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 · 45% saltStream 2ṁ₂ = ? · 20% saltProduct40% salt

How much of stream 2 must be added each hour to stream 1 so that the blended product contains 40% salt? Find ṁ2.

ṁ1 = 750 kg/hx1 = 45 %x2 = 20 %x3 = 40 %
ṁ2 = kg/h
2.Mass balanceHard
EvaporatorFeed F5000 kg/h · 7% 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 = 5000 kg/hxF = 7 %xP = 65 %
W = kg/h
3.Energy balanceMedium
hot sidecold sidelight oil · 150 °C80 °Cwater · 20 °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: 150 → 80 °Cṁc = 1.5 kg/scp,c = 4.18 kJ/(kg·K)tin = 20 °C
tout = °C
4.Energy balanceHard
hot sidecold sidelight oil · 140 °C80 °Cwater · 15 °C45 °C · ṁ = ?5% lost

Hot light oil 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 = 0.5 kg/scp,h = 2.1 kJ/(kg·K)Th: 140 → 80 °Ct: 15 → 45 °Ccp,c = 4.18 kJ/(kg·K)loss = 5 %
ṁc = kg/s
5.Ideal gas lawMedium
140 g of CH₄P = 750 kPaT = 60 °CV = ?

What volume V does 140 g of CH₄ occupy at 750 kPa and 60 °C? Treat it as an ideal gas.

m = 140 gM = 16.04 g/molP = 750 kPaT = 60 °CR = 8.314 J/(mol·K)
V = L
6.Ideal gas lawHard
Ar, M = 39.95 g/molP = 550 kPaT = 60 °Cρ = ?

Find the density ρ of Ar at 550 kPa and 60 °C, treating it as an ideal gas.

M = 39.95 g/molP = 550 kPaT = 60 °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)

12 g of CH4 and 16 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(CH4​)n(CH4​)n(H2​O)m(H2​O)÷ M(CH4​)× 2/1× M(H2​O)

In the reaction CH4 + 2O2 → CO2 + 2H2O, 24 g of CH4 reacts completely (the other reactants are in excess) and 44.22 g of H2O is collected. Find the percent yield Y.

M(CH4) = 16.04 g/molM(H2O) = 18.02 g/molactual: 44.22 g
Y = %

3Solutions

1.Mass balance

Answer: ṁ2 = 187.5 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) = 750·(45 − 40)/(40 − 20) = 187.5 kg/h

Answer

ṁ2 = 187.5 kg/h
2.Mass balance

Answer: W = 4462 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 = 5000·7/65 = 538.5 kg/h

Total balance

W = F − P = 5000 − 538.5 = 4462 kg/h

Answer

W = 4462 kg/h
3.Energy balance

Answer: tout = 43.44 °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·(150 − 80) = 147 kW

Heat picked up by the water

tout = tin + Q/(ṁccp,c) = 20 + 147/(1.5·4.18) = 43.44 °C

Answer

tout = 43.44 °C
4.Energy balance

Answer: ṁc = 0.4773 kg/s

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

Heat released

Qh = 0.5·2.1·(140 − 80) = 63 kW

Heat that reaches the water

Qc = (1 − 0.05)·63 = 59.85 kW

Water flow rate

ṁc = Qc/(cp,c·Δt) = 59.85/(4.18·30) = 0.4773 kg/s

Answer

ṁc = 0.4773 kg/s
5.Ideal gas law

Answer: V = 32.23 L

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

Moles

n = 140/16.04 = 8.728 mol
T = 60 + 273.15 = 333.1 K

Volume

V = nRT/P = 8.728·8.314·333.1/750 = 32.23 L

Answer

V = 32.23 L
6.Ideal gas law

Answer: ρ = 7.933 kg/m³

Ideal gas density: ρ = PM/(RT).

Rearrange the gas law

PV = (m/M)RT → ρ = m/V = PM/(RT)
T = 60 + 273.15 = 333.1 K

Density

ρ = 550·39.95/(8.314·333.1) = 7.933 kg/m³

Answer

ρ = 7.933 kg/m³
7.Stoichiometry

Answer: m = 9.01 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) = 12/16.04 = 0.7481 mol
n(O2) = 16/32 = 0.5 mol

Limiting reactant

0.7481/1 = 0.7481, 0.5/2 = 0.25 → O2

Mass of product

n(H2O) = 0.25·2 = 0.5 mol
m = 0.5·18.02 = 9.01 g

Answer

m = 9.01 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) = 24/16.04 = 1.496 mol
m = 1.496·2/1·18.02 = 53.93 g

Percent yield

Y = 44.22/53.93·100 = 82%

Answer

Y = 82%

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