Question

The burning of gasoline in a car releases about\({\bf{3}}{\bf{.0}} \times {\bf{1}}{{\bf{0}}{\bf{4}}}{\bf{ kcal/gal}}\).If a car averages\({\bf{41}}{\rm{ }}{\bf{km/gal}}\)when driving at a speed of\({\bf{110}}{\rm{ }}{\bf{km/h}}\),which requires 25 hp, what is the efficiency of the engine under those conditions?

Short answer

The efficiency of the car engine is \(19.86\% \).

Step-by-step solution

Understanding the efficiency of the engine

The efficiency of the car engine is dependent on the output power and input power. The output power can be obtained by the power developed by the car. The input power can be obtained by the heat released by the gasoline per unit time.

For this question, evaluate the heat energy released by the gasoline with the help of unit conversions and then estimate the car's efficiency.

Identification of the given data

The given data can be listed as

• The heat released by the gasoline is

\({Q_{\rm{H}}} = 3 \times {104}{\rm{ kcal/gal}}\left( {\frac{{4186{\rm{ J}}}}{{1{\rm{ kcal}}}}} \right) = 1.26 \times {108}{\rm{ J/gal}}\).

• The useful power required is\(P = 25{\rm{ hp}}\left( {\frac{{746{\rm{ W}}}}{{1{\rm{ hp}}}}} \right) = 18650{\rm{ W}}\).
• The average of the car is\(A = {\rm{41 km/gal}}\).
• The speed of the car is \(v = {\rm{110 km/h}}\).

Determination of the heat released by the gasoline

The heat released per unit time can be expressed as

\(\begin{array}{c}\frac{{{Q_{\rm{H}}}}}{t} = {P_{\rm{H}}}\\{P_{\rm{H}}} = \frac{{{Q_{\rm{H}}}}}{A} \times v\end{array}\).

Here, the heat released per unit time\(\left( {{Q_{\rm{H}}}{\rm{/}}t} \right)\)is the input power of the car, which is denoted as\({P_{\rm{H}}}\).

Substituting the values in the above equation,

\(\begin{array}{c}{P_{\rm{H}}} = \frac{{1.26 \times {{10}8}{\rm{ J/gal}}}}{{{\rm{41 km/gal}}}} \times {\rm{110 km/h}}\\ = 3.38 \times {108}{\rm{ J/h}}\left( {\frac{{1{\rm{ h}}}}{{3600{\rm{ s}}}}} \right)\left( {\frac{{1{\rm{ W}}}}{{1{\rm{ J/s}}}}} \right)\\ = 93888.89{\rm{ W}}\end{array}\).

Determination of the efficiency of the car engine

The efficiency of the car engine can be expressed as

\(e = \left( {\frac{P}{{{P_{\rm{H}}}}}} \right)\).

Substituting the values in the above equation,

\(\begin{array}{c}e = \left( {\frac{{18650{\rm{ W}}}}{{93888.89{\rm{ W}}}}} \right)\\e = 0.1986\\e = 0.1986 \times 100\% \\e = 19.86\% \end{array}\).

Thus, the efficiency of the car engine is \(19.86\% \).