12.4 Filtration, Crystallization, and Solid-Fluid Separations

12.4 Filtration, Crystallization, and Solid-Fluid Separations

The NCEES specification lists Other separations (e.g., liquid-liquid, liquid-solid, gas-solid, extraction, drying, adsorption, filtration, membrane separations, crystallization) as a single subtopic under Mass Transfer Applications. Sections 12.2 and 12.3 covered extraction, adsorption, membranes, and drying. This section covers the solid-handling operations that complete the list, and which appear on the exam as short, highly numerical problems.

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1. Cake Filtration Theory

Slurry flows through a septum; solids build a cake of increasing thickness whose resistance grows with the volume already filtered. Combining Darcy's law through the cake and medium gives the general filtration rate equation:

$\frac{dt}{dV} = \frac{\mu}{A \Delta P} \left( \frac{\alpha c V}{A} + R_m \right)$

Constant-pressure operation. Holding (\Delta P) fixed and integrating from (V = 0):

$\boxed{\ \frac{t}{V} = \left( \frac{\mu \alpha c}{2 A^2 \Delta P} \right) V + \frac{\mu R_m}{A \Delta P}\ }$

This is the workhorse. Plot (t/V) versus (V); the data fall on a straight line. The slope gives (\alpha) and the intercept gives (R_m). The exam typically hands you the fitted line and asks for a time, a required area, or a scale-up.

Constant-rate operation. If instead a positive-displacement pump holds (dV/dt) constant, (V = \dot{V}t) and the required pressure rises linearly with time. Most industrial filtrations begin at constant rate (pump-limited) and switch to constant pressure once the pump reaches its discharge limit.

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2. Cake Compressibility

Real cakes consolidate under load. The specific resistance follows:

$\alpha = \alpha_0 (\Delta P)^{s}$

where (s) is the compressibility index.

Substituting into the cake-dominated limit, the filtration rate scales as (\Delta P^{1-s}). At (s = 1) the exponent is zero: squeezing harder simply collapses the pore structure by exactly as much as the extra driving force gains. The industrial answer for a highly compressible cake is a filter aid (diatomaceous earth precoat or body feed), which provides a rigid incompressible matrix, not a bigger pump.

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3. Worked Example: Constant-Pressure Filtration

Problem. A plate-and-frame press operating at constant pressure yields the fitted relation (t/V = 3.00 \times 10^4 V + 1.20 \times 10^3), with (t) in seconds and (V) in (\text{m}^3). (a) How long to collect (0.50\text{ m}^3) of filtrate? (b) The cake is essentially incompressible; how long if (\Delta P) is doubled?

(a) Multiply through by (V): $t = V \left[ 3.00 \times 10^4 V + 1.20 \times 10^3 \right] = 0.50 \left[ (3.00\times10^4)(0.50) + 1.20\times10^3 \right]$ $t = 0.50 \left[ 15{,}000 + 1{,}200 \right] = 0.50(16{,}200) = 8{,}100\text{ s} = 2.25\text{ h}$

(b) Both the slope and the intercept carry (\Delta P) in the denominator, and for an incompressible cake (\alpha) does not change. Doubling (\Delta P) halves both coefficients: $t = 0.50 \left[ (1.50\times10^4)(0.50) + 600 \right] = 0.50(8{,}100) = 4{,}050\text{ s} = 1.125\text{ h}$

Exactly half. Had the cake been fully compressible ((s = 1)), (\alpha) would have doubled along with (\Delta P), the slope would have been unchanged, and the gain would have been confined to the small medium-resistance term.

Equipment note. A plate-and-frame press is batch, high-pressure, labor-intensive, and suited to low-throughput high-value solids. A rotary vacuum drum filter is continuous but limited to roughly (80\text{ kPa}) of driving force by the vacuum; its capacity is set by the submergence fraction (fraction of the drum immersed, typically (0.2) to (0.35)) and the cycle time, since each drum revolution filters, washes, dries, and discharges.

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4. Crystallization

Driving force. Crystallization requires supersaturation, generated by cooling, by evaporating solvent, by adding an antisolvent, or by reaction. Between the solubility curve and the metastable limit lies the metastable zone, where existing crystals grow but spontaneous nucleation does not occur. Good crystallizer operation stays inside that zone: nucleate once, then grow. Blow through the metastable limit and you get a shower of fines and an unfilterable product.

Yield with a hydrate — the classic exam problem.

Problem. (2{,}000\text{ kg}) of a (30.0\text{ wt}%) (\text{MgSO}_4) solution is cooled to (20^\circ\text{C}), where solubility is (25.0\text{ kg}) anhydrous (\text{MgSO}_4) per (100\text{ kg}) water. Crystals form as (\text{MgSO}_4 \cdot 7\text{H}_2\text{O}) ((M = 246.5); anhydrous (M = 120.4)). Evaporation is negligible. Find the crystal yield.

Solution. Feed contains (600\text{ kg}) (\text{MgSO}_4) and (1{,}400\text{ kg}) water. Let (C) be kg of heptahydrate. The anhydrous fraction of the hydrate is (120.4/246.5 = 0.4884), so each kg of crystal removes (0.4884\text{ kg}) (\text{MgSO}_4) and (0.5116\text{ kg}) water. Saturating the mother liquor:

$\frac{600 - 0.4884C}{1{,}400 - 0.5116C} = 0.250$ $600 - 0.4884C = 350 - 0.1279C \quad \Longrightarrow \quad 250 = 0.3606C \quad \Longrightarrow \quad C = 693\text{ kg}$

The trap. Ignoring water of crystallization gives (600 - C = 0.250(1{,}400)), so (C = 250\text{ kg}) — low by a factor of (2.8). Removing water with the crystals concentrates the remaining liquor and drives far more solute out of solution than the anhydrous balance predicts. Always check whether the crystal is a hydrate.

MSMPR population balance. For a mixed-suspension, mixed-product-removal crystallizer at steady state with size-independent growth rate (G) and residence time (\tau):

$n(L) = n^0 \exp\!\left( -\frac{L}{G\tau} \right)$

A semilog plot of population density (n) against size (L) is a straight line of slope (-1/(G\tau)) and intercept (n^0) (the nuclei population density). Consequences worth memorizing:

• Dominant (mass-weighted modal) size: (L_D = 3 G \tau). To grow bigger crystals, increase residence time or growth rate — nothing else in the ideal model matters.
• Coefficient of variation: a fixed 50% for an ideal MSMPR, independent of (G), (\tau), and supersaturation. A narrow distribution therefore cannot be achieved by tuning an MSMPR; it requires fines destruction, classified product removal, or seeded batch operation.

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5. Leaching and Gas-Solid Separation

Leaching (solid-liquid extraction) recovers a solute from a solid matrix with a solvent — oilseed extraction with hexane, metal recovery from ore, sugar from beets. Stage calculations mirror liquid-liquid extraction, with one addition: each stage's underflow carries retained solution with the solids. The usual simplifying assumption is constant underflow, meaning each stage's exiting solids retain a fixed mass of solution per mass of inert solid, which makes the operating line straight and permits a McCabe-Thiele or Kremser construction.

Cyclones separate solids from gas by centrifugal action. The Lapple cut diameter — the size collected at 50% efficiency — is:

$d_{pc} = \sqrt{\frac{9 \mu W}{2 \pi N_e v_i (\rho_p - \rho)}}$

with (W) the inlet width, (v_i) the inlet velocity, and (N_e) the effective number of gas turns (conventionally (N_e \approx 5) for a standard high-efficiency cyclone).

Example. (\mu = 1.8\times10^{-5}\text{ Pa}\cdot\text{s}), (W = 0.10\text{ m}), (v_i = 15\text{ m/s}), (N_e = 5), (\rho_p = 2{,}000\text{ kg/m}^3), (\rho = 1.2\text{ kg/m}^3):

$d_{pc} = \sqrt{\frac{9(1.8\times10^{-5})(0.10)}{2\pi(5)(15)(1{,}998.8)}} = \sqrt{\frac{1.62\times10^{-5}}{9.42\times10^{5}}} = \sqrt{1.72\times10^{-11}} = 4.1\times10^{-6}\text{ m} = 4.1\ \mu\text{m}$

Because (d_{pc} \propto \sqrt{W/v_i}), a smaller, faster cyclone collects finer particles — which is why high-efficiency service uses several small parallel cyclones rather than one large one, at the cost of higher pressure drop. Below roughly (5\ \mu\text{m}), cyclones become ineffective and a baghouse or electrostatic precipitator is required.