User:Hansmuller/Car mathematical physics

The energy consumption of a car covering a given distance for locomotion depends on the speed of the vehicle. The faster a car drives, the more energy is required for propulsion. At speeds below 40km/h, energy for a given distance is roughly constant with speed; at higher speed the energy consumption of a car per unit distance increases proportional to the square of the speed. This is caused by the air drag which dominates over the tyre rolling resistance at high speeds.

Driving twice as fast therefore costs approximately four times as much fuel per unit of distance. If we neglect the speed dependence of the heat production, the required power increases with the cube of the speed, so driving twice as fast costs theoretically eight times as much fuel per unit of time.

Mechanisms for energy loss at low speed
An average passenger car dissipates fuel energy through various mechanisms. For a low speed of 60 km/h, Holmberg et al. (2012) found the following breakup by percentages from the literature: MacKay's simulation shown in Graph 1 largly concurs with slightly different parameters, while not considering the heat losses.
 * 1) 33% (range, 30–37%): heat loss in exhaust gases
 * 2) 29% (25–33%): heat loss in the engine, the radiator, and the car’s heating system
 * 3) 38% (33–40%): air drag and friction losses, subdivided into:
 * 4) 5% (3–12%): air drag
 * 5) 33%: friction in the car, including rolling resistance.

Energy loss by acceleration, air drag and rolling resistance at various speeds
The energy $$E_\text{motor}$$ for propulsion supplied by a car's engine is spent on: so that
 * 1) accelerating the car (abbreviated to ac),
 * 2) overcoming the air resistance (aerodynamic drag, dr), and
 * 3) overcoming the rolling resistance (rr).
 * $$E_\text{motor}=E_\text{ac} + E_\text{dr} +E_\text{rr}$$

The energy effects of wind, ascent and descent are not considered here, nor the repeated braking and accelerating when driving in a city.

Acceleration
The kinetic energy $$E_\text{kin}$$ of a moving car with mass $$m_{car}$$ and speed $$v_{car}$$ can be calculated theoretically as
 * $$E_\text{kin} = \tfrac12 m_{car} \cdot v^2_{car}$$

so for a car with a mass of 1,000 kg at a speed of 120 km/h (33.3 m/s)
 * $$E_\text{kin} = 0{.}5 \times 1000 \times 33{.}3^2 = 5{.}56 \times 10^5\, \text{J} = 0{. }15\, \text{kWh}$$

(One kilowatt hour (kWh) equals 3.6 million joules (J). )

The required kinetic energy is supplied by the motor when the car accelerates starting from rest. It is lost when the driver brakes (unless the available kinetic energy is stored in, for example, a flywheel). Part of this energy can also be recovered from electric cars via recuperative braking and can be stored in a accumulator.

In practice, the real energy consumption of a car depends on many additional factors, including the traffic situation and personal driving style. In order to be able to make comparisons, standardized consumption tests have been devised, driving cycles, such as the New European Driving Cycle (NEDC) and the newer Worldwide harmonized Light vehicles Test Procedures (WLTP).

Without any resistance or friction, by Newton's First Law a car with speed $$v$$ would continue at this speed forever. However, air drag and rolling resistance cause energy losses which must be overcome to drive at a constant speed.

Air drag
Depending on its surface area $$A_{car}$$, streamlining and speed $$v$$, a car moves a lot of air, which causes energy loss by the aerodynamic drag force. We can imagine the moving air as a tube (trunk) of air behind the car, with a volume of the cross-sectional area of that tube times the length, which equals the distance traveled by the car. This length is then equal to the car speed times the elapsed time. The effective cross-section of the air tube is less than the front surface of the car $$A_{car}$$, for example due to streamlining. This can be taken in account by introducing a drag coefficient $$c_{drag}$$ for the air resistance of about 1/3. Suppose the surface area of the wind shield of a car is
 * $$A_{car} = 2 m \times 1.5 m = 3 m^2$$

then the effective volume $$V_{eff-air}$$ of that tube of swirling air behind the car, after a period of driving $$t$$ at speed $$v$$ equals
 * $$V_{eff-air} = A_{car} \cdot tube length = A_{car} \cdot c_{drag} \cdot v_{car} \cdot t$$

with $$v_{car} \cdot t$$ = $$d$$ the distance traveled. The mass of the displaced air over a journey of $$d$$ meters is equal to the volume of the tube times the specific mass $$\rho_{air}$$ of air, so
 * $$m_{air} = V_{eff-air} \cdot \rho_{air} = 3 \times 1/3 \cdot d \cdot \rho_{air}$$.

The energy loss due to air resistance is equal to the kinetic energy that the car imparts to the air. Using the value of $$1.3 kgm^{-3}$$ for the density of air $$\rho_{air}$$ results in
 * $$E_{drag} = \tfrac12 m_{air} \cdot v_{car}^2 = \tfrac12 \rho_{air} \cdot A_{car} \cdot d \cdot c_{drag} \cdot v_{car}^2 \cong 0.65 \cdot v^2_{car} \cdot d$$

Rolling resistance
The mathematical formula for the force due to the rolling resistance of the car tyres is:
 * $$ F_{rr} = C_{rr} \cdot m \cdot g$$

with
 * $$C_{rr}$$ the roll coefficient which is around 0.01.
 * $$m$$ the mass of the car, and
 * $$g$$ the gravitational acceleration.

The corresponding energy loss caused by the rolling resistance is (the work done by the rolling resistance)


 * $$ E_{rr} = F_{rr} \cdot d = constant \cdot d$$,

with $$d$$ the distance covered, so the energy spent on overcoming the rolling resistance per unit distance is a constant.

Total energy consumption to overcome friction
Adding up the energy consumption required to overcome both air drag and tyre rolling resistance, we have


 * $$E_\text{total car over distance d} = E_\text{air resistance} + E_\text{rolling resistance} $$


 * $$E_{drag} \cong 0.65 \cdot v^2_{car} \cdot d + constant \cdot d $$

From the graphs it is seen that per unit distance the drag resistance dominates over the rolling resistance at higher car speeds.
 * $$E_\text{tot} \cong 0{.}7 \cdot v^2 \cdot d$$

The energy consumption over a distance and the air pollution are proportional to the speed squared
The energy consumption of a car at higher speeds is seen to be proportional to its speed $$v$$ squared, and of course also to the distance $$d$$ traveled. So if the air resistance on the car dominates – and acceleration/braking and rolling resistance are of less significance for energy consumption, such as on the highway – the energy consumption is proportional to the square of the car speed. If air pollution increases in line with petrol consumption, this means that traveling at 120 km/h over the same distance pollutes 15% less than at 130 km/hour, because (120/130)2 = 0.85 = 85% and 100% - 85% = 15%.

Because a car mainly produces heat and only uses 25% of the energy from the fuel for movement (an electric car uses about 90% for movement), the petrol must provide much more energy than the car uses for movement:
 * $$E_\text{supplied by gasoline} = 4 \times E_{tot} \cong 2{,}8 \cdot v^2 \cdot d$$

Car power is proportional to the cube of the speed
Precise details of a car's energy consumption depend on the construction etc. of a car, but in general this quadratic relationship with speed is correct for speeds above 60 km/h, but refer to the accompanying figure. This can also be seen in the measured relationship between the power (energy per time, so energy times speed divided by distance) $$P_{car}$$ of cars and the cube of their top speeds $$v_ \text{top}$$
 * $$P_{top, car} = E_{top, car} / t = E_{top, car} / (d / v_\text{top}) = E_{top, car} \cdot v_\text{top} / d \cong v^3_\text{top}$$, so


 * $$P_{top, car} \cong v^3_\text{top}$$

Energy is proportional to the square of the speed and must be multiplied again by the speed to obtain the expended power. So the car power scales with the third power of the car speed. Neglecting the speed dependence of the heat production, the consumption of energy per second would increase eightfold (23) if you drive twice faster.

Friction dominates at low speed
At 60km/h, the global average speed, energy loss due to air drag in fossil fuel cars is approximately 5% of the total energy loss. Friction (33%), exhaust (29%), and cooling the engine (33%) account for the rest. Electric vehicles are estimated to have about half as much loss due to friction. At high speeds air drag losses are overwhelming the other car energy losses, due to the cubic dependence on speed.

City driving
In driving over short distances with lots of starting and stopping, as in cities, the energy is mainly spent on speeding up the car but also on air resistance. To consume less energy, the car should weigh less and travel further between stops. Going slower saves energy and regenerative braking would be optimal. For the case without this advanced braking, an estimate can be made which energy loss is more important. If $$d_{stop}$$ is the distance between stops, then $$d_{stop}$$ = $$v_{car} \cdot t$$ and the time between stops equals $$\frac{d_{stop}}{v_{car}}$$ so that the energy loss rate is $$R_\text{loss rate by braking} = \frac{\text{kinetic energy car}}{\text{time between stops}} = \frac{\tfrac12 m_{car} \cdot v^2_{car}}{\frac{d_{stop}}{v_{car}} } = \frac{ \tfrac12 m_{car} \cdot v^3_{car} } {d_{stop}}$$ From the analysis above for the tube of swirling air set in motion by a passing car: $$R_\text{loss rate by air drag} = \frac{\text{kinetic energy swirling air}}{\text{time between stops}} = \frac{ \tfrac12 \rho_{air} \cdot A_{car} \cdot d \cdot c_{drag} \cdot v_{car}^2 }{\frac{d_{stop}}{v_{car}} } =  \tfrac12 \rho_{air} \cdot A_{car} \cdot v^3_{car}  $$ so the total rate of energy loss is the sum $$R_\text{loss rate by braking + air drag} = \tfrac12 \cdot v_{car}^3 \left(\frac{m}{d_{stop}} + \rho_{air} \cdot A_{car} \right) $$ Braking losses are higher than air drag losses if $$ \frac{m_{car}}{d_{stop}} > \rho_{air} \cdot A_{car} $$, so for a car mass $$m_{car}$$ of 1000 kg, a windscreen $$A_{car}$$ of 3 m^2, air density $$ \rho_{air}$$of 1.3 kg/m3, the critical distance $$d_{stop}$$ under which braking is more important than air drag is 750 m and regenerative braking is extremely helpful. With a longer distance between stops, air drag losses dominate and the streamlining of the car becomes vital. As the cubic dependence on speed is the same for both braking and drag losses, driving slowly is best for energy saving in both regimes.

Literature

 * Fundamentals of Physics:
 * On the related topic of internal friction: