Rate controlling step: halving the particle moves nothing, doubling the velocity does

Type three resistances on one basis. Halving the particle moves only the pore term and doubling the velocity only the film term, and two of the worked terms end too close to separate.

Live result

At the typed reference, the largest resistance is consistent with Surface kinetics. Halving size gives 1.14 times the rate and doubling velocity gives 1.09 times.

Typed resistances and response sweep

Sweep rail

Try a scenario

Purpose

The mechanism request is outside this instrument and stops the finding.

Typed resistance case

Use one consistent resistance basis throughout. No unit conversion is applied.

Surface resistance

Pore resistance at reference

Film resistance at reference

Reference particle size

Reference external velocity

Particle size exponent

Velocity exponent

Driving force

Tie tolerance

Particle size sweep

Start

End

Step

External velocity sweep

Start

End

Step

Tracked answer and response fingerprint

Resistance	Typed value
External film transport	1
Diffusion inside pores	1
Surface kinetics	4

Observed rate: 1
Rate after halving particle size: 1.143
Rate after doubling velocity: 1.091

Candidates within tolerance: Surface kinetics
Other typed terms: External film transport, Diffusion inside pores

First checked changes

Particle size rail: First checked value: 2, No single resistance
Velocity rail: No change on the checked grid

The boundary is three typed resistance terms on one consistent basis, their reference size and velocity, two declared scaling exponents, and checked sweep grids. First changes are sampled values, not interpolated boundaries.

Here is the maths

The formula, in words

Pore resistance scales with particle size raised to its exponent. Film resistance scales with the inverse velocity ratio raised to its exponent. Surface resistance stays fixed, and observed rate equals driving force divided by the sum of all three.

R_p = R_{p,\mathrm{ref}} \left(\frac{d}{d_{\mathrm{ref}}}\right)^p

where $R_p$ is pore resistance, $R_{p,\mathrm{ref}}$ its reference value, $d$ the particle size, $d_{\mathrm{ref}}$ the reference size, and $p$ its exponent.

R_f = R_{f,\mathrm{ref}} \left(\frac{v_{\mathrm{ref}}}{v}\right)^q

where $R_f$ is film resistance, $R_{f,\mathrm{ref}}$ its reference value, $v_{\mathrm{ref}}$ the reference velocity, $v$ the velocity, and $q$ its exponent.

r = \frac{F}{R_f + R_p + R_s}

where $r$ is the observed rate, $F$ the driving force, and $R_s$ surface resistance.

A worked example

At reference size 1 and velocity 1, type surface resistance 4, pore resistance 1, film resistance 1, and zero tie tolerance. Surface is the largest term.

At size 1.5, pore resistance is 1 times 1.5 squared, or 2.25, so surface remains largest.

At size 2, pore resistance reaches 4 and ties surface resistance. A 0.5 step therefore names size 2 as the first checked change and returns no single resistance.

What this does not model

The terms are typed on one basis. The model does not convert units, fit kinetic constants, represent every transport effect, or establish a physical mechanism.

Check it yourself

Halving particle size changes only the pore term. Doubling external velocity changes only the film term. Their two rate ratios are the response fingerprint.

Sources

Model version

1.0.0

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A response pattern is consistent with a typed resistance regime and establishes no mechanism. One sensitivity cannot settle rate control, and no resistance is fitted from plant data here.

Resistance response record