Specific Energy Consumption Evaporator Guide: kWh per Tonne

Last updated: 2026-10-22

The specific energy consumption evaporator designers should plan for ranges from about 620–700 kWh per tonne of water evaporated for a single-effect unit (quoted as steam-equivalent thermal energy), through roughly 200–225 kWh/t for a four-effect multiple effect evaporator, down to 15–40 kWh/t of electricity for a mechanical vapor recompression (MVR) system. This guide gives the formulas, the unit conversions, and three fully worked arithmetic examples behind those numbers, so you can audit any vendor quote. For the underlying equipment choices, start with our evaporation and crystallization hub.

Key Takeaways

• Single-effect evaporators consume roughly 620–700 kWh of steam-equivalent energy per tonne of water evaporated.
• A four-effect MEE cuts steam SEC to about 200–225 kWh/t by reusing vapor heat several times.
• MVR evaporators use only 15–40 kWh/t of electricity and usually have the lowest running cost.
• Steam economy, latent heat of about 2,257 kJ/kg, and boiler efficiency drive every evaporator energy calculation.
• Feed preheat, boiling point elevation, and fouling can add 10–30% to the theoretical SEC.

What Is Specific Energy Consumption in Evaporation?

Specific energy consumption (SEC) is a performance metric that states how much energy an evaporator must supply in order to evaporate one tonne of water from the feed. It is the single most useful number for comparing evaporation flowsheets, because it collapses feed rate, concentration, and equipment type into one figure that can be priced directly against your utility bill.

how MVR Evaporator Selection is decided

SEC is normally expressed in kWh per tonne of water evaporated (kWh/t). Because 1 kWh equals 3,600 kJ, an SEC of 627 kWh/t is identical to 2,257 kJ/kg — which is no coincidence: 2,257 kJ/kg is the latent heat of vaporization of water at 100 °C and atmospheric pressure. That figure is the thermodynamic floor for evaporation at atmospheric conditions; no evaporator can beat it, and real units exceed it because of feed preheating, heat losses, and non-ideal vapor use. Precise enthalpy values at other pressures come from the IAPWS industrial formulation for water and steam, which is the basis of every reputable steam table.

Steam economy is a ratio that compares kilograms of water evaporated to kilograms of live steam consumed. It is the inverse of SEC for steam-heated units: an economy of 0.9 kg/kg corresponds to an SEC of 627/0.9 ≈ 697 kWh/t, while an economy of 3.0 corresponds to 209 kWh/t. Mechanical vapor recompression replaces the economy logic with electricity, so a second basis is needed there.

Latent heat of vaporization is a thermodynamic property that varies gently with pressure — roughly 2,360 kJ/kg near 40 °C to 2,257 kJ/kg at 100 °C — so evaporators running under vacuum show slightly higher latent heat per kilogram, even though the whole train gains from lower operating temperatures.

One caution that process engineers repeat constantly: whenever a vendor quotes a specific energy consumption evaporator figure, ask what basis it uses. Thermal SEC (steam), electrical SEC (compressor power), and primary-fuel SEC (after boiler or grid efficiency) differ by factors large enough to invert a purchasing decision. Every number in this article states its basis explicitly, and you should demand the same discipline from any proposal you receive.

How Do You Calculate kWh per Tonne of Water Evaporated?

There are two practical routes to an evaporator energy calculation: the steam-economy method, which is fast and accurate enough for screening, and the enthalpy method, which is the rigorous version used for detailed design. Both give the same answer when the same assumptions are used.

Steam-economy method (screening). For a steam-heated evaporator:

SEC (kWh/t) = 627 / E

where E is the steam economy in kg of water evaporated per kg of live steam, and 627 kWh/t is the latent-heat constant derived from 2,257 kJ/kg (2,257 ÷ 3,600 = 0.627 kWh per kg of water). The economy E must already include allowances for feed preheat and losses; the worked examples below show how to choose it honestly.

Enthalpy method (rigorous). Sum every heat duty the evaporator must carry, then divide by the evaporation rate:

1. Evaporation duty: Q_evap = V × λ, where V is the water evaporation rate and λ the latent heat at the boiling pressure.
2. Feed preheat duty: Q_preheat = F × cp × (T_boil − T_feed), where F is feed flow and cp the feed specific heat.
3. Losses and BPE correction: add radiation losses and the extra heat needed where boiling point elevation suppresses vaporization.
4. Total: Q_total = Q_evap + Q_preheat + Q_losses, and SEC = Q_total ÷ V, converted to kWh/t.

The variables you need, with typical values for hydrometallurgical liquors:

A quick mass-balance relationship underpins both methods: for a feed of F kg/h at feed concentration x_F (mass fraction of dissolved solids) concentrated to x_P, the evaporation rate is V = F × (1 − x_F/x_P). Get this number right first; every downstream energy figure inherits its errors.

How Does Steam Economy Work in a Multiple Effect Evaporator?

A multiple effect evaporator (MEE) chains several vessels so that the vapor boiled off one effect becomes the heating steam for the next, which operates at a lower pressure. Each effect reuses the same kilogram of primary steam once more, so the specific energy consumption evaporator trains achieve improves almost in proportion to the number of effects — approximately E ≈ 0.75 × N, where N is the effect count.

The 0.75 factor (rather than 1.0 per effect) is where practical reality enters the steam economy calculation. Every effect must operate at a lower temperature than the previous one, and the total available temperature drop between steam and condenser is finite — commonly 60–90 K for a whole train. Each effect also has to absorb boiling point elevation, feed preheating between stages, and small radiation losses, all of which shave the achievable economy. That is why a four-effect unit delivers an economy near 3.0 rather than 4.0, and why a sixth effect adds less than the fifth did.

Thermal vapor recompression (TVR) is a middle path: a steam-jet ejector entrains low-pressure vapor from the boiloff and mixes it with high-pressure motive steam, effectively raising the economy of a single effect to roughly 1.3–1.7 kg/kg. TVR is cheap and nearly maintenance-free, but it needs a motive-steam supply at 6–12 bar and its economy is fixed by ejector geometry, so it cannot track varying duties the way a mechanical compressor can.

The economics of the MEE route are developed further in our MVR vs multiple effect comparison, but the headline is simple: doubling the effects roughly halves the steam bill while adding vessel count, height, and capital cost.

How Is MVR Energy Consumption Calculated?

Mechanical vapor recompression treats the evaporator as a heat pump. Water vapor boiled off the liquor is compressed slightly — typically by 5–15 K of saturation temperature lift — so it can condense inside the same unit’s heating surface, releasing its latent heat back into the boiling liquid. The only continuous energy input is the compressor’s electrical drive.

Evaporator Crystallizer Systems for Mining Brines in practice

Coefficient of performance (COP) is an efficiency ratio that compares the latent heat recycled inside the unit to the electrical or shaft work the compressor consumes. Because the temperature lift in evaporation service is small, MVR COP values of 15–35 are routine — far above the 3–5 typical of industrial heat pumps with larger lifts. Two equivalent formulas describe the energy consumption:

1. Via COP: SEC_electric (kWh/t) ≈ 627 ÷ COP. A COP of 32 gives about 19.6 kWh/t.
2. Via compressor work: SEC_electric = w × 1,000 ÷ 3,600,000 per tonne, where w is the specific compressor work in kJ per kg of vapor. More conveniently, SEC = shaft power (kW) ÷ evaporation rate (t/h).

The specific compressor work comes from the isentropic relation for steam:

w_s = [k ÷ (k − 1)] × R × T_1 × [(PR)^((k−1)/k) − 1]

where k ≈ 1.33 for steam, R = 0.4615 kJ/(kg·K), T_1 is the suction temperature in kelvin, and PR the pressure ratio. Dividing w_s by the isentropic efficiency (0.70–0.80) gives the actual shaft work. Worked Example 3 below runs this arithmetic end to end.

Two practical corrections belong in any honest MVR energy consumption estimate. First, motor and drive losses add roughly 3–7% between shaft and meter. Second, boiling point elevation raises the lift: if the liquor boils 4 K above pure-water temperature, the compressor must produce a condensing temperature at least 4 K higher for the same heat-transfer ΔT, and compressor work rises roughly in proportion to the lift. Concentrated sodium hydroxide or salt liquors with BPE of 5–10 K therefore push real-world MVR consumption from the low teens toward 25–40 kWh/t. The MVR working principle guide covers the mechanical layout, and our MVR evaporator systems page shows the equipment options at each lift.

Worked Example 1: Single-Effect Evaporator Steam SEC

All three worked examples use the same illustrative duty so the configurations can be compared directly. Assumptions: feed 10 t/h at 5 wt% dissolved solids, concentrated to 20 wt%; feed preheated to its boiling point of 100 °C by external streams (so preheat is not charged to the evaporator here); radiation losses ignored in the ideal case, then corrected through the economy figure. These values are illustrative, not a design basis.

1. Mass balance. Dissolved solids in = 10,000 kg/h × 0.05 = 500 kg/h. Product flow at 20 wt% = 500 ÷ 0.20 = 2,500 kg/h. Water evaporated V = 10,000 − 2,500 = 7,500 kg/h = 7.5 t/h.
2. Steam demand. Take a realistic single-effect economy of 0.9 kg/kg, which already allows for feed flash, venting, and losses. Live steam = 7,500 ÷ 0.9 = 8,333 kg/h.
3. Thermal power. Steam energy = 8,333 kg/h × 2,257 kJ/kg = 18.8 GJ/h ≈ 5,224 kW of condensing steam.
4. SEC (thermal basis). 5,224 kW ÷ 7.5 t/h = 697 kWh per tonne of water evaporated. Cross-check: 627 ÷ 0.9 = 697 kWh/t — the two routes agree.
5. Primary-fuel basis. If the boiler delivering that steam runs at 90% efficiency, fuel-side SEC = 697 ÷ 0.90 ≈ 774 kWh/t of fuel energy.
6. Interpretation. At a steam price of $30/t, this unit burns 8.33 t of steam per operating hour — about $250/h before any other operating cost. That is what a single-effect installation must justify through low capital cost or small duty.

The theoretical minimum for this duty is 7,500 × 2,257 = 16.9 GJ/h = 4,702 kW, or 627 kWh/t; the gap between 627 and 697 is the price of the 0.9 economy.

Worked Example 2: Four-Effect MEE Steam SEC

Same duty and same assumptions, now evaporated in a four-effect train in forward feed, with steam economy estimated by the 0.75 × N rule.

1. Duty. From Example 1, V = 7.5 t/h of water evaporated, unchanged by the flowsheet.
2. Economy estimate. E ≈ 0.75 × 4 = 3.0 kg of water per kg of live steam. (A vendor might promise 3.2–3.4 for a large train with extensive condensate flashing; 3.0 is the defensible screening value.)
3. Steam demand. Live steam = 7,500 ÷ 3.0 = 2,500 kg/h — exactly one third of the single-effect figure.
4. Thermal power. 2,500 kg/h × 2,257 kJ/kg = 5.64 GJ/h ≈ 1,567 kW.
5. SEC (thermal basis). 1,567 ÷ 7.5 = 209 kWh/t. Cross-check: 627 ÷ 3.0 = 209 kWh/t.
6. Saving. 697 − 209 = 488 kWh/t saved, a 70% reduction in steam consumption versus the single effect. At $30/t steam that is worth roughly (8,333 − 2,500) kg/h = 5.83 t/h of steam, about $175/h, every operating hour.

The catch is capital and temperature: the four effects must share the same 60–90 K total ΔT that the single effect enjoyed, so each effect sees only 15–22 K, which demands more heat-transfer area and vacuum equipment. Whether the steam saving pays for the extra vessels is precisely the trade-off our MVR vs multiple effect comparison quantifies.

Worked Example 3: MVR Evaporator Electricity SEC

Same duty again, now in a single-vessel MVR evaporator with a centrifugal compressor. Assumptions: vapor leaves the boiling surface as saturated steam at 100 °C (101.3 kPa); the compressor discharge condenses at 108 °C, so the saturation pressure there is about 134 kPa; steam k = 1.33; compressor isentropic efficiency 75%; motor and drive efficiency 95%. Values are illustrative.

1. Vapor flow. V = 7,500 kg/h of water vapor to be compressed (from the same mass balance).
2. Pressure ratio. PR = 134 ÷ 101.3 ≈ 1.32. This corresponds to an 8 K saturation-temperature lift, enough to drive heat transfer through the heating surface at this scale.
3. Isentropic specific work. w_s = [k ÷ (k − 1)] × R × T_1 × [(PR)^((k−1)/k) − 1] = 4.03 × 0.4615 × 373 × [(1.32)^(0.248) − 1]. Now (1.32)^0.248 ≈ 1.071, so the bracket term is 0.071, and w_s ≈ 4.03 × 172.2 × 0.071 ≈ 49.5 kJ/kg of vapor.
4. Actual shaft work. w = 49.5 ÷ 0.75 ≈ 66.0 kJ/kg. Compressor shaft power = 7,500 kg/h × 66.0 kJ/kg = 495,000 kJ/h ≈ 137.5 kW.
5. Electrical power and SEC. Allowing 5% motor/drive losses: 137.5 ÷ 0.95 ≈ 144.7 kW at the meter. SEC = 144.7 ÷ 7.5 ≈ 19.3 kWh of electricity per tonne of water evaporated.
6. Cross-check via COP. COP = λ ÷ w = 2,257 ÷ 66.0 ≈ 34.2 on a shaft basis, and 627 ÷ 34.2 ≈ 18.3 kWh/t at the shaft, rising to 19.3 kWh/t after drive losses — consistent with step 5.

A 145 kW machine evaporating 7.5 t/h with no live steam at all is the reason MVR dominates mid-size duties wherever electricity is reasonably priced. Add 5–10 K of boiling point elevation — concentrated brines, caustic liquors — and the lift grows to 13–18 K, roughly doubling compressor work toward 30–40 kWh/t. Size the compressor for the end-of-run BPE, not the clean feed.

How Do You Compare Specific Energy Consumption Evaporator Options?

A fair specific energy consumption evaporator comparison needs a single basis, a stated set of assumptions, and the utility prices of your site — not the vendor’s demo plant. The table below collects the screening ranges developed in this article, all per tonne of water evaporated and all self-consistent with the 627 kWh/t latent-heat constant.

Every range in this table is an original screening estimate derived from the formulas above; treat detailed design values as vendor-specific. When a proposal lands on your desk, rebuild these numbers for the quoted duty before comparing prices. If your flowsheet targets zero liquid discharge, the concentration step interacts with crystallization duty, as described in our zero liquid discharge applications overview and the brine concentrator equipment page.

What Increases Specific Energy Consumption Evaporator Duty?

The headline SEC of any flowsheet is set by configuration, but four site-level factors move the real number by 10–30% or more. Each factor in the table below can be estimated with the enthalpy method from earlier in this guide.

The feed-preheat row deserves the arithmetic spelled out, because it is the most frequently forgotten term. Heating 10 t/h of feed by 80 K requires 10,000 kg/h × 4.18 kJ/(kg·K) × 80 K ≈ 3.34 GJ/h ≈ 929 kW. Spread over the 7.5 t/h evaporated, that is 929 ÷ 7.5 ≈ 124 kWh/t — nearly 18% of the single-effect SEC in Example 1, recovered almost free if condensate and concentrate are flashed and counter-currently exchanged against the incoming feed.

Fouling is the silent inflation of every energy guarantee. As the overall heat-transfer coefficient U decays, the unit either loses evaporation rate or demands a higher ΔT, and in an MVR unit higher ΔT means higher lift and more kilowatt-hours. Specifying a forced-circulation evaporator for scaling liquors, and following the evaporator scaling control practices, keeps U near its clean value so the guaranteed SEC survives contact with your liquor.

Converting Between Steam and Electricity Cost Bases

Because steam-heated and MVR units draw on different utilities, the final comparison is financial, not thermodynamic. Three conversion rules make the arithmetic transparent:

1. One tonne of steam delivers ≈2,257 MJ ≈ 627 kWh of latent heat when condensing.
2. Steam price per tonne ÷ 627 = steam cost per kWh thermal. At $30/t: 30 ÷ 627 ≈ 4.8 ¢/kWh.
3. Electricity cost per kWh is simply the tariff — say $0.08/kWh in this illustration.

Reading the table: MVR beats the four-effect train in this illustration unless electricity exceeds roughly (209 × 0.0478) ÷ 19.3 ≈ $0.52/kWh — a tariff almost nobody pays. The reverse sensitivity matters too: at a $4/t steam price (waste-heat boiler, flare gas), the four-effect train drops to about $1.33/t and becomes competitive with MVR. Run the conversion with your tariffs before locking a flowsheet; the ranking of configurations is a property of the site, not of the technology.

Frequently Asked Questions

What is a good specific energy consumption evaporator figure?
A good figure is one stated on a declared basis and consistent with the physics: steam-heated units fall between 620–700 kWh/t (single effect) and about 130–225 kWh/t (four to six effects), while MVR units consume 15–40 kWh/t of electricity depending on lift. Query any quote outside these bands or lacking a basis.

MVR vs Multi-Effect Evaporation, step by step

How many kWh does it take to evaporate one tonne of water?
The thermodynamic minimum at atmospheric pressure is 2,257 kJ/kg × 1,000 kg ÷ 3,600 = 627 kWh of thermal energy. Real single-effect evaporators need roughly 10% more because of preheat and losses, while multi-effect and MVR designs recycle latent heat so effectively that their billed energy is far lower — as little as 15–25 kWh of electricity per tonne in favorable MVR cases.

Why does steam economy fall short of the number of effects?
Each effect must run colder than the previous one, and the total temperature drop from steam to condenser is finite. Feed preheating between effects, boiling point elevation, flashing losses, and radiation all consume part of that drop, so an N-effect train typically achieves only about 0.75 × N kilograms evaporated per kilogram of steam instead of the ideal N.

Is MVR always cheaper than steam heating?
Not always. MVR converts cheap latent-heat duty into electricity, so it wins when power costs $0.05–0.10/kWh, as the worked examples show (≈$1.55/t versus ≈$10/t for a four-effect train). Where steam is essentially free — waste heat, low-pressure surplus — a six-effect train can match MVR, and MVR adds compressor maintenance and electrical infrastructure costs that small plants may not want.

How does boiling point elevation affect energy consumption?
BPE raises the boiling temperature of the liquor above that of pure water at the same pressure. In steam-heated trains it shrinks the usable ΔT per effect, trimming economy by roughly 3–8%. In MVR units the compressor must add the BPE to its lift, and compressor work rises roughly in proportion — concentrated caustic or salt liquors can push MVR consumption from 15–25 up to 25–40 kWh/t.

Should SEC be quoted on a thermal or electrical basis?
Quote both, and state which utilities each configuration uses. Thermal SEC (kWh/t of steam) suits MEE and TVR plants; electrical SEC (kWh/t) suits MVR. Converting to a common cost basis using your steam and power tariffs — as shown in the conversion table above — is the only fair way to rank configurations, because a kWh of steam and a kWh of electricity are not worth the same.

Does boiler efficiency change the SEC calculation?
It changes the primary-energy SEC but not the process SEC. An evaporator consuming 697 kWh/t of steam draws 697 ÷ 0.90 ≈ 774 kWh/t of fuel if the boiler is 90% efficient. Guarantees are written on the process basis; fuel budgets and carbon accounting belong on the primary basis.

Conclusion

The specific energy consumption evaporator projects should target is not a catalogue constant — it is a calculation you can now reproduce: 627 kWh/t divided by steam economy for steam-heated trains, or compressor work times vapor flow for MVR. Run the three worked examples against your own feed analysis, concentrations, and tariffs, and the cheapest flowsheet usually declares itself within an afternoon of arithmetic.

When the screening numbers are close, or your liquor’s boiling point elevation and scaling behavior are uncertain, the deciding data comes from a pilot run on your actual feed. Talk to us about pilot testing with Lexin-tech’s evaporation and crystallization team, or browse the resource library for more engineering guides in this series.

References

• Perry, R. H. and Green, D. W., Perry’s Chemical Engineers’ Handbook, 9th ed., McGraw-Hill (sections on evaporation and vapor recompression).
• McCabe, W. L., Smith, J. C. and Harriott, P., Unit Operations of Chemical Engineering, 7th ed., McGraw-Hill.
• Geankoplis, C. J., Transport Processes and Separation Process Principles, 4th ed., Pearson.
• IAPWS, Industrial Formulation 1997 for the Thermodynamic Properties of Water and Steam, International Association for the Properties of Water and Steam.
• Mersmann, A. (ed.), Crystallization Technology Handbook, 2nd ed., Marcel Dekker.
• Desalination (Elsevier journal; peer-reviewed studies on evaporator and vapor recompression energy consumption).
• U.S. Geological Survey, Mineral Commodity Summaries (annual; hydrometallurgical production and processing context).

Changelog

• 2026-10-22 — First published.

Related reading

• Crystallization in practice