Solids Handling practice questions

Answer: A) Particle size, density and shape

Dry blends segregate mainly because their particles differ in size, density and shape. Size is usually the strongest factor, because fines percolate through the gaps between coarser particles.

• ACorrect. Mixed solids tend to segregate because their particles differ in size, density and shape. Size is usually the dominant factor, because fines sift (percolate) through the gaps between coarser particles.
• BIncorrect. Moisture affects how cohesive and free-flowing a powder is, but it is not the main mechanism that separates the components of a free-flowing blend.
• CIncorrect. These chemical properties do not drive the mechanical separation of particles during flow, vibration or conveying.
• DIncorrect. Thermal properties have little effect on segregation during ambient handling. Physical differences between particles drive it.

Answer: A) Energy for crushing and grinding with a work index

• ACorrect. The work index comes from standard tests.
• BIncorrect. That uses filtration equations.
• CIncorrect. That's Stokes' law.
• DIncorrect. That uses the Ergun equation.

Answer: A) 1

• ACorrect. Other shapes have sphericity below 1.
• BIncorrect. A sphere has the maximum value, 1.
• CIncorrect. Sphericity is a ratio at most 1.
• DIncorrect. Sphericity is a ratio at most 1.

Answer: C) Volume-to-surface-area ratio as the whole population

The Sauter mean diameter D32 = Σn·d³/Σn·d² is the diameter of a sphere with the same volume-to-surface ratio as the whole population. It is different from the number (D10), surface (D20) and volume (D30) mean diameters.

• AIncorrect. That describes the arithmetic (number-length) mean diameter, D10 = Σn·d/Σn. It does not weight particles by surface or volume.
• BIncorrect. That describes the surface mean diameter, D20 = (Σn·d²/Σn)^1/2.
• CCorrect. D32 = Σn·d³/Σn·d². It is used where specific surface controls the process, such as reactions, mass transfer, drying and sprays.
• DIncorrect. That describes the volume mean diameter, D30 = (Σn·d³/Σn)^1/3.

Answer: A) 1,500 kg/m³

Bulk density equals particle density times the solids volume fraction: ρbulk = ρp(1 − ε) = 2,500 × 0.6 = 1,500 kg/m³.

• ACorrect. Bulk density = ρp(1 − ε) = 2,500 × 0.6 = 1,500 kg/m³. This ignores the negligible mass of the gas in the voids.
• BIncorrect. This multiplies by the void fraction (0.4). The solids take up the fraction 1 − ε of the volume, not ε.
• CIncorrect. This is the particle (true) density. The voids make the bulk density lower than the particle density.
• DIncorrect. This divides by (1 − ε) instead of multiplying. Bulk density can never be higher than the particle density.

Answer: A) Undersize, or fines

• ACorrect. Retained particles are oversize.
• BIncorrect. Those are retained.
• CIncorrect. That's a different process.
• DIncorrect. That isn't screening terminology.

Answer: A) Centrifugal force

• ACorrect. Swirling flow throws particles to the wall.
• BIncorrect. That's an electrostatic precipitator.
• CIncorrect. That's a baghouse.
• DIncorrect. It's a mechanical separator.

Answer: A) The new surface area created by grinding

Rittinger's law (E ∝ 1/D2 − 1/D1) bases grinding energy on the new surface created and suits fine grinding. Kick's law uses the reduction ratio (coarse crushing), and Bond's law uses 1/√D (intermediate sizes).

• ACorrect. Rittinger's law gives E = K_R(1/D2 − 1/D1), where D1 is the feed size and D2 is the product size. This is proportional to the new specific surface created. It applies best to fine grinding.
• BIncorrect. This is Kick's law, E = K_K·ln(D1/D2). Under Kick's law, equal reduction ratios need equal energy. It applies best to coarse crushing.
• CIncorrect. This is the basis of Bond's law, W = Wi·(10/√P − 10/√F). Bond's law is the intermediate case between Kick's and Rittinger's laws.
• DIncorrect. Total power does scale with feed rate, but every comminution law makes energy per unit mass depend on the feed and product sizes.

Answer: C) It replaces g with a centrifugal acceleration, ω²r, that is many times g

Settling velocity is proportional to the acceleration acting on the particle. A centrifuge replaces g with ω²r, which is many times larger, so particles separate much faster.

• AIncorrect. Spinning the slurry does not change the particle or liquid densities, so the density difference (ρp − ρ) stays the same.
• BIncorrect. Any shear heating is small and incidental. A centrifuge is not designed to work by changing viscosity.
• CCorrect. In the settling-velocity relation (for example, Stokes' law), velocity is proportional to the acceleration. Replacing g with ω²r, which can be hundreds or thousands of times g, makes settling that much faster.
• DIncorrect. Flocculation comes from chemical additives, not from the centrifuge itself. High shear in a centrifuge can actually break up flocs.

Answer: D) Smaller openings

• AIncorrect. It means smaller openings.
• BIncorrect. Higher mesh means finer screens.
• CIncorrect. It means smaller openings.
• DCorrect. More wires per inch leave smaller openings.