Energy Use in Vehicles

Physics textbooks often talk about static friction force and kinetic friction force, for objects that are either stationary with respect to the surface they sit on, or sliding on that surface. Neither of these concepts are appropriate for the motion of wheels on a surface. The wheel does not slide on the road or rail surface unless something is going wrong. The wheel at the point of contact with the road or rail is stationary, so the static friction force is what determines the limits of the vehicle’s acceleration or deceleration before skidding occurs. However, something else is going on: the wheel (and road/rail) is constantly deforming and relaxing slightly at the point of contact and this process is not perfectly elastic, which leads to energy loss.

The force exerted by this rolling friction depends on the weight of the vehicle supported by the wheels, and the quality of the contact point, which is expressed as a dimensionless coefficient of rolling resistance $\mu_r$. Hard, unyielding surface pairs like steel-on-steel have the lowest $\mu_r$ (see table 1), which is why trains and trams run on rails. Tires on roads are about an order of magnitude worse, and highly dependent on how well inflated the tires are. Anyone who has tried to ride a bicycle with soft tires will understand that.

Force due to Rolling Resistance $F_r$
= Coefficient of Rolling Resistance $\mu_r$
$\times$ Mass of vehicle
$\times$ Acceleration of gravity $g$ $\tag{1}$

The Coefficient of Rolling Resistance is usually written as $\mu_{RR}$, and it has different values for different types of vehicles. Some example values of rolling resistance are given in the table below[note]A Discovery Company. How Tires Work (online). http://auto.howstuffworks.com/tire4.htm [25 August 2009].[/note].

\begin{array}{|c c|}
\hline
\textbf{Tire Type} & \textbf{Coefficient of Rolling Friction} \\ \hline
\text{Low rolling resistance car tire} & 0.006\, – 0.01 \\
\text{Ordinary car tire} & 0.015 \\
\text{Truck tire} & 0.006\, – 0.01 \\
\text{Train wheel} & 0.001 \\
\hline
\end{array}

So what does this tell us? In order to figure out how this force impacts our fuel economy we need to figure out how much energy is required to overcome it. For this we use the Work-Energy principle, which tells us how much energy a force will add to a system.

$\text{Work = (Force)}\times\text{(Distance)}$

Because the rolling friction opposes the motion of the car, it actually subtracts energy from the car. This energy needs to be made up by burning more fuel.

A typical sedan has a mass of around 1200 kg. For this car, plus a single driver (70 kg) the force of rolling resistance will be:

\begin{eqnarray}
F_{RR} & =& \mu_{RR} m g \nonumber \tag{2} \\
& =& (0.015)(1270 \text{ kg})(9.8 \text{ m/s}^2) \nonumber \\
& =& 187 \text{ Newtons} \nonumber
\end{eqnarray}

Over the course of driving one kilometre, this will require extra energy given by:

\begin{eqnarray}
W & =& (F_{RR})\times(\text{Distance}) \nonumber \tag{3} \\
& =& (187 \text {N}) \times (1000 \text{ m}) \nonumber \\
& =& 187,000 \text{ N m} \nonumber \\
& =& 187 \text{ kJ for each kilometre driven} \nonumber
\end{eqnarray}

We can figure out how much fuel is required to drive one kilometre by using the efficiency formula:

\begin{eqnarray}
\text{Efficiency} & =& \dfrac{\text{Work Output}}{\text{Work Input}} \nonumber \tag{3}\\
& =& \dfrac{\text{Work Output}}{\text{Fuel Energy Input}} \nonumber \tag{4} \end{eqnarray}

\begin{eqnarray}
\text{Fuel Energy Input} &=& \dfrac{\text{Work Output}}{\text{Efficiency}} \nonumber \tag{5}\\
& =& \dfrac{187 \text{ kJ}} {25 \%} \nonumber \\
& =& 748 \text{ kJ} \nonumber
\end{eqnarray}

And to provide this amount of energy we need to use

\begin{eqnarray}
\text{Energy per litre} & =& \dfrac{\text{# of Joules}}{\text{# of litres}} \nonumber \tag{6}\\
\text{# of litres} & =& \dfrac{\text{# of Joules}}{\text{Energy per litre}} \nonumber \tag{7}\\
& =& \dfrac{748 \text{ kJ}}{32 \text{ MJ/L}} \nonumber \\
& =& 0.023 \text{ L} \nonumber
\end{eqnarray}

So, 0.023 L of fuel is required to drive 1 km.

Remember, this result is just to overcome the rolling friction. If we add this to the 0.064 L/km highway mileage we calculated in the Constant Speed Cruising section (taking air drag into account) this comes out to a total of 0.087 L/km. This is a little bit higher than the reported average of 0.076 L/km[note]MacKay DJC. Sustainable Energy – Without the Hot Air (Online). UIT Cambridge. p.31. http://www.inference.phy.cam.ac.uk/sustainable/book/tex/ps/1.112.pdf [25 August 2009].[/note], which seems reasonable as the highway mileage was calculated at a speed of 100 km/h which is perhaps a bit fast.

So, now what would be the impact of having low air pressure in our tires? Let’s imagine that having your air pressure reduced by 5% would result in a 5% increase in the coefficient of rolling resistance. A typical car’s tires are inflated to around 40 psi, so this would correspond to being 2 psi lower than average. A 5% increase in the Coefficient of Rolling Resistance would bring it up to 0.01575, and the associated fuel consumption would increase to 0.024 L/km. This is an extra 0.01 L/km, or approximately an extra 1% of fuel mileage. This corresponds closely with the guidelines published by Natural Resources Canada[note value=1][/note].

This extra drag starts to add up when your tires are really low on air. If they are 10 psi low, that would correspond to an extra 5% fuel cost!

Updated (CEW) 2019-10-17

At cruising speed on a flat road, the forces that the car’s engine and transmission have to overcome are rolling resistance and air resistance. The force from rolling resistance is approximately constant with speed, while that from air resistance rises steeply with increasing speed. The basic physics is well described in ref.[note]MacKay D. J. C., Sustainable Energy – Without the Hot Air (Online). UIT Cambridge. p.257. http://www.inference.phy.cam.ac.uk/sustainable/book/tex/ps/253.326.pdf [2019-09-12].[/note]. Here, let us do a simple analysis of a real situation. The author has family in Vancouver and north-eastern BC, and has made many trips across the province, over the last decade and half, in a 2005 Honda Civic[note]US DoE/EPA fuel economy data,  https://www.fueleconomy.gov/feg/noframes/20709.shtml [2019-09-16].[/note]. These long trips involve long uninterrupted stretches on mostly reasonably flat road at an average 90 km/h. The fuel economy of the car has remained constant at $R$ = 5.8 L/100 km. We will use these data to estimate the mechanical efficiency of the car engine and transmission, and to find how rolling and air resistance contribute to the total drag force.

Other details about the car we will need are the mass, rolling resistance and drag coefficient. Assuming two occupants, the mass of the vehicle $m$ = 1500 kg. The rolling resistance coefficient $\mu_r$ we take to be a typical 0.012[note]Rolling resistance https://en.wikipedia.org/wiki/Rolling_resistance#Pneumatic_tires_.28load.29 [2019-09-12][/note]. For vehicles, the drag coefficient and frontal area are usually combined into the product $C_DA$[note] Automobile drag coefficient https://en.wikipedia.org/wiki/Automobile_drag_coefficient [2019-09-12][/note], which is roughly the area of a flat sheet facing the air flow that would produce the same drag at the same speed. For the 2005 Civic, $C_DA$ is 0.654 m$^2$, or about 1/3 of the actual frontal area of the car.

The force from rolling resistance $F_r$ is straightforward to calculate from Eqn. 1.

\begin{eqnarray}
F_r = \mu_r mg = 176 \text{ N} \tag{1}
\end{eqnarray}

The force from air resistance $F_a$ requires the speed, $v$, which is 25 m/s, and the density of air, $\rho$ = 1.2 kg/m$^3$. To see where this Eqn. 2 comes from see ref.[note]Drag, https://en.wikipedia.org/wiki/Drag_(physics)#Types_of_drag [2019-09-13][/note] and/or check the dimensions (the factor of 1/2 is conventional).

\begin{eqnarray}
F_a = \dfrac{1}{2} \rho C_DA v^2 = 245 \text{ N} \tag{2}
\end{eqnarray}

The mechanical power $P_{mech}$ is the total drag force $F_r + F_a$ times the velocity (Eqn. 3).

\begin{eqnarray}
P_{mech} = (F_r + F_a)v  = (425 \text{ N})(25 \text{ m/s}) = 10,500 \text{ W} = 10.5 \text{ kW} \tag{3}
\end{eqnarray}

To find the mechanical efficiency of the engine and transmission, we need the thermal power $P_{therm}$; this we can obtain from the fuel economy $R$, and the enthalpy of combustion (“energy content”) of the fuel, $E$ = 36 MJ/L. The easiest way to see where Eqn. 4 comes from is to check the dimensions.

\begin{eqnarray} P_{therm} &=& R E v \tag{4} \\ &=& (5.8 \times 10^{-5} \text{ L/m})(36 \times 10^6 \text{ J/L})(25 \text{ m/s}) = 52,200 \text{ W} \\ &=& 52.2 \text{ kW} \end{eqnarray}

Now the mechanical efficiency of the engine and transmission $\epsilon$ is just the ratio of these two powers.

\begin{eqnarray} \epsilon &=& \dfrac{P_{mech}} {P_{therm}} \tag{5}\\ &=& 10.5/52.2 = 0.202 \\ &\approx& 20\% \end{eqnarray}
Not bad for an old car. The real steady cruise efficiency will be somewhat higher, as there were some stops on the 1200 km run.

Updated (CEW) 2019-10-17

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,  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 25¢ 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 3¢, currently.

Updated (CEW) 2019-09-16