green electric energy lecture 13

8/14/2019 Green Electric Energy Lecture 13

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Announcements

Be reading Chapter 4, test covers through Section 4.5

First exam is Oct 8 in class (as specified on syllabus). Note that the old Exam 1from last semester is on Compass.

After test be reading Chapter 5

Homework 6 is due Oct 15. It is 4.9, 5.2, 5.4, 5.6, 5.11


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In the News: Prof. Chapman,Harnessing Human Power


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Wave Power

The potential energy available waves is quite high, withsome estimates up to 2 TW (2000 GW) worldwide.

The potential wave power per meter varies with the square of the wave height and linearly with the period.

A 3 meter wave with an 8 second period produces about 36 kW/m, while a 15 meter wave with a 15 second period

produces about 1.7 MW/m.

Worlds largest wave park is

in Portugal, with capacity of

2.25 MW; each of three devices is142m long, and has a diameter of

3.5 m, and uses 700 metric tons of steel
http://upload.wikimedia.org/wikipedia/commons/c/cc/Pelamis_bursts_out_of_a_wave.JPG


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Pelamis Wave Power

Source: www.pelamiswave.com


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Tidal Power

There is tremendous potential energy in harnessing theearths tides and is predictable, but like wave energy it

is very difficult to economically achieve.

The largest tidal power plant in the world is the 240MW (max) La Rance in France, built in the 1960s. Average power generation is 68 MW

A 330m dam contains a 22 square km basin, with average

tides of 8m.


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Tidal Power

No new tidal plants have been constructed since 1960s

The two main approaches to tidal power are 1) dams (barrages) across a tidalestuary (harnessing a height difference like conventional hydro), 2) tidal stream

systems which work similar to wind turbines

Tidal stream systems are seen as

more viable since they do notrequire construction of a dam.

Verdant Power has recently

installed such a system in the

NYC East River

Each turbine has about 35 kW Max Capacity


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Fuel Cells

Convert chemical energy contained in a fuel directlyinto electrical power

Skip conversion to mechanical energy, not

constrained by Carnot limitsChemical energy

Heat

Mechanical energy

Electrical energy

Chemical energy

Electrical energy

Conventional

Combustion

Fuel Cells


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Fuel Cells

Up to ~65% efficiencies

No combustion products (SOX,CO) although there maybe NOX at high temperatures

Vibration free, almost silent can be located close to theload

Waste heat can be used for cogeneration

Byproduct is water

Modular in nature


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Fuel Cells - History

Developed more than 150 years ago Used in NASAs Gemini earth-orbiting missions,

1960s

tp://scienceservice.si.edu/pages/059017.htm

For more information on thehistory of fuel cells, see the

Smithsonian project-

http://americanhistory.si.edu

/fuelcells/

http://americanhistory.si.edu/fuelcells/pem/pem3.htm


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Fuel Cells - History

http://americanhistory.si.edu/fuelcells/pem/pem5.htm

http://www.fuelcells.org/basics/apps.html


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Fuel Cells- Basic Operation

Protons diffuse though electrolyte so cathode is positive with respect to anode

Anode CathodeElectrolyte

2 2 2 H H e+

+2 2

12 2

2O H e H O+ + +

2H+

I

Electrical Load

Catalyst


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Fuel Cells- Basic Operation

Combined anode and cathode reactions:

This reaction is exothermic- it releases heat

A single cell only produces ~0.5V under normal operatingconditions, so multiple cells are stacked to build up the voltage

2 2 2 H H e+

+

2 2

12 2

2O H e H O+ + +

2 2 21 (4.20)2

H O H O+


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Fuel Cells

How much energy is liberated and how much can be converted toelectricity?

Need to talk about enthalpy, entropy, and free energy

Enthalpy - Sum of internal energy U and its volume V and

pressure P

U = internal energy, microscopic properties

PV = observable, macroscopic energies

Units kJ/mole

H U PV = +


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Enthalpy

A measure of the energy it takes to form the substance outof its constituent elements

For fuel cells, changes in chemical energy are of interestand those are best described in terms of enthalpy changes

As with potential energy, we describe enthalpy with respectto a reference (it is the change that matters)

At Standard Temperature and Pressure (STP)= 25C, 1 atm,stable form of element has zero enthalpy


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Enthalpy of Formation

Enthalpy of formation - difference between enthalpy of thesubstance and enthalpies of its constituent elements

Exothermic heat is liberated, enthalpy in final products isless than reactants, enthalpy of formation is negative,

chemical energy of substance is less than that of itsconstituents

Endothermic heat is absorbed

Depends on state (liquid, solid, gas), see Table 4.6


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Entropy and Fuel Cells

How much of the energy can be converted directlyinto electricity?

Well use entropy concepts to develop the

maximum efficiency of a fuel cellEnthalpy in

H

Enthalpy outWe

Rejected heat

Q

Fuel Cell

Fig. 4.28- Energy Balance for a Fuel Cell


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Assume isothermal

Q = heat released

Net entropy must increase

QS

T =

2 2 2

1

2 H O H O Q+ +

Enthalpy in

H

Enthalpy out

We

Rejected heat

Q

Fuel Cell

products reactants

QS S

T+

Entropy gain Entropy loss


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From

the minimum heat we can release is

Now find maximum efficiency

products reactants(4.28)Q

S ST

+

( )min reactants productsQ T S S =

Enthalpy in

H

Enthalpy out

We

Rejected heat

Q

Fuel Cell

(4.30)e H W Q= +1 (4.31)e

W Q

H H = =

minmax 1

Q

H =

Theoretical maximum

can be quite high

(> 80%)


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Gibbs Free Energy and Fuel Cells

Consider the chemical energy released in a reaction as havingtwo parts:

1) entropy-free part (the work) called free energy G

2) heat Q

G = H-Q=H-TS G is the maximum entropy-free output (work) from a chemical

reaction

Then products reactantsG G G =

eW

H = max

G

H

=


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Output of an Ideal Fuel Cell

Equal to the magnitude of G

From Table 4.6,

Maximum electrical output at STP is

n = rate of hydrogen flow into the cell, mol/sec

2 2 2

1

2 H O H O+ 237.2 kJ/molG =

237.2 kJ/moleW G= =

[ ] [ ] [ ] [ ]237.2 kJ/mol mol/sec 1000 J/kJ 237, 200 P W n n= =


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Types of Fuel Cells

Proton Exchange Membrane Fuel Cells (PEMFC) Direct Methanol Fuel Cells (DMFC)

Phosphoric Acid Fuel Cells (PAFC)

Alkaline Fuel Cells (AFC) Molten-Carbonate Fuel Cells (MCFC)

Solid Oxide Fuel Cells (SOFC)


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Types of Fuel Cells


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Hydrogen Production

Obtaining a supply of hydrogen of sufficient quality and at areasonable cost is difficult

Critical to solve this before fuel cells can become widelydeployed

Main technologies are steam reforming of methane (SMR) andpartial oxidation (POX)

Generation IV nuclear reactors could be used as well (whenthey become available)

E E i C t


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The economic evaluation of a renewable energy resourcerequires a meaningful quantification of cost elements

fixed costs

variable costs

We use engineering economics notions for this purposesince they provide the means to compare on a consistent

basis

two different projects; or, the costs with and without a given project

Energy Economic Concepts(From Prof. Gross)


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Basic notion: a dollar today is not the same as adollar in a year

We represent the time value of money by the

standard approach of discounted cash flows The notation isP = principal

i = interest value

The convention we use is that payments occur at theend of each period

Time Value of Money


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loanP for 1 yearrepayP+ iP = P(1 + i) at the end of 1 year

year0 P

year 1 P(1 + i)

loanP forn years

year0 P

year 1 (1 + i) P repay/reborrow

year 2 (

1 + i

)2

P repay/reborrowyear 3 (1 + i)3P repay/reborrow

.

yearn (1 + i)nP repay

Simple Example

M M M


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Compound Interest

e.o.p. amount owed interest for

next period

amount owed for next period

0 P Pi P + Pi = P(1+i)

1 P(1+i) P(1+i) i P(1+i) + P(1+i) i = P(1+i)2

2 P(1+i)2

P(1+i)2

i P(1+i)2

+ P(1+i)2

i = P(1+i)3

3 P(1+i)3 P(1+i)3 i P(1+i)3 + P(1+i)3 i = P(1+i)4

n-1 P(1+i)n-1 P(1+i)n-1 i P(1+i)n-1 + P(1+i)n-1 i = P(1+i)n

n P(1+i)n

M M

The value in the last column for the e.o.p. (k-1) provides the value in

the first column for the e.o.p.k(e.o.p. is end of period)


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Terminology

( )1 nF P i+compoundinterest

Lump sum repayment at theend of n periods. F is

called the future worth, while

P is called the present worth

Need not be integer-valued


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Terminology

We call (1+ i)n the single payment compound amountfactor

We define

and

is the single payment present worth factor Fis called the future worth;Pis called the present worth

or present value at interest i of a future sumF

( ) 11 i +@( )1 nn i = +


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Example 1

Consider a loan of$4,000 at 8% interest to be repaid intwo installments

$ 1,000 and interest at the e.o.y. (end of year) 1

$ 3,000 and interest at the e.o.y. 4

The cash flows are e.o.y. 1: 1000 + 4000 (.08) = $ 1,320

e.o.y. 2: 3000 (1 + .08 )3

= $ 3,779.14 Note that the loan is made in year0 presentdollars,

but the repayments are in year 1 and

year 4future dollars


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Example 2

Given that

Then we say that for the cost of money of 12%,PandFare equivalent in the sense that $1,000 todayhas the same worth as $1,762.34 in 5 years

1, 000 .12 P $ i =and

( ) ( )5 5

1 1, 000 1 .12 1, 762.34 P i $ $ F = + = =


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Example 3

Consider an investment that returns

$1,000 at the e.o.y. 1

$2,000 at the e.o.y. 2

i = 10%

We evaluateP

rate at which

money can befreely lent or

borrowed

( ) ( )1 22

1, 000 1 .1 2, 000 1 .1

909.9 1, 652.09 2, 561.98

P $ $

$ $ $

= + + +

= + =142 43 142 43


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Cash Flow Diagram for Example 3

We can illustrate this with a cash flow diagram

0 1 2

$ 2,561.98

$ 1,000

$ 2,000

year


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Net Present Value (NPV) for Example 3

Next, suppose that this investment requires $ 2,400now. So at 10% we say that the investment has a net

present value (NPV) of

NPV = $ 2,561.98 $ 2,400 = $ 161.98

0 1 2

$ 2,561.98

$ 1,000

$ 2,000

year

NPV

$ 161.98


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Cash Flows

A cash flow is a transfer of an amount At from one

entity to another at e.o.p. t

We consider a cash-flow set

corresponding to the set of times The convention for cash flows is

+ inflow

outflow Each cash flow requires the specification of:

amount;

time; and,

sign

{ }1 2, , , ...,0 n A A A A{ },1,2,...,0 n


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Cash Flows, cont.

Given a cash-flow set we

define the future worthFnof the cash flow set at e.o.y.

n as

{ }1 2, , , ...,0 n A A A A

( )1n n tn tt 0

F A i =

= +

0 1 2 t n 2 nn 2

A0 A1 A2 At An-2 An-1 An

. . . . . .


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Cash Flows, cont.

Note that each cash flow At in the set contributesdifferently toF

n

( )( )( )( )

1

1 1

2

2 2

1

1

1

1

n

0 0

n

n

n t

t t

n n

A A i

A A i

A A i

A A i

A A

+ + + +

M M

M M


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Cash Flows, cont.

We define the present worthPof the cash flow set as

Note that

( )1n n ttt tt 0 t 0

P A A i

= == = +

( )

( ) ( ) ( )

1

1 1 1

1

nt

t

t 0

n

t n nt

t 0

P A i

A i i i

=

=

= +

= + + +

1 4 4 2 4 4 3


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Cash Flows, cont.

or equivalently

( ) ( )1 1n

n

nn n t

t

t 0

n

n

F

i A i

F

=

= + +

=

142 431 4 42 4 43

( )1 nnF i P+


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Uniform Cash Flow Set

Consider the cash-flow set with

Such a set is called an equal payment cash flow set We compute the present worth

1,2,...,t A A t n={ }1 2, , , ...,0 n A A A A

2 1

1 1

1 ...n n

t t n

t

t t

P A A A = =

= = + + + +


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Uniform Cash Flow Set, cont.

Therefore

But

and so( )

1

1 d

= +

1

1

n

P A

=


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Uniform Cash Flow Set, cont.

We write

and we call the equal payment series present

value function

1n

P Ad

=

11 1

1 1

dd

d d = = =+ +

1

n

d


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Present Value Function (PVF)

1 1 (1 ) (1 ) 1

(1 )

n n n

n

d d

d d d d

+ + = =+


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Equivalence

We consider two cash-flow sets

under a given discount rate d

We say are equivalent cash-flow sets if

their future worths are identical.

{ } { }: 0,1, 2,..., : 0,1, 2,...,a bt t A t n A t n=and

{ } { }a bt tA Aand


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Equivalence, Example

Consider the two cash-flow setsunder d = 7%

0 1 2 3

a

4 5 6 7

2000 2000 2000 2000 2000

0 1 2

b

8,200.40


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Equivalence, cont.

We compute

and

Therefore, are equivalent cash flow setsunder d= 7%

7

3

2000 7162.33a tt

P == =2

8200.40 7162.33bP =

{ } { }a bt tA Aand


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Example

Consider the set of cash flows illustrated below

0 1 2

3

4 5 6 7 8

$ 300

$ 300

$ 200

$ 400

$ 200

d = 6%


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Discount Rate

The interest rate i is typically referred to as thediscount rate and is denoted by d

In converting a future amount F to a present worthP we can view the discount rate as the interest rate

that can be earned from the best investment

alternative

A postulated savings of $10,000 in a project in 5

years is worth at present

( ) 555 10,000 1 P F d = = +

Di t R t


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Discount Rate

Ford = 0.1, P = $6,201,while ford = 0.2, P = $4,019

In general, the lower the discount factor, the

higher the present worth The present worth of a set of costs under a given

discount rate is called the life-cycle costs

M t P h E l


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Motor Purchase Example

We consider the purchase of two 100-hp motors aand b to be used over a 20-year period; the discount

rate is 10%

The relative merits ofa and b are

The motor is used 1,600 hours per year and electricity

costs are constant at 0.08 $/kWh

motor costs ($) load(kW)

a 2,400 79.0

b 2,900 77.5

M t P h E l t


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Motor Purchase Example, cont.

We evaluate yearly energy costs fora and b

We next evaluate the present worth ofa and b

( ) ( ) ( )

( ) ( ) ( )

79.0 1600 .08 / 10,112

1, 2, .. . , 20

77.5 1600 .08 / 9, 920

a

t

b

t

A kW h $ kWh $

t

A kW h $ kWh $

= ==

= =

( )201

2, 400 10,112 1.1

88,489

ta

t

P

$

=

= +=

M t P h E l t


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Motor Purchase Example, cont.

Now, we evaluate

Therefore, the purchase of motorb results in thesavings of$1,135 due to the use of the smaller loadmotor under the specified 10% discount rate

88, 489 87, 354 1,135a b P P $= =

( )201

2, 900 9, 920 1.1

87,354

tb

t

P

$

=

= +=

I fi it H i C h Fl S t


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Infinite Horizon Cash-Flow Sets

Consider a uniform cash-flow set with

Then,

For an infinite horizon uniform cash-flow set

{ }: , 1, 2, ...t A A t 0=n

( )1 1n P A And d = A dP

=

I fi it H i C h Fl S t t


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Infinite Horizon Cash-Flow Sets, cont.

We may view das the capital recovery factor with thefollowing interpretation:

For an initial investment ofP,

dP = A

is the annual amount recovered in terms of

returns on investment

I t l R t f R t


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Internal Rate of Return

We consider a cash-flow set

The value of d for which

is called the internal rate of return (IRR)

TheIRR is a measure of how fast we recover aninvestment or stated differently, the speed with whichthe returns recover an investment

{ }: , 1, 2, ...t A A t 0=n t

t

t 0

P A 0=

= =

I t l R t f R t E l


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Internal Rate of Return Example

8

Consider the following cash-flow set

0

1 2

$30,000

3 4

$6,000 $6,000 $6,000 $6,000 $6,000

I t l R t f R t


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The present value

has the (non-obvious) solution of d equal to about 12%.

The interpretation is that under a 12% discount rate, the presentvalue of the cash flow set is 0 and so 12% is theIRR for the given

cash- flow set

The investment makes sense as long as other investments yield less than

12%.

81

30, 000 6,000P 0d

= + =

Internal Rate of Ret rn


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Consider an infinite horizon simple investment

Therefore

ForI= $1,000 andA = $200, d= 20% and we

interpret that the returns capture 20% of the investmenteach year or equivalently that we have a simple

payback period of 5 years

I

AdI

= ratio of annual return toinitial investment

A A A

0 1 2

. . .n

Efficient Refrigerator Example


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Efficient Refrigerator Example

A more efficient refrigerator incurs an investment ofadditional $1,000 but provides $200 of energy savings

annually

For a lifetime of 10 years, theIRR is computed from thesolution of

or

101

1,000 2000d

= +10

15

d

= The solution of this equationrequires either an iterativeapproach or a value looked

up from a table

Efficient Refrigerator Example cont


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Efficient Refrigerator Example, cont.

IRR tables show that

and so theIRR is approximately 15%

If the refrigerator has an expected lifetime of 15 years this

value becomes

10

15

15.02

d %d

=

=

15

18.4

1 5.00

d %d = =

s was mentioned earlier, the value is 20% if it lasts forever

Impacts of Inflation


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Impacts of Inflation

Inflation is a general increase in the level of prices in aneconomy; equivalently, we may view inflation as a general

decline in the value of the purchasing power of money

Inflation is measured using prices: different products may

have distinct escalation rates Typically, indices such as the CPI the consumer price index

use a market basket of goods and services as a proxy for the

entire U.S. economy

reference basis is the year 1967 with the price of$100 for the basket(L0); in the year 1990, the same basket cost $374 (L23 )

Figuring Average Rate of Inflation


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Figuring Average Rate of Inflation

Calculate average inflation rate e from 1967 to 1990

( ) 23 3741 3.74100

e = =( ) ( )ln 3.74ln 1 0.059%

23e e= =

Current

(1/2009)basket

value is

about 632.

green electric energy lecture 13

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