Subtopic: I.C. engine systems – fuel, cooling, lubrication, ignition, electrical, intake and exhaust
Question Statement:
A four stroke cycle engine has the following valve events: inlet valve opens at 8° before HDC; inlet valve closes at 55° after CDC; exhaust valve opens at 47° before CDC; exhaust valve closes at 12° after HDC. If the engine runs at 2000 rpm, the time in milli-seconds during which inlet and exhaust valves remain closed simultaneously is
Options:
(A) 19.67
(B) 21.50
(C) 40.58
(D) 80.67
Official Answer Key:
B
Step-by-Step Solution & Explanation
Taking HDC (start of intake) as 0°: IVC = 180°(BDC)+55° = 235°. EVO = 540°(BDC of exhaust stroke)−47° = 493°. Both valves remain closed together from IVC (235°) until EVO (493°): duration = 493−235 = 258° of crank rotation. Time = (258/360)×(60/2000)×1000 = 21.5 ms.
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