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 1600 kg/h · 35% NaOHStream 2ṁ₂ = ? · 5% NaOHProduct30% NaOH

How much of stream 2 must be added each hour to stream 1 so that the blended product contains 30% NaOH? Find ṁ2.

ṁ1 = 600 kg/hx1 = 35 %x2 = 5 %x3 = 30 %
ṁ2 = kg/h
2.Mass balanceHard
EvaporatorFeed F1500 kg/h · 15% NaOHProduct P50% NaOHVapour W (pure water)W = ? kg/h

An evaporator concentrates a NaOH solution by boiling off pure water. Find the water evaporated W in kg/h.

F = 1500 kg/hxF = 15 %xP = 50 %
W = kg/h
3.Energy balanceMedium
hot sidecold sidelight oil · 120 °C90 °Cwater · 15 °Ct = ? °C

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

ṁh = 1.5 kg/scp,h = 2.1 kJ/(kg·K)Th: 120 → 90 °Cṁc = 1 kg/scp,c = 4.18 kJ/(kg·K)tin = 15 °C
tout = °C
4.Energy balanceHard
hot sidecold sideglycol mix · 140 °C90 °Cwater · 30 °C80 °C · ṁ = ?15% lost

Hot glycol mix 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 = 2 kg/scp,h = 3.6 kJ/(kg·K)Th: 140 → 90 °Ct: 30 → 80 °Ccp,c = 4.18 kJ/(kg·K)loss = 15 %
ṁc = kg/s
5.Ideal gas lawMedium
400 g of O₂V = 20 LT = 75 °CP = ?

A tank of 20 L contains 400 g of O₂ at 75 °C. Treat it as an ideal gas and find the pressure P.

m = 400 gM = 32 g/molV = 20 LT = 75 °CR = 8.314 J/(mol·K)
P = kPa
6.Ideal gas lawHard
O₂, M = 32 g/molP = 450 kPaT = 30 °Cρ = ?

Find the density ρ of O₂ at 450 kPa and 30 °C, treating it as an ideal gas.

M = 32 g/molP = 450 kPaT = 30 °CR = 8.314 J/(mol·K)
ρ = kg/m³
7.StoichiometryMedium
m(X)n(X)n(H2​O)m(H2​O)÷ M(X)× 4/ν× M(H2​O)

16 g of C3H8 and 48 g of O2 react according to C3H8 + 5O2 → 3CO2 + 4H2O. What mass m of H2O can form?

M(C3H8) = 44.1M(O2) = 32M(H2O) = 18.02 g/mol
m = g
8.StoichiometryHard
m(Na)n(Na)n(NaCl)m(NaCl)÷ M(Na)× 2/2× M(NaCl)

In the reaction 2Na + Cl2 → 2NaCl, 8 g of Na reacts completely (the other reactants are in excess) and 16.68 g of NaCl is collected. Find the percent yield Y.

M(Na) = 22.99 g/molM(NaCl) = 58.44 g/molactual: 16.68 g
Y = %

3Solutions

1.Mass balance

Answer: ṁ2 = 120 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) = 600·(35 − 30)/(30 − 5) = 120 kg/h

Answer

ṁ2 = 120 kg/h
2.Mass balance

Answer: W = 1050 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 = 1500·15/50 = 450 kg/h

Total balance

W = F − P = 1500 − 450 = 1050 kg/h

Answer

W = 1050 kg/h
3.Energy balance

Answer: tout = 37.61 °C

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

Heat released by the hot stream

Q = 1.5·2.1·(120 − 90) = 94.5 kW

Heat picked up by the water

tout = tin + Q/(ṁccp,c) = 15 + 94.5/(1·4.18) = 37.61 °C

Answer

tout = 37.61 °C
4.Energy balance

Answer: ṁc = 1.464 kg/s

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

Heat released

Qh = 2·3.6·(140 − 90) = 360 kW

Heat that reaches the water

Qc = (1 − 0.15)·360 = 306 kW

Water flow rate

ṁc = Qc/(cp,c·Δt) = 306/(4.18·50) = 1.464 kg/s

Answer

ṁc = 1.464 kg/s
5.Ideal gas law

Answer: P = 1809 kPa

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

Moles

n = m/M = 400/32 = 12.5 mol
T = 75 + 273.15 = 348.1 K

Pressure

P = nRT/V = 12.5·8.314·348.1/0.02 = 1809 kPa

Answer

P = 1809 kPa
6.Ideal gas law

Answer: ρ = 5.713 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 = 450·32/(8.314·303.1) = 5.713 kg/m³

Answer

ρ = 5.713 kg/m³
7.Stoichiometry

Answer: m = 21.62 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(C3H8) = 16/44.1 = 0.3628 mol
n(O2) = 48/32 = 1.5 mol

Limiting reactant

0.3628/1 = 0.3628, 1.5/5 = 0.3 → O2

Mass of product

n(H2O) = 0.3·4 = 1.2 mol
m = 1.2·18.02 = 21.62 g

Answer

m = 21.62 g
8.Stoichiometry

Answer: Y = 82.02 %

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(Na) = 8/22.99 = 0.348 mol
m = 0.348·2/2·58.44 = 20.34 g

Percent yield

Y = 16.68/20.34·100 = 82.02%

Answer

Y = 82.02%

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