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 1550 kg/h · 50% NaOHStream 2ṁ₂ = ? · 10% NaOHProduct20% NaOH

How much of stream 2 must be added each hour to stream 1 so that the blended product contains 20% NaOH? Find ṁ2.

ṁ1 = 550 kg/hx1 = 50 %x2 = 10 %x3 = 20 %
ṁ2 = kg/h
2.Mass balanceHard
EvaporatorFeed F600 kg/h · 20% NaOHProduct P30% NaOHVapour W (pure water)W = ? kg/h

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

F = 600 kg/hxF = 20 %xP = 30 %
W = kg/h
3.Energy balanceMedium
hot sidecold sideethanol · 140 °C100 °Cwater · 15 °Ct = ? °C

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

ṁh = 0.5 kg/scp,h = 2.44 kJ/(kg·K)Th: 140 → 100 °Cṁc = 1 kg/scp,c = 4.18 kJ/(kg·K)tin = 15 °C
tout = °C
4.Energy balanceHard
hot sidecold sideglycol mix · 180 °C130 °Cwater · 10 °C45 °C · ṁ = ?5% lost

Hot glycol mix 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 = 3.6 kJ/(kg·K)Th: 180 → 130 °Ct: 10 → 45 °Ccp,c = 4.18 kJ/(kg·K)loss = 5 %
ṁc = kg/s
5.Ideal gas lawMedium
30 g of O₂P = 400 kPaT = 80 °CV = ?

What volume V does 30 g of O₂ occupy at 400 kPa and 80 °C? Treat it as an ideal gas.

m = 30 gM = 32 g/molP = 400 kPaT = 80 °CR = 8.314 J/(mol·K)
V = L
6.Ideal gas lawHard
O₂, M = 32 g/molP = 900 kPaT = 90 °Cρ = ?

Find the density ρ of O₂ at 900 kPa and 90 °C, treating it as an ideal gas.

M = 32 g/molP = 900 kPaT = 90 °CR = 8.314 J/(mol·K)
ρ = kg/m³
7.StoichiometryMedium
m(X)n(X)n(NaCl)m(NaCl)÷ M(X)× 2/ν× M(NaCl)

20 g of Na and 32 g of Cl2 react according to 2Na + Cl2 → 2NaCl. What mass m of NaCl can form?

M(Na) = 22.99M(Cl2) = 70.9M(NaCl) = 58.44 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, 30 g of Fe2O3 reacts completely (the other reactants are in excess) and 13.01 g of Fe is collected. Find the percent yield Y.

M(Fe2O3) = 159.69 g/molM(Fe) = 55.85 g/molactual: 13.01 g
Y = %

3Solutions

1.Mass balance

Answer: ṁ2 = 1650 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) = 550·(50 − 20)/(20 − 10) = 1650 kg/h

Answer

ṁ2 = 1650 kg/h
2.Mass balance

Answer: W = 200 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 = 600·20/30 = 400 kg/h

Total balance

W = F − P = 600 − 400 = 200 kg/h

Answer

W = 200 kg/h
3.Energy balance

Answer: tout = 26.67 °C

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

Heat released by the hot stream

Q = 0.5·2.44·(140 − 100) = 48.8 kW

Heat picked up by the water

tout = tin + Q/(ṁccp,c) = 15 + 48.8/(1·4.18) = 26.67 °C

Answer

tout = 26.67 °C
4.Energy balance

Answer: ṁc = 1.753 kg/s

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

Heat released

Qh = 1.5·3.6·(180 − 130) = 270 kW

Heat that reaches the water

Qc = (1 − 0.05)·270 = 256.5 kW

Water flow rate

ṁc = Qc/(cp,c·Δt) = 256.5/(4.18·35) = 1.753 kg/s

Answer

ṁc = 1.753 kg/s
5.Ideal gas law

Answer: V = 6.881 L

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

Moles

n = 30/32 = 0.9375 mol
T = 80 + 273.15 = 353.1 K

Volume

V = nRT/P = 0.9375·8.314·353.1/400 = 6.881 L

Answer

V = 6.881 L
6.Ideal gas law

Answer: ρ = 9.539 kg/m³

Ideal gas density: ρ = PM/(RT).

Rearrange the gas law

PV = (m/M)RT → ρ = m/V = PM/(RT)
T = 90 + 273.15 = 363.1 K

Density

ρ = 900·32/(8.314·363.1) = 9.539 kg/m³

Answer

ρ = 9.539 kg/m³
7.Stoichiometry

Answer: m = 50.84 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(Na) = 20/22.99 = 0.8699 mol
n(Cl2) = 32/70.9 = 0.4513 mol

Limiting reactant

0.8699/2 = 0.435, 0.4513/1 = 0.4513 → Na

Mass of product

n(NaCl) = 0.435·2 = 0.8699 mol
m = 0.8699·58.44 = 50.84 g

Answer

m = 50.84 g
8.Stoichiometry

Answer: Y = 62 %

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) = 30/159.69 = 0.1879 mol
m = 0.1879·2/1·55.85 = 20.98 g

Percent yield

Y = 13.01/20.98·100 = 62%

Answer

Y = 62%

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