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 1650 kg/h · 15% saltStream 2ṁ₂ = ? · 40% saltProduct35% salt

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

ṁ1 = 650 kg/hx1 = 15 %x2 = 40 %x3 = 35 %
ṁ2 = kg/h
2.Mass balanceHard
EvaporatorFeed F800 kg/h · 20% saltProduct PP = ? · 30%Vapour W (pure water)0% salt

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

F = 800 kg/hxF = 20 %xP = 30 %
P = kg/h
3.Energy balanceMedium
hot sidecold sidelight oil · 160 °C70 °Cwater · 15 °Ct = ? °C

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

ṁh = 2 kg/scp,h = 2.1 kJ/(kg·K)Th: 160 → 70 °Cṁc = 2 kg/scp,c = 4.18 kJ/(kg·K)tin = 15 °C
tout = °C
4.Energy balanceHard
hot sidecold sidelight oil · 120 °C60 °Cwater · 10 °C60 °C · ṁ = ?5% lost

Hot light oil heats water in an exchanger that loses 5% of the heat released by the hot stream to the surroundings. Find the water flow rate ṁc needed.

ṁh = 1.5 kg/scp,h = 2.1 kJ/(kg·K)Th: 120 → 60 °Ct: 10 → 60 °Ccp,c = 4.18 kJ/(kg·K)loss = 5 %
ṁc = kg/s
5.Ideal gas lawMedium
290 g of HeP = 200 kPaT = 10 °CV = ?

What volume V does 290 g of He occupy at 200 kPa and 10 °C? Treat it as an ideal gas.

m = 290 gM = 4.003 g/molP = 200 kPaT = 10 °CR = 8.314 J/(mol·K)
V = L
6.Ideal gas lawHard
N₂, M = 28.01 g/molP = 200 kPaT = 150 °Cρ = ?

Find the density ρ of N₂ at 200 kPa and 150 °C, treating it as an ideal gas.

M = 28.01 g/molP = 200 kPaT = 150 °CR = 8.314 J/(mol·K)
ρ = kg/m³
7.StoichiometryMedium
m(X)n(X)n(AlCl3​)m(AlCl3​)÷ M(X)× 2/ν× M(AlCl3​)

6 g of Al and 32 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(Fe2​O3​)n(Fe2​O3​)n(Fe)m(Fe)÷ M(Fe2​O3​)× 2/1× M(Fe)

In the reaction Fe2O3 + 3CO → 2Fe + 3CO2, 10 g of Fe2O3 reacts completely (the other reactants are in excess) and 5.46 g of Fe is collected. Find the percent yield Y.

M(Fe2O3) = 159.69 g/molM(Fe) = 55.85 g/molactual: 5.46 g
Y = %

3Solutions

1.Mass balance

Answer: ṁ2 = 2600 kg/h

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

Balances

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

Solve for ṁ2

ṁ2 = ṁ1(x1 − x3)/(x3 − x2) = 650·(15 − 35)/(35 − 40) = 2600 kg/h

Answer

ṁ2 = 2600 kg/h
2.Mass balance

Answer: P = 533.3 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 = 800·20/30 = 533.3 kg/h

Answer

P = 533.3 kg/h
3.Energy balance

Answer: tout = 60.22 °C

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

Heat released by the hot stream

Q = 2·2.1·(160 − 70) = 378 kW

Heat picked up by the water

tout = tin + Q/(ṁccp,c) = 15 + 378/(2·4.18) = 60.22 °C

Answer

tout = 60.22 °C
4.Energy balance

Answer: ṁc = 0.8591 kg/s

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

Heat released

Qh = 1.5·2.1·(120 − 60) = 189 kW

Heat that reaches the water

Qc = (1 − 0.05)·189 = 179.6 kW

Water flow rate

ṁc = Qc/(cp,c·Δt) = 179.6/(4.18·50) = 0.8591 kg/s

Answer

ṁc = 0.8591 kg/s
5.Ideal gas law

Answer: V = 852.7 L

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

Moles

n = 290/4.003 = 72.45 mol
T = 10 + 273.15 = 283.1 K

Volume

V = nRT/P = 72.45·8.314·283.1/200 = 852.7 L

Answer

V = 852.7 L
6.Ideal gas law

Answer: ρ = 1.592 kg/m³

Ideal gas density: ρ = PM/(RT).

Rearrange the gas law

PV = (m/M)RT → ρ = m/V = PM/(RT)
T = 150 + 273.15 = 423.1 K

Density

ρ = 200·28.01/(8.314·423.1) = 1.592 kg/m³

Answer

ρ = 1.592 kg/m³
7.Stoichiometry

Answer: m = 29.65 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) = 6/26.98 = 0.2224 mol
n(Cl2) = 32/70.9 = 0.4513 mol

Limiting reactant

0.2224/2 = 0.1112, 0.4513/3 = 0.1504 → Al

Mass of product

n(AlCl3) = 0.1112·2 = 0.2224 mol
m = 0.2224·133.33 = 29.65 g

Answer

m = 29.65 g
8.Stoichiometry

Answer: Y = 78.06 %

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(Fe2O3) = 10/159.69 = 0.06262 mol
m = 0.06262·2/1·55.85 = 6.995 g

Percent yield

Y = 5.46/6.995·100 = 78.06%

Answer

Y = 78.06%

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