Green Electric Energy: Wave/tidal Power, Fuel Cells, Economics

ECE 333 Green Electric Energy Lecture 13 Wave/Tidal Power, Fuel Cells, Economics

Professor Tom Overbye Department of Electrical and Computer Engineering

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 1 • •

from 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

In the News: Prof. Chapman, Harnessing Human Power

Wave Power The potential energy available waves is quite high, with

• some 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.

World’s largest wave park” is

• in Portugal, with capacity of

2.25 MW; each of three devices is 142m long, and has a diameter of 3.5 m, and uses 700 metric tons of steel

Pelamis Wave Power

Source: www.pelamiswave.com

Tidal Power • •

There is tremendous potential energy in harnessing the earth’s 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 240 MW (max) La Rance in France, built in the 1960’s. – –

Average power generation is 68 MW A 330m dam contains a 22 square km basin, with average tides of 8m.

Tidal Power • No new tidal plants have been constructed since 1960’s • The two main approaches to tidal power are 1) dams (barrages) across a tidal •

estuary (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 not require 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

Fuel Cells • •

Convert chemical energy contained in a fuel directly into electrical power Skip conversion to mechanical energy, not constrained by Carnot limits

Conventional Combustion

Chemical energy

Chemical energy Fuel Cells

Heat

Electrical energy

Mechanical energy Electrical energy

Fuel Cells • Up to ~65% efficiencies • No combustion products (SOX,CO) although there may be NOX at high temperatures

• Vibration free, almost silent – can be located close to the • • •

load Waste heat can be used for cogeneration Byproduct is water Modular in nature

Fuel Cells - History • •

Developed more than 150 years ago Used in NASA’s Gemini earth-orbiting missions, 1960s For more information on the history of fuel cells, see the Smithsonian projecthttp://americanhistory.si.edu /fuelcells/

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

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

Fuel Cells - History

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

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

Fuel Cells- Basic Operation Electrical Load I

2H +

Anode H 2 → 2 H + + 2e −

Electrolyte Catalyst

Cathode 1 O2 + 2 H + + 2e − → H 2O 2

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

Fuel Cells- Basic Operation •

Combined anode and cathode reactions: H 2 → 2 H + + 2e − 1 O2 + 2 H + + 2e− → H 2O 2 1 H 2 + O2 → H 2 O (4.20) 2

• •

This reaction is exothermic- it releases heat A single cell only produces ~0.5V under normal operating conditions, so multiple cells are stacked to build up the voltage

Fuel Cells • How much energy is liberated and how much can be converted to • •

electricity? 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 = U + PVenergies • PV = observable,Hmacroscopic • Units – kJ/mole

Enthalpy • • • •

A measure of the energy it takes to form the substance out of its constituent elements For fuel cells, changes in chemical energy are of interest and those are best described in terms of enthalpy changes As with potential energy, we describe enthalpy with respect to a reference (it is the change that matters) At Standard Temperature and Pressure (STP)= 25˚C, 1 atm, stable form of element has zero enthalpy

Enthalpy of Formation • •

• •

Enthalpy of formation - difference between enthalpy of the substance and enthalpies of its constituent elements Exothermic – heat is liberated, enthalpy in final products is less than reactants, enthalpy of formation is negative, chemical energy of substance is less than that of its constituents Endothermic – heat is absorbed Depends on state (liquid, solid, gas), see Table 4.6

Enthalpy of Formation • Liquid Water

1 H 2 + O2 → H 2 O ( l) 2 ∆H = −285.8 kJ/mol

• Water Vapor

1 H 2 + O2 → H 2 O ( g ) 2 ∆H = −241.8 kJ/mol

• Latent heat of vaporization of water

44 kJ/mol

Entropy and Fuel Cells • •

How much of the energy can be converted directly into electricity? We’ll use entropy concepts to develop the maximum efficiency of a fuel cell Enthalpy in H Fuel Cell

Enthalpy out We

Rejected heat Q Fig. 4.28- Energy Balance for a Fuel Cell

Entropy and Fuel Cells •

Assume isothermal ∆S =

Q T

1 H 2 + O2 → H 2 O + Q 2

• •

Q = heat released Net entropy must increase Entropy gain ≥ Entropy loss

Q + ∑ S products ≥∑ Sreactants T

Enthalpy in H Fuel Cell

Enthalpy out We

Rejected heat Q

Entropy and Fuel Cells • From

Q heat we can release is the minimum T

+ ∑ S products ≥∑ Sreactants

(4.28)

Qmin = T ( ∑ S reactants − ∑ S products )

• Now find maximum efficiency

H = We + Q (4.30) We Q η= =1− (4.31) H H Qmin ηmax = 1 − H

Enthalpy in H Fuel Cell

Enthalpy out We

Rejected heat Q

Theoretical maximum can be quite high (> 80%)

Gibbs Free Energy and Fuel Cells • Consider the chemical energy released in a reaction as having two parts: – –

1) entropy-free part (the work) called free energy ΔG 2) heat Q

• G = H-Q=H-TΔS • ΔG is the maximum entropy-free output (work) from a chemical reaction

• Then

∆G = ∑ Gproducts We η= H

− ∑ Greactants

ηmax =

∆G ∆H

Output of an Ideal Fuel Cell • •

Equal to the magnitude of ΔG From Table 4.6,

1 H 2 + O2 → H 2 O Maximum electrical output 2 at STP is •

∆G = −237.2 kJ/mol

We = ∆G = 237.2 kJ/mol •

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

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

Electrical Characteristics • Activation losses • Fuel crossover losses

• Mass transport losses

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)

Types of Fuel Cells

Hydrogen Production • Obtaining a supply of hydrogen of sufficient quality and at a • • •

reasonable cost is difficult Critical to solve this before fuel cells can become widely deployed Main technologies are steam reforming of methane (SMR) and partial oxidation (POX) Generation IV nuclear reactors could be used as well (when they become available)

Energy Economic Concepts (From Prof. Gross) •

The economic evaluation of a renewable energy resource requires a meaningful quantification of cost elements – –

•

fixed costs variable costs

We use engineering economics notions for this purpose since they provide the means to compare on a consistent basis – –

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

Time Value of Money • • • •

Basic notion: a dollar today is not the same as a dollar in a year We represent the time value of money by the standard approach of discounted cash flows The notation is P = principal i = interest value The convention we use is that payments occur at the end of each period

Simple Example loan P for 1 year repay P + iP = P ( 1 + i ) at the end of 1 year year 0 P year 1 P (1 + i) loan P for n years year 0 P year 1 ( 1 + i ) P repay/reborrow year 2 ( 1 + i )2 P repay/reborrow year 3 ( 1 + i )3 P repay/reborrow .M M M year n ( 1 + i )n P repay

Compound Interest e.o.p.

amount owed

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

P(1+i ) n-1 i

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

M

interest for next period

amount owed for next period

M

n-1

P(1+i ) n-1

n

P(1+i ) n

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)

Terminology

F = P ( 1 + i)

n

compound interest Lump sum repayment at the end of n periods. F is called the future worth, while P is called the present worth

Need not be integer-valued

Terminology • •

We call (1 + i) n the single payment compound amount factor −1 We define β @( 1 + i ) and

•

β

n

= ( 1+ i)

−n

is the single payment present worth factor F is called the future worth; P is called the present worth or present value at interest i of a future sum F

Example 1 •

• •

Consider a loan of $4,000 at 8% interest to be repaid in two 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 year 0 present dollars, but the repayments are in year 1 and year 4 future dollars

Example 2 •

Given that

P = $1,000

•

and

i = .12

Then we say that for the cost of money of 12%, P and F are equivalent in the sense that $1,000 today has the same worth as $1,762.34 in 5 years

P ( 1 + i)

5

= $1,000( 1 + .12)

5

= $1,762.34 = F

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 evaluate P

rate at which money can be freely lent or borrowed

P = $ 1,000 ( 1 + .1) + $ 2,000( 1 + .1) 142 43 142 43 2 β β = $ 909.9 + $ 1,652.09 = $ 2,561.98 −1

−2

Cash Flow Diagram for Example 3 •

We can illustrate this with a cash flow diagram $ 2,000 $ 1,000 0

1

2 year

$ 2,561.98

Net Present Value (NPV) for Example 3 •

Next, suppose that this investment requires $ 2,400 now. So at 10% we say that the investment has a net present value (NPV ) of NPV = $ 2,561.98 – $ 2,400 = $ 161.98 NPV $ 161.98 0

$ 2,000 $ 1,000 1

2 year

$ 2,561.98

Cash Flows • • • •

A cash flow is a transfer of an amount A t from one entity to another at e.o.p. t We consider a cash-flow set { A0 , A1 , A2 ,..., An} corresponding to the set of times { 0 ,1, 2,..., n} The convention for cash flows is + inflow − outflow Each cash flow requires the specification of: – – –

amount; time; and, sign

Cash Flows, cont. •

{ A , A , A ,..., A }

Given a cash-flow set

0

1

2

we

n

define the future worth F n of the cash flow set at e.o.y. n as

A0

n

∑ A ( 1 + i)

Fn = A1

t =0

1

2

t

At

A2

... 0

n− t

A n-2

A n-1

An

n–2

n–2

n

... t

Cash Flows, cont. •

Note that each cash flow A t in the set contributes differently to F n

A0 A1

→ →

A2 → M At → M An →

A0 ( 1 + i )

A1 ( 1 + i )

n n−1

A2 ( 1 + i ) M n− t At ( 1 + i ) M An

n− 2

Cash Flows, cont. •

We define the present worth P of the cash flow set as

P =

•

n

∑A β t =0

Note that

P =

n

=

∑ A ( 1 + i) t =0

=

t

t

n

∑ A ( 1 + i) t =0

−t

t

−t

t

n

∑ A ( 1 + i ) (11 4+ i4) 2( 14+4i) 3 t =0

−t

−n

n

t

1

Cash Flows, cont. n

= ( 1 + i ) ∑ At ( 1 + i ) 142 43 t =0 1 4 42 4 43 n β F −n

n −t

n

= β Fn n

or equivalently

Fn = ( 1 + i ) P n

Uniform Cash Flow Set • • •

Consider the cash-flow set { A0 , A1 , A2 ,..., An}

At = A

with

t = 1, 2,..., n

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

n

n

t =1

t =1

t t 2 n− 1   A β = A β = A β 1 + β + β + ... + β ∑ t ∑ 

Uniform Cash Flow Set, cont. •

Now, for 0 < β < 1 , we have the identity ∞

∑β •

It follows that

j

j =0

1 = 1−β

∞

∑β

j

j=0

1 + β + ... + β n−1 =

∞

j n 2 n −1  β − β 1 + β + β + ... + β +... ∑  j =0

∞

= ( 1−β n ) ∑ β 1− β = 1− β

n

j =0

j

Uniform Cash Flow Set, cont. •

Therefore

1− β P = Aβ 1−β

•

But

β = (1 + d) and so

−1

n

Uniform Cash Flow Set, cont.

•

We write

1 d 1−β = 1− = = βd 1+ d 1+d

1−β P = A d

and we call

n

1− β value function d

n

the equal payment series present

Present Value Function (PVF)

1−β d

n

1 − (1 + d ) = d

−n

(1 + d ) − 1 = d (1 + d )n n

Equivalence •

We consider two cash-flow sets

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

and

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

under a given discount rate d

•

We say

{A }

are equivalent cash-flow sets if

and { A } their future worths are identical. a t

b t

Equivalence, Example •

Consider the two cash-flow sets under d = 7%

8,200.40

2000 2000 2000 2000 2000

0

1

2

3

4 a

5

6

7

0

1 b

2

Equivalence, cont. •

We compute P a = 2000

7

∑β

t

= 7162.33

t =3

•

and P b = 8200.40 β 2 = 7162.33 Therefore, A a and A b are equivalent cash flow sets { t} { t} under d = 7%

Example •

Consider the set of cash flows illustrated below $ 400 $ 300

$ 200

$ 200 3 0

1

4

2 $ 300

5

6

7

8

d = 6%

Example, cont. •

We compute F 8 at t = 8 for d = 6% F8 = 300 ( 1 + .06)

7

− 300( 1 + .06)

200 ( 1 + .06) + 400( 1 + .06) = $ 951.56 4

•

We next compute P

P = 300 ( 1 + .06)

−1

+ 200 ( 1 + .06 ) = $ 597.04

•

2

− 300( 1 + .06)

−4

5

+ + 200

−3

+ 400( 1 + .06)

−6

+ 200( 1 + .06)

We check that for d = 6% 8 F8 = 597.04 ( 1 + .06) = $ 951.56

−8

Discount Rate • •

•

The interest rate i is typically referred to as the discount rate and is denoted by d In converting a future amount F to a present worth P 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

P = F5 β

5

= 10,000 ( 1 + d )

−5

Discount Rate • • •

For d = 0.1, P = $ 6,201, while for d = 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

Motor Purchase Example • •

•

We consider the purchase of two 100-hp motors – a and b – to be used over a 20-year period; the discount rate is 10% The relative merits of a and b are motor

costs ( $ )

load ( kW )

a

2,400

79.0

b

2,900

77.5

The motor is used 1,600 hours per year and electricity costs are constant at 0.08 $/kWh

Motor Purchase Example, cont. •

We evaluate yearly energy costs for a and b A at = ( 79.0 kW ) ( 1600 h) ( .08 $ / kWh) = $ 10,112

t = 1, 2, ... , 20 A bt = ( 77.5 kW ) ( 1600 h) ( .08 $ / kWh)

•

= $ 9,920

We next evaluate the present worth of a and b

P

a

20

= 2,400 + 10,112 ∑ ( 1.1) t =1

= $ 88,489

−t

Motor Purchase Example, cont. P

b

20

= 2,900 + 9,920 ∑ ( 1.1) = $ 87, 354

•

−t

t =1

Now, we evaluate

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

•

Therefore, the purchase of motor b results in the savings of $ 1,135 due to the use of the smaller load motor under the specified 10% discount rate

Infinite Horizon Cash-Flow Sets •

Consider a uniform cash-flow set with n → ∞

{ At = A : t = 0, 1, 2, ... } •

Then,

P = A

n 1 − β ( )

d

n →∞

1 A d

For an infinite horizon uniform cash-flow set

A = d P

Infinite Horizon Cash-Flow Sets, cont. •

We may view d as the capital recovery factor with the following interpretation: For an initial investment of P, dP = A is the annual amount recovered in terms of returns on investment

Internal Rate of Return • •

We consider a cash-flow set

{A

t

= A : t = 0, 1, 2, ...

}

The value of d for which

P =

n

t A β ∑ t =0 t =0

•

is called the internal rate of return (IRR) The IRR is a measure of how fast we recover an investment or stated differently, the speed with which the returns recover an investment

Internal Rate of Return Example

•

Consider the following cash-flow set $6,000 $6,000 $6,000 $6,000

$6,000

0 1

$30,000

2

3

4

8

Internal Rate of Return • The present value 1− β 8 P = − 30,000 + 6,000 =0 d

•

has the (non-obvious) solution of d equal to about 12%. The interpretation is that under a 12% discount rate, the present value of the cash flow set is 0 and so 12% is the IRR for the given cash- flow set –

The investment makes sense as long as other investments yield less than 12%.

Internal Rate of Return •

Consider an infinite horizon simple investment A 0

I

A

A 1

2

A • Therefore d = I

•

... n

ratio of annual return to initial investment

For I = $ 1,000 and A = $ 200, d = 20% and we interpret that the returns capture 20% of the investment each year or equivalently that we have a simple payback period of 5 years

Efficient Refrigerator Example • •

A more efficient refrigerator incurs an investment of additional $ 1,000 but provides $ 200 of energy savings annually For a lifetime of 10 years, the IRR is computed from the solution of

1− β 0 = − 1,000 + 200 d or

1− β d

10

= 5

10

The solution of this equation requires either an iterative approach or a value looked up from a table

Efficient Refrigerator Example, cont. •IRR tables show that

1 − β 10 = 5.02 d d = 15%

and so the IRR is approximately 15% If the refrigerator has an expected lifetime of 15 years this value becomes

1 − β 15 = 5.00 d d = 18.4%

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

Impacts of Inflation • Inflation is a general increase in the level of prices in an • •

economy; 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 (L 0); in the year 1990, the same basket cost $ 374 (L 23 )

Figuring Average Rate of Inflation •

Calculate average inflation rate e from 1967 to 1990

374 = 3.74 ( 1 + e) = 100 ln ( 3.74 ) ln ( 1 + e ) = → e = 0.059% 23 23

Current (1/2009) basket value is about 632.

Source: http://en.wikipedia.org/wiki/File:US_Historical_Inflation.svg