Particle size descriptors: which mean diameter corresponds to the “volume–surface mean” (often denoted D[3,2]) used when correlating surface-related phenomena?

Difficulty: Medium

Correct Answer: Volume surface

Explanation:


Introduction / Context:
Particle size distributions can be represented by many different “means,” each emphasizing a particular physical effect. For heat transfer, reaction, dissolution, or any phenomenon linked to particle surface, the volume–surface mean diameter (D[3,2]) is frequently used because it weights particles by volume in the numerator and by surface in the denominator.


Given Data / Assumptions:

  • We seek the name of the mean diameter whose definition involves volume in the numerator and surface in the denominator.
  • No detailed formula is asked for, only the correct descriptor.
  • Standard nomenclature D[3,2] corresponds to volume–surface mean.


Concept / Approach:
The family D[p,q] uses p for the power in the numerator and q for the denominator. For spheres: volume ∝ d^3 and surface ∝ d^2. Hence D[3,2] = (Σ n_i d_i^3 / Σ n_i d_i^2). It is particularly meaningful when processes are controlled by surface area but inventory or feed measurements relate to mass/volume.


Step-by-Step Solution:
Identify target mean: volume–surface (D[3,2]).Match to option text: “Volume surface”.Exclude alternatives: arithmetic mean (D[1,0]), “mass” or “volume” alone are ambiguous labels here.


Verification / Alternative check:
In dissolution or reaction-rate correlations, surface-specific rates scale well with D[3,2], validating its selection when surface area is the controlling factor.


Why Other Options Are Wrong:
Mass/Volume (alone): imprecise and do not distinguish how powers weight the distribution.Arithmetic: emphasizes number averaging and is rarely appropriate for surface-controlled phenomena.


Common Pitfalls:
Using the Sauter mean incorrectly; note that “Sauter mean diameter” is another name for D[3,2] in many texts—do not confuse it with simple volumetric or arithmetic means.


Final Answer:
Volume surface

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