Crushing rolls: derive the theoretical capacity (tons per hour) in terms of peripheral velocity V (m/s), roll width W (m), roll gap Dr (m), and solid density ρ (kg/m^3). Use V = π N D, where N is speed in rps and D is roll diameter.

Introduction / Context:
In particle technology, predicting the theoretical throughput of crushing rolls helps with preliminary sizing and capacity checks. The rolls drag a bed of solids through the nip; capacity scales with the cross-sectional opening, the machine width, and the bed velocity, then with material density to obtain a mass rate.

Given Data / Assumptions:

• Peripheral velocity V = π · N · D (m/s).
• Effective cross-sectional opening is W (width) by Dr (gap) in metres.
• Bulk density of the solid entering the nip is ρ (kg/m^3).
• Theoretical (no slip, full loading), medium resistance neglected.

Concept / Approach:
Volumetric flow = velocity * area. Mass flow = volumetric flow * density. Convert kg/s to tons/h with factor 3.6.

Step-by-Step Solution:
Volumetric rate, Q_v = V * (W * Dr) [m^3/s].Mass rate, ṁ = Q_v * ρ = V * W * Dr * ρ [kg/s].Throughput, T = 3.6 * ṁ = 3.6 * V * W * Dr * ρ [t/h].

Verification / Alternative check:
Dimensional analysis: [V]=m/s, [W·Dr]=m^2 → m^3/s; multiply by ρ (kg/m^3) → kg/s; multiply by 3.6 → t/h. Units are consistent.

Why Other Options Are Wrong:
3.6 V · W · ρ: ignores the gap Dr; missing area term.3.6 W · Dr · ρ: ignores velocity; no transport without V.3.6 V · W · Dr / ρ: divides by density, which is unphysical for mass rate.

Common Pitfalls:
Confusing roll diameter D with gap Dr; only the opening Dr contributes to area. Also, forgetting the 3.6 factor when converting kg/s to t/h.

Final Answer:
3.6 V · W · Dr · ρ