Material balances on multiple-unit processes

Choosing subsystems, counting independent balances, overall-first strategy, tie components through series units and solving a two-column separation train.

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Why it matters

Real plants are trains of units – evaporator effects in series, distillation columns in sequence, a reactor followed by separators. Balances must be written so that the whole flowsheet closes, and the order in which you choose subsystems decides whether the problem takes five minutes or an hour. This topic is the bridge from single-unit balances to recycle, bypass and purge problems.

Key ideas

Subsystems. A balance can be written around any boundary:

  • each individual unit;
  • a mixing point (several streams combine) or a splitting point (one stream divides);
  • any group of units;
  • the overall process, whose only streams are the fresh feeds and final products. Internal streams that cross no chosen boundary do not appear in that balance. The overall balance is often the best starting point because it eliminates internal (tie) streams.

Independent balances. For N components, each subsystem gives at most N independent material balances. But the balances on all the individual units together already contain the overall balance – the overall balance is not independent of the full set of unit balances. Use it as a convenient replacement for one unit, not as an extra equation.

Strategy.

  1. Draw and fully label the flowchart; number the streams.
  2. Do a DOF analysis for the overall process and for each unit or mixing/splitting point.
  3. Start with a subsystem with DOF = 0, solve it, and carry the results to the next subsystem (its DOF falls as streams become known).
  4. Finish with a check balance that was not used.

Units in series. For a series of concentrating steps (multiple-effect evaporators, a train of dryers), a tie component (dissolved solid, dry solid) passes through every unit unchanged, so each intermediate flow follows from its concentration directly.

Separation trains. In a sequence of distillation columns, each column takes one key component overhead. Component recoveries (fraction of feed component sent to a given product) are often specified instead of compositions.

Mixing and splitting points. A mixing point has N balances and no equipment. A splitter has the same composition in all outlets and only one independent balance. These points are where recycle and bypass streams join and leave – the next topic.

Formulas

  • Overall: Σ (fresh feeds) = Σ (final products) – steady state, no reaction, each component and total.
  • Unit k: Σ ṁ_in,k = Σ ṁ_out,k – one set of N balances per unit.
  • DOF_process = Σ DOF_units − number of tie-stream variables – a bookkeeping check (tie variables are counted in two units).
  • Recovery of i in stream j = ṅ_j·x_i,j / (ṅ_F·x_i,F) – fraction of feed component i leaving in stream j.
  • Tie component in series: ṁ_k = ṁ_solid / w_k – flow leaving step k when the solid mass fraction there is w_k.

Worked examples

Example 1 (standard). 10 000 kg/h of 10 wt % NaOH solution is concentrated in a double-effect evaporator: to 20 % in the first effect and to 50 % in the second. Find the water evaporated in each effect.

  1. Tie component: NaOH = 0.10 × 10 000 = 1000 kg/h through both effects.
  2. Leaving effect 1: 1000 / 0.20 = 5000 kg/h. Water evaporated in effect 1 = 10 000 − 5000 = 5000 kg/h.
  3. Leaving effect 2: 1000 / 0.50 = 2000 kg/h. Water evaporated in effect 2 = 5000 − 2000 = 3000 kg/h.
  4. Overall check: total water evaporated = 10 000 − 2000 = 8000 = 5000 + 3000 ✓.

Answer: 5000 kg/h from effect 1; 3000 kg/h from effect 2

Example 2 (GATE level). 100 kmol/h of a mixture of 30 % benzene (B), 40 % toluene (T) and 30 % xylene (X) is separated in two columns. Column 1 gives distillate D1 (95 % B, 5 % T, no X); its bottoms B1 feeds column 2. Column 2 gives distillate D2 (3 % B, 95 % T, 2 % X) and bottoms B2 (2 % T, 98 % X, no B). Find D1, D2, B2 and the flow and composition of B1.

  1. DOF: the overall process has three unknown product flows and three component balances – DOF = 0. Start with the overall balance (the tie stream B1 drops out).
  2. B balance: 30 = 0.95·D1 + 0.03·D2, so D1 = (30 − 0.03·D2) / 0.95.
  3. X balance: 30 = 0.02·D2 + 0.98·B2, so B2 = (30 − 0.02·D2) / 0.98.
  4. T balance: 40 = 0.05·D1 + 0.95·D2 + 0.02·B2. Substituting: 1.579 + 0.612 + (0.95 − 0.00158 − 0.00041)·D2 = 40, so 0.94801·D2 = 37.809 and D2 = 39.88 kmol/h.
  5. Then D1 = (30 − 1.196) / 0.95 = 30.32 kmol/h and B2 = (30 − 0.798) / 0.98 = 29.80 kmol/h. Check: 30.32 + 39.88 + 29.80 = 100.0 ✓.
  6. Column 2 balance gives B1 = D2 + B2 = 69.68 kmol/h, with B = 0.03 × 39.88 / 69.68 = 0.017, T = (0.95 × 39.88 + 0.02 × 29.80) / 69.68 = 0.552, X = 0.431 (sum 1.000).
  7. Recoveries: benzene in D1 = 0.95 × 30.32 / 30 = 96.0 %; toluene in D2 = 0.95 × 39.88 / 40 = 94.7 %.

Answer: D1 = 30.3, D2 = 39.9, B2 = 29.8 kmol/h; B1 = 69.7 kmol/h (1.7 % B, 55.2 % T, 43.1 % X)

Common mistakes

  • Writing balances on every unit and the overall process and counting them all as independent.
  • Starting with a unit that has DOF > 0 and getting stuck in simultaneous equations.
  • Forgetting internal streams when balancing a single unit, or including them in the overall balance.
  • Mixing mass and mole fractions between units.
  • Not checking that compositions in each stream sum to one.

For GATE CH

Expect two- or three-unit flowsheets (evaporator trains, column sequences, absorber plus stripper) with NAT answers for an intermediate stream flow or composition, and recovery-based specifications. Practise choosing the overall balance first and then the unit that closes the remaining unknowns.

Quick check

  1. Which streams appear in the overall balance of a multi-unit process?
  2. A 3-unit process handles 2 components. How many independent material balances exist in total?
  3. 1000 kg/h of 5 % sugar solution is concentrated to 25 % in two stages. How much water is removed in total?
  4. Why is the overall balance often solved first?
  5. In Example 2, why is B1 not needed for the overall balance?

Answers: 1. only fresh feeds and final products; 2. six (2 per unit); 3. 800 kg/h; 4. internal streams cancel, reducing unknowns; 5. it is an internal (tie) stream.

Interview questions

All Process Calculations interview questions

Try answering each one aloud before you open it.

  1. 1.Explain the difference between a single-unit and a multiple-unit process in material balances.Concept

    A single-unit process involves only one processing step or unit, where material balances are calculated for that specific unit. In contrast, a multiple-unit process involves several interconnected units, and material balances must be calculated for each unit as well as the overall process to ensure mass conservation throughout the entire system.

  2. 2.Why is it important to perform material balances on multiple-unit processes?Application

    Performing material balances on multiple-unit processes is crucial because it helps identify inefficiencies, losses, or accumulation within the system. It ensures that all units are operating optimally and that the overall process is balanced, which is essential for process control, optimization, and safety.

  3. 3.What happens if there is an accumulation of material in a unit within a multiple-unit process?Application

    If there is an accumulation of material in a unit, it indicates that the input exceeds the output, which can lead to operational issues such as overpressure, equipment damage, or process inefficiencies. It is essential to identify and rectify the cause of accumulation to maintain a balanced and safe process.

  4. 4.How do you approach solving a material balance problem for a multiple-unit process?Application

    To solve a material balance problem for a multiple-unit process, follow these steps: 1) Define the system boundaries and identify all units involved. 2) Write material balance equations for each unit and the overall process. 3) Use known process data and relationships to solve the equations. 4) Check the consistency of the results to ensure mass conservation.

  5. 5.Explain the role of recycle streams in multiple-unit processes.Application

    Recycle streams are used in multiple-unit processes to improve efficiency by returning unreacted materials or by-products back to an earlier stage in the process. This reduces waste, increases yield, and can lower raw material costs. Material balances must account for these streams to ensure accurate calculations.

  6. 6.What is the impact of a purge stream in a multiple-unit process?Application

    A purge stream is used to remove unwanted by-products or inert materials from a process, preventing their accumulation. While it helps maintain process efficiency and product quality, it also represents a loss of material that must be accounted for in material balances.

  7. 7.In a two-unit process, Unit A receives 200 kg/h and sends out 150 kg/h, while Unit B receives 150 kg/h and sends out 180 kg/h (no reaction). What is happening in each unit and in the whole system?Numerical

    Unit A accumulates 200 − 150 = +50 kg/h, so its hold-up is increasing. Unit B has 150 − 180 = −30 kg/h, i.e. it is being emptied – its inventory is falling by 30 kg/h (it cannot do this for ever). The whole system accumulates 50 − 30 = +20 kg/h, which matches the overall balance 200 − 180 = 20 kg/h. Neither unit, nor the system, is at steady state.

  8. 8.Describe how you would handle a situation where the material balance equations for a multiple-unit process do not initially balance.Application

    If the material balance equations do not initially balance, first check for errors in data or assumptions. Verify all inputs, outputs, and process conditions. Consider if there are unaccounted streams, leaks, or measurement errors. Adjust the equations or process data as necessary and re-calculate to achieve a balanced system.

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