Question

An engine is operating at 600 rpm. The inlet valve opens just at TDC and closes \(20^\circ\) after BDC. Calculate the time for which inlet valve remains open

• A. \(1/12\text{ second}\)
• B. \(1/16\text{ second}\)
• C. \(1/18\text{ second}\)
• D. \(1/20\text{ second}\)

Hint:

Formula shortcut: $t = \frac{\theta}{6N}$. Here $t = \frac{200}{6 \times 600} = \frac{200}{3600} = \frac{1}{18}\text{ s}$.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
Valve opening duration in an IC engine corresponds to a specific angular rotation of the crankshaft at a given rotational engine speed.
Key Formula or Approach:
\[ \text{Crankshaft angular velocity } (\omega) = N \times \frac{360^\circ}{60\text{ s}} = 6 N \, (\text{degrees/second}) \]
\[ \text{Time } (t) = \frac{\theta_{\text{open}}}{\omega} = \frac{\theta_{\text{open}}}{6N} \]

Step 2: Detailed Explanation:
Total crankshaft angle during which the inlet valve is open:
- From Top Dead Center (TDC) to Bottom Dead Center (BDC) = \(180^\circ\)
- Valve remains open until \(20^\circ\) after BDC:
\[ \theta = 180^\circ + 20^\circ = 200^\circ \]
Engine speed \(N = 600\text{ rpm}\).
Angular speed in degrees per second:
\[ \omega = 6 \times 600 = 3600^\circ\text{/second} \]
Time duration for which the valve remains open:
\[ t = \frac{200^\circ}{3600^\circ\text{/s}} = \frac{2}{36} = \frac{1}{18}\text{ second} \]

Step 3: Final Answer:
Thus, the time for which the inlet valve remains open is \(1/18\text{ second}\), matching option (C).