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 · 45% saltStream 2ṁ₂ = ? · 20% saltProduct30% salt

How much of stream 2 must be added each hour to stream 1 so that the blended product contains 30% salt? Find ṁ2.

ṁ1 = 600 kg/hx1 = 45 %x2 = 20 %x3 = 30 %
ṁ2 = kg/h
2.Mass balanceHard
EvaporatorFeed F4100 kg/h · 12% sugarProduct P35% sugarVapour W (pure water)W = ? kg/h

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

F = 4100 kg/hxF = 12 %xP = 35 %
W = kg/h
3.Energy balanceMedium
hot sidecold sideglycol mix · 140 °C90 °Cwater · 10 °Ct = ? °C

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

ṁh = 1.5 kg/scp,h = 3.6 kJ/(kg·K)Th: 140 → 90 °Cṁc = 0.5 kg/scp,c = 4.18 kJ/(kg·K)tin = 10 °C
tout = °C
4.Energy balanceHard
hot sidecold sideglycol mix · 120 °C60 °Cwater · 10 °C25 °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 kg/scp,h = 3.6 kJ/(kg·K)Th: 120 → 60 °Ct: 10 → 25 °Ccp,c = 4.18 kJ/(kg·K)loss = 5 %
ṁc = kg/s
5.Ideal gas lawMedium
480 g of O₂P = 450 kPaT = 20 °CV = ?

What volume V does 480 g of O₂ occupy at 450 kPa and 20 °C? Treat it as an ideal gas.

m = 480 gM = 32 g/molP = 450 kPaT = 20 °CR = 8.314 J/(mol·K)
V = L
6.Ideal gas lawHard
O₂, M = 32 g/molP = 750 kPaT = 10 °Cρ = ?

Find the density ρ of O₂ at 750 kPa and 10 °C, treating it as an ideal gas.

M = 32 g/molP = 750 kPaT = 10 °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)

10 g of Na and 48 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(N2​)n(N2​)n(NH3​)m(NH3​)÷ M(N2​)× 2/1× M(NH3​)

In the reaction N2 + 3H2 → 2NH3, 24 g of N2 reacts completely (the other reactants are in excess) and 16.05 g of NH3 is collected. Find the percent yield Y.

M(N2) = 28.02 g/molM(NH3) = 17.03 g/molactual: 16.05 g
Y = %

3Solutions

1.Mass balance

Answer: ṁ2 = 900 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) = 600·(45 − 30)/(30 − 20) = 900 kg/h

Answer

ṁ2 = 900 kg/h
2.Mass balance

Answer: W = 2694 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 = 4100·12/35 = 1406 kg/h

Total balance

W = F − P = 4100 − 1406 = 2694 kg/h

Answer

W = 2694 kg/h
3.Energy balance

Answer: tout = 139.2 °C

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

Heat released by the hot stream

Q = 1.5·3.6·(140 − 90) = 270 kW

Heat picked up by the water

tout = tin + Q/(ṁccp,c) = 10 + 270/(0.5·4.18) = 139.2 °C

Answer

tout = 139.2 °C
4.Energy balance

Answer: ṁc = 3.273 kg/s

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

Heat released

Qh = 1·3.6·(120 − 60) = 216 kW

Heat that reaches the water

Qc = (1 − 0.05)·216 = 205.2 kW

Water flow rate

ṁc = Qc/(cp,c·Δt) = 205.2/(4.18·15) = 3.273 kg/s

Answer

ṁc = 3.273 kg/s
5.Ideal gas law

Answer: V = 81.24 L

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

Moles

n = 480/32 = 15 mol
T = 20 + 273.15 = 293.1 K

Volume

V = nRT/P = 15·8.314·293.1/450 = 81.24 L

Answer

V = 81.24 L
6.Ideal gas law

Answer: ρ = 10.19 kg/m³

Ideal gas density: ρ = PM/(RT).

Rearrange the gas law

PV = (m/M)RT → ρ = m/V = PM/(RT)
T = 10 + 273.15 = 283.1 K

Density

ρ = 750·32/(8.314·283.1) = 10.19 kg/m³

Answer

ρ = 10.19 kg/m³
7.Stoichiometry

Answer: m = 25.42 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) = 10/22.99 = 0.435 mol
n(Cl2) = 48/70.9 = 0.677 mol

Limiting reactant

0.435/2 = 0.2175, 0.677/1 = 0.677 → Na

Mass of product

n(NaCl) = 0.2175·2 = 0.435 mol
m = 0.435·58.44 = 25.42 g

Answer

m = 25.42 g
8.Stoichiometry

Answer: Y = 55.02 %

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(N2) = 24/28.02 = 0.8565 mol
m = 0.8565·2/1·17.03 = 29.17 g

Percent yield

Y = 16.05/29.17·100 = 55.02%

Answer

Y = 55.02%

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