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The burning of gasoline in a car releases about $\SI{3.0e4}{kcal \per gal}$. If a car averages $\SI{41}{\kilo\meter\per gal}$ when driving $\SI{90}{\kilo\meter\per\hour}$, which requires $\SI{2}{hp}$, what is the efficiency of the engine under those conditions? (One gallon equals \SI{3.785}{\liter} and one imperial horsepower equals \SI{746}{W}.)
2\%
In SIunits, the burning of gasoline releases: \begin{align} H &= \SI{3.0e4}{kcal/gal}\\ &= \numprint{3.0e4}\cdot \frac{\SI{4182}{J}}{\SI{3.785}{\liter}}\\ &= \pq{3.31e7}{\joule\per\liter} \end{align} In one hour, the car drives $\SI{90}{km}$, and its intake is therefore \begin{align} V = \frac{\SI{90}{km}}{\SI{41}{km \per gal}}=\SI{2.19}{gal}=\SI{8.29}{\liter}, \end{align} which release \begin{align} E &= V \cdot H\\ &= \SI{8.29}{\liter}\cdot \SI{3.31e7}{\joule\per\liter}\\ &= \SI{2.743e8}{J}. \end{align} The engines power of $\SI{2}{hp} (two horsepower, one imperial horsepower is $\pq{746}{W}$) is in SIunits $\SI{1492}{W}$. In one hour, this corresponds to the energy \begin{align} E' &= P\cdot t\\ &= \SI{5.371e6}{J}. \end{align} The efficiency under those conditions is: \begin{align} \eta &= \frac{\SI{5.371e6}{J}}{\SI{2.743e8}{J}}\\ &= 0.01958 \end{align} That's around 2\%.
19:32, 12. June 2017  si  Urs Zellweger (urs)  Current Version 