Question

An elutriator is used to separate particles based on their sizes in flowing air as shown in the figure. Assuming spherical particles, the diameter (\(D_{50}\)) of the suspended particles which have 50% chance to report to overflow by turbulent air flow is expressed as 

• A. \( D_{50} = \frac{3 f v^2 \rho_a}{4 g (\rho_s - \rho_a)} \)
• B. \( D_{50} = \left( \frac{18 \mu v}{g f (\rho_s - \rho_a)} \right)^{0.5} \)
• C. \( D_{50} = \left( \frac{9 \mu v}{2 g (\rho_s - \rho_a)} \right)^2 \)
• D. \( D_{50} = \frac{3 f v \rho_a}{8 g (\rho_s - \rho_a)} \)

Hint:

For particle suspension problems, the core concept is always the balance of forces: Drag Force vs. Gravitational Force. Identify the flow regime (laminar/Stokes or turbulent/Newton) to choose the correct expression for the drag force. Turbulent drag is proportional to \(v^2\), while laminar drag is proportional to \(v\). The presence of \(v^2\) in the correct option confirms a turbulent regime.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
An elutriator separates particles by size using an upward-flowing fluid (air in this case). Smaller particles are carried upwards (overflow) while larger particles settle downwards (underflow). The separation occurs when the upward drag force exerted by the fluid on a particle balances the downward gravitational force. The \(D_{50}\) is the "cut size," where a particle has an equal chance of going up or down.
Step 2: Key Formula or Approach:
The condition for a particle to be suspended is that the upward drag force (\(F_D\)) equals the net downward gravitational force (buoyancy-corrected weight, \(F_g\)). \[ F_D = F_g \] The forces are defined as:
- Drag Force (\(F_D\)): For a turbulent flow regime, the drag force is given by \( F_D = C_D A_p \frac{\rho_a v^2}{2} \), where \(C_D\) is the drag coefficient, \(A_p\) is the projected area of the particle, \(\rho_a\) is the air density, and \(v\) is the air velocity. The question gives an "air/solid friction factor \(f\)", which relates to the drag coefficient. Often, \(f = C_D/4\), but here it seems to be used directly in the context of the drag force equation developed for turbulent pipe flow, which is adapted here. A common form for turbulent drag relates force to \(f\). The standard drag equation is \(F_D = C_D \frac{\pi D^2}{4} \frac{\rho_a v^2}{2}\).
- Gravitational Force (\(F_g\)): The net gravitational force is the weight of the particle minus the buoyant force: \( F_g = V_p (\rho_s - \rho_a) g \), where \(V_p\) is the particle volume, \(\rho_s\) is the particle density, and \(g\) is the acceleration due to gravity.
Step 3: Detailed Calculation:
For a spherical particle of diameter D:
- Volume \(V_p = \frac{\pi D^3}{6}\)
- Projected Area \(A_p = \frac{\pi D^2}{4}\)
- Net Gravitational Force: \(F_g = \frac{\pi D^3}{6} (\rho_s - \rho_a) g\)
- Drag Force: The options suggest a relationship involving \(f\). A common definition for drag in turbulent regimes, analogous to pipe flow friction, gives the drag stress as \( \tau = f \frac{\rho_a v^2}{2} \). The force is this stress times the area. Let's use the standard drag formula and see how \(f\) relates. The drag force in turbulent flow is often expressed using a friction factor. Let's assume the drag force is given in a form consistent with the options: \(F_D = \frac{\pi D^2}{4} . f . \frac{\rho_a v^2}{2}\). This is a non-standard form. A more likely form is \(F_D = (\text{some constant}) \times f \times A_p \times \rho_a v^2\). Let's try to work from the answer. Option (A) is \( D_{50} = \frac{3 f v^2 \rho_a}{4 g (\rho_s - \rho_a)} \).
Let's rearrange it: \( D_{50} g (\rho_s - \rho_a) = \frac{3 f v^2 \rho_a}{4} \).
The left side is proportional to force per unit volume. The right side is proportional to pressure. This doesn't seem to balance forces directly.
Let's retry the force balance \(F_D = F_g\).
\[ C_D \frac{\pi D^2}{4} \frac{\rho_a v^2}{2} = \frac{\pi D^3}{6} (\rho_s - \rho_a) g \] Let's simplify:
\[ \frac{C_D \rho_a v^2}{8} = \frac{D g (\rho_s - \rho_a)}{6} \] \[ D = \frac{6 C_D \rho_a v^2}{8 g (\rho_s - \rho_a)} = \frac{3 C_D \rho_a v^2}{4 g (\rho_s - \rho_a)} \] Step 4: Final Answer:
By equating the turbulent drag force with the buoyant weight of the particle and assuming the friction factor \(f\) is equivalent to the drag coefficient \(C_D\), we arrive at the expression in option (A).
Step 5: Why This is Correct:
The derivation is based on a fundamental force balance that governs the suspension of particles in a fluid stream. The resulting formula matches option (A) under the reasonable assumption that \(f\) represents the particle drag coefficient. The other options correspond to different flow regimes (e.g., Stokes' law for laminar flow) or are dimensionally inconsistent.