Sodium Sulfate Crystallization: Yield & Heat Balance

Last updated: 2026-10-29

Sodium sulfate crystallization is decided by a single temperature: 32.4 °C. Below it, the stable solid is mirabilite (Na2SO4·10H2O) and you cool the liquor; above it, the stable solid is thenardite (Na2SO4) and you evaporate. For 10.0 m³/h of liquor saturated at 40 °C, cooling to 5 °C closes at 9.31 t/h of mirabilite — a 94.1% recovery — against a 1,034 kW cooling duty. Every intermediate number is worked out below; swap in your own assay and rerun the balance.

This page is the calculation companion to our evaporator-crystallizer pillar page, which covers the hardware. Here we stay on paper: the solubility curve, three worked calculations, and the heat balance that fixes evaporator sizing, steam demand, and cooling capacity.

Key Takeaways

• The 32.4 °C mirabilite–thenardite transition decides the route: cool below it, evaporate above it.
• Cooling 10 m³/h of 40 °C saturated liquor to 5 °C yields 9.31 t/h of decahydrate crystals.
• The decahydrate locks 55.9% of its mass in hydration water, lifting crystal tonnage 2.27-fold.
• Evaporating the same feed at 60 °C to a 90% yield needs 7.98 t/h of water removed.
• The heat balance adds roughly 2.3 GJ/h of crystallization heat to the cooling duty.

What Is Sodium Sulfate Crystallization and Why Does 32.4 °C Decide the Route?

Sodium sulfate crystallization is a solid–liquid separation step that recovers Na2SO4 from a liquor by pushing its concentration past the solubility curve — by cooling, by evaporating water, or by combining the two. What makes it unusual is that the curve bends the wrong way halfway up, and that bend is set by a phase change in the solid.

The mirabilite–thenardite transition is a reversible hydration change that swaps a ten-water hydrate for an anhydrous salt at 32.4 °C. Below the line, crystals are Na2SO4·10H2O (mirabilite, Glauber’s salt); above it, anhydrous Na2SO4 (thenardite). The hydrate holds 10 mol of water per mol of salt — 55.9% of its own crystal mass — and that fact reshapes crystal tonnage, handling temperature, drying duty, and the yield arithmetic itself.

A mother liquor is a saturated solution that remains after crystals form, holding whatever salt the curve still allows at the final temperature. Below the transition, solubility collapses as temperature falls — 40.8 g Na2SO4 per 100 g water at 30 °C, roughly 5 g at 0 °C — so a modest cooling duty throws down a huge crystal crop. Above it, the curve is nearly flat and slightly retrograde: heating from 40 °C to 100 °C lowers solubility from 48.8 to about 42.5 g per 100 g water. A thenardite-saturated liquor yields almost nothing on cooling; the only lever is water removal.

The route decision is mechanical, not preferential. A feed below 32.4 °C, or one coolable there economically, suits a cooling crystallizer producing mirabilite cheaply; a hot feed, or a product that must be anhydrous for dry handling, suits an evaporative circuit at 50–70 °C. Combined circuits — evaporate to concentrate, then cool — appear when a zero-liquid-discharge obligation or a mother-liquor bleed forces the issue.

The Solubility Curve: Numbers to Work With

Every number downstream inherits from this curve, so the data belongs on one page. The values below are typical literature values for the Na2SO4–H2O system at atmospheric pressure — to be confirmed against a validated data source for your system before they anchor a design.

Between 0 °C and 32.4 °C the curve rises from 4.8 to about 49.7 g per 100 g water, and that steepness is the economic case for cooling crystallization of mirabilite: each degree of cooling strands a large amount of salt on the wrong side of the curve. At the transition the curve peaks and reverses, falling gently with temperature. Sodium sulfate itself is a commodity tracked annually by the USGS Mineral Commodity Summaries.

One caution: real plant liquors are never pure Na2SO4–H2O. Dissolved NaCl, Na2CO3 or borates shift both the transition temperature and the curve itself, so confirm solubility on your own liquor with simple isothermal tests — and treat the 30–35 °C band as a no-man’s-land, for reasons the scaling section returns to.

How Much Yield Does a Cooling Duty Deliver?

The duty. 10.0 m³/h of liquor at 40 °C, saturated in Na2SO4, is cooled to 5 °C in a continuous cooling crystallizer. All figures are per hour; salt is on an anhydrous Na2SO4 basis unless stated; crystals are dry solid.

Step 1 — feed mass and composition. Saturated sodium sulfate liquor at 40 °C has a density of about 1.33 t/m³ (typical literature value, to be confirmed). Feed mass: 10.0 m³/h × 1,330 kg/m³ = 13,300 kg/h. At 40 °C the liquor holds 48.8 g Na2SO4 per 100 g water, so the salt mass fraction is 48.8 / 148.8 = 0.328:

Na2SO4 in feed: 13,300 × 0.328 = 4,362 kg/h
Water in feed: 13,300 × 0.672 = 8,938 kg/h

Step 2 — what the 5 °C liquor can still hold. From the mirabilite branch (4.8 at 0 °C, 9.1 at 10 °C), interpolating at 5 °C gives about 6.9 g Na2SO4 per 100 g water. A first pass ignoring hydration water would say: dissolved salt remaining = 0.069 × 8,938 = 617 kg/h, so anhydrous-equivalent crystals = 4,362 − 617 = 3,745 kg/h. That pass is wrong — instructively so.

Step 3 — the hydration water correction. Each kilogram of decahydrate drags 0.559 kg of water out of the liquid phase: the hydrate is 142.04 / 322.19 = 0.4409 Na2SO4 and 180.15 / 322.19 = 0.5591 water by mass. Let H be the hydrate rate. Two balances must close simultaneously:

Na2SO4: 4,362 = 0.4409·H + dissolved salt
Water: 8,938 = 0.5591·H + free water
with dissolved salt = 0.069 × free water.

Substituting: 4,362 − 0.4409·H = 0.069 × (8,938 − 0.5591·H) = 616.7 − 0.0386·H. Collecting terms: 3,745.3 = 0.4023·H, so:

H = 9,308 kg/h of Na2SO4·10H2O ≈ 9.31 t/h ≈ 223 t/d

Step 4 — close the balance. Salt in crystals: 0.4409 × 9,308 = 4,104 kg/h (anhydrous equivalent, 98.5 t/d). Water locked in crystals: 0.5591 × 9,308 = 5,204 kg/h. Mother liquor: 8,938 − 5,204 = 3,734 kg/h water and 4,362 − 4,104 = 258 kg/h salt. Check: 258 / 3,734 = 0.069, exactly the 5 °C solubility. The balance closes.

What the hydration water penalty really means. This iteration is what every sodium sulfate crystallization yield calculation on a hydrate product reduces to. Three consequences. First, the mass uplift: 9,308 kg/h of wet crystal carries 4,104 kg/h of actual Na2SO4 — a factor of 2.27 on every conveyor, centrifuge and freight bill. Second, the recovery paradox: withdrawing hydration water shrinks the mother liquor, so the mirabilite route recovers 94.1% of feed salt where a hypothetical anhydrous product at 5 °C would cap out at 3,745 / 4,362 = 85.9%. Third, the bleed: 3,992 kg/h of saturated mother liquor will redissolve or rescale on cooling or drying, which is why mother liquor recovery deserves its own design pass.

What Evaporation Duty Does an Evaporative Route Demand at 60 °C?

The duty. The same 13,300 kg/h feed — 4,362 kg/h Na2SO4 and 8,938 kg/h water — goes to a forced-circulation evaporative crystallizer at 60 °C, where the stable solid is thenardite. Target: 90.0% recovery of feed salt as dry anhydrous Na2SO4.

Step 1 — crystallization target. 0.90 × 4,362 = 3,926 kg/h of thenardite — 3.93 t/h or 94.2 t/d of dry product. Dissolved salt remaining: 4,362 − 3,926 = 436 kg/h.

Step 2 — water balance. At 60 °C the liquor holds 45.3 g Na2SO4 per 100 g water. Free water needed to keep 436 kg/h dissolved: 436 / 0.453 = 963 kg/h. Since thenardite carries no water of crystallization, everything else must leave as vapor:

Evaporation rate = 8,938 − 963 = 7,975 kg/h ≈ 8.0 t/h of water vapor

That is the number an evaporator vendor sizes against. Magma check: the liquor phase is 963 + 436 = 1,399 kg/h at 436 / 1,399 = 31.2 wt% Na2SO4, matching 45.3 / 145.3 = 31.2% at saturation — the magma closes on both curves.

Step 3 — heat demand. Saturated Na2SO4 liquor at 60 °C boils with a boiling point elevation of roughly 5 K (typical literature value, to be confirmed), so the magma boils at about 65 °C at atmospheric pressure, where water’s latent heat is about 2,345 kJ/kg (typical literature value). Evaporation duty: 7,975 × 2,345 = 18.70 GJ/h. Sensible heat to lift the 40 °C feed to 65 °C at cp ≈ 3.0 kJ/(kg·K) (typical literature value): 13,300 × 3.0 × 25 = 998 MJ/h ≈ 1.00 GJ/h. Total evaporator duty ≈ 19.7 GJ/h, or 5,472 kW.

Step 4 — energy carriers. With live steam at 2.0 bar(g) (latent heat ≈ 2,200 kJ/kg), single effect: 19,700,000 / 2,200 ≈ 8,950 kg/h ≈ 9.0 t/h of steam — 2.28 t of steam per tonne of product. With mechanical vapor recompression the compressor does the same job electrically: at 25–35 kWh per tonne of water evaporated (typical literature range for MVR on inorganic salts, to be confirmed by vendor data), 30 kWh/t gives 7.975 × 30 ≈ 239 kW — about 61 kWh per tonne of product. That gap, 1,394 vs 61 kWh/t, is why MVR is the default where grid power is dependable; see MVR working principle before shortlisting.

How Does the Heat Balance Close for the Cooling Route?

Cooling crystallization looks energy-trivial because no vapor is produced, but the balance has two terms — and ignoring the second undersizes the cooling system by about 60% here.

Term 1 — sensible heat. Cool 13,300 kg/h of liquor from 40 °C to 5 °C at cp = 3.0 kJ/(kg·K):

Q1 = 13,300 × 3.0 × (40 − 5) = 1,396,500 kJ/h ≈ 1.40 GJ/h removed

Term 2 — heat of crystallization. Crystallization is exothermic: the salt releases heat as it organizes into the hydrate lattice. For Na2SO4·10H2O the crystallization enthalpy is approximately 250 kJ per kg of hydrate formed (literature value on hydrate basis, typically quoted at 243–255 kJ/kg — to be confirmed against a validated data source for your system). Using 250:

Q2 = 9,308 kg/h × 250 kJ/kg = 2,327,000 kJ/h ≈ 2.33 GJ/h released

Total duty. Heat to be removed: Q1 + Q2 = 1.40 + 2.33 = 3.72 GJ/h = 1,034 kW (≈ 3.5 MMBtu/h). Nearly two-thirds of that is not cooling the liquor at all — it is removing the latent heat the crystals release. A cooler sized on the 40 K drop alone installs 1.40 GJ/h against a 3.72 GJ/h reality, and the crystallizer runs hot while the yield slides down the solubility curve.

Turning duty into equipment. Plant cooling water at 25–30 °C cannot close the gap at a 5 °C product; you need chilled water at roughly 0–2 °C or a refrigeration loop. At a 5 K coolant rise, flow is 3,723,500 / (4.18 × 5) ≈ 178,000 kg/h ≈ 178 m³/h (≈ 785 US gal/min). With refrigeration at a coefficient of performance of 3.0, the electrical draw is 1,034 / 3.0 ≈ 345 kW — 37 kWh per tonne of hydrate, or about 84 kWh per tonne on the anhydrous-equivalent basis (4.10 t/h). Stating the basis keeps the numbers honest.

Which Route Wins on Energy: Cooling, Evaporation, or Both?

The routes are not competing on the same product, so the comparison is a basis exercise.

Three conclusions survive the basis-checking. First, cooling is the cheapest crystallization per tonne of salt recovered — but only while the mirabilite form is acceptable and a cold sink exists on site. Second, live steam evaporation at 2.28 t steam per tonne of product is almost always the most expensive carrier; proper comparisons belong in our evaporator energy consumption guide. Third, MVR converts the evaporation duty into roughly 61 kWh/t of electricity — competitive with cooling per tonne, without the hydrate’s 2.27× mass uplift, which is why mechanical recompression now anchors most new sodium sulfate crystallization plants. Under a zero liquid discharge obligation the combined route is not optional, and the same logic extends to the concentrates covered in brine concentrator sizing and selection.

Two decision sentences follow from the table. For a site with a 5 °C cooling sink and a buyer for the decahydrate, cooling crystallization is preferred because 37 kWh/t of refrigeration sits an order of magnitude below the 1,394 kWh/t carried by live-steam evaporation. For a site that must ship a dry, free-flowing anhydrous product, evaporative crystallization above 32.4 °C is preferred because thenardite leaves the centrifuge as dry salt and needs no dehydration step behind it.

What Controls Crystal Size Distribution — and Why Recycle Fines?

Yield and heat balance decide whether the plant works; crystal size distribution decides whether it works well. An MSMPR crystallizer is a mixed-suspension, mixed-product-removal vessel that sets its product size through the competition between nucleation, which multiplies small crystals, and growth, which coarsens the population. Supersaturation feeds both — nucleation far more violently — and mirabilite nucleates readily, producing fine, hard-to-dewater magma.

The levers, in the order they pay back:

• Moderate the driving force. Spread the 35 K cooling over staged vessels or a longer residence rather than quenching; a gentler supersaturation profile shifts the balance from nucleation to growth.
• Seed. Sized seed crystals give the duty a place to land without spawning new nuclei.
• Destroy fines. Dissolve the fines fraction (typically everything below a cut size of a few tens of microns) in a heated or diluted trap stream and return it as clear liquor. The surviving crystals then grow on the dissolved salt, and dewatering improves accordingly.
• Classify the product removal. Withdrawing product at coarse sizes extends residence time for the fraction that carries the mass.

Fines recycling matters most in mirabilite service because the hydrate dewateres unforgivingly: fine crystals blind centrifuge cloths and carry mother liquor into the dryer, eroding the very yield the balance promised. Circuits compared in continuous vs batch crystallizer service trade these levers differently — batch vessels seed and cool cyclically, continuous vessels run fines traps permanently — and the choice follows from tonnage and dryer demands. Hardware implications are covered in crystallizer design.

Why Does Thenardite Scale So Hard Above the Transition?

Above 32.4 °C the system has a property that turns ordinary fouling into a maintenance problem: retrograde solubility. Because dissolving thenardite is weakly exothermic, warmer liquor holds less salt — so the hottest surface, not the coldest, is where crystals preferentially deposit. Add boiling at the wall, which concentrates the boundary layer past saturation, and you get dense, adherent anhydrous scale exactly where the heat flux is highest. The scale does not slough off with thermal cycling: the wall-to-liquor ΔT that created it keeps it baked on.

Mitigation is thermal and hydraulic, not chemical. Keep the wall-to-boiling-liquid ΔT low — typically below about 10–15 K in sodium sulfate service (field practice, to be confirmed for your liquor) — hold circulation velocities high so the magma scrubs the surfaces and supplies growth sites in the bulk, and avoid dead zones where liquor can linger past the transition band and flip phase. Equipment straddling 30–35 °C sees repeated mirabilite–thenardite cycling, which builds a layered scale worse than either phase alone. Scaling-prone liquors deserve the countermeasures in evaporator scaling control — and in our project experience, the most common avoidable error is a high-ΔT evaporator body chosen to save surface area, with the savings spent on descaling shutdowns.

Calculation Sequence: Seven Steps Before You Specify Equipment

Run every duty through this sequence before contacting a vendor; each step feeds the next.

1. Fix the basis. State feed flow, temperature, saturation state, and the product form — mirabilite or thenardite — plus wet/dry and hydrate/anhydrous bases for every number.
2. Pull the solubility data at both temperatures. Take curve values at feed and product temperatures, confirmed on your own liquor.
3. Balance the water. Split the feed into salt and water; for evaporative routes, solve the free water left at the product solubility to get the evaporation rate.
4. Iterate for hydration water. For mirabilite duty, solve the coupled balances — 0.4409·H salt, 0.5591·H water — because hydrate growth shrinks the mother liquor and changes its own yield.
5. Compute the sensible heat. Feed mass × cp × ΔT, on the temperature span the chosen route actually imposes.
6. Add the heat of crystallization. Roughly 250 kJ per kg of hydrate (literature value, to be confirmed); it enlarges a cooling duty and slightly relieves an evaporative one.
7. Convert to equipment duty. Close with evaporation rate (t/h vapor), steam or MVR demand, and cooling capacity in kW — the three numbers a vendor quote must be written against.

Steps 4 and 6 are the two most often skipped, and both err on the small side of equipment.

If you want your own duty list independently checked — or turned into a sized evaporator-crystallizer proposal — request a quote with your balance sheet attached, and our process team will review the arithmetic against your assay before sizing anything.

Conclusion: Closing the Sodium Sulfate Crystallization Balance

The whole discipline of sodium sulfate crystallization reduces to one transition and three closures. The transition at 32.4 °C picks the route; the yield closure fixes the tonnage through the centrifuges; the heat closure — sensible plus roughly 250 kJ/kg of crystallization heat — fixes the cooling plant or steam balance behind it. On the worked duty, cooling to 5 °C recovers 94.1% of the feed salt as 223 t/d of hydrate against a 1,034 kW duty, while the evaporative route delivers 94.2 t/d of dry thenardite for 8.0 t/h of vapor. Close those numbers on your own liquor and equipment selection becomes arithmetic, not argument. The resource library carries the supporting guides for each piece of that flowsheet.

Frequently Asked Questions

Why is 32.4 °C the dividing line in sodium sulfate crystallization?

Below 32.4 °C the stable solid is mirabilite (Na2SO4·10H2O); above it, the decahydrate gives up its water and thenardite (Na2SO4) is stable. The solubility curve rises steeply to the transition — about 49.7 g Na2SO4 per 100 g water at 32.4 °C (typical literature value, to be confirmed) — then falls gently above it. That shape is why cooling works below the line and evaporation above it.

Should the product be mirabilite or thenardite?

Follow the liquor temperature and the downstream use. Mirabilite forms below 32.4 °C mostly by cooling, so its energy bill is small, but it carries 55.9% crystal water — 2.27 kg of wet crystal per kilogram of anhydrous salt — and must be melted or dried before most uses. Thenardite from a 60 °C evaporator is anhydrous and transport-efficient, at the price of a larger evaporation duty.

How much yield does the hydration water cost?

In the worked cooling example, the decahydrate locks 5,204 kg/h of water inside 9,308 kg/h of crystals. Paradoxically, that raises recovery: anhydrous crystals at 5 °C would leave 617 kg/h dissolved and cap the yield at 85.9%, while mirabilite reaches 94.1% because withdrawing hydration water shrinks the mother liquor. The real cost is logistics — every tonne of anhydrous salt leaves the plant as about 2.27 t of hydrate.

Why does sodium sulfate solubility fall as temperature rises above 32.4 °C?

The curve is retrograde above the transition because dissolving thenardite releases heat: the hydration energy of the ions outweighs the lattice energy broken, so warming the liquor pushes the equilibrium back toward the solid. A liquor saturated at 40 °C (48.8 g per 100 g water) holds only about 42.5 g at 100 °C (typical literature values). This is also why hot heat-transfer surfaces are the preferred scale sites.

What causes hard scale in sodium sulfate evaporators?

Three mechanisms stack up: retrograde solubility, which precipitates thenardite directly on the hottest surfaces; boiling at the wall, which concentrates the boundary layer past saturation; and the hardness of anhydrous scale, which bonds tightly to metal. Mitigation is thermal, not chemical — keep the wall-to-liquid temperature difference low, typically below about 10–15 K, hold circulation velocities high, and seed the magma so growth happens on suspended crystals rather than tubes.

How is crystal size distribution controlled in a mirabilite circuit?

Through the supersaturation balance in an MSMPR crystallizer: nucleation multiplies fine crystals while growth coarsens the population, and supersaturation feeds both. The levers are moderate, staged cooling rates; seeding with sized crystals; fines destruction, which dissolves the sub-cut-size fraction and returns it as clear liquor; and classified product removal. Fines recycling matters most in mirabilite service, where fine crystals blind centrifuge media and carry mother liquor into the dryer.

Can mechanical vapor recompression be used on sodium sulfate liquors?

Yes, and it is the usual choice in sodium sulfate crystallization where electricity is cheaper than steam. Saturated Na2SO4 liquor boils with a boiling point elevation of roughly 4–6 K (typical literature value, to be confirmed), which the compressor must bridge on top of the pressure lift. In the worked example, an MVR evaporator draws about 239 kW to remove 7.98 t/h of water — roughly 61 kWh per tonne of anhydrous product.

References

1. USGS, Mineral Commodity Summaries — Sodium Sulfate, U.S. Geological Survey, annual edition. https://pubs.usgs.gov/periodicals/mcs2024/mcs2024-sodium-sulfate.pdf
2. Perry, R.H. and Green, D.W. (eds.), Perry’s Chemical Engineers’ Handbook, 9th ed., McGraw-Hill — physical and chemical data tables (sodium sulfate solubility) and liquid–solid operations and equipment.
3. Lide, D.R. (ed.), CRC Handbook of Chemistry and Physics, CRC Press — solubility of inorganic salts in water, Na2SO4–H2O system.
4. Mullin, J.W., Crystallization, 4th ed., Butterworth-Heinemann — crystallization kinetics, MSMPR population balance, heat of crystallization.
5. Myerson, A.S. (ed.), Handbook of Industrial Crystallization, 3rd ed., Butterworth-Heinemann — crystallizer design, CSD control, fines destruction.
6. Fuerstenau, M.C., Han, K.N. et al. (eds.), SME extractive metallurgy references, Society for Mining, Metallurgy & Exploration — hydrometallurgical crystallization and evaporation practice.
7. Linke, W.F. and Seidell, A., Solubilities of Inorganic and Metal-Organic Compounds, 4th ed., American Chemical Society — compiled Na2SO4 solubility data.

Changelog

• 2026-10-29: Initial publication with three fully worked examples (cooling yield, evaporative duty, heat balance), solubility data table, energy comparison, CSD and scaling sections, and seven FAQs.