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 1400 kg/h · 50% saltStream 2ṁ₂ = ? · 5% 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 = 400 kg/hx1 = 50 %x2 = 5 %x3 = 20 %
ṁ2 = kg/h
2.Mass balanceHard
EvaporatorFeed F1600 kg/h · 10% sugarProduct P65% 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 = 1600 kg/hxF = 10 %xP = 65 %
W = kg/h
3.Energy balanceMedium
hot sidecold sidelight oil · 160 °C90 °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: 160 → 90 °Cṁc = 3 kg/scp,c = 4.18 kJ/(kg·K)tin = 25 °C
tout = °C
4.Energy balanceHard
hot sidecold sideglycol mix · 150 °C80 °Cwater · 15 °C35 °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: 150 → 80 °Ct: 15 → 35 °Ccp,c = 4.18 kJ/(kg·K)loss = 10 %
ṁc = kg/s
5.Ideal gas lawMedium
330 g of O₂P = 550 kPaT = 80 °CV = ?

What volume V does 330 g of O₂ occupy at 550 kPa and 80 °C? Treat it as an ideal gas.

m = 330 gM = 32 g/molP = 550 kPaT = 80 °CR = 8.314 J/(mol·K)
V = L
6.Ideal gas lawHard
He, M = 4.003 g/molP = 650 kPaT = 30 °Cρ = ?

Find the density ρ of He at 650 kPa and 30 °C, treating it as an ideal gas.

M = 4.003 g/molP = 650 kPaT = 30 °CR = 8.314 J/(mol·K)
ρ = kg/m³
7.StoichiometryMedium
m(X)n(X)n(CO2​)m(CO2​)÷ M(X)× 1/ν× M(CO2​)

4 g of CH4 and 64 g of O2 react according to CH4 + 2O2 → CO2 + 2H2O. What mass m of CO2 can form?

M(CH4) = 16.04M(O2) = 32M(CO2) = 44.01 g/mol
m = g
8.StoichiometryHard
m(CH4​)n(CH4​)n(CO2​)m(CO2​)÷ M(CH4​)× 1/1× M(CO2​)

In the reaction CH4 + 2O2 → CO2 + 2H2O, 12 g of CH4 reacts completely (the other reactants are in excess) and 27 g of CO2 is collected. Find the percent yield Y.

M(CH4) = 16.04 g/molM(CO2) = 44.01 g/molactual: 27 g
Y = %

3Solutions

1.Mass balance

Answer: ṁ2 = 800 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) = 400·(50 − 20)/(20 − 5) = 800 kg/h

Answer

ṁ2 = 800 kg/h
2.Mass balance

Answer: W = 1354 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 = 1600·10/65 = 246.2 kg/h

Total balance

W = F − P = 1600 − 246.2 = 1354 kg/h

Answer

W = 1354 kg/h
3.Energy balance

Answer: tout = 36.72 °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·(160 − 90) = 147 kW

Heat picked up by the water

tout = tin + Q/(ṁccp,c) = 25 + 147/(3·4.18) = 36.72 °C

Answer

tout = 36.72 °C
4.Energy balance

Answer: ṁc = 2.713 kg/s

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

Heat released

Qh = 1·3.6·(150 − 80) = 252 kW

Heat that reaches the water

Qc = (1 − 0.1)·252 = 226.8 kW

Water flow rate

ṁc = Qc/(cp,c·Δt) = 226.8/(4.18·20) = 2.713 kg/s

Answer

ṁc = 2.713 kg/s
5.Ideal gas law

Answer: V = 55.05 L

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

Moles

n = 330/32 = 10.31 mol
T = 80 + 273.15 = 353.1 K

Volume

V = nRT/P = 10.31·8.314·353.1/550 = 55.05 L

Answer

V = 55.05 L
6.Ideal gas law

Answer: ρ = 1.032 kg/m³

Ideal gas density: ρ = PM/(RT).

Rearrange the gas law

PV = (m/M)RT → ρ = m/V = PM/(RT)
T = 30 + 273.15 = 303.1 K

Density

ρ = 650·4.003/(8.314·303.1) = 1.032 kg/m³

Answer

ρ = 1.032 kg/m³
7.Stoichiometry

Answer: m = 10.98 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) = 64/32 = 2 mol

Limiting reactant

0.2494/1 = 0.2494, 2/2 = 1 → CH4

Mass of product

n(CO2) = 0.2494·1 = 0.2494 mol
m = 0.2494·44.01 = 10.98 g

Answer

m = 10.98 g
8.Stoichiometry

Answer: Y = 82 %

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(CH4) = 12/16.04 = 0.7481 mol
m = 0.7481·1/1·44.01 = 32.93 g

Percent yield

Y = 27/32.93·100 = 82%

Answer

Y = 82%

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