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 1750 kg/h · 15% NaOHStream 2ṁ₂ = ? · 30% 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 = 750 kg/hx1 = 15 %x2 = 30 %x3 = 20 %
ṁ2 = kg/h
2.Mass balanceHard
EvaporatorFeed F1300 kg/h · 8% saltProduct P50% 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 = 1300 kg/hxF = 8 %xP = 50 %
W = kg/h
3.Energy balanceMedium
hot sidecold sidelight oil · 140 °C110 °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 = 2.5 kg/scp,h = 2.1 kJ/(kg·K)Th: 140 → 110 °Cṁc = 0.5 kg/scp,c = 4.18 kJ/(kg·K)tin = 25 °C
tout = °C
4.Energy balanceHard
hot sidecold sidelight oil · 120 °C70 °Cwater · 30 °C90 °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 kg/scp,h = 2.1 kJ/(kg·K)Th: 120 → 70 °Ct: 30 → 90 °Ccp,c = 4.18 kJ/(kg·K)loss = 15 %
ṁc = kg/s
5.Ideal gas lawMedium
160 g of HeP = 200 kPaT = 80 °CV = ?

What volume V does 160 g of He occupy at 200 kPa and 80 °C? Treat it as an ideal gas.

m = 160 gM = 4.003 g/molP = 200 kPaT = 80 °CR = 8.314 J/(mol·K)
V = L
6.Ideal gas lawHard
Ar, M = 39.95 g/molP = 850 kPaT = 90 °Cρ = ?

Find the density ρ of Ar at 850 kPa and 90 °C, treating it as an ideal gas.

M = 39.95 g/molP = 850 kPaT = 90 °CR = 8.314 J/(mol·K)
ρ = kg/m³
7.StoichiometryMedium
m(X)n(X)n(H2​O)m(H2​O)÷ M(X)× 2/ν× M(H2​O)

4 g of CH4 and 10 g of O2 react according to CH4 + 2O2 → CO2 + 2H2O. What mass m of H2O can form?

M(CH4) = 16.04M(O2) = 32M(H2O) = 18.02 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, 4 g of H2 reacts completely (the other reactants are in excess) and 33.61 g of H2O is collected. Find the percent yield Y.

M(H2) = 2.016 g/molM(H2O) = 18.02 g/molactual: 33.61 g
Y = %

3Solutions

1.Mass balance

Answer: ṁ2 = 375 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) = 750·(15 − 20)/(20 − 30) = 375 kg/h

Answer

ṁ2 = 375 kg/h
2.Mass balance

Answer: W = 1092 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 = 1300·8/50 = 208 kg/h

Total balance

W = F − P = 1300 − 208 = 1092 kg/h

Answer

W = 1092 kg/h
3.Energy balance

Answer: tout = 100.4 °C

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

Heat released by the hot stream

Q = 2.5·2.1·(140 − 110) = 157.5 kW

Heat picked up by the water

tout = tin + Q/(ṁccp,c) = 25 + 157.5/(0.5·4.18) = 100.4 °C

Answer

tout = 100.4 °C
4.Energy balance

Answer: ṁc = 0.3559 kg/s

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

Heat released

Qh = 1·2.1·(120 − 70) = 105 kW

Heat that reaches the water

Qc = (1 − 0.15)·105 = 89.25 kW

Water flow rate

ṁc = Qc/(cp,c·Δt) = 89.25/(4.18·60) = 0.3559 kg/s

Answer

ṁc = 0.3559 kg/s
5.Ideal gas law

Answer: V = 586.8 L

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

Moles

n = 160/4.003 = 39.97 mol
T = 80 + 273.15 = 353.1 K

Volume

V = nRT/P = 39.97·8.314·353.1/200 = 586.8 L

Answer

V = 586.8 L
6.Ideal gas law

Answer: ρ = 11.25 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 = 850·39.95/(8.314·363.1) = 11.25 kg/m³

Answer

ρ = 11.25 kg/m³
7.Stoichiometry

Answer: m = 5.631 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(CH4) = 4/16.04 = 0.2494 mol
n(O2) = 10/32 = 0.3125 mol

Limiting reactant

0.2494/1 = 0.2494, 0.3125/2 = 0.1563 → O2

Mass of product

n(H2O) = 0.1563·2 = 0.3125 mol
m = 0.3125·18.02 = 5.631 g

Answer

m = 5.631 g
8.Stoichiometry

Answer: Y = 94 %

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) = 4/2.016 = 1.984 mol
m = 1.984·2/2·18.02 = 35.75 g

Percent yield

Y = 33.61/35.75·100 = 94%

Answer

Y = 94%

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