Micro-turbines,Fuel Cells and Energy Storage Schemes

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Energy Storage Schemes

Distributed Generation Technologies

Microturbines

2Photovoltaics

Microturbines

Fuel Cells

Advanced TurbinesReciprocating Engines

Wind turbines

Basic Gas Turbine

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• Fresh air is drawn into a compressor where spinning rotor blades compress the air, elevating its temperature and pressure

• This hot, compressed air is mixed with fuel (natural gas, LPG, kerosene) and burnt in the combustion chamber

• The hot exhaust gases are expanded in a turbine and released to the atmosphere

• The compressor and turbine shares a common shaft, so that a portion, typically more than half, of the rotational energy created by the turbine is used to power the turbine

Combined-Cycle Power Plants

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• Carnot efficiency reveals that the maximum possible efficiency of a heat engine is limited by a low-temperature sink as well as by a high-temperature source

• With exhaust gases leaving a gas turbine at temperatures frequently above 5000 C, overall efficiencies are low, in the 30% range

• A heat recovery steam generator (HRSG) can capture some of that waste heat for process steam whereby the gas turbine waste heat can be used to power a second stage steam turbine

• Working together such combined cycle power plants have achieved efficiencies above 50%

Microturbines• Recently, a new generation of

very small gas turbines entered the marketplace. Often referred to as microturbines, these units generate anywhere from about 500 W to several hundred kW.

• Incoming air is compressed and sent through a heat exchanger called recuperartor, where its temperature is

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where its temperature is elevated by the hot exhaust gases. By preheating the compressed incoming air, the recuperator helps boost the efficiency

��The hot, compressed air is mixed with fuel in the combustion chamber The hot, compressed air is mixed with fuel in the combustion chamber and is burnt. The expansion of hot gases through the turbine spins and is burnt. The expansion of hot gases through the turbine spins the compressor and the generatorthe compressor and the generator��The exhaust gas is released to the atmosphere after The exhaust gas is released to the atmosphere after transferring much of its heat to the incoming compressed air in the transferring much of its heat to the incoming compressed air in the recuperatorrecuperator

ExampleThe Elliott TA 100A microturbine at its full 105 kW output burns 1.24 X 106 Btu/hr ( 1Btu = 1055J ) of natural gas. Its waste heat is used to supplement a boiler used for water and space heating in anapartment house. The design calls for water from the boiler at 1200 F to be heated to 1400 F and returned to the boiler. Thesystem operates in this mode for 8000 hours per year.a) If 47% of the fuel energy is transferred to the boiler water, what

should the water flow rate be ?

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should the water flow rate be ?b) If the boiler is 75% efficient, and it is fueled with natural gas

costing $6 per million Btu, how much money will the microturbine save in displaced boiler fuel ?

c) If utility electricity costs $0.08/kWh, how much will the microturbine save in avoided utility electricity ?

d) If O&M is $1500/yr, what is the net annual savings for the microturbine ?

e) If the microturbine costs $220,000, what is the ratio of annual savings to initial investment (called the initial rate of return) ?

Solutiona) The heat Q required to raise a substance with specific

heat c and mass flow rate by a temperature difference of ∆T is . Since it takes 1 Btu to raise 1 lb of water by 10 F, and one gallon of water weighs 8.34 lb, we can write,Water flow rate = 0.47X 1.24 X 106 Btu/h X 1/60 h/min

m&

TCmQ ∆= &

m&

1 Btu/lb0F X 200F X 8.34 lb/gal

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= 58 gpm

b) The fuel displaced by not using the 75% efficient boiler is worth,Fuel savings = 0.47 X 1.24 X 106 Btu/h X $6 X 8000 h/yr

= $37,300/yr

1 Btu/lb F X 20 F X 8.34 lb/gal

0.75 106 Btu

Solution

c) The utility electricity savings = 105 kW X 8000 h/yr X

$0.08/kWh = $67,200/yr

d) The cost of fuel for the microturbine is

= 1.24 X 106Btu/h X ($6/106 Btu) X 8000 h/yr= $59,520/yr

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= 1.24 X 10 Btu/h X ($6/10 Btu) X 8000 h/yr= $59,520/yr

So the net annual savings of the microturbine, including $1500/yr in O&M, is = ($37,300 + $67,200) - $59,520 -$1500 = $43,480/yr

e) The initial rate of return on this investment would be

= Annual savings = $43,480/yr = 0.198 = 19.8%/yr

Initial investment $220,000

Fuel Cells

• Why the Hydrogen hype ?

Energy density of some materials

kWh/kg

Gasoline 14

Lead/acid batteries

0.04•Hydrogen is the only non-polluting fuel available

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Flywheel, steel 0.05

Flywheel, Carbon fiber

0.2

Flywheel, fused Silica

0.9

Hydrogen 38

Compressed air 2/m3

non-polluting fuel available

• High energy density

• Hydrogen can be produced either from Methane or by splitting water

Fuel Cells

In the archetypal example of a hydrogen/oxygen proton exchange membrane fuel cell (PEMFC), a proton-conducting polymer membrane, (the electrolyte), separates the anode and cathode sides.

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anode and cathode sides.On the anode side, hydrogen diffuses to the anode catalyst where it dissociates into protons and electrons.

The protons are conducted through the membrane to the cathode, but the electrons are forced to travel in an external circuit (supplying power) because the membrane is electrically insulating. On the cathode catalyst, oxygen molecules react with the electrons (which have traveled through the external circuit) and protons to form water. In this example, the only waste product is water vapor and/or liquid water.

Parts of a fuel cellThere are four basic elements of a PEMFC:

The anode, the negative post of the fuel cell, has several jobs. It conducts the electrons that are freed from the hydrogen molecules so that they can be used in an external circuit. It has

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be used in an external circuit. It has channels etched into it that disperse the hydrogen gas equally over the surface of the catalyst.

The cathode, the positive post of the fuel cell, has channels etched into it that distribute the oxygen to the surface of the catalyst. It also conducts the electrons back from the external circuit to the catalyst, where they can recombine with the hydrogen ions and oxygen to form water.

Parts of a fuel cell

• The electrolyte is the proton exchange membrane. This specially treated material, which looks something like ordinary kitchen plastic wrap, only conducts positively charged ions. The membrane blocks electrons.

• The catalyst is a special material that facilitates the

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• The catalyst is a special material that facilitates the reaction of oxygen and hydrogen. It is usually made of platinum powder very thinly coated onto carbon paper or cloth. The catalyst is rough and porous so that the maximum surface area of the platinum can be exposed to the hydrogen or oxygen.

Parts of a fuel cell

Figure on page 10 shows the pressurized hydrogen gas (H2) enteringthe fuel cell on the anode side. This gas is forced through the catalystby the pressure. When an H2 molecule comes in contact with the platinum on the catalyst, it splits into two H+ ions and two electrons (e-). The electrons are conducted through the anode, where they make their way through the external circuit (doing useful work such as turning

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a motor) and return to the cathode side of the fuel cell.

Meanwhile, on the cathode side of the fuel cell, oxygen gas (O2) isbeing forced through the catalyst, where it forms two oxygen atoms. Each of these atoms has a strong negative charge. This negative charge attracts the two H+ ions through the membrane, where they combine with an oxygen atom and two of the electrons from the external circuit to form a water molecule (H2O).

Phosphoric Acid (PAFC).Phosphoric acid fuel cells are commercially available today. Hundreds offuel cell systems have been installed in 19 nations - in hospitals, nursing homes, hotels, office buildings, schools, utility power plants, landfills and waste water treatment plants. PAFCs generate electricity at more than 40% efficiency –and nearly 85% of the steam this fuel cell

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produces is used for cogeneration - this compares to about 35% for the utility power grid in the United States. Phosphoric acid fuel cells use liquid phosphoric acid as the electrolyte and operate at about 450°F.

One of the main advantages to this type of fuel cell, besides the nearly 85% One of the main advantages to this type of fuel cell, besides the nearly 85% cogeneration efficiency, is that it can use impure hydrogen as fuel. PAFCs cogeneration efficiency, is that it can use impure hydrogen as fuel. PAFCs can tolerate a CO concentration of about 1.5 percent, which broadens the can tolerate a CO concentration of about 1.5 percent, which broadens the choice of fuels they can use. If gasoline is used, the sulfur must be removed.choice of fuels they can use. If gasoline is used, the sulfur must be removed.

Proton Exchange Membrane (PEM)

• These fuel cells operate at relatively low temperatures (about 175°F), have high power density, can vary their output quickly to meet shifts in power demand, and are suited for applications, such as in automobiles, where quick startup is required.

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where quick startup is required. According to the U.S. Department of Energy (DOE), "they are the primary candidates for light-duty vehicles, for buildings, and potentially for much smaller applications such as replacements for rechargeable batteries." This type of fuel cell is sensitive to fuel impurities. Cell outputs generally range from 50 watts to 75 kW.

Solid Oxide Fuel Cells(SOFC)

• Solid oxide fuel cells use a hard, non-porous ceramic compound as the electrolyte, and operate at very high temperatures - around 1800°F. One type of SOFC uses an array of meter-long tubes, and other variations include a compressed disc that resembles

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other variations include a compressed disc that resembles the top of a soup can. Tubular SOFC designs are closer to commercialization and are being produced by several companies around the world. SOFCs are suitable for stationary applications as well as for auxiliary power units (APUs) used in vehicles to power electronics.

Molten Carbonate (MCFC)Molten carbonate fuel cells use anelectrolyte composed of a moltencarbonate salt mixture suspended in aporous, chemically inert matrix, and operate at high temperatures –approximately 1,200ºF. They requirecarbon dioxide and oxygen to be

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carbon dioxide and oxygen to be delivered to the cathode. To date, MCFCshave been operated on hydrogen, carbonmonoxide, natural gas, propane, landfillgas, marine diesel, and simulated coalgasification products. 10 kW to 2 MWMCFCs have been tested on a variety offuels and are primarily targeted to electricutility applications.

Summary of Fuel Cell Types

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Fuel Cell Thermodynamics –Enthalpy of Formation

The fuel cell shown on page 10 is described by the following pair of Reactions:

Anode: H2 � 2H+ + 2e-

Cathode: (1/2)O2 + 2H+ + 2e- � H2O

When combined, the above equations result in the equation for

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When combined, the above equations result in the equation for combustion of hydrogen:

H2 + (1/2)O2 � H2O

One way to think about enthalpy is that it is a measure of the energy that it takes to form that substance out of its constituent elements. The difference between the enthalpy of the substance and the enthalpies of its elements is called the enthalpy of formation (∆H).

Enthalpy of formation for selected substances (At 250C and 1 atm)

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Electrochemical reactions in fuel cells

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Enthalpy of formation of water

Analyzing the reactions on page 21, the enthalpies of H2 and O2 are zero so the enthalpy of formation is simply the enthalpy of the resulting H2O. Noticing table on page 20, the enthalpy of H2O depends on whether it is liquid water or gaseous water vapor. When the result is liquid water:

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H2 + (1/2)O2 � H2O(l) ∆H= -285.8 kJ

When the resulting product is water vapor:

H2 + (1/2)O2 � H2O(g) ∆H= -241.8 kJ

The negative signs for the enthalpy changes in above tells us these reactions are exothermic: that is, heat is released.

ExampleFind the HHV of methane (CH4) in kJ/mol and kJ/kg when it is oxidized to CO2

and liquid H2O.

SOLUTIONThe reaction is written below, and beneath it are enthalpies. Notice that we must balance the equation so that we know how many moles of each constituent are involved.CH4(g) + 2O2(g) CO2(g) + 2 H2O(l)(-74.9) 2X(0) (-393.5) 2X(-285.8)

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(-74.9) 2X(0) (-393.5) 2X(-285.8)

Notice, too, that we have used the enthalpy of liquid water to find HHV.The difference between the total enthalpy of the reaction products and the reactants is

∆H = [(-393.5) + 2 X (-285.8)] – [(-74.9) + 2 X (0)] = -890.2 kJ/mol of CH4

Since the result is negative, heat is released during combustion; that is, it is exothermic. The HHV is the absolute value of ∆H which is 890.2 kJ/mol.Since there are 12.011 + 4X1.008 = 16.043 g/mol of CH4, the HHV can be Written as

HHV = (890.2 kJ/mol) / (16.403 g/mol) X1000 g/kg = 55,490 kJ/kg

Entropy and Theoretical Efficiency of Fuel Cells

The reactions H2 + 1/2O2 � H2O(l) and H2 + 1/2O2 � H2O(g) act as a source of enthalpy Hthat can be converted to heat and work. The cell generates an amount of electricity We and rejects an amount of thermal energy Q to its environment. Since there is heat transfer and it is a real system,

FuelCell

Enthalpyin H

We

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Since there is heat transfer and it is a real system, there must be an increase in entropy. We can write the entropy appearing in the rejected heat as ∆S=Q/T. However there is no entropy associated with the work done We as there is no heat transfer. According to 2nd law of thermodynamics,Entropy gain ≥ Entropy loss

Q/T + ΣSproducts ≥ ΣSreactants which leads to

Q ≥ T(ΣSreactants - ΣSproducts )

Cell

RejectedHeat Q

EnthalpyOutput

Entropy and Theoretical Efficiency of Fuel Cells

Above equation tells us that the minimum amount of heat that must appear in the fuel cell. That is we cannot convert all of the fuel’s energy into electricity – we are stuck with some thermal losses. We can now easily determine the maximum efficiency of the fuel cell. From the figure in page 24, the enthalpy

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the fuel cell. From the figure in page 24, the enthalpy supplied by the chemical reaction ∆H equals the electricity produced We plus the heat rejected Q:∆H = We + QSince it is the electrical output that we want, we can writethe fuel cell’s efficiency asη= We/∆H = (∆H – Q)/∆H = 1 – Q/∆H

Example

Suppose a fuel cell that operates at 250C (298K) and 1 atm

forms liquid water (that is we are working with the HHV of

the hydrogen fuel):

H2 + (½ )O2 H2O(l) ∆H = -285.8 kJ/mol of H2

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a) Find the minimum amount of heat rejected per mole of H2.

b) What is the maximum efficiency of the fuel cell ?

Solutiona) From the reaction, 1 mole of H2 reacts with ½ mole of O2 to

produce 1 mole of liquid H2O. The loss of entropy by the reactants per

mole of H2 is found as follows:

ΣSreactants = 0.13 kJ/mol-K X 1mol H2

+ 0.205 kJ/mol-K X (1/2) mol O2 = 0.2325 kJ/K

The gain in entropy in the product water is

ΣSproduct = 0.0699 kJ/mol-K X 1mol H2O = 0.0699 kJ/K

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ΣSproduct = 0.0699 kJ/mol-K X 1mol H2O = 0.0699 kJ/K

The minimum amount of heat released during the reaction

is therefore

Qmin = T( ΣSreactants - ΣSproduct ) = 298 K (0.2325 – 0.0699) kJ/K

= 48.45 kJ per mole H2

b) The enthalpy made available during the formation of liquid water from

H2 and O2 is H = 285.8 kJ/mol of H2.

The maximum efficiency possible occurs when Q is a minimum; thus

ηmax = 1 – Qmin/H = 1 – 48.45/285.8 = 0.83 = 83%

Gibbs Free Energy and Fuel Cell Efficiency

The chemical energy released in a reaction can be thought of as consisting of two parts: an entropy-free part, called free energy ∆G, thatcan be converted directly into electrical or mechanical work, plus a partthat must appear as heat Q. The free energy G is the enthalpy H created by the chemical reaction, minus the heat that must be liberated, Q=T∆S, to satisfy the 2nd law.

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The Gibbs free energy ∆G corresponds to the maximum possible, entropy-free, electrical output from a chemical reaction. It can be foundat STP using Table on page 20 by taking the difference between the sum of the Gibbs energies of the reactants and the products:∆G = ΣGproducts - Σ Greactants

This means that the maximum possible efficiency is just the ratio of theGibbs free energy to the enthalpy change in the chemical reaction.

ExampleWhat is the maximum efficiency at STP of a proton-

exchange –membrane (PEM) fuel cell based on the higher

heating value (HHV) of hydrogen ?

SOLUTION

The HHV corresponds to liberated water in the liquid state

so that the appropriate reaction is

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so that the appropriate reaction is

H2 + (½ )O2 H2O(l) ∆H = -285.8 kJ/mol of H2

The Gibbs free energy of the reactants H2 and O2 are both

zero, and that of the product, liquid water, is -237.2 kJ.

Therefore, ∆G = -237.2 –(0+0) = -237.2 kJ/mol.

So, ηmax = ∆G / ∆H = -237.2/(-285.8) = 0.83 = 83%

This is the same answer that we found in the previous example.

Electrical Output of an Ideal Cell

The Gibbs free energy ∆G is the maximum possible amount of work orelectricity that a fuel cell can deliver. For an ideal hydrogen fuel cell,the maximum possible electrical output is therefore equal to the magnitude of ∆G. For a fuel cell producing liquid water, this makes themaximum electrical output at STP equal to We= | ∆G | = 237.2 kJ/mol of H2.

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2

To use above equation we just have to adjust the units so that the electrical output We will have the conventional electrical units of volts, amps and watts. For each mole of H2 into an ideal fuel cell, two electrons will pass through the electrical load (see figure on page 10).

Electrical Output of an Ideal Cell

If n is the rate of flow of hydrogen into the cell (in mol/s), we can therefore write that the current flowing through the load will beI(A) = n (mol/s). 6.022X1023(molecules of H2/mol).

2 (electrons/molecules H2) . 1.602X10-19(coulombs/electron)= 192945n A

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The ideal power (watts) delivered to the load will be 237.2 kJ/mol of H2

times the rate of hydrogen use:P(W) = 237.2 (kJ/mol) . n (mol/s) . 1000

= 237200n W

And the reversible voltage produced across the terminals of this ideal fuel cell will be,VR = P(W)/ I(A) = 237200n/192945n

= 1.229 V

Electrical Characteristics of Fuel Cells

Usually real fuel cells do not deliver the full Gibbs free energy. Activation losses result from the energy required by the catalysts to initiate the reactions. The relatively slow speed of reactions at the cathode, where oxygen combines with protons and electrons to form water, tends to limit the fuel cell power. Ohmic losses result from current passing through the internal resistance posed by the electrolyte

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membrane, electrodes, and various interconnections in the cell. Another loss, referred to as fuel crossover, results from fuel passing through the electrolyte without releasing its electrons to the external circuit. And finally, mass transport losses result when hydrogen andoxygen gasses have difficulty reaching the electrodes. This is especially true at the cathode if water is allowed to build up, clogging the catalyst. For these and other reasons, fuel cells, in general, generate only about 60-70% of the theoretical maximum.

Electrical Characteristics of Fuel Cells

Figure shows the relationship between current and voltage for a typical fuel cell. Notice that the voltage at zero current, called the open circuit voltage, is a little less than 1V, which is about 25%lower than the theoretical value

0.9V

1.229 V

0.85V

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lower than the theoretical valueof 1.229 V. Over most of the length of the fuel cell I-V graph,voltage drops linearly as current increases. This suggests a Simple equivalent circuit consistingof a voltage source in series with some internal resistance.

Fitting theFitting the VV--I curve in the ohmic region for the fuel cell, I curve in the ohmic region for the fuel cell, V = 0.85 V = 0.85 –– 0.25J = 0.85 0.25J = 0.85 –– (0.25/A)I (0.25/A)I where A is the cell area (cmwhere A is the cell area (cm22), I is current (A) and J is current density(A/cm), I is current (A) and J is current density(A/cm22).).

0.25 V/(A/cm2)

Energy storage

• Why energy storage?

• Utility load levelling: to improve load factors, reduce pollution in populated urban areas and to make better use of available plants and fuels

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make better use of available plants and fuels

• Utilization of renewable energy in its various forms to relieve the burden on finite fossil fuel resources and to improve the environment

• Storage for remote users

• Uninterruptible power supplies

Pumped Storage SystemsPumped Storage balances supply and demand by moving water between reservoirs (70% efficient for water) construct a reservoir above a source of H2O.In times of slack demand, use electricity to pump water to the reservoir. When you need

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reservoir. When you need electricity, run water through turbines to generate it. It can beused to increase base load, decrease peak load....is this good?

Pumped water storagePumped water storage is used in Michigan (2000 MWe plant in Ludington) is used in Michigan (2000 MWe plant in Ludington) Pennsylvania (870 MWe). Pennsylvania (870 MWe). Can also do this for Can also do this for airair at about 90% efficiency. In 2000 the US had about 20GW of at about 90% efficiency. In 2000 the US had about 20GW of pumped storage capability.pumped storage capability.

Superconductors• What is a superconductor?

Materials in which the electrical resistance disappears at some critical temperature (Mercury Tc = 4.2K; Niobium-Tin Tc = 23K).

• Why superconductors? Joule losses (I2R)! True workable room temperature superconductors would revolutionize technology.

• Efficiency of electric generators

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• Efficiency of electric generators-Conventional electric generators are 98.6% efficient.-Superconducting generators will be ~99.5% efficient. Is this gain worth it?

• Room-temperature superconducting transmission lines would save10-15% for short-distance transmission2% for long-distance transmission.

• What does this mean?-1% improvement in transmission $1 billion in savings.

Superconducting Magnetic Energy Storage (SMES)

A typical SMES configuration comprisesof two 6-pulse thyristor bridges series connected to the superconducting coil on the DC part of the bridge and coupled through an AC transformer to a power system on the AC side

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Superconducting

winding

to a power system on the AC side of the bridge.

The energy stored in the coil is magnetic energy given by (1/2)LI2.The current can be increased to a very high value due to the superconductivenature of the coil to provide superiorenergy storage capacity.

Flywheels• Flywheels are an old technology. Storing energy in a

slow moving heavy object has been around for a long time.

• Example: The car engine

So what is new about flywheels?....

• It is the material....strong plastics and epoxy allow edge

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• It is the material....strong plastics and epoxy allow edge speed of 1400 m/s!

• It is now feasible to create cars which store energy in flywheels. These could be powered by electricity, natural gas, etc.

• The presently available flywheels for cars rotate at 30,000 rpm and can deliver 0.55 kWhe. These may be considered flywheel batteries. The Prius uses flywheels to store energy!

Flywheels

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Flywheels

• A wheel winds up through some system of gears and then delivers rotational energy until friction dissipates it

• Stored energy = sum of kinetic energy of individual mass elements that comprise the flywheel

• Kinetic Energy = (1/2)*Iω2

• I = moment of inertia ability of an object to resist changes in its rotational velocity

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changes in its rotational velocity • ω = rotational velocity • To optimize the energy-to-mass ratio, the flywheel needs

to spin at the maximum possible speed. This is because kinetic energy only increases linearly with mass but goes as the square of the rotation speed.

Flywheels

• Rapidly rotating objects are subject to centrifugal forces that can rip them apart. Centrifugal force for a rotating object goes as: MRω2

• Thus, while dense material can store more energy it is also subject to higher centrifugal force and thus fails at lower rotation speeds than low density material.

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lower rotation speeds than low density material. • Tensile Strength is More important than density of

material. • Long rundown times are also required. Frictionless

bearings and a vacuum to minimize air resistance can result in rundown times of 6 months steady supply of energy.

• Flywheels are about 80% efficient (like hydro) • Flywheels do take up much less land than pumped hydro

systems

Batteries

• The oldest electrical storage device is the battery. The lead-acid battery in a car is capable of producing 50-100W/kg for a total stored energy of 25-35 Wh/kg.

• However lead is expensive. Other alternatives include:

• - sodium-sulfur battery (100 kWh - GE)- zinc-chloride prototype (50 Wh - Gulf)- lithium-aluminum (100 Wh/kg - ANL)

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- lithium-aluminum (100 Wh/kg - ANL)- lithium-water (55 Wh/kg - Canada)

• Batteries comprise two electrode systems and an electrolyte, placed together in a special container and connected to an external circuit or load. These two electrodes, fitted on both sides of an electrolyte and exchanging ions with the electrolyte and electrons with the external circuit, are called anode and cathode respectively.

• Today batteries are mainly used in emergency standby supplies duty for DC auxiliaries.