ABE TUTORIALS

AREA 1 PROBLEM 53: ENGINE VALVE TIMING

THE INTAKE VALVE OF AN ENGINE STARTS TO OPEN AT 12 DEGREES BEFORE TOP DEAD CENTER POSITION OF THE CRANKSHAFT AND CLOSES AT 43 DEGREES AFTER BOTTOM DEAD CENTER. IF THE TIME DURING EACH INLET VALVE OPENING IS 0.0208 SECOND, ESTIMATE THE CRANKSHAFT SPEED IN REVOLUTIONS PER MINUTE.

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a. 1,883 rpm

b. 1,553 rpm

c. 1,223 rpm

d. 993 rpm

CORRECT ANSWER: A. 1,883 RPM

Solution:
• Intake Opens: 12° BTDC
• Intake Closes: 43° ABDC
• Opening Time (t) = 0.0208 second

Step 1: Calculate Total Degrees of Opening (θ)
θ = (Degrees BTDC) + 180° + (Degrees ABDC)
θ = 12° + 180° + 43° = 235°

Step 2: Calculate Angular Speed (ω)
ω = θ / t = 235° / 0.0208 s = 11,298.08 degrees/second

Step 3: Convert to RPM
Speed (RPM) = (ω × 60 s/min) / 360 degrees/rev
Speed (RPM) = (11,298.08 × 60) / 360
Speed (RPM) = 677,884.8 / 360 = 1,883 rpm

The crankshaft must rotate through the sum of the lead, the standard stroke (180°), and the lag.