Mass balance in food processing: milk, evaporator and dryer examples

A mango drying cooperative in Burkina Faso loads 1,000 kg of fresh slices at 85% moisture and wants dried mango at 18%. The supervisor subtracts the percentages, expects to remove 670 kg of water and sets the dryer time to match. The batch comes out soft and sticky, because the real figure is about 817 kg. The arithmetic was fine; the method was wrong. The fix is the simplest mass balance food processing engineers use: anchor the sum on the dry solids, which never leave the dryer.

In short

  • A mass balance applies conservation of mass: input = output + accumulation. At steady state, everything that goes in comes out.
  • Write one total balance and one balance per component. A tie component, such as the dry solids in drying, turns most problems into a two-line calculation.
  • Milk standardisation is a fat balance; the Pearson square gives the parts of whole and skimmed milk for a target fat.
  • Evaporation and drying are solids balances: the dry solids stay constant while water leaves.
  • Never subtract wet-basis moisture percentages. Work through the dry solids instead.

What is a mass balance in food processing?

A mass balance is an account of every kilogram that enters and leaves a process step. Mass is not created or destroyed, so for any system:

input = output + accumulation

The system can be one machine or a whole factory. You define it with a system boundary, an imaginary line around the part of the process you are studying. Every stream that crosses the line is an input or an output.

A process at steady state is one whose flows and compositions do not change with time, so nothing accumulates and input = output. A batch process, such as a cheese vat, is balanced from start to finish: initial contents plus additions equal final contents plus removals.

You can write one balance for total mass and one for each component, such as fat, water or dry solids. A tie component is a component that passes through a step unchanged and appears in only one input and one output. Dry solids are the tie in evaporation and drying. In blending, the key balance is on the component you are standardising, such as fat.

How do you build a mass balance food processing teams can trust?

Follow the same six steps every time. Most errors come from a missed stream or mixed units, not from the arithmetic.

  1. Sketch and bound the process. Label every stream, including rinse water, laboratory samples, start-up and shut-down flushes and waste.
  2. Choose a basis, such as 100 kg of feed, one hour of operation or one batch. Every flow then refers to it.
  3. Convert everything to mass. Litres become kilograms through density; whole milk is about 1.03 kg/L at 20 Β°C.
  4. Express compositions as mass fractions on one consistent basis, wet or dry, never a mix.
  5. Write the balances: the tie component first, then the total, then any other component.
  6. Solve and check. Put the answers back into every balance, then ask whether the product mass and composition look plausible.

The Introduction to Food Process Engineering course drills this method on blending, standardisation, evaporation and drying problems before moving on to energy balances.

How do you standardise milk fat with a mass balance?

Milk standardisation is the adjustment of milk fat, and sometimes protein, to a fixed target by blending streams or removing cream. Raw milk fat varies with breed, season and feed, but every pack must match its label. Minimum fat contents for each milk category are set nationally, so check your national legislation.

When two streams are blended, the fat balance is:

m₁ Γ— x₁ + mβ‚‚ Γ— xβ‚‚ = (m₁ + mβ‚‚) Γ— x_mix

Here m is mass (kg) and x is the fat mass fraction. The Pearson square is a graphical shortcut for two ingredients and one component. Write each ingredient’s fat content on the left of a square and the target in the centre, then subtract diagonally, ignoring signs, to get the parts of each ingredient. The target must lie between the two ingredient values.

Worked example

A dairy needs 6,000 kg of reduced-fat milk at 2.0% fat, blended from whole milk at 3.8% fat and skimmed milk at 0.05% fat.

Parts of whole milk = 2.0 βˆ’ 0.05 = 1.95. Parts of skimmed milk = 3.8 βˆ’ 2.0 = 1.80. Total = 3.75 parts.

Whole milk = 6,000 kg Γ— 1.95 Γ· 3.75 = 3,120 kg. Skimmed milk = 6,000 kg Γ— 1.80 Γ· 3.75 = 2,880 kg.

Check: fat = 3,120 Γ— 0.038 + 2,880 Γ— 0.0005 = 118.56 + 1.44 = 120.0 kg, and 120.0 Γ· 6,000 = 2.00%.

Large dairies usually standardise in-line: a separator splits milk into skimmed milk and cream, and a control system remixes just enough cream to hit the target. The surplus cream comes from the same fat balance.

Worked example

20,000 kg of raw milk at 4.1% fat is standardised to 3.5% fat by removing cream at 40% fat. How much cream leaves?

Let C be the mass of cream. Fat balance: 20,000 Γ— 0.041 = 0.40 Γ— C + 0.035 Γ— (20,000 βˆ’ C).

820 = 0.40C + 700 βˆ’ 0.035C, so 120 = 0.365C and C = 328.8 kg of cream.

Standardised milk = 20,000 βˆ’ 328.8 = 19,671.2 kg. Check: 19,671.2 Γ— 0.035 + 328.8 Γ— 0.40 = 688.5 + 131.5 = 820.0 kg of fat, as in the feed.

The same arithmetic makes juice from concentrate, brine and syrup. With a fixed batch size, you need one more ingredient than the number of component targets, so hitting both fat and protein takes three ingredients and simultaneous equations instead of a Pearson square.

How do you calculate evaporation with a solids balance?

In evaporation, the solids balance fixes the concentrate flow and the total balance gives the water removed, because only water leaves as vapour:

F Γ— x_F = P Γ— x_P and V = F βˆ’ P

F, P and V are the feed, product (concentrate) and vapour flows in kg/h, and x is the solids mass fraction.

Worked example

An orange juice plant evaporates 10,000 kg/h of juice at 12 Β°Brix to concentrate at 65 Β°Brix. Treat Β°Brix as per cent soluble solids by mass, which is what it approximates.

Solids in: 10,000 Γ— 0.12 = 1,200 kg/h.

Concentrate: P = 1,200 Γ· 0.65 = 1,846 kg/h.

Water evaporated: V = 10,000 βˆ’ 1,846 = 8,154 kg/h, or 81.5% of the feed.

Check: 1,846 kg/h of concentrate holds 1,200 kg/h of solids and 646 kg/h of water, which is 65% solids.

Over four-fifths of the feed leaves as vapour, and each kilogram needs roughly 2,250 to 2,400 kJ of latent heat, depending on the boiling temperature. That is why a single-effect evaporator uses roughly 1 kg of steam per kilogram of water removed, and why large plants add effects or vapour recompression. Sizing the evaporator and its steam supply is the next step, covered in Intermediate Food Process Engineering.

What is the difference between wet-basis and dry-basis moisture?

Wet-basis moisture is water as a fraction of total mass; dry-basis moisture is water per unit mass of dry solids. Labels, specifications and laboratory reports use wet basis. Drying engineers prefer dry basis, because the dry solids stay constant while water leaves.

X_wb = water Γ· (water + dry solids) X_db = water Γ· dry solids

X_db = X_wb Γ· (1 βˆ’ X_wb) X_wb = X_db Γ· (1 + X_db)

Wet basis (kg water per kg product)Dry basis (kg water per kg dry solids)What it means
0.90 (90%)9.009 kg of water per kg of solids
0.85 (85%)5.67Many fresh fruits
0.50 (50%)1.00Equal water and solids
0.18 (18%)0.220Bases start to converge
0.12 (12%)0.136Dried pasta or grain range
0.04 (4%)0.042Milk powder range

At low moisture the two figures are close; at high moisture they diverge sharply.

How much water must a dryer remove?

The water a dryer removes equals the dry solids multiplied by the fall in dry-basis moisture, as the mango batch shows.

Worked example

Basis: 1,000 kg of fresh mango slices at 85% moisture (wet basis), dried to 18% moisture.

Dry solids = 1,000 Γ— (1 βˆ’ 0.85) = 150 kg. These do not leave the dryer.

In the dried product, 150 kg is 82% of the mass, so dried mango = 150 Γ· 0.82 = 182.9 kg.

Water removed = 1,000 βˆ’ 182.9 = 817.1 kg.

Dry-basis check: initial 0.85 Γ· 0.15 = 5.667; final 0.18 Γ· 0.82 = 0.2195; water removed = 150 Γ— (5.667 βˆ’ 0.2195) = 817.1 kg.

Subtracting percentages (85 βˆ’ 18 = 67%, so 670 kg) understates the load by 147 kg, about 18%.

The yield follows directly: 182.9 kg of dried mango per tonne of fresh slices, or about 5.5 kg of fresh fruit per kilogram of product. That figure drives raw material buying and product costing.

The error also has a food safety side. Dried fruit is kept stable by its low water activity (aw), a measure of how available its water is to microorganisms. Product left wetter than specified has a higher water activity and a greater risk of mould and yeast growth.

Where do mass balances go wrong in real plants?

  • Subtracting wet-basis percentages. Work through the dry solids or convert to dry basis.
  • Mixing litres and kilograms. Fat in milk is a mass percentage; some instruments report grams per litre. Convert with density before blending.
  • Missing streams. Line flushes, start-up and shut-down mixes of water and product, laboratory samples, residues in tanks and pipes, and spills all leave the system.
  • Trusting drifting instruments. Flow meters and in-line fat or Brix sensors need regular checks against laboratory results. When a balance will not close, question the data before the arithmetic.
  • Rounding too early. Keep three or four significant figures until the final answer.

Plants run daily balances to track yield and losses: comparing milk received with milk packed exposes leaks, overfilling and poor product recovery. Balances also support traceability, because when rework is blended into a batch, the record of how much went where matters for allergen control and any recall.

Frequently asked questions

What is a tie component in a mass balance?

A tie component is a substance that passes through a process step unchanged and appears in only one input and one output stream. Because its mass is the same on both sides, it links the flows directly. Dry solids are the classic tie in evaporation and drying, because only water leaves. Fat in milk blending is not a strict tie, since every stream contains some, but a fat balance still solves it.

Can you subtract moisture percentages to find the water removed?

No. Wet-basis percentages refer to a total mass that shrinks as water leaves, so subtracting them understates the water removed. Drying 1,000 kg of fruit from 85% to 18% moisture removes about 817 kg of water, not 670 kg. Calculate the dry solids first, find the final mass from the target moisture, then subtract.

Is the Pearson square accurate for milk standardisation?

Yes, for two ingredients and one component, provided every composition is a mass fraction and the target lies between the two ingredient values. The Pearson square is a graphical form of the fat balance, so it gives exactly the same answer as the algebra. For three or more ingredients or two targets, such as fat and protein, write one balance per component and solve them together.

What does steady state mean in a mass balance?

Steady state means that flows, temperatures and compositions at every point stay constant over time. Nothing accumulates inside the system, so total input equals total output. Continuous operations such as pasteurisers and evaporators run close to steady state once started. Start-up, shut-down and batch operations do not, so they are balanced over a whole batch or a defined period.

Why does a dairy milk balance never close exactly?

Real plants lose small amounts of product to line flushes, start-up and shut-down mixes, laboratory samples, residues in tanks and pipes, and leaks, and flow meters drift. A small, stable gap between milk received and milk packed is normal. A gap that grows over several days points to a fault, such as a leaking valve, poor product recovery or a meter that needs calibration.

What is the difference between Brix and total solids?

Degrees Brix measures soluble solids, by refractometer or density, expressed as the equivalent percentage of sucrose by mass. Total solids is everything left after all the water is removed, including insoluble fibre and pulp. For clear juices and syrups the two are close. For tomato paste and pulpy juices, total solids exceed Brix, so use one measure consistently in every balance.

Next step. The Introduction to Food Process Engineering course teaches moisture bases and mass and energy balances for blending, milk standardisation, evaporation and drying, then heat exchangers, pumps and hygienic design. It ends with a proctored final assessment and an ASC certificate. You can also see all eleven food science and technology courses.

Sources. R. Paul Singh, Dennis R. Heldman and Ferruh Erdogdu, Introduction to Food Engineering, 6th edn (Academic Press, 2024). P. J. Fellows, Food Processing Technology: Principles and Practice, 5th edn (Woodhead Publishing, 2022). Romeo T. Toledo, Rakesh K. Singh and Fanbin Kong, Fundamentals of Food Process Engineering, 4th edn (Springer, 2018). Codex Alimentarius Commission, General Standard for the Use of Dairy Terms (CXS 206-1999), available from the Codex Alimentarius website.

This article is general guidance on food process calculations and is not a substitute for the applicable standard, your national legislation or the advice of a qualified food engineer.

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