Concept:
A spark-ignition engine works upon Otto cycle
Air standard, efficiency for an Otto cycle is given by \({\eta = 1 - \frac{1}{{{r^{\gamma - 1}}}}} \)
Where ‘r’ is compression ratio, γ = 1.4 for air
Efficiency is also given by
\(\eta = \frac{{Power\;output}}{{Heat\;input}}\)
Heat inpur = Q = (mass flow rate of fuel) × (calorific value of fuel) = ṁ × cv
Calculation:
Given, ṁ = 10.3 kg/hr, cv = 44,000 kJ/kg, Power output = 70 kW
\(Q = \left( {\frac{{10.3}}{{3600}} × {44,000}} \right)kW = \;125.89\;KW\)
\(\eta = \frac{{70}}{{125.89}} = 0.556\)
Now,
\(0.556 = 1 - \frac{1}{{{r^{\left( {1.4 - 1} \right)}}}} \Rightarrow \frac{1}{{{{\text{r}}^{0.4}}}} = 0.444\)
⇒ r = 7.61