\marginfig{
\begin{center}
\begin{tabular}{@{}c@{}}
\lowres%
{\mbox{\epsfxsize=53mm\epsfbox{../../images/Peugeot_206_WRCS.jpg.eps}}}%
{\mbox{\epsfxsize=53mm\epsfbox{../../images/Peugeot_206_WRC.jpg.eps}}}
\\
\end{tabular}
\end{center}
\caption[a]{A Peugot 206 has a \ind{drag coefficient} of 0.33. Photo by Christopher Batt.
}\label{Peugot206}
}
We estimated that a \ind{car}\index{transport} driven
100\,\km\ uses about 80\,\kWh\ of energy.
Where does this energy go? How does it depend on
properties of the car?
Could we make cars that are 100 times more efficient?
\marginfignocaption{
\fbox{
\begin{minipage}{50mm}
The key formula for most of the calculations in this book is:
\[
\mbox{kinetic energy} = \frac{1}{2} m v^2 .
\]
For example, a car of mass $m = 1000\,\kg$
moving at 100\,\km\ per hour or $v = 28$\,m/s
%% 27.777 etres per second
has an energy of
\[
\frac{1}{2} m v^2 \simeq 390\,000 \,\J \simeq 0.1 \,\kWh .
\]
\end{minipage}
}
}%
\marginfig{
% \begin{figure} \figuremargin{
\begin{center}
\mbox{\mono{\epsfbox{metapost/sign.101}}{\epsfbox{metapost/sign.1}}}
\end{center}
%}{
\caption[a]{Our cartoon:\index{cartoon!car}
a car moves at speed $v$ between stops separated
by a distance $d$.}
}%
% \end{figure}
Let's make a simple cartoon of car-driving, to
describe where the energy goes.
The energy in a typical fossil-fuel car goes to four main destinations, all
of which we will explore:
\ben
\item
speeding up then slowing down using the brakes;
\item
air resistance;
\item
rolling resistance;
\item
heat -- 75\% of the energy is thrown away as heat, because the
energy-conversion chain is inefficient.
% to the radiator
\een
Initially our cartoon will ignore rolling resistance; we'll
add in this effect later in the chapter.
Assume the driver accelerates rapidly up to
a cruising speed $v$, and maintains that speed
for a distance $d$, which is the distance between traffic lights,
stop signs, or congestion events.
%% incidents.
At this point, he slams on the brakes and
turns all his kinetic energy into heat in the brakes.
(This vehicle doesn't have fancy regenerative braking.)
Once he's able to move again, he accelerates back up
to his cruising speed, $v$.
This acceleration gives the car kinetic energy; braking
throws that \ind{kinetic energy} away.
Energy goes not only into the brakes:
while the car is moving, it makes air swirl around.
A car leaves behind it a tube of swirling
air, moving at a speed similar to $v$.
Which of these two forms of energy is
the bigger: kinetic energy of
the swirling air, or heat in the brakes?
Let's work it out.
\begin{itemize}
\item
The car speeds up and slows down once in each duration
$d/v$.
The rate at which energy pours into the brakes is:\index{brakes}
\begin{equation}
\frac{\mbox{kinetic energy}}{\mbox{time between braking events} }
=
\frac{ \frac{1}{2} m_{\rm c} v^2 }{ d/v } =
\frac{ \frac{1}{2} m_{\rm c} v^3 }{ d } ,
\end{equation}
where $m_{\rm c}$ is the mass of the car.
\begin{figure}
\figuremargin{
\begin{center}
\mbox{\epsfbox{crosspad/cars4.ps}}
\end{center}
}{
\caption[a]{A car moving at speed $v$ creates
behind it a tube of swirling air;
the cross-sectional
area of the tube is similar to the
frontal area of the car, and the speed at which air in the tube
swirls is roughly $v$.}
}
\end{figure}
\item
The tube of air created in a time $t$ has a volume $A v t$,
where $A$ is the cross-sectional area of the tube, which
is similar to the area of the front view of the car.
(For a \ind{streamlined} \ind{car},
$A$ is usually a little smaller than the \ind{frontal area}
$A_{\rm car}$, and the ratio
% $\cd$
of the tube's effective cross-sectional area to the
car area is called the \ind{drag coefficient}
$\cd$. Throughout the following equations,
$A$ means the effective area of the
car, $\cd A_{\rm car}$.)%
\marginpar{
\fbox{\small
\begin{minipage}{50mm}
I'm using this formula:
\[
\mbox{mass} = \mbox{density} \times \mbox{volume}
\]
The symbol $\rho$ (Greek letter `rho') denotes the density.
\end{minipage}
}
\label{miniformula}
}
The tube has mass $m_{\rm air} = \rho A v t$
(where $\rho$ is the density of air) and
swirls at speed $v$, so its \ind{kinetic energy} is:
\[
\frac{1}{2} m_{\rm air} v^2 =
\frac{1}{2} \rho
%c_{\rm d}
A v t \, v^2,
\]
and the rate of generation of kinetic energy in \ind{swirling air} is:
\[
\frac{ \frac{1}{2} \rho
% c_{\rm d}
A v t v^2}{t} =
\frac{1}{2} \rho
% c_{\rm d}
A v^3 .
\]
\label{pageAvcubed}
% This is very similar to a formula that we had for windmills!
% (Page \pageref{eq.windpower}.)
\end{itemize}
% SPACE TOO BIG ***
So the total rate of energy production by the car is:
\beq
\begin{array}{rrcl}
& \mbox{power going into brakes} & + &
\mbox{power going into swirling air} \\
&
= \frac{1}{2} m_{\rm c} v^3 /d
& + & \frac{1}{2} \rho A v^3 .
\end{array}
\label{eq.totpow}
\eeq
Both forms of energy dissipation scale as $v^3$.
So this cartoon predicts that a driver who halves his speed $v$ makes
his power consumption $8$ times smaller.\index{energy saving!transport}
If he ends up driving the same total distance, his journey
will take twice as long, but the total energy consumed by
his journey will be four times smaller.
Which of the two forms of energy dissipation -- brakes or air-swirling --
is the bigger?
It depends on the ratio of
\[
(m_{\rm c}/d) \left/ (\rho A) \right. .
\]
If this ratio is much bigger than 1, then more power is
going into brakes; if it is smaller,
more power is going into swirling air.%
\marginfig{
% \begin{figure} \figuremargin{ STOP sign
\begin{center}
\mbox{\mono{\epsfbox{metapost/sign.102}}{\epsfbox{metapost/sign.2}}}
\end{center}
%}{
\caption[a]{To know whether energy consumption is braking-dominated\index{brakes}
or air-swirling-dominated, we compare the mass of the car with
the mass of the \ind{tube of air} between \ind{stop-sign}s.}\label{fig.tubesize}
}% \end{figure}
\marginfig{
%\begin{figure}
%\figuremargin{
\begin{center}
\begin{tabular}{@{}c@{}}
% {\mbox{\epsfxsize=53mm\epsfbox{../../images/cam/carsFordSmart2.eps}}} \\
%{\mbox{\epsfxsize=43mm\epsfbox{../../images/cam/carsTankMini.eps}}} \\
\lowres{\mbox{\epsfxsize=53mm\epsfbox{../../images/cam/carsVWHondaS.jpg.eps}}}%
{\mbox{\epsfxsize=53mm\epsfbox{../../images/cam/carsVWHonda.eps}}} \\
\end{tabular}
\end{center}
%}{
\caption[a]{Power consumed by a car
is proportional to its cross-sectional area,
during \ind{motorway driving}, and to its mass, during
town driving. Guess which gets better mileage --
the \index{Volkswagen}VW on the left, or the \ind{spaceship}?
}
}
% \end{figure}
Rearranging this ratio,
it is bigger than 1 if
\[
m_{\rm c} > \rho A d .
\]
Now, $A d$ is the volume of the tube of air swept out from
one stop sign to the next. And $\rho A d$ is the mass of that
tube of air. So we have a very simple situation:
energy dissipation is dominated by kinetic-energy-being-dumped-into-the-brakes if the mass of the car is {\em{bigger}\/}
than the mass
of the tube of air from one stop sign to the next;
and energy dissipation is dominated by making-air-swirl
if the mass of the car is {\em smaller} (\figref{fig.tubesize}).
% that tube of air is bigger than the car's mass.
Let's work out the special distance $d^*$ between stop signs,
below which the dissipation is braking-dominated and above which
it is \ind{air-swirling} dominated (also known as \ind{drag}-dominated).
If the frontal area of the car is:
\[
A_{\rm car} = 2\,\m\:\mbox{wide} \times 1.5\,\m\:\mbox{high}
= 3 \, \m^2
\]
and the \ind{drag coefficient} is $\cd = 1/3$
and the mass is $m_{\rm c} = 1000 \,\kg$
then the special distance is:
%% given by
\[
d^* = \frac{m_{\rm c} }{ \rho \cd A_{\rm car}}
= \frac{ 1000 \,\kg }{1.3\,\kg/ \m^3 \times \frac{1}{3} \times 3 \,\m^2} = 750 \,\m .
\]
So ``city-driving'' is dominated by kinetic energy and braking\index{city driving}\index{driving!city driving}
if the distance between stops is less than 750\,m.
Under these conditions, it's a good idea, if you want to save
energy:
\begin{enumerate}
\item
to reduce the mass of your car;
\item
to get a car with \ind{regenerative brakes}\index{brakes!regenerative} (which roughly
halve the energy lost in
braking -- see \chref{ch.transport}); and
\label{pRegenEff}
\item to drive more slowly.
\end{enumerate}
When the stops are significantly more than 750\,m\ apart, energy dissipation
is drag-dominated.\index{driving!motorway driving}
Under these conditions, it doesn't much matter what your
car weighs.
% how much mass you are transporting in the car.
Energy dissipation will
be much the same whether the car contains one person or six.
% It's all about area and speed.
Energy dissipation can be reduced:
\begin{enumerate}
\item
by reducing the car's \ind{drag coefficient};
\item
by reducing its cross-sectional area; or
\item by driving more slowly.
\end{enumerate}
% (Should I include a drag coefficient? Typical values are $1/3$ or $1/2$.)
The actual energy consumption of the car will be the energy dissipation
in \eqref{eq.totpow}, cranked up by a factor related to the
inefficiency of the engine and the transmission.
Typical petrol engines are about 25\% efficient,\label{pCarEng25}
%%% GET REFERENCE
so of the chemical energy that a car guzzles,\index{engine!exhaust}\index{car!engine efficiency}\index{car!radiator}
three quarters
is wasted in making the car's \ind{exhaust}, \ind{engine} and \ind{radiator} hot, and just one quarter
goes into ``useful'' energy:
\margintab{ \small
% \begin{figure}
\begin{center}
\begin{tabular}{c@{\,$\leftrightarrow$\,}c} \toprule
\multicolumn{2}{c}{\sc Energy-per-distance }\\ \midrule
\begin{tabular}{@{}c@{}}Car \\at 110\,km/h\\
\end{tabular}
& 80\,kWh/(100 km) \\
\begin{tabular}{@{}c@{}}
\index{bicycle}Bicycle \\ at 21\,km/h\\
\end{tabular}
& ${2.4}$\,kWh/(100 km) \\
\bottomrule
\end{tabular}
\medskip
\medskip
\begin{tabular}{ll} \toprule
\multicolumn{2}{c}{{\sc{Planes at 900 km/h }}} \\ \midrule
\ind{A380}\index{airbus A380} & 27\,kWh/100 seat-km\\ \bottomrule
\end{tabular}
\end{center}
% }{
\caption[a]{Facts worth remembering:
car energy consumption.
% (In an earlier chapter the car was
% doing 0.7\,kWh/km, but that was at lower speed.)
}
}
\[
\mbox{total power of car} \simeq 4 \left[
\frac{1}{2} m_{\rm c} v^3 /d
+ \frac{1}{2} \rho A v^3
\right] .
\]
Let's check this theory of cars by
plugging in plausible numbers for
\uk{motorway}{freeway} driving.
Let $v= 70\,\miles \, \per \, \hour = 110 \,\km/\h = 31\,\m/\s$
and
% assume the drag coefficient $\cd=1$, so that $A=\cd A_{\rm car}$ is the
% frontal area of the car.
$A=\cd A_{\rm car} = 1\,$m$^2$.
The power consumed by the engine is estimated to be roughly\label{pageDragCar}
\[
4 \times \frac{1}{2} \rho A v^3 = 2 \times {1.3\,\kg/ \m^3 \times 1 \,\m^2}
\times {( 31\,\m/\s)^3}
= 80 \,\kW.
\label{Poweris80}
\]
If you drive the car at this speed for one hour every day, then you
travel 110\,\km\ and use \Red{80\,kWh} of energy per day. If you drove
at half this speed for two hours per day instead, you would travel
the same distance and use
up \Red{20\,\kWh} of energy. This simple theory seems consistent
with the mileage figures for cars quoted in \chref{ch.car}.
% the
% previous chapter (73\,kWh/day for travelling 100\,km at an
% unspecified typical mix of speeds).
Moreover, the theory gives insight into how
the energy consumed by your car could be reduced.
The theory has a couple of flaws which we'll explore in a moment.
Could we make a new car that consumes 100 times less energy
and still goes at 70\,mph?
{\bf{No}}. Not if the car has the same shape.
On the \uk{motorway}{freeway} at 70\,mph,
the energy is going mainly into making air swirl.
Changing the materials the car is made from makes no difference
to that. A miraculous improvement to the fossil-fuel engine
could perhaps boost its efficiency from
25\% to 50\%, bringing the energy consumption of a fossil-fuelled
car down to
roughly 40\,kWh per 100\,km.
% (Could put electric vehicle information here.)
%% see action.tex
Electric vehicles have some wins:
while the weight of the energy store, per useful kWh stored,
is about 25 times bigger than that of petrol, the weight of an electric engine
can be about 8 times smaller.\nlabel{pElecEnginPowRat} And the energy-chain in an electric car is much more
efficient: electric motors can be 90\% efficient.
{% begin troublesomepage hack
% this should be *before* the start of troublesome page
\renewcommand{\floatpagefraction}{0.8}
We'll come back to electric cars in more detail towards the end of this
chapter.
%
\margintab{
\begin{center}
\begin{tabular}{lc}\toprule
\multicolumn{2}{c}{ \sc Drag coefficients } \\ \midrule
\multicolumn{1}{c}{\sc Cars}\\
% Typical drag coefficient: $\cd=0.3$ for most cars where effort has
% been taken to do a little streamlining.
% For example Renault~25 &% , very ordinary car, has 0.28
Honda Insight& 0.25\\
Prius& 0.26\\
Renault~25 & 0.28\\
Honda Civic (2006)& 0.31\\
VW Polo GTi & 0.32\\%% frontal area 2.04m**2 -- C_x = 0.65
Peugeot 206 & 0.33\\
Ford Sierra & 0.34\\
Audi TT& 0.35\\
Honda Civic (2001)& 0.36\\
Citro\"en 2CV& 0.51 \\
% Audi A2& 0.25--0.28\\
\midrule
\multicolumn{1}{l}{Cyclist}& 0.9 \\ % $\cd=0.9$
\midrule
\multicolumn{1}{l}{Long-distance coach} & 0.425 \\
\midrule
\multicolumn{1}{c}{\sc Planes}\\
Cessna & 0.027 \\ % $\cd=0.027$; learjet $\cd=0.022$;
Learjet & 0.022 \\
Boeing 747 & 0.031 \\ % $\cd=0.031$.
\bottomrule
\end{tabular}
\medskip
\medskip
\begin{tabular}{ll}\toprule
%Typical full-size passenger cars have a drag-area of roughly
\multicolumn{2}{c}{ {\sc Drag-areas} (m$^2$)} \\ \midrule
Land Rover Discovery &
% 17.4 sq ft
1.6\\%\,m$^2$ \\
%% in ft**2
Volvo 740 & 0.81 \\ %% 8.70 1990 Volvo 740 Turbo
{\bf Typical car} & {\bf 0.8} \\
Honda Civic & 0.68\\ %% 7.34 2001 Honda Civic
%% 7.02 1992 BMW 325I
VW Polo GTi & 0.65\\%% frontal area 2.04m**2 -- C_x = 0.6528
%% http://www.carfolio.com/specifications/models/car/?car=143110
Honda Insight & 0.47\\% m$^2$ \\
\bottomrule
\end{tabular}
\end{center}
\caption[a]{Drag coefficients and drag areas.}
}
\subsection{Bicycles and the scaling trick}
Here's a fun question:
what's the energy consumption of a bicycle, in kWh per 100\,km?
Pushing yourself along on a bicycle requires energy for the same
reason as a car: you're making air swirl around.
Now, we could do all the calculations from scratch, replacing
car-numbers by bike-numbers.
But there's a simple trick we can use to get
the answer for the bike from the answer for the car.
The energy consumed by a car, per distance travelled, is
the power-consumption associated with air-swirling,
\[
4 \times \frac{1}{2} \rho A v^3,
\]
divided by the speed, $v$; that is,
\[
\mbox{ energy per distance } =
4 \times \frac{1}{2} \rho A v^2 .
\]
% [Another name for `energy per distance' is 'force'.
The ``$4$'' came from engine inefficiency;
$\rho$ is the density of air;
the area $A = \cd A_{\rm car}$ is the effective frontal area of a car;
and $v$ is its speed.
Now, we can compare a bicycle
with a car by dividing $ 4 \times \frac{1}{2} \rho A v^2$ for the bicycle
by
$4 \times \frac{1}{2} \rho A v^2 $ for the car.
All the fractions and $\rho$s cancel, if
the efficiency of the carbon-powered
bicyclist's engine is similar
to the efficiency of the carbon-powered
car engine (which it is).
% -- that'd be a fun thing to double-check later, but let's assume it is so.
The ratio is:
\[
\frac{\mbox{energy per distance of bike}}
{\mbox{energy per distance of car}}
=
\frac{ \cd^{\rm bike} A_{\rm{bike}} v^2_{\rm{bike}} }
{ \cd^{\rm car} A_{\rm{car}} v^2_{\rm{car}} } .
\]
The trick we are using is called ``scaling.'' If we know how
energy consumption scales with speed and area, then
we can predict energy consumption of objects with completely
different speeds and areas.
Specifically, let's assume that the area ratio is
\[
\frac{ A_{\rm{bike}} }
{ A_{\rm{car}} } = \frac{1}{4} .
\]
(Four cyclists can sit shoulder to shoulder in the width of one car.)
Let's assume the bike is not very well streamlined:
\[
\frac{ \cd^{\rm bike}}
{ \cd^{\rm car}} = \frac{1}{1/3}
\]
And let's assume the speed of the bike is 21\,km/h (13 miles per hour),
so
\[
\frac{ v_{\rm{bike}} }
{ v_{\rm{car}} } = \frac{1}{5} .
\]
Then
\beqa
\frac{\mbox{energy-per-distance of bike}}
{\mbox{energy-per-distance of car}}
&=&
\left(
\frac{ \cd^{\rm bike}}
{ \cd^{\rm car}}
\frac{ A_{\rm{bike}} }
{ A_{\rm{car}} }
\right)
\left( \frac{ v_{\rm{bike}} }
{ v_{\rm{car}} } \right)^2
\\
&=&
\left( \frac{3}{4} \right) \times
\left( \frac{1}{5} \right)^2\\
&=&
\frac{3}{100}
\eeqa
So a cyclist at 21\,km/h consumes about 3\% of the energy per kilometre of
a lone car-driver on the motorway -- about {\bf{2.4\,kWh per 100\,km}}.
% (Put this fact into the main text too.)
If you would like a vehicle whose fuel efficiency is
30 times better than a car's, it's simple: ride a
% well-streamlined
bike.
\begin{table}[!b]
%% bikes
%% http://www.recumbents.com/mars/pages/proj/tetz/other/Crr.html
%% trains and cars
%% http://www.lafn.org/~dave/trans/energy/rail_vs_auto_EE.html#s5
%% typical car tyre: C = 0.010
%% typical truck tyre: C = 0.007
%% rail : C = 0.002
\figuremarginwidecap{%%% sizes determined in cft.sty specialwidtha,b
\begin{tabular}{ll} \toprule
wheel & $C_{\rm{rr}}$ \\
\midrule
train\index{rail}\index{train} (steel on steel)&0.002 \\
\ind{bicycle} \ind{tyre}&0.005 \\
truck rubber tyres &0.007 \\%on smooth road \\
car \ind{rubber} tyres &0.010 \\%on smooth road \\
%% see _cars.tex for notes
\bottomrule
\end{tabular}}{
\caption[a]{% Rolling resistance --
The \ind{rolling resistance}\index{data!rolling resistance}
is equal to the weight multiplied
by the coefficient of rolling resistance, $C_{\rm{rr}}$.
The rolling resistance includes the force due to
wheel flex, friction losses in the wheel bearings, shaking and
vibration of both the roadbed and the vehicle (including energy
absorbed by the vehicle's shock absorbers), and sliding of the
wheels on the road or rail.
The coefficient varies with the quality of the road,
with the material the wheel is made from, and with temperature.
The numbers given here assume smooth roads.
\tinyurl{2bhu35}{http://www.lafn.org/~dave/trans/energy/rail_vs_auto_EE.html}
}
\label{tab.Crr}
}
\end{table}
%
\begin{myfloat}
\begin{center}{\small
\threefigures{
\begin{center}
\mbox{\rotatebox{90}{\footnotesize\sf Energy consumption (kWh/100\,km)}\hspace*{8mm}%
{\epsfxsize=40mm\epsfbox{../data/carsTheory.eps}}%
}%
\\[0.2in]
{\sf{speed (km/h)}}\\
\end{center}
}{
\caption[a]{Simple theory of car
fuel consumption (energy per distance)
when driving at steady speed.
% Horizontal axis is speed in km/h.
% Vertical axis is fuel consumption in kWh per 100\,km.
Assumptions: the car's engine uses energy with
an efficiency of 0.25, whatever the
speed; $\cd A_{\rm car} = 1$\,m$^2$;
$m_{\rm car} = 1000\,\kg$; and $C_{\rm{rr}}=0.01$.
% the air resistance has an energy demand that grows
% as the square of the speed,
% and the rolling resistance makes a constant
% energy demand, whatever the speed.
} \label{fig.carsTheory}
}
%% figure 2
{
\begin{center}
\mbox{\ \ \epsfxsize=40mm\mono%
{\epsfbox{../data/mono/bikeTheory.eps}}%
{\epsfbox{../data/bikeTheory.eps}}%
}\\[0.2in]
{\sf{speed (km/h)}}\\
\end{center}
\label{fig.bikeTheory}
}{
\caption[a]{
Simple theory of bike\index{transport!efficiency!bicycle}
fuel consumption (energy per distance).\index{cartoon!bicycle}
% Horizontal axis is speed in km/h.
Vertical axis is energy consumption in
kWh per 100\,km.
Assumptions: the bike's engine (that's you!)
uses energy with an efficiency of 0.25,\nlabel{pBikeCrr};
% source \cite{Prampero}
% whatever the speed;
the drag-area of the cyclist is 0.75\,m$^2$;
the cyclist+bike's mass is
$90\,\kg$; and
$C_{\rm{rr}}=0.005$.
}
}
% figure 3
{
\begin{center}
\mbox{\rotatebox{90}{\footnotesize\sf Energy consumption (kWh/100\,pkm)}\hspace*{8mm}%
{\epsfxsize=40mm\epsfbox{../data/trainTheory8.eps}}%
}%
\\[0.2in]
{\sf{speed (km/h)}}\\
\end{center}
}{
\caption[a]{Simple theory of train\index{rail}\index{train}
energy consumption, {\em{per passenger}},
for an eight-carriage train carrying 584 passengers.
% Horizontal axis is speed in km/h.
Vertical axis is energy consumption in kWh per 100\,\pkm.
Assumptions: the train's engine uses energy with
an efficiency of 0.90;
% the drag-area of the train is
$\cd A_{\rm train} = 11$\,m$^2$;
$m_{\rm train}=400\,000\,\kg$; and $C_{\rm{rr}}=0.002$.
} \label{fig.trainTheory8}
}
}
\end{center}
\end{myfloat}
%\section{Queries}
\subsection{What about rolling resistance?}
Some things we've completely ignored so far are
the energy consumed in the tyres and bearings of the car,
the energy that goes into the noise of wheels against
asphalt, the energy that goes into grinding
rubber off the tyres, and the energy that vehicles put into
shaking the ground.
Collectively, these forms of energy consumption
are called {\em\ind{rolling resistance}}. The standard model
of rolling resistance
asserts that the force of rolling resistance
is simply proportional to the weight of the vehicle,
independent of the speed.
The constant of proportionality is called the
coefficient of rolling resistance, $C_{\rm{rr}}$.
\Tabref{tab.Crr} gives some typical values.
The coefficient of rolling resistance for a car is about 0.01.
The effect of rolling resistance is just like perpetually
driving up a hill with a slope of one in a hundred.
% If set moving, a car will roll at steady speed
% down a hill with a slope of one in a hundred.
So rolling friction is about 100 newtons per \tonne, independent of
speed. You can confirm this by pushing a typical one-\tonne\ car
along a flat road. Once you've got it moving, you'll find you can
keep it moving with one hand. (100 newtons is the weight of 100
apples.)\index{newton}\index{apple}\index{weight}
So at a speed of
% 25\,m/s (55\,mph), the power required to overcome
31\,m/s (70\,mph), the power required to overcome
rolling resistance, for a one-\tonne\ vehicle, is
\[
% (\mbox{10\,Newtons}) \times (25\,\m/\s) = 250\,\W ;
\mbox{force} \times \mbox{velocity} \:=\:
(\mbox{100 newtons}) \times (31\,\m/\s) \:=\: 3100\,\W ;
\]
which, allowing for an engine efficiency of 25\%,
requires 12\,kW of power to go into the engine;
whereas the power required to overcome drag was estimated
on \pref{Poweris80}
% {pageDragCar}
to be 80\,\kW\@.
So, at high speed, about 15\% of the power is required for
rolling resistance.
\Figref{fig.carsTheory} shows the theory of
fuel consumption (energy per unit distance)
as a function of steady speed, when we add together the
air resistance and rolling resistance.
The speed at which a car's rolling resistance is equal to
its air resistance is
given by
\[
C_{\rm{rr}} m_{\rm c} g = \frac{1}{2} \rho \cd A v^2 ,
\]
that is,
\[
v = \sqrt{
2 \frac{C_{\rm{rr}} m_{\rm c} g}{ \rho \cd A }
} =
% corrected error feb 2010 !!!!!!!
% sqrt( 2 * 0.01 * 1e3 * 9.81 / 1.3 / 1.0 )
13\,\m/\s
= 29\,\mbox{miles per hour.}
\]
% or 16\,mph.
% At 32\,mph, one fifth of the power is required for
% rolling resistance. At 48\,mph, one tenth.
}% end troublesoem page hack
\subsubsection{Bicycles}
%For truly sustainable transport look to human powered vehicles (HPVs)
% this includes bicycles with the aid of a trailer
% http://www.bikesandtrailers.com/bike-trailers/index.html for example
% or with a load carrying wheelbase extension
% http://www.xtracycle.com/html/home.php. If you must ride in all
% weathers then you may either use good quality waterproof clothing or
% a velomobile eg http://www.letra.dk ,http://www.velomobiel.nl or
% http://www.cab-bike.com to name but a few.
%% from http://www.bbc.co.uk/dna/actionnetwork/A3651932
%% Alexander Rice
For a bicycle ($m=90\,\kg$, $A=0.75\,\m^2$),
% the speed above which rolling resistance is less than air resistance is
%\[
% v =
%% sqrt( 2 * 0.005 * 90 * 9.81 / 1.3 / 0.75 )
% 3\,\m/\s
%\]
% or 7\,mph.
%% 6.71
the transition%
\amarginfig{b}{\footnotesize
\begin{center}
\mbox{\rotatebox{90}{\footnotesize\sf Energy consumption (kWh/100 km)}\hspace*{8mm}%
{\epsfxsize=40mm\epsfbox{../data/cars.eps}}%
}\\[0.2in]
{\sf{speed (km/h)}}
\\
\end{center}
% COLON
\caption[a]{Current cars' fuel consumptions
do not vary as speed squared.
% Vertical axis: Energy consumption in kWh per 100\,km.
% Horizontal axis: Speed in km/h.
Prius data from B.Z.\ Wilson;
BMW data from Phil C.\ Stuart.
% The Prius figures are from three separate experiments
%
% with slightly different conditions, but in all
% cases the car moved at constant speed under cruise control.
The smooth curve shows what a speed-squared curve would look like,
assuming a drag-area of 0.6\,m$^2$.
}
\label{fig.cars0}
% this figure deliberately placed early *** (check whether it can go later)
}
from rolling-res\-ist\-ance-dom\-in\-ated
cycling to air-resistance-dominated cycling takes place
at a speed of about 12\,km/h.
At a steady speed of 20\,km/h,
cycling costs about \eccol{2.2\,kWh per 100\,km}.
By adopting an aerodynamic posture, you can reduce your
drag area and cut the energy consumption down to
about 1.6\,kWh per 100\,km.
% Electric bicycle (according to a friendly user):
% uses 1\,kWh(e) per 100\,km.
% (I think he pedals too.)
% For a bike, {\em \ldots move bike here?}
\subsubsection{Trains}
For an eight-carriage
% high-speed
train as depicted in \figref{fig.stoptrain}
($m=400\,000\,\kg$, $A=11\,\m^2$), the speed
above which\index{cartoon!train}\index{rail}\index{train}
air resistance
is greater than rolling resistance
is
\[
v =
% sqrt( 2 * 0.002 * 400000 * 9.81 / 1.3 / 11.0 )
33\,\m/\s = 74\,\mbox{miles per hour}.
\]
% or 74\,mph.
For a single-carriage
train ($m=50\,000\,\kg$, $A=11\,\m^2$) , the speed
above which
air resistance
is greater than rolling resistance
is
\[
v =
% sqrt( 2 * 0.002 * 50000 * 9.81 / 1.3 / 11.0 )
12\,\m/\s = 26\,\mbox{miles per hour}.
\]
% or 26\,mph.
\subsection{Dependence of power on speed}
When I say that halving your driving speed should reduce
fuel consumption (in miles per gallon) to {\em one quarter\/} of current levels,
some people feel sceptical. They have a point: most cars'
engines have an optimum revolution rate,
\amarginfig{b}{
\begin{center}
\mbox{\epsfxsize=43mm%
\mono{\epsfbox{../data/mono/EnginePower.eps}}%
{\epsfbox{../data/EnginePower.eps}}%
}%
% \\[0.16in]
\end{center}
\caption[a]{Powers of cars (kW) versus their top speeds (km/h).
Both scales are logarithmic.
The power increases as the third power of the speed.
To go twice as fast requires eight times as much engine power.
From \protect\cite{flight}.}
\label{fig.cars}
}%
and the choice of gears of the car determines a range
of speeds at which the optimum engine efficiency can be delivered.
If my suggested experiment of halving the car's speed takes the car out
of this designed range of speeds, the consumption might not fall by
as much as four-fold.
My tacit assumption that the engine's efficiency is the
same at all speeds and all loads led to the conclusion that it's
always good (in terms of miles per gallon) to travel slower;
but if the engine's efficiency drops off at low speeds, then
the most fuel-efficient speed might be at an intermediate speed that
makes a compromise between going slow and keeping the engine efficient.
For the BMW\,318ti in \index{BMW}\figref{fig.cars0}, for example, the optimum
speed is about 60\,km/h.
But if society were to decide that car speeds should be reduced,
there is nothing to stop engines and gears being redesigned
so that the peak engine efficiency was found at the right speed.
As further evidence that the power a car requires
really does increase as the cube of speed, \figref{fig.cars} shows the
engine power versus the top speeds of a range of cars. The line shows
the relationship ``power proportional to $v^3$.''
\subsection{Electric cars: is range a problem?}
\label{ch.cars2Range}
\index{electric car!range}People
often say that the range of electric cars is not big enough.
Electric car advocates say ``no problem, we can just put in more batteries'' --
and that's true, but we need to work out what effect the extra batteries
have on the energy consumption.
The answer depends sensitively on
what energy density we assume the batteries deliver:
for an energy density of 40\,Wh/kg (typical
of lead-acid batteries), we'll see that
it's hard to push the range
beyond 200 or 300\,km; but for an energy density
of 120\,Wh/kg (typical of various lithium-based
batteries), a range of 500\,km is easily achievable.
\amarginfig{b}{
\begin{center}
\mbox{\epsfxsize=53mm%
{\epsfbox{../data/carselec.eps}}%
}%\\[0.1in]
\end{center}
\caption[a]{Theory of electric car range (horizontal
axis) and transport cost (vertical axis) as a function
of battery mass, for two battery technologies.\index{electric car!theory}
A car with 500\,kg of
old batteries, with an energy density of 40\,Wh per kg,
has a range of 180\,km.
With the same weight of
modern batteries, delivering 120\,Wh per kg,
an electric car can have a range of more than 500\,km.
Both cars would have an energy cost of about 13\,kWh per 100\,km.
These numbers allow for a
battery charging efficiency of 85\%.
}
\label{fig.carselec}
}
% So, let's work out what our cartoon says about the range of an electric car.
Let's assume that the mass of the car and occupants\index{electric car!theory}
is 740\,kg, {\em without\/} any batteries.\index{range!electric car}
In due course we'll add 100\,kg, 200\,kg, 500\,kg, or
perhaps 1000\,kg of batteries.\index{electric car!range}
Let's assume a typical speed of 50\,km/h (30\,mph);
a drag-area of 0.8\,m$^2$; a rolling resistance of 0.01;
a distance between stops of 500\,m; an engine efficiency of 85\%;
and that
during stops and starts, regenerative braking recovers half
of the kinetic energy of the car.
Charging up the car from the mains is assumed to be 85\% efficient.
\Figref{fig.carselec} shows the transport cost of the
car versus its range, as we vary the amount of
battery on board.
The upper curve shows the result for a battery whose energy density
is 40\,Wh/kg (old-style lead-acid batteries).
The range is limited by a wall at about
500\,km. To get close to this maximum range, we
have to take along comically large batteries:
for a range of 400\,km, for example, 2000\,kg of
batteries are required, and the transport cost is above 25\,kWh per 100\,km.
If we are content with a range of
180\,km, however, we can get by with 500\,kg of batteries.
Things get much better when we switch to lighter lithium-ion batteries.
At an energy density of 120\,Wh/kg, electric cars with 500\,kg
of batteries can easily deliver a range of 500\,km.
The transport cost is predicted to be about 13\,kWh per 100\,km.
It thus seems to me that the range problem\index{electric car!range}
has been solved by the advent of modern batteries.
It would be nice to have even better batteries, but
an energy density of 120\,Wh per kg is already
good enough, as long as we're happy for the
batteries in a car to weigh up to 500\,kg.\index{battery!energy density}
In practice I imagine most people would be
content to have a range of 300\,km, which can be
delivered by 250\,kg of batteries. If these batteries
were divided into ten 25\,kg chunks, separately
unpluggable, then a car user could keep just
four of the ten chunks on board when he's doing regular
commuting (100\,kg gives a range of 140\,km);
and collect an extra six chunks from a\index{car!recharging}
battery-recharging station when he wants to
make longer-range trips. During long-range trips, he would
exchange his batteries for a fresh set\index{electric car!recharging}\index{recharging}
at a \index{battery exchange}%
battery-exchange station\index{filling station!role in electric transport}
every 300\,km or so.
% BP would be responsible for recharging the batteries
\small
\section*{Notes and further reading}
\beforenotelist
\begin{notelist}
\item[page no.]
\item[\npageref{pCarEng25}]
{\nqs{
Typical petrol engines are about 25\% efficient}.
}
Encarta \tinyurl{6by8x}{http://encarta.msn.com/encyclopedia_761553622/Internal-Combustion_Engine.html}
says ``The efficiencies of good modern Otto-cycle engines range between 20 and 25\%.''
% This rough statistic is backed up by this webpage
% \myurl{http://www.fueleconomy.gov/feg/atv.shtml},
% 17\% of the energy idle
% 62.4 engine losses
% 18.2 comes out of the engine and goes somewhere.
% 2.2 to other things
% which shows that engine losses
% are three times the power going to the driveline
% and other accessories.
% Wikipedia
% http://en.wikipedia.org/wiki/Internal_combustion_engine#Engine_Efficiency
% says ``as high as 37\% at the optimum operating point'' but
% normally about 20\%, because car engines are usually not at their
% sweet spot, sadly.
% I wonder if they are using net or gross 5% difference!
The petrol engine
of a Toyota Prius, famously one of the most efficient
car engines, uses the Atkinson cycle instead of the
Otto cycle; it has a peak power output of 52\,kW
% 70 hp
and has an efficiency of 34\%
when delivering 10\,kW
\tinyurl{348whs}{http://www.cleangreencar.co.nz/page/prius-petrol-engine}.
% Octane engines reach 32\%
% http://ecen.com/content/eee7/motoref.htm
%
%
% From Toyota:
% http://www.cleangreencar.co.nz/page/toyota-prius-iii-hybrid-car-technical-information
% prius electric motor 1 is 18 kW
% Motore 2 maximum power of 33 kW in the Generation II Prius and 50 kW in the Generation III Prius.
% The Generation II Prius high-voltage battery pack consists of 228 cells of 1.2 volts each for a total nominal voltage of 273.6 volts. The cells are arranged in 38 modules of 6 cells each and the whole lot is assembled into a unit that is fixed behind the rear seat. You can see where it is from the bump at the bottom rear of the boot. The maximum current of the battery is 80 amps discharge and 50 amps charge. This is remarkable, since each cell is similar in size to an ordinary D-cell such as you would use in a large flashlight.
% Maximum efficiency generally occurs at around half of the engine's peak power output.
% Battery capacity is 1.5kWh - delivers 1.35kWh?
%
The most efficient diesel engine in the world is
52\%-efficient, but it's not suitable for cars
as it weighs 2300 \tonnes:
the \ind{Wartsila--Sulzer} RTA96-C turbocharged diesel
engine (\figref{Wart}) is intended for container ships and has a power output of 80\,MW\@.
\marginfig{
\begin{center}
\begin{tabular}{@{}c@{}}
\lowres%
{\mbox{\epsfxsize=53mm\epsfbox{../../images/wartsila400.eps}}}%
{\mbox{\epsfxsize=53mm\epsfbox{../../images/wartsila800.eps}}}
\\
\end{tabular}
\end{center}
\caption[a]{The Wartsila-Sulzer RTA96-C 14-cylinder two-stroke diesel engine.
27\,m long and 13.5\,m high.
\myurlb{www.wartsila.com}{http://www.wartsila.com/}
}\label{Wart}
}
% image from http://www.wartsila.com/ press release site (youfigure.mgp)
% the most powerful and most efficient diesel engine - Japan's Diesel United
\item[\npageref{pRegenEff}]
{\nqs{Regenerative brakes roughly
halve the energy lost in braking}.}
Source: \cite{E4tech}.
%The e4 document in refs/transport says current hybrids recover about half the braking energy.
\item[\npageref{pElecEnginPowRat}]
{\nqs{Electric engines can be about 8 times lighter than petrol engines}.}
% {\bf Engine power:}
\par
% this par added to prevent an ORB
A 4-stroke petrol engine has a power-to-mass ratio of
% aero-engine power rating delivers
% ~ 1HP/kg =
roughly 0.75\,kW$\!$/kg.
The best electric motors have an efficiency of 90\% and a power-to-mass ratio of 6\,kW$\!$/kg.
So replacing a 75\,kW petrol engine with a 75\,kW electric motor saves 85\,kg in weight.
% , but this
% corresponds to only 13\,kWh of battery capacity = 4\,kg of petrol
Sadly, the power to weight ratio of batteries is about 1\,kW per kg, so
what the electric vehicle gained on the motor, it loses on the batteries.
\item[\npageref{pBikeCrr}]
{\nqs{The bike's engine
uses energy with an efficiency of 0.25}.}
This and the other assumptions about cycling
are confirmed by \citet{Prampero}.
The drag-area of a cyclist in racing posture is
% cf the 0.75 that I assumed
$\cd A = 0.3\,\m^2$. The rolling resistance of
a cyclist on a high-quality
racing cycle (total weight 73\,kg) is 3.2\,N.\label{bikeref}
% \cite{Prampero}
% (Di Prampero et al 1979).
% from Alan Cummings ``Cycling in the Wind''
% J Appl Physiology Prampero
% (1.6 kWh) per (100 km) = 57.6 newtons
\item[\npageref{fig.cars0}] {\nqs{\Figref{fig.cars0}.}}
Prius data from B.\ Z.\ Wilson
[\myurlb{home.hiwaay.net/ ~bzwilson/prius/}{http://home.hiwaay.net/ ~bzwilson/prius/}].
BMW data from Phil C.\ Stuart
[\myurlb{www.randomuseless.info/318ti/economy.html}{http://www.randomuseless.info/318ti/economy.html}].
\item[Further reading:]
\cite{WhatPriceSpeed}.
\end{notelist}
\normalsize