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Solids Handling

3% of exam

Particle properties and size distributions, surface and bulk forces, solids processing (crushing, grinding, and crystallization), and transportation and storage (belt and pneumatic conveying, slurries, tanks, and hoppers).

2 concepts

A. Particle properties

Particle Size, Distributions and Comminution

How to describe a particle-size distribution, compute the right mean diameter, relate surface area to size, and apply the Rittinger/Kick/Bond grinding-energy laws and crystallization basics.

A powder is never one size — it is a ·distribution· of sizes, and almost every property you care about (how fast it dissolves, how it settles, how much it costs to grind, how a catalyst performs) depends on which feature of that distribution you weight by. The single most common FE error in solids work is quoting an arithmetic average diameter when the physics demands a surface- or volume-weighted mean, and they can differ by tens of percent. This concept builds the vocabulary the exam expects: density functions

f_N,f_V

and cumulative functions

F_N,F_V

, the family of mean diameters, the inverse-size scaling of surface area, and the energy it costs to make small particles out of big ones. The handbook gives the distribution and mean-diameter machinery in the ·Chemical Engineering — Solids Processing· section (particle-size distributions), and it does print a crushing/grinding equipment-selection table and a hydrate-formation phase diagram for crystallization. What it does ·not· tabulate are the specific comminution energy-law equations (Rittinger/Kick/Bond and the work index) and the crystallization supersaturation/yield relations, so memorize those from references such as Perry's or McCabe.

Density and cumulative distribution functions

Two equivalent pictures describe a PSD. The **density function**

f

tells you the fraction of the sample falling in a narrow band of size around

x

; the **cumulative function**

F

tells you the fraction smaller than

x

, so

F

rises monotonically from 0 to 1 and

f = dF/dx

. Crucially, the answer depends on the ·basis·: a number basis (

f_N,F_N

, counting particles) and a volume or mass basis (

f_V,F_V

, weighting by particle volume) give very different curves, because one large particle carries the mass of thousands of fines. In discrete form the size range is split into

m

sub-ranges, each with a mean size

x_i

and a fraction

\Delta F_{Ni}

(number) or

\Delta F_{Vi}

(volume); every sum of fractions is 1. The descriptors

d_{10},d_{50},d_{60}

— the sizes at which 10%, 50%, 60% of the sample is smaller — come straight off the cumulative curve, and

d_{60}/d_{10}

is the uniformity coefficient.

f = \dfrac{dF}{dx}, \qquad F(x)=\int_0^{x} f\,dx, \qquad \sum_{i=1}^{m}\Delta F_i = 1

Mean diameters depend on what you weight by

There is no single 'average' particle size — there is a family of means, each correct for a different purpose. The **number-length (mean-length) diameter**

X_{ML}=\sum x_i\,\Delta F_{Ni}

is the plain number average. The **Sauter (surface-mean) diameter**

X_{SM}

is the diameter of a sphere with the same surface-area-to-volume ratio as the whole sample; it is the right size for any rate process governed by surface — dissolution, drying, catalysis, two-film mass transfer — and it is the one examiners most often want. The **volume (mass-mean) diameter**

X_{MV}

weights toward the coarse end and governs how mass is distributed. Because

X_{SM}

and

X_{MV}

over-weight large particles relative to

X_{ML}

— and the Sauter mean leans on the coarse tail most heavily of all — you will always find

X_{ML}\le X_{MV}\le X_{SM}

for the same powder (a power-mean/moment-ratio result, with equality only for a perfectly monodisperse powder); quoting the wrong one is a classic trap.

X_{ML}=\sum_i x_i\,\Delta F_{Ni}, \qquad X_{SM}=\dfrac{\sum_i x_i^{3}\,\Delta F_{Ni}}{\sum_i x_i^{2}\,\Delta F_{Ni}}=\dfrac{1}{\sum_i \frac{1}{x_i}\Delta F_{Vi}}, \qquad X_{MV}=\left(\sum_i x_i^{3}\,\Delta F_{Ni}\right)^{1/3}

Specific surface area scales as 1/d

For a sphere of diameter

d

, surface area is

\pi d^{2}

and volume is

\pi d^{3}/6

, so the surface area per unit volume of solid is

a_v = 6/d

and per unit mass

a_m = 6/(\rho_p d)

. The key qualitative fact for the exam: **halve the diameter and you double the specific surface area**. Real particles are not spheres, so we fold in the **sphericity**

\Phi_s

— the ratio of the surface area of the equal-volume sphere to the actual particle surface area (

\Phi_s=1

for a sphere,

\approx 0.6

–

0.8

for crushed solids). The surface area per volume becomes

a_v = 6/(\Phi_s d)

, which is exactly the

\Phi_s

that appears in the Ergun packed-bed equation. This

1/d

scaling is why grinding to a finer product is so valuable for reaction and dissolution rates — and so expensive in energy, as the next section shows.

a_v=\dfrac{6}{\Phi_s\,d}, \qquad a_m=\dfrac{6}{\Phi_s\,\rho_p\,d}, \qquad \Phi_s=\dfrac{\text{surface area of equal-volume sphere}}{\text{actual particle surface area}}

Comminution and the energy laws

Size reduction (crushing, grinding, milling) is notoriously inefficient — a few percent of the input energy actually creates new surface. The three classical laws all come from the same differential statement that the energy to make a small reduction

dx

scales as a power of size,

dE/dx=-C\,x^{-n}

, integrated between feed and product sizes. **Rittinger's law** (

n=2

) says energy is proportional to the ·new surface created· and works best for fine grinding. **Kick's law** (

n=1

) says energy depends only on the size ·ratio· and suits coarse crushing of large pieces. **Bond's law** (

n=1.5

) sits between them and, with its empirical work index, is the workhorse for the rod- and ball-mill range that dominates industrial practice. Notice all three predict that pushing to ever-finer products costs disproportionately more energy.

\dfrac{dE}{dx}=-C\,x^{-n}\;\Longrightarrow\; E_{Ritt}=K_R\!\left(\dfrac{1}{x_2}-\dfrac{1}{x_1}\right),\quad E_{Kick}=K_K\ln\dfrac{x_1}{x_2},\quad E_{Bond}=K_B\!\left(\dfrac{1}{\sqrt{x_2}}-\dfrac{1}{\sqrt{x_1}}\right)

Bond work index in practice

Bond cast his law in a form built around a measurable material property, the **work index**

W_i

— defined as the energy (in kWh per short ton) needed to reduce a very large feed to a product that is 80% finer than 100 micrometers. With the feed and product characterized by their 80%-passing sizes

F_{80}

and

P_{80}

in micrometers, the specific grinding energy follows the equation below; multiplying by the mass throughput gives the required mill power. Typical work indices run from about 6–7 kWh/short ton for genuinely soft materials like gypsum or clay, ~11.6–12.7 for limestone, to 15+ for hard rocks like granite or trap rock. Watch the units — Bond's classic constants assume micrometers and short tons (2000 lb), so an SI throughput in t/h must be converted, or the result will be off by the ton-conversion factor.

W = 10\,W_i\!\left(\dfrac{1}{\sqrt{P_{80}}}-\dfrac{1}{\sqrt{F_{80}}}\right)\qquad [\text{kWh/short ton},\ P_{80},F_{80}\ \text{in }\mu\text{m}]

Crystallization: supersaturation, nucleation, growth, and yield

Crystallization makes particles instead of breaking them, and its driving force is **supersaturation** — a solution holding more solute than its equilibrium solubility

C^{*}

. The supersaturation ratio

S=C/C^{*}

(or the difference

\Delta C=C-C^{*}

) drives two competing rate processes: **nucleation** (birth of new crystals, very steep in

S

) and **growth** (enlargement of existing crystals). Operating inside the ·metastable zone· — modest supersaturation, reached by cooling or evaporation — favors growth over nucleation and yields large, pure, easily filtered crystals; blow past it and you get a shower of fines. The recoverable **yield** comes from a mass balance: solute in feed minus solute left dissolved at the final solubility equals solute crystallized, and if the solid is a hydrate you must charge the crystal phase with its water of crystallization (the hydrate-to-anhydrous molar-mass ratio) and credit any evaporated solvent. That mass balance, not a memorized formula, is what the exam tests.

S=\dfrac{C}{C^{*}}, \qquad \Delta C = C-C^{*}, \qquad m_{\text{crystal}}=\dfrac{(\text{solute fed})-(\text{solute in mother liquor})}{\text{anhydrous mass fraction of crystal}}

Exam strategy

First decide the **basis** the question wants: number, surface, or volume/mass. If the process is rate- or surface-limited, reach for the Sauter mean

X_{SM}

; if it is about mass holdup, use

X_{MV}

. Build a small table of

x_i,\,\Delta F_{Ni},\,x_i^{2}\Delta F_{Ni},\,x_i^{3}\Delta F_{Ni}

and the means fall out of column sums — do not average the diameters directly. For specific surface area, remember the lone fact

a=6/(\Phi_s d)

and that it scales as

1/d

. For comminution, identify the regime (fine

\to

Rittinger, coarse

\to

Kick, mid-range/work-index data

\to

Bond) and keep Bond's micrometre-and-short-ton units straight. For crystallization, draw the box, label solute and water on every stream, and don't forget the water of crystallization. Scope note: this concept is filed under ·A. Particle properties·, but its comminution (crushing/grinding) and crystallization halves belong to the NCEES Solids spec sub-area ·B. Processing· — treat this single entry as spanning both A and B so the B-area outline is not left uncovered.

Key equations

Number-length (mean-length) diameter

X_{ML}=\int_0^1 x\,dF_N=\sum_{i=1}^{m} x_i\,\Delta F_{Ni}

Plain number-average particle size. $x_i$ = mean size of sub-range $i$ ( $\mu$ m or m); $\Delta F_{Ni}$

Sauter (surface-mean) diameter

X_{SM}=\dfrac{\sum_i x_i^{3}\,\Delta F_{Ni}}{\sum_i x_i^{2}\,\Delta F_{Ni}}=\dfrac{1}{\sum_i \frac{1}{x_i}\,\Delta F_{Vi}}

Mean-volume diameter

X_{MV}=\left(\sum_i x_i^{3}\,\Delta F_{Ni}\right)^{1/3}

Specific surface area of a sphere

a_v=\dfrac{6}{d}, \qquad a_m=\dfrac{6}{\rho_p\,d}

Specific surface area with sphericity

a_v=\dfrac{6}{\Phi_s\,d}

Non-spherical correction.

General comminution law

\dfrac{dE}{dx}=-C\,x^{-n}

Energy per unit mass to reduce size; $n=2$

Rittinger's law

E=K_R\left(\dfrac{1}{x_2}-\dfrac{1}{x_1}\right)

Kick's law

E=K_K\,\ln\dfrac{x_1}{x_2}

Bond's law (size form)

E=K_B\left(\dfrac{1}{\sqrt{x_2}}-\dfrac{1}{\sqrt{x_1}}\right)

Bond work-index equation

W = 10\,W_i\left(\dfrac{1}{\sqrt{P_{80}}}-\dfrac{1}{\sqrt{F_{80}}}\right)

Supersaturation

S=\dfrac{C}{C^{*}}, \qquad \Delta C=C-C^{*}

Stokes terminal settling velocity

v_t=\dfrac{g\,(\rho_p-\rho_f)\,d^{2}}{18\,\mu}

Worked examples

Three mean diameters from a discrete PSD

Problem. A powder is split into four number sub-ranges with mean sizes and number fractions: $x_i=50,100,150,200\ \mu\text{m}$ and $\Delta F_{Ni}=0.40,0.30,0.20,0.10$ . Find the number-length, Sauter, and mean-volume diameters.

Solution. Build the column sums.

\sum x_i\Delta F_{Ni}=50(0.40)+100(0.30)+150(0.20)+200(0.10)=20+30+30+20=100

X_{SM}=\dfrac{\sum x_i^{3}\Delta F_{Ni}}{\sum x_i^{2}\Delta F_{Ni}}=\dfrac{1{,}825{,}000}{12{,}500}=146\ \mu\text{m}

Specific surface area of a ground quartz

Problem. Quartz ( $\rho_p=2650\ \text{kg/m}^3$ ) is ground to a Sauter mean diameter of $50\ \mu\text{m}$ . Estimate the specific surface area per unit mass treating particles (a) as spheres and (b) with sphericity $\Phi_s=0.80$

Grinding power from Bond's work index

Problem. Limestone with a Bond work index $W_i=11.6\ \text{kWh/short ton}$ is reduced from a feed $F_{80}=6500\ \mu\text{m}$

Yield from cooling crystallization of a hydrate

Problem. $1000\ \text{kg}$ of solution containing $30\,\text{wt\%}$ Na $_2$ CO $_3$

Common pitfalls

•Averaging diameters directly. The Sauter and mean-volume diameters are ratios of higher moments ( $\sum x^3/\sum x^2$ , etc.) — never just $\sum x_i\Delta F_i$ . Build moment columns and divide.
•Confusing number and volume/mass basis. $f_N,F_N$ count particles; $f_V,F_V$
•Using the wrong mean for the physics. Surface-rate processes (dissolution, catalysis, two-film mass transfer) need the surface (Sauter) mean; mass holdup needs the volume mean. The exam picks the diameter on purpose.
•Forgetting sphericity. Real crushed solids have $\Phi_s<1$ , so the specific surface area is $6/(\Phi_s d)$ , larger than the equal-volume sphere — dividing by $\Phi_s$
•Mixing Bond units. The classic Bond equation uses micrometres for $F_{80},P_{80}$ and kWh per short ton; an SI metric-tonne throughput must be converted, or the power is wrong by the ton factor (1 short ton = 0.907 t).
•Picking the wrong comminution law. Rittinger is for fine grinding (new-surface energy), Kick for coarse crushing (size-ratio energy), Bond for the mid/work-index range — using one regime's law in another's regime gives a misleading energy.
•Crystallization mass balance ignoring water of crystallization or evaporation. A hydrate locks up solvent in the crystal, and any evaporated water leaves the mother liquor — both shift the yield substantially if omitted.

References

NCEES FE Reference Handbook — Chemical Engineering (Solids Processing: particle-size distributions and mean diameters)
NCEES FE Reference Handbook — Fluid Mechanics (sphericity, packed beds, drag/terminal velocity)
Perry's Chemical Engineers' Handbook — particle-size statistics, comminution energy laws and work index, crystallization
McCabe, Smith & Harriott, Unit Operations of Chemical Engineering — size reduction and crystallization fundamentals

C. Transportation and storage

Solids Conveying, Slurries and Storage

Belt and pneumatic conveying, slurry critical-deposition velocity, hopper mass- versus funnel-flow, arching and rat-holing, and the angle of repose that ties bulk-solid behavior together.

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= number fraction in that sub-range (dimensionless, summing to 1).

Diameter of a sphere with the same surface-to-volume ratio as the sample; use for surface-rate processes (dissolution, drying, catalysis). $\Delta F_{Vi}$ = volume/mass fraction in sub-range $i$ .

Volume-weighted mean; governs mass distribution. Always $X_{ML}\le X_{MV}\le X_{SM}$ for the same powder.

Surface area per unit volume ( $a_v$ , $\text{m}^2/\text{m}^3$ ) or per unit mass ( $a_m$ , $\text{m}^2/\text{kg}$ ). Scales as $1/d$ : halving size doubles area. $\rho_p$ = particle density.

\Phi_s

= sphericity (0–1): area of equal-volume sphere divided by actual surface area; the same

\Phi_s

used in the Ergun equation.

Rittinger,

n=1

Kick,

n=1.5

Bond.

x

= particle size;

C

= material constant.

Energy $\propto$ new surface created; best for fine grinding. $x_1$ feed, $x_2$ product size; $K_R$ Rittinger constant.

Energy depends only on size reduction ratio; best for coarse crushing. $K_K$ = Kick constant.

Intermediate law for rod/ball mills. $K_B$ = Bond constant tied to the work index.

Specific energy in kWh/short ton; $W_i$ = Bond work index (kWh/short ton), $F_{80},P_{80}$ = 80%-passing feed and product sizes in $\mu$ m. Mill power $=W\times$ mass throughput.

Crystallization driving force. $C$ = actual concentration, $C^{*}$ = equilibrium solubility at the operating temperature. $S>1$ required for nucleation and growth.

Laminar ( $Re\le1$ ) settling/classification of a particle of diameter $d$ ; $\rho_p,\rho_f$ = particle and fluid density, $\mu$ = fluid viscosity. Ties particle size to separation.

.

\sum x_i^{2}\Delta F_{Ni}=2500(0.40)+10000(0.30)+22500(0.20)+40000(0.10)=1000+3000+4500+4000=12{,}500

.

\sum x_i^{3}\Delta F_{Ni}=125{,}000(0.40)+1{,}000{,}000(0.30)+3{,}375{,}000(0.20)+8{,}000{,}000(0.10)=1{,}825{,}000

. Number-length:

X_{ML}=\sum x_i\Delta F_{Ni}=100\ \mu\text{m}

. Sauter:

X_{SM}=1{,}825{,}000/12{,}500=146\ \mu\text{m}

. Mean-volume:

X_{MV}=(1{,}825{,}000)^{1/3}=122\ \mu\text{m}

. Answers:

X_{ML}=100\ \mu\text{m}

,

X_{SM}=146\ \mu\text{m}

,

X_{MV}=122\ \mu\text{m}

. Sanity check: the ordering

X_{ML}<X_{MV}<X_{SM}

here reflects the surface mean over-weighting the coarse tail; all three lie inside the 50–200

\mu

m range, as they must.

.

Solution. (a) Sphere:

a_m=\dfrac{6}{\rho_p d}=\dfrac{6}{2650\times 50\times10^{-6}}=\dfrac{6}{0.1325}=45.3\ \text{m}^2/\text{kg}

. (b) With sphericity:

a_m=\dfrac{6}{\Phi_s\rho_p d}=\dfrac{45.3}{0.80}=56.6\ \text{m}^2/\text{kg}

. Answers:

a_m\approx 45.3\ \text{m}^2/\text{kg}

(spheres),

56.6\ \text{m}^2/\text{kg}

(

\Phi_s=0.80

). Sanity check: rough crushed grains expose more surface than the equal-volume sphere, so dividing by

\Phi_s<1

correctly raises the area;

\sim

45 m

^2

/kg (

\approx 0.045

m

^2

/g) is typical for a fine sand.

a_m=\dfrac{6}{\Phi_s\,\rho_p\,d}=\dfrac{6}{0.80(2650)(50\times10^{-6})}=56.6\ \text{m}^2/\text{kg}

to a product

P_{80}=100\ \mu\text{m}

at a throughput of

50\ \text{short ton/h}

. Find the specific energy and the mill power.

Solution. Specific energy from Bond:

W=10(11.6)\left(\dfrac{1}{\sqrt{100}}-\dfrac{1}{\sqrt{6500}}\right)=116\left(\dfrac{1}{10}-\dfrac{1}{80.6}\right)=116(0.1000-0.01240)

.

W=116(0.08760)=10.16\ \text{kWh/short ton}

. Power

=W\times\dot m=10.16\ \dfrac{\text{kWh}}{\text{ton}}\times 50\ \dfrac{\text{ton}}{\text{h}}=508\ \text{kW}

. Answers:

W=10.16\ \text{kWh/short ton}

(the full-precision value;

W

rounds to

10.2

, but carrying

10.16

is what gives

10.16\times 50=508

kW),

P=508\ \text{kW}

. Sanity check: units kWh/ton

\times

ton/h = kW; the feed term contributes little (

1/\sqrt{6500}

is small), so almost all the energy is spent making the fine product — exactly the inefficiency the energy laws predict.

W=10\,W_i\!\left(\dfrac{1}{\sqrt{P_{80}}}-\dfrac{1}{\sqrt{F_{80}}}\right)=10.16\ \text{kWh/short ton}

is cooled so the salt crystallizes as the decahydrate Na

_2

CO

_3\cdot

10H

_2

O. The final solubility is

12.5\ \text{g}

anhydrous Na

_2

CO

_3

per

100\ \text{g}

water, and

5\%

of the original water evaporates. Find the mass of crystals. (Molar masses: anhydrous

=106

, decahydrate

=286\ \text{g/mol}

.)

Solution. Feed:

0.30(1000)=300\ \text{kg}

anhydrous Na

_2

CO

_3

and

700\ \text{kg}

water. Water lost to evaporation

=0.05(700)=35\ \text{kg}

, leaving

665\ \text{kg}

water in the system. Let

C

= mass of decahydrate crystals. Anhydrous fraction of the crystal

=106/286=0.3706

; water of crystallization fraction

=180/286=0.6294

. Anhydrous left in mother liquor

=300-0.3706\,C

; free water in mother liquor

=665-0.6294\,C

. The solubility constraint is

\dfrac{300-0.3706\,C}{665-0.6294\,C}=\dfrac{12.5}{100}=0.125

.

300-0.3706\,C=0.125(665)-0.125(0.6294)\,C=83.13-0.07868\,C

.

300-83.13=(0.3706-0.07868)\,C\Rightarrow 216.9=0.2919\,C\Rightarrow C=743\ \text{kg}

. Answer:

C\approx 743\ \text{kg}

of Na

_2

CO

_3\cdot

10H

_2

O. Sanity check: anhydrous in crystals

=0.3706(743)=275\ \text{kg}

, leaving

25\ \text{kg}

in

197\ \text{kg}

of mother-liquor water, ratio

0.125

— matches the solubility exactly.

\dfrac{300-0.3706\,C}{665-0.6294\,C}=0.125 \;\Rightarrow\; C=743\ \text{kg}

weight by volume. The two cumulative curves and their means differ greatly because one coarse particle equals thousands of fines by mass.

raises area, never lowers it.