Laws of Conservation of Mass & Energy — Material and Energy Balances — Free Notes for APPSC Assistant Environmental Engineer

Last reviewed 4 Nov 2026 · 3 min read

1. The general balance equation

For any quantity (total mass, a component, energy) in a system:

FormulaGeneral balance

Input + Generation - Output - Consumption = Accumulation

Case	Simplified form
Steady state (nothing changes with time)	Accumulation = 0
Total mass (no nuclear reactions)	Generation = Consumption = 0 → Input − Output = Accumulation
Steady state, no reaction	Input = Output
Unsteady (transient) process	Accumulation ≠ 0 — e.g. filling or draining a tank

Mass is conserved in a chemical reaction, but moles of a component are not (they are generated or consumed). Atoms (elements) are conserved.

2. Solving material balance problems — steps

Draw and label a flowsheet with all streams, flow rates and compositions.
Choose a basis — e.g. 100 kg of feed or 1 hour of operation.
List unknowns and write independent balances (total + component, or atom balances).
Check the degrees of freedom (unknowns − equations = 0).
Solve, starting with a "tie" component that passes unchanged between two streams, and check with an unused balance.

Worked Example1 — Mixing (no reaction)

100 kg/h of a 20% (by mass) salt solution is diluted with pure water to give a 5% solution. Find the water added W and the product P. Salt balance: $0.20 \times 100 = 0.05 P \Rightarrow P = 400$ kg/h. Total balance: $100 + W = 400 \Rightarrow W = 300$ kg/h.

3. Balances with reaction

Term	Meaning
Stoichiometric ratio	Moles in the balanced equation
Limiting reactant	The one present in the smallest stoichiometric proportion — it would be used up first
Excess reactant	% excess = (fed − theoretical) ÷ theoretical × 100
Fractional conversion	Moles reacted ÷ moles fed (of a reactant)
Yield	Moles of desired product formed ÷ moles that could have formed (or per mole reactant fed/consumed — definitions vary)
Selectivity	Moles of desired product ÷ moles of undesired product

Worked Example2 — Combustion of methane

CH₄ + 2O₂ → CO₂ + 2H₂O. 10 mol/h of CH₄ is burnt with 20% excess air, completely. Theoretical O₂ = 20 mol/h; O₂ supplied = 24 mol/h; N₂ with it = 24 × 79/21 = 90.3 mol/h. Flue gas: CO₂ 10, H₂O 20, O₂ 4 (excess), N₂ 90.3 mol/h — total 124.3 mol/h.

4. Energy balance

Closed system (no mass crosses the boundary)

FormulaFirst law, closed system

Δ U + Δ E_{k} + Δ E_{p} = Q - W

Q = heat added to the system; W = work done by the system.

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1. The general balance equation

2. Solving material balance problems — steps

3. Balances with reaction

4. Energy balance

Closed system (no mass crosses the boundary)

Practise this chapter for your exam