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 1650 kg/h · 15% NaOHStream 2ṁ₂ = ? · 40% NaOHProduct20% NaOH

How much of stream 2 must be added each hour to stream 1 so that the blended product contains 20% NaOH? Find ṁ2.

ṁ1 = 650 kg/hx1 = 15 %x2 = 40 %x3 = 20 %
ṁ2 = kg/h
2.Mass balanceHard
EvaporatorFeed F4700 kg/h · 8% 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 = 4700 kg/hxF = 8 %xP = 65 %
W = kg/h
3.Energy balanceMedium
hot sidecold sidelight oil · 130 °C100 °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 = 0.5 kg/scp,h = 2.1 kJ/(kg·K)Th: 130 → 100 °Cṁc = 1.5 kg/scp,c = 4.18 kJ/(kg·K)tin = 20 °C
tout = °C
4.Energy balanceHard
hot sidecold sidelight oil · 180 °C130 °Cwater · 20 °C65 °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 = 1 kg/scp,h = 2.1 kJ/(kg·K)Th: 180 → 130 °Ct: 20 → 65 °Ccp,c = 4.18 kJ/(kg·K)loss = 5 %
ṁc = kg/s
5.Ideal gas lawMedium
260 g of HeV = 20 LT = 65 °CP = ?

A tank of 20 L contains 260 g of He at 65 °C. Treat it as an ideal gas and find the pressure P.

m = 260 gM = 4.003 g/molV = 20 LT = 65 °CR = 8.314 J/(mol·K)
P = kPa
6.Ideal gas lawHard
O₂, M = 32 g/molP = 650 kPaT = 160 °Cρ = ?

Find the density ρ of O₂ at 650 kPa and 160 °C, treating it as an ideal gas.

M = 32 g/molP = 650 kPaT = 160 °CR = 8.314 J/(mol·K)
ρ = kg/m³
7.StoichiometryMedium
m(X)n(X)n(MgO)m(MgO)÷ M(X)× 2/ν× M(MgO)

4 g of Mg and 10 g of O2 react according to 2Mg + O2 → 2MgO. What mass m of MgO can form?

M(Mg) = 24.31M(O2) = 32M(MgO) = 40.3 g/mol
m = g
8.StoichiometryHard
m(Al)n(Al)n(AlCl3​)m(AlCl3​)÷ M(Al)× 2/2× M(AlCl3​)

In the reaction 2Al + 3Cl2 → 2AlCl3, 16 g of Al reacts completely (the other reactants are in excess) and 71.95 g of AlCl3 is collected. Find the percent yield Y.

M(Al) = 26.98 g/molM(AlCl3) = 133.33 g/molactual: 71.95 g
Y = %

3Solutions

1.Mass balance

Answer: ṁ2 = 162.5 kg/h

Steady-state mass balance: in = out for the total and for each component. Blending problems combine both.

Balances

Total: ṁ3 = ṁ1 + ṁ2
NaOH: ṁ1x1 + ṁ2x2 = (ṁ1 + ṁ2)x3

Solve for ṁ2

ṁ2 = ṁ1(x1 − x3)/(x3 − x2) = 650·(15 − 20)/(20 − 40) = 162.5 kg/h

Answer

ṁ2 = 162.5 kg/h
2.Mass balance

Answer: W = 4122 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 = 4700·8/65 = 578.5 kg/h

Total balance

W = F − P = 4700 − 578.5 = 4122 kg/h

Answer

W = 4122 kg/h
3.Energy balance

Answer: tout = 25.02 °C

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

Heat released by the hot stream

Q = 0.5·2.1·(130 − 100) = 31.5 kW

Heat picked up by the water

tout = tin + Q/(ṁccp,c) = 20 + 31.5/(1.5·4.18) = 25.02 °C

Answer

tout = 25.02 °C
4.Energy balance

Answer: ṁc = 0.5303 kg/s

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

Heat released

Qh = 1·2.1·(180 − 130) = 105 kW

Heat that reaches the water

Qc = (1 − 0.05)·105 = 99.75 kW

Water flow rate

ṁc = Qc/(cp,c·Δt) = 99.75/(4.18·45) = 0.5303 kg/s

Answer

ṁc = 0.5303 kg/s
5.Ideal gas law

Answer: P = 9130 kPa

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

Moles

n = m/M = 260/4.003 = 64.95 mol
T = 65 + 273.15 = 338.1 K

Pressure

P = nRT/V = 64.95·8.314·338.1/0.02 = 9130 kPa

Answer

P = 9130 kPa
6.Ideal gas law

Answer: ρ = 5.776 kg/m³

Ideal gas density: ρ = PM/(RT).

Rearrange the gas law

PV = (m/M)RT → ρ = m/V = PM/(RT)
T = 160 + 273.15 = 433.1 K

Density

ρ = 650·32/(8.314·433.1) = 5.776 kg/m³

Answer

ρ = 5.776 kg/m³
7.Stoichiometry

Answer: m = 6.631 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(Mg) = 4/24.31 = 0.1645 mol
n(O2) = 10/32 = 0.3125 mol

Limiting reactant

0.1645/2 = 0.08227, 0.3125/1 = 0.3125 → Mg

Mass of product

n(MgO) = 0.08227·2 = 0.1645 mol
m = 0.1645·40.3 = 6.631 g

Answer

m = 6.631 g
8.Stoichiometry

Answer: Y = 91 %

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(Al) = 16/26.98 = 0.593 mol
m = 0.593·2/2·133.33 = 79.07 g

Percent yield

Y = 71.95/79.07·100 = 91%

Answer

Y = 91%

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