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 · 30% NaOHStream 2ṁ₂ = ? · 15% NaOHProduct25% NaOH

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

ṁ1 = 750 kg/hx1 = 30 %x2 = 15 %x3 = 25 %
ṁ2 = kg/h
2.Mass balanceHard
EvaporatorFeed F600 kg/h · 18% saltProduct PP = ? · 60%Vapour W (pure water)0% salt

An evaporator concentrates a salt solution by boiling off pure water. Find the product flow P in kg/h.

F = 600 kg/hxF = 18 %xP = 60 %
P = kg/h
3.Energy balanceMedium
hot sidecold sideglycol mix · 180 °C120 °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 = 2.5 kg/scp,h = 3.6 kJ/(kg·K)Th: 180 → 120 °Cṁc = 3 kg/scp,c = 4.18 kJ/(kg·K)tin = 15 °C
tout = °C
4.Energy balanceHard
hot sidecold sideglycol mix · 120 °C80 °Cwater · 25 °C45 °C · ṁ = ?5% lost

Hot glycol mix 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 = 1.5 kg/scp,h = 3.6 kJ/(kg·K)Th: 120 → 80 °Ct: 25 → 45 °Ccp,c = 4.18 kJ/(kg·K)loss = 5 %
ṁc = kg/s
5.Ideal gas lawMedium
330 g of CO₂V = 20 LT = 30 °CP = ?

A tank of 20 L contains 330 g of CO₂ at 30 °C. Treat it as an ideal gas and find the pressure P.

m = 330 gM = 44.01 g/molV = 20 LT = 30 °CR = 8.314 J/(mol·K)
P = kPa
6.Ideal gas lawHard
O₂, M = 32 g/molP = 650 kPaT = 80 °Cρ = ?

Find the density ρ of O₂ at 650 kPa and 80 °C, treating it as an ideal gas.

M = 32 g/molP = 650 kPaT = 80 °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​)

8 g of N2 and 80 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(Fe)m(Fe)÷ M(Fe2​O3​)× 2/1× M(Fe)

In the reaction Fe2O3 + 3CO → 2Fe + 3CO2, 12 g of Fe2O3 reacts completely (the other reactants are in excess) and 6.88 g of Fe is collected. Find the percent yield Y.

M(Fe2O3) = 159.69 g/molM(Fe) = 55.85 g/molactual: 6.88 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·(30 − 25)/(25 − 15) = 375 kg/h

Answer

ṁ2 = 375 kg/h
2.Mass balance

Answer: P = 180 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 = 600·18/60 = 180 kg/h

Answer

P = 180 kg/h
3.Energy balance

Answer: tout = 58.06 °C

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

Heat released by the hot stream

Q = 2.5·3.6·(180 − 120) = 540 kW

Heat picked up by the water

tout = tin + Q/(ṁccp,c) = 15 + 540/(3·4.18) = 58.06 °C

Answer

tout = 58.06 °C
4.Energy balance

Answer: ṁc = 2.455 kg/s

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

Heat released

Qh = 1.5·3.6·(120 − 80) = 216 kW

Heat that reaches the water

Qc = (1 − 0.05)·216 = 205.2 kW

Water flow rate

ṁc = Qc/(cp,c·Δt) = 205.2/(4.18·20) = 2.455 kg/s

Answer

ṁc = 2.455 kg/s
5.Ideal gas law

Answer: P = 944.9 kPa

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

Moles

n = m/M = 330/44.01 = 7.498 mol
T = 30 + 273.15 = 303.1 K

Pressure

P = nRT/V = 7.498·8.314·303.1/0.02 = 944.9 kPa

Answer

P = 944.9 kPa
6.Ideal gas law

Answer: ρ = 7.084 kg/m³

Ideal gas density: ρ = PM/(RT).

Rearrange the gas law

PV = (m/M)RT → ρ = m/V = PM/(RT)
T = 80 + 273.15 = 353.1 K

Density

ρ = 650·32/(8.314·353.1) = 7.084 kg/m³

Answer

ρ = 7.084 kg/m³
7.Stoichiometry

Answer: m = 9.724 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) = 8/28.02 = 0.2855 mol
n(H2) = 80/2.016 = 39.68 mol

Limiting reactant

0.2855/1 = 0.2855, 39.68/3 = 13.23 → N2

Mass of product

n(NH3) = 0.2855·2 = 0.571 mol
m = 0.571·17.03 = 9.724 g

Answer

m = 9.724 g
8.Stoichiometry

Answer: Y = 81.97 %

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) = 12/159.69 = 0.07515 mol
m = 0.07515·2/1·55.85 = 8.394 g

Percent yield

Y = 6.88/8.394·100 = 81.97%

Answer

Y = 81.97%

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