Flight conveyor sizing: derive the capacity in tons/hour in terms of width W (m), depth D (m), belt/flight speed V (m/s), and bulk density ρ (kg/m^3). Which option matches the standard form?

Introduction / Context:
Designing a flight (scraper) conveyor requires a quick estimate of its mass throughput. The capacity depends on the cross-sectional area of material carried, its bulk density, and the conveying velocity. This question reinforces the basic capacity relation and the unit conversion factor used in plant calculations.

Given Data / Assumptions:

• Cross-section approximated as W * D for filled section.
• Speed V is uniform along the conveyor.
• Bulk density ρ is constant and expressed in kg/m^3.
• Desired capacity unit: tons/hour.

Concept / Approach:
Mass flow rate m_dot equals volumetric flow rate times bulk density. Volumetric flow rate for a prismatic cross-section is area * velocity. With SI units, m_dot is in kg/s. Converting kg/s to tons/hour introduces the factor 3.6 because 1 kg/s equals 3.6 t/h (1000 kg per ton and 3600 s per hour).

Step-by-Step Solution:
Cross-sectional area A = W * D (m^2).Volumetric rate Q = A * V = W * D * V (m^3/s).Mass rate m_dot = Q * ρ = W * D * V * ρ (kg/s).Convert to tons/hour: Capacity = 3.6 * W * D * V * ρ (t/h).

Verification / Alternative check:
Dimensional analysis: [W][D] gives m^2; multiplied by V (m/s) → m^3/s; times ρ (kg/m^3) → kg/s. Multiply by 3.6 → t/h. The expression matches plant datasheets for flight/drag conveyors under full cross-section loading.

Why Other Options Are Wrong:
Options omitting ρ cannot yield mass capacity; those omitting either W or D do not represent area.

Common Pitfalls:
Using loose fill depth rather than effective cross-section; forgetting de-rating for slope or return flights; confusing metric ton with short ton (use consistent units).

Final Answer:
3.6 W.D.V.ρ