How can I find the mechanical efficiency of a car engine from easily obtained or measured data?
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,ย 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)vย = (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.