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 1850 kg/h · 5% sugarStream 2ṁ₂ = ? · 30% sugarProduct15% sugar

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

ṁ1 = 850 kg/hx1 = 5 %x2 = 30 %x3 = 15 %
ṁ2 = kg/h
2.Mass balanceHard
EvaporatorFeed F3900 kg/h · 8% saltProduct P55% 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 = 3900 kg/hxF = 8 %xP = 55 %
W = kg/h
3.Energy balanceMedium
hot sidecold sidelight oil · 130 °C80 °Cwater · 25 °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 kg/scp,h = 2.1 kJ/(kg·K)Th: 130 → 80 °Cṁc = 3 kg/scp,c = 4.18 kJ/(kg·K)tin = 25 °C
tout = °C
4.Energy balanceHard
hot sidecold sidelight oil · 180 °C60 °Cwater · 10 °C65 °C · ṁ = ?15% lost

Hot light oil heats water in an exchanger that loses 15% 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: 180 → 60 °Ct: 10 → 65 °Ccp,c = 4.18 kJ/(kg·K)loss = 15 %
ṁc = kg/s
5.Ideal gas lawMedium
30 g of CO₂P = 150 kPaT = 75 °CV = ?

What volume V does 30 g of CO₂ occupy at 150 kPa and 75 °C? Treat it as an ideal gas.

m = 30 gM = 44.01 g/molP = 150 kPaT = 75 °CR = 8.314 J/(mol·K)
V = L
6.Ideal gas lawHard
N₂, M = 28.01 g/molP = 200 kPaT = 50 °Cρ = ?

Find the density ρ of N₂ at 200 kPa and 50 °C, treating it as an ideal gas.

M = 28.01 g/molP = 200 kPaT = 50 °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​)

12 g of Al and 10 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(Na)n(Na)n(NaCl)m(NaCl)÷ M(Na)× 2/2× M(NaCl)

In the reaction 2Na + Cl2 → 2NaCl, 20 g of Na reacts completely (the other reactants are in excess) and 35.59 g of NaCl is collected. Find the percent yield Y.

M(Na) = 22.99 g/molM(NaCl) = 58.44 g/molactual: 35.59 g
Y = %

3Solutions

1.Mass balance

Answer: ṁ2 = 566.7 kg/h

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

Balances

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

Solve for ṁ2

ṁ2 = ṁ1(x1 − x3)/(x3 − x2) = 850·(5 − 15)/(15 − 30) = 566.7 kg/h

Answer

ṁ2 = 566.7 kg/h
2.Mass balance

Answer: W = 3333 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 = 3900·8/55 = 567.3 kg/h

Total balance

W = F − P = 3900 − 567.3 = 3333 kg/h

Answer

W = 3333 kg/h
3.Energy balance

Answer: tout = 33.37 °C

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

Heat released by the hot stream

Q = 1·2.1·(130 − 80) = 105 kW

Heat picked up by the water

tout = tin + Q/(ṁccp,c) = 25 + 105/(3·4.18) = 33.37 °C

Answer

tout = 33.37 °C
4.Energy balance

Answer: ṁc = 1.398 kg/s

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

Heat released

Qh = 1.5·2.1·(180 − 60) = 378 kW

Heat that reaches the water

Qc = (1 − 0.15)·378 = 321.3 kW

Water flow rate

ṁc = Qc/(cp,c·Δt) = 321.3/(4.18·55) = 1.398 kg/s

Answer

ṁc = 1.398 kg/s
5.Ideal gas law

Answer: V = 13.15 L

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

Moles

n = 30/44.01 = 0.6817 mol
T = 75 + 273.15 = 348.1 K

Volume

V = nRT/P = 0.6817·8.314·348.1/150 = 13.15 L

Answer

V = 13.15 L
6.Ideal gas law

Answer: ρ = 2.085 kg/m³

Ideal gas density: ρ = PM/(RT).

Rearrange the gas law

PV = (m/M)RT → ρ = m/V = PM/(RT)
T = 50 + 273.15 = 323.1 K

Density

ρ = 200·28.01/(8.314·323.1) = 2.085 kg/m³

Answer

ρ = 2.085 kg/m³
7.Stoichiometry

Answer: m = 12.54 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) = 12/26.98 = 0.4448 mol
n(Cl2) = 10/70.9 = 0.141 mol

Limiting reactant

0.4448/2 = 0.2224, 0.141/3 = 0.04701 → Cl2

Mass of product

n(AlCl3) = 0.04701·2 = 0.09403 mol
m = 0.09403·133.33 = 12.54 g

Answer

m = 12.54 g
8.Stoichiometry

Answer: Y = 70 %

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(Na) = 20/22.99 = 0.8699 mol
m = 0.8699·2/2·58.44 = 50.84 g

Percent yield

Y = 35.59/50.84·100 = 70%

Answer

Y = 70%

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