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 1450 kg/h · 15% saltStream 2ṁ₂ = ? · 50% saltProduct20% salt

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

ṁ1 = 450 kg/hx1 = 15 %x2 = 50 %x3 = 20 %
ṁ2 = kg/h
2.Mass balanceHard
EvaporatorFeed F1800 kg/h · 18% sugarProduct P30% 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 = 1800 kg/hxF = 18 %xP = 30 %
W = kg/h
3.Energy balanceMedium
hot sidecold sideglycol mix · 130 °C70 °Cwater · 15 °Ct = ? °C

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

ṁh = 2 kg/scp,h = 3.6 kJ/(kg·K)Th: 130 → 70 °Cṁc = 1 kg/scp,c = 4.18 kJ/(kg·K)tin = 15 °C
tout = °C
4.Energy balanceHard
hot sidecold sideglycol mix · 130 °C90 °Cwater · 20 °C40 °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 = 1 kg/scp,h = 3.6 kJ/(kg·K)Th: 130 → 90 °Ct: 20 → 40 °Ccp,c = 4.18 kJ/(kg·K)loss = 10 %
ṁc = kg/s
5.Ideal gas lawMedium
80 g of N₂P = 450 kPaT = 70 °CV = ?

What volume V does 80 g of N₂ occupy at 450 kPa and 70 °C? Treat it as an ideal gas.

m = 80 gM = 28.01 g/molP = 450 kPaT = 70 °CR = 8.314 J/(mol·K)
V = L
6.Ideal gas lawHard
CH₄, M = 16.04 g/molP = 1000 kPaT = 190 °Cρ = ?

Find the density ρ of CH₄ at 1000 kPa and 190 °C, treating it as an ideal gas.

M = 16.04 g/molP = 1000 kPaT = 190 °CR = 8.314 J/(mol·K)
ρ = kg/m³
7.StoichiometryMedium
m(X)n(X)n(NH3​)m(NH3​)÷ M(X)× 2/ν× M(NH3​)

6 g of N2 and 20 g of H2 react according to N2 + 3H2 → 2NH3. What mass m of NH3 can form?

M(N2) = 28.02M(H2) = 2.016M(NH3) = 17.03 g/mol
m = g
8.StoichiometryHard
m(H2​)n(H2​)n(H2​O)m(H2​O)÷ M(H2​)× 2/2× M(H2​O)

In the reaction 2H2 + O2 → 2H2O, 16 g of H2 reacts completely (the other reactants are in excess) and 111.5 g of H2O is collected. Find the percent yield Y.

M(H2) = 2.016 g/molM(H2O) = 18.02 g/molactual: 111.5 g
Y = %

3Solutions

1.Mass balance

Answer: ṁ2 = 75 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) = 450·(15 − 20)/(20 − 50) = 75 kg/h

Answer

ṁ2 = 75 kg/h
2.Mass balance

Answer: W = 720 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 = 1800·18/30 = 1080 kg/h

Total balance

W = F − P = 1800 − 1080 = 720 kg/h

Answer

W = 720 kg/h
3.Energy balance

Answer: tout = 118.3 °C

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

Heat released by the hot stream

Q = 2·3.6·(130 − 70) = 432 kW

Heat picked up by the water

tout = tin + Q/(ṁccp,c) = 15 + 432/(1·4.18) = 118.3 °C

Answer

tout = 118.3 °C
4.Energy balance

Answer: ṁc = 1.55 kg/s

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

Heat released

Qh = 1·3.6·(130 − 90) = 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/(4.18·20) = 1.55 kg/s

Answer

ṁc = 1.55 kg/s
5.Ideal gas law

Answer: V = 18.11 L

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

Moles

n = 80/28.01 = 2.856 mol
T = 70 + 273.15 = 343.1 K

Volume

V = nRT/P = 2.856·8.314·343.1/450 = 18.11 L

Answer

V = 18.11 L
6.Ideal gas law

Answer: ρ = 4.166 kg/m³

Ideal gas density: ρ = PM/(RT).

Rearrange the gas law

PV = (m/M)RT → ρ = m/V = PM/(RT)
T = 190 + 273.15 = 463.1 K

Density

ρ = 1000·16.04/(8.314·463.1) = 4.166 kg/m³

Answer

ρ = 4.166 kg/m³
7.Stoichiometry

Answer: m = 7.293 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(N2) = 6/28.02 = 0.2141 mol
n(H2) = 20/2.016 = 9.921 mol

Limiting reactant

0.2141/1 = 0.2141, 9.921/3 = 3.307 → N2

Mass of product

n(NH3) = 0.2141·2 = 0.4283 mol
m = 0.4283·17.03 = 7.293 g

Answer

m = 7.293 g
8.Stoichiometry

Answer: Y = 78 %

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(H2) = 16/2.016 = 7.937 mol
m = 7.937·2/2·18.02 = 143 g

Percent yield

Y = 111.5/143·100 = 78%

Answer

Y = 78%

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