# Transport

Transport accounts for 32% of the energy we use and 34% of the emissions we generate (Figure 2.4). Cars contribute a large part of both. The primary eco-objective in car design is to provide transport at minimum environ­mental impact, which we will measure here by the CO2 rating in grams per kilometer (g/km). The audits of Chapter 7 confirmed what we already knew: that the energy consumed during the life phase of a car exceeds that of all the other phases put together. If we are going to reduce it, we first need to know how it depends on vehicle weight and propulsion system.

Energy, carbon, and cars. The fuel consumption and CO2 emission of cars increase with their weight. Figures 9.11 and 9.12 show the evidence for pet­rol, diesel, LPG, and hybrid-engine vehicles. The first shows energy plotted against mass on log scales, allowing the power-law fit for the energy con­sumption, Hkm, in MJ/km and carbon emission, CO2/km in g/km, as a func­tion of the vehicle mass, m, in kg, shown in the second column of Table 9.6. The second, Figure 9.12, shows the carbon rating in g/km as a func­tion of energy per km (MJ/km) on linear scales. The two are proportional, with the constants of proportionality marked on the figure. The CO2 rating (g/km) as a function of mass (kg) in the third column of the table is found by multiplying this by the energy/km in column 2.

From these we calculate the energy penalty associated with one kilogram of increased weight, evaluated here for a car of weight 1000 leg, by differentiating the expressions for Hm in Table 9.6. The results are listed in the last column of the table. We now have the inputs we need for modeling and selection.

But first, let us look at another aspect of car data by performing the selection that was set up in Chapter 8, in Section 8.2 and Figure 8.1: selec­tion of a car to meet constraints on power, fuel type, and number of doors and subject to two objectives, one environmental (CO2 rating), the other economic (cost of ownership). Figure 8.2 was a schematic of the tradeoff between the objectives. Figure 9.13a is real. It shows carbon rating and cost of ownership for 2600 cars plotted on the same axes as the schematic.2 We are in luck: the tradeoff line has a sharp nose; the cars with the lowest CO2 rating also have the lowest cost of ownership. The best combinations 2Data from What Car magazine (2005).

 Table 9.6 The energy and CO2 rating of cars as a function of their mass Fuel Type Energy per km: mass (Hkm in MJ/km, m in kg) CO2 per km: mass (COzfkm in g/km, m in kg) dHkm/dm MJ/km. kg (m = 1000 kg) Petrol power Hkm – 3.7 x 10~3 m0 93 CO2/km – 0.25 m0 93 2.1 x 10~3 Diesel power Hkm – 2.8 x 10~3 m0 93 CO2/km – 0.21 m0 93 1.6 x 10~3 LPG power Hkm – 3.7 x 10~3 m0 93 CO2/km – 0.17 m0 93 2.2 x 10~3 Hybrid power Hkm – 2.3 x 10~3 m0 93 CO2/km – 0.16 m0 93 1.3 x 10~3

are the cars that lie nearest to the tradeoff line, at the lower left of the plot. Not surprisingly, they are all very small.

That is the picture before the constraints are applied. If we now screen out all models that are not gas powered, have less than 150 hp, and fewer than four doors, the tradeoff plot looks like Figure 9.13b. Again, the trade­off line has a sharp nose. The cars that lie there have the best combination of cost and carbon and meet all the constraints; we don’t need a penalty function to find them. The Honda Civic 2.0 is a winner, but the Toyota Corolla 1.8 and the Renault Laguna 2.0 lie close. It is worth exploring these in more depth. What are the service intervals? How close is the nearest ser­vice center? What do consumer magazines say about them? It is this docu­mentation that allows a final choice to be reached.

Not all tradeoffs are so straightforward. Performance, to many car own­ers, is a matter of importance. For someone who wants to be responsible about carbon yet drive a high-performance vehicle, the compromise is more difficult: high performance means high carbon. Figure 9.14 shows the trad­eoff, using acceleration, measured by the time from 0 to 60 mph, as one objective (the shorter the time, the greater the performance) and carbon rating as the second. Now the tradeoff line is a broad curve. As before, the best choices are those that lie on or near this line; all others (and there are many) can be rejected. To get further, it is necessary to assign relative val­ues to performance and carbon rating. If the first is the most highly valued, it is the cars at the upper left that become the prime candidates. If it is carbon, it is those at the lower right. The compromise is a harsh one; any choice with low carbon has poor performance.

Modeling: where Does the Energy Go? Energy is dissipated in transport in three ways: as the energy needed to accelerate the vehicle up to its cruising

four doors, gas fuel, and 150+ hp have not yet been applied. Bottom: the same tradeoff after applying the constraints. The figure identifies the cars that meet the constraints and minimize the objectives.

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Time, 0-60 mph (s)

A tradeoff plot performance, measured by time to accelerate from 0 to 60 mph (100 km/hour), against CO2 rating for 1700 cars.

speed, giving it kinetic energy that is lost on braking; as drag exerted by the air or water through which it is passing; and as rolling friction in bear­ings and the contact between wheels and road or track. Imagine (following MacKay, 2008) that a vehicle with mass m accelerates to a cruising velocity v acquiring kinetic energy:

Eke = 2 mv2. (9.191

It continues over a distance d for a time d/v before stopping, losing this kinetic energy as heat in the brakes, thus dissipating energy per unit time (power) of

while cruising the vehicle drags behind it a column of air with a cross­section proportional to its frontal area A. The column created in time t has a volume cd Avt where cd is the drag coefficient, typically about 0.3 for a car, 0.4 for a bus or truck. This column has a mass paircd Avt and it moves with velocity v, so the kinetic energy imparted to the air is

Edrag = 1 mair v2 = 1 PairCdA v31 (9.21)

where pair is the density of air. The drag is the rate of change of this energy,

%- = 2PairCdA V3. I9-22!

If this cycle is repeated over and over again, the power dissipated is the sum of these two:

Rolling resistance adds another small term that is proportional to the mass, which we will ignore.

The first term in the brackets is proportional to the mass of the vehicle and the distance it moves between stops. The second depends only on the frontal area and the drag coefficient. Thus for short haul, stop-and-go, or city driving, the way to save fuel is to make the vehicle as light as possible. For long-haul, steady cruising, or motorway driving, mass is less important and minimizing drag (meaning frontal area A and drag coefficient cd) is key. The data for average energy consumption of Table 9.6 shows a near-linear dependence of energy per km on mass, meaning that, in normal use, it is the first term that dominates. Thus design to minimize the energy and CO2 of vehicle use must focus on material selection to minimize mass.

Selection Making cars lighter means replacing heavy steel and cast iron com­ponents by those made of lighter materials: light alloys based on aluminum, magnesium, or titanium and composites reinforced by glass or carbon fibers. All these materials have greater embodied energy per kg than steel, intro – ducing a new sort of tradeoff: that between the competing energy demands of different life phases. There is a net saving of energy and CO2 only if that saved by weight reduction exceeds that invested as extra embodied energy. Figure 9.15 illustrates the problem. It shows what happens when an existing material (which we take to be steel for this example) for a vehicle

FIGURE 9.15

The diagonal lines are contours of constant (negative) sum.

component is replaced with a substitute. The horizontal axis plots the change in embodied energy, AHemb. The vertical one plots the change in use energy, AHuse. The black circle at (0,0) is the steel; the other circles, enclosed by a tradeoff line, represent substitutes. The diagonal contours show the penalty function, which, in this instance, is particularly simple:

Z = AHemb + AHuse

The best choice of material is that with the lowest (most negative) value of Z. Any substitute that lies on the contour of Z = 0 passing through steel offers no reduction or increase in life energy. Those lying in the blue "No win" zone give an increase. Those in the white "Win" zone offer a saving. The best choice of all is that nearest the point that a penalty contour is tangent to the tradeoff line; it is indicated on the figure.

All very straightforward. Well, not quite. Material substitution in a com­ponent that performs a mechanical function requires that the component be rescaled to have the same stiffness or strength (whichever is design limit­ing) as the original. In making the substitution, the mass of the component changes both because the density of the new material differs from that of the old and because the scaling changes the volume of material that is used.

The scaling rules are known; they are given by material indices developed in Chapter 8. They allow the change in mass Am and in embodied energy AHemb of the component resulting from the change of material to be calculated. Multiplying Am by the energy per kg. km from Table 9.6 and the distance traveled over life, which we will take as 200,000 km, gives the change in use-phase energy AHuse resulting from substitution.

Consider, then, the replacement of the pressed steel bumper set of a car by one made of a lighter material. The function and weight scaling of the bumper were described earlier: it is a beam of given bending strength with a mass, for a given bending strength, that scales as p/a^73, where p is the density and ay is the yield strength of the material of which it is made. The weight change of the bumper set, on replacing one made of steel (subscript 0) with one made of a lighter material (subscript 1) , is thus: where B is a constant. If the mass mo of the steel bumper set is 20 kg, then

mo = B-pfi = 20kg (9.y5)

ay,0

defining B. Substituting this value for B into the previous equation gives

 2/3 p a y,0 – 1 2/3 Po a y, l

The property group in the square brackets is that for the steel of the original bumper; we will use data for an AISI 1022 rolled steel with 0.18% C 0.7% Mn, a yield strength of 295 MPa and a density of 7,900 kg/m3, giving the quantity in square brackets the value, based on these units, of

5.6 X 10-3. From Table 9.6, a gasoline-engine vehicle weighing 1000 kg requires 2.1 X 10-3 MJ/km. kg. Thus the change in use energy, AHuse, found by multiplying this by Am and the distance traveled over a life of 200,000 km, is

The change in embodied energy AHemb is found in a similar way. The embodied energy of the initial steel bumper set is:

Hemb = mHmo = CHmo^O – = 20 x 33MJ (9.28)

a y,0

where m = 20 kg is the mass and Hmo = 33 MJ/kg the embodied energy of the original steel bumper set. The constant C is defined by this equation. The change on switching to a new material is then:

h p H p

ah = C nm1p ___________

emb 2/3 2/3

. ay,1 ay,0

As before, the property group in the square brackets is that for the steel of the original bumper. Its value, using steel data given previously, is

1.7 x 10 _4, giving the final expression for change in embodied energy:

The tradeoff between AHuse and AHemb is plotted in Figure 9.16 . It explores the total energy saving (or lack of it) when the carbon steel bum­per set is replaced by one made of a low alloy steel, a 6000 series aluminum alloy, a wrought magnesium alloy, or a composite. Steel lies at the origin (0,0). The other bubbles, labeled, show the changes in AHuse and AHemb calculated from Equations 9.27 and 9.30. The "No win" area is shaded. For titanium alloys, the energy saving is slight. For magnesium and aluminum alloys, it is considerable. The greatest saving is made possible using com­posites. The tradeoff line is sketched in. The penalty contours show the total energy saved over 200,000 km. The contour that is tangent to the trad­eoff line has (by interpolation) a value of about 7 GJ. It identifies the best choice: here, the glass-epoxy laminate.

Before a final decision is reached, it is helpful to have a feeling for the likely change in cost. The change in use cost ACuse is the use energy change AHuse (Equation 9.27) multiplied by the price of energy; gasoline at \$0.8/liter (\$3 per U. S. gallon) gives an energy price of approximately 0.025\$/MJ. Thus:

 ACuse = 210 5.6 x 10_3 P1 2/3 _ 1 a y,1

The change in material cost ACmat might seem (in parallel with that of embodied energy) to be the change in the product of component mass m and the price per unit mass Cm, and we shall use this:

Taking the cost Cmo of steel to be \$0.8/kg makes the value of the steel property group in the square brackets equal to 7.0 x 10-3 and the equation becomes:

But this ignores manufacture. Cost has other contributions; at least half the cost of a component is that of manufacturing. The dominant feature in this is time. If forming, joining, and finishing with the new material are slower than with the old, there is an additional cost penalty. So the mate­rial cost of Equation 9.13 must be regarded as approximate only.

With this simplifying approximation, the cost tradeoff (following the pattern used for energy) appears as in Figure 9.17. Titanium alloys lie well inside the "No win" zone. Low alloy steel, aluminum alloys, and magne­sium alloys are the most attractive from an economic point of view, though epoxy glass is almost as good. The most striking result is that the total sum saved is so small.

Postscript. We have calculated the primary mass saving that material sub­stitution brings. In reality the saving is—or can be—even larger because the lighter vehicle can get away with less heavy suspension, lighter tires, and less powerful brakes, allowing a secondary weight saving. In current prac­tice, however, aluminum cars are not much lighter than the steel ones they replace, because manufacturers tend to load them with more extras (such as air conditioning as standard), adding mass back on.

9.6 Summary and conclusion

Rational selection of materials to meet environmental objectives starts by identifying the phase of product life that causes greatest concern: production, manufacture, transport, use, or disposal. Dealing with all these requires data not only for the obvious eco-attributes (energy, emissions, toxicity,
ability to be recycled, and the like) but also data for mechanical, thermal, electrical, and chemical properties. Thus if material production is the phase of concern, selection is based on minimizing embodied energy or the associated emissions (CO2 release, for example). But if it is the use phase that is of concern, selection is based instead on light weight or excellence as a thermal insulator or electrical conductor. These define the objective; the idea is to minimize this while meeting all the other constraints of the design: adequate stiffness, strength, durability, and the like.

Almost always there is more than one objective, and almost always they conflict. They arise in more than one way. One is the obvious conflict between eco-objectives and cost, illustrated both in Chapter 8 and here, for which trade­off methods offer a way forward, provided a value can be assigned to the eco­objective, something that is not always easy. Another is the conflict between the energy demands and emissions of different phases of life: the conflict between increased embodied energy of material and reduced energy of use, for example. Tradeoff methods work particularly well for this type of problem because both energies can be quantified. The case studies of this chapter illustrate how such problems are tackled. The exercises of Section 9.9 present more.