Material Balances
Mass is conserved, so everything that enters a process must leave, react or accumulate. Learn the general balance equation, flowcharts, degrees-of-freedom analysis, balances on mixers, splitters, evaporators and distillation columns, and systems with recycle, bypass and purge.
Material balances are the bread and butter of chemical engineering: they size every stream in a plant, find unknown flows from a few measurements, and reveal losses.
The method is always the same — draw a flowchart, label knowns and unknowns, count degrees of freedom, write balances, solve.
An analogy
A material balance is like your bank account: deposits − withdrawals + interest − fees = change in balance. For steady plants, the change is zero.
The balance equation
Choose a system boundary. Total mass is always conserved; component moles are conserved unless reactions generate or consume them. At steady state nothing accumulates.
Solving procedure
Draw and label the flowchart; choose a basis (e.g. 100 kmol feed); convert units; count DOF; write total and component balances; solve (simultaneous equations); scale to actual flows; check with an unused balance.
Multiple units
Balances can be written around the whole plant, each unit or mixing points. Start where DOF = 0. Tie components (inert in one stream) simplify calculations.
Recycle, bypass and purge
Recycle returns unreacted feed so overall conversion exceeds single-pass conversion. Bypass mixes part of the feed around a unit. Purge removes inerts that would otherwise build up in a recycle loop. Large flowsheets are solved by tearing streams and iterating (sequential modular) or solving all equations together (equation-oriented).
Formulas
- General balance:
In − Out + Generation − Consumption = Accumulation - Steady state, no reaction:
In = Out (total and each component) - Binary column:
D = F (z_F − x_B)/(x_D − x_B), B = F − D - Degrees of freedom:
DOF = unknowns − independent equations (must be 0 to solve)
Worked example: Evaporator
Given: 1000 kg/h brine, 10 wt % salt; Concentrated to 40 wt %
Find: Product and water evaporated.
- Salt balance: 100 = 0.40 P → P = 250 kg/h
- Water evaporated = 1000 − 250 = 750 kg/h
Answer: 250 kg/h product, 750 kg/h vapour
Worked example: Distillation split
Given: F = 100 kmol/h, z = 0.40; x_D = 0.95, x_B = 0.05
Find: D and B.
- D = 100 × (0.40 − 0.05)/(0.95 − 0.05) = 38.9 kmol/h
- B = 61.1 kmol/h
- Check: 0.95 × 38.9 + 0.05 × 61.1 = 40 ✓
Answer: D = 38.9, B = 61.1 kmol/h
Common mistakes
- Wrong: Balancing total moles in reacting systems. — Right: Moles aren't conserved with reaction — use mass, atoms or extents.
- Wrong: Mixing mass and mole fractions. — Right: Pick one basis and convert everything consistently.
Applications
- Plant design: Stream tables for every flowsheet start from balances.
- Operations: Data reconciliation finds meter errors and losses.
- Environmental reporting: Emission inventories from mass balances.
Key takeaways
- In − Out + Gen − Cons = Acc.
- Flowchart → basis → DOF → balances → check.
- Recycle raises overall conversion; purge removes inerts.
- D = F(z − x_B)/(x_D − x_B) for binary splits.
Going further
- Data reconciliation: Least-squares adjustment of measured flows to satisfy balances.
- Sequential modular vs equation-oriented: Two strategies used by process simulators.
- Transient balances: Accumulation terms give ODEs for tanks and batch processes.