Eco Solution Manual

Materials and the Environment

Solution manual for exercises

Solutions to exercise:

Eco text: solution manual

Page

Chapter 1

2

Chapter 2

4

Chapter 3

12

Chapter 4

16

Chapter 5

20

Chapter 6

23

Chapter 7

30

Chapter 8

41

Chapter 9

53

Chapter 10

66

Chapter 11

69

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Chapter 1 E 1.1. Use Google to research the history and uses of one of the following materials 

Tin



Glass



Cement



Bakelite



Titanium

Carbon fiber Present the result as a short report of about 100 100 - 200 words (roughly half a page). Imagine that you are preparing it for school school children. Who used it first? Why? What is exciting about the material? material? Do we now depend on it or could we, with no loss of engineering performance of great increase in cost, live without it? 

Specimen answer: tin. Tin (symbol Sn), a silver-white metal, has a long history. It was traded in the civilizations of the Mediterranean as early as 1500 BC (the Old Testament of the Christian bible contains many references to it). Its importance at that time lay in its ability to harden copper to give bronze (copper containing about 10% tin), the key material for weapons, tools and statuary of the Bronze age (1500 BC – 500 BC). Today tin is still used to make bronze, for solders and as as a corrosion resistant coating on steel sheet (“tin plate” ) for food and drink containers – a “tinnie”, to an Australian, is a can of beer. Plate glass is made by floating molten glass glass on a bed of liquid tin (the Pilkington process). Thin deposits of tin compounds on glass give transparent, electrically conducting coatings used for frost-free windshields and for panel lighting. Most of the applications of tin could be filled by other materials – polymer coatings for food containers, aluminum instead of tin to make bronzes, indium for transparent coatings (though at an increased cost). Finding a replacement for tin in solders is more difficult.

E 1.2. There is international agreement that it is desirable (essential, in the view of some) to reduce global energy consumption. Producing materials from ores and feedstocks requires energy (its “embodied energy”). The table lists the energy per kg and the annual consumption of 4 materials of engineering. If consumption of each could be reduced by 10%, which material offers the the greatest global energy saving? Which the least?

Material

Embodied energy MJ/kg

Annual global consumption (tonnes/yr)

Annual energy commitment (MJ)

Steels

29

1.1 x 109

3.2 x 1010

Aluminum alloys

200

3.2 x 107

6.4 x 109

Polyethylene

80

6.8 x 107

5.4 x 109

Concrete

1.2

1.5 x 1010

1.8 x 1010

Approximately 2000

5 x 103

1.0 x 107

Device-grade silicon

Answer. An additional column has been added to the table above: it shows the annual energy commitment associated with the production of each material (the product of the numbers in the two columns to its left). Reducing consumption of steel and of concrete – by more efficient design of structures perhaps – have by far the greatest potential for global energy saving. Doing the same for device grade silicon has the least, least, by a large factor. Although the embodied energies of materials materials differ considerably, it is the much greater differences in annual consumption that dominate the total energy commitment and the carbon burden they generate. This is one reason that much of the discussion discussion of this book focuses on the materials used in the greatest quantities.


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E1.3. The ultimate limits of most resources are difficult to assess precisely, although estimates can be made. One resource, however, has a well-defined limit: that of usable land area. The surface area of the globe is 511 million square km, or 5.11 x 1010 hectares (a hectare is 0.01 sq. km.) Only a fraction of this is land, and only part of that land is useful – the best estimate is that 1.1 x 10 10 hectares of the earth’s surface is biologically productive. Industrial countries require 6 hectares of biologically productive land per head of population to support current levels of consumption. The current (2008) global population is close to 6.7 billion (6.7 x 10 9). What conclusions can you draw from these facts?

Answer. The current biologically productive area per person is 1.1 x 10 10 / 6.7 x 109 = 1.64 hectares/person, less than 1/3 of that currently needed to support a person in a developed country. Thus a large fraction of the world’s population can never reach the same level of consumption currently enjoyed by developed countries unless the global population declines, or new forms of intensive agriculture, perhaps based on genetically modified crops and animals, enable a dramatic increase (factor 3) in the productivity of the areas of the earth’s surface that are biologically productive.


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Chapter 2

E2.1 Explain the distinction between reserves and the resource base.

Answer. A mineral reserve is defined as that part of a known mineral deposit that can be extracted legally and economically at the time it is determined. Reserves are an economic construct, which grow and shrink under varying economic, technical and legal conditions. Improved extraction technology can enlarge it, but environmental legislation or changing political climate may make it shrink. Demand stimulates prospecting, with the consequence that reserves tend to grow in line with consumption.

The resource base (or just resource) of a mineral is the r eal total, and it is much larger than the reserve but much less certain. It includes not only the current reserves but also all usable deposits that might be revealed by future prospecting and that, by various extrapolation techniques, can be estimated. It inclu des, too, kn own and unkn own depos its that cann ot be mined profi tably now but which – due to higher prices, better technology or improved tr anspor tatio n – may become available in the future. The answer should include a sketch of the diagram illustrating reserves and resources is (Figur e 2.10 of the book).

E2.2 The world consumption rate of CFRP is rising at 8 % per year. How long does it take to double?

Answer. Equation 20.3 of the text gives the doubling time t D as

t D =

100 r

log e ( 2 ) ≈

70 r

where r is the percentage fractional rate of growth per year. Thus a growth rate of 8% means that consumption doubles every 8.7 years.

E2.3

Derive the dynamic index t ex ,d =

r R + 1 ) log e ( r 100 P o

100

starting with equation (2.2) of the text.

Answer. Equation (2.2) of the text describes the production rate of a material, with P 0 = production

rate at time t = t o and r = annual growth rate in % per year.

⎧ r (t − t o ) ⎫ ⎬ ⎩ 100 ⎭

P = P 0 exp⎨

Integrating this over time to give the cumulative production Qt * up to time t = t * and equating the result to the reserve R gives

⎞ ⎧⎪ r ( t * − t ) ⎫⎪ 100 P o ⎛ ⎜ ⎟ o Qt * = ∫ P dt = exp ⎨ ⎬ − 1⎟ = R ⎜ r ⎜ ⎟ ⎪ 100 ⎪ t t *

o


⎝

⎩

⎭

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⎠

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Solving for the time interval ( t * − t o ) gives the desired result t * − t o = t ex ,d =

r R log e ( + 1 ) r 100 P o

100

E2.4 A total of 5 million cars were sold in China in 2007; in 2008 the sale was 6.6 million. What is the annual growth rate of car sales, expressed as % per year? If there were 15 million cars already on Chinese roads by the end of 2007 and this growth rate continues, how many cars will there be in 2020, assuming that the number that are removed from the roads in this time interval can be neglected?

Answer. Starting with equation (2.2)

⎧ r (t − t o ) ⎫ ⎬ ⎩ 100 ⎭

P = P 0 exp⎨

we enter P = 6 .6 x10 6 and P o = 5 x10 6 and the time interval (t − t o ) = 1 year, and solve for r . The result is r = 27.8% per year. The cumulative number of cars entering use in the subsequent 13 years is found from the integral of this equation over time

⎞ ⎧⎪ r ( t * − t ) ⎫⎪ 100 P o ⎛ ⎜ ⎟ o Qt * = ∫ P dt = exp ⎨ ⎬ − 1⎟ ⎜ r ⎜ ⎟ ⎪ 100 ⎪ t t *

⎝

o

⎩

⎭

⎠

Entering P o = 5 x10 6 (the number 2007), r = 27.8% per year and (t * −t o ) = 13 gives the additional number of cars by 2020 at Qt * = 650 x10 6 . To this (if we are picky) we must add the number already there in 2007, giving a final total of 655 x10 6 , larger than the number in 2007 by a factor of 100. (This colossal number is still only equivalent to 1 car per 3-person family, l ess than the current car ownership per family in the US,)

E2.5 Prove the statement made in the text that, “at a global growth rate of just 3% per year we will mine, process and dispose of more “stuff” in the next 25 years than in the entire 300 years since the start of the industrial revolution.”. . Answer. Exponential growth has a number of alarming features, among them, the doubling time both of consumption and of the total quantity consumed. The consumption rate C of a resource grows by follows the law

(1)

C = C 0 exp r (t – t 0 )

here C 0 is the consumption rate when t = t 0. The doubling time for the consumption rate is tDC, meaning that the rate will have doubled to 2C 0 in the time tD = t – t0. It is calculated by equating C in equation 1 to 2C0. 2C 0 = C 0 exp (r t D)

giving

tD =

1

r

ln 2

the quantity ln 2 = 0.69, so that, at a growth rate of 3% per year ( r = 0.03 ) , consumption doubles in 23 years.


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We are interested here not in the consumption rate but in the total quantity, Q, consumed since consumption began. It is the integral of C over time. Q = ∫ 0t Cdτ = C o ∫ 0t exp r ( τ − t o ) dτ

giving

Q=

Co r

{ exp (r (t - t o )) − exp( −r t o )}

(2)

Q0 is the total amount consumed prior to the present day (when t = t 0), thus, C (3) Q0 = o { 1 − exp( − r t o )} r We want the time tQ (the time to double the total quantity consumed) where tQ = t – t0 is the time from the present day that Q = 2 Q 0, i.e. when (from equations 2 and 3) Co 2 Co { 1 − exp( − r t o )} exp r t Q − exp( − r t o ) = r r solving for tQ gives exp (r t Q ) = 2 – exp (– r t 0 ) Now note that if t 0 = 100 years (meaning that consumption started 100 years ago) and r = 3% per year (0.03), then exp (- r t0 ) = 0.05. If t0 = 1000 years, it is roughly 10 -13. Compared with 2 both numbers are negligible. Thus, the result for the doubling time of consumption is the same as that for consumption rate, namely.

{

tQ =

1

r

( )

ln

}

2

(The underlying reason both tD and tQ are the same is simply that ∫ e x dx = e x )

That means that the total quantity of a given resource that will be consumed in the next 23 years (given the modest rate of growth of consumption of 3%/year) is equal to the total quantity consumed over the history of industrial development. Not a happy thought.

E2.6 Understanding reserves: copper. The table lists the world production and reported reserves of copper over the last 20 years





Year

Price (US$/kg)

World production (million tonnes/yr)

Reserves (million tonnes)

Reserves/World production (yrs)

Growth rate (%/yr)

1995

2.93

9.8

310

31.6

1996

2.25

10.7

310

29.0

8.8

1997

2.27

11.3

320

28.3

5.5

1998

1.65

12.2

340

27.9

7.7

1999

1.56

12.6

340

27.0

3.2

2000

1.81

13.2

340

25.8

4.7

2001

1.67

13.7

340

24.8

3.7

2002

1.59

13.4

440

32.8

-2.2

2003

1.78

13.9

470

33.8

3.7

2004

2.86

14.6

470

32.2

4.9

2005

3.7

14.9

470

31.5

2.0

2006

6.81

15.3

480

31.4

2.6

Examine trends (plot price, production and reserves against time) – what do you conclude? Tabulate the Reserves / World production to give the static index of exhaustion. What does the result suggest about reserves?


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Answer. The figure shows the data. The rates of growth of production are listed in the last column of the table above: world production and reserves have increased steadily over the 12 year period 1995 - 2006. The ratio of reserves to production (second last column of the table) is steady at around 30 years. All this suggests a well-balanced market, with no indicators of supply problems. Only the price, which was steady or falling up to 2003 shows a sudden increase in 2005 / 2006 because of a surge in demand from China. All the other indicators suggest that the supply chain is capable of adapting to meet it.

E2.7 The table shows the production rate and the reserves of 5 metals over a period of 10 years. What has been the growth rate of production? What is that of the reserves? What can you conclusions can you draw about the criticality of the material? Metal

Year

Production rate, tonnes/year

Reserves, tonnes

Growth rate of production % per year

Growth rate of reserves % per year

Platinum

2005 1995

217 145

71 x 103 56 x 103

4.03

2.37

64 x 106 47 x 106

3.6

3.09

1995

2005 1.49 x 106 1.04 x 106

Lead

2005 1995

3.27 x 106 2.71 x 106

67 x 106 69 x 106

1.88

1.97

Copper

2005 1995

15.0 x 106 10.0x 106

480 x 106 310 x 106

4.05

4.37

Cobalt

2005 1995

57.5 x 103 22.1 x 103

7 x 106 4 x 106

9.56

5.60

Nickel

Answer. The last two columns of the table above show the growth rates, calculated from equation 2.2 of the text

⎧ r (t − t o ) ⎫ ⎬ which, on inversion, gives ⎩ 100 ⎭

P = P o exp⎨

r =

⎛ P ⎞ ⎟ ln ⎜⎜ (t − t o ) ⎝ P o ⎠⎟ 100

Inserting the 2005 value of production or reserves for P , the 1995 value for P o and (t − t o ) = 10 years gives the average growth rate over the 10 year interval. For two of the metals – lead and copper – the rate of growth of reserves is more or less in balance with the growth in production. For nickel, reserves lag behind production but not by much. The two metals that are cause for concern are platinum and cobalt. Both are used for critical functions for which there is no other easy substitute: platinum for catalysts and cobalt for high temperature alloys. For both the growth of reserves – meaning the rate at which new, exploitable ore bodies are proven – lags well behind the rate of production.


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E2.8 Tabulate the annual world production in tonnes/year and the densities (kg/m 3) of carbon steel, PE, softwood and concrete (you will find the data in data sheets for these materials in Chapter 12 of this book – use an average of the ranges given in the data sheets). Calculate, for each, the annual world production measured in m3/year. How does the ranking change?

Answer. The first three columns of the table below list data for world production in tonnes/year,

P m ,and density, ρ , in kg/m3, drawn (as averages) from the data sheets of Chapter 12. The last

column lists the world production in m3/year , P v calculated as P P v = 1000 m ρ

(The 1000 is to convert the density into tonnes/m 3.) Concrete and steel dominated production when measured in tonnes per year . But for at least two of these materials – softwood and concrete – it is the volume that is used that is important, not the weight. By this measure wood is second only to concrete in the quantity used. Wood remains one of the central materials of structural engineering.

Material

World production tonnes/year

Density 3 kg/m

World production 3 m /year

Carbon steel

1.15 x 109

7850

1.46 x 108

Polyethylene

6.85 x 107

950

7.2 x 107

Softwood

9.65 x 108

520

1.86 x 109

Concrete

1.53 x 1010

2450

6.25 x 109

E2.9 The price of cobalt, copper and nickel have fluctuated wildly in the past decade. Those of aluminum, magnesium and iron have remained much more stable. Why? Research this by examining uses (which metal are used in high value-added products?) and the localization of the producing mines. The USGS web site listed under Further Reading is a good starting point. Answer. Take cobalt as an example. A web search (and a search of CES) gives the following information. World production of cobalt has increased steadily year-on-year, and has almost trebled from around 20,000 tonnes in 1995 to 54,000 tonnes in 2005, an average rate of growth at times exceeding 12% per year, far higher than that of most materials. The ores of cobalt are localized in relatively few countries. Since 2002, the strongest growth in production of cobalt has come in China, where output grew at over 60% per year in 2005 as a result of expansion by domestic producers. Other recent increases have come from new projects including Voisey's Bay in Canada (Inco), Coral Bay in the Philippines (Sumitomo) and Sally Malay in Australia (Sally Malay and Sumitomo). Cobalt costs about $30/kg – far more than common structural materials. It is used for catalysts; medical implants, cermets (tungsten carbide – cobalt) cutting tools, alloying of steels, hightemperature cobalt based superalloys, alloys for high field magnets, and as a pigment in glass and paints. These are high value-added applications, as you might anticipate from the high price of the metal. When the supply chain for metals with unique applications, like cobalt, is unable to meet demand, the price rises steeply because there are no substitutes. When, by contrast, substitutes are readily available (plastics substituting for steel or aluminum in many applications, the existence and availability of the substituted dampens price fluctuations.


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E2.10 The production of zinc over the period 1992 – 2006 increased more or less steadily at a rate of 3.1 % per year. The reserves, over the same period, increased by 3.5%. What conclusions can you draw from this about the criticality of zinc supply? Answer. The symptoms of criticality are

• • • •

The rate of growth of discovery falls below the rate of growth of production. The production rate curve peaks and starts to decline The minimum economic ore grade falls The price starts to rise sustainable.

We are told in the question that growth in production has been steady – no sign of a flattening out suggesting a peak in production is near. The growth rate of reserves is comfortably larger than that of production indicating no shortage of supply. Taken together these indicators point to a market for zinc that is well balanced.

E2.11 The production of platinum, vital for catalysts and catalytic converters, has risen from 145 to 217 tonnes per year over the last 10 years. The ores are highly localized in South Africa, Russia and Canada. The reserves have risen from 56,000 to 71,000 tonnes in the same time interval. Would you classify platinum as a critical material? Base your judgment on the these facts, making use of the relative growth rates of production and reserves and on the dynamic index (equation 2.6 of the text) calculated using 2005 data. Answer. Platinum has applications for which there is no present substitute. The growth rates of production and of reserves, calculated as described in the solution to Exercise E2.7, are listed in the table below. The dynamic index of exploitation, calculated from these data, is 20.9 years.

Platinum

Year

Production rate, tonnes/year

Reserves, tonnes

2005 1995

217 145

71 x 103 56 x 103

Growth rate of Growth rate of production reserves % per year % per year 4.03

2.37

Those are the facts. What deductions? There are many danger signs here. Sources are localized and only one is in a country with a long history of economic and political stability. The growth rate of reserves has, over a 10 year period, lagged behind that of exploitation. The dynamic index – not a reliable measure, but another indicator none the less – is low when compared with those of most other metals. E2.12 Global water consumption has tripled in the last 50 years. What is the growth rate, r %, in consumption assuming exponential growth? By what factor will water consumption increase between now (2008) and 2050?

Answer. Write consumption as

⎧ r (t − t o ) ⎫ ⎬ ⎩ 100 ⎭

C = C o exp⎨

which, on inversion, gives

r =

100

(t − t o )

⎛ C ⎞ ⎟ ⎟ ⎝ C o ⎠

ln ⎜⎜

The ratio C / C o = 3 for the time interval (t − t o ) = 50 years, giving r = 2.2% / year . If this growth rate continues, the consumption of water will increase over the next 42 years by the factor


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⎧ r (t − t o ) ⎫ = exp⎨ ⎬ = 2.52 C o ⎩ 100 ⎭ C

that is, it will more than double. Water supply in many parts of the world is already barely able to meet demand. It is not clear how this need will be met.

E2.13 Plot the Annual world production of metals against their Price , using mean values from the data sheets in Chapter 12 of this book. What trend is visible? Answer. See the answer to Exercise E2.14, in which the annual world production for materials is plotted against price as a CES exercise. Metals are labeled. The trend described in the solution to E2.14 appears also when metal data alone areplotted.

Exercises using CES Eco Level 2. E2.14 Use CES to plot the Annual world production of materials against their Price. What trend is visible? Answer. The figure shows the plot. There is a correlation – the lower the price, the larger is the annual production. The contours have a slope -2, which, with log scales (as here) means that, very approximately,

Annual world production ∝

1

( Pr ice)2

-2


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E2.15 Make a plot of the static index of exhaustion for metals (the reserves in tonnes divided by the annual production rate in tonnes per year), using the “Advanced” facility in CES to plot the ratio. Which metals have the longest apparent resource life? Why is this index not a reliable measure of the true resource life of the metal?

Answer. The figure shows the plot of the static index, created with the CES Eco Level 2. The ores of magnesium, aluminum and titanium are plentiful and very widely distributed; for these there is no concern about depletion. Those of nickel, copper, zinc and lead are more localized and already significantly depleted, with the result that the ore grade currently mined is much leaner than it was 25 years ago.

The static index, however, is not a true indicator of resource life. This is because the reserves are an economic construct, which grow and shrink under varying economic, technical and legal conditions. Improved extraction technology can enlarge it, but environmental legislation or changing political climate may make it shrink. Demand stimulates prospecting, with the consequence that reserves tend to grow in line with consumption. The figure shows that the reserves of the more critical metals are typically between 20 to 50 times the annual production. This is a comfort zone. If the reserves fall below this zone, there is growing incentive to prospect for and develop new mines; if they rise above it, there is little incentive to do so.


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Chapter 3 E3.1 Which phase of life would you expect to be the most energy intensive (in the sense of consuming fossil fuel) for the following products? Pick one and list the resources and emissions associated with each phase of its life along the lines of Figure 3.3 (template provided).

Answer. At this stage we can only guess at the energy-intensive phase – later, when we get to ecoaudits, the question can be answered properly. As a general rule, if a product requires energy to perform the use-phase of its life, then it is the cumulative energy of use that dominates. When the product does not use energy, it is the material production phase that dominates. Product

Energy intensive life phase

A toaster

Probably the use phase

A two car garage

The material production phase if garage is unheated; if heated, then the use phase

A bicycle

The material production phase of life

A motorbike

The use phase

A refrigerator

The use phase

A coffee maker

Perhaps material production if the coffee maker is rarely used, otherwise use.

A LPG fired patio heater

No need to ask.

Take the coffee maker as an example.

Resources Energy: polymer molding Energy: metal forming Energy: glass molding

Resources Resources Oil for polymers Minerals for metals, glass Energy for both

Emissions Cut-offs (recycled) Low grade heat

Water Paper for filters Coffee Energy to heat water

Resources Emissions CO2, NOx, SOx Slag, tailings Chemical waste Low grade heat

Energy for collection Energy to disassemble Energy to recycle

Emissions Low grade heat Coffee grounds Waste filters

Emissions Materials (recycled) Waste to landfill


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E3.2 Functional units. Think of the basic need the product provides – it is this that determines use – and list what you would choose, thinking of all from an environmental standpoint Answer. The table lists the suggested functional units

Product

Functional unit

Washing machines

Energy (kW.hr or MJ) or liters of water per kg of clothes washed

Refrigerators

Energy (kW.hr or MJ) per m3 of cooled space per year

Home heating systems

Energy (kW.hr or MJ) per unit volume of heated space per unit time (per m3 per year)

Air conditioners

Energy (kW.hr or MJ) per unit volume of cooled space per unit time (per m3 per year)

Lighting

Power (Watts) per lumen (the measure of light intensity)

Home coffee maker

Energy (kW.hr or MJ) per cup of coffee

Public transport

Energy (kW.hr or MJ) per passenger.mile

Hand-held hair dryers

Energy (kW.hr or MJ) per drying episode (difficult: long hair takes longer than short)

E3.3 What is meant by “externalized” costs and costs that are “internalized” in an environmental context? Now a moment of introspection. List three internalized costs associated with your life style. Now list three that are externalized. If your life is so pure that you have less than three, then list some of other people you know. Answer. Many human activities incur costs, some obvious, some hidden. Industrial emissions, in particular, incur costs though damage to health and property via air and water pollution, acid rain, ozone layer depletion and climate change. When the damage is local and the creator of the emissions accepts the responsibility and the cost for containing and remediating it, the environmental cost is said to be internalized . When the damage not attributable, and particularly when it is global in impact, it is much more difficult to establish creator-responsibility. In these circumstances the environmental cost becomes a burden on society as a whole and is said to be externalized . Much current negotiation and legislation aims to internalize environmental costs that, at present, are externalized.

Internalized Cost of mobile phone (assuming microwave radiation is harmless) Cost of use of personal computer Cost of broadband connection Cost of private health care Cost of home ownership

Externalized Dropping cigarette butts, gum, hamburger wrappers in the street Disposing of television sets or personal computers by dumping Travel by plane (the traveler does not yet pay for the eco-damage of the emissions) Dumping, rather than recycling, nickel-cadmium batteries


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E3.4 What, in the context of life-cycle assessment, is meant by “system boundaries”? How are they set? Answer. To assess the energy demand, carbon footprint and emissions-profile of a product it is necessary to sum the contributions it makes to each of these over its entire life cycle. Deciding what to include and what exclude in the life-cycle (the “system”),however, is not simple. The first step in lifecycle assessment (LCA) is to decide on the system boundary. The boundary should normally enclose all four phases of life: material production, product manufacture, product use and product disposal. For the LCA to meet ISO standards, the LCA expert must define and state what aspects of each phase of life are included within the system boundary and which aspects remain outside it.

It is sometimes desirable to conduct a partial LCA, setting the system boundary around only one phase of life. Thus the embodied energy and carbon footprint of a material is found by enclosing only the material production within the system boundary.

E3.5 Describe briefly the steps prescribed by the ISO 14040 Standard for life-cycle assessment of products. Answer. The four steps in conducting an LCA, elaborated in the ISO 14040 set of standards, are: 

Setting goals and scope. Why do the assessment? What is the subject and which bit(s) of its life are to be assessed (setting system boundaries)?



Inventory compilation: what resources are consumed, what emissions excreted?



Impact assessment: what do these do to the environment – particularly, what bad things?



Interpretation: what do the results mean and how are they to be used?

E3.6 What are the difficulties with a full LCA? Why would a simpler, if approximate, technique be helpful? Answer.

•

A full LCA is a time-consuming task requiring days or weeks of experts’ time, and thus is expensive.

•

Despite the formalism that attaches to LCA methods, the results are subject to considerable uncertainty.

•

The output of an LCA that meets the ISO 14040 Standard is complex and detailed – it is of little help for design.

A simpler, approximate approach would, by reducing the cost and speeding the process, allow much wider application.

E3.7 Pick two of the products listed in Exercise E.3.1 and, using your judgment, attempt to fill out the simplified streamlined LCA matrix below to give an environmentally responsible product rating. Make your own assumptions (and report them) about where the product was made and thus how far it has to be transported, and whether it will be recycled. Assign an integer between 0 (highest impact) and 4 (least impact) to each box and then sum to give an environmental rating, providing a comparison. Try the protocol


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•

Material: is it energy-intensive? Does it create excessive emissions? Is it difficult or impossible to recycle? Is the material toxic? If the answer to these questions is yes, score 4. If the reverse, score 0. Use the intermediate integers for other combinations.

•

Manufacture: is the process one that uses much energy? Is it wasteful (meaning cut-offs and rejects are high)? Does it produce toxic or hazardous waste? Does make use of volatile organic solvents? If yes, score 4. If no, score 0, etc.

•

Transport: is the product manufactured far from its ultimate market? Is it shipped by air freight? If yes, score 4. If no, score 0.

•

Use: does the product use energy during its life? Is the energy derived from fossil fuels? Are any emissions toxic? Is it possible to provide the use-function in a less energyintensive way? Scoring as above.

•

Disposal: Will the product be sent to land-fill at end of life? Does disposal involve toxic or long-lived residues? Scoring as above.

What difficulties did you have? Do you feel confident that the results are meaningful?

Answer. Example: the toaster.

A toaster uses about 1 kg of materials, largely metals. Manufacture, almost certainly in South Asia, requires only simple processes, though chromium plating is a particularly toxic process. Most toasters used in the US and Europe are made in South-east Asia, so considerable transport is involved. A toaster consumes electrical energy during use, most of which at present derives from fossil fuels, with associated emissions. The toaster itself produces no emissions (unless you burn the toast) and is simple to recycle at end of life. The figure shows an attempt to fill out the matrix, based on this information.

It is not all easy to fill out the matrix with confidence unless you have considerable experience of assessing products in this way – the rankings chosen here are, quite frankly, guesses. This is just one example of the matrix approach – there are many variants of the row and column categories. Some are better adapted for particular products than others, but it is evident that any one of them requires a degree of experience. The present attempt (correctly) identifies the use and the transport phase (assuming the product is shipped over large distances) as the most damaging . The final total, 34 out of a possible maximum of 80 suggests the product is not a particularly harmful one.


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Chapter 4 E4.1 Many products are thrown away and enter the waste stream even though they still work. What are the reasons for this? Answer. A product reaches the end of its life when it’s no longer valued, even though it may still work well The life expectancy is the least of

•

The physical life, meaning the time in which the product breaks down beyond economic repair;

•

The functional life, meaning the time when the need for it ceases to exist;

•

The technical life, meaning the time at which advances in technology have made the product unacceptably obsolete;

•

The economical life, meaning the time at which advances in design and technology offer the same functionality at significantly lower operating cost;

•

The legal life – the time at which new standards, directives, legislation or restrictions make the use of the product illegal

•

And finally the loss of desirability – the time at which changes in taste, fashion, or aesthetic preference render the product unattractive.

E4.2 Do you think manufacture without waste is possible? “Waste”, here, includes waste (i.e. lowgrade) heat, emissions and solid and liquid residues that cannot be put to a useful purpose. If not, why not? Answer. All manufacture involves conversion: conversion of ores and feedstock into raw materials, conversion of raw materials into products, the use of the products and their ultimate disposal. All of these steps involve energy, sometimes in its primary form as oil or gas, but more usually in a converted form as electricity or mechanical power.

All conversions involve conversion efficiencies. Some conversion efficiencies are high – conversion of electric to mechanical power, for instance, can be achieved with an efficiency of 90% . Others, for fundamental thermodynamic reasons, are not – the conversion of fossil fuels to electric power is, at best, only 45% efficient. The losses, small or large, constitute one sort of waste – waste heat – and, if fossil fuels are involved, another: waste emissions. Just as there are conversion efficiencies for energy, there are conversion efficiencies for the transformation of materials during manufacture, and for the recovery of materials at end of life. Some are economic in origin – salvaging and sorting 100% of the scrap produced during manufacture is not economic, though salvaging 90% might be. Recovering materials at end of life has lower efficiencies because the dispersion of materials in products makes full recovery uneconomic, and because contamination of one material by another cannot, in some products, be prevented. All these factors combine to make some waste unavoidable.

E4.3 What options are available for coping with the waste-stream generated by modern industrial society? Answer. Materials have a life-cycle. They are extracted and refined, manufactured into products, used, and at end of first life, rejected as “waste”. But what is waste to some markets is a resource to others, creating a number of alternative channels down which the materials continue to flow. The sketch of Figure 4.1 of the text introduced the options: commit to landfill, combust for heat recovery, recycle (or downcycle), re-engineer (refurbish or recondition) and reuse.


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E4.4 Recycling has the attraction of returning materials into the use-stream. What are the obstacles to recycling? Answer. The recycling market is like any other, with prices that fluctuate according to the balance of supply and demand. In a free market the materials that are recycled are those from which a profit can be made. Recycling requires energy, and this energy carries its burden of emissions. The recycle energy is generally small compared to the initial embodied energy, making recycling – when it is possible at all – an energy-efficient propositions. But there are difficulties.

•

Recycling may not be cost-efficient; that depends on the degree to which the material has become dispersed. In-house scrap, generated at the point of production or manufacture is localized and is already recycled efficiently (near 100% recovery). Widely distributed “scrap” – material contained in discarded products – is a much more expensive proposition to collect, separate and clean.

•

Many materials cannot recycled, although they may still be reused in a lower-grade activity; continuous-fiber composites, for instance, cannot be re-separated economically into fiber and polymer in order to reuse them, though they can be chopped and used as fillers.

•

Most materials require an input of virgin material to avoid build-up of uncontrollable impurities. Thus the fraction of a material production that can ultimately re-enter the usestream depends on the material itself, on the product into which has been incorporated and on the price of virgin material with which the recycled material must compete.

E4.5 Car tyres create a major waste problem. Use the internet to research ways in which the materials contained in car tyres can be used, either in the form of the tyre or in some decomposition of it.

Answer. The following sites are among many that provide information about used tyres.

1.

www.ct.gov/dep/

2.

www.defra.gov.uk/environment/waste/topics/tyres.htm

3.

www.ni-environment.gov.uk/usedtyresleaflet.pdf

4.

www.tyredisposal.co.uk/cms/index.php

The first describes regulations about disposal of tyres, the second lists ways to use them, and the third provides statistics for collection, reuse and disposal. From them we learn that 3.9 million tons of used tyres are produced in the US every year, of which 18.7% are recycled. Tyres do not compost. The main channels for reusing used tyres are:

•

Retreading almost doubles the life of tyres. Most can only be retreaded once, however.

•

Rubber crumb is made by shredding tyres. It is used for playground and sports track surfaces.

•

Whole or part tyres can be used in wall structures for earth retention.

•

Sea defences can be constructed using floating arrays of tyres

•

Fuel: used tyres are burnt for heat, particularly in cement making.


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E4.6 List three important functions of packaging.

Answer. Packaging performs at least five functions

•

Protection: packaging extends product life by protecting foodstuffs and controlling the atmosphere that surrounds them.

•

Security: tamper-proof packaging protects the consumer.

•

Information: pack information identifies the product, its sell-by date (if it has one) and gives instructions for use.

•

Affiliation: brands are defined by their packaging.

•

Presentation: packaging “presents” the product in much the same way that clothing presents the wearer.

E4.7 As a member of a brain-storming group you are asked to devise ways of reusing polystyrene foam packaging – the sort that encases TV sets, computers, appliances and much else when transported – most of which at present is sent to landfill. Use free thinking: no suggestion is too ridiculous.

Answer. The idea here is to encourage free thinking. Here are just a few possible uses

•

Break up for bean bags.

•

Drainage for patio garden pots.

•

Flotation for fish-farm enclosures.

•

Shred to make artificial snow for theatre productions.

•

Send to Art schools as raw material for hot-wire sculptures.

•

Pack in the attic to improve the thermal insulation of your home / garage / outhouse.

E4.8 You are employed to recycle German washing machines, separating the materials for recycling. You encounter components with the following recycle marks:

How do you interpret them?

Answer. Chapter 4 and its Appendix (Section 4.8) gives the information needed to crack the codes. (a) Polypropylene. (b) Polyamide 6 (Nylon 6) with 10% glass fiber. (c) Polypropylene with 20% talc. (d) Polystyrene with 15% glass fines (powdered glass).


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E4.9 The metal lead has a number of uses: principally as electrodes in vehicle batteries, in architecture for roofing and pipe-work (particularly on churches) and as pigment for paints. The first two of these allow recycling, the third does not. Batteries consume 38% of all lead, have an average life of 4 years, a growth rate of 4% per year and the lead they contain is recycled with an efficiency of 80%. Architectural lead accounts for 16 % of total consumption. The lead on buildings has an average life of 70 years, the same growth rate (4% per year) as batteries, and 95% of it is recycled. What is the fractional contribution of recycled lead from each source to current supply?

Answer. Inserting these data into equation (4.4) show that lead from batteries contributes a fraction 0.26 that from buildings only 0.009 to current supply. The short life of the batteries means that the lead they contain makes a much larger contribution to the secondary scrap stream, even though the fraction recycled is smaller than that of architectural lead.

E4.10 A material M is imported into a country principally to manufacture one family of products with

an average life of 5 years and a growth rate of r c % per year. The material is not at present recycled at end of life but it could be. The government is concerned that imports should not grow. What level of recycling is necessary to make this possible?

Answer. Consumption C is growing at a rate r c % per year . Over the life of the product, Δt , it

grows from C o to r C = C o exp( c Δt ) 100

an increase of r c ΔC = C o exp( Δt ) − C o 100

For zero growth in imports recycling must feed back into the “Material” phase of life a quantity of material equal to the growth in consumption over the product life time. This requires a recycle fraction f crit where r f crit C o = ΔC = C o exp( c Δt ) − C o 100

Thus r f crit = exp( c Δt ) − 1 100

Clearly f crit cannot be greater than 1. Thus the scheme will only work if r exp( c Δt ) ≤ 2 100

or

Δt ≤

100 r c

ln( 2 )


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Chapter 5 E5.1 What is a Protocol? What do the Montreal Protocol and the Kyoto Protocol commit the signatories to do? Answer. The word “protocol” crops up most frequently in the language of medicine: there it means a detailed plan, or set of steps, to be followed in a study, an investigation or an intervention. The meaning of “protocol” in the context of this Chapter is a treaty – an agreement under international law entered into by the actors, namely states and international organizations.

•

The Montreal Protocol of 1987 commits the nations that signed it to phase out the use of chemicals that deplete ozone in the stratosphere.

•

The Kyoto Protocol of 1998 commits the nations that signed it to reduce the emissions of gases that, through the greenhouse effect, cause climate change.

E5.2 What is meant by “internalized” and “externalized” environmental costs? What legislative steps can be taken to pressure polluters to internalize environmental costs?

Answer. When environmental damage can be attributed to an particular industry or activity that accepts both the responsibility and the expense of containing and remediating it, the environmental cost is said to be internalized . When the damage is not attributable, the environmental cost becomes a burden on society as a whole and is said to be externalized .

Much current negotiation and legislation aims to internalize environmental costs that, at present, are externalized (the “Polluter pays” policy). This can take the form of outright bans, fines, penalties or taxes, all of which require a mechanism to insure compliance, and many of which are unpopular. The alternative is to use market forces to initiate change, through trading schemes, subsidies and other incentives to make environmental protection economically attractive.

E5.3. What is the difference between command and control methods and the use of economic instruments to protect the environment? Answer. Command and control environmental legislation makes it illegal to use specific resources or allow specific emissions – examples are bans on dumping of toxic waste, the use of lead in petrol, the release of certain chemicals that pollute water or those that damage the ozone layer. There is a growing recognition that this can lead to perverse effects where action to fix one isolated problem just shifts the burden elsewhere and may even increase it. For this reason there has been a shift from command and control legislation towards the use of economic instruments – green taxes, subsidies, trading schemes – that seek to use market forces to encourage the efficient use of materials and energy. Instead of outright bans, economic instruments seek to target environmental burdens that have costs that are not paid for by the provider or user ( externalized costs), transferring the costs back to the activity creating it, thereby internalising them.

E5.4 How does emissions trading work? Answer. Emissions trading is a market-based scheme that allows participants to buy and sell permits for emissions, or credits for reduction in emissions in certain pollutants. Taking carbon as an example, the regulator first decides on a total acceptable emissions level and divides this into tradable units


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called permits. These are allocated to the participants, based on their actual carbon emissions at a chosen point in time. The actual carbon emissions of any one participant change with time, falling if they develop more efficient production technology or rising if they increase capacity. A company that emits more than its allocated allowances must purchase allowances from the market, while a company that emits less than its allocations can sell its surplus. The buyer is paying a charge for polluting while the seller is rewarded for having reduced emissions. Emissions trading has another dimension – that of off-setting carbon release by buying credits in activities that absorb or sequester carbon or that replace the use of fossil fuels by energy sources that are carbon-free: tree planting, solar, wind or wave power for example. By purchasing sufficient credits the generator of CO2 can claim to be “carbon neutral”.

E5.5. Carbon trading sounds like the perfect control mechanism to enable emissions reduction. But nothing in this world is perfect. Use the internet to research the imperfections in the system and report your findings.

Answer. The criticisms of Carbon trading and offsetting are as follows.

•

It provides an excuse for enterprises to continue to pollute as before by buying credits and passing the cost on to the consumer.

•

The scheme only achieves it aim if the mitigating project runs for its planned life, and this is often very long. Trees, for instance, have to grow for 50 to 80 years to capture the carbon with which they are credited – fell them sooner for quick profit and the off-set has not been achieved. Wind turbines and wave power, similarly, achieved their claimed offset only at the end of their design life, typically 20 to 25 years.

•

It is hard to verify that the credit payments actually reach the mitigating projects – the tree planters or wind turbine builders – for which they were sold; too much of it gets absorbed in administrative costs.

E5.6. What are the merits and difficulties associated with (a) taxation and (b) trading schemes as economic instruments to control pollution?

Answer.

Historically, environmental legislation has targeted individual, obvious problems – dumping of toxic waste, lead in petrol, water pollution, ozone depletion – taking a command and control approach based on outright bans, fines, penalties or direct taxes. There is a growing recognition that this can lead to perverse effects where action to fix one isolated problem just shifts the burden elsewhere and may even increase it. Trading schemes and other economic instruments – green taxes, subsidies, carbon credits – seek to use market forces to encourage the efficient use of materials and energy. The motive is to transfer externalized costs back to the activity creating them. The difficulties are those of ensuring that the trading schemes function as a well-balanced market and are not distorted by high administrative costs or misuse.


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E5.7. Your neighbour with a large 4 x 4 boasts that his car, despite its size, is carbon-neutral. What on earth does he mean (or thinks he means)?

Answer. He means that he, or the company from which he bought the vehicle, has purchased carbon reduction certificates equal to the expected carbon emission of the vehicle. He thinks he means that these pay for projects that absorb carbon (like tree planting) or replace carbon-using energy sources with those that do not, but it is not always easy to verify how much of the payment reaches the project and how much is absorbed in administrative costs on the way.

E5.8. In December 2007 Saab posted advertisements urging consumers to “switch to carbon neutral motoring” claiming that “every Saab is green”. In a press release the company said they planned to plant 17 native trees for each car bought. The company claimed that its purchase of offsets for each car sold made Saab the first car brand to make its entire range carbon free.

What is misleading about this statement? (The company have since withdrawn it). Answer. The scheme only achieves it aim if the mitigating project runs for its planned life, and this is often very long. Trees, for instance, have to grow for 50 to 80 years to capture the carbon with which they are credited – fell them sooner for quick profit and the off-set has not been achieved. Wind turbines and wave power, similarly, achieved their claimed offset only at the end of their design life, typically 25 years. Even Saabs only last for 15 years, so the mitigation claimed for them is not even half-achieved by the time the vehicle dies.

E5.9. What tools are available to help companies meet the European Union VOC (volatile organic compounds) regulations? Carry out a Web search to find out and report your findings.

Answer. Numerous sites offer advice, help and new disposal methods for dealing with volatile organic compounds. Among them are

1.

en.wikipedia.org/wiki/Volatile_organic_compound

2.

www.parish-supply.com/volatile_organic_compounds.htm

3.

www.processingtalk.com

4.

www.coatings.org.uk/

5.

http://www.epd.gov.hk/epd

From them we learn that VOCs (volatile organic compounds) are found in wide variety of everyday products such as solvent-based paints, printing inks, many consumer products, organic solvents and petroleum products. The US, the European Union, Australia and New Zealand all are in progress of introducing labelling, restrictions on use, and a ban on sales without permit to limit the release of VOCs into the atmosphere. At present most sites focus on explaining the regulation; a few offer technical solutions for dealing with VOCs to render them harmless (generally by controlled combustion).


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E5.10. What tools are available to help companies meet the European Union EuP (Energy-using Products) regulations? Carry out a Web search to find out and report your findings.

Answer. Numerous sites report, offer advice and consultancy on the EuP Directive. Among them are:

1.

www.infoworld.com/article/06/10/31/45OPreality_1.html

2.

www.ee.sgs.com/eup_compliance_eup_ee

3.

www.synapsistech.com/solutions/compliance/eup.html

4.

www.era.co.uk/Services/ecodesign.asp

From these we learn the following

•

The Directive. The EuP Directive will require manufacturers to calculate the energy used to produce, transport, sell, use, and dispose of almost every one of its products. It will require that the manufacturer go all the way back to the energy used when extracting the raw materials to make its product, including all subassemblies and components. And in time, it will set limits on a product-by-product basis of how much energy can be used in a product’s entire lifecycle (Source (1), above).

•

EuP compliance in brief. Manufacturers can choose to perform the conformity assessment either by ecological product profiles or within a management system like ISO 14001 or EMAS. Companies will be required to produce a document file for each of their products and should therefore start by collecting data on the mentioned environmental aspects in the different life cycle phases (Source (2)).

•

Services and tools to help with EuP compliance. A number of web sites (Source (3) is an example) offer in-house training and planning assistance to help companies set up procedures to comply with the Directive. Others (e.g. Source (4) go further, offering to design and put it place compliance procedures. It appears that many organizations with environmental knowledge are seeking to capture what maybe a large market in establishing procedures that are practical and affordable, but none have yet emerged as clear market leader.


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Chapter 6 E 6.1. What is meant by embodied energy per kilogram of a metal? Why does it differ from the free energy of formation of the oxide, carbonate or sulfide from the which it was extracted? Answer. The embodied energy of a metal is the output of an energy audit of the resources and processes need to extract, refine or synthesize it. Energy is consumed in mining or extracting the ores and feedstock, concentrating the ore and reducing it to metal, all with energy losses. To convert an oxide, sulfide, or carbonate ore to metal, the free energy of formation of those compounds must, of course, be provided. The embodied energy includes this, but it also includes the energy to mine, transport and concentrate the ore, and the energy “loss” that results from the inefficiencies of the processes.

E 6.2 What is meant by the process energy per kilogram for casting a metal? Why does it differ from the latent heat of melting of the metal? Answer. In order to cast a metal, energy must be provided to heat the metal to its melting point, to melt it (requiring the latent heat of melting) and to raise the temperature of the liquid a little higher to ensure that, when cast, it does not solidify before it has filled the mold. The kiln itself has to be heated to the same temperature, requiring additional energy. The heat itself is provided by the combustion of fossil fuel (efficiency about 70%) or by electric heating (oil-equivalent efficiency about 40%). The thermal energy of both kiln and metal is lost as low-grade heat when the casting cools to room temperature. The casting energy is the sum of all the energies involved.

E 6.3 Make a bar chart of CO2 footprint divided by Embodied energy using data from the data sheets of Chapter 12. Which material has the highest ratio? Why? Answer. The Figure shows the important features. For most materials the ratio is about0.04 - 0.06. For cement and concrete it is far higher. This is because, making cement (a component of concrete) involves “calcining” calcium carbonate, driving off carbon dioxide to leave lime. This CO2 contribution adds to that derived from the fuel that is used to heat the kiln to give the high total.


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E 6.4 The embodied energies and CO 2 footprints for woods, plywood and paper do not include a credit for the energy and carbon stored in the wood itself, for the reasons explained in the text. Recalculate these, crediting them with sequestering energy and carbon by subtracting out the stored contributions (take them to be 25 MJ/kg and 2.8 kg CO 2 per kg). Is there a net saving?

Answer. The table lists the embodied energies and carbon footprints for wood, plywood and paper, taking the means of the ranges shown on the data sheets. The last two columns show the values adjusted as described in the question. Crediting the materials with storing energy and carbon gives all of them a negative carbon footprint and all but one (paper) a negative embodied energy. The reasons we use the “uncorrected” values are explained in the text.

Material

Embodied energy (MJ/kg)

CO2 footprint (kg/kg)

Embodied energy, adjusted (MJ/kg)

CO2 footprint, adjusted (kg/kg)

Wood

7.4

0.43

-17.6

-2.37

Plywood

15

0.75

-10

-2.05

Paper

27

1.44

2

-1.36

E6.5 Rank the three common commodity materials Low carbon steel, Aluminum alloy and

Polyethylene by embodied energy/ kg, H m ,and embodied energy/m3, H m ρ , where ρ is the density,

using data drawn from the data sheets of Chapter 12 (use the means of the ranges given in the databases). Finally rank them by embodied energy per unit stiffness (measured by H m ρ / E where E is Young’s modulus).

Answer. The table shows the data and the calculated information. Column 2 gives the embodied

energy per kg, H m . Steel has the lowest. Column 5 lists the embodied energy per unit volume, H m ρ – polyethylene has by far the lowest. The last column is the embodied energy per unit stiffness (note the inclusion of ρ in the ratio so that top and bottom are both in the same units). Now steel is the lowest, and by a large margin. If you want a material that is stiff and has low embodied energy, steel is the best bet.

Material

MJ/kg

kg/m

Modulus E GPa

Low carbon steel

32

7850

207

2.51 x 105

1.2

Aluminum alloys

220

2700

75

5.94 x 105

7.9

Polyethylene

81

950

0.74

7.7 x 104

104

Energy H m

Density ρ 3

H m ρ MJ/m

(Dimensionless)

3

H m ρ / E

E6.6 Iron is made by the reduction of iron oxide, Fe 2O3, with carbon, aluminum by the electrochemical reduction of Bauxite, basically Al2O3. The enthalpy of oxidation of iron is 5.5 MJ/kg, that of aluminum to its oxide is 20.5 MJ/kg. Compare these with the embodied energies of cast iron and of aluminum, retrieved from the data sheets of Chapter 12 (use means of the ranges given there). What conclusions do you draw?


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Answer. The table shows the data. The embodied energies are larger, by a factor of 3 to 10, than the enthalpy of oxidation. This is because of the inherent irreversibility of the reduction process, heat and other energy losses, and the energy required for mining transporting and concentrating the ores before reduction.

Material

Enthalpy of oxidation

Embodied energy

(MJ/kg)

(MJ/kg) from Chapter 12

Cast iron

5.5

17

Aluminum

20.5

220

E6.7 Estimate the energy to mold PET by assuming it to be equal to the energy required to heat PET from room temperature to its melting temperature, T m. Compare this with the actual molding energy. You will find the molding energy, the specific heat and the melting temperature in the data sheet for PET in Chapter 12 (use means of the ranges). What conclusions do you draw?

Answer. PET melts at about 238 C, about 200 C above room temperature. Its specific heat is 1445 J/kg.K, so the energy to raise one kg of PET from room temperature (20 o C) to the melting point is about 0.31 MJ/kg. The extrusion energy is 3.8 MJ/kg, the molding energy 9.6 MJ/kg. Both are more than ten times greater than the estimate based simply on heat the polymer.

Where does the extra energy go? Part in the relatively low conversion efficiency of fossil-fuel energy to electric power (about 38%), part in heating the extrusion or molding press, in power dissipated by the mechanism of the press, and as incidentals – the general energy overhead of the plant.

E6.8. The data sheets of Chapter 12 list eco-indicator values where these are available. As explained in the text, the eco-indicator value is a normalized, weighted sum involving resource consumption, emissions and estimates of impact factors. Plot eco-indicator values against embodied energy (a much simpler measure of impact). Is there a correlation?

Answer. See the chart in the answer to exercise E6.11 (below). It is a CES plot of eco-indicator values against embodied energy. There is a clear correlation, but with scatter. Is the scatter significant? Think back to the inherent uncertainty in determining embodied energy (Figure 6.2) and in the arbitrary nature of the weight factors used to calculate eco-indictors (Figure 3.11 and its discussion). Given these, you would expect some scatter. We conclude the embodied energy is an approximate but still useful proxy for the eco-indicator.


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Exercises using the CES software. E6.9. Figures 6.8 and 6.9 of the text are plots of the embodied energy of materials per kg and per m 3. Use CES to make similar plots for the carbon footprint. Use the “Advanced” facility in the axis selection window to make the one for kg CO 2/m3 by multiplying kg CO2/kg by the density in kg/m3.

Answer. The chart is shown below. Not surprisingly, it looks very like the embodied energy per m3, shown in Figure 6.9 of the text. (The Metals, Polymers, Ceramics and Hybrids are segregated to separate columns by selecting them for the x-axis using the “Trees” option in the Advanced facility.)

E6.10 Plot a bar chart for the embodied energies of metals and compare it with one for polymers, on a a “per unit yield strength” basis, using CES. You will need to use the “Advanced” facility in the axisselection window to make the function

H m ρ σ y

,

where H m is the embodied energy per kg, ρ is the density and σ y is the yield strength. Do this by using the “Advanced” facility in the Axis selection box to form (Embodied energy * Density) / Yield strength (elastic limit) Which materials are attractive by this measure?


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MFA, 08/12/2008

Answer. The figure shows the CES output. (The Metals and Polymers are segregated to separate columns by selecting them for the x-axis using the “Trees” option in the Advanced facility.) Carbon steels and cast irons have a lower embodied energy per unit of strength – than any other metal or any polymer.

E6.11 Compare the eco-indicator values of materials with their embodied energy. To do so, make a chart with (Embodied energy x Density) on the x-axis and Eco-indicator value on the y-axis. (Ignore the data for foams since these have an artificially-inflated volume). Is there a correlation between the two? Is it linear? Given that the precision of both could be in error by 10 % are they significantly different measures? Does this give a way of estimating, approximately, eco-indicator values where none are available?

Answer. The plot of eco-indicator values against embodied energy is shown below. There is a clear correlation, but with some scatter. See the answer to Exercise 6.9 for a commentary on this.


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Eco – indicator values

E6.12 Plot material price against annual production in tonnes per year – is there a correlation?

Answer. The plot shows that there is a correlation. The grid of broken lines has a slope of -2, meaning

Annual world production ∝

1 (Pr ice ) 2

It is no surprise that production falls as price rises. The origin of the power of -2 is obscure.

Slo e -2


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Chapter 7 E7.1 If the embodied energies and CO 2 used in the Alpure water case study in the chapter are uncertain by a factor of ± 25%, do the conclusion change? Mark ± 25% margins onto each bar in a copy of Figure 7.3 (you are free to copy it) and then state your case. Answer. The error margins are marked on the figures below. Even when the most extreme values are taken, the dominant contributions to energy and carbon are those of the material of the bottle. The conclusions reached in the text still stand.

E7.2 Alpure water has proved to be popular. The importers now wish to move up-market. To do so they plan to market their water in 1 liter glass bottles of appealing design instead of the rather-ordinary PET bottles with which we are familiar from the case-study in the chapter. A single 1-liter glass bottles weigh 430 grams, much more than the 40 grams of those made of PET. Critics argue that this marketing-strategy is irresponsible because of the increased weight. The importers respond that glass has lower embodied energy than PET.

Use the methods of this chapter and the data available in Chapter 12 to analyze this situation. What do you conclude? Answer. The energy audit for 100 glass bottles, carried out in the same way as the PET bottles described in the text and assuming the glass is recycled at end of life, gives the breakdown shown below. (Data are taken from the record for soda-glass in Chapter 12, using mean values of the ranges of embodied and process energies). The total life energies without recycling are summarized in the table below. The choice of glass is almost 3 times more energy intensive than PET.

If both the PET and the glass are recycled, energy is “recovered”: it is the difference between the embodied energy and the recycle energy, also listed in the data sheets of Chapter 12. The last column of the table lists the life-energy with recycling; the bar chart shows the contributions. The glass bottle remains almost three more energy intensive than the one made of PET.

Material (100 bottles)

Life energy without recycling (MJ)

Life energy with recycling (MJ)

PET bottles

490

300

Glass bottles

1200

830


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Eco-audit for 100 glass Alpure bottles.

E7.3 If the glass container of the coffee maker audited in Section 7.4 of the text is replace by a double walled stainless steel one weighing twice as much, how much does the total embodied energy of the product change? If this reduces the electric power consumed over life by 10%, does the energy balance favor the substitution? Answer. The table lists the data and results, using information about the mass of the container and the energy consumption over life given in the question. The stainless steel jug is twice as heavy as the glass one but saves 10% of the use energy. Using data for the embodied energy of and borosilicate glass and stainless steel from Chapter 12 gives the results in the table. The stainless steel jug carries 45 MJ more embodied energy but saves more than three times this energy over life. In addition, borosilicate glass is difficult to recycle (it cannot be mixed with ordinary glass) whereas stainless steel is readily recycled.

Material

Mass (kg)


Energy per container (MJ)

Electrical energy over 5 years (MJ)

Borosilicate glass

0.33

25

8.25

1380

Stainless steel

0.66

81

53.5

1240

Difference:

+ 45 MJ

- 140 MJ

E7.4 The figure shows a 1700 Watt steam iron. It weighs 1.3 kg, 98% of which is accounted for by the 7 components listed in the table. The iron heats up on full power in 4 minutes, is then used, typically, for 20 minutes on half power. At end of life the iron is dumped as landfill. Create an eco-audit for the iron assuming that it is used once per week over a life of 5 years, using data from the data sheets of Chapter 12. Use PVC as a proxy for polyurethane.

What conclusions can you draw? How might the energy be reduced?


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Steam iron: bill of materials

Component

Mass (kg)

Material

Body

0.15

Polypropylene

Heating element

0.03

Base

Shaping process

Material energy* MJ/kg

Process energy* MJ/kg

Molded

97

8.6

Nichrome

Drawn

134

2.65

0.80

Stainless steel

Cast

82

3.35

Cable sheath, 3 meter

0.18

Polyurethane

Molded

82

10.1

Cable core, 3 meter

0.05

Copper

Drawn

71

2.0

Plug body

0.037

Phenolic

Molded

90

12.7

Plug pins

0.03

Brass

Rolled

71

2.0

* From the data sheets of Chapter 12

Answer. The material energy is found by multiplying the mass of each component by its embodied energy/kg (listed, using data from Chapter 12, on the right of the table) and summing. The result is 108 MJ. The process energy is found in the same way; here the result is 6.5 MJ. To calculate the use phase energy multiply the power (1.7 kW) by the effective use-time on full power (4 minutes on full power, 20 on half power, equivalent to 14 minutes per day, 52 days per year, 5 years) converted to seconds, then divided by 1000 (to convert kJ to MJ), giving 371 MJ. But this is electrical energy, generated from fossil fuel with an efficiency of around 38%, so to get the “oil equivalent” energy this electrical energy must be divided by 0.38, giving 977 MJ.

The bar chart shows this distribution of energy commitment over the phases of life. The use phase dominates, accounting for 90% of the total energy. About 29% of the use-phase energy is used to heat the iron itself up to the working temperature, requiring energy Q = m C p ΔT

Where m is the mass of the base of the iron (0.8 kg), C p is its specific heat and ΔT is the temperature interval through which it is heated. Thus energy could be saved by reducing the mass of the base of the iron or by selecting a material for it with a lower specific heat, or both.


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E7.5 The figure shows a 970 Watt toaster. It weighs 1.2 kg including 0.75 m of cable and plug. It takes 2 minutes 15 seconds to toast a pair of slices. It is used to toast, on average, 8 slices per day, so it draws its full electrical power for 9 minutes (540 seconds) per day over its design life of 3 years. The toasters are made locally – transport energy and CO 2 are negligible. At end of life it is dumped. Create an eco-audit for the toaster using data from the data sheets of Chapter 12. Toaster: bill of materials

Component

Mass (kg)

Body

0.24

Heating element

Material

Shaping process

Material energy* MJ/kg

Process energy* MJ/kg

Polypropylene

Molded

97

8.6

0.03

Nichrome

Drawn

134

2.65

Inner frame

0.93

Low carbon steel

Rolled

32

2.4

Cable sheath, 0.75 meter

0.045

Polyurethane

Molded

82

10.1

Cable core, 0.75 meter

0.011

Copper

Drawn

71

2.0

Plug body

0.037

Phenolic

Molded

90

12.7

Plug pins

0.03

Brass

Rolled

71

2.0

* From the data sheets of Chapter12

Answer. The material energy is found by multiplying the mass of each component by its embodied energy/kg (listed, using data from Chapter 12, on the right of the table) and summing. The result is 67 MJ. The process energy is found in the same way; here the result is 5.4 MJ. To calculate the use-phase energy multiply the power (0.97 kW) by the use-time on full power (9 minutes per day, 365 days per year, 3 years) converted to seconds, then divided by 1000 (to convert kJ to MJ), giving 574 MJ. The result is the MJ of electrical energy consumed over life. But this is electrical energy, generated from fossil fuel with an efficiency of around 38%, so to get the “oil equivalent” energy this electrical energy must be divided by 0.38, giving 1510 MJ.

The bar chart shows this distribution of energy commitment over the phases of life. The use phase dominates, accounting for 90% of the total energy. How could the energy efficiency be increased? Air circulates through the toaster during use, necessary to carry off the moisture distilled from the bread, but carrying off energy too. Optimizing this air flow and the use of infra-red lamp elements instead of the resistance coils could focus the heat more precisely where it is needed. When toasting a single slice in a 2-slice toaster half the heat is wasted. Provision of a switch or sensor to limit heating to the slot containing bread would significantly reduce energy consumption.


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E7.6 It is proposed to replace the low-alloy steel bumper set of Case Study 7.6 in the text by one made of CFRP. It is anticipated that the CFRP set will weigh 7 kg. Following the procedure of the text, drawing data from the data sheets of Chapter 12, to estimate whether, over the life pattern used in the text, there is a net energy saving.

Answer. The table below lists mean values for the embodied energies of the two materials. Using the energy consumption of a gasoline-powered car as 2.1 MJ/tonne.km (Table 6.7 of the text) we find the use-energy over 250,000 km to have the values shown in the second last column. The two bar charts display the relative magnitudes of material and use energies. Based on this very crude comparison, the substitution gives a substantial energy saving if the vehicle is driven the full 250,000 km. The breakeven distance is found by equating the total energy associated with the steel bumper-set to that of the CFRP set for a driven distance of X km, and solving for X. The result is 98,500 km.

Mass kg

Material energy, MJ/kg*

Material energy, MJ

Use energy, MJ

Total: material plus use, MJ

Low alloy steel

14

35

490

7210

7700

CFRP

7

273

1910

3610

5520

Material of fender

* From the data sheets of Chapter 12.

E7.7 It is reported that the production of a small car (mass 1000 kg) requires materials with a total embodied energy of 70 GJ, and a further 15 GJ for the manufacturing phase. The car is manufactured in Germany and delivered to the US show room by sea freight (distance 10,000 km) followed by delivery by heavy truck over a further 250 km (Table 6.7 of the text gives the energy per tonne.km for both). The car has a useful life of 10 years, and will be driven on average 25,000 km per year consuming 2MJ/km. Assume that recycling at end of life recovers 25 GJ per vehicle.

Make an energy-audit bar chart for the car with bars for material, manufacture, distribution, use and disposal. Which phase of life consumes most energy? The inherent uncertainty of current data for embodied and processing energies are considerable – if both of these were in error by a factor of 2 either way can you still draw firm conclusions from the data? If so, what steps would do most to reduced lifeenergy requirements?


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MFA, 08/12/2008

Answer. The figure shows the bar chart based on the data with the factor of 2 uncertainty indicated by T bars showing the ranges on the Material and the Manufacturing bars. Even with this colossal allowance for the imprecision of the base-line data the conclusion is inescapable: the use-phase of the vehicle remains the dominant contribution to life-energy consumption (more on this in Chapter 9). Material choice should focus first on reducing vehicle mass since it is this that most directly correlates with fuel consumption. It is essential to then check that the consequent change in embodied energy of the material choice has not negated the gain resulting from the lower mass, using the method of exercise E7.6.

E7.8 The table below lists one European automaker’s summary of the material content of a mid-sized family car. Material proxies for the vague material descriptions are given in brackets and italicized. The vehicle is gasoline-powered and weighs 1800 kg. The data sheets of Chapter 12 provide the embodied energies of the materials: mean values are listed in the table. Table 6.7 gives the energy of use: 2.1 MJ/tonne.km, equating to 3.8 MJ/km for a car of this weight. Use this information here to make an approximate comparison of embodied and use energies of the car assuming it is driven 25,000 km per year for 10 years.

Material content of a family car, total weight 1800 kg Material

Mass (kg)

Material energy, MJ/kg*

Energy per car GJ

Steel (Low alloy steel)

950

32

30.4

Aluminum (Cast aluminum alloy)

438

220

96.4

Thermoplastic polymers (PU, PVC)

148

80

11.8

Thermosetting polymers (Polyester)

93

88

8.2

Elastomers (Butyl rubber)

40

110

4.4

Glass (Borosilicate glass)

40

15

0.6

Other metals (Copper)

61

72

4.4

Textiles (Polyester)

47

47

2.2

Total mass

1800

Total energy

158.4



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MFA, 08/12/2008

Answer. The material energy of the car is found by multiplying the mass of each material used in it by its embodied energy. The column added to the right of the table above lists the contributions and their sum, 158 GJ per car. The bar chart below shows the comparison. The input data are of the most approximate nature, but it would take very large discrepancies to change the conclusion: the energy consumed in the use phase (here about 85%) greatly exceeds that embodied in the materials of the vehicle. Fuel consumption scales with weight (more detail on this in Chapter 9). The answer here is to make the car from lighter materials – provided, of course, that the embodied energy of the replacement is not so high that it offsets the gain made by light-weighting.

E7.10 Conduct a CO2 eco-audit for the patio heater shown here. It is manufactured in SE Asia and shipped 8,000 km to the US where it is sold and used. It weighs 24 kg, of which 17 kg is rolled stainless steel, 6 kg is rolled carbon steel, 0.6 kg is cast brass and 0.4 kg is unidentified injection-molded plastic (so use a proxy of your own choosing for this). In use it delivers 14 kW of heat (“enough to keep 8 people warm”) consuming 0.9 kg of propane gas (LPG) per hour, releasing 0.059 kg of CO 2 /MJ. The heater is used for 3 hours per day for 30 days per year, over 5 years, at which time the owner tires of it and takes it to the recycling depot (only 6 miles / 10 km away, so neglect the transport CO 2) where the stainless steel, carbon steel and brass are sent for recycling. Use data from the text and data sheets of Chapter 12 to construct a bar-chart for CO2 emission over life.

Answer. The table below lists the carbon footprints of the materials and manufacturing processes for the patio heater. The bar chart plots the totals. Sea transport over 8000km, consuming 0.015 kg CO2/ tonne.km (from Table 6.7 of the text) releases 2.9 kg of carbon dioxide per unit, so small as to be invisible on the bar chart. The use of the heater, delivering 14 kW, or 50.4 MJ/hour with the release of 0.059 kg CO2/MJ, emits 1338 kg of CO2.

Recycling saves the difference between the carbon footprint of virgin and that of recycled material. The second table lists these differences, using data for virgin and recycled material from the data sheets of Chapter 12. Not surprisingly, the carbon emission during use (over 1 tonne of CO 2 over the course of life) dominates. There is only one way to reduce the carbon footprint of a device like this: turn it off.


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MFA, 08/12/2008

Material

Mass (kg)

Material CO2 kg/kg*

Material CO2 Process CO2 per unit, kg kg/kg*

Process CO2 per unit, kg

Stainless steel, rolled

17

5.05

85.9

0.285

4.8

Carbon steel, rolled

6

2.5

15

0.195

1.2

Brass, cast

0.6

5.25

3.2

0.16

0.1

Thermoplastic polymer molded (polypropylene)

0.4

2.7

1.1

0.68

0.27

Totals

24

105

6.4


Recycled material

Mass (kg)

Difference between material CO2 and recycling CO2 kg/kg*

CO2 saved by recycling per unit, kg

Stainless steel, rolled

17

3.6

61.2

Carbon steel, rolled

6

1.8

10.8

Brass, cast

0.6

3.95

2.4

Thermoplastic polymer molded (polypropylene)

0.4

1.55

0.6

Totals


24

75

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Exercises using the CES eco-audit tool These exercises repeat or extend those of the text or of the preceding exercises, using the ecoaudit tool of the CES Edu software (found in the pull-down menu under “Tools”). The output and bar charts in some cases differ slightly from those of the earlier exercises because of the way the software creates means of the property ranges (it uses geometric rather than arithmetic means, more logical when ranges are large) and because it calculates the oil equivalent of electrical energy using a countrydependent energy mix (the mix of fossil fuel, hydro, wind and nuclear in the grid supply).

E7.11 Repeat the analysis of the PET bottle for the Alpure water – the first case study in the Chapter – entering data from Table 7.1 the CES eco-audit tool. Then repeat, replacing the PET bottle by a glass one, using the additional information in exercise E7.2.

Answer. The resulting bar charts and report reproduce those already presented in Exercises E7.1 and 7.2.

E7.12 Carry out the eco-audit for the jug kettle of Section 7.3, now using the fact that the PP kettle body and the cardboard packaging are recycled.

Answer. Enter the bill of materials for the jug kettle, listed in the first four columns of Table 7.5 of the text into the eco-audit tool, selecting “Recycle” for end of life for the PP body and the packaging and “Landfill” for all the others. The report summary and bar chart delivered by the software are show below, here using the electrical energy mix for Australia. The recycling offers a potential energy saving of 51 MJ, 2.4% of the energy total.

Summary: jug kettle with recycling

Life phase

Energy, MJ

Material

120

Manufacture

10

Transport

130

Use

1900

End of life

Jug kettle

-51 Total


2110

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E7.13 Carry out the eco-audit for the coffee maker of Section 7.4, now using the fact that it is made entirely from recycled materials (select the grade “100% recycled”) and that the PP housing and the steel parts are themselves recycled at end of life.

Answer. Enter the bill of materials for the coffee maker, listed in the first four columns of Table 7.6 of the text into the eco-audit tool, selecting “Recycle” for end of life for the PP body (as in the original case study of the text) and “Landfill” for all the others. The report summary delivered by the software is show below, here using the average electrical energy mix for Europe, and selecting “Virgin materials” for all. The resulting bar-chart is shown below – it reproduces that shown as Figure 7.7 of the text. The material energy of 147 MJ is 8% if the total.

When all the material inputs are entered instead as “100% recycled” the material energy falls to 59 MJ, now only 4 % of the total.

Coffee maker

Summary: Coffee maker, virgin materials

Life phase

Energy, MJ

Material

147

Manufacture

13

Transport

6

Use

1480

End of life

-52 1594

Total

E7.14 Carry out the eco-audit for CO 2 for the portable space heater using the bill of materials from Section 7.5, Table 7.7 of the text, assuming that at end of first life the fan and heat shield are reengineered to incorporate them into a new product.

Answer. Enter the bill of materials for the space heater, listed in the first four columns of Table 7.7 of the text into the eco-audit tool, selecting “Re-engineer” for end of life for the heat shield and all the components of the fan, and “Landfill” for all the others. The report summary and bar-chart are shown below.

Re-engineering (and then reusing) the fan and heat shield saves 3.3 kg of CO 2.

Space heater

Summary: Space heater

Life phase

CO2, kg

Material

19

Manufacture

1.7

Transport

1.3

Use

190

End of life

-3.3 Total


209

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E7.15 Carry out Exercise E7.5 (the toaster) using the CES eco-audit tool, using the bill of materials listed there. Make bar charts both for energy and for CO2. Then select the grade “Standard” in the calculation, and repeat (“Standard” grade is material with a recycle content equal to the recycle fraction in current supply, data for which is included in the data sheets). Compare the result with using virgin material for making the toaster .

Answer. Enter the bill of materials for the toaster, from Exercise E7.5 into the eco-audit tool. The report summaries and bar-charts for energy and CO 2 when virgin materials are used are shown below. The same audit, using “Standard” grades for all materials reduces the Material energy from 68 to 56 MJ and the Material CO2 from 3.7 to 2.9 kg.

Summary: Toaster - energy

Life phase

Toaster

Energy, MJ

Material

68

Manufacture

5.4

Transport

0

Use

1250

End of life

0 1323

Total

Summary: Toaster – CO 2

Life phase

CO2, kg

Material

3.7

Manufacture

0.4

Transport

0

Use

164

End of life

0 Total


168

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Chapter 8 E 8.1. What is meant by an objective and what by a constraint in the requirements for a design? How do they differ?

Answer. A constraint is an essential condition that must be met, usually expressed as an upper or lower limit on a material property. An objective is a quantity for which an extreme value (a maximum or minimum) is sought, frequently the minimization of cost, mass, volume or – of particular relevance here – environmental impact. The distinction is brought out by comparing common constraints and objectives, as in the table below. A constraint is a “Go / No go” criterion – if the constraint is not met, the material is rejected. An objective allows the materials that meet all the constraints to be ranked: the one that minimizes the objective is the best choice.

Common constraints Must meet a target value of Stiffness 









Strength Fracture toughness Thermal conductivity Service temperature

Common objectives











Minimize Cost

Mass Volume Energy consumption Carbon emissions

E 8.2. Describe and illustrate the “Translation” step of the material selection strategy. Materials are required to make safe, eco-friendly swings and climbing frames for a children’s’ playground. How would you translate these design requirements into a specification for selecting materials?

Answer. The design requirements for a component of a product specify what it should do but not what properties its materials should have. Translation is the step of converting the design requirements into constraints and objectives that can be applied to a materials database. Thus the design requirement “Materials are required to make safe, eco-friendly swings and climbing frames for a children’s’ playground” might translate as shown in the table.

Design requirements

Materials are required to make safe, eco-friendly swings and climbing frames for a children’s’ playground

Translation

Constraints: seek materials that meet specified levels of Stiffness 





Strength Fracture toughness

Objective: minimize Carbon footprint of material 


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E8.3. Bicycles come in many forms, each aimed at a particular sector of the market: 

Sprint bikes



Touring bikes



Mountain bikes



Shopping bikes



Children’s bikes



Folding bikes

Use your judgment to identify the primary objective and the constraints that must be met for each of these.

Answer. Suggested constraints and objectives are listed in the table below. The balance between cost and other objectives (like mass) depend on the consumer-group at which the bike is aimed.

Constraints

Bike type

Objective(s)

Spring bikes

Meet target values of stiffness and strength.

Minimize mass

Touring bikes

Meet target value strength and mass.

Minimize cost

Mountain bikes

Meet target values of stiffness and strength

Maximize damage tolerance (toughness, fatigue strength)

Shopping bikes

Meet target values of strength and mass.

Minimize cost

Children’s bikes

Meet target values of strength and toughness, ease of molding or forming.

Minimize cost

Folding bikes

Meet target values of stiffness and strength.

Minimize mass (for portability) and volume (for compactness).

E 8.4. You are asked to design a fuel-saving cooking pan with the goal of wasting as little heat as possible while cooking. What objective would you choose, and what constraints would you recommend should be met?

Answer. The table lists constraints and an objective Constraints

• Must be non-toxic • Meet an upper limit on • •

Objective: minimize

• Thermal resistance (thus maximize thermal conductivity)

acceptable cost Meet constraints of formability Meet FDA* standards for contact with food.

• Must be recyclable (to save embodied energy) *The US Food and Drug Administration approve s materials for contact with foods.


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E 8.5. Formulate the constraints and objective you would associate with the choice of material to make the forks of a racing bicycle. Answer. The important constraints are those of meeting target values of stiffness (for control and comfort) and strength (for safety). Cost, in highly competitive sport, is irrelevant. Mass, however, carries a heavy performance penalty. The central objective is to minimize mass.

E8.6. What is meant by a material index?

Answer. The performance of a component or product is limited by the properties of the materials of which it is made. Sometimes it is limited by a single property, sometimes by a combination of them. The property or property-group that limits performance for a given design is called its material index. Performance is maximized by seeking the materials with the most extreme (biggest or smallest) value of the index, usually the smallest.

E8.7. The objective in selecting a material for a panel of given in-plane dimensions for the lid-casing of

an ultra-thin portable computer is that of minimizing the panel thickness h while meeting a constraint on bending stiffness, S * to prevent damage to the screen. What is the appropriate material index?

Answer. The index, read from Table 8.3 of the text, is that for a stiffness limited panel of minimum

volume: minimize 1 / E 1 / 3 , where E is Young’s modulus (or, more strictly, the flexural modulus, equal to E for isotropic materials but not for those, like fiber composites, that are not isotropic).

E8.8. Plot the index for a light, stiff panel on a copy of the Modulus-Density chart of Figure 8.11, positioning the line such that six materials are left above it, excluding ceramics because of their brittleness. Which six to you find? What material classes do they belong to? 1 / 3 Answer. The index for selecting materials for light stiff panels, read from Table 8.3, is ρ / E .

Construct a line by hand, or on a scanned image of the chart in WORD, with the slope indicated by the guide line on the figure, then position it so that a few materials lie to its left, as shown. These materials have the lowest (best) values of the index. Omitting ceramics and glasses because of brittleness, we find

•

Woods

•

CFRP (carbon fiber reinforced polymers)

•

Magnesium (Mg) alloys

•

Three density-grades of rigid polymer foams.

Polymer foams have the problem that their modulus is low, so although they are light, the panel has to be thick to achieve much stiffness. It is possible to apply a constraint on thickness (it translates into a lower limit for modulus); applying it then removes any materials that can only meet the stiffness constraint if they are too thick.


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MFA, 08/12/2008

Index ρ / E 1 / 3

The Modulus – Density chart: the one for stiffness at minimum weight.

E8.9. Panels are needed to board-up the windows of an unused building. The panels should have the lowest possible embodied energy but be strong enough to deter an intruder who, in attempting to break in, will load the panels in bending. Which index would you choose to guide choice?

Plot the index on the Strength-Embodied energy chart of Figure 8.14, positioning the line to find the best choice, excluding ceramics because of their brittleness. Which six to you find? What material classes do they belong to?

Answer. The index for a strength limited panel of minimum embodied energy, read from Table 8.3 of

the text, is H m ρ / σ y1 / 2 , where H m is the embodied energy per kg of the material, ρ is its density and σ y its yield strength. The index is plotted on the Strength – Embodied energy chart using the procedure described in the answer to example E8.8, reading the slope from the appropriate guide line in the lower right of the chart. It is position to leave a few materials to its left, as shown. Ignoring ceramics because or brittleness, the best choices are

•

Woods

•

Cast irons

•

Carbon steels

•

Three density-grades of rigid polymer foams.

Polymer foams are unsuitable for other reasons (too easily cut, and too thick) leaving the others as the most environmentally friendly choices.


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MFA, 08/12/2008

Index H m ρ / σ y1 / 2

The Strength – Embodied energy chart: the one for strength at minimum embodied energy.

E8.10. A material is required for disposable forks for a fast-food chain. List the objective and the constraints that you would see as important in this application.

Answer. The primary considerations in choosing materials for disposable applications are health, cost and the environmental damage that disposal might cause. The table list the resulting constraints and objectives.

Constraints: the material must be

• • • • •

Non toxic Cheap Cheap also to mold or shape Low embodied energy Easily (and profitably) recyclable


Objective: minimize

• Embodied energy or • Carbon footprint (they lead to essentially the same choice)

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MFA, 08/12/2008

E8.12. Show that the index for selecting materials for a strong panel with the dimensions shown here, loaded in bending, with minimum embodied energy content is

M =

ρ H m 1 / 2 σ y

where H m is the embodied energy of the material, ρ its density and σ y its yield strength. To do so, rework the panel derivation in Section 8.10 replacing the stiffness constraint with a constraint on failure load F requiring that it exceed a chosen value F * where F = C 2

I σ y h L

> F *

where C2 is a constant and I is the second moment of area of the panel : I =

b h3 12

.

Answer. The objective function for the embodied energy of the panel, H p , is the volume Lbh times

the embodied energy of the material per unit volume, ρ H m H p = b h L ρ H m

Its failure load F must be at least F * , where F = C 2

I σ y h L

≥ F *

Here C 2 is a constant that depends only on the distribution of the loads and I is the second moment of area, which, for a rectangular section, is I =

bh

3

12

We can reduce the energy H p by reducing h , but only so far that the stiffness constraint is still met. Combining the last two equations and solving for h gives

⎛ 12 F * L ⎞ ⎟ h ≥⎜ ⎜ C 2 b σ y ⎟ ⎝ ⎠

1. / 2

Using this to eliminate h in the objective function gives

H p

⎛ 12 F * L3 b ⎞ ⎟ ≥ ⎜ ⎜ C 2 ⎟ ⎝ ⎠

1 / 2

ρ H m 1 / 2 σ y

The quantities F * , L , b and C 2 are all specified; the only freedom of choice left is that of the material. The best materials for a strong panel with the lowest embodied energy are those with the smallest values of M p =

ρ H m 1 / 2 σ y


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MFA, 08/12/2008

E8.13. Use the chart E − H m ρ chart of Figure 8.13 to find the metal with a modulus E greater than

100 GPa and the lowest embodied energy per unit volume. Answer. The construction is shown below. The metal with a modulus greater than 100 GPa and the lowest embodied energy is cast iron.

Lower embodied energy

Modulus E = 100 GPa

The Modulus – Embodied energy chart: the one for stiffness at minimum embodied energy.

8.14. A maker of polypropylene (PP) garden furniture is concerned that the competition is stealing part of his market by claiming that the “traditional” material for garden furniture, cast iron, is much less energy and CO2 intensive than the PP. A typical PP chair weighs 1.6 kg; one made of cast iron weighs 11 kg. Use the data sheets for these two materials in Chapter 12 of the book to find out who is right – are the differences significant? (Remember the warning about precision at the start of Chapter 12).

If the PP chair lasts 5 years and the cast iron chair lasts 25 years, does the conclusion change? Answer. The table lists mean values of embodied energy and carbon footprint, per kg, for the two materials. The last two columns show the values per chair. If the difference in lifetime is ignored, the two chairs do not differ significantly in embodied energy, but they do in carbon release. If the longer life of the cast iron chair is recognized by dividing the values by the life in years (to give energy and carbon per chair.year) the cast iron chair wins easily.

Material

Embodied energy*

CO2

MJ/kg

Cast iron Polypropylene

*

Embodied energy

CO2

kg/kg

MJ/chair

kg/chair

17

1.05

187

11.6

103

2.7

165

4.32

* From the data sheets of Chapter12.


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Exploring design using CES Edu Level 2 Eco.

E8.15. Use a “Limit” stage to find materials with modulus E > 180 GPa and embodied energy H m <

30 MJ/kg. Answer. Apply the two constraints using a “Limit” stage in CES Edu Level 2 with Eco. Four materials meet them. They are

Selected materials Cast iron, ductile (nodular) High carbon steel Low carbon steel Medium carbon steel

E8.16. Use a “Limit” stage to find materials with yield strength σ y > 100 MPa and a carbon footprint CO 2 < 1 kg/kg.

Answer. Two materials meet the constraints. They are

Selected materials Cast iron, ductile (nodular) Cast iron, gray

E8.17. Make a bar chart of embodied energy H m . Add a “Tree” “Tree” stage to limit the selection to

polymers alone. Which polymers have the lowest embodied energy? Answer. The figure below shows the bar-chart created by CES. It was made using a “Graph” stage with Embodied energy plotted on the y-axis. The selection was limited to Polymers and elastomers by selecting, for the x-axis, this subset using the “Tree” option found under Advanced in the axis choice dialog box. The four polymers with the lowest embodied energy are listed below. They are all bio polymers – polymers made from natural feedstock, not oil.

Selected materials Natural Rubber (NR) Polyhydroxyalkanoates (PHA, PHB) Polylactide (PLA) Starch-based thermoplastics (TPS)


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E8.18.

Make a chart showing modulus E and density ρ .

Apply a selection selection line of slope 1,

corresponding to the index ρ / E positioning positioning the line such that six materials materials are left above it. What families do they belong to?

Answer. The chart generated by CES is shown below with a selection line of slope 1 positioned to leave six materials above the line. These are the materials with the lowest values of the index. They are listed in the table. Five are ceramics, ceramics, one is a composite.

Selected materials Alumina Aluminum nitride Boron carbide CFRP, epoxy matrix (isotropic) Silicon carbide Silicon nitride


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E8.19. A material is required for a tensile tie to link the front and back walls of a barn to stabilize both. It must meet a constraint on strength and have as low an embodied energy as possible. To be safe the material of the tie must have fracture toughness K 1c > 18 MPa.m1/2. The relevant index is M = H m ρ / σ y

Construct a chart of σ y plotted against H m ρ . Add the constraint of adequate fracture toughness, meaning K 1c > 18 MPa.m1/2, using a “Limit” stage. Then plot an appropriate selection selecti on line on the chart and report the three materials that are the best choices for the tie. Answer. The chart is shown below with a selection selecti on line leaving just three materials exposed. They are listed in the table.

Selected materials Cast iron, ductile (nodular) High carbon steel Low alloy steel


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E8.20. A company wishes to enhance its image by replacing oil-based plastics in its products by polymers based on natural materials. Use the “Search” facility in CES to find biopolymers. List the material you find. Are their embodied energies and CO2 footprints less than those of conventional plastics? Make bar charts of embodied energy and CO2 footprint to find out. Answer. A search on “Biopolymer” delivers the 5 materials listed in the table. The plot of embodied energies for polymers, shown in the answer to exercise E8.17 shows that the first four have lower embodied energies than any other polymer or elastomers. The equivalent chart of carbon footprint, below, reveals that CO2 release associated with their production is at the lower end of that for polymers – it is about the same as that polyethylene or PVC.

Selected materials Natural Rubber (NR) Polylactide (PLA) Polyhydroxyalkanoates (PHA, PHB) Starch-based thermoplastics (TPS) Cellulose polymers (CA)


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Chapter 9 E9.1. The materials of the drink containers of Figure 9.1 are recycled to different degrees. How does the ranking of Table 9.2 change if the contribution of recycling is included? To do so, multiply the energy per liter in the last column of the table by the factor

⎛ H ⎞ 1 − f rc ⎜⎜ 1 − rc ⎟⎟ H ⎝

m ⎠

where f rc is the recycle fraction in current supply, H m is the embodied energy for primary material production and H rc is that for recycling of the material. You will find data for all three attributes in the data sheets of Chapter 12.

Answer. The first table below lists the materials of the drink containers and the embodied energies, energies to recycle and typical recycled fraction in current supply. The second table reproduces Table 9.2 of the text with two additional columns added. The first lists the correction factor , the second lists the energy per liter multiplied by the correction factor.

The inclusion of recycled material significantly reduces all the energies. Steel still emerges as the best choice, but glass rather than aluminium now becomes the most energy-intensive.

Container type

Material

Embodied energy* MJ/kg

Recycle energy* (MJ/kg)

Recycle fr. in current supply*

PET 400 ml bottle

PET

84

38.5

0.21

PE 1 liter milk bottle

High density PE

81

34

0.085

Glass 750 ml bottle

Soda glass

15.5

6.8

0.24

Al 440 ml can

5000 series Al alloy

208

19.5

0.44

Steel 440 ml can

Plain carbon steel

32

9.0

0.42


Container type

Material

Mass, grams

Energy/liter MJ/liter

Correction factor

Corrected energy MJ/liter

PET 400 ml bottle

PET

25

5.3

0.89

4.7

PE 1 liter milk bottle

High density PE

38

3.8

0.95

3.6

Glass 750 ml bottle

Soda glass

325

6.7

0.87

5.8

Al 440 ml can

5000 series Al alloy

20

9.5

0.60

5.7

Steel 440 ml can

Plain carbon steel

45

3.3

0.70

2.3

E9.2. Derive the correction factor to allow for recycle content cited in Exercise E9.1

Answer. If the drink containers are made of virgin material, each carries an embodied energy of H m

per kg. If instead they are made of material containing a fraction f rc of recycled material that had required the recycle energy H rc to produce it, the energy per kg falls to ( 1 − f rc ) H m + f rc H rc . The factor we seek is the ratio of these two: ( 1 − f rc ) H m + f rc H rc H m


⎛ H ⎞ = 1 − f rc ⎜⎜ 1 − rc ⎟⎟ ⎝ H m ⎠

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E9.4. Use the indices for the crash barriers (equations 9.1 and 9.2) with the charts for strength and density (Figure 8.12) and strength and embodied energy (Figure 8.14) to select materials for each of the barriers. Position your selection line to include one metal for each. Reject ceramics and glass on the grounds of brittleness. List what you find for each barrier.

ρ

Answer.

σ y2 / 3

H m ρ σ y2 / 3


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The figure above shows the two charts with the indices M 1 =

H m ρ σ y2 / 3

and M 2 =

ρ σ y2 / 3

marked.

Each is position to leave one class of metal exposed. The upper selection is that for the mobile barrier with minimizing mass as the objective. The best choice is CFRP (carbon fiber reinforced polymers); among metals, magnesium alloys offer the lightest solution . The lower selection is that for the static barrier with minimizing embodied energy as the objective. The selection is wood; among , metals cast irons offer the solution with lowest embodied energy. The exercise brings out how strongly the selection to minimize environmental burden depends on the application, as did the selection method used in the text.

Selected materials: mobile barrier

Selected materials: static barrier

Carbon-fiber reinforced polymers (CFRP)

Wood

Magnesium alloys

Cast irons

(Aluminum alloys)

(Carbon steels)

(Wood)

E9.5. Complete the selection of materials for light, stiff shells of Section 9.4 by plotting the stiffness index

M 1 =

ρ E 1 / 2

onto a copy of the modulus-density chart of Figure 8.11. Reject ceramics and glass on the grounds of brittleness, and foams on the grounds that the shell would have to be very thick. Which materials do you find? Which of these would be practical for a real shell?

Answer. The selection line with slope 2 (picked up from the guide line on the lower right) is position here such that CFRP, magnesium alloys, aluminium alloys and wood lie above it (ignoring the ceramics and foams). All are practical for shells. The earliest light-weight shells were made of plywood (some still are), aluminium sheet is routinely used for shell structures; magnesium sheet can be used in the same way but its has more limited ductility, making it harder to shape; finally CFRP is now the material of choice for light weight shells.


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ρ E 1 / 2

The Modulus – Density chart: the one for stiffness at minimum weight. .

E9.6. In a far-away land, fridges cost that same as they do here but electrical energy costs 10 times more than here – that is, it costs $2/kW.hr. Make a copy of the trade-off plot for fridges (Figure 9.8) and plot a new set of penalty lines onto it, using this value for the exchange constant, α e . If you had to

choose just one fridge to use in this land, which would it be?

Answer. We wish to minimize life-cost, which we take to be the sum of the initial cost and the cost of the energy used over life. So define the penalty function * * Z * = C f t + α e H f * * where C f is the initial cost and H f is the energy per year (both per cubic meter of cold space), α e ,

the exchange constant, is the cost of energy per kW.hr and t is the service life of the fridge in years, making Z * the life-cost of the fridge per cubic meter of cold space. Inserting a life t of 10 years and an energy cost α e of $2/kW.hr gives * * Z * = C f + 20 H f

Rearranging gives * H f =


1 20

Z * −

1 20

* C f

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* * So the penalty function plots as a set of lines of slope 1/20 on the trade of plot of H f against C f .

They are shown in black on the figure. The choice that minimizes total life-cost when electrical power costs $2/kW.hr is that nearest the point where the new Z * contours are tangent to the trade-off surface. It is ringed in red: the Hotpoint RLA 175.

Contours of Z* for an exchange constant of $2/kW.hr

The trade-off plot for fridges with contours of the new penalty function Z *.

E9.7. You are asked to design a large heated work-space in a cold climate, making it as eco-friendly as possible by using straw-bale insulation. Straw, when compressed, has a density of 600 kg/m 3, a specific heat capacity of 1670 J/kg.K and a thermal conductivity of 0.09 W/m.K. The space will be heated during the day (12 hours) but not at night. What is the optimum thickness of straw to minimize the energy loss?

Answer. The required insulation wall-thickness, w, is given by equation 9.16 of the text:

⎛ 2 λ t ⎞ ⎟ w = ⎜ ⎜ C p ρ ⎟ ⎝ ⎠

1/2

where λ is the thermal conductivity, C p is the specific heat of the wall per unit mass (so C p ρ is the specific heat per unit volume) and t is the time interval between heating and cooling. Inserting the data given in the question (and remembering to convert the 12 hour cycle time into seconds) gives the result w = 0.09 meters = 9 centimeters.


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E9.8. The makers of a small electric car wish to make bumpers out of a molded thermoplastic. Which index is the one to guide this selection if the aim is to maximize the range for a given battery storage capacity? Plot it on the appropriate chart from the set shown as Figures 8.11 – 8.14, and make a selection.

Answer. The use-energy of any vehicle, be it gasoline or electric powered, increases with the mass of the vehicle. The bumper contributes to this mass. Maximizing range for a given battery capacity means minimizing mass. The function of the bumper is to sustain bending loads. The required index (equation 9.2 of the text) is

M 2 =

ρ σ y2 / 3

This is plotted on the strength – density chart below, picking up the slope of 1.5 from the guide line at the lower right. The selection line is position such that a few polymers remain above it – they are the best choice. They are:

•

Polycarbonate, PC

•

Polyamide (nylon) PA and

•

Polyetheretherketone PEEK

In practice bumpers are made of blends of Polycarbonate with other polymers such as Polypropylene or Polyamide, or of fiber reinforced thermosets such as Polyester. PEEK is too expensive for application such as this. ρ σ y2 / 3

The Strength – Density chart: the one for strength at minimum weight. . Eco text: solution manual

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E9.9. Car bumpers used to be made of steel. Most cars now have extruded aluminium or glassreinforced polymer bumpers. Both materials have a much higher embodied energy than steel. Take the weight of a steel bumper-set to be 20 kg, and that of an aluminum one to be 14 kg. Table 9.6 gives the equation for the energy consumption in MJ/km as a function of weight for petrol engine cars based on the data plotted in Figure 9.11 of the text.

(a) Work out how much energy is saved by changing the bumper-set of a 1500 kg car from steel to aluminum, over an assumed life of 200,000 km (b) Calculate whether the switch from steel to aluminum has saved energy over life. You will find the embodied energies of steel and aluminum in the datasheets of Chapter 12. Ignore the differences in energy in manufacturing the two bumpers – it is small. (c) The switch from steel to aluminum increases the price of the car by $60. Using current pump prices for gasoline, work out whether, over the assumed life, it is cheaper to have the aluminum bumper or the steel one.

Answer. (a) The energy in MJ/km, H km , as a function of vehicle mass m is listed in Table 9.6:

H km = 3.7 x10 −3 m 0.93

The marginal change in this energy resulting from a small change is mass when bumpers of one material are replace by those of another is found by differentiating this: Δ H km =

3.44 x10 −3 m 0.07

Δm

Thus a weight saving of 6 kg made possible by the change from steel to aluminum bumpers on a car of mass 1500 kg saves, over 200,000 km, the energy 2,470 MJ. (b) The table below lists the embodied energies of steel and aluminum; they are the means of the ranges given in data sheets of Chapter 12. From these the material energy per bumper-set is calculated: the difference is 2,440 MJ. Thus the switch from steel to aluminum bumpers has saved energy, but not much. The difference between the material and the use energy over 200,000 for a vehicle of this mass is only 30 MJ. Material

Bumper mass (kg)


Material energy per bumper (MJ)

Steel

20

32

640

Aluminum

14

220

3080

(c) One liter of gasoline provides 35 MJ of energy (see Table 6.5 of the text). Thus the change of material has saved the equivalent of one liter of fuel, and has cost $60. Fuel cost depends on country; it varies at present between $1/liter and $2/liter. Thus from an economic point of view,, and assuming the conditions of use described here (200,000 km on a vehicle of mass 1500 kg) the change of material is not justified on cost grounds. If the life-distance is increased to 400,000 km the picture changes. The energy of use then becomes 4,940 MJ, the difference between material and use energy becomes 2,500 MJ, equivalent to 71 liters of gasoline, and then, even with cheap gasoline ($1 per liter) the change becomes economically attractive.


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Exercises using the CES Edu software.

E9.10. Refine the selection for shells (Chapter 9, Section 9.4) using Level 3 of the CES software. Make a chart with the two shell indices

M 1 =

ρ E 1 / 2

and

M 2 =

ρ 1 / 2 σ y

as axes, using the “Advanced” facility to make the combination of properties. Then add a “Tree” stage, selecting only metals, polymers and composites and natural materials. Which ones emerge as the best choice? Why?

Answer. The figure below shows the CES Level 2 output. The materials that have the most attractive values of both indices are those at the bottom left. They are listed in the table below. All have low

densities. Density ρ enters the two indicies with a power of 1; modulus E and strength σ y enter only as the square root. Thus density plays a larger role in determining the value of the index than the other two properties, which in part explains the selection. Selected materials Age-hardening wrought Al-alloys CFRP, epoxy matrix (isotropic) Softwood: pine, along grain Titanium alloys Wrought magnesium alloys

The interesting bit


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E9.11. Tackle the crash barrier case study using CES Level 2 following the requirements set out in

Table 9.3. Use a Limit stage to apply the constraints on fracture toughness K 1c ≥ 18 MPa.m and the requirement of recyclability. Then make a chart with density ρ on the x-axis and yield strength σ y on the y-axis and apply a selection line with the appropriate slope to represent the index for the mobile barrier: M 2 =

ρ σ y2 / 3

List the best candidates. Then replace Level 2 with Level 3 data and explore what you find.

Answer. The two figures below show the Level 2 and the Level 3 charts on exactly the same axes, and with a selection line with the appropriate slope (1.5) in the same position on both. Switching from Level 2 to Level 3 has populated the chart in much more detail. There are no polymer matrix composites on either plot – they are eliminated by the requirement of recyclability. The selection at Level 2 with the selection line in the position shown is

Selected materials, Level 2 Age-hardening wrought Al-alloys Titanium alloys Wrought magnesium alloys

The same selection at Level 3 gives 76 alloys of magnesium, aluminum, titanium and steels. Displacing the line upwards to refine the search until just six are left gives a much more specific selection listed below. ρ σ y2 / 3

1.5

Level 2 data


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ρ σ y2 / 3

1.5

Level Level 3 3 data data

Selected materials, Level 3 Titanium alpha-beta alloy, Ti-6Al-2Sn-2Zr-2Mo, Solution Treated and Aged Titanium alpha-beta alloy, Ti-6Al-2Sn-2Zr-2Mo, annealed Wrought aluminum alloy, 7055, T77511 Wrought aluminum alloy, 7150, T61511 Wrought magnesium alloy (EA55RS) Wrought magnesium alloy (ZC71)


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E9.12. Repeat the procedure of Exercise E9.11, but this time make a chart using CES Level 2 on which the index for the static barrier

M 1 =

H m ρ σ y2 / 3

can be plotted – you will need the Advanced facility to make the product H m ρ . List what you find to be the best candidates. Then, as before, dump in Level 3 data and explore what you find.

Answer. The two figures below show the Level 2 and the Level 3 charts on exactly the same axes, and with a selection line with the appropriate slope (1.5) in the same position on both. Switching from Level 2 to Level 3 has populated the chart in much more detail. As in the preceding exercise there are no polymer matrix composites on either plot – they are eliminated by the requirement of recyclability. The selection at Level 2 with the selection line in the position shown is

Selected materials, Level 2 Cast iron, ductile (nodular) Cast iron, gray High carbon steel Low alloy steel Low carbon steel Medium carbon steel

The same selection at Level 3 gives 379 cast irons and steels, many of them sophisticated tool steels. Displacing the line upwards to refine the search until just six are left, for instance, gives a much more specific selection shown, below.

Selected materials, Level 3 Low-carbon mold tool steels, AISI P2 Low-carbon mold tool steels, AISI P3 Low-carbon mold tool steels, AISI P5 Oil-hardening cold work tool steel, AISI O6 Water-hardening tool steel, AISI W2 (Annealed) Water-hardening tool steel, AISI W5 (Annealed)

These steels are expensive and an inappropriate choice for a low-grade application such as a static crash barrier. The problem is overcome by applying a further constraint, using a “Limit” stage, requiring that the material cost less than $1/kg. The result is shown in the third chart and table below.


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H m ρ σ y2 / 3

Level 2 data

1.5

H m ρ σ y2 / 3

Level 3 data

1.5


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Selected materials, Level 3 with limit on price of £1/kg Carbon steel, AISI 1340 (tempered @ 205 C, oil quenched) Low alloy steel, AISI 9255 (tempered @ 205 C, oil quenched) Nodular graphite cast iron (BS 700/2, Hardened & Tempered) Nodular graphite cast iron (BS 900/2, Hardened & Tempered) Pearlitic malleable cast iron (BS grade P 70-02) Pearlitic malleable cast iron (former BS P 690/2-OQ)

H m ρ σ y2 / 3

1.5

Level 3 data with price < $1/kg


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Chapter 10 E10.1. Distinguish pollution control and prevention (PCP) from design for the environment (DFE). When would you use the first? When the second? Answer. Pollution control and prevention (PC and P) is intervention to mitigate environmental damage caused by existing products or processes without the burden of redesigning them. It is basically a clean-up or “end-or-pipe” measure. Taking transport as an example, it is the addition of catalytic converters to cars, a step to mitigate an identified problem with an existing product or system. Taking water-borne pollution of beaches as an example, it is extending the exit pipe to carry effluent further out to sea.

Design for the environment (DFE) is the strategy to foresee and minimize the damaging effects of product families at the design stage, balancing them against the conflicting objectives of performance, reliability, quality and cost. Returning to the example of the car: it is to redesign the vehicle, giving emphasis to the objectives of minimizing emissions by reducing weight or adopting an alternative propulsion system – hybrid, perhaps, or electric. Using the example of water-borne waste again, it is redesigning the process generating the effluent so that, through filtration, chemical or biological treatment, the water leaving the plant is no longer polluted. E10.2. What is meant by the ecological metaphor ? What does it suggest about ways to use materials in a sustainable way?

Answer. The ecological metaphor , a concept emerging from industrial ecology, is based on the precept that we must see human activities as part of the global eco-system, and that there is, in some senses, a parallel between the functioning of the natural and the industrial systems. It has sometimes led to the idea is that a study of the processes and balances that have evolved in nature might suggest ways reconcile the imbalance between the industrial and the natural systems.

E10.3 What are the potential sources of renewable energy? What are the positive and negative aspects of converting to an economy based wholly on renewable sources? Answer. There are five potential sources of renewable energy:.

• • • • •

Wind Wave Tidal Geothermal Solar, both as direct heat and, via photocells, as electricity

It is these that offer the possibility of power generation without atmospheric carbon. Positive aspects of converting to renewable energy sources. All renewable energy sources generate electrical power without significant emissions. All offer independence from the international trade in fossil fuels and the uncertainty of price and supply that this involves. And all require a large number capture devices distributed over a large area of land or sea, making the system as a whole difficult to disrupt. Negative aspects of converting to renewable energy sources. Extracting significant power for renewable resources involves committing large areas of land or sea because the energy is so dispersed. Establishing a renewable-energy based system requires very large capital investment but so, too, does the building of conventional power generating plant. The dispersion and reliance on (and thus lack of protection from) the forces of nature creates maintenance problems.


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E10.4. The land area of the Netherlands (Holland) is 41,526 km 2. Its population is 16.5 million and the average power consumed per capita there is 6.7 kW. If the average wind power is 2 W/m2 of land area and wind turbines operate at a load factor of 0.5, what fractional of the area of the country would be taken up by turbines in order to meet the energy needs?

Answer. The average power demand of the Netherlands is

P = Population x power per person = 1.11 x10 8 kW

The wind power that can be generated from an area A km2 is P w = A x 10 6 x 2 x 0.5 / 1000 = 1000 A

kW

Here the factor of 106 is to convert km2 to m2 and the factor of 1000 is to convert W to kW . Equating this to the power demand gives A = 1.11 x10 5 km 2

This is 2.67 times the area of the Netherlands. There is no way wind power can supply all the country’s needs.

E10.5. The land area of the New York State is 131,255 km2. Its population is 19.5 million and the average power consumed per capita there is 10.5 kW. If the average wind power is 2 W/m2 of land area and wind turbines operate at a load factor of 0.25, what fractional of the area of the country would be taken up by turbines in order to meet the energy needs? Answer. The average power demand of the State of New York is


The wind power that can be generated from an area A km2 is P w = A x 10 6 x 2 x 0.25 / 1000 = 500 A

kW

Here the factor of 106 is to convert km2 to m2 and the factor of 1000 is to convert W to kW . Equating this to the power demand gives A = 14.1 x10 5 km 2

This is 3.12 times the area of the State of New York. There is no way wind power from within the State can supply all the energy it needs. Even with a probably unrealistic load factor of 0.5, the area is insufficient.


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E10.6. The combined land area of the State of New Mexico (337,367 km 2) and Nevada (286,367 km 2) is 623,734 km2. The population if the United States is 301 million and the average power consumed per capita there is 10.2 kW. Mass-produced solar cells can capture 10% of the energy that falls on them, which, in New Mexico, is roughly 50W/m 2. Is the combined area of the two States large enough to provide solar power that would meet the current needs of the United States? Answer. The average power demand of the US is


The solar power that can be generated from an area A km2 is P w = A x 10 6 x 50 x 0.1 / 1000 = 5000 A

kW

Here the factor of 106 is to convert km2 to m2 and the factor of 1000 is to convert W to kW . Equating this to the power demand gives A = 6 .14 x10 5 km 2

This is 0.98 times the combined area of the two States. The power needs could just be met if both States were completely paved with solar cells.


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Chapter 11 E11.1. An aluminum saucepan weighs 1.2 kg, and costs $10. The embodied energy of aluminum is 220 MJ/kg. If the cost of industrial electric power doubles from 0.0125 $/MJ to 0.025 $/MJ, how much will it change the cost of the saucepan?

Answer. The cost of the electric power at 0.0125 $/MJ to make the aluminum for one 1.2 kg saucepan is $ 3.30. If the price of energy doubles to 0.025 $/MJ this rises to $ 6.6. The cost of the saucepan will rise to $ 13.3, an increase of 33%.

E11.2. An MP3 player weighs 100 grams, and costs $120. The embodied energy of assembled integrated electronics of this sort is about 2000 MJ/kg. If the cost on industrial electric power doubles from 0.0125 $/MJ to 0.025 $/MJ, how much will it change the cost of the MP3 player?

Answer. The cost of the electric power at 0.0125 $/MJ to make one MP3 player is $2.5. If the price of energy doubles this rises to $ 5. The cost of the player will rise to $ 125, an increase of just 4%.

E 11.3. Cars in Cuba are repaired and continue in use when 25 years old. The average life of a car in the US is 13 years. What is the underlying reason for this?

Answer. In affluent economies such as those of the US and much of Europe consumers are sufficiently wealthy to purchase products as much for their character, associations and styling as for their functionality. New models often appear on an annual basis, stimulating replacement of previously purchased products before the end of their functional life. High labor costs make maintenance and repair uneconomic so that even minor breakdown causes rejection and replacement.

In less affluent economies the capital cost of products represents a much larger fraction of disposable income. To avoid the need for more capital investment products are kept for their full functional life, which is extended by maintenance and repair enabled by low labor costs.

E11.4. Use the Worldwide Web to research the meaning and history of “The precautionary principle”. Select and report the definition that, in your view, best sums up the meaning.

Answer. Numerous sites comment on or criticize the precautionary principle, which has a long history. Among them are:

1.

en.wikipedia.org/wiki/Precautionary_principle

2.

www.jncc.gov.uk/page-1575

3.

www.sirc.org/articles/beware

4.

www.i-sis.org.uk/prec.php

From these we learn the following. The precautionary principle states that when there is reasonable suspicion of harm, lack of scientific certainty or consensus must not be used to postpone preventative action. Originating in 1930s


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Eco Solution Manual

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