Subtopic: Calculation of power, torque, fuel consumption, heat load and power losses
Question Statement:
A four-stroke single cylinder diesel engine with $120\text{ mm}$ bore and $150\text{ mm}$ stroke operates at $1500\text{ rpm}$. The engine takes in $0.012\text{ kg/s}$ of air. If ambient air density is $1.18\text{ kg/m}^3$, calculate the volumetric efficiency of the engine in percentage.
Answer Key:
80.3
Step-by-Step Solution & Explanation
Swept volume of cylinder:
$V_s = \frac{\pi}{4} D^2 L = \frac{\pi}{4} (0.12)^2 (0.15) = 0.0016965\text{ m}^3$
Theoretical volume displaced per second for four-stroke ($N/2$ cycles/min):
$V_{\text{theoretical}} = V_s \times \frac{N}{2 \times 60} = 0.0016965 \times \frac{1500}{120} = 0.0016965 \times 12.5 = 0.021206\text{ m}^3/\text{s}$
Actual volume of air ingested per second:
$V_{\text{actual}} = \frac{\text{Mass rate}}{\text{density}} = \frac{0.012\text{ kg/s}}{1.18\text{ kg/m}^3} = 0.0101695\text{ m}^3/\text{s}$
Volumetric efficiency:
$\eta_v = \frac{V_{\text{actual}}}{V_{\text{theoretical}}} \times 100 = \frac{0.0101695}{0.021206} \times 100 \approx 80.3\%$