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 · 35% saltStream 2ṁ₂ = ? · 15% saltProduct25% salt

How much of stream 2 must be added each hour to stream 1 so that the blended product contains 25% salt? Find ṁ2.

ṁ1 = 400 kg/hx1 = 35 %x2 = 15 %x3 = 25 %
ṁ2 = kg/h
2.Mass balanceHard
EvaporatorFeed F4400 kg/h · 18% sugarProduct P50% 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 = 4400 kg/hxF = 18 %xP = 50 %
W = kg/h
3.Energy balanceMedium
hot sidecold sideglycol mix · 160 °C110 °Cwater · 25 °Ct = ? °C

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

ṁh = 0.5 kg/scp,h = 3.6 kJ/(kg·K)Th: 160 → 110 °Cṁc = 1.5 kg/scp,c = 4.18 kJ/(kg·K)tin = 25 °C
tout = °C
4.Energy balanceHard
hot sidecold sideethanol · 140 °C110 °Cwater · 25 °C40 °C · ṁ = ?15% lost

Hot ethanol 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 = 2.44 kJ/(kg·K)Th: 140 → 110 °Ct: 25 → 40 °Ccp,c = 4.18 kJ/(kg·K)loss = 15 %
ṁc = kg/s
5.Ideal gas lawMedium
130 g of CH₄P = 100 kPaT = 25 °CV = ?

What volume V does 130 g of CH₄ occupy at 100 kPa and 25 °C? Treat it as an ideal gas.

m = 130 gM = 16.04 g/molP = 100 kPaT = 25 °CR = 8.314 J/(mol·K)
V = L
6.Ideal gas lawHard
O₂, M = 32 g/molP = 100 kPaT = 130 °Cρ = ?

Find the density ρ of O₂ at 100 kPa and 130 °C, treating it as an ideal gas.

M = 32 g/molP = 100 kPaT = 130 °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​)

16 g of N2 and 64 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(Fe2​O3​)n(Fe2​O3​)n(CO2​)m(CO2​)÷ M(Fe2​O3​)× 3/1× M(CO2​)

In the reaction Fe2O3 + 3CO → 2Fe + 3CO2, 8 g of Fe2O3 reacts completely (the other reactants are in excess) and 6.22 g of CO2 is collected. Find the percent yield Y.

M(Fe2O3) = 159.69 g/molM(CO2) = 44.01 g/molactual: 6.22 g
Y = %

3Solutions

1.Mass balance

Answer: ṁ2 = 400 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·(35 − 25)/(25 − 15) = 400 kg/h

Answer

ṁ2 = 400 kg/h
2.Mass balance

Answer: W = 2816 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 = 4400·18/50 = 1584 kg/h

Total balance

W = F − P = 4400 − 1584 = 2816 kg/h

Answer

W = 2816 kg/h
3.Energy balance

Answer: tout = 39.35 °C

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

Heat released by the hot stream

Q = 0.5·3.6·(160 − 110) = 90 kW

Heat picked up by the water

tout = tin + Q/(ṁccp,c) = 25 + 90/(1.5·4.18) = 39.35 °C

Answer

tout = 39.35 °C
4.Energy balance

Answer: ṁc = 0.9923 kg/s

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

Heat released

Qh = 1·2.44·(140 − 110) = 73.2 kW

Heat that reaches the water

Qc = (1 − 0.15)·73.2 = 62.22 kW

Water flow rate

ṁc = Qc/(cp,c·Δt) = 62.22/(4.18·15) = 0.9923 kg/s

Answer

ṁc = 0.9923 kg/s
5.Ideal gas law

Answer: V = 200.9 L

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

Moles

n = 130/16.04 = 8.105 mol
T = 25 + 273.15 = 298.1 K

Volume

V = nRT/P = 8.105·8.314·298.1/100 = 200.9 L

Answer

V = 200.9 L
6.Ideal gas law

Answer: ρ = 0.9547 kg/m³

Ideal gas density: ρ = PM/(RT).

Rearrange the gas law

PV = (m/M)RT → ρ = m/V = PM/(RT)
T = 130 + 273.15 = 403.1 K

Density

ρ = 100·32/(8.314·403.1) = 0.9547 kg/m³

Answer

ρ = 0.9547 kg/m³
7.Stoichiometry

Answer: m = 19.45 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) = 16/28.02 = 0.571 mol
n(H2) = 64/2.016 = 31.75 mol

Limiting reactant

0.571/1 = 0.571, 31.75/3 = 10.58 → N2

Mass of product

n(NH3) = 0.571·2 = 1.142 mol
m = 1.142·17.03 = 19.45 g

Answer

m = 19.45 g
8.Stoichiometry

Answer: Y = 94.04 %

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(Fe2O3) = 8/159.69 = 0.0501 mol
m = 0.0501·3/1·44.01 = 6.614 g

Percent yield

Y = 6.22/6.614·100 = 94.04%

Answer

Y = 94.04%

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