumina" ceramic - Surrounding (cathode) outer cylinder (orange) Typically: Sulfur / Sodium Sulfide (Na 2 S x ) In the central anode:At the outer cathode: 2 Na => 2 Na + + 2 e - x S + 2 Na + + 2 e - => Na 2 S x With Na + ions formed in anode migrating through beta alumina toward cathode

Yielding Li ion battery structure and behavior: During CHARGING, Li is actually transferred from inside cathode to inside anode DISCHARGE reverses this: Li transferred from inside the anode to inside cathode: - Anode: Li desorbing and ionizing Cathode: Li absorbing and deionizing + + - + - Anode: Li absorbing and deionizing Cathode: Li dissolving and ionizing + + - + - -

1) http://www.sc.ehu.es/sbweb/energias-renovables/temas/almacenamiento/almacenamiento.html Comparative battery energy density data on all of these: The most complete data set I found (from the University del Pais Vasco, Spain): 1 Ion flow battery volume energy density: Up to 30 kW-h / m 3 Molten sodium battery volume energy density: Up to 270 kW-h / m 3 Li Ion battery volume energy density: Up to 350 kW-h / m 3 Li Ion Molten Sodium Flow

Using those battery numbers: As before, total storage goal is: 887 GW-h daily 887 GW-h daily Peak Power Midnight Noon Midnight 80% Most effective power plants, always at full power Stored energy 887 GW-h Li ion battery volume: 887 GW-h/(350 kW-h/m 3 ) = 2,534,000 m 3 = (1m)(1.6 km)(1.6 km) Molten Na battery volume: 887 GW-h/(270 kW-h/m 3 ) = 3,285,000 m 3 = (1m)(1.8 km)(1.8 km) Flow battery volume: 887 GW-h/(30 kW-h/m 3 ) = 29,566,000 m 3 = (1m)(5.4 km)(5.4 km) Volumes (or area of 1m tall piles) have shrunk! But those exotic batteries have still got to be affordable But those exotic batteries have still got to be affordable

Scenario 2d: Energy storage via Molten Salts In the earlier lecture on solar energy, I noted that: Solar Thermal plants naturally require a heat transfer fluid Now usually water but, more desirably, high temperature oils But I also noted the possibility of using molten salts, offering advantages that: They are non-toxic and non-flammable They melt at low temperature (~130 °C for Na / K / Ca salt mixture) Large THERMAL MASS => heating sucks in a huge amount of energy With that energy then available for return during the night Making molten salt use more plausible: It's been used in past nuclear reactors And would be used in some of the most promising "NextGen reactors" An Introduction to Sustainable Energy Systems: www.virlab.virginia.edu/Energy_class/Energy_class.htm

Here the important point is that almost all power plants = heaters So to store energy in the morning (for use in the evening): We'd just switch over what some of the power plants were heating To produce power for NOW:To store energy for LATER: To LATER convert stored energy to power:

How MUCH salt would U.S. would need for power storage? Batteries return more than 90% of electrical energy That is, electrical power in => electrical storage => electrical power out But molten salts return only about 70% of heat energy So here, heat in => heat storage => heat power out A 20% drop would seem to be a huge handicap If it were not for the fact that batteries are expensive, but salt is cheap We might even mine AND store heated salt in... SALT MINES! So where super-capacitors and super-batteries invoke high-tech breakthroughs Molten salt heat storage is almost quintessentially low-tech And thus likely to be both cheaper, and available sooner An Introduction to Sustainable Energy Systems: www.virlab.virginia.edu/Energy_class/Energy_class.htm

Digging up the details: As found in the U.S. National Renewable Energy Lab report on the Halotechnics Molten Salt Energy Storage Test Project, Emeryville CA: 35.45 kw-h of energy was stored in 0.19 m 3 of salt, returning 25.1 kw-h (71%) 1 Giving an energy storage capacity of 25.1 kW-h/0.19 m 3 = 132 kW/m 3 And above we found that, ideally, in US we wanted to store/shift 887 GW-h daily So need this volume of molten salt: 887 GW-h / (132 kW-h/m 3 ) => 6,719,696 m 3 = (1m)(2.6 km)(2.6 km) Larger than with some batteries But this is just a tank of salt! But this is just a tank of salt! 1) http://www.nrel.gov/docs/fy13osti/58595.pdf, page 61 Peak Power Midnight Noon Midnight 80% Most effective power plants, always at full power Stored energy 887 GW-h

Comparing our 2a-d energy storage scenarios: These scenarios were all for "best" power plants operating at continuous full power Storing 8% of off-peak power for later => 887 GW-h daily U.S. storage Taking what follows with a large "grain of salt," especially for my super capacitor/battery predictions: a) Pumped Hydro would require: (230 km) 2 of reservoir area For upper reservoirs 385 meters above lower reservoirs (as at Bath) Required area increases with decrease in vertical separation b) Capacitors might require: 1.26 x 10 8 m 3 = (1 m)(11.2 km) 2 c) Batteries would require: 2.5 to 30 x 10 6 m 3 = (1m){(1.5 km) 2 to (5 km) 2 } d) Molten Salt would require: 6,719,696 m 3 = (1m)(2.6 km) 2 An Introduction to Sustainable Energy Systems: www.virlab.virginia.edu/Energy_class/Energy_class.htm

1) http://www.nytimes.com/gwire/2010/10/15/15greenwire-doe-promotes-pumped-hydro-as-option-for-renewa- 51805.html?partner=rss&emc=rss&pagewanted=print Leading to the surprising result: While the results have stretched plausibility a bit, pumped storage hydro, super-capacitors, super-batteries and molten salt all appear to be semi-viable ways of implementing scenario #2 But why then does PSH now account for ~ 99% of large scale energy storage? And why is there currently so little energy storage of any type? Best guess is that demonstrated PSH is cheaper, but not yet cheap enough Which is supported New York Times 1 quote of then US Energy Secretary Chu: "Using pumped hydro to store electricity costs less than \$100 per kilowatt-hour and is highly efficient, Chu told his energy advisory board during a recent meeting. By contrast, he said, using sodium ion flow batteries, another option for storing large amounts of power, would cost \$400 per kWh and have less than 1 percent of pumped hydro’s capacity."

Confirming cost comparison data: U.S. National Renewable Energy Lab: "Hydrogen Energy Storage Overview" http://www.nrel.gov/hydrogen/pdfs/48360.pdf

An Introduction to Sustainable Energy Systems: www.virlab.virginia.edu/Energy_class/Energy_class.htm Scenario 3: Use of sustainable power sources Most renewable energy sources can't produce energy all of the time Examples include solar power, wind power, tidal power And when they DO produce, it's not when we most need their production! This is how we LIKE to consume energy (EIA data from second slide): www.eia.gov/todayinenergy/detail.cfm?id=12711 YELLOW midnight lines added

TEXAS: www.seco.cpa.state.tx.us/publications/renewenergy /windenergy.php These are data on actual wind speeds vs. time of day ONTARIO CANADA: www.omafra.gov.on.ca/english/engineer/facts/03- 047.htm NETHERLANDS: www.ekopower.nl/wind-energy- monitoring-data-logger-data- processing.htm WISCONSIN: www.windpowerweather.com/history?date=last2days

An Introduction to Sustainable Energy Systems: www.virlab.virginia.edu/Energy_class/Energy_class.htm Which, recalling that wind energy goes as wind velocity cubed: Would lead to a typical wind energy versus time of day plot something like this: That is, wind energy would peak sharply in the late afternoon 100% Midnight Noon Midnight

An Introduction to Sustainable Energy Systems: www.virlab.virginia.edu/Energy_class/Energy_class.htm Solar energy peaks midday, but also varies by season: Starts at sunrise, peaks at about noon, ends at sunset And lasts longer and is more intense in the summer: 100% Midnight Noon Midnight Summer Winter

What we now need/want in the way of daily power: What we'd get from a wind plant:What we'd get from a solar plant: So use of renewables => BIG NEED for storage: 100% Midnight Noon Midnight 100% Midnight Noon Midnight 100% Midnight Noon Midnight

Let me (somewhat crudely) approximate either as half sine wave: Area under half sinusoid = Amplitude x day / Pi = P peak_renewable x day / Pi But if this to be all our power, this must = 80% P peak_use x day Implying P peak_renewable = 0.8 Pi P peak_use = 2.5 P peak_use An Introduction to Sustainable Energy Systems: www.virlab.virginia.edu/Energy_class/Energy_class.htm Peak Power Use Midnight Noon Midnight 80% Power of Renewable at Peak

Whoops, so figure should be more like: But I now need to store and shift ALL energy above green "use" line! Yellow Base ~ (0.8 P peak_use )(d/2) = (0.4 d)(P peak_renewable /2.5) = 0.16 P peak_renewable d Fraction that must be stored and shifted = 1 - Yellow base / All = 1 – (0.16 P peak_renewable d) / (P peak_renewable d day / Pi) = 1 – 0.16 Pi = 50% An Introduction to Sustainable Energy Systems: www.virlab.virginia.edu/Energy_class/Energy_class.htm Peak Power Use Midnight Noon Midnight Power of Renewable at Peak 80% of Peak Power Use Midnight Noon Midnight Power of Renewable at Peak ≈

Whoops squared! Scenario #2: Flat power production: Must store and shift 8% of energy to accommodate our power use curve Scenario #3: Renewable production, approximated as ½ day long half sine wave: Must store and shift 50% of energy to accommodate our power use curve If we use a single such renewable, storage need increases by ~ 6X 6X all of my earlier calculations which already stretched plausibility limits Two renewables, peaking at different times (e.g. solar and wind) help a little Two renewables, coming from different time zones, help a little more But it really suggests that we will have to retain some 24/7 power sources Producing a good fraction of the Scenario #1 base load It's going to be really hard to abandon hydro & nuclear power!

Credits / Acknowledgements Some materials used in this class were developed under a National Science Foundation "Research Initiation Grant in Engineering Education" (RIGEE). Other materials, including the "UVA Virtual Lab" science education website, were developed under even earlier NSF "Course, Curriculum and Laboratory Improvement" (CCLI) and "Nanoscience Undergraduate Education" (NUE) awards. This set of notes was authored by John C. Bean who also created all figures not explicitly credited above. Copyright John C. Bean (2015) (However, permission is granted for use by individual instructors in non-profit academic institutions) An Introduction to Sustainable Energy Systems: www.virlab.virginia.edu/Energy_class/Energy_class.htm

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