\label{ch.transport2}
% \newpage
\marginfig{
\begin{center}
\begin{tabular}{@{}c@{}}
\lowres{\mbox{\epsfxsize=53mm\epsfbox{../../images/MPV2S.jpg.eps}}}%
{\mbox{\epsfxsize=53mm\epsfbox{../../images/MPV2.jpg.eps}}}\\
\lowres{\mbox{\epsfxsize=53mm\epsfbox{../../images/Lexus1S.jpg.eps}}}%
{\mbox{\epsfxsize=53mm\epsfbox{../../images/Lexus1.jpg.eps}}}\\
{\mbox{\epsfxsize=53mm\epsfbox{../../images/tandem.eps}}}\\
%\lowres{\mbox{\epsfxsize=53mm\epsfbox{../../images/Lexus2S.jpg.eps}}}%
%{\mbox{\epsfxsize=53mm\epsfbox{../../images/Lexus2.jpg.eps}}}\\
%% Lexus RX300
\end{tabular}
\end{center}
\caption[a]{
Multi-passenger vehicles.
}
}
\section{Short-range high-speed trains}
One carriage has weight 50 \tonnes,
length 20m, width 2.82 m, and contains 79 seats.
(six windows and 2 doors, WAGN stock).
Height is I think 4m.
What is the best that a high-speed train's energy-per-distance
could possibly be?
Just like a fast-moving car, a train needs energy because it
makes air swirl.
The energy-per-distance (or ``force'') needed to make
air swirl is
$$\frac{1}{2} c_d \rho A v^2,$$
where $c_d$ is a drag coefficient, which I'll take to be $1$,
$\rho$ is the density of air, $A$ is the frontal area of the train,
and $v$ is its speed. This is the unavoidable energy that a
train travelling at speed $v$ must guzzle.
Let's plug in the figures for a 100\,mph train from London to Cambridge
($v = 45$\,m/s).
The train has width 2.8\,m, height 4\,m, weighs 400\,\tonnes,
and seats 630 passengers (who themselves weigh an extra 50\,\tonnes).
% rho = 1.3
The energy per distance associated with air resistance
comes out to
\[
15\,000\,\J/\m
\]
or
\[
400\,\kWh\,\mbox{per 100\,\km}
\]
which, shared between 630 is\index{transport!efficiency!train}
\[
0.65\,\kWh\,\mbox{per 100\,seat-km} .
\]
This is the smallest that the energy consumption
could ever be. In practice, it's about five times bigger
because of the inefficiency of energy conversion, and the
other things that energy goes into.
Is the London-to-Cambridge train's energy consumption really
drag-dominated?
Let's check what the distance between stops needs to be
for drag-power to exceed the starting-and-stopping power.
The two powers are equal if
$\rho A D = m_{\rm{train}}$;
which means
$D \simeq 40\,\km$.
So if the train comes to a stop a couple of times between
London and Cambridge (a distance of 100\,km),
the train would be putting equal energy into
making its own kinetic energy as into air resistance.
So to get an accurate estimate, I can't neglect the mass of the train.
What about rolling friction?
If rolling friction is about 100\,Newtons per \tonne,
the force for rolling friction is about 45\,000\,Newtons.
This is three times the force I estimated for air resistance.
Train mpg?
Assume drag dominated;
energy per distance = force = $\dfrac{1}{2} c_d rho A v^2$
Diesel or electric?
Diesel:
Scale compared to car (70mph).
v = 100mph so $v^2$ is 4 times bigger;
A = 4 times bigger; (8m**2)
$c_d$ much same as car;
so energy per distance is 16 times bigger; but occupancy
is 16*4 per carriage, times 8 carriages -> could be 500 per train.
So pmpg is up to 500/16 which is 31 times better.
Compute distance between stops for energy to indeed be drag
dominated:
$\rho A D = m_{\rm{train}}$;
$m_{\rm{train}} \simeq 50$ tons per carriage, 8 carriages -> 400,000 kg
D = 400,000/ 1 / 8
= 50,000 metres = 50km. (400 times bigger than car by weight; 4 times smaller
by area).
Stopping train: kinetic energy dominated.
energy cost per distance is (1/2)mv**2 / D
where D is the dist between stops, say 3km (so always within
a mile of a station, if you are adjacent to the line).
v = 30 mph = 15 m/s
m (single carriage) = 20,000 kg; actually 50,000.
64 people max.
-> energy cost per distance is 750 J per m (*5/2)
or 750 kJ per km (for all 64 people) (*5/2)
or 12 kJ per km per person; (*5/2)
which is 0.3 units per 100\,km; (300 km per unit) (*5/2)
[what did I do here about internal combustion efficiency? Did I assume electric train?]
or in mpg, 300km per (1/56 gall); /(5/2)
which is 17000 km per gallon. /(5/2)
10\,000 passenger miles per gallon /(5/2)
Don't use mpg, it doesn't translate to America or Europe.
Let's use kWh per 100 seat-km instead.
%Cost of new trains:
%250 million pounds for 51 trains each 3 carriages and diesel powered.
% source Railwatch Number 110. January 2007 (Railfuture)
% ie 30 million per train.