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% sugarStream 2ṁ₂ = ? · 30% sugarProduct20% sugar

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

ṁ1 = 650 kg/hx1 = 15 %x2 = 30 %x3 = 20 %
ṁ2 = kg/h
2.Mass balanceHard
EvaporatorFeed F4200 kg/h · 13% saltProduct P30% saltVapour W (pure water)W = ? kg/h

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

F = 4200 kg/hxF = 13 %xP = 30 %
W = kg/h
3.Energy balanceMedium
hot sidecold sidelight oil · 160 °C100 °Cwater · 25 °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: 160 → 100 °Cṁc = 0.5 kg/scp,c = 4.18 kJ/(kg·K)tin = 25 °C
tout = °C
4.Energy balanceHard
hot sidecold sideethanol · 170 °C130 °Cwater · 10 °C50 °C · ṁ = ?5% lost

Hot ethanol 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 = 2 kg/scp,h = 2.44 kJ/(kg·K)Th: 170 → 130 °Ct: 10 → 50 °Ccp,c = 4.18 kJ/(kg·K)loss = 5 %
ṁc = kg/s
5.Ideal gas lawMedium
400 g of HeP = 200 kPaT = 100 °CV = ?

What volume V does 400 g of He occupy at 200 kPa and 100 °C? Treat it as an ideal gas.

m = 400 gM = 4.003 g/molP = 200 kPaT = 100 °CR = 8.314 J/(mol·K)
V = L
6.Ideal gas lawHard
Ar, M = 39.95 g/molP = 500 kPaT = 20 °Cρ = ?

Find the density ρ of Ar at 500 kPa and 20 °C, treating it as an ideal gas.

M = 39.95 g/molP = 500 kPaT = 20 °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)

6 g of Mg and 80 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(Mg)n(Mg)n(MgO)m(MgO)÷ M(Mg)× 2/2× M(MgO)

In the reaction 2Mg + O2 → 2MgO, 4 g of Mg reacts completely (the other reactants are in excess) and 5.84 g of MgO is collected. Find the percent yield Y.

M(Mg) = 24.31 g/molM(MgO) = 40.3 g/molactual: 5.84 g
Y = %

3Solutions

1.Mass balance

Answer: ṁ2 = 325 kg/h

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

Balances

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

Solve for ṁ2

ṁ2 = ṁ1(x1 − x3)/(x3 − x2) = 650·(15 − 20)/(20 − 30) = 325 kg/h

Answer

ṁ2 = 325 kg/h
2.Mass balance

Answer: W = 2380 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 = 4200·13/30 = 1820 kg/h

Total balance

W = F − P = 4200 − 1820 = 2380 kg/h

Answer

W = 2380 kg/h
3.Energy balance

Answer: tout = 115.4 °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·(160 − 100) = 189 kW

Heat picked up by the water

tout = tin + Q/(ṁccp,c) = 25 + 189/(0.5·4.18) = 115.4 °C

Answer

tout = 115.4 °C
4.Energy balance

Answer: ṁc = 1.109 kg/s

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

Heat released

Qh = 2·2.44·(170 − 130) = 195.2 kW

Heat that reaches the water

Qc = (1 − 0.05)·195.2 = 185.4 kW

Water flow rate

ṁc = Qc/(cp,c·Δt) = 185.4/(4.18·40) = 1.109 kg/s

Answer

ṁc = 1.109 kg/s
5.Ideal gas law

Answer: V = 1550 L

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

Moles

n = 400/4.003 = 99.93 mol
T = 100 + 273.15 = 373.1 K

Volume

V = nRT/P = 99.93·8.314·373.1/200 = 1550 L

Answer

V = 1550 L
6.Ideal gas law

Answer: ρ = 8.196 kg/m³

Ideal gas density: ρ = PM/(RT).

Rearrange the gas law

PV = (m/M)RT → ρ = m/V = PM/(RT)
T = 20 + 273.15 = 293.1 K

Density

ρ = 500·39.95/(8.314·293.1) = 8.196 kg/m³

Answer

ρ = 8.196 kg/m³
7.Stoichiometry

Answer: m = 9.947 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) = 6/24.31 = 0.2468 mol
n(O2) = 80/32 = 2.5 mol

Limiting reactant

0.2468/2 = 0.1234, 2.5/1 = 2.5 → Mg

Mass of product

n(MgO) = 0.1234·2 = 0.2468 mol
m = 0.2468·40.3 = 9.947 g

Answer

m = 9.947 g
8.Stoichiometry

Answer: Y = 88.07 %

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(Mg) = 4/24.31 = 0.1645 mol
m = 0.1645·2/2·40.3 = 6.631 g

Percent yield

Y = 5.84/6.631·100 = 88.07%

Answer

Y = 88.07%

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