Solution manual for exercises - MAELabs UCSD
Solution manual for exercises - MAELabs UCSD
Solution manual for exercises - MAELabs UCSD
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Materials and the Environment<br />
<strong>Solution</strong> <strong>manual</strong> <strong>for</strong> <strong>exercises</strong><br />
<strong>Solution</strong>s to exercise: Page<br />
Chapter 1 2<br />
Chapter 2 4<br />
Chapter 3 12<br />
Chapter 4 16<br />
Chapter 5 20<br />
Chapter 6 23<br />
Chapter 7 30<br />
Chapter 8 41<br />
Chapter 9 53<br />
Chapter 10 66<br />
Chapter 11 69<br />
Eco text: solution <strong>manual</strong> 0 MFA, 08/12/2008
Eco text: solution <strong>manual</strong> 1 MFA, 08/12/2008
Chapter 1<br />
E 1.1. Use Google to research the history and uses of one of the following materials<br />
� Tin<br />
� Glass<br />
� Cement<br />
� Bakelite<br />
� Titanium<br />
� Carbon fiber<br />
Present the result as a short report of about 100 - 200 words (roughly half a page). Imagine that you<br />
are preparing it <strong>for</strong> school children. Who used it first? Why? What is exciting about the material? Do<br />
we now depend on it or could we, with no loss of engineering per<strong>for</strong>mance of great increase in cost,<br />
live without it?<br />
Specimen answer: tin. Tin (symbol Sn), a silver-white metal, has a long history. It was traded in the<br />
civilizations of the Mediterranean as early as 1500 BC (the Old Testament of the Christian bible<br />
contains many references to it). Its importance at that time lay in its ability to harden copper to give<br />
bronze (copper containing about 10% tin), the key material <strong>for</strong> weapons, tools and statuary of the<br />
Bronze age (1500 BC – 500 BC). Today tin is still used to make bronze, <strong>for</strong> solders and as a corrosion<br />
resistant coating on steel sheet (“tin plate” ) <strong>for</strong> food and drink containers – a “tinnie”, to an Australian,<br />
is a can of beer. Plate glass is made by floating molten glass on a bed of liquid tin (the Pilkington<br />
process). Thin deposits of tin compounds on glass give transparent, electrically conducting coatings<br />
used <strong>for</strong> frost-free windshields and <strong>for</strong> panel lighting.<br />
Most of the applications of tin could be filled by other materials – polymer coatings <strong>for</strong> food<br />
containers, aluminum instead of tin to make bronzes, indium <strong>for</strong> transparent coatings (though at an<br />
increased cost). Finding a replacement <strong>for</strong> tin in solders is more difficult.<br />
E 1.2. There is international agreement that it is desirable (essential, in the view of some) to reduce<br />
global energy consumption. Producing materials from ores and feedstocks requires energy (its<br />
“embodied energy”). The table lists the energy per kg and the annual consumption of 4 materials of<br />
engineering. If consumption of each could be reduced by 10%, which material offers the greatest<br />
global energy saving? Which the least?<br />
Material Embodied energy<br />
MJ/kg<br />
Annual global<br />
consumption (tonnes/yr)<br />
Annual energy<br />
commitment (MJ)<br />
Steels 29 1.1 x 10 9 3.2 x 10 10<br />
Aluminum alloys 200 3.2 x 10 7 6.4 x 10 9<br />
Polyethylene 80 6.8 x 10 7 5.4 x 10 9<br />
Concrete 1.2 1.5 x 10 10<br />
1.8 x 10 10<br />
Device-grade silicon Approximately 2000 5 x 10 3 1.0 x 10 7<br />
Answer. An additional column has been added to the table above: it shows the annual energy<br />
commitment associated with the production of each material (the product of the numbers in the two<br />
columns to its left). Reducing consumption of steel and of concrete – by more efficient design of<br />
structures perhaps – have by far the greatest potential <strong>for</strong> global energy saving. Doing the same <strong>for</strong><br />
device grade silicon has the least, by a large factor. Although the embodied energies of materials differ<br />
considerably, it is the much greater differences in annual consumption that dominate the total energy<br />
commitment and the carbon burden they generate. This is one reason that much of the discussion of<br />
this book focuses on the materials used in the greatest quantities.<br />
Eco text: solution <strong>manual</strong> 2 MFA, 08/12/2008
E1.3. The ultimate limits of most resources are difficult to assess precisely, although estimates can be<br />
made. One resource, however, has a well-defined limit: that of usable land area. The surface area of<br />
the globe is 511 million square km, or 5.11 x 10 10 hectares (a hectare is 0.01 sq. km.) Only a fraction<br />
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<br />
earth’s surface is biologically productive. Industrial countries require 6 hectares of biologically<br />
productive land per head of population to support current levels of consumption. The current (2008)<br />
global population is close to 6.7 billion (6.7 x 10 9 ). What conclusions can you draw from these facts?<br />
Answer. The current biologically productive area per person is 1.1 x 10 10 / 6.7 x 10 9 = 1.64<br />
hectares/person, less than 1/3 of that currently needed to support a person in a developed country. Thus<br />
a large fraction of the world’s population can never reach the same level of consumption currently<br />
enjoyed by developed countries unless the global population declines, or new <strong>for</strong>ms of intensive<br />
agriculture, perhaps based on genetically modified crops and animals, enable a dramatic increase<br />
(factor 3) in the productivity of the areas of the earth’s surface that are biologically productive.<br />
Eco text: solution <strong>manual</strong> 3 MFA, 08/12/2008
Chapter 2<br />
E2.1 Explain the distinction between reserves and the resource base.<br />
Answer. A mineral reserve is defined as that part of a known mineral deposit that can be extracted<br />
legally and economically at the time it is determined. Reserves are an economic construct, which grow<br />
and shrink under varying economic, technical and legal conditions. Improved extraction technology<br />
can enlarge it, but environmental legislation or changing political climate may make it shrink. Demand<br />
stimulates prospecting, with the consequence that reserves tend to grow in line with consumption.<br />
The resource base (or just resource) of a mineral is the real total, and it is much larger<br />
than the reserve but much less certain. It includes not only the current reserves but also all<br />
usable deposits that might be revealed by future prospecting and that, by various extrapolation<br />
techniques, can be estimated. It includes, too, known and unknown deposits that cannot be mined<br />
profitably now but which – due to higher prices, better technology or improved transportation – may<br />
become available in the future.<br />
The answer should include a sketch of the diagram illustrating reserves and resources is<br />
(Figure 2.10 of the book).<br />
E2.2 The world consumption rate of CFRP is rising at 8 % per year. How long does it take to double?<br />
Answer. Equation 20.3 of the text gives the doubling time t D as<br />
100<br />
r<br />
tD = loge<br />
( 2 )<br />
≈<br />
70<br />
r<br />
where r is the percentage fractional rate of growth per year. Thus a growth rate of 8% means that<br />
consumption doubles every 8.7 years.<br />
E2.3 Derive the dynamic index<br />
100 r R<br />
tex<br />
, d = loge<br />
(<br />
r 100 Po<br />
starting with equation (2.2) of the text.<br />
+<br />
1)<br />
Answer. Equation (2.2) of the text describes the production rate of a material, with P 0 = production<br />
rate at time t = to<br />
and r = annual growth rate in % per year.<br />
( t t )<br />
⎧r − ⎫<br />
P = P exp<br />
o<br />
0 ⎨ ⎬<br />
⎩ 100 ⎭<br />
Integrating this over time to give the cumulative production Q t*<br />
up to time<br />
result to the reserve R gives<br />
t<br />
*<br />
Qt<br />
* =<br />
∫ P dt<br />
to<br />
=<br />
100 P<br />
⎛ *<br />
r(<br />
t t )<br />
⎞<br />
o ⎜ ⎪<br />
⎧ − o ⎪<br />
⎫<br />
exp<br />
1<br />
⎟<br />
=<br />
r ⎜ ⎨ ⎬ −<br />
100 ⎟<br />
⎝ ⎪⎩ ⎪⎭ ⎠<br />
t = t<br />
*<br />
and equating the<br />
Eco text: solution <strong>manual</strong> 4 MFA, 08/12/2008<br />
R
Solving <strong>for</strong> the time interval ( t<br />
*<br />
− to<br />
) gives the desired result<br />
t*<br />
− to<br />
= tex,<br />
d =<br />
100 r R<br />
loge<br />
(<br />
r 100 Po<br />
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<br />
annual growth rate of car sales, expressed as % per year? If there were 15 million cars already on<br />
Chinese roads by the end of 2007 and this growth rate continues, how many cars will there be in 2020,<br />
assuming that the number that are removed from the roads in this time interval can be neglected?<br />
Answer. Starting with equation (2.2)<br />
( t t )<br />
⎧r − ⎫<br />
P = P exp<br />
o<br />
0 ⎨ ⎬<br />
⎩ 100 ⎭<br />
6<br />
we enter P = 6.<br />
6 x10<br />
6<br />
and P o = 5x10<br />
and the time interval ( t − to<br />
) = 1 year, and solve <strong>for</strong> r . The<br />
result is r = 27.8% per year.<br />
The cumulative number of cars entering use in the subsequent 13 years is found from the integral<br />
of this equation over time<br />
Eco text: solution <strong>manual</strong> 5 MFA, 08/12/2008<br />
+<br />
1)<br />
t<br />
*<br />
⎛<br />
⎞<br />
⎜ ⎪<br />
⎧ *<br />
100 P<br />
⎪<br />
⎫<br />
o r(<br />
t − to<br />
)<br />
Q = =<br />
⎟<br />
t * ∫ P dt<br />
⎜<br />
exp ⎨ ⎬ − 1<br />
r<br />
⎟<br />
t<br />
⎝ ⎪⎩<br />
100<br />
o<br />
⎪⎭ ⎠<br />
Entering P<br />
6<br />
o = 5x10<br />
(the number 2007), r = 27.8% per year and ( t* −to ) = 13 gives the additional<br />
number of cars by 2020 at Q<br />
6<br />
t*<br />
= 650x10<br />
. To this (if we are picky) we must add the number already<br />
6<br />
there in 2007, giving a final total of 655 x10<br />
, larger than the number in 2007 by a factor of 100.<br />
(This colossal number is still only equivalent to 1 car per 3-person family, less than the current car<br />
ownership per family in the US,)<br />
E2.5 Prove the statement made in the text that, “at a global growth rate of just 3% per year we will<br />
mine, process and dispose of more “stuff” in the next 25 years than in the entire 300 years since the start<br />
of the industrial revolution.”.<br />
.<br />
Answer. Exponential growth has a number of alarming features, among them, the doubling time both<br />
of consumption and of the total quantity consumed. The consumption rate C of a resource grows by<br />
follows the law<br />
C = C0 exp r (t – t0) (1)<br />
here C0 is the consumption rate when t = t0. The doubling time <strong>for</strong> the consumption rate is tDC, meaning<br />
that the rate will have doubled to 2C0 in the time tD = t – t0. It is calculated by equating C in equation 1<br />
to 2C0.<br />
1<br />
2C0 = C0 exp (r tD) giving tD = n 2<br />
r l<br />
the quantity l n 2 = 0.69, so that, at a growth rate of 3% per year ( r = 0.<br />
03 ) , consumption doubles in 23<br />
years.
We are interested here not in the consumption rate but in the total quantity, Q, consumed since<br />
consumption began. It is the integral of C over time.<br />
Q =<br />
t<br />
Cdτ C<br />
t<br />
∫ 0<br />
= o ∫0<br />
exp r ( τ − t o ) dτ<br />
C<br />
giving Q =<br />
o { ( r (t - t ) ) exp( −rt<br />
) }<br />
r<br />
exp o − o<br />
(2)<br />
Q0 is the total amount consumed prior to the present day (when t = t0), thus,<br />
C<br />
Q0 =<br />
o { 1− exp( −r<br />
t o ) }<br />
(3)<br />
r<br />
We want the time tQ (the time to double the total quantity consumed) where tQ = t – t0 is the time from<br />
the present day that Q = 2 Q0, i.e. when (from equations 2 and 3)<br />
Co<br />
2 C<br />
{ exp ( r tQ<br />
) − exp( −rt<br />
) } o<br />
o = { 1−<br />
exp( − r t o ) }<br />
r<br />
r<br />
solving <strong>for</strong> tQ gives<br />
exp (r tQ) = 2 – exp (– r t0)<br />
Now note that if t0 = 100 years (meaning that consumption started 100 years ago) and r = 3% per year<br />
(0.03), then exp (- r t0 ) = 0.05. If t0 = 1000 years, it is roughly 10 -13 . Compared with 2 both numbers<br />
are negligible. Thus, the result <strong>for</strong> the doubling time of consumption is the same as that <strong>for</strong><br />
consumption rate, namely.<br />
1<br />
tQ = n 2<br />
r l<br />
(The underlying reason both tD and tQ are the same is simply that<br />
e<br />
x<br />
= e<br />
x<br />
∫ dx )<br />
That means that the total quantity of a given resource that will be consumed in the next 23 years<br />
(given the modest rate of growth of consumption of 3%/year) is equal to the total quantity consumed<br />
over the history of industrial development. Not a happy thought.<br />
E2.6 Understanding reserves: copper. The table lists the world production and reported reserves of<br />
copper over the last 20 years<br />
Year Price World production Reserves Reserves/World Growth rate<br />
(US$/kg) (million tonnes/yr) (million tonnes) production (yrs) (%/yr)<br />
1995 2.93 9.8 310 31.6<br />
1996 2.25 10.7 310 29.0 8.8<br />
1997 2.27 11.3 320 28.3 5.5<br />
1998 1.65 12.2 340 27.9 7.7<br />
1999 1.56 12.6 340 27.0 3.2<br />
2000 1.81 13.2 340 25.8 4.7<br />
2001 1.67 13.7 340 24.8 3.7<br />
2002 1.59 13.4 440 32.8 -2.2<br />
2003 1.78 13.9 470 33.8 3.7<br />
2004 2.86 14.6 470 32.2 4.9<br />
2005 3.7 14.9 470 31.5 2.0<br />
2006 6.81 15.3 480 31.4 2.6<br />
� Examine trends (plot price, production and reserves against time) – what do you conclude?<br />
� Tabulate the Reserves / World production to give the static index of exhaustion. What does the<br />
result suggest about reserves?<br />
Eco text: solution <strong>manual</strong> 6 MFA, 08/12/2008
Answer. The figure shows the data. The rates<br />
of growth of production are listed in the last<br />
column of the table above: world production<br />
and reserves have increased steadily over the<br />
12 year period 1995 - 2006. The ratio of<br />
reserves to production (second last column of<br />
the table) is steady at around 30 years. All this<br />
suggests a well-balanced market, with no<br />
indicators of supply problems. Only the price,<br />
which was steady or falling up to 2003 shows a<br />
sudden increase in 2005 / 2006 because of a<br />
surge in demand from China. All the other<br />
indicators suggest that the supply chain is<br />
capable of adapting to meet it.<br />
E2.7 The table shows the production rate and the reserves of 5 metals over a period of 10 years. What<br />
has been the growth rate of production? What is that of the reserves? What can you conclusions can<br />
you draw about the criticality of the material?<br />
Metal Year Production rate,<br />
tonnes/year<br />
Platinum 2005<br />
1995<br />
Nickel 2005<br />
1995<br />
Lead 2005<br />
1995<br />
Copper 2005<br />
1995<br />
Cobalt 2005<br />
1995<br />
217<br />
145<br />
1.49 x 10 6<br />
1.04 x 10 6<br />
3.27 x 10 6<br />
2.71 x 10 6<br />
15.0 x 10 6<br />
10.0x 10 6<br />
57.5 x 10 3<br />
22.1 x 10 3<br />
Reserves,<br />
tonnes<br />
71 x 10 3<br />
56 x 10 3<br />
64 x 10 6<br />
47 x 10 6<br />
67 x 10 6<br />
69 x 10 6<br />
480 x 10 6<br />
310 x 10 6<br />
7 x 10 6<br />
4 x 10 6<br />
Growth rate<br />
of production<br />
% per year<br />
Growth rate<br />
of reserves<br />
% per year<br />
4.03 2.37<br />
3.6 3.09<br />
1.88 1.97<br />
4.05 4.37<br />
9.56 5.60<br />
Answer. The last two columns of the table above show the growth rates, calculated from equation 2.2<br />
of the text<br />
( t t )<br />
⎧r − ⎫<br />
P = P exp<br />
o<br />
o ⎨ ⎬ which, on inversion, gives<br />
⎩ 100 ⎭<br />
⎛ P ⎞<br />
⎝ Po<br />
⎠<br />
( ) ⎟ ln ⎜<br />
t − t ⎜<br />
Inserting the 2005 value of production or reserves <strong>for</strong> P , the 1995 value <strong>for</strong> o<br />
years gives the average growth rate over the 10 year interval.<br />
Eco text: solution <strong>manual</strong> 7 MFA, 08/12/2008<br />
r<br />
=<br />
100<br />
o<br />
P and ( t )<br />
t − o = 10<br />
For two of the metals – lead and copper – the rate of growth of reserves is more or less in balance<br />
with the growth in production. For nickel, reserves lag behind production but not by much. The two<br />
metals that are cause <strong>for</strong> concern are platinum and cobalt. Both are used <strong>for</strong> critical functions <strong>for</strong><br />
which there is no other easy substitute: platinum <strong>for</strong> catalysts and cobalt <strong>for</strong> high temperature alloys.<br />
For both the growth of reserves – meaning the rate at which new, exploitable ore bodies are proven –<br />
lags well behind the rate of production.
E2.8 Tabulate the annual world production in tonnes/year and the densities (kg/m 3 ) of carbon steel,<br />
PE, softwood and concrete (you will find the data in data sheets <strong>for</strong> these materials in Chapter 12 of<br />
this book – use an average of the ranges given in the data sheets). Calculate, <strong>for</strong> each, the annual world<br />
production measured in m 3 /year. How does the ranking change?<br />
Answer. The first three columns of the table below list data <strong>for</strong> world production in tonnes/year,<br />
P m ,and density, ρ , in kg/m 3 , drawn (as averages) from the data sheets of Chapter 12. The last<br />
column lists the world production in m 3 /year , Pv calculated as<br />
Pm<br />
P v = 1000<br />
ρ<br />
(The 1000 is to convert the density into tonnes/m 3 .) Concrete and steel dominated production when<br />
measured in tonnes per year . But <strong>for</strong> at least two of these materials – softwood and concrete – it is the<br />
volume that is used that is important, not the weight. By this measure wood is second only to concrete<br />
in the quantity used. Wood remains one of the central materials of structural engineering.<br />
Material World production Density<br />
tonnes/year<br />
kg/m 3<br />
World production<br />
m 3 /year<br />
Carbon steel 1.15 x 10 9 7850 1.46 x 10 8<br />
Polyethylene 6.85 x 10 7<br />
950 7.2 x 10 7<br />
Softwood 9.65 x 10 8 520 1.86 x 10 9<br />
Concrete 1.53 x 10 10 2450 6.25 x 10 9<br />
E2.9 The price of cobalt, copper and nickel have fluctuated wildly in the past decade. Those of<br />
aluminum, magnesium and iron have remained much more stable. Why? Research this by examining<br />
uses (which metal are used in high value-added products?) and the localization of the producing mines.<br />
The USGS web site listed under Further Reading is a good starting point.<br />
Answer. Take cobalt as an example. A web search (and a search of CES) gives the following<br />
in<strong>for</strong>mation.<br />
World production of cobalt has increased steadily year-on-year, and has almost trebled from<br />
around 20,000 tonnes in 1995 to 54,000 tonnes in 2005, an average rate of growth at times exceeding<br />
12% per year, far higher than that of most materials. The ores of cobalt are localized in relatively few<br />
countries. Since 2002, the strongest growth in production of cobalt has come in China, where output<br />
grew at over 60% per year in 2005 as a result of expansion by domestic producers. Other recent<br />
increases have come from new projects including Voisey's Bay in Canada (Inco), Coral Bay in the<br />
Philippines (Sumitomo) and Sally Malay in Australia (Sally Malay and Sumitomo).<br />
Cobalt costs about $30/kg – far more than common structural materials. It is used <strong>for</strong> catalysts;<br />
medical implants, cermets (tungsten carbide – cobalt) cutting tools, alloying of steels, hightemperature<br />
cobalt based superalloys, alloys <strong>for</strong> high field magnets, and as a pigment in glass and<br />
paints. These are high value-added applications, as you might anticipate from the high price of the<br />
metal.<br />
When the supply chain <strong>for</strong> metals with unique applications, like cobalt, is unable to meet demand,<br />
the price rises steeply because there are no substitutes. When, by contrast, substitutes are readily<br />
available (plastics substituting <strong>for</strong> steel or aluminum in many applications, the existence and<br />
availability of the substituted dampens price fluctuations.<br />
Eco text: solution <strong>manual</strong> 8 MFA, 08/12/2008
E2.10 The production of zinc over the period 1992 – 2006 increased more or less steadily at a rate of<br />
3.1 % per year. The reserves, over the same period, increased by 3.5%. What conclusions can you<br />
draw from this about the criticality of zinc supply?<br />
Answer. The symptoms of criticality are<br />
• The rate of growth of discovery falls below the rate of growth of production.<br />
• The production rate curve peaks and starts to decline<br />
• The minimum economic ore grade falls<br />
• The price starts to rise sustainable.<br />
We are told in the question that growth in production has been steady – no sign of a flattening out<br />
suggesting a peak in production is near. The growth rate of reserves is com<strong>for</strong>tably larger than that of<br />
production indicating no shortage of supply. Taken together these indicators point to a market <strong>for</strong> zinc<br />
that is well balanced.<br />
E2.11 The production of platinum, vital <strong>for</strong> catalysts and catalytic converters, has risen from 145 to<br />
217 tonnes per year over the last 10 years. The ores are highly localized in South Africa, Russia and<br />
Canada. The reserves have risen from 56,000 to 71,000 tonnes in the same time interval. Would you<br />
classify platinum as a critical material? Base your judgment on the these facts, making use of the<br />
relative growth rates of production and reserves and on the dynamic index (equation 2.6 of the text)<br />
calculated using 2005 data.<br />
Answer. Platinum has applications <strong>for</strong> which there is no present substitute. The growth rates of<br />
production and of reserves, calculated as described in the solution to Exercise E2.7, are listed in the<br />
table below. The dynamic index of exploitation, calculated from these data, is 20.9 years.<br />
Platinum 2005<br />
1995<br />
Year Production rate,<br />
tonnes/year<br />
217<br />
145<br />
Reserves, tonnes Growth rate of<br />
production<br />
% per year<br />
71 x 10 3<br />
56 x 10 3<br />
Growth rate of<br />
reserves<br />
% per year<br />
4.03 2.37<br />
Those are the facts. What deductions? There are many danger signs here. Sources are localized<br />
and only one is in a country with a long history of economic and political stability. The growth rate of<br />
reserves has, over a 10 year period, lagged behind that of exploitation. The dynamic index – not a<br />
reliable measure, but another indicator none the less – is low when compared with those of most other<br />
metals.<br />
E2.12 Global water consumption has tripled in the last 50 years. What is the growth rate, r %, in<br />
consumption assuming exponential growth? By what factor will water consumption increase between<br />
now (2008) and 2050?<br />
Answer. Write consumption as<br />
( t t )<br />
⎧r − ⎫<br />
C = C exp<br />
o<br />
o ⎨ ⎬ which, on inversion, gives<br />
⎩ 100 ⎭<br />
( ) ⎟ 100 ⎛ C ⎞<br />
r = ln ⎜<br />
t − t ⎜<br />
o ⎝ Co<br />
⎠<br />
The ratio C / Co<br />
= 3 <strong>for</strong> the time interval ( t − to<br />
) = 50 years, giving r = 2.<br />
2%<br />
/ year .<br />
If this growth rate continues, the consumption of water will increase over the next 42 years by the<br />
factor<br />
Eco text: solution <strong>manual</strong> 9 MFA, 08/12/2008
( t t )<br />
C ⎧r −<br />
exp<br />
o ⎫<br />
= ⎨ ⎬ = 2.<br />
52<br />
Co<br />
⎩ 100 ⎭<br />
that is, it will more than double. Water supply in many parts of the world is already barely able to meet<br />
demand. It is not clear how this need will be met.<br />
E2.13 Plot the Annual world production of metals against their Price, using mean values from the<br />
data sheets in Chapter 12 of this book. What trend is visible?<br />
Answer. See the answer to Exercise E2.14, in which the annual world production <strong>for</strong> materials is<br />
plotted against price as a CES exercise. Metals are labeled. The trend described in the solution to<br />
E2.14 appears also when metal data alone areplotted.<br />
Exercises using CES Eco Level 2.<br />
E2.14 Use CES to plot the Annual world production of materials against their Price. What trend is<br />
visible?<br />
Answer. The figure shows the plot. There is a correlation – the lower the price, the larger is the<br />
annual production. The contours have a slope -2, which, with log scales (as here) means that, very<br />
approximately,<br />
Annual world production ∝<br />
-2<br />
( ) 2<br />
1<br />
Pr ice<br />
Eco text: solution <strong>manual</strong> 10 MFA, 08/12/2008
E2.15 Make a plot of the static index of exhaustion <strong>for</strong> metals (the reserves in tonnes divided by the<br />
annual production rate in tonnes per year), using the “Advanced” facility in CES to plot the ratio.<br />
Which metals have the longest apparent resource life? Why is this index not a reliable measure of the<br />
true resource life of the metal?<br />
Answer. The figure shows the plot of the static index, created with the CES Eco Level 2. The ores of<br />
magnesium, aluminum and titanium are plentiful and very widely distributed; <strong>for</strong> these there is no<br />
concern about depletion. Those of nickel, copper, zinc and lead are more localized and already<br />
significantly depleted, with the result that the ore grade currently mined is much leaner than it was 25<br />
years ago.<br />
The static index, however, is not a true indicator of resource life. This is because the reserves are<br />
an economic construct, which grow and shrink under varying economic, technical and legal conditions.<br />
Improved extraction technology can enlarge it, but environmental legislation or changing political<br />
climate may make it shrink. Demand stimulates prospecting, with the consequence that reserves tend<br />
to grow in line with consumption. The figure shows that the reserves of the more critical metals are<br />
typically between 20 to 50 times the annual production. This is a com<strong>for</strong>t zone. If the reserves fall<br />
below this zone, there is growing incentive to prospect <strong>for</strong> and develop new mines; if they rise above<br />
it, there is little incentive to do so.<br />
Eco text: solution <strong>manual</strong> 11 MFA, 08/12/2008
Chapter 3<br />
E3.1 Which phase of life would you expect to be the most energy intensive (in the sense of consuming<br />
fossil fuel) <strong>for</strong> the following products? Pick one and list the resources and emissions associated with<br />
each phase of its life along the lines of Figure 3.3 (template provided).<br />
Answer. At this stage we can only guess at the energy-intensive phase – later, when we get to ecoaudits,<br />
the question can be answered properly. As a general rule, if a product requires energy to<br />
per<strong>for</strong>m the use-phase of its life, then it is the cumulative energy of use that dominates. When the<br />
product does not use energy, it is the material production phase that dominates.<br />
Product Energy intensive life phase<br />
A toaster Probably the use phase<br />
A two car garage The material production phase if garage is unheated; if heated, then the use phase<br />
A bicycle The material production phase of life<br />
A motorbike The use phase<br />
A refrigerator The use phase<br />
A coffee maker Perhaps material production if the coffee maker is rarely used, otherwise use.<br />
A LPG fired patio heater No need to ask.<br />
Take the coffee maker as an example.<br />
Resources<br />
Oil <strong>for</strong> polymers<br />
Minerals <strong>for</strong> metals, glass<br />
Energy <strong>for</strong> both<br />
Emissions<br />
CO2, NOx, SOx<br />
Slag, tailings<br />
Chemical waste<br />
Low grade heat<br />
Resources<br />
Energy: polymer molding<br />
Energy: metal <strong>for</strong>ming<br />
Energy: glass molding<br />
Emissions<br />
Cut-offs (recycled)<br />
Low grade heat<br />
Resources<br />
Energy <strong>for</strong> collection<br />
Energy to disassemble<br />
Energy to recycle<br />
Emissions<br />
Materials (recycled)<br />
Waste to landfill<br />
Resources<br />
Water<br />
Paper <strong>for</strong> filters<br />
Coffee<br />
Energy to heat water<br />
Emissions<br />
Low grade heat<br />
Coffee grounds<br />
Waste filters<br />
Eco text: solution <strong>manual</strong> 12 MFA, 08/12/2008
E3.2 Functional units. Think of the basic need the product provides – it is this that determines use –<br />
and list what you would choose, thinking of all from an environmental standpoint<br />
Answer. The table lists the suggested functional units<br />
Product Functional unit<br />
Washing machines Energy (kW.hr or MJ) or liters of water per kg of clothes washed<br />
Refrigerators Energy (kW.hr or MJ) per m 3 of cooled space per year<br />
Home heating systems Energy (kW.hr or MJ) per unit volume of heated space per unit time (per m 3 per<br />
year)<br />
Air conditioners Energy (kW.hr or MJ) per unit volume of cooled space per unit time (per m 3 per<br />
year)<br />
Lighting Power (Watts) per lumen (the measure of light intensity)<br />
Home coffee maker Energy (kW.hr or MJ) per cup of coffee<br />
Public transport Energy (kW.hr or MJ) per passenger.mile<br />
Hand-held hair dryers Energy (kW.hr or MJ) per drying episode (difficult: long hair takes longer than<br />
short)<br />
E3.3 What is meant by “externalized” costs and costs that are “internalized” in an environmental<br />
context? Now a moment of introspection. List three internalized costs associated with your life style.<br />
Now list three that are externalized. If your life is so pure that you have less than three, then list some<br />
of other people you know.<br />
Answer. Many human activities incur costs, some obvious, some hidden. Industrial emissions, in<br />
particular, incur costs though damage to health and property via air and water pollution, acid rain,<br />
ozone layer depletion and climate change. When the damage is local and the creator of the emissions<br />
accepts the responsibility and the cost <strong>for</strong> containing and remediating it, the environmental cost is said<br />
to be internalized. When the damage not attributable, and particularly when it is global in impact, it is<br />
much more difficult to establish creator-responsibility. In these circumstances the environmental cost<br />
becomes a burden on society as a whole and is said to be externalized. Much current negotiation and<br />
legislation aims to internalize environmental costs that, at present, are externalized.<br />
Internalized<br />
Cost of mobile phone (assuming microwave radiation is harmless)<br />
Cost of use of personal computer<br />
Cost of broadband connection<br />
Cost of private health care<br />
Cost of home ownership<br />
Externalized<br />
Dropping cigarette butts, gum, hamburger wrappers in the street<br />
Disposing of television sets or personal computers by dumping<br />
Travel by plane (the traveler does not yet pay <strong>for</strong> the eco-damage of the emissions)<br />
Dumping, rather than recycling, nickel-cadmium batteries<br />
Eco text: solution <strong>manual</strong> 13 MFA, 08/12/2008
E3.4 What, in the context of life-cycle assessment, is meant by “system boundaries”? How are they<br />
set?<br />
Answer. To assess the energy demand, carbon footprint and emissions-profile of a product it is<br />
necessary to sum the contributions it makes to each of these over its entire life cycle. Deciding what to<br />
include and what exclude in the life-cycle (the “system”),however, is not simple. The first step in lifecycle<br />
assessment (LCA) is to decide on the system boundary. The boundary should normally enclose<br />
all four phases of life: material production, product manufacture, product use and product disposal. For<br />
the LCA to meet ISO standards, the LCA expert must define and state what aspects of each phase of<br />
life are included within the system boundary and which aspects remain outside it.<br />
It is sometimes desirable to conduct a partial LCA, setting the system boundary around only one<br />
phase of life. Thus the embodied energy and carbon footprint of a material is found by enclosing only<br />
the material production within the system boundary.<br />
E3.5 Describe briefly the steps prescribed by the ISO 14040 Standard <strong>for</strong> life-cycle assessment of<br />
products.<br />
Answer. The four steps in conducting an LCA, elaborated in the ISO 14040 set of standards, are:<br />
� Setting goals and scope. Why do the assessment? What is the subject and which bit(s) of its life<br />
are to be assessed (setting system boundaries)?<br />
� Inventory compilation: what resources are consumed, what emissions excreted?<br />
� Impact assessment: what do these do to the environment – particularly, what bad things?<br />
� Interpretation: what do the results mean and how are they to be used?<br />
E3.6 What are the difficulties with a full LCA? Why would a simpler, if approximate, technique be<br />
helpful?<br />
Answer.<br />
• A full LCA is a time-consuming task requiring days or weeks of experts’ time, and thus is<br />
expensive.<br />
• Despite the <strong>for</strong>malism that attaches to LCA methods, the results are subject to considerable<br />
uncertainty.<br />
• The output of an LCA that meets the ISO 14040 Standard is complex and detailed – it is of<br />
little help <strong>for</strong> design.<br />
A simpler, approximate approach would, by reducing the cost and speeding the process, allow much<br />
wider application.<br />
E3.7 Pick two of the products listed in Exercise E.3.1 and, using your judgment, attempt to fill out the<br />
simplified streamlined LCA matrix below to give an environmentally responsible product rating. Make<br />
your own assumptions (and report them) about where the product was made and thus how far it has to<br />
be transported, and whether it will be recycled. Assign an integer between 0 (highest impact) and 4<br />
(least impact) to each box and then sum to give an environmental rating, providing a comparison. Try<br />
the protocol<br />
Eco text: solution <strong>manual</strong> 14 MFA, 08/12/2008
• Material: is it energy-intensive? Does it create excessive emissions? Is it difficult or<br />
impossible to recycle? Is the material toxic? If the answer to these questions is yes, score<br />
4. If the reverse, score 0. Use the intermediate integers <strong>for</strong> other combinations.<br />
• Manufacture: is the process one that uses much energy? Is it wasteful (meaning cut-offs<br />
and rejects are high)? Does it produce toxic or hazardous waste? Does make use of<br />
volatile organic solvents? If yes, score 4. If no, score 0, etc.<br />
• Transport: is the product manufactured far from its ultimate market? Is it shipped by air<br />
freight? If yes, score 4. If no, score 0.<br />
• Use: does the product use energy during its life? Is the energy derived from fossil fuels?<br />
Are any emissions toxic? Is it possible to provide the use-function in a less energyintensive<br />
way? Scoring as above.<br />
• Disposal: Will the product be sent to land-fill at end of life? Does disposal involve toxic or<br />
long-lived residues? Scoring as above.<br />
What difficulties did you have? Do you feel confident that the results are meaningful?<br />
Answer. Example: the toaster.<br />
A toaster uses about 1 kg of materials, largely metals. Manufacture, almost certainly in South<br />
Asia, requires only simple processes, though chromium plating is a particularly toxic process. Most<br />
toasters used in the US and Europe are made in South-east Asia, so considerable transport is involved.<br />
A toaster consumes electrical energy during use, most of which at present derives from fossil fuels,<br />
with associated emissions. The toaster itself produces no emissions (unless you burn the toast) and is<br />
simple to recycle at end of life. The figure shows an attempt to fill out the matrix, based on this<br />
in<strong>for</strong>mation.<br />
It is not all easy to fill out the matrix with confidence unless you have considerable experience of<br />
assessing products in this way – the rankings chosen here are, quite frankly, guesses. This is just one<br />
example of the matrix approach – there are many variants of the row and column categories. Some are<br />
better adapted <strong>for</strong> particular products than others, but it is evident that any one of them requires a<br />
degree of experience. The present attempt (correctly) identifies the use and the transport phase<br />
(assuming the product is shipped over large distances) as the most damaging . The final total, 34 out of<br />
a possible maximum of 80 suggests the product is not a particularly harmful one.<br />
Eco text: solution <strong>manual</strong> 15 MFA, 08/12/2008
Chapter 4<br />
E4.1 Many products are thrown away and enter the waste stream even though they still work. What<br />
are the reasons <strong>for</strong> this?<br />
Answer. A product reaches the end of its life when it’s no longer valued, even though it may still<br />
work well The life expectancy is the least of<br />
• The physical life, meaning the time in which the product breaks down beyond economic<br />
repair;<br />
• The functional life, meaning the time when the need <strong>for</strong> it ceases to exist;<br />
• The technical life, meaning the time at which advances in technology have made the product<br />
unacceptably obsolete;<br />
• The economical life, meaning the time at which advances in design and technology offer the<br />
same functionality at significantly lower operating cost;<br />
• The legal life – the time at which new standards, directives, legislation or restrictions make the<br />
use of the product illegal<br />
• And finally the loss of desirability – the time at which changes in taste, fashion, or aesthetic<br />
preference render the product unattractive.<br />
E4.2 Do you think manufacture without waste is possible? “Waste”, here, includes waste (i.e. lowgrade)<br />
heat, emissions and solid and liquid residues that cannot be put to a useful purpose. If not, why<br />
not?<br />
Answer. All manufacture involves conversion: conversion of ores and feedstock into raw materials,<br />
conversion of raw materials into products, the use of the products and their ultimate disposal. All of<br />
these steps involve energy, sometimes in its primary <strong>for</strong>m as oil or gas, but more usually in a converted<br />
<strong>for</strong>m as electricity or mechanical power.<br />
All conversions involve conversion efficiencies. Some conversion efficiencies are high –<br />
conversion of electric to mechanical power, <strong>for</strong> instance, can be achieved with an efficiency of 90% .<br />
Others, <strong>for</strong> fundamental thermodynamic reasons, are not – the conversion of fossil fuels to electric<br />
power is, at best, only 45% efficient. The losses, small or large, constitute one sort of waste – waste<br />
heat – and, if fossil fuels are involved, another: waste emissions.<br />
Just as there are conversion efficiencies <strong>for</strong> energy, there are conversion efficiencies <strong>for</strong> the<br />
trans<strong>for</strong>mation of materials during manufacture, and <strong>for</strong> the recovery of materials at end of life. Some<br />
are economic in origin – salvaging and sorting 100% of the scrap produced during manufacture is not<br />
economic, though salvaging 90% might be. Recovering materials at end of life has lower efficiencies<br />
because the dispersion of materials in products makes full recovery uneconomic, and because<br />
contamination of one material by another cannot, in some products, be prevented. All these factors<br />
combine to make some waste unavoidable.<br />
E4.3 What options are available <strong>for</strong> coping with the waste-stream generated by modern industrial<br />
society?<br />
Answer. Materials have a life-cycle. They are extracted and refined, manufactured into products,<br />
used, and at end of first life, rejected as “waste”. But what is waste to some markets is a resource to<br />
others, creating a number of alternative channels down which the materials continue to flow. The<br />
sketch of Figure 4.1 of the text introduced the options: commit to landfill, combust <strong>for</strong> heat recovery,<br />
recycle (or downcycle), re-engineer (refurbish or recondition) and reuse.<br />
Eco text: solution <strong>manual</strong> 16 MFA, 08/12/2008
E4.4 Recycling has the attraction of returning materials into the use-stream. What are the obstacles to<br />
recycling?<br />
Answer. The recycling market is like any other, with prices that fluctuate according to the balance of<br />
supply and demand. In a free market the materials that are recycled are those from which a profit can<br />
be made. Recycling requires energy, and this energy carries its burden of emissions. The recycle<br />
energy is generally small compared to the initial embodied energy, making recycling – when it is<br />
possible at all – an energy-efficient propositions. But there are difficulties.<br />
• Recycling may not be cost-efficient; that depends on the degree to which the material has<br />
become dispersed. In-house scrap, generated at the point of production or manufacture is<br />
localized and is already recycled efficiently (near 100% recovery). Widely distributed<br />
“scrap” – material contained in discarded products – is a much more expensive proposition to<br />
collect, separate and clean.<br />
• Many materials cannot recycled, although they may still be reused in a lower-grade activity;<br />
continuous-fiber composites, <strong>for</strong> instance, cannot be re-separated economically into fiber and<br />
polymer in order to reuse them, though they can be chopped and used as fillers.<br />
• Most materials require an input of virgin material to avoid build-up of uncontrollable<br />
impurities. Thus the fraction of a material production that can ultimately re-enter the usestream<br />
depends on the material itself, on the product into which has been incorporated and on<br />
the price of virgin material with which the recycled material must compete.<br />
E4.5 Car tyres create a major waste problem. Use the internet to research ways in which the materials<br />
contained in car tyres can be used, either in the <strong>for</strong>m of the tyre or in some decomposition of it.<br />
Answer. The following sites are among many that provide in<strong>for</strong>mation about used tyres.<br />
1. www.ct.gov/dep/<br />
2. www.defra.gov.uk/environment/waste/topics/tyres.htm<br />
3. www.ni-environment.gov.uk/usedtyresleaflet.pdf<br />
4. www.tyredisposal.co.uk/cms/index.php<br />
The first describes regulations about disposal of tyres, the second lists ways to use them, and the third<br />
provides statistics <strong>for</strong> collection, reuse and disposal. From them we learn that 3.9 million tons of used<br />
tyres are produced in the US every year, of which 18.7% are recycled.<br />
Tyres do not compost. The main channels <strong>for</strong> reusing used tyres are:<br />
• Retreading almost doubles the life of tyres. Most can only be retreaded once, however.<br />
• Rubber crumb is made by shredding tyres. It is used <strong>for</strong> playground and sports track surfaces.<br />
• Whole or part tyres can be used in wall structures <strong>for</strong> earth retention.<br />
• Sea defences can be constructed using floating arrays of tyres<br />
• Fuel: used tyres are burnt <strong>for</strong> heat, particularly in cement making.<br />
Eco text: solution <strong>manual</strong> 17 MFA, 08/12/2008
E4.6 List three important functions of packaging.<br />
Answer. Packaging per<strong>for</strong>ms at least five functions<br />
• Protection: packaging extends product life by protecting foodstuffs and controlling the<br />
atmosphere that surrounds them.<br />
• Security: tamper-proof packaging protects the consumer.<br />
• In<strong>for</strong>mation: pack in<strong>for</strong>mation identifies the product, its sell-by date (if it has one) and gives<br />
instructions <strong>for</strong> use.<br />
• Affiliation: brands are defined by their packaging.<br />
• Presentation: packaging “presents” the product in much the same way that clothing presents<br />
the wearer.<br />
E4.7 As a member of a brain-storming group you are asked to devise ways of reusing polystyrene<br />
foam packaging – the sort that encases TV sets, computers, appliances and much else when transported<br />
– most of which at present is sent to landfill. Use free thinking: no suggestion is too ridiculous.<br />
Answer. The idea here is to encourage free thinking. Here are just a few possible uses<br />
• Break up <strong>for</strong> bean bags.<br />
• Drainage <strong>for</strong> patio garden pots.<br />
• Flotation <strong>for</strong> fish-farm enclosures.<br />
• Shred to make artificial snow <strong>for</strong> theatre productions.<br />
• Send to Art schools as raw material <strong>for</strong> hot-wire sculptures.<br />
• Pack in the attic to improve the thermal insulation of your home / garage / outhouse.<br />
E4.8 You are employed to recycle German washing machines, separating the materials <strong>for</strong> recycling.<br />
You encounter components with the following recycle marks:<br />
How do you interpret them?<br />
Answer. Chapter 4 and its Appendix (Section 4.8) gives the in<strong>for</strong>mation needed to crack the codes.<br />
(a) Polypropylene. (b) Polyamide 6 (Nylon 6) with 10% glass fiber. (c) Polypropylene with 20% talc.<br />
(d) Polystyrene with 15% glass fines (powdered glass).<br />
Eco text: solution <strong>manual</strong> 18 MFA, 08/12/2008
E4.9 The metal lead has a number of uses: principally as electrodes in vehicle batteries, in architecture<br />
<strong>for</strong> roofing and pipe-work (particularly on churches) and as pigment <strong>for</strong> paints. The first two of these<br />
allow recycling, the third does not. Batteries consume 38% of all lead, have an average life of 4 years,<br />
a growth rate of 4% per year and the lead they contain is recycled with an efficiency of 80%.<br />
Architectural lead accounts <strong>for</strong> 16 % of total consumption. The lead on buildings has an average life of<br />
70 years, the same growth rate (4% per year) as batteries, and 95% of it is recycled. What is the<br />
fractional contribution of recycled lead from each source to current supply?<br />
Answer. Inserting these data into equation (4.4) show that lead from batteries contributes a fraction<br />
0.26 that from buildings only 0.009 to current supply. The short life of the batteries means that the lead<br />
they contain makes a much larger contribution to the secondary scrap stream, even though the fraction<br />
recycled is smaller than that of architectural lead.<br />
E4.10 A material M is imported into a country principally to manufacture one family of products with<br />
an average life of 5 years and a growth rate of r c % per year. The material is not at present recycled at<br />
end of life but it could be. The government is concerned that imports should not grow. What level of<br />
recycling is necessary to make this possible?<br />
Answer. Consumption C is growing at a rate r c % per year . Over the life of the product, Δ t , it<br />
grows from C o to<br />
r<br />
C = C exp(<br />
c<br />
o Δt<br />
)<br />
100<br />
an increase of<br />
r<br />
Δ C = C<br />
c<br />
o exp( Δt<br />
) − Co<br />
100<br />
For zero growth in imports recycling must feed back into the “Material” phase of life a quantity of<br />
material equal to the growth in consumption over the product life time. This requires a recycle fraction<br />
f crit where<br />
Thus<br />
r<br />
f<br />
c<br />
crit Co<br />
= Δ C = Co<br />
exp( Δt<br />
) − Co<br />
100<br />
r<br />
f exp(<br />
c<br />
crit = Δ t ) − 1<br />
100<br />
Clearly fcrit cannot be greater than 1. Thus the scheme will only work if<br />
r<br />
exp(<br />
c<br />
Δ t ) ≤ 2<br />
100<br />
100<br />
or Δ<br />
t ≤ ln( 2 )<br />
rc<br />
Eco text: solution <strong>manual</strong> 19 MFA, 08/12/2008
Chapter 5<br />
E5.1 What is a Protocol? What do the Montreal Protocol and the Kyoto Protocol commit the<br />
signatories to do?<br />
Answer. The word “protocol” crops up most frequently in the language of medicine: there it means a<br />
detailed plan, or set of steps, to be followed in a study, an investigation or an intervention. The<br />
meaning of “protocol” in the context of this Chapter is a treaty – an agreement under international<br />
law entered into by the actors, namely states and international organizations.<br />
• The Montreal Protocol of 1987 commits the nations that signed it to phase out the use of<br />
chemicals that deplete ozone in the stratosphere.<br />
• The Kyoto Protocol of 1998 commits the nations that signed it to reduce the emissions of<br />
gases that, through the greenhouse effect, cause climate change.<br />
E5.2 What is meant by “internalized” and “externalized” environmental costs? What legislative steps<br />
can be taken to pressure polluters to internalize environmental costs?<br />
Answer. When environmental damage can be attributed to an particular industry or activity that<br />
accepts both the responsibility and the expense of containing and remediating it, the environmental<br />
cost is said to be internalized. When the damage is not attributable, the environmental cost becomes a<br />
burden on society as a whole and is said to be externalized.<br />
Much current negotiation and legislation aims to internalize environmental costs that, at present, are<br />
externalized (the “Polluter pays” policy). This can take the <strong>for</strong>m of outright bans, fines, penalties or<br />
taxes, all of which require a mechanism to insure compliance, and many of which are unpopular. The<br />
alternative is to use market <strong>for</strong>ces to initiate change, through trading schemes, subsidies and other<br />
incentives to make environmental protection economically attractive.<br />
E5.3. What is the difference between command and control methods and the use of economic<br />
instruments to protect the environment?<br />
Answer. Command and control environmental legislation makes it illegal to use specific resources or<br />
allow specific emissions – examples are bans on dumping of toxic waste, the use of lead in petrol, the<br />
release of certain chemicals that pollute water or those that damage the ozone layer. There is a growing<br />
recognition that this can lead to perverse effects where action to fix one isolated problem just shifts the<br />
burden elsewhere and may even increase it. For this reason there has been a shift from command and<br />
control legislation towards the use of economic instruments – green taxes, subsidies, trading schemes –<br />
that seek to use market <strong>for</strong>ces to encourage the efficient use of materials and energy. Instead of<br />
outright bans, economic instruments seek to target environmental burdens that have costs that are not<br />
paid <strong>for</strong> by the provider or user (externalized costs), transferring the costs back to the activity creating<br />
it, thereby internalising them.<br />
E5.4 How does emissions trading work?<br />
Answer. Emissions trading is a market-based scheme that allows participants to buy and sell permits<br />
<strong>for</strong> emissions, or credits <strong>for</strong> reduction in emissions in certain pollutants. Taking carbon as an example,<br />
the regulator first decides on a total acceptable emissions level and divides this into tradable units<br />
Eco text: solution <strong>manual</strong> 20 MFA, 08/12/2008
called permits. These are allocated to the participants, based on their actual carbon emissions at a<br />
chosen point in time. The actual carbon emissions of any one participant change with time, falling if<br />
they develop more efficient production technology or rising if they increase capacity. A company that<br />
emits more than its allocated allowances must purchase allowances from the market, while a company<br />
that emits less than its allocations can sell its surplus. The buyer is paying a charge <strong>for</strong> polluting while<br />
the seller is rewarded <strong>for</strong> having reduced emissions.<br />
Emissions trading has another dimension – that of off-setting carbon release by buying credits in<br />
activities that absorb or sequester carbon or that replace the use of fossil fuels by energy sources that<br />
are carbon-free: tree planting, solar, wind or wave power <strong>for</strong> example. By purchasing sufficient credits<br />
the generator of CO2 can claim to be “carbon neutral”.<br />
E5.5. Carbon trading sounds like the perfect control mechanism to enable emissions reduction. But<br />
nothing in this world is perfect. Use the internet to research the imperfections in the system and report<br />
your findings.<br />
Answer. The criticisms of Carbon trading and offsetting are as follows.<br />
• It provides an excuse <strong>for</strong> enterprises to continue to pollute as be<strong>for</strong>e by buying credits and<br />
passing the cost on to the consumer.<br />
• The scheme only achieves it aim if the mitigating project runs <strong>for</strong> its planned life, and this is<br />
often very long. Trees, <strong>for</strong> instance, have to grow <strong>for</strong> 50 to 80 years to capture the carbon<br />
with which they are credited – fell them sooner <strong>for</strong> quick profit and the off-set has not been<br />
achieved. Wind turbines and wave power, similarly, achieved their claimed offset only at the<br />
end of their design life, typically 20 to 25 years.<br />
• It is hard to verify that the credit payments actually reach the mitigating projects – the treeplanters<br />
or wind turbine builders – <strong>for</strong> which they were sold; too much of it gets absorbed in<br />
administrative costs.<br />
E5.6. What are the merits and difficulties associated with (a) taxation and (b) trading schemes as<br />
economic instruments to control pollution?<br />
Answer.<br />
Historically, environmental legislation has targeted individual, obvious problems – dumping of toxic<br />
waste, lead in petrol, water pollution, ozone depletion – taking a command and control approach based<br />
on outright bans, fines, penalties or direct taxes. There is a growing recognition that this can lead to<br />
perverse effects where action to fix one isolated problem just shifts the burden elsewhere and may even<br />
increase it.<br />
Trading schemes and other economic instruments – green taxes, subsidies, carbon credits – seek to use<br />
market <strong>for</strong>ces to encourage the efficient use of materials and energy. The motive is to transfer<br />
externalized costs back to the activity creating them. The difficulties are those of ensuring that the<br />
trading schemes function as a well-balanced market and are not distorted by high administrative costs<br />
or misuse.<br />
Eco text: solution <strong>manual</strong> 21 MFA, 08/12/2008
E5.7. Your neighbour with a large 4 x 4 boasts that his car, despite its size, is carbon-neutral. What on<br />
earth does he mean (or thinks he means)?<br />
Answer. He means that he, or the company from which he bought the vehicle, has purchased carbon<br />
reduction certificates equal to the expected carbon emission of the vehicle. He thinks he means that<br />
these pay <strong>for</strong> projects that absorb carbon (like tree planting) or replace carbon-using energy sources<br />
with those that do not, but it is not always easy to verify how much of the payment reaches the project<br />
and how much is absorbed in administrative costs on the way.<br />
E5.8. In December 2007 Saab posted advertisements urging consumers to “switch to carbon neutral<br />
motoring” claiming that “every Saab is green”. In a press release the company said they planned to<br />
plant 17 native trees <strong>for</strong> each car bought. The company claimed that its purchase of offsets <strong>for</strong> each car<br />
sold made Saab the first car brand to make its entire range carbon free.<br />
What is misleading about this statement? (The company have since withdrawn it).<br />
Answer. The scheme only achieves it aim if the mitigating project runs <strong>for</strong> its planned life, and this is<br />
often very long. Trees, <strong>for</strong> instance, have to grow <strong>for</strong> 50 to 80 years to capture the carbon with which<br />
they are credited – fell them sooner <strong>for</strong> quick profit and the off-set has not been achieved. Wind<br />
turbines and wave power, similarly, achieved their claimed offset only at the end of their design life,<br />
typically 25 years. Even Saabs only last <strong>for</strong> 15 years, so the mitigation claimed <strong>for</strong> them is not even<br />
half-achieved by the time the vehicle dies.<br />
E5.9. What tools are available to help companies meet the European Union VOC (volatile organic<br />
compounds) regulations? Carry out a Web search to find out and report your findings.<br />
Answer. Numerous sites offer advice, help and new disposal methods <strong>for</strong> dealing with volatile organic<br />
compounds. Among them are<br />
1. en.wikipedia.org/wiki/Volatile_organic_compound<br />
2. www.parish-supply.com/volatile_organic_compounds.htm<br />
3. www.processingtalk.com<br />
4. www.coatings.org.uk/<br />
5. http://www.epd.gov.hk/epd<br />
From them we learn that VOCs (volatile organic compounds) are found in wide variety of<br />
everyday products such as solvent-based paints, printing inks, many consumer products, organic<br />
solvents and petroleum products. The US, the European Union, Australia and New Zealand all are in<br />
progress of introducing labelling, restrictions on use, and a ban on sales without permit to limit the<br />
release of VOCs into the atmosphere. At present most sites focus on explaining the regulation; a few<br />
offer technical solutions <strong>for</strong> dealing with VOCs to render them harmless (generally by controlled<br />
combustion).<br />
Eco text: solution <strong>manual</strong> 22 MFA, 08/12/2008
E5.10. What tools are available to help companies meet the European Union EuP (Energy-using<br />
Products) regulations? Carry out a Web search to find out and report your findings.<br />
Answer. Numerous sites report, offer advice and consultancy on the EuP Directive. Among them are:<br />
1. www.infoworld.com/article/06/10/31/45OPreality_1.html<br />
2. www.ee.sgs.com/eup_compliance_eup_ee<br />
3. www.synapsistech.com/solutions/compliance/eup.html<br />
4. www.era.co.uk/Services/ecodesign.asp<br />
From these we learn the following<br />
• The Directive. The EuP Directive will require manufacturers to calculate the energy used to<br />
produce, transport, sell, use, and dispose of almost every one of its products. It will require<br />
that the manufacturer go all the way back to the energy used when extracting the raw materials<br />
to make its product, including all subassemblies and components. And in time, it will set<br />
limits on a product-by-product basis of how much energy can be used in a product’s entire<br />
lifecycle (Source (1), above).<br />
• EuP compliance in brief. Manufacturers can choose to per<strong>for</strong>m the con<strong>for</strong>mity assessment<br />
either by ecological product profiles or within a management system like ISO 14001 or<br />
EMAS. Companies will be required to produce a document file <strong>for</strong> each of their products and<br />
should there<strong>for</strong>e start by collecting data on the mentioned environmental aspects in the<br />
different life cycle phases (Source (2)).<br />
• Services and tools to help with EuP compliance. A number of web sites (Source (3) is an<br />
example) offer in-house training and planning assistance to help companies set up procedures<br />
to comply with the Directive. Others (e.g. Source (4) go further, offering to design and put it<br />
place compliance procedures. It appears that many organizations with environmental<br />
knowledge are seeking to capture what maybe a large market in establishing procedures that<br />
are practical and af<strong>for</strong>dable, but none have yet emerged as clear market leader.<br />
Eco text: solution <strong>manual</strong> 23 MFA, 08/12/2008
Chapter 6<br />
E 6.1. What is meant by embodied energy per kilogram of a metal? Why does it differ from the free<br />
energy of <strong>for</strong>mation of the oxide, carbonate or sulfide from the which it was extracted?<br />
Answer. The embodied energy of a metal is the output of an energy audit of the resources and<br />
processes need to extract, refine or synthesize it. Energy is consumed in mining or extracting the ores<br />
and feedstock, concentrating the ore and reducing it to metal, all with energy losses. To convert an<br />
oxide, sulfide, or carbonate ore to metal, the free energy of <strong>for</strong>mation of those compounds must, of<br />
course, be provided. The embodied energy includes this, but it also includes the energy to mine,<br />
transport and concentrate the ore, and the energy “loss” that results from the inefficiencies of the<br />
processes.<br />
E 6.2 What is meant by the process energy per kilogram <strong>for</strong> casting a metal? Why does it differ<br />
from the latent heat of melting of the metal?<br />
Answer. In order to cast a metal, energy must be provided to heat the metal to its melting point, to<br />
melt it (requiring the latent heat of melting) and to raise the temperature of the liquid a little higher to<br />
ensure that, when cast, it does not solidify be<strong>for</strong>e it has filled the mold. The kiln itself has to be heated<br />
to the same temperature, requiring additional energy. The heat itself is provided by the combustion of<br />
fossil fuel (efficiency about 70%) or by electric heating (oil-equivalent efficiency about 40%). The<br />
thermal energy of both kiln and metal is lost as low-grade heat when the casting cools to room<br />
temperature. The casting energy is the sum of all the energies involved.<br />
E 6.3 Make a bar chart of CO2 footprint divided by Embodied energy using data from the data sheets<br />
of Chapter 12. Which material has the highest ratio? Why?<br />
Answer. The Figure shows the important features. For most materials the ratio is about0.04 - 0.06.<br />
For cement and concrete it is far higher. This is because, making cement (a component of concrete)<br />
involves “calcining” calcium carbonate, driving off carbon dioxide to leave lime. This CO2<br />
contribution adds to that derived from the fuel that is used to heat the kiln to give the high total.<br />
Eco text: solution <strong>manual</strong> 24 MFA, 08/12/2008
E 6.4 The embodied energies and CO2 footprints <strong>for</strong> woods, plywood and paper do not include a credit<br />
<strong>for</strong> the energy and carbon stored in the wood itself, <strong>for</strong> the reasons explained in the text. Recalculate<br />
these, crediting them with sequestering energy and carbon by subtracting out the stored contributions<br />
(take them to be 25 MJ/kg and 2.8 kg CO2 per kg). Is there a net saving?<br />
Answer. The table lists the embodied energies and carbon footprints <strong>for</strong> wood, plywood and paper,<br />
taking the means of the ranges shown on the data sheets. The last two columns show the values<br />
adjusted as described in the question. Crediting the materials with storing energy and carbon gives all<br />
of them a negative carbon footprint and all but one (paper) a negative embodied energy. The reasons<br />
we use the “uncorrected” values are explained in the text.<br />
Material Embodied<br />
energy (MJ/kg)<br />
CO2 footprint<br />
(kg/kg)<br />
Embodied energy,<br />
adjusted (MJ/kg)<br />
CO2 footprint,<br />
adjusted (kg/kg)<br />
Wood 7.4 0.43 -17.6 -2.37<br />
Plywood 15 0.75 -10 -2.05<br />
Paper 27 1.44 2 -1.36<br />
E6.5 Rank the three common commodity materials Low carbon steel, Aluminum alloy and<br />
Polyethylene by embodied energy/ kg, H m ,and embodied energy/m 3 , H m ρ , where ρ is the density,<br />
using data drawn from the data sheets of Chapter 12 (use the means of the ranges given in the<br />
databases). Finally rank them by embodied energy per unit stiffness (measured by H m ρ / E where<br />
E is Young’s modulus).<br />
Answer. The table shows the data and the calculated in<strong>for</strong>mation. Column 2 gives the embodied<br />
energy per kg, H m . Steel has the lowest. Column 5 lists the embodied energy per unit volume, H m ρ<br />
– polyethylene has by far the lowest. The last column is the embodied energy per unit stiffness (note<br />
the inclusion of ρ in the ratio so that top and bottom are both in the same units). Now steel is the<br />
lowest, and by a large margin. If you want a material that is stiff and has low embodied energy, steel is<br />
the best bet.<br />
Material Energy H m<br />
MJ/kg<br />
Density ρ<br />
kg/m 3<br />
Modulus E<br />
GPa<br />
Eco text: solution <strong>manual</strong> 25 MFA, 08/12/2008<br />
H m<br />
ρ<br />
MJ/m 3<br />
Low carbon steel 32 7850 207 2.51 x 10 5<br />
H m ρ / E<br />
(Dimensionless)<br />
Aluminum alloys 220 2700 75 5.94 x 10 5 7.9<br />
Polyethylene 81 950 0.74 7.7 x 10 4 104<br />
E6.6 Iron is made by the reduction of iron oxide, Fe2O3, with carbon, aluminum by the electrochemical<br />
reduction of Bauxite, basically Al2O3. The enthalpy of oxidation of iron is 5.5 MJ/kg, that of<br />
aluminum to its oxide is 20.5 MJ/kg. Compare these with the embodied energies of cast iron and of<br />
aluminum, retrieved from the data sheets of Chapter 12 (use means of the ranges given there). What<br />
conclusions do you draw?<br />
1.2
Answer. The table shows the data. The embodied energies are larger, by a factor of 3 to 10, than the<br />
enthalpy of oxidation. This is because of the inherent irreversibility of the reduction process, heat and<br />
other energy losses, and the energy required <strong>for</strong> mining transporting and concentrating the ores be<strong>for</strong>e<br />
reduction.<br />
Material Enthalpy of oxidation<br />
Embodied energy<br />
(MJ/kg)<br />
(MJ/kg) from Chapter 12<br />
Cast iron 5.5 17<br />
Aluminum 20.5 220<br />
E6.7 Estimate the energy to mold PET by assuming it to be equal to the energy required to heat PET<br />
from room temperature to its melting temperature, Tm. Compare this with the actual molding energy.<br />
You will find the molding energy, the specific heat and the melting temperature in the data sheet <strong>for</strong><br />
PET in Chapter 12 (use means of the ranges). What conclusions do you draw?<br />
Answer. PET melts at about 238 C, about 200 C above room temperature. Its specific heat is 1445<br />
J/kg.K, so the energy to raise one kg of PET from room temperature (20 o C) to the melting point is<br />
about 0.31 MJ/kg. The extrusion energy is 3.8 MJ/kg, the molding energy 9.6 MJ/kg. Both are more<br />
than ten times greater than the estimate based simply on heat the polymer.<br />
Where does the extra energy go? Part in the relatively low conversion efficiency of fossil-fuel<br />
energy to electric power (about 38%), part in heating the extrusion or molding press, in power<br />
dissipated by the mechanism of the press, and as incidentals – the general energy overhead of the plant.<br />
E6.8. The data sheets of Chapter 12 list eco-indicator values where these are available. As explained<br />
in the text, the eco-indicator value is a normalized, weighted sum involving resource consumption,<br />
emissions and estimates of impact factors. Plot eco-indicator values against embodied energy (a much<br />
simpler measure of impact). Is there a correlation?<br />
Answer. See the chart in the answer to exercise E6.11 (below). It is a CES plot of eco-indicator<br />
values against embodied energy. There is a clear correlation, but with scatter. Is the scatter<br />
significant? Think back to the inherent uncertainty in determining embodied energy (Figure 6.2) and in<br />
the arbitrary nature of the weight factors used to calculate eco-indictors (Figure 3.11 and its<br />
discussion). Given these, you would expect some scatter. We conclude the embodied energy is an<br />
approximate but still useful proxy <strong>for</strong> the eco-indicator.<br />
Eco text: solution <strong>manual</strong> 26 MFA, 08/12/2008
Exercises using the CES software.<br />
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 .<br />
Use CES to make similar plots <strong>for</strong> the carbon footprint. Use the “Advanced” facility in the axis<br />
selection window to make the one <strong>for</strong> kg CO2/m 3 by multiplying kg CO2/kg by the density in kg/m 3 .<br />
Answer. The chart is shown below. Not surprisingly, it looks very like the embodied energy per m 3 ,<br />
shown in Figure 6.9 of the text. (The Metals, Polymers, Ceramics and Hybrids are segregated to<br />
separate columns by selecting them <strong>for</strong> the x-axis using the “Trees” option in the Advanced facility.)<br />
E6.10 Plot a bar chart <strong>for</strong> the embodied energies of metals and compare it with one <strong>for</strong> polymers, on a<br />
a “per unit yield strength” basis, using CES. You will need to use the “Advanced” facility in the axisselection<br />
window to make the function<br />
H m ρ<br />
,<br />
σ y<br />
where H m is the embodied energy per kg, ρ is the density and σ y is the yield strength. Do this by<br />
using the “Advanced” facility in the Axis selection box to <strong>for</strong>m<br />
(Embodied energy * Density) / Yield strength (elastic limit)<br />
Which materials are attractive by this measure?<br />
Eco text: solution <strong>manual</strong> 27 MFA, 08/12/2008
Answer. The figure shows the CES output. (The Metals and Polymers are segregated to separate<br />
columns by selecting them <strong>for</strong> the x-axis using the “Trees” option in the Advanced facility.) Carbon<br />
steels and cast irons have a lower embodied energy per unit of strength – than any other metal or any<br />
polymer.<br />
E6.11 Compare the eco-indicator values of materials with their embodied energy. To do so, make a<br />
chart with (Embodied energy x Density) on the x-axis and Eco-indicator value on the y-axis. (Ignore<br />
the data <strong>for</strong> foams since these have an artificially-inflated volume). Is there a correlation between the<br />
two? Is it linear? Given that the precision of both could be in error by 10 % are they significantly<br />
different measures? Does this give a way of estimating, approximately, eco-indicator values where<br />
none are available?<br />
Answer. The plot of eco-indicator values against embodied energy is shown below. There is a clear<br />
correlation, but with some scatter. See the answer to Exercise 6.9 <strong>for</strong> a commentary on this.<br />
Eco text: solution <strong>manual</strong> 28 MFA, 08/12/2008
Eco – indicator values<br />
E6.12 Plot material price against annual production in tonnes per year – is there a correlation?<br />
Answer. The plot shows that there is a correlation. The grid of broken lines has a slope of -2, meaning<br />
Annual world production ∝<br />
1<br />
2<br />
(Pr ice )<br />
It is no surprise that production falls as price rises. The origin of the power of -2 is obscure.<br />
Slope -2<br />
Eco text: solution <strong>manual</strong> 29 MFA, 08/12/2008
Chapter 7<br />
E7.1 If the embodied energies and CO2 used in the Alpure water case study in the chapter are<br />
uncertain by a factor of ± 25%, do the conclusion change? Mark ± 25% margins onto each bar in a<br />
copy of Figure 7.3 (you are free to copy it) and then state your case.<br />
Answer. The error margins are marked on the figures below. Even when the most extreme values are<br />
taken, the dominant contributions to energy and carbon are those of the material of the bottle. The<br />
conclusions reached in the text still stand.<br />
E7.2 Alpure water has proved to be popular. The importers now wish to move up-market. To do so<br />
they plan to market their water in 1 liter glass bottles of appealing design instead of the rather-ordinary<br />
PET bottles with which we are familiar from the case-study in the chapter. A single 1-liter glass bottles<br />
weigh 430 grams, much more than the 40 grams of those made of PET. Critics argue that this<br />
marketing-strategy is irresponsible because of the increased weight. The importers respond that glass<br />
has lower embodied energy than PET.<br />
Use the methods of this chapter and the data available in Chapter 12 to analyze this situation.<br />
What do you conclude?<br />
Answer. The energy audit <strong>for</strong> 100 glass bottles, carried out in the same way as the PET bottles<br />
described in the text and assuming the glass is recycled at end of life, gives the breakdown shown<br />
below. (Data are taken from the record <strong>for</strong> soda-glass in Chapter 12, using mean values of the ranges<br />
of embodied and process energies). The total life energies without recycling are summarized in the<br />
table below. The choice of glass is almost 3 times more energy intensive than PET.<br />
If both the PET and the glass are recycled, energy is “recovered”: it is the difference between the<br />
embodied energy and the recycle energy, also listed in the data sheets of Chapter 12. The last column<br />
of the table lists the life-energy with recycling; the bar chart shows the contributions. The glass bottle<br />
remains almost three more energy intensive than the one made of PET.<br />
Material<br />
(100 bottles)<br />
Life energy without<br />
recycling (MJ)<br />
Life energy with<br />
recycling (MJ)<br />
PET bottles 490 300<br />
Glass bottles 1200 830<br />
Eco text: solution <strong>manual</strong> 30 MFA, 08/12/2008
Eco-audit <strong>for</strong> 100 glass Alpure bottles.<br />
E7.3 If the glass container of the coffee maker audited in Section 7.4 of the text is replace by a double<br />
walled stainless steel one weighing twice as much, how much does the total embodied energy of the<br />
product change? If this reduces the electric power consumed over life by 10%, does the energy balance<br />
favor the substitution?<br />
Answer. The table lists the data and results, using in<strong>for</strong>mation about the mass of the container and the<br />
energy consumption over life given in the question. The stainless steel jug is twice as heavy as the<br />
glass one but saves 10% of the use energy. Using data <strong>for</strong> the embodied energy of and borosilicate<br />
glass and stainless steel from Chapter 12 gives the results in the table. The stainless steel jug carries 45<br />
MJ more embodied energy but saves more than three times this energy over life. In addition,<br />
borosilicate glass is difficult to recycle (it cannot be mixed with ordinary glass) whereas stainless steel<br />
is readily recycled.<br />
Material Mass Embodied energy Energy per Electrical energy<br />
(kg)<br />
(MJ/kg) container (MJ) over 5 years (MJ)<br />
Borosilicate glass 0.33 25 8.25 1380<br />
Stainless steel 0.66 81 53.5 1240<br />
Difference: + 45 MJ - 140 MJ<br />
E7.4 The figure shows a 1700 Watt steam iron. It weighs 1.3 kg, 98% of<br />
which is accounted <strong>for</strong> by the 7 components listed in the table. The iron<br />
heats up on full power in 4 minutes, is then used, typically, <strong>for</strong> 20<br />
minutes on half power. At end of life the iron is dumped as landfill.<br />
Create an eco-audit <strong>for</strong> the iron assuming that it is used once per week<br />
over a life of 5 years, using data from the data sheets of Chapter 12. Use<br />
PVC as a proxy <strong>for</strong> polyurethane.<br />
What conclusions can you draw? How might the energy be reduced?<br />
Eco text: solution <strong>manual</strong> 31 MFA, 08/12/2008
Steam iron: bill of materials<br />
Component Mass (kg) Material Shaping<br />
Material Process<br />
process energy* MJ/kg energy* MJ/kg<br />
Body 0.15 Polypropylene Molded 97 8.6<br />
Heating element 0.03 Nichrome Drawn 134 2.65<br />
Base 0.80 Stainless steel Cast 82 3.35<br />
Cable sheath, 3 meter 0.18 Polyurethane Molded 82 10.1<br />
Cable core, 3 meter 0.05 Copper Drawn 71 2.0<br />
Plug body 0.037 Phenolic Molded 90 12.7<br />
Plug pins 0.03 Brass Rolled 71 2.0<br />
* From the data sheets of Chapter 12<br />
Answer. The material energy is found by multiplying the mass of each component by its embodied<br />
energy/kg (listed, using data from Chapter 12, on the right of the table) and summing. The result is<br />
108 MJ. The process energy is found in the same way; here the result is 6.5 MJ. To calculate the usephase<br />
energy multiply the power (1.7 kW) by the effective use-time on full power (4 minutes on full<br />
power, 20 on half power, equivalent to 14 minutes per day, 52 days per year, 5 years) converted to<br />
seconds, then divided by 1000 (to convert kJ to MJ), giving 371 MJ. But this is electrical energy,<br />
generated from fossil fuel with an efficiency of around 38%, so to get the “oil equivalent” energy this<br />
electrical energy must be divided by 0.38, giving 977 MJ.<br />
The bar chart shows this distribution of energy commitment over the phases of life. The use<br />
phase dominates, accounting <strong>for</strong> 90% of the total energy. About 29% of the use-phase energy is used<br />
to heat the iron itself up to the working temperature, requiring energy<br />
Q = m C p ΔT<br />
Where m is the mass of the base of the iron (0.8 kg), C p is its specific heat and ΔT is the<br />
temperature interval through which it is heated. Thus energy could be saved by reducing the mass of<br />
the base of the iron or by selecting a material <strong>for</strong> it with a lower specific heat, or both.<br />
Eco text: solution <strong>manual</strong> 32 MFA, 08/12/2008
E7.5 The figure shows a 970 Watt toaster. It weighs 1.2 kg<br />
including 0.75 m of cable and plug. It takes 2 minutes 15 seconds to<br />
toast a pair of slices. It is used to toast, on average, 8 slices per day,<br />
so it draws its full electrical power <strong>for</strong> 9 minutes (540 seconds) per<br />
day over its design life of 3 years. The toasters are made locally –<br />
transport energy and CO2 are negligible. At end of life it is dumped.<br />
Create an eco-audit <strong>for</strong> the toaster using data from the data sheets of<br />
Chapter 12.<br />
Toaster: bill of materials<br />
Component Mass Material Shaping<br />
Material Process<br />
(kg)<br />
process energy* MJ/kg energy* MJ/kg<br />
Body 0.24 Polypropylene Molded 97 8.6<br />
Heating element 0.03 Nichrome Drawn 134 2.65<br />
Inner frame 0.93 Low carbon steel Rolled 32 2.4<br />
Cable sheath, 0.75 meter 0.045 Polyurethane Molded 82 10.1<br />
Cable core, 0.75 meter 0.011 Copper Drawn 71 2.0<br />
Plug body 0.037 Phenolic Molded 90 12.7<br />
Plug pins 0.03 Brass Rolled 71 2.0<br />
* From the data sheets of Chapter12<br />
Answer. The material energy is found by multiplying the mass of each component by its embodied<br />
energy/kg (listed, using data from Chapter 12, on the right of the table) and summing. The result is 67<br />
MJ. The process energy is found in the same way; here the result is 5.4 MJ. To calculate the use-phase<br />
energy multiply the power (0.97 kW) by the use-time on full power (9 minutes per day, 365 days per<br />
year, 3 years) converted to seconds, then divided by 1000 (to convert kJ to MJ), giving 574 MJ. The<br />
result is the MJ of electrical energy consumed over life. But this is electrical energy, generated from<br />
fossil fuel with an efficiency of around 38%, so to get the “oil equivalent” energy this electrical energy<br />
must be divided by 0.38, giving 1510 MJ.<br />
The bar chart shows this distribution of energy commitment over the phases of life. The use<br />
phase dominates, accounting <strong>for</strong> 90% of the total energy. How could the energy efficiency be<br />
increased? Air circulates through the toaster during use, necessary to carry off the moisture distilled<br />
from the bread, but carrying off energy too. Optimizing this air flow and the use of infra-red lamp<br />
elements instead of the resistance coils could focus the heat more precisely where it is needed. When<br />
toasting a single slice in a 2-slice toaster half the heat is wasted. Provision of a switch or sensor to<br />
limit heating to the slot containing bread would significantly reduce energy consumption.<br />
Eco text: solution <strong>manual</strong> 33 MFA, 08/12/2008
E7.6 It is proposed to replace the low-alloy steel bumper set of Case Study 7.6 in the text by one<br />
made of CFRP. It is anticipated that the CFRP set will weigh 7 kg. Following the procedure of the<br />
text, drawing data from the data sheets of Chapter 12, to estimate whether, over the life pattern<br />
used in the text, there is a net energy saving.<br />
Answer. The table below lists mean values <strong>for</strong> the embodied energies of the two materials. Using the<br />
energy consumption of a gasoline-powered car as 2.1 MJ/tonne.km (Table 6.7 of the text) we find the<br />
use-energy over 250,000 km to have the values shown in the second last column. The two bar charts<br />
display the relative magnitudes of material and use energies. Based on this very crude comparison, the<br />
substitution gives a substantial energy saving if the vehicle is driven the full 250,000 km. The breakeven<br />
distance is found by equating the total energy associated with the steel bumper-set to that of the<br />
CFRP set <strong>for</strong> a driven distance of X km, and solving <strong>for</strong> X. The result is 98,500 km.<br />
Material of<br />
fender<br />
Mass<br />
kg<br />
Material<br />
energy, MJ/kg*<br />
Material<br />
energy, MJ<br />
Use energy,<br />
MJ<br />
Total: material<br />
plus use, MJ<br />
Low alloy steel 14 35 490 7210 7700<br />
CFRP 7 273 1910 3610 5520<br />
* From the data sheets of Chapter 12.<br />
E7.7 It is reported that the production of a small car (mass<br />
1000 kg) requires materials with a total embodied energy of<br />
70 GJ, and a further 15 GJ <strong>for</strong> the manufacturing phase. The<br />
car is manufactured in Germany and delivered to the US show<br />
room by sea freight (distance 10,000 km) followed by delivery<br />
by heavy truck over a further 250 km (Table 6.7 of the text<br />
gives the energy per tonne.km <strong>for</strong> both). The car has a useful<br />
life of 10 years, and will be driven on average 25,000 km per<br />
year consuming 2MJ/km. Assume that recycling at end of life<br />
recovers 25 GJ per vehicle.<br />
Make an energy-audit bar chart <strong>for</strong> the car with bars <strong>for</strong> material, manufacture, distribution, use and<br />
disposal. Which phase of life consumes most energy? The inherent uncertainty of current data <strong>for</strong><br />
embodied and processing energies are considerable – if both of these were in error by a factor of 2 either<br />
way can you still draw firm conclusions from the data? If so, what steps would do most to reduced lifeenergy<br />
requirements?<br />
Eco text: solution <strong>manual</strong> 34 MFA, 08/12/2008
Answer. The figure shows the bar chart based on the data with the factor of 2 uncertainty indicated by Tbars<br />
showing the ranges on the Material and the Manufacturing bars. Even with this colossal allowance<br />
<strong>for</strong> the imprecision of the base-line data the conclusion is inescapable: the use-phase of the vehicle<br />
remains the dominant contribution to life-energy consumption (more on this in Chapter 9). Material<br />
choice should focus first on reducing vehicle mass since it is this that most directly correlates with fuel<br />
consumption. It is essential to then check that the consequent change in embodied energy of the material<br />
choice has not negated the gain resulting from the lower mass, using the method of exercise E7.6.<br />
E7.8 The table below lists one European automaker’s summary of the material content of a mid-sized<br />
family car. Material proxies <strong>for</strong> the vague material descriptions are given in brackets and italicized.<br />
The vehicle is gasoline-powered and weighs 1800 kg. The data sheets of Chapter 12 provide the<br />
embodied energies of the materials: mean values are listed in the table. Table 6.7 gives the energy of<br />
use: 2.1 MJ/tonne.km, equating to 3.8 MJ/km <strong>for</strong> a car of this weight. Use this in<strong>for</strong>mation here to<br />
make an approximate comparison of embodied and use energies of the car assuming it is driven 25,000<br />
km per year <strong>for</strong> 10 years.<br />
Material content of a family car, total weight 1800 kg<br />
Material Mass (kg) Material energy,<br />
MJ/kg*<br />
Energy per car<br />
GJ<br />
Steel (Low alloy steel) 950 32 30.4<br />
Aluminum (Cast aluminum alloy) 438 220 96.4<br />
Thermoplastic polymers (PU, PVC) 148 80 11.8<br />
Thermosetting polymers (Polyester) 93 88 8.2<br />
Elastomers (Butyl rubber) 40 110 4.4<br />
Glass (Borosilicate glass) 40 15 0.6<br />
Other metals (Copper) 61 72 4.4<br />
Textiles (Polyester) 47 47 2.2<br />
Total mass 1800 Total energy 158.4<br />
* From the data sheets of Chapter 12<br />
Eco text: solution <strong>manual</strong> 35 MFA, 08/12/2008
Answer. The material energy of the car is found by multiplying the mass of each material used in it by<br />
its embodied energy. The column added to the right of the table above lists the contributions and their<br />
sum, 158 GJ per car. The bar chart below shows the comparison. The input data are of the most<br />
approximate nature, but it would take very large discrepancies to change the conclusion: the energy<br />
consumed in the use phase (here about 85%) greatly exceeds that embodied in the materials of the<br />
vehicle. Fuel consumption scales with weight (more detail on this in Chapter 9). The answer here is to<br />
make the car from lighter materials – provided, of course, that the embodied energy of the replacement<br />
is not so high that it offsets the gain made by light-weighting.<br />
E7.10 Conduct a CO2 eco-audit <strong>for</strong> the patio heater shown here. It is<br />
manufactured in SE Asia and shipped 8,000 km to the US where it is sold and<br />
used. It weighs 24 kg, of which 17 kg is rolled stainless steel, 6 kg is rolled<br />
carbon steel, 0.6 kg is cast brass and 0.4 kg is unidentified injection-molded<br />
plastic (so use a proxy of your own choosing <strong>for</strong> this). In use it delivers 14 kW of<br />
heat (“enough to keep 8 people warm”) consuming 0.9 kg of propane gas (LPG)<br />
per hour, releasing 0.059 kg of CO2 /MJ. The heater is used <strong>for</strong> 3 hours per day<br />
<strong>for</strong> 30 days per year, over 5 years, at which time the owner tires of it and takes it<br />
to the recycling depot (only 6 miles / 10 km away, so neglect the transport CO2)<br />
where the stainless steel, carbon steel and brass are sent <strong>for</strong> recycling. Use data<br />
from the text and data sheets of Chapter 12 to construct a bar-chart <strong>for</strong> CO2<br />
emission over life.<br />
Answer. The table below lists the carbon footprints of the materials and manufacturing processes <strong>for</strong><br />
the patio heater. The bar chart plots the totals. Sea transport over 8000km, consuming 0.015 kg CO2/<br />
tonne.km (from Table 6.7 of the text) releases 2.9 kg of carbon dioxide per unit, so small as to be<br />
invisible on the bar chart. The use of the heater, delivering 14 kW, or 50.4 MJ/hour with the release of<br />
0.059 kg CO2/MJ, emits 1338 kg of CO2.<br />
Recycling saves the difference between the carbon footprint of virgin and that of recycled<br />
material. The second table lists these differences, using data <strong>for</strong> virgin and recycled material from the<br />
data sheets of Chapter 12.<br />
Not surprisingly, the carbon emission during use (over 1 tonne of CO2 over the course of life)<br />
dominates. There is only one way to reduce the carbon footprint of a device like this: turn it off.<br />
Eco text: solution <strong>manual</strong> 36 MFA, 08/12/2008
Material Mass (kg) Material CO2<br />
kg/kg*<br />
Material CO 2<br />
per unit, kg<br />
Process CO 2<br />
kg/kg*<br />
Process CO 2<br />
per unit, kg<br />
Stainless steel, rolled 17 5.05 85.9 0.285 4.8<br />
Carbon steel, rolled 6 2.5 15 0.195 1.2<br />
Brass, cast 0.6 5.25 3.2 0.16 0.1<br />
Thermoplastic polymer molded<br />
(polypropylene)<br />
0.4 2.7 1.1 0.68 0.27<br />
Totals 24 105 6.4<br />
* From the data sheets of Chapter 12<br />
Recycled material Mass (kg) Difference between material CO2<br />
and recycling CO2 kg/kg*<br />
CO 2 saved by recycling<br />
per unit, kg<br />
Stainless steel, rolled 17 3.6 61.2<br />
Carbon steel, rolled 6 1.8 10.8<br />
Brass, cast 0.6 3.95 2.4<br />
Thermoplastic polymer molded<br />
(polypropylene)<br />
0.4 1.55 0.6<br />
Totals 24 75<br />
Eco text: solution <strong>manual</strong> 37 MFA, 08/12/2008
Exercises using the CES eco-audit tool<br />
These <strong>exercises</strong> repeat or extend those of the text or of the preceding <strong>exercises</strong>, using the ecoaudit<br />
tool of the CES Edu software (found in the pull-down menu under “Tools”). The output and bar<br />
charts in some cases differ slightly from those of the earlier <strong>exercises</strong> because of the way the software<br />
creates means of the property ranges (it uses geometric rather than arithmetic means, more logical<br />
when ranges are large) and because it calculates the oil equivalent of electrical energy using a countrydependent<br />
energy mix (the mix of fossil fuel, hydro, wind and nuclear in the grid supply).<br />
E7.11 Repeat the analysis of the PET bottle <strong>for</strong> the Alpure water – the first case study in the Chapter –<br />
entering data from Table 7.1 the CES eco-audit tool. Then repeat, replacing the PET bottle by a glass<br />
one, using the additional in<strong>for</strong>mation in exercise E7.2.<br />
Answer. The resulting bar charts and report reproduce those already presented in Exercises E7.1 and<br />
7.2.<br />
E7.12 Carry out the eco-audit <strong>for</strong> the jug kettle of Section 7.3, now using the fact that the PP kettle<br />
body and the cardboard packaging are recycled.<br />
Answer. Enter the bill of materials <strong>for</strong> the jug kettle, listed in the first four columns of Table 7.5 of the<br />
text into the eco-audit tool, selecting “Recycle” <strong>for</strong> end of life <strong>for</strong> the PP body and the packaging and<br />
“Landfill” <strong>for</strong> all the others. The report summary and bar chart delivered by the software are show<br />
below, here using the electrical energy mix <strong>for</strong> Australia. The recycling offers a potential energy<br />
saving of 51 MJ, 2.4% of the energy total.<br />
Summary: jug kettle with recycling<br />
Life phase Energy, MJ<br />
Material 120<br />
Manufacture 10<br />
Transport 130<br />
Use 1900<br />
End of life -51<br />
Total 2110<br />
Jug kettle<br />
Eco text: solution <strong>manual</strong> 38 MFA, 08/12/2008
E7.13 Carry out the eco-audit <strong>for</strong> the coffee maker of Section 7.4, now using the fact that it is made<br />
entirely from recycled materials (select the grade “100% recycled”) and that the PP housing and the<br />
steel parts are themselves recycled at end of life.<br />
Answer. Enter the bill of materials <strong>for</strong> the coffee maker, listed in the first four columns of Table 7.6 of<br />
the text into the eco-audit tool, selecting “Recycle” <strong>for</strong> end of life <strong>for</strong> the PP body (as in the original<br />
case study of the text) and “Landfill” <strong>for</strong> all the others. The report summary delivered by the software<br />
is show below, here using the average electrical energy mix <strong>for</strong> Europe, and selecting “Virgin<br />
materials” <strong>for</strong> all. The resulting bar-chart is shown below – it reproduces that shown as Figure 7.7 of<br />
the text. The material energy of 147 MJ is 8% if the total.<br />
When all the material inputs are entered instead as “100% recycled” the material energy falls to<br />
59 MJ, now only 4 % of the total.<br />
Summary: Coffee maker, virgin materials<br />
Life phase Energy, MJ<br />
Material 147<br />
Manufacture 13<br />
Transport 6<br />
Use 1480<br />
End of life -52<br />
Total 1594<br />
E7.14 Carry out the eco-audit <strong>for</strong> CO2 <strong>for</strong> the portable space heater using the bill of materials from<br />
Section 7.5, Table 7.7 of the text, assuming that at end of first life the fan and heat shield are reengineered<br />
to incorporate them into a new product.<br />
Answer. Enter the bill of materials <strong>for</strong> the space heater, listed in the first four columns of Table 7.7 of<br />
the text into the eco-audit tool, selecting “Re-engineer” <strong>for</strong> end of life <strong>for</strong> the heat shield and all the<br />
components of the fan, and “Landfill” <strong>for</strong> all the others. The report summary and bar-chart are shown<br />
below.<br />
Re-engineering (and then reusing) the fan and heat shield saves 3.3 kg of CO2.<br />
Summary: Space heater<br />
Life phase CO2, kg<br />
Material 19<br />
Manufacture 1.7<br />
Transport 1.3<br />
Use 190<br />
End of life -3.3<br />
Total 209<br />
Coffee maker<br />
Space heater<br />
Eco text: solution <strong>manual</strong> 39 MFA, 08/12/2008
E7.15 Carry out Exercise E7.5 (the toaster) using the CES eco-audit tool, using the bill of materials<br />
listed there. Make bar charts both <strong>for</strong> energy and <strong>for</strong> CO2. Then select the grade “Standard” in the<br />
calculation, and repeat (“Standard” grade is material with a recycle content equal to the recycle fraction<br />
in current supply, data <strong>for</strong> which is included in the data sheets). Compare the result with using virgin<br />
material <strong>for</strong> making the toaster .<br />
Answer. Enter the bill of materials <strong>for</strong> the toaster, from Exercise E7.5 into the eco-audit tool. The<br />
report summaries and bar-charts <strong>for</strong> energy and CO2 when virgin materials are used are shown below.<br />
The same audit, using “Standard” grades <strong>for</strong> all materials reduces the Material energy from 68 to 56<br />
MJ and the Material CO2 from 3.7 to 2.9 kg.<br />
Summary: Toaster - energy<br />
Life phase Energy, MJ<br />
Material 68<br />
Manufacture 5.4<br />
Transport 0<br />
Use 1250<br />
End of life 0<br />
Total 1323<br />
Summary: Toaster – CO2<br />
Life phase CO2, kg<br />
Material 3.7<br />
Manufacture 0.4<br />
Transport 0<br />
Use 164<br />
End of life 0<br />
Total 168<br />
Toaster<br />
Eco text: solution <strong>manual</strong> 40 MFA, 08/12/2008
Chapter 8<br />
E 8.1. What is meant by an objective and what by a constraint in the requirements <strong>for</strong> a design? How<br />
do they differ?<br />
Answer. A constraint is an essential condition that must be met, usually expressed as an upper or<br />
lower limit on a material property. An objective is a quantity <strong>for</strong> which an extreme value (a maximum<br />
or minimum) is sought, frequently the minimization of cost, mass, volume or – of particular relevance<br />
here – environmental impact. The distinction is brought out by comparing common constraints and<br />
objectives, as in the table below. A constraint is a “Go / No go” criterion – if the constraint is not met,<br />
the material is rejected. An objective allows the materials that meet all the constraints to be ranked: the<br />
one that minimizes the objective is the best choice.<br />
Common constraints Common objectives<br />
Must meet a target value of<br />
� Stiffness<br />
� Strength<br />
� Fracture toughness<br />
� Thermal conductivity<br />
� Service temperature<br />
Minimize<br />
� Cost<br />
� Mass<br />
� Volume<br />
� Energy consumption<br />
� Carbon emissions<br />
E 8.2. Describe and illustrate the “Translation” step of the material selection strategy. Materials are<br />
required to make safe, eco-friendly swings and climbing frames <strong>for</strong> a children’s’ playground. How<br />
would you translate these design requirements into a specification <strong>for</strong> selecting materials?<br />
Answer. The design requirements <strong>for</strong> a component of a product specify what it should do but not what<br />
properties its materials should have. Translation is the step of converting the design requirements into<br />
constraints and objectives that can be applied to a materials database. Thus the design requirement<br />
“Materials are required to make safe, eco-friendly swings and climbing frames <strong>for</strong> a children’s’<br />
playground” might translate as shown in the table.<br />
Design requirements Translation<br />
Materials are required to make safe,<br />
eco-friendly swings and climbing<br />
frames <strong>for</strong> a children’s’ playground<br />
Constraints: seek materials that meet<br />
specified levels of<br />
� Stiffness<br />
� Strength<br />
� Fracture toughness<br />
Objective: minimize<br />
� Carbon footprint of material<br />
Eco text: solution <strong>manual</strong> 41 MFA, 08/12/2008
E8.3. Bicycles come in many <strong>for</strong>ms, each aimed at a particular sector of the market:<br />
� Sprint bikes<br />
� Touring bikes<br />
� Mountain bikes<br />
� Shopping bikes<br />
� Children’s bikes<br />
� Folding bikes<br />
Use your judgment to identify the primary objective and the constraints that must be met <strong>for</strong> each of<br />
these.<br />
Answer. Suggested constraints and objectives are listed in the table below. The balance between cost<br />
and other objectives (like mass) depend on the consumer-group at which the bike is aimed.<br />
Spring bikes<br />
Bike type Constraints Objective(s)<br />
Meet target values of stiffness and<br />
strength.<br />
Touring bikes Meet target value strength and<br />
mass.<br />
Mountain bikes Meet target values of stiffness and<br />
strength<br />
Shopping bikes Meet target values of strength and<br />
mass.<br />
Children’s bikes Meet target values of strength and<br />
toughness, ease of molding or<br />
<strong>for</strong>ming.<br />
Folding bikes Meet target values of stiffness and<br />
strength.<br />
Minimize mass<br />
Minimize cost<br />
Maximize damage tolerance<br />
(toughness, fatigue strength)<br />
Minimize cost<br />
Minimize cost<br />
Minimize mass (<strong>for</strong> portability)<br />
and volume (<strong>for</strong> compactness).<br />
E 8.4. You are asked to design a fuel-saving cooking pan with the goal of wasting as little heat as<br />
possible while cooking. What objective would you choose, and what constraints would you<br />
recommend should be met?<br />
Answer. The table lists constraints and an objective<br />
Constraints Objective: minimize<br />
• Must be non-toxic<br />
• Meet an upper limit on<br />
acceptable cost<br />
• Meet constraints of <strong>for</strong>mability<br />
• Meet FDA* standards <strong>for</strong><br />
contact with food.<br />
• Must be recyclable (to save<br />
embodied energy)<br />
• Thermal resistance (thus maximize<br />
thermal conductivity)<br />
*The US Food and Drug Administration approves materials <strong>for</strong> contact with foods.<br />
Eco text: solution <strong>manual</strong> 42 MFA, 08/12/2008
E 8.5. Formulate the constraints and objective you would associate with the choice of material to make<br />
the <strong>for</strong>ks of a racing bicycle.<br />
Answer. The important constraints are those of meeting target values of stiffness (<strong>for</strong> control and<br />
com<strong>for</strong>t) and strength (<strong>for</strong> safety). Cost, in highly competitive sport, is irrelevant. Mass, however,<br />
carries a heavy per<strong>for</strong>mance penalty. The central objective is to minimize mass.<br />
E8.6. What is meant by a material index?<br />
Answer. The per<strong>for</strong>mance of a component or product is limited by the properties of the materials of<br />
which it is made. Sometimes it is limited by a single property, sometimes by a combination of them.<br />
The property or property-group that limits per<strong>for</strong>mance <strong>for</strong> a given design is called its material index.<br />
Per<strong>for</strong>mance is maximized by seeking the materials with the most extreme (biggest or smallest) value<br />
of the index, usually the smallest.<br />
E8.7. The objective in selecting a material <strong>for</strong> a panel of given in-plane dimensions <strong>for</strong> the lid-casing of<br />
an ultra-thin portable computer is that of minimizing the panel thickness h while meeting a constraint<br />
on bending stiffness, *<br />
S to prevent damage to the screen. What is the appropriate material index?<br />
Answer. The index, read from Table 8.3 of the text, is that <strong>for</strong> a stiffness limited panel of minimum<br />
1/<br />
3<br />
volume: minimize 1 / E , where E is Young’s modulus (or, more strictly, the flexural modulus,<br />
equal to E <strong>for</strong> isotropic materials but not <strong>for</strong> those, like fiber composites, that are not isotropic).<br />
E8.8. Plot the index <strong>for</strong> a light, stiff panel on a copy of the Modulus-Density chart of Figure 8.11,<br />
positioning the line such that six materials are left above it, excluding ceramics because of their<br />
brittleness. Which six to you find? What material classes do they belong to?<br />
Answer. The index <strong>for</strong> selecting materials <strong>for</strong> light stiff panels, read from Table 8.3, is ρ / E<br />
1/<br />
3<br />
.<br />
Construct a line by hand, or on a scanned image of the chart in WORD, with the slope indicated by the<br />
guide line on the figure, then position it so that a few materials lie to its left, as shown. These materials<br />
have the lowest (best) values of the index. Omitting ceramics and glasses because of brittleness, we<br />
find<br />
• Woods<br />
• CFRP (carbon fiber rein<strong>for</strong>ced polymers)<br />
• Magnesium (Mg) alloys<br />
• Three density-grades of rigid polymer foams.<br />
Polymer foams have the problem that their modulus is low, so although they are light, the panel has to<br />
be thick to achieve much stiffness. It is possible to apply a constraint on thickness (it translates into a<br />
lower limit <strong>for</strong> modulus); applying it then removes any materials that can only meet the stiffness<br />
constraint if they are too thick.<br />
Eco text: solution <strong>manual</strong> 43 MFA, 08/12/2008
The Modulus – Density chart: the one <strong>for</strong> stiffness at minimum weight.<br />
E8.9. Panels are needed to board-up the windows of an unused building. The panels should have the<br />
lowest possible embodied energy but be strong enough to deter an intruder who, in attempting to break<br />
in, will load the panels in bending. Which index would you choose to guide choice?<br />
Plot the index on the Strength-Embodied energy chart of Figure 8.14, positioning the line to find<br />
the best choice, excluding ceramics because of their brittleness. Which six to you find? What material<br />
classes do they belong to?<br />
Answer. The index <strong>for</strong> a strength limited panel of minimum embodied energy, read from Table 8.3 of<br />
the text, is H<br />
1/ 2<br />
m ρ / σ y , where H m is the embodied energy per kg of the material, ρ is its density<br />
and σ y its yield strength. The index is plotted on the Strength – Embodied energy chart using the<br />
procedure described in the answer to example E8.8, reading the slope from the appropriate guide line in<br />
the lower right of the chart. It is position to leave a few materials to its left, as shown. Ignoring<br />
ceramics because or brittleness, the best choices are<br />
• Woods<br />
• Cast irons<br />
• Carbon steels<br />
Index<br />
ρ<br />
/ E<br />
1 / 3<br />
• Three density-grades of rigid polymer foams.<br />
Polymer foams are unsuitable <strong>for</strong> other reasons (too easily cut, and too thick) leaving the others as the<br />
most environmentally friendly choices.<br />
Eco text: solution <strong>manual</strong> 44 MFA, 08/12/2008
The Strength – Embodied energy chart: the one <strong>for</strong> strength at minimum embodied<br />
energy.<br />
E8.10. A material is required <strong>for</strong> disposable <strong>for</strong>ks <strong>for</strong> a fast-food chain. List the objective and the<br />
constraints that you would see as important in this application.<br />
Answer. The primary considerations in choosing materials <strong>for</strong> disposable applications are health, cost<br />
and the environmental damage that disposal might cause. The table list the resulting constraints and<br />
objectives.<br />
Constraints: the material must be Objective: minimize<br />
• Non toxic<br />
• Cheap<br />
• Cheap also to mold or shape<br />
• Low embodied energy<br />
• Easily (and profitably) recyclable<br />
Index<br />
H<br />
1/ 2<br />
m ρ<br />
/ σ y<br />
• Embodied energy or<br />
• Carbon footprint<br />
(they lead to essentially the same<br />
choice)<br />
Eco text: solution <strong>manual</strong> 45 MFA, 08/12/2008
E8.12. Show that the index <strong>for</strong> selecting materials <strong>for</strong> a<br />
strong panel with the dimensions shown here, loaded in<br />
bending, with minimum embodied energy content is<br />
M<br />
ρ H m<br />
=<br />
1/ 2<br />
σ y<br />
where H m is the embodied energy of the material, ρ its density and σ y its yield strength. To do so,<br />
rework the panel derivation in Section 8.10 replacing the stiffness constraint with a constraint on<br />
failure load F requiring that it exceed a chosen value *<br />
F where<br />
I σ y<br />
F = C<br />
*<br />
2 > F<br />
h L<br />
where C2 is a constant and I is the second moment of area of the panel :<br />
Answer. The objective function <strong>for</strong> the embodied energy of the panel, p H ,<br />
the embodied energy of the material per unit volume, ρ H m<br />
H p = b h L ρ H m<br />
Its failure load F must be at least<br />
I σ y<br />
F = C2<br />
≥<br />
h L<br />
F<br />
*<br />
*<br />
F , where<br />
Eco text: solution <strong>manual</strong> 46 MFA, 08/12/2008<br />
I<br />
=<br />
b h<br />
12<br />
3<br />
.<br />
is the volume Lbh times<br />
Here C 2 is a constant that depends only on the distribution of the loads and I is the second moment of<br />
area, which, <strong>for</strong> a rectangular section, is<br />
3<br />
b h<br />
I =<br />
12<br />
We can reduce the energy H by reducing h , but only so far that the stiffness constraint is still met.<br />
p<br />
Combining the last two equations and solving <strong>for</strong> h gives<br />
1.<br />
/ 2<br />
⎛ *<br />
12 F L ⎞<br />
h ≥ ⎜ ⎟<br />
⎜ C2b<br />
σ y<br />
⎟<br />
⎝ ⎠<br />
Using this to eliminate h in the objective function gives<br />
H p<br />
1/<br />
2<br />
⎛ * 3<br />
12 F L b ⎞ ρ H m<br />
≥ ⎜ ⎟<br />
⎜ C2<br />
⎟ 1/ 2<br />
⎝ ⎠ σ y<br />
*<br />
The quantities F , L , b and C 2 are all specified; the only freedom of choice left is that of the<br />
material. The best materials <strong>for</strong> a strong panel with the lowest embodied energy are those with the<br />
smallest values of<br />
ρ H m<br />
M p =<br />
1/ 2<br />
σ y
E8.13. Use the chart E − H m ρ chart of Figure 8.13 to find the metal with a modulus E greater than<br />
100 GPa and the lowest embodied energy per unit volume.<br />
Answer. The construction is shown below. The metal with a modulus greater than 100 GPa and the<br />
lowest embodied energy is cast iron.<br />
The Modulus – Embodied energy chart: the one <strong>for</strong> stiffness at minimum embodied energy.<br />
8.14. A maker of polypropylene (PP) garden furniture is concerned that the competition is stealing part<br />
of his market by claiming that the “traditional” material <strong>for</strong> garden furniture, cast iron, is much less<br />
energy and CO2 intensive than the PP. A typical PP chair weighs 1.6 kg; one made of cast iron weighs<br />
11 kg. Use the data sheets <strong>for</strong> these two materials in Chapter 12 of the book to find out who is right –<br />
are the differences significant? (Remember the warning about precision at the start of Chapter 12).<br />
If the PP chair lasts 5 years and the cast iron chair lasts 25 years, does the conclusion change?<br />
Answer. The table lists mean values of embodied energy and carbon footprint, per kg, <strong>for</strong> the two<br />
materials. The last two columns show the values per chair. If the difference in lifetime is ignored, the<br />
two chairs do not differ significantly in embodied energy, but they do in carbon release. If the longer<br />
life of the cast iron chair is recognized by dividing the values by the life in years (to give energy and<br />
carbon per chair.year) the cast iron chair wins easily.<br />
Material Embodied energy*<br />
MJ/kg<br />
CO 2 *<br />
kg/kg<br />
Embodied energy<br />
MJ/chair<br />
Eco text: solution <strong>manual</strong> 47 MFA, 08/12/2008<br />
CO 2<br />
kg/chair<br />
Cast iron 17 1.05 187 11.6<br />
Polypropylene 103 2.7 165 4.32<br />
* From the data sheets of Chapter12.<br />
Lower embodied<br />
energy<br />
Modulus<br />
E = 100 GPa
Exploring design using CES Edu Level 2 Eco.<br />
E8.15. Use a “Limit” stage to find materials with modulus E > 180 GPa and embodied energy H m <<br />
30 MJ/kg.<br />
Answer. Apply the two constraints using a “Limit” stage in CES Edu Level 2 with Eco. Four materials<br />
meet them. They are<br />
Selected materials<br />
Cast iron, ductile (nodular)<br />
High carbon steel<br />
Low carbon steel<br />
Medium carbon steel<br />
E8.16. Use a “Limit” stage to find materials with yield strength σ y > 100 MPa and a carbon footprint<br />
CO 2 < 1 kg/kg.<br />
Answer. Two materials meet the constraints. They are<br />
Selected materials<br />
Cast iron, ductile (nodular)<br />
Cast iron, gray<br />
E8.17. Make a bar chart of embodied energy H m . Add a “Tree” stage to limit the selection to<br />
polymers alone. Which polymers have the lowest embodied energy?<br />
Answer. The figure below shows the bar-chart created by CES. It was made using a “Graph” stage<br />
with Embodied energy plotted on the y-axis. The selection was limited to Polymers and elastomers by<br />
selecting, <strong>for</strong> the x-axis, this subset using the “Tree” option found under Advanced in the axis choice<br />
dialog box. The four polymers with the lowest embodied energy are listed below. They are all biopolymers<br />
– polymers made from natural feedstock, not oil.<br />
Selected materials<br />
Natural Rubber (NR)<br />
Polyhydroxyalkanoates (PHA, PHB)<br />
Polylactide (PLA)<br />
Starch-based thermoplastics (TPS)<br />
Eco text: solution <strong>manual</strong> 48 MFA, 08/12/2008
E8.18. Make a chart showing modulus E and density ρ . Apply a selection line of slope 1,<br />
corresponding to the index ρ / E positioning the line such that six materials are left above it. What<br />
families do they belong to?<br />
Answer. The chart generated by CES is shown below with a selection line of slope 1 positioned to<br />
leave six materials above the line. These are the materials with the lowest values of the index. They<br />
are listed in the table. Five are ceramics, one is a composite.<br />
Selected materials<br />
Alumina<br />
Aluminum nitride<br />
Boron carbide<br />
CFRP, epoxy matrix (isotropic)<br />
Silicon carbide<br />
Silicon nitride<br />
Eco text: solution <strong>manual</strong> 49 MFA, 08/12/2008
E8.19. A material is required <strong>for</strong> a tensile tie to link the front and back walls of a barn to stabilize both.<br />
It must meet a constraint on strength and have as low an embodied energy as possible. To be safe the<br />
material of the tie must have fracture toughness K 1c<br />
> 18 MPa.m 1/2 . The relevant index is<br />
M = H m ρ / σ y<br />
Construct a chart of σ y plotted against H m ρ . Add the constraint of adequate fracture toughness,<br />
meaning K 1c<br />
> 18 MPa.m 1/2 , using a “Limit” stage. Then plot an appropriate selection line on the<br />
chart and report the three materials that are the best choices <strong>for</strong> the tie.<br />
Answer. The chart is shown below with a selection line leaving just three materials exposed. They are<br />
listed in the table.<br />
Selected materials<br />
Cast iron, ductile (nodular)<br />
High carbon steel<br />
Low alloy steel<br />
Eco text: solution <strong>manual</strong> 50 MFA, 08/12/2008
Eco text: solution <strong>manual</strong> 51 MFA, 08/12/2008
E8.20. A company wishes to enhance its image by replacing oil-based plastics in its products by<br />
polymers based on natural materials. Use the “Search” facility in CES to find biopolymers. List the<br />
material you find. Are their embodied energies and CO2 footprints less than those of conventional<br />
plastics? Make bar charts of embodied energy and CO2 footprint to find out.<br />
Answer. A search on “Biopolymer” delivers the 5 materials listed in the table. The plot of embodied<br />
energies <strong>for</strong> polymers, shown in the answer to exercise E8.17 shows that the first four have lower<br />
embodied energies than any other polymer or elastomers. The equivalent chart of carbon footprint,<br />
below, reveals that CO2 release associated with their production is at the lower end of that <strong>for</strong><br />
polymers – it is about the same as that polyethylene or PVC.<br />
Selected materials<br />
Natural Rubber (NR)<br />
Polylactide (PLA)<br />
Polyhydroxyalkanoates (PHA, PHB)<br />
Starch-based thermoplastics (TPS)<br />
Cellulose polymers (CA)<br />
Eco text: solution <strong>manual</strong> 52 MFA, 08/12/2008
Chapter 9<br />
E9.1. The materials of the drink containers of Figure 9.1 are recycled to different degrees. How does<br />
the ranking of Table 9.2 change if the contribution of recycling is included? To do so, multiply the<br />
energy per liter in the last column of the table by the factor<br />
⎛ H ⎞<br />
1 − f ⎜ ⎟<br />
⎜<br />
−<br />
rc<br />
rc 1<br />
⎟<br />
⎝ H m ⎠<br />
where f rc is the recycle fraction in current supply, H m is the embodied energy <strong>for</strong> primary material<br />
production and H rc is that <strong>for</strong> recycling of the material. You will find data <strong>for</strong> all three attributes in<br />
the data sheets of Chapter 12.<br />
Answer. The first table below lists the materials of the drink containers and the embodied energies,<br />
energies to recycle and typical recycled fraction in current supply. The second table reproduces Table<br />
9.2 of the text with two additional columns added. The first lists the correction factor , the second lists<br />
the energy per liter multiplied by the correction factor.<br />
The inclusion of recycled material significantly reduces all the energies. Steel still emerges as the<br />
best choice, but glass rather than aluminium now becomes the most energy-intensive.<br />
Container type Material Embodied<br />
energy* MJ/kg<br />
PET 400 ml bottle PET<br />
84<br />
PE 1 liter milk bottle High density PE<br />
81<br />
Glass 750 ml bottle Soda glass<br />
15.5<br />
Al 440 ml can 5000 series Al alloy 208<br />
Steel 440 ml can Plain carbon steel<br />
32<br />
* From the data sheets of Chapter 12<br />
Container type Material Mass,<br />
grams<br />
PET 400 ml bottle PET<br />
25<br />
PE 1 liter milk bottle High density PE 38<br />
Glass 750 ml bottle Soda glass<br />
325<br />
Al 440 ml can 5000 series Al alloy 20<br />
Steel 440 ml can Plain carbon steel 45<br />
Energy/liter<br />
MJ/liter<br />
5.3<br />
3.8<br />
6.7<br />
9.5<br />
3.3<br />
Recycle energy*<br />
(MJ/kg)<br />
38.5<br />
34<br />
6.8<br />
19.5<br />
9.0<br />
Correction<br />
factor<br />
0.89<br />
0.95<br />
0.87<br />
0.60<br />
0.70<br />
E9.2. Derive the correction factor to allow <strong>for</strong> recycle content cited in Exercise E9.1<br />
Recycle fr. in<br />
current supply*<br />
0.21<br />
0.085<br />
0.24<br />
0.44<br />
0.42<br />
Corrected<br />
energy MJ/liter<br />
4.7<br />
3.6<br />
5.8<br />
5.7<br />
2.3<br />
Answer. If the drink containers are made of virgin material, each carries an embodied energy of H m<br />
per kg. If instead they are made of material containing a fraction f rc of recycled material that had<br />
required the recycle energy H rc to produce it, the energy per kg falls to ( 1−<br />
f rc ) H m<br />
The factor we seek is the ratio of these two:<br />
+ f rc H rc .<br />
(<br />
1−<br />
frc<br />
) H<br />
m<br />
H m<br />
+<br />
f rc H rc<br />
⎛ H ⎞<br />
= 1 − f ⎜ ⎟<br />
⎜<br />
−<br />
rc<br />
rc 1<br />
⎟<br />
⎝ H m ⎠<br />
Eco text: solution <strong>manual</strong> 53 MFA, 08/12/2008
E9.4. Use the indices <strong>for</strong> the crash barriers (equations 9.1 and 9.2) with the charts <strong>for</strong> strength<br />
and density (Figure 8.12) and strength and embodied energy (Figure 8.14) to select materials <strong>for</strong> each<br />
of the barriers. Position your selection line to include one metal <strong>for</strong> each. Reject ceramics and glass on<br />
the grounds of brittleness. List what you find <strong>for</strong> each barrier.<br />
Answer.<br />
ρ<br />
σ<br />
2 / 3<br />
y<br />
H m ρ<br />
σ<br />
2 / 3<br />
y<br />
Eco text: solution <strong>manual</strong> 54 MFA, 08/12/2008
H m ρ<br />
ρ<br />
The figure above shows the two charts with the indices M 1 = and M<br />
2 / 3<br />
2 = marked.<br />
2 / 3<br />
σ y<br />
σ y<br />
Each is position to leave one class of metal exposed. The upper selection is that <strong>for</strong> the mobile barrier<br />
with minimizing mass as the objective. The best choice is CFRP (carbon fiber rein<strong>for</strong>ced polymers);<br />
among metals, magnesium alloys offer the lightest solution . The lower selection is that <strong>for</strong> the static<br />
barrier with minimizing embodied energy as the objective. The selection is wood; among , metals cast<br />
irons offer the solution with lowest embodied energy.<br />
The exercise brings out how strongly the selection to minimize environmental burden depends on<br />
the application, as did the selection method used in the text.<br />
Selected materials: mobile barrier<br />
Carbon-fiber rein<strong>for</strong>ced polymers (CFRP)<br />
Magnesium alloys<br />
(Aluminum alloys)<br />
(Wood)<br />
Selected materials: static barrier<br />
Wood<br />
Cast irons<br />
(Carbon steels)<br />
E9.5. Complete the selection of materials <strong>for</strong> light, stiff shells of Section 9.4 by plotting the stiffness<br />
index<br />
ρ<br />
M 1 =<br />
E<br />
1/<br />
2<br />
onto a copy of the modulus-density chart of Figure 8.11. Reject ceramics and glass on the grounds of<br />
brittleness, and foams on the grounds that the shell would have to be very thick. Which materials do<br />
you find? Which of these would be practical <strong>for</strong> a real shell?<br />
Answer. The selection line with slope 2 (picked up from the guide line on the lower right) is position<br />
here such that CFRP, magnesium alloys, aluminium alloys and wood lie above it (ignoring the<br />
ceramics and foams). All are practical <strong>for</strong> shells. The earliest light-weight shells were made of<br />
plywood (some still are), aluminium sheet is routinely used <strong>for</strong> shell structures; magnesium sheet can<br />
be used in the same way but its has more limited ductility, making it harder to shape; finally CFRP is<br />
now the material of choice <strong>for</strong> light weight shells.<br />
Eco text: solution <strong>manual</strong> 55 MFA, 08/12/2008
The Modulus – Density chart: the one <strong>for</strong> stiffness at minimum weight.<br />
.<br />
E9.6. In a far-away land, fridges cost that same as they do here but electrical energy costs 10 times<br />
more than here – that is, it costs $2/kW.hr. Make a copy of the trade-off plot <strong>for</strong> fridges (Figure 9.8)<br />
and plot a new set of penalty lines onto it, using this value <strong>for</strong> the exchange constant, α e . If you had to<br />
choose just one fridge to use in this land, which would it be?<br />
Answer. We wish to minimize life-cost, which we take to be the sum of the initial cost and the cost<br />
of the energy used over life. So define the penalty function<br />
* * *<br />
Z = C f + αe<br />
H f t<br />
where * C f is the initial cost and<br />
*<br />
H f is the energy per year (both per cubic meter of cold space), α e ,<br />
the exchange constant, is the cost of energy per kW.hr and t is the service life of the fridge in years,<br />
making *<br />
Z the life-cost of the fridge per cubic meter of cold space. Inserting a life t of 10 years and<br />
an energy cost α e of $2/kW.hr gives<br />
Rearranging gives<br />
Z<br />
*<br />
= C<br />
*<br />
f +<br />
* 1<br />
H<br />
*<br />
f = Z −<br />
20<br />
20H<br />
1<br />
20<br />
* f<br />
C<br />
*<br />
f<br />
ρ<br />
E<br />
1/<br />
2<br />
Eco text: solution <strong>manual</strong> 56 MFA, 08/12/2008
So the penalty function plots as a set of lines of slope 1/20 on the trade of plot of<br />
They are shown in black on the figure.<br />
*<br />
H f against * C f .<br />
The choice that minimizes total life-cost when electrical power costs $2/kW.hr is that nearest the<br />
point where the new *<br />
Z contours are tangent to the trade-off surface. It is ringed in red: the Hotpoint<br />
RLA 175.<br />
The trade-off plot <strong>for</strong> fridges with contours of the new penalty function Z * .<br />
E9.7. You are asked to design a large heated work-space in a cold climate, making it as eco-friendly<br />
as possible by using straw-bale insulation. Straw, when compressed, has a density of 600 kg/m 3 , a<br />
specific heat capacity of 1670 J/kg.K and a thermal conductivity of 0.09 W/m.K. The space will be<br />
heated during the day (12 hours) but not at night. What is the optimum thickness of straw to minimize<br />
the energy loss?<br />
Answer. The required insulation wall-thickness, w, is given by equation 9.16 of the text:<br />
w<br />
1/2<br />
⎛ 2 t ⎞<br />
⎜ λ<br />
= ⎟<br />
⎜ C p ρ ⎟<br />
⎝ ⎠<br />
where λ is the thermal conductivity, C p is the specific heat of the wall per unit mass (so C pρ<br />
is the<br />
specific heat per unit volume) and t is the time interval between heating and cooling. Inserting the<br />
data given in the question (and remembering to convert the 12 hour cycle time into seconds) gives the<br />
result<br />
w = 0.09 meters = 9 centimeters.<br />
Contours of Z* <strong>for</strong> an<br />
exchange constant of<br />
$2/kW.hr<br />
Eco text: solution <strong>manual</strong> 57 MFA, 08/12/2008
E9.8. The makers of a small electric car wish to make bumpers out of a molded thermoplastic. Which<br />
index is the one to guide this selection if the aim is to maximize the range <strong>for</strong> a given battery storage<br />
capacity? Plot it on the appropriate chart from the set shown as Figures 8.11 – 8.14, and make a<br />
selection.<br />
Answer. The use-energy of any vehicle, be it gasoline or electric powered, increases with the mass of<br />
the vehicle. The bumper contributes to this mass. Maximizing range <strong>for</strong> a given battery capacity<br />
means minimizing mass. The function of the bumper is to sustain bending loads. The required index<br />
(equation 9.2 of the text) is<br />
M 2<br />
ρ<br />
=<br />
σ<br />
2 / 3<br />
y<br />
This is plotted on the strength – density chart below, picking up the slope of 1.5 from the guide line at<br />
the lower right. The selection line is position such that a few polymers remain above it – they are the<br />
best choice. They are:<br />
• Polycarbonate, PC<br />
• Polyamide (nylon) PA and<br />
• Polyetheretherketone PEEK<br />
In practice bumpers are made of blends of Polycarbonate with other polymers such as Polypropylene or<br />
Polyamide, or of fiber rein<strong>for</strong>ced thermosets such as Polyester. PEEK is too expensive <strong>for</strong> application<br />
such as this.<br />
The Strength – Density chart: the one <strong>for</strong> strength at minimum weight.<br />
.<br />
ρ<br />
2 / 3<br />
σ y<br />
Eco text: solution <strong>manual</strong> 58 MFA, 08/12/2008
E9.9. Car bumpers used to be made of steel. Most cars now have extruded aluminium or glassrein<strong>for</strong>ced<br />
polymer bumpers. Both materials have a much higher embodied energy than steel. Take<br />
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<br />
the equation <strong>for</strong> the energy consumption in MJ/km as a function of weight <strong>for</strong> petrol engine cars based<br />
on the data plotted in Figure 9.11 of the text.<br />
(a) Work out how much energy is saved by changing the bumper-set of a 1500 kg car from steel to<br />
aluminum, over an assumed life of 200,000 km<br />
(b) Calculate whether the switch from steel to aluminum has saved energy over life. You will find the<br />
embodied energies of steel and aluminum in the datasheets of Chapter 12. Ignore the differences<br />
in energy in manufacturing the two bumpers – it is small.<br />
(c) The switch from steel to aluminum increases the price of the car by $60. Using current pump<br />
prices <strong>for</strong> gasoline, work out whether, over the assumed life, it is cheaper to have the aluminum<br />
bumper or the steel one.<br />
Answer. (a) The energy in MJ/km, H km , as a function of vehicle mass m is listed in Table 9.6:<br />
H km<br />
= 3.<br />
7x10<br />
−3<br />
m<br />
0.<br />
93<br />
The marginal change in this energy resulting from a small change is mass when bumpers of one<br />
material are replace by those of another is found by differentiating this:<br />
−<br />
3.<br />
44x10<br />
3<br />
ΔH<br />
km =<br />
Δm<br />
m<br />
0.<br />
07<br />
Thus a weight saving of 6 kg made possible by the change from steel to aluminum bumpers on a car of<br />
mass 1500 kg saves, over 200,000 km, the energy 2,470 MJ.<br />
(b) The table below lists the embodied energies of steel and aluminum; they are the means of the<br />
ranges given in data sheets of Chapter 12. From these the material energy per bumper-set is calculated:<br />
the difference is 2,440 MJ. Thus the switch from steel to aluminum bumpers has saved energy, but not<br />
much. The difference between the material and the use energy over 200,000 <strong>for</strong> a vehicle of this mass<br />
is only 30 MJ.<br />
Material Bumper mass<br />
(kg)<br />
Embodied energy<br />
(MJ/kg)<br />
Material energy per<br />
bumper (MJ)<br />
Steel 20 32 640<br />
Aluminum 14 220 3080<br />
(c) One liter of gasoline provides 35 MJ of energy (see Table 6.5 of the text). Thus the change of<br />
material has saved the equivalent of one liter of fuel, and has cost $60. Fuel cost depends on country; it<br />
varies at present between $1/liter and $2/liter. Thus from an economic point of view,, and assuming<br />
the conditions of use described here (200,000 km on a vehicle of mass 1500 kg) the change of material<br />
is not justified on cost grounds.<br />
If the life-distance is increased to 400,000 km the picture changes. The energy of use then<br />
becomes 4,940 MJ, the difference between material and use energy becomes 2,500 MJ, equivalent to<br />
71 liters of gasoline, and then, even with cheap gasoline ($1 per liter) the change becomes<br />
economically attractive.<br />
Eco text: solution <strong>manual</strong> 59 MFA, 08/12/2008
Exercises using the CES Edu software.<br />
E9.10. Refine the selection <strong>for</strong> shells (Chapter 9, Section 9.4) using Level 3 of the CES software.<br />
Make a chart with the two shell indices<br />
M 1<br />
ρ<br />
= and<br />
E<br />
1/<br />
2<br />
M 2<br />
=<br />
ρ<br />
σ<br />
1/ 2<br />
y<br />
as axes, using the “Advanced” facility to make the combination of properties. Then add a “Tree” stage,<br />
selecting only metals, polymers and composites and natural materials. Which ones emerge as the best<br />
choice? Why?<br />
Answer. The figure below shows the CES Level 2 output. The materials that have the most attractive<br />
values of both indices are those at the bottom left. They are listed in the table below. All have low<br />
densities. Density ρ enters the two indicies with a power of 1; modulus E and strength σ y enter<br />
only as the square root. Thus density plays a larger role in determining the value of the index than the<br />
other two properties, which in part explains the selection.<br />
The interesting<br />
bit<br />
Selected materials<br />
Age-hardening wrought Al-alloys<br />
CFRP, epoxy matrix (isotropic)<br />
Softwood: pine, along grain<br />
Titanium alloys<br />
Wrought magnesium alloys<br />
Eco text: solution <strong>manual</strong> 60 MFA, 08/12/2008
E9.11. Tackle the crash barrier case study using CES Level 2 following the requirements set out in<br />
Table 9.3. Use a Limit stage to apply the constraints on fracture toughness K1c ≥ 18MPa.<br />
m and the<br />
requirement of recyclability. Then make a chart with density ρ on the x-axis and yield strength σ y<br />
on the y-axis and apply a selection line with the appropriate slope to represent the index <strong>for</strong> the mobile<br />
barrier:<br />
M 2<br />
ρ<br />
=<br />
2 / 3<br />
σ y<br />
List the best candidates. Then replace Level 2 with Level 3 data and explore what you find.<br />
Answer. The two figures below show the Level 2 and the Level 3 charts on exactly the same axes,<br />
and with a selection line with the appropriate slope (1.5) in the same position on both. Switching from<br />
Level 2 to Level 3 has populated the chart in much more detail. There are no polymer matrix<br />
composites on either plot – they are eliminated by the requirement of recyclability. The selection at<br />
Level 2 with the selection line in the position shown is<br />
Level 2 data<br />
Selected materials, Level 2<br />
Age-hardening wrought Al-alloys<br />
Titanium alloys<br />
Wrought magnesium alloys<br />
The same selection at Level 3 gives 76 alloys of magnesium, aluminum, titanium and steels.<br />
Displacing the line upwards to refine the search until just six are left gives a much more specific<br />
selection listed below.<br />
ρ<br />
2 / 3<br />
σ y<br />
1.5<br />
Eco text: solution <strong>manual</strong> 61 MFA, 08/12/2008
1.5<br />
Level<br />
Level<br />
3<br />
data<br />
data<br />
Selected materials, Level 3<br />
Titanium alpha-beta alloy, Ti-6Al-2Sn-2Zr-2Mo, <strong>Solution</strong> Treated and Aged<br />
Titanium alpha-beta alloy, Ti-6Al-2Sn-2Zr-2Mo, annealed<br />
Wrought aluminum alloy, 7055, T77511<br />
Wrought aluminum alloy, 7150, T61511<br />
Wrought magnesium alloy (EA55RS)<br />
Wrought magnesium alloy (ZC71)<br />
ρ<br />
2 / 3<br />
σ y<br />
Eco text: solution <strong>manual</strong> 62 MFA, 08/12/2008
E9.12. Repeat the procedure of Exercise E9.11, but this time make a chart using CES Level 2 on<br />
which the index <strong>for</strong> the static barrier<br />
H m ρ<br />
M 1 =<br />
2 / 3<br />
σ y<br />
can be plotted – you will need the Advanced facility to make the product H mρ<br />
. List what you find to<br />
be the best candidates. Then, as be<strong>for</strong>e, dump in Level 3 data and explore what you find.<br />
Answer. The two figures below show the Level 2 and the Level 3 charts on exactly the same axes,<br />
and with a selection line with the appropriate slope (1.5) in the same position on both. Switching from<br />
Level 2 to Level 3 has populated the chart in much more detail. As in the preceding exercise there are<br />
no polymer matrix composites on either plot – they are eliminated by the requirement of recyclability.<br />
The selection at Level 2 with the selection line in the position shown is<br />
Selected materials, Level 2<br />
Cast iron, ductile (nodular)<br />
Cast iron, gray<br />
High carbon steel<br />
Low alloy steel<br />
Low carbon steel<br />
Medium carbon steel<br />
The same selection at Level 3 gives 379 cast irons and steels, many of them sophisticated tool<br />
steels. Displacing the line upwards to refine the search until just six are left, <strong>for</strong> instance, gives a much<br />
more specific selection shown, below.<br />
Selected materials, Level 3<br />
Low-carbon mold tool steels, AISI P2<br />
Low-carbon mold tool steels, AISI P3<br />
Low-carbon mold tool steels, AISI P5<br />
Oil-hardening cold work tool steel, AISI O6<br />
Water-hardening tool steel, AISI W2 (Annealed)<br />
Water-hardening tool steel, AISI W5 (Annealed)<br />
These steels are expensive and an inappropriate choice <strong>for</strong> a low-grade application such as a static<br />
crash barrier. The problem is overcome by applying a further constraint, using a “Limit” stage,<br />
requiring that the material cost less than $1/kg. The result is shown in the third chart and table below.<br />
Eco text: solution <strong>manual</strong> 63 MFA, 08/12/2008
Level 2 data<br />
1.5<br />
Level 3 data<br />
1.5<br />
H m ρ<br />
σ<br />
2 / 3<br />
y<br />
H m ρ<br />
2 / 3<br />
σ y<br />
Eco text: solution <strong>manual</strong> 64 MFA, 08/12/2008
Selected materials, Level 3 with limit on price of £1/kg<br />
Carbon steel, AISI 1340 (tempered @ 205 C, oil quenched)<br />
Low alloy steel, AISI 9255 (tempered @ 205 C, oil quenched)<br />
Nodular graphite cast iron (BS 700/2, Hardened & Tempered)<br />
Nodular graphite cast iron (BS 900/2, Hardened & Tempered)<br />
Pearlitic malleable cast iron (BS grade P 70-02)<br />
Pearlitic malleable cast iron (<strong>for</strong>mer BS P 690/2-OQ)<br />
1.5<br />
H m ρ<br />
σ<br />
2 / 3<br />
y<br />
Level 3 data<br />
with price < $1/kg<br />
Eco text: solution <strong>manual</strong> 65 MFA, 08/12/2008
Chapter 10<br />
E10.1. Distinguish pollution control and prevention (PCP) from design <strong>for</strong> the environment (DFE).<br />
When would you use the first? When the second?<br />
Answer. Pollution control and prevention (PC and P) is intervention to mitigate environmental<br />
damage caused by existing products or processes without the burden of redesigning them. It is<br />
basically a clean-up or “end-or-pipe” measure. Taking transport as an example, it is the addition of<br />
catalytic converters to cars, a step to mitigate an identified problem with an existing product or system.<br />
Taking water-borne pollution of beaches as an example, it is extending the exit pipe to carry effluent<br />
further out to sea.<br />
Design <strong>for</strong> the environment (DFE) is the strategy to <strong>for</strong>esee and minimize the damaging effects of<br />
product families at the design stage, balancing them against the conflicting objectives of per<strong>for</strong>mance,<br />
reliability, quality and cost. Returning to the example of the car: it is to redesign the vehicle, giving<br />
emphasis to the objectives of minimizing emissions by reducing weight or adopting an alternative<br />
propulsion system – hybrid, perhaps, or electric. Using the example of water-borne waste again, it is<br />
redesigning the process generating the effluent so that, through filtration, chemical or biological<br />
treatment, the water leaving the plant is no longer polluted.<br />
E10.2. What is meant by the ecological metaphor? What does it suggest about ways to use materials<br />
in a sustainable way?<br />
Answer. The ecological metaphor, a concept emerging from industrial ecology, is based on the<br />
precept that we must see human activities as part of the global eco-system, and that there is, in some<br />
senses, a parallel between the functioning of the natural and the industrial systems. It has sometimes<br />
led to the idea is that a study of the processes and balances that have evolved in nature might suggest<br />
ways reconcile the imbalance between the industrial and the natural systems.<br />
E10.3 What are the potential sources of renewable energy? What are the positive and negative aspects<br />
of converting to an economy based wholly on renewable sources?<br />
Answer. There are five potential sources of renewable energy:.<br />
• Wind<br />
• Wave<br />
• Tidal<br />
• Geothermal<br />
• Solar, both as direct heat and, via photocells, as electricity<br />
It is these that offer the possibility of power generation without atmospheric carbon.<br />
Positive aspects of converting to renewable energy sources. All renewable energy sources generate<br />
electrical power without significant emissions. All offer independence from the international trade in<br />
fossil fuels and the uncertainty of price and supply that this involves. And all require a large number<br />
capture devices distributed over a large area of land or sea, making the system as a whole difficult to<br />
disrupt.<br />
Negative aspects of converting to renewable energy sources. Extracting significant power <strong>for</strong><br />
renewable resources involves committing large areas of land or sea because the energy is so dispersed.<br />
Establishing a renewable-energy based system requires very large capital investment but so, too, does<br />
the building of conventional power generating plant. The dispersion and reliance on (and thus lack of<br />
protection from) the <strong>for</strong>ces of nature creates maintenance problems.<br />
Eco text: solution <strong>manual</strong> 66 MFA, 08/12/2008
E10.4. The land area of the Netherlands (Holland) is 41,526 km 2 . Its population is 16.5 million and<br />
the average power consumed per capita there is 6.7 kW. If the average wind power is 2 W/m 2 of land<br />
area and wind turbines operate at a load factor of 0.5, what fractional of the area of the country would<br />
be taken up by turbines in order to meet the energy needs?<br />
Answer. The average power demand of the Netherlands is<br />
8<br />
P = Population x power per person = 1.<br />
11x10<br />
kW<br />
The wind power that can be generated from an area A km 2 is<br />
Pw<br />
=<br />
6<br />
A x10<br />
x 2 x0.<br />
5<br />
/ 1000<br />
=<br />
1000 A<br />
kW<br />
Here the factor of 10 6 is to convert km 2 to m 2 and the factor of 1000 is to convert W to kW. Equating<br />
this to the power demand gives<br />
A =<br />
1.<br />
11x10<br />
5<br />
km<br />
2<br />
This is 2.67 times the area of the Netherlands. There is no way wind power can supply all the<br />
country’s needs.<br />
E10.5. The land area of the New York State is 131,255 km 2 . Its population is 19.5 million and the<br />
average power consumed per capita there is 10.5 kW. If the average wind power is 2 W/m 2 of land<br />
area and wind turbines operate at a load factor of 0.25, what fractional of the area of the country would<br />
be taken up by turbines in order to meet the energy needs?<br />
Answer. The average power demand of the State of New York is<br />
8<br />
P = Population x power per person = 2.<br />
05x10<br />
kW<br />
The wind power that can be generated from an area A km 2 is<br />
6<br />
Pw<br />
= A x10<br />
x 2 x0.<br />
25 / 1000 =<br />
500 A<br />
kW<br />
Here the factor of 10 6 is to convert km 2 to m 2 and the factor of 1000 is to convert W to kW. Equating<br />
this to the power demand gives<br />
A =<br />
5 2<br />
14.<br />
1x10<br />
km<br />
This is 3.12 times the area of the State of New York. There is no way wind power from within the<br />
State can supply all the energy it needs. Even with a probably unrealistic load factor of 0.5, the area is<br />
insufficient.<br />
Eco text: solution <strong>manual</strong> 67 MFA, 08/12/2008
E10.6. The combined land area of the State of New Mexico (337,367 km 2 ) and Nevada (286,367 km 2 )<br />
is 623,734 km 2 . The population if the United States is 301 million and the average power consumed<br />
per capita there is 10.2 kW. Mass-produced solar cells can capture 10% of the energy that falls on<br />
them, which, in New Mexico, is roughly 50W/m 2 . Is the combined area of the two States large enough<br />
to provide solar power that would meet the current needs of the United States?<br />
Answer. The average power demand of the US is<br />
9<br />
P = Population x power per person = 3.<br />
07x10<br />
kW<br />
The solar power that can be generated from an area A km 2 is<br />
6<br />
Pw<br />
= A x10<br />
x 50 x0.<br />
1 / 1000 =<br />
5000 A<br />
kW<br />
Here the factor of 10 6 is to convert km 2 to m 2 and the factor of 1000 is to convert W to kW. Equating<br />
this to the power demand gives<br />
A =<br />
6.<br />
14<br />
5 2<br />
x10<br />
km<br />
This is 0.98 times the combined area of the two States. The power needs could just be met if both<br />
States were completely paved with solar cells.<br />
Eco text: solution <strong>manual</strong> 68 MFA, 08/12/2008
Chapter 11<br />
E11.1. An aluminum saucepan weighs 1.2 kg, and costs $10. The embodied energy of aluminum is 220<br />
MJ/kg. If the cost of industrial electric power doubles from 0.0125 $/MJ to 0.025 $/MJ, how much will it<br />
change the cost of the saucepan?<br />
Answer. The cost of the electric power at 0.0125 $/MJ to make the aluminum <strong>for</strong> one 1.2 kg saucepan is<br />
$ 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<br />
$ 13.3, an increase of 33%.<br />
E11.2. An MP3 player weighs 100 grams, and costs $120. The embodied energy of assembled integrated<br />
electronics of this sort is about 2000 MJ/kg. If the cost on industrial electric power doubles from 0.0125<br />
$/MJ to 0.025 $/MJ, how much will it change the cost of the MP3 player?<br />
Answer. The cost of the electric power at 0.0125 $/MJ to make one MP3 player is $2.5. If the price of<br />
energy doubles this rises to $ 5. The cost of the player will rise to $ 125, an increase of just 4%.<br />
E 11.3. Cars in Cuba are repaired and continue in use when 25 years old. The average life of a car in the<br />
US is 13 years. What is the underlying reason <strong>for</strong> this?<br />
Answer. In affluent economies such as those of the US and much of Europe consumers are sufficiently<br />
wealthy to purchase products as much <strong>for</strong> their character, associations and styling as <strong>for</strong> their functionality.<br />
New models often appear on an annual basis, stimulating replacement of previously purchased products<br />
be<strong>for</strong>e the end of their functional life. High labor costs make maintenance and repair uneconomic so that<br />
even minor breakdown causes rejection and replacement.<br />
In less affluent economies the capital cost of products represents a much larger fraction of disposable<br />
income. To avoid the need <strong>for</strong> more capital investment products are kept <strong>for</strong> their full functional life, which<br />
is extended by maintenance and repair enabled by low labor costs.<br />
E11.4. Use the Worldwide Web to research the meaning and history of “The precautionary principle”.<br />
Select and report the definition that, in your view, best sums up the meaning.<br />
Answer. Numerous sites comment on or criticize the precautionary principle, which has a long<br />
history. Among them are:<br />
1. en.wikipedia.org/wiki/Precautionary_principle<br />
2. www.jncc.gov.uk/page-1575<br />
3. www.sirc.org/articles/beware<br />
4. www.i-sis.org.uk/prec.php<br />
From these we learn the following.<br />
The precautionary principle states that when there is reasonable suspicion of harm, lack of scientific<br />
certainty or consensus must not be used to postpone preventative action. Originating in 1930s<br />
Eco text: solution <strong>manual</strong> 69 MFA, 08/12/2008
Germany as Vorsorgeprinzip (literally <strong>for</strong>esight principle), it has become one of the guiding principles<br />
in dealing with issues of loss of biodiversity, the deployment of genetically modified crops, the<br />
uncontrolled development of nanotechnology and, above all, to issues relating to climate change. Since<br />
the 1970s the principle has sometimes been invoked by green activists and those opposed to<br />
technological change, often with little scientific justification, thereby ensnaring what is, at bottom,<br />
sound common sense in an emotional net. Despite this it remains at the heart of discussions of mans’<br />
influence on the environment.<br />
E11.5. Use the Worldwide Web to research examples of problems that are approached by predictive<br />
modeling.<br />
Answer. A number of sites discuss and give examples of predictive modelling. Among them are:<br />
1. en.wikipedia.org/wiki/Predictive_modeling<br />
2. searchdatamanagement.techtarget.com<br />
3. www.dmreview.com/specialreports/20050215/1019956-1.html<br />
From these we learn the following.<br />
Predictive modeling is the use of mathematical models based on known or measured behavior to<br />
predict the behavior of systems under conditions <strong>for</strong> which measurements do not exist.<br />
Physically-based models are based on the principles of physics, often expressed as differential<br />
equations that can then be calibrated against known response and subsequently integrated with new<br />
boundary conditions to explore response under different circumstances. The design of a new aircraft,<br />
behavior of systems in space and the explosion of nuclear devices are examples; in all three examples<br />
response to service or use conditions cannot easily be measured because of the expense of doing so or<br />
because international treaties <strong>for</strong>bid it. The behavior must be predicted. Physically-based models<br />
enable this.<br />
Statistically-based models use predictive analytics to create a statistical model of future behavior.<br />
Predictive analytics is the area of data mining concerned with <strong>for</strong>ecasting probabilities and trends. A<br />
predictive model is made up of a number of predictors, variable factors that are likely to influence<br />
future behavior or results. In marketing, <strong>for</strong> example, a customer's gender, age, and purchase history<br />
might predict the likelihood of a future sale.<br />
To create a predictive model, data is collected <strong>for</strong> the relevant predictors, a statistical model is<br />
<strong>for</strong>mulated, predictions are made and the model is validated (or revised) as additional data becomes<br />
available. The model may employ a simple linear equation or a complex neural network, mapped out<br />
by sophisticated software. The Data Mining Group, a vendor consortium, uses a XML-based language<br />
called the Predictive Model Markup Language (PMML) to enable the definition and sharing of<br />
predictive models between applications.<br />
Eco text: solution <strong>manual</strong> 70 MFA, 08/12/2008