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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.JPG8/14/2019 Green Electric Energy Lecture 13
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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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Entropy and Fuel Cells
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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Entropy and Fuel Cells
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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Internal Rate of Return
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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Internal Rate of Return
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.
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