Material and Energy Balances: practice problems and exam questions

The core of the first year in chemical engineering: mass balances on mixers, separators and reactors, energy balances, gas behaviour and stoichiometry.

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Cram pack (PDF)Formula sheets, method steps and practice problems with full solutions, ready to print.

Topics in this course

  1. Mass balancemixing two streams · blending to a target · evaporatorcalculations
  2. Energy balanceheating a stream · heat exchanger · exchanger with heat losscalculations
  3. Ideal gas lawP from n, V, T · from mass · state change or densitycalculations
  4. Stoichiometrymass to mass · limiting reactant · percent yieldcalculations

Exam tips for Material and Energy Balances

  • Draw the flowsheet and label every stream before you write a balance.
  • Count the unknowns and the independent equations before you start solving.
  • Choose a basis, such as 100 kg/h, when no flow rate is given.

Worked examples

Mass balanceEasy
MixerStream 1300 kg/h · 30% saltStream 2250 kg/h · 40% saltStream 3x₃ = ? %

Two salt solutions are mixed at steady state. Find the mass fraction of salt in the outlet x3.

ṁ1 = 300 kg/hx1 = 30 %ṁ2 = 250 kg/hx2 = 40 %

Find x3 (%)

Show hints
  1. At steady state, what goes in comes out: write a total balance and a balance for the dissolved component.
  2. Total: ṁ₁ + ṁ₂ = ṁ₃. Component: ṁ₁x₁ + ṁ₂x₂ = ṁ₃x₃.
  3. ṁ3 = 550 kg/h.
Show the answer and the step-by-step solution

Steady-state mass balance: in = out, for the total mass and separately for each component (no reaction).

Total balance

ṁ3 = ṁ1 + ṁ2 = 300 + 250 = 550 kg/h

Salt balance

ṁ1x1 + ṁ2x2 = ṁ3x3
x3 = (300·30 + 250·40)/550 = 34.55 %

Answer

x3 = 34.55 %
Energy balanceMedium
hot sidecold sideglycol mix · 150 °C120 °Cwater · 15 °Ct = ? °C

Hot glycol mix cools water in a counter-flow heat exchanger with no heat loss. Find the water outlet temperature tout.

ṁh = 1 kg/scp,h = 3.6 kJ/(kg·K)Th: 150 → 120 °Cṁc = 1 kg/scp,c = 4.18 kJ/(kg·K)tin = 15 °C

Find tout (°C)

Show hints
  1. Heat lost by the hot stream equals heat gained by the cold stream.
  2. Find Q from the hot side first, since you know both of its temperatures.
  3. Q = 108 kW.
Show the answer and the step-by-step solution

Exchanger energy balance: with no losses, ṁhcp,h(Th,in − Th,out) = ṁccp,c(tout − tin).

Heat released by the hot stream

Q = 1·3.6·(150 − 120) = 108 kW

Heat picked up by the water

tout = tin + Q/(ṁccp,c) = 15 + 108/(1·4.18) = 40.84 °C

Answer

tout = 40.84 °C

How practising works

  • “Practise this course” mixes problems from all its topics, like a real exam.
  • Every problem has new values, so you learn the method instead of an answer.
  • Stuck? Open a small hint first, and the full explanation only when you need it.
  • Mock exams are timed and have no hints, and you get your score and a full review at the end.

Calculators for this course

Ideal gas law calculatorPercent yield calculatorStoichiometry calculator (grams to grams and limiting reactant)Molar mass calculatorCombined gas law calculator (Boyle, Charles and Gay-Lussac)Empirical formula calculator (and molecular formula)

Related courses

FE Chemical exam practiceThermodynamics 1General ChemistryChemical Reaction EngineeringPhysics for Engineers
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