Material and Energy Balances: cram pack

Formula sheets, the key method steps for every topic and 8 practice problems, with full solutions at the back.

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1Formulas and methods

Mass balance

Steady state
in = out
Total balance
ṁ1 + ṁ2 = ṁ3
Component balance
ṁ1x1 + ṁ2x2 = ṁ3x3
Evaporator
F·xF = P·xP, W = F − P

Energy balance

Sensible heat
Q = ṁ·cp·ΔT
Latent heat
Q = ṁ·hfg
Mixing
ΣṁcpTin = ΣṁcpTout
Water
cp ≈ 4.18 kJ/(kg·K)

Ideal gas law

Ideal gas law
pV = nRT
Gas constant
R = 8.314 J/(mol·K)
Combined gas law
p1V1/T1 = p2V2/T2
Moles
n = m/M
Temperature
T[K] = T[°C] + 273.15

Stoichiometry

Moles
n = m/M
Concentration
c = n/V
Mole ratio
use the coefficients of the balanced equation
Limiting reactant
the one that runs out first
Percent yield
actual / theoretical × 100%

Method

  1. Write and balance the chemical equation.
  2. Convert the given mass to moles: n = m/M.
  3. Use the coefficients as the mole ratio to find the moles of the substance you want.
  4. Convert back to mass with m = n·M. With two reactants, find the limiting reactant first.

2Practice problems

1.Mass balanceMedium
MixerStream 1550 kg/h · 15% NaOHStream 2ṁ₂ = ? · 30% NaOHProduct20% NaOH

How much of stream 2 must be added each hour to stream 1 so that the blended product contains 20% NaOH? Find ṁ2.

ṁ1 = 550 kg/hx1 = 15 %x2 = 30 %x3 = 20 %
ṁ2 = kg/h
2.Mass balanceHard
EvaporatorFeed F2400 kg/h · 18% sugarProduct P45% sugarVapour W (pure water)W = ? kg/h

An evaporator concentrates a sugar solution by boiling off pure water. Find the water evaporated W in kg/h.

F = 2400 kg/hxF = 18 %xP = 45 %
W = kg/h
3.Energy balanceMedium
hot sidecold sideethanol · 130 °C100 °Cwater · 15 °Ct = ? °C

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

ṁh = 2 kg/scp,h = 2.44 kJ/(kg·K)Th: 130 → 100 °Cṁc = 0.5 kg/scp,c = 4.18 kJ/(kg·K)tin = 15 °C
tout = °C
4.Energy balanceHard
hot sidecold sideethanol · 130 °C90 °Cwater · 15 °C45 °C · ṁ = ?10% lost

Hot ethanol heats water in an exchanger that loses 10% of the heat released by the hot stream to the surroundings. Find the water flow rate ṁc needed.

ṁh = 3 kg/scp,h = 2.44 kJ/(kg·K)Th: 130 → 90 °Ct: 15 → 45 °Ccp,c = 4.18 kJ/(kg·K)loss = 10 %
ṁc = kg/s
5.Ideal gas lawMedium
130 g of CO₂P = 100 kPaT = 90 °CV = ?

What volume V does 130 g of CO₂ occupy at 100 kPa and 90 °C? Treat it as an ideal gas.

m = 130 gM = 44.01 g/molP = 100 kPaT = 90 °CR = 8.314 J/(mol·K)
V = L
6.Ideal gas lawHard
CO₂, M = 44.01 g/molP = 200 kPaT = 180 °Cρ = ?

Find the density ρ of CO₂ at 200 kPa and 180 °C, treating it as an ideal gas.

M = 44.01 g/molP = 200 kPaT = 180 °CR = 8.314 J/(mol·K)
ρ = kg/m³
7.StoichiometryMedium
m(X)n(X)n(H2​O)m(H2​O)÷ M(X)× 2/ν× M(H2​O)

8 g of H2 and 80 g of O2 react according to 2H2 + O2 → 2H2O. What mass m of H2O can form?

M(H2) = 2.016M(O2) = 32M(H2O) = 18.02 g/mol
m = g
8.StoichiometryHard
m(CaCO3​)n(CaCO3​)n(CO2​)m(CO2​)÷ M(CaCO3​)× 1/1× M(CO2​)

In the reaction CaCO3 → CaO + CO2, 16 g of CaCO3 reacts completely (the other reactants are in excess) and 5.77 g of CO2 is collected. Find the percent yield Y.

M(CaCO3) = 100.09 g/molM(CO2) = 44.01 g/molactual: 5.77 g
Y = %

3Solutions

1.Mass balance

Answer: ṁ2 = 275 kg/h

Steady-state mass balance: in = out for the total and for each component. Blending problems combine both.

Balances

Total: ṁ3 = ṁ1 + ṁ2
NaOH: ṁ1x1 + ṁ2x2 = (ṁ1 + ṁ2)x3

Solve for ṁ2

ṁ2 = ṁ1(x1 − x3)/(x3 − x2) = 550·(15 − 20)/(20 − 30) = 275 kg/h

Answer

ṁ2 = 275 kg/h
2.Mass balance

Answer: W = 1440 kg/h

Tie component: a component that passes through unchanged in one stream (here the solute) links the flows directly. Then the total balance gives the rest.

Solute balance

F·xF = P·xP → P = 2400·18/45 = 960 kg/h

Total balance

W = F − P = 2400 − 960 = 1440 kg/h

Answer

W = 1440 kg/h
3.Energy balance

Answer: tout = 85.05 °C

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

Heat released by the hot stream

Q = 2·2.44·(130 − 100) = 146.4 kW

Heat picked up by the water

tout = tin + Q/(ṁccp,c) = 15 + 146.4/(0.5·4.18) = 85.05 °C

Answer

tout = 85.05 °C
4.Energy balance

Answer: ṁc = 2.101 kg/s

Energy balance with losses: heat released = heat absorbed + heat lost.

Heat released

Qh = 3·2.44·(130 − 90) = 292.8 kW

Heat that reaches the water

Qc = (1 − 0.1)·292.8 = 263.5 kW

Water flow rate

ṁc = Qc/(cp,c·Δt) = 263.5/(4.18·30) = 2.101 kg/s

Answer

ṁc = 2.101 kg/s
5.Ideal gas law

Answer: V = 89.18 L

Ideal gas law: PV = nRT, with n = m/M.

Moles

n = 130/44.01 = 2.954 mol
T = 90 + 273.15 = 363.1 K

Volume

V = nRT/P = 2.954·8.314·363.1/100 = 89.18 L

Answer

V = 89.18 L
6.Ideal gas law

Answer: ρ = 2.336 kg/m³

Ideal gas density: ρ = PM/(RT).

Rearrange the gas law

PV = (m/M)RT → ρ = m/V = PM/(RT)
T = 180 + 273.15 = 453.1 K

Density

ρ = 200·44.01/(8.314·453.1) = 2.336 kg/m³

Answer

ρ = 2.336 kg/m³
7.Stoichiometry

Answer: m = 71.51 g

Stoichiometry: a balanced equation counts particles, not grams. Convert mass to moles with n = m/M, use the coefficients of the equation as the mole ratio, then convert back with m = n·M. With two reactants, the limiting reactant is the one that runs out first and fixes how much product forms.

Moles of each reactant

n(H2) = 8/2.016 = 3.968 mol
n(O2) = 80/32 = 2.5 mol

Limiting reactant

3.968/2 = 1.984, 2.5/1 = 2.5 → H2

Mass of product

n(H2O) = 1.984·2 = 3.968 mol
m = 3.968·18.02 = 71.51 g

Answer

m = 71.51 g
8.Stoichiometry

Answer: Y = 82.02 %

Stoichiometry: a balanced equation counts particles, not grams. Convert mass to moles with n = m/M, use the coefficients of the equation as the mole ratio, then convert back with m = n·M. With two reactants, the limiting reactant is the one that runs out first and fixes how much product forms.

Theoretical yield

n(CaCO3) = 16/100.09 = 0.1599 mol
m = 0.1599·1/1·44.01 = 7.035 g

Percent yield

Y = 5.77/7.035·100 = 82.02%

Answer

Y = 82.02%

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