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.

New problemsBack to the course

In the print window, choose “Save as PDF” to keep a copy on your device.

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 1450 kg/h · 45% NaOHStream 2ṁ₂ = ? · 15% NaOHProduct35% NaOH

How much of stream 2 must be added each hour to stream 1 so that the blended product contains 35% NaOH? Find ṁ2.

ṁ1 = 450 kg/hx1 = 45 %x2 = 15 %x3 = 35 %
ṁ2 = kg/h
2.Mass balanceHard
EvaporatorFeed F4100 kg/h · 14% sugarProduct P40% 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 = 4100 kg/hxF = 14 %xP = 40 %
W = kg/h
3.Energy balanceMedium
hot sidecold sideethanol · 180 °C60 °Cwater · 20 °Ct = ? °C

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

ṁh = 3 kg/scp,h = 2.44 kJ/(kg·K)Th: 180 → 60 °Cṁc = 1.5 kg/scp,c = 4.18 kJ/(kg·K)tin = 20 °C
tout = °C
4.Energy balanceHard
hot sidecold sideethanol · 170 °C70 °Cwater · 15 °C65 °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 = 1 kg/scp,h = 2.44 kJ/(kg·K)Th: 170 → 70 °Ct: 15 → 65 °Ccp,c = 4.18 kJ/(kg·K)loss = 10 %
ṁc = kg/s
5.Ideal gas lawMedium
470 g of CO₂P = 550 kPaT = 75 °CV = ?

What volume V does 470 g of CO₂ occupy at 550 kPa and 75 °C? Treat it as an ideal gas.

m = 470 gM = 44.01 g/molP = 550 kPaT = 75 °CR = 8.314 J/(mol·K)
V = L
6.Ideal gas lawHard
CO₂, M = 44.01 g/molP = 150 kPaT = 0 °Cρ = ?

Find the density ρ of CO₂ at 150 kPa and 0 °C, treating it as an ideal gas.

M = 44.01 g/molP = 150 kPaT = 0 °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)

30 g of H2 and 20 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(Al)n(Al)n(AlCl3​)m(AlCl3​)÷ M(Al)× 2/2× M(AlCl3​)

In the reaction 2Al + 3Cl2 → 2AlCl3, 6 g of Al reacts completely (the other reactants are in excess) and 26.09 g of AlCl3 is collected. Find the percent yield Y.

M(Al) = 26.98 g/molM(AlCl3) = 133.33 g/molactual: 26.09 g
Y = %

3Solutions

1.Mass balance

Answer: ṁ2 = 225 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) = 450·(45 − 35)/(35 − 15) = 225 kg/h

Answer

ṁ2 = 225 kg/h
2.Mass balance

Answer: W = 2665 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 = 4100·14/40 = 1435 kg/h

Total balance

W = F − P = 4100 − 1435 = 2665 kg/h

Answer

W = 2665 kg/h
3.Energy balance

Answer: tout = 160.1 °C

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

Heat released by the hot stream

Q = 3·2.44·(180 − 60) = 878.4 kW

Heat picked up by the water

tout = tin + Q/(ṁccp,c) = 20 + 878.4/(1.5·4.18) = 160.1 °C

Answer

tout = 160.1 °C
4.Energy balance

Answer: ṁc = 1.051 kg/s

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

Heat released

Qh = 1·2.44·(170 − 70) = 244 kW

Heat that reaches the water

Qc = (1 − 0.1)·244 = 219.6 kW

Water flow rate

ṁc = Qc/(cp,c·Δt) = 219.6/(4.18·50) = 1.051 kg/s

Answer

ṁc = 1.051 kg/s
5.Ideal gas law

Answer: V = 56.2 L

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

Moles

n = 470/44.01 = 10.68 mol
T = 75 + 273.15 = 348.1 K

Volume

V = nRT/P = 10.68·8.314·348.1/550 = 56.2 L

Answer

V = 56.2 L
6.Ideal gas law

Answer: ρ = 2.907 kg/m³

Ideal gas density: ρ = PM/(RT).

Rearrange the gas law

PV = (m/M)RT → ρ = m/V = PM/(RT)
T = 0 + 273.15 = 273.1 K

Density

ρ = 150·44.01/(8.314·273.1) = 2.907 kg/m³

Answer

ρ = 2.907 kg/m³
7.Stoichiometry

Answer: m = 22.52 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) = 30/2.016 = 14.88 mol
n(O2) = 20/32 = 0.625 mol

Limiting reactant

14.88/2 = 7.44, 0.625/1 = 0.625 → O2

Mass of product

n(H2O) = 0.625·2 = 1.25 mol
m = 1.25·18.02 = 22.52 g

Answer

m = 22.52 g
8.Stoichiometry

Answer: Y = 87.99 %

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(Al) = 6/26.98 = 0.2224 mol
m = 0.2224·2/2·133.33 = 29.65 g

Percent yield

Y = 26.09/29.65·100 = 87.99%

Answer

Y = 87.99%

ExamProblems · examproblems.com/courses/material-and-energy-balances