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 1600 kg/h · 10% NaOHStream 2ṁ₂ = ? · 40% NaOHProduct15% NaOH

How much of stream 2 must be added each hour to stream 1 so that the blended product contains 15% NaOH? Find ṁ2.

ṁ1 = 600 kg/hx1 = 10 %x2 = 40 %x3 = 15 %
ṁ2 = kg/h
2.Mass balanceHard
EvaporatorFeed F4300 kg/h · 19% NaOHProduct PP = ? · 50%Vapour W (pure water)0% NaOH

An evaporator concentrates a NaOH solution by boiling off pure water. Find the product flow P in kg/h.

F = 4300 kg/hxF = 19 %xP = 50 %
P = kg/h
3.Energy balanceMedium
hot sidecold sidelight oil · 140 °C100 °Cwater · 10 °Ct = ? °C

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

ṁh = 1.5 kg/scp,h = 2.1 kJ/(kg·K)Th: 140 → 100 °Cṁc = 3 kg/scp,c = 4.18 kJ/(kg·K)tin = 10 °C
tout = °C
4.Energy balanceHard
hot sidecold sideglycol mix · 150 °C60 °Cwater · 30 °C50 °C · ṁ = ?10% lost

Hot glycol mix 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 = 0.5 kg/scp,h = 3.6 kJ/(kg·K)Th: 150 → 60 °Ct: 30 → 50 °Ccp,c = 4.18 kJ/(kg·K)loss = 10 %
ṁc = kg/s
5.Ideal gas lawMedium
270 g of N₂P = 800 kPaT = 20 °CV = ?

What volume V does 270 g of N₂ occupy at 800 kPa and 20 °C? Treat it as an ideal gas.

m = 270 gM = 28.01 g/molP = 800 kPaT = 20 °CR = 8.314 J/(mol·K)
V = L
6.Ideal gas lawHard
State 1State 2

A fixed amount of CO₂ goes from 200 kPa, 2 L, 30 °C to 750 kPa and 200 °C. Find the new volume V2.

P1 = 200 kPaV1 = 2 LT1 = 30 °CP2 = 750 kPaT2 = 200 °C
V2 = L
7.StoichiometryMedium
m(X)n(X)n(AlCl3​)m(AlCl3​)÷ M(X)× 2/ν× M(AlCl3​)

10 g of Al and 20 g of Cl2 react according to 2Al + 3Cl2 → 2AlCl3. What mass m of AlCl3 can form?

M(Al) = 26.98M(Cl2) = 70.9M(AlCl3) = 133.33 g/mol
m = g
8.StoichiometryHard
m(H2​)n(H2​)n(H2​O)m(H2​O)÷ M(H2​)× 2/2× M(H2​O)

In the reaction 2H2 + O2 → 2H2O, 20 g of H2 reacts completely (the other reactants are in excess) and 146.6 g of H2O is collected. Find the percent yield Y.

M(H2) = 2.016 g/molM(H2O) = 18.02 g/molactual: 146.6 g
Y = %

3Solutions

1.Mass balance

Answer: ṁ2 = 120 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) = 600·(10 − 15)/(15 − 40) = 120 kg/h

Answer

ṁ2 = 120 kg/h
2.Mass balance

Answer: P = 1634 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 = 4300·19/50 = 1634 kg/h

Answer

P = 1634 kg/h
3.Energy balance

Answer: tout = 20.05 °C

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

Heat released by the hot stream

Q = 1.5·2.1·(140 − 100) = 126 kW

Heat picked up by the water

tout = tin + Q/(ṁccp,c) = 10 + 126/(3·4.18) = 20.05 °C

Answer

tout = 20.05 °C
4.Energy balance

Answer: ṁc = 1.744 kg/s

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

Heat released

Qh = 0.5·3.6·(150 − 60) = 162 kW

Heat that reaches the water

Qc = (1 − 0.1)·162 = 145.8 kW

Water flow rate

ṁc = Qc/(cp,c·Δt) = 145.8/(4.18·20) = 1.744 kg/s

Answer

ṁc = 1.744 kg/s
5.Ideal gas law

Answer: V = 29.37 L

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

Moles

n = 270/28.01 = 9.639 mol
T = 20 + 273.15 = 293.1 K

Volume

V = nRT/P = 9.639·8.314·293.1/800 = 29.37 L

Answer

V = 29.37 L
6.Ideal gas law

Answer: V2 = 0.8324 L

Combined gas law: for a fixed amount of ideal gas, P1V1/T1 = P2V2/T2.

Absolute temperatures

T1 = 303.1 K, T2 = 473.1 K

Combined gas law

V2 = V1·(P1/P2)·(T2/T1) = 2·(200/750)·(473.1/303.1) = 0.8324 L

Answer

V2 = 0.8324 L
7.Stoichiometry

Answer: m = 25.07 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(Al) = 10/26.98 = 0.3706 mol
n(Cl2) = 20/70.9 = 0.2821 mol

Limiting reactant

0.3706/2 = 0.1853, 0.2821/3 = 0.09403 → Cl2

Mass of product

n(AlCl3) = 0.09403·2 = 0.1881 mol
m = 0.1881·133.33 = 25.07 g

Answer

m = 25.07 g
8.Stoichiometry

Answer: Y = 82 %

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(H2) = 20/2.016 = 9.921 mol
m = 9.921·2/2·18.02 = 178.8 g

Percent yield

Y = 146.6/178.8·100 = 82%

Answer

Y = 82%

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