Question

An engine has a swept volume of \(300\ cm^3\), clearance volume of \(25\ cm^3\). Its volumetric efficiency is 0.80 and mechanical efficiency is 0.90. The volume of mixture taken in per stroke is

• A. \(325\ cm^3\)
• B. \(275\ cm^3\)
• C. \(240\ cm^3\)
• D. \(270\ cm^3\)

Hint:

Volumetric efficiency links only the actual charge drawn in to the swept volume, not every given number is needed.

Answer 1

The Correct Option is C

Step 1: Volumetric efficiency is defined as the ratio of the actual volume of fresh charge drawn into the cylinder per stroke to the swept volume: \(\eta_v = \dfrac{V_{actual}}{V_s}\).

Step 2: Here \(V_s = 300\ cm^3\) and \(\eta_v = 0.80\), so \(V_{actual} = 0.80 \times 300 = 240\ cm^3\).

Step 3: The clearance volume of 25 cm3 is only needed to work out the compression ratio, and the mechanical efficiency of 0.90 relates brake power to indicated power, not the charge drawn in. Both values are distractors and not needed for this calculation.

So the volume of mixture taken in per stroke is 240 cm3, option 3.