# A six-cylinder 4-stroke engine has bore diameter 11 cm and stroke length 12.2 cm running at 2800 rev/min consumes 455m^2 of air per hour. What would be the volumetric efficiency of the engine?

A six-cylinder 4-stroke engine has bore diameter 11 cm and stroke length 12.2 cm running at 2800 rev/min consumes 455m^2 of air per hour. What would be the volumetric efficiency of the engine?

(a) 75%

(b) 78.7%

(c) 77.8%

(d) 76%

The question was asked during an interview.

I'm obligated to ask this question of Diesel Engine Vs. Petrol Engine topic in portion Internal Combustion Engine of Farm Machinery

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The correct option is (c) 77.8%

The best I can explain: Swept volume per hour = $\frac{π}{4}$*D^2*Number of cycles per hour*Number of cylinder

= $\frac{π}{4}$[11/100]^2*[12.2/100]*2800/2*60*6

= 584.33 m^2

Volumetric efficiency = $\frac{Actual \,volume \,of \,air \,taken}{Swept \,Volume}$*100

= $\frac{455}{584.33}$*100

= 77.8%.

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