Material balances without reaction on single units

Why it matters

Mixers, dryers, evaporators, crystallisers, filters, distillation and extraction columns change compositions without chemical reaction. A material balance on one such unit fixes its product flows, tells you how much water an evaporator must remove or how much product a crystalliser gives, and is the building block for every larger flowsheet.

Key ideas

General balance. For any system and any conserved quantity: Input + Generation − Output − Consumption = Accumulation. Without chemical reaction, generation and consumption of every species are zero. At steady state accumulation is zero, so input = output, both for the total mass and for each component.

Types of process. Continuous processes run with streams flowing in and out; at steady state, balances are written on rates (kg/h). Batch processes load material, process it, then unload; the balance is written on amounts (kg) between the initial and final states (input = initial contents, output = final contents). Semi-batch processes have flows in or out but not both and are unsteady.

Mass vs moles. Without reaction, both total mass and total moles are conserved, and each component is conserved in either unit. Use whichever matches the data (mass for solids and liquids, moles for gases).

Independent equations. For N components, N independent balances exist (N component balances, or N − 1 component balances plus the total). The total balance is not an extra equation.

Tie (key) component. A component that passes from one stream to a single other stream unchanged – dry solid in a dryer, salt in an evaporator, an inert gas in an absorber – lets you link two streams in one step. Look for it first; it often gives the answer without simultaneous equations.

Procedure.

1. Draw and label the flowchart with all known and unknown flows and compositions.
2. Choose a basis (a given flow, or 100 kg/h).
3. Do a degree-of-freedom check.
4. Write balances, starting with those that involve only one unknown (tie component).
5. Solve, then check with a balance you did not use.

Phase-equilibrium data – solubility in crystallisation, vapour–liquid or liquid–liquid equilibrium – often provide the extra relation needed. Take such data from your data book; do not assume them.

Formulas

• Input + Generation − Output − Consumption = Accumulation – general balance.
• Σ ṁ_in = Σ ṁ_out – steady state, no reaction (kg/h).
• Σ ṁ_in·w_i,in = Σ ṁ_out·w_i,out – component i balance, w mass fraction.
• F = D + B and F·x_F = D·x_D + B·x_B – two-product separator.
• D / F = (x_F − x_B) / (x_D − x_B) – from the two equations above.
• Recovery of i in D = D·x_D / (F·x_F)
• ṁ_tie,in = ṁ_tie,out – tie component, e.g. dry solid: F·(1 − w_F) = P·(1 − w_P).

Worked examples

Example 1 (standard). A dryer receives 1000 kg/h of wet solid containing 60 % moisture (wet basis) and dries it to 10 % moisture. Find the dried product rate and the water evaporated.

1. Tie component – dry solid: 1000 × (1 − 0.60) = 400 kg/h.
2. Dried product: P × (1 − 0.10) = 400, so P = 400 / 0.90 = 444.4 kg/h.
3. Total balance: water evaporated = 1000 − 444.4 = 555.6 kg/h.
4. Check with a water balance: in 600 kg/h; out 0.10 × 444.4 + 555.6 = 44.4 + 555.6 = 600 kg/h ✓.

Answer: P = 444.4 kg/h; water evaporated = 555.6 kg/h

Example 2 (GATE level). 1000 kg/h of an aqueous solution containing 50 wt % KNO₃ enters a crystalliser. 100 kg/h of water is evaporated and the remainder is cooled to 20 °C, where the solubility of KNO₃ is 0.316 kg per kg water (take from data book). Pure anhydrous crystals are separated from saturated mother liquor. Find the crystal yield and the mother-liquor flow.

1. Feed: KNO₃ = 500 kg/h, water = 500 kg/h.
2. Water balance: water in mother liquor = 500 − 100 = 400 kg/h (crystals are anhydrous).
3. KNO₃ dissolved in mother liquor = 0.316 × 400 = 126.4 kg/h.
4. KNO₃ balance: crystals = 500 − 126.4 = 373.6 kg/h.
5. Mother liquor = 400 + 126.4 = 526.4 kg/h.
6. Check (total balance): 100 + 373.6 + 526.4 = 1000 kg/h ✓. Fraction of KNO₃ recovered = 373.6 / 500 = 74.7 %.

Answer: Crystals ≈ 373.6 kg/h; mother liquor ≈ 526.4 kg/h

Common mistakes

• Using wet-basis and dry-basis moisture interchangeably (60 % wet basis = 1.5 kg water/kg dry solid).
• Assuming the dried product is the dry solid only; it still holds residual moisture.
• Writing N component balances and the total balance and treating them as N + 1 independent equations.
• Forgetting water of crystallisation when crystals are hydrates (e.g. Na₂SO₄·10H₂O).
• Using solubility per 100 g of solution when the data are per 100 g of water.
• Solving for a product flow from one component only, when the other product also carries that component.

For GATE CH

Expect NAT questions on dryers, evaporators, crystallisers, mixers and two-product separators, often with a recovery or purity specification. Practise identifying the tie component, converting wet/dry basis, and using solubility data with hydrates.

Quick check

1. At steady state without reaction, what is the accumulation term?
2. 100 kg/h of 70 % moisture solid is dried to 10 % moisture. What is the dried product rate?
3. Feed 200 kg/h with 60 % A gives a distillate with 90 % A and bottoms with 5 % A. What is D?
4. What is 20 % moisture (wet basis) on a dry basis?
5. Mixing 100 kg/h water and 50 kg/h ethanol gives what ethanol mass fraction?

Answers: 1. zero; 2. 33.3 kg/h; 3. 129.4 kg/h; 4. 0.25 kg water/kg dry solid; 5. 0.333.