What does it cost in fuel to stop a bus at a pedestrian light?
The bus is stopped at a light. It has lost all its kinetic energy. How much fuel does it need to start up again? Where does the chemical energy in the fuel go?
Let us consider the 99-B Line in Vancouver BC, which is the busiest bus route in North America. The articulated Xcelsior buses on this route are built by New Flyer Industries of Winnipeg, Manitoba, and their diesel engines have a typical mechanical efficiency of 40%[note]Some diesel engines have an efficiency as high as 54% but here we use a more conservative number to reflect real operating conditions. https://en.wikipedia.org/wiki/Engine_efficiency [2019-09-11] [/note]. Although this figure is much higher than that of automobile gasoline engines, it still means that most of the chemical energy in the fuel – 60% – goes out of the exhaust pipe as heat.
$$\text{Engine Efficiency} = \dfrac{ \text{Mechanical Energy Out}}{\text{Fuel Energy In}}$$
Imagine the bus was initially moving at 50 km/h and now needs to accelerate back to that speed.
The mass of an empty 99 bus is 20 tonnes[note]New Flyer Xcelsior,ย https://en.wikipedia.org/wiki/New_Flyer_Xcelsior [2019-09-11][/note]. The full load of 120 passengers have a mean mass of 70 kg each[note]Human mass data, https://en.wikipedia.org/wiki/Human_body_weight#By_country. The US mean is 82 kg rather than 70 kg for Canada.[2019-09-11][/note].
So the total mass of the bus is
$$20,000 \text{ kg} + (120 \text{ passengers})\times\left(\dfrac{70 \text{ kg}} {\text{passenger}}\right) = 28,400 \text{ kg}$$
Now we need to know how much energy it takes to get this mass moving again at 50 km/h (14 m/s). The kinetic energy ($KE$) of a moving object is given by Eqn. 1.
$$\begin{eqnarray}KE &=& \dfrac{1}{2} m v^2 \nonumber \\
&=& \dfrac{1}{2} (28,400 \text{ kg})\times(50 \text{ km/h})^2 \nonumber \\
&=& \dfrac{1}{2} (28,400 \text{ kg})\times(13.9 \text{ m/s})^2 \nonumber \\
&=& 2.74 \times 10^6 \text{ kg} \text{ m}^2 \text{/s}^2 \nonumber \\
&=& 2.74 \text{ MJ} \nonumber
\end{eqnarray}$$
So far we have ignored bullet points 2 and 3. We assume stopping the bus does not cause it to travel any further distance that it would have anyway. The energy used to overcome air resistance depends on speed and the distance travelled, in fact it rises sharply with increasing speed. The interrupted bus journey will have a lower mean speed than an uninterrupted journey, and so the losses due to air resistance will actually be slightly less. The force due to rolling resistance is fairly independent of speed and so the losses incurred by that mechanism will be the same regardless of whether the journey is interrupted or not. Thus the differences are small and for simplicity we will only consider bullet points 1 and 4, kinetic energy and mechanical efficiency.
In order to create this much mechanical energy, we need to burn fuel. The enthalpy of combustion of diesel oil (i.e. “energy content”) is around 38 MJ/L[note]C21 article: Useful Numbers, https://c21-wp.phas.ubc.ca/article/useful-numbers[/note]. Now we can calculate how much fuel energy is required.
$$\begin{eqnarray}
\text{Fuel Energy} &=& \dfrac{\text{Mechanical Energy}}{\text{Mechanical Efficiency}} \nonumber \\
&=& \dfrac{2.77 \text{ MJ}}{0.40} \nonumber \\
&=& 6.9 \text{ MJ}\nonumber
\end{eqnarray}$$
To determine how much fuel this is, we calculate
$$\begin{eqnarray}
\text{Volume of diesel oil} &=& \dfrac{6.9 \text{ MJ}}{38 \text{ MJ/L}} \nonumber \\
&=& 0.18 \text{ L} \nonumber
\end{eqnarray}$$
In Vancouver, this amount of fuel costs about 25ยข and its combustion will emit about 400 g of CO$_2$ (using a conversion factor of 2.3 kg CO$_2$/L oil)[note]C21 article: GHG Calculations, https://c21-wp.phas.ubc.ca/article/energy-and-greenhouse-gas-ghg-calculations [/note] plus smaller amounts for each of all the other vehicles that were also stopped by the light, multiplied by the number of times that it happens. One way of avoiding this cost and major annoyance to passengers is to have transponders in the buses that make sure that traffic lights up ahead stay green for them.
A similar calculation for a small family car goes as follows. The author’s 2005 Honda Civic with two passengers has a mass of 1440 kg. At 50 km/h, the kinetic energy is 140 kJ. The enthalpy of combustion of gasoline is 34 MJ/L and the car engine’s mechanical efficiency is 20%, so the kinetic energy at 50 km/h represents 0.02 L of fuel, i.e. stopping the car and then resuming speed costs about 3ยข, currently.
Updated 2019-09-16
