Question

The limiting particle diameter to which the cyclone separator can separate in terms of volumetric flow (V), exhaust flow velocity (v1), loss of pressure (\(\Delta_p\)), height (L) below the outlet and the viscosity of air (\(\mu\)) can be written as

• A. (18 \(\mu\) V v1^2 \(\Delta_p\) 2\(\pi\) L)
• B. 1 (18 \(\mu\) V v1^2 \(\Delta_p\) 2\(\pi\) L)
• C. \(\sqrt{18 \mu V / v_1^2 \Delta_p 2\pi L}\)
• D. (18 \(\mu\) V v1^2 \(\Delta_p\) 2\(\pi\) L)^2

Hint:

Cyclone Limiting Diameter: All analytical cut-diameter formulas feature a SQUARE ROOT term: $d_{\text{limit}} = \mathbf{\sqrt{\frac{18 \mu V}{2\pi L v_1^2 \Delta p}}}$.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
Barth and Muschelknautz cyclone separation mechanics: balancing Stokes drag force against centrifugal force on a limiting cut-size powder particle yields the critical cut diameter expressed with square root dependence ($\sqrt{ rac{18 \mu V}{2\pi L v_1^2 \Delta p}}$).
Key Formula or Approach:
\[ \text{Centrifugal Balance: } 3 \pi \mu d_p v_r = \left( \frac{\pi d_p^3}{6} \right) \frac{v_t^2}{r} \rho_p \implies \mathbf{d_{\text{limit}}} = \mathbf{\sqrt{\frac{18 \mu V}{v_1^2 \Delta_p \cdot 2\pi L}}} \]

Step 2: Detailed Explanation:
In aerosol physics and cyclone dust collector analytical modeling:
- A particle of limiting diameter $d_p$ suspended in a swirling vortex experiences:
1. Radial outward centrifugal force: $F_c = \frac{\pi d_p^3}{6} \rho_p \frac{v_t^2}{r}$.
2. Inward fluid drag force (Stokes' Law): $F_d = 3 \pi \mu d_p v_r$.
- Equating centrifugal force and drag force at the boundary of the inner vortex core of height $L$ and flow area $2\pi r L$:
- Substituting volumetric air flow rate $V$, exhaust exit velocity $v_1$, pressure loss $\Delta p$, and gas viscosity $\mu$ yields the classical critical diameter relationship:
\[ \mathbf{d_{pc} = \sqrt{\frac{18 \; \mu \; V}{v_1^2 \; \Delta_p \; 2\pi L}}} \]
- Notice the fundamental square root functional dependency characteristic of all aerodynamic cut-size formulas.

Step 3: Final Answer:
Therefore, the limiting particle diameter is written as \(\sqrt{18 \mu V / v_1^2 \Delta_p 2\pi L}\), corresponding to option (C).