Classifier capacity for solids throughput: which expression (in tons of solids per hour) is dimensionally and practically correct? (A = cross-sectional area, m^2; V = rising fluid velocity, m/s; S = volume fraction solids; ρ = solids density, kg/m^3.)

Introduction / Context:
Classifier capacity relates volumetric flux through cross-sectional area to solid content and density to yield mass rate. Converting SI to tons/hour introduces the factor 3.6 (since 1 kg/s = 3.6 t/h).

Concept / Approach:
Volumetric flow of mixture through area A at velocity V is Q = A * V (m^3/s). The solids volume flow is Q_s = S * Q = A * V * S (m^3/s). Multiplying by solids density ρ (kg/m^3) gives mass flow m_s = A * V * S * ρ (kg/s). Converting to tons/hour multiplies by 3.6, giving 3.6 * A * V * S * ρ (t/h).

Step-by-Step Solution:
Compute Q = A V.Compute solids volumetric flux: S * Q.Compute solids mass rate: multiply by ρ.Convert kg/s to t/h with factor 3.6.

Why Other Options Are Wrong:
Omitting S ignores the fraction of solids.Dividing by ρ or dropping V produces incorrect dimensions.

Final Answer:
3.6 A V S ρ