A six-cylinder 4-stroke engine has bore diameter 11 cm and stroke length 12.2 cm running at 2800 rev/min consumes 455m² of air per hour. What would be the volumetric efficiency of the engine?
a) 75%
b) 78.7%
c) 77.8%
d) 76%

Answer: c
Explanation: Swept volume per hour = \(\frac{π}{4}\)*D²*Number of cycles per hour*Number of cylinder
= \(\frac{π}{4}\)[11/100]²*[12.2/100]*2800/2*60*6
= 584.33 m²
Volumetric efficiency = \(\frac{Actual \,volume \,of \,air \,taken}{Swept \,Volume}\)*100
= \(\frac{455}{584.33}\)*100
= 77.8%.