Difficulty: Medium
Correct Answer: 230
Explanation:
Introduction / Context:
In gas–solid separation, the cyclone's ability to drive particles toward the wall is often expressed using a separation factor (also termed the centrifugal acceleration ratio). It compares the centrifugal acceleration acting on particles at a given radius to gravitational acceleration. This dimensionless number helps relate operating conditions to expected cut size and collection efficiency for a given particle density and size range.
Given Data / Assumptions:
Concept / Approach:
Centrifugal acceleration at radius r is a_c = v_t^2 / r. The separation factor S compares a_c to g: S = a_c / g = v_t^2 / (r * g). Larger S generally implies stronger outward particle forcing and potentially smaller achievable cut size, all else equal (geometry, inlet design, and dust properties).
Step-by-Step Solution:
Compute radius: r = 0.2 / 2 = 0.1 m.Compute v_t^2: 15^2 = 225 m^2/s^2.Compute denominator: r * g = 0.1 * 9.81 = 0.981.Evaluate S = 225 / 0.981 ≈ 229.46 ≈ 230 (rounded).
Verification / Alternative check:
The order of magnitude is typical for small industrial cyclones at moderate velocities (hundreds of g's). Using a representative wall radius is standard practice for quick estimates; more detailed designs consider inlet configuration, vortex finder, and actual velocity profiles.
Why Other Options Are Wrong:
2250, 1125, 460: These values are one or more orders too large for the given v_t and diameter. They would imply unrealistically high accelerations for the stated conditions.
Common Pitfalls:
Using cyclone body diameter instead of radius in the denominator; forgetting to square the velocity; or mixing units (e.g., cm vs m) can inflate S by factors of 10–100 erroneously.
Final Answer:
230
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