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 1250 kg/h · 35% NaOHStream 2ṁ₂ = ? · 20% NaOHProduct25% NaOH

How much of stream 2 must be added each hour to stream 1 so that the blended product contains 25% NaOH? Find ṁ2.

ṁ1 = 250 kg/hx1 = 35 %x2 = 20 %x3 = 25 %
ṁ2 = kg/h
2.Mass balanceHard
EvaporatorFeed F3700 kg/h · 7% saltProduct P30% saltVapour W (pure water)W = ? kg/h

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

F = 3700 kg/hxF = 7 %xP = 30 %
W = kg/h
3.Energy balanceMedium
hot sidecold sideethanol · 180 °C150 °Cwater · 10 °Ct = ? °C

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

ṁh = 2.5 kg/scp,h = 2.44 kJ/(kg·K)Th: 180 → 150 °Cṁc = 2 kg/scp,c = 4.18 kJ/(kg·K)tin = 10 °C
tout = °C
4.Energy balanceHard
hot sidecold sideglycol mix · 160 °C80 °Cwater · 25 °C60 °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: 160 → 80 °Ct: 25 → 60 °Ccp,c = 4.18 kJ/(kg·K)loss = 10 %
ṁc = kg/s
5.Ideal gas lawMedium
340 g of O₂P = 100 kPaT = 80 °CV = ?

What volume V does 340 g of O₂ occupy at 100 kPa and 80 °C? Treat it as an ideal gas.

m = 340 gM = 32 g/molP = 100 kPaT = 80 °CR = 8.314 J/(mol·K)
V = L
6.Ideal gas lawHard
O₂, M = 32 g/molP = 600 kPaT = 100 °Cρ = ?

Find the density ρ of O₂ at 600 kPa and 100 °C, treating it as an ideal gas.

M = 32 g/molP = 600 kPaT = 100 °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)

12 g of CH4 and 8 g of O2 react according to CH4 + 2O2 → CO2 + 2H2O. What mass m of H2O can form?

M(CH4) = 16.04M(O2) = 32M(H2O) = 18.02 g/mol
m = g
8.StoichiometryHard
m(Mg)n(Mg)n(MgO)m(MgO)÷ M(Mg)× 2/2× M(MgO)

In the reaction 2Mg + O2 → 2MgO, 30 g of Mg reacts completely (the other reactants are in excess) and 42.27 g of MgO is collected. Find the percent yield Y.

M(Mg) = 24.31 g/molM(MgO) = 40.3 g/molactual: 42.27 g
Y = %

3Solutions

1.Mass balance

Answer: ṁ2 = 500 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) = 250·(35 − 25)/(25 − 20) = 500 kg/h

Answer

ṁ2 = 500 kg/h
2.Mass balance

Answer: W = 2837 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 = 3700·7/30 = 863.3 kg/h

Total balance

W = F − P = 3700 − 863.3 = 2837 kg/h

Answer

W = 2837 kg/h
3.Energy balance

Answer: tout = 31.89 °C

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

Heat released by the hot stream

Q = 2.5·2.44·(180 − 150) = 183 kW

Heat picked up by the water

tout = tin + Q/(ṁccp,c) = 10 + 183/(2·4.18) = 31.89 °C

Answer

tout = 31.89 °C
4.Energy balance

Answer: ṁc = 0.8859 kg/s

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

Heat released

Qh = 0.5·3.6·(160 − 80) = 144 kW

Heat that reaches the water

Qc = (1 − 0.1)·144 = 129.6 kW

Water flow rate

ṁc = Qc/(cp,c·Δt) = 129.6/(4.18·35) = 0.8859 kg/s

Answer

ṁc = 0.8859 kg/s
5.Ideal gas law

Answer: V = 312 L

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

Moles

n = 340/32 = 10.63 mol
T = 80 + 273.15 = 353.1 K

Volume

V = nRT/P = 10.63·8.314·353.1/100 = 312 L

Answer

V = 312 L
6.Ideal gas law

Answer: ρ = 6.189 kg/m³

Ideal gas density: ρ = PM/(RT).

Rearrange the gas law

PV = (m/M)RT → ρ = m/V = PM/(RT)
T = 100 + 273.15 = 373.1 K

Density

ρ = 600·32/(8.314·373.1) = 6.189 kg/m³

Answer

ρ = 6.189 kg/m³
7.Stoichiometry

Answer: m = 4.505 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(CH4) = 12/16.04 = 0.7481 mol
n(O2) = 8/32 = 0.25 mol

Limiting reactant

0.7481/1 = 0.7481, 0.25/2 = 0.125 → O2

Mass of product

n(H2O) = 0.125·2 = 0.25 mol
m = 0.25·18.02 = 4.505 g

Answer

m = 4.505 g
8.Stoichiometry

Answer: Y = 84.99 %

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(Mg) = 30/24.31 = 1.234 mol
m = 1.234·2/2·40.3 = 49.73 g

Percent yield

Y = 42.27/49.73·100 = 84.99%

Answer

Y = 84.99%

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