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 1500 kg/h · 20% saltStream 2ṁ₂ = ? · 45% 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 = 500 kg/hx1 = 20 %x2 = 45 %x3 = 25 %
ṁ2 = kg/h
2.Mass balanceHard
EvaporatorFeed F700 kg/h · 19% 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 = 700 kg/hxF = 19 %xP = 30 %
W = kg/h
3.Energy balanceMedium
hot sidecold sidelight oil · 140 °C110 °Cwater · 30 °Ct = ? °C

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

ṁh = 2.5 kg/scp,h = 2.1 kJ/(kg·K)Th: 140 → 110 °Cṁc = 2 kg/scp,c = 4.18 kJ/(kg·K)tin = 30 °C
tout = °C
4.Energy balanceHard
hot sidecold sideethanol · 160 °C80 °Cwater · 20 °C45 °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 = 1.5 kg/scp,h = 2.44 kJ/(kg·K)Th: 160 → 80 °Ct: 20 → 45 °Ccp,c = 4.18 kJ/(kg·K)loss = 5 %
ṁc = kg/s
5.Ideal gas lawMedium
410 g of ArP = 700 kPaT = 50 °CV = ?

What volume V does 410 g of Ar occupy at 700 kPa and 50 °C? Treat it as an ideal gas.

m = 410 gM = 39.95 g/molP = 700 kPaT = 50 °CR = 8.314 J/(mol·K)
V = L
6.Ideal gas lawHard
N₂, M = 28.01 g/molP = 500 kPaT = 110 °Cρ = ?

Find the density ρ of N₂ at 500 kPa and 110 °C, treating it as an ideal gas.

M = 28.01 g/molP = 500 kPaT = 110 °CR = 8.314 J/(mol·K)
ρ = kg/m³
7.StoichiometryMedium
m(X)n(X)n(NaCl)m(NaCl)÷ M(X)× 2/ν× M(NaCl)

24 g of Na and 40 g of Cl2 react according to 2Na + Cl2 → 2NaCl. What mass m of NaCl can form?

M(Na) = 22.99M(Cl2) = 70.9M(NaCl) = 58.44 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, 20 g of Na reacts completely (the other reactants are in excess) and 46.26 g of NaCl is collected. Find the percent yield Y.

M(Na) = 22.99 g/molM(NaCl) = 58.44 g/molactual: 46.26 g
Y = %

3Solutions

1.Mass balance

Answer: ṁ2 = 125 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) = 500·(20 − 25)/(25 − 45) = 125 kg/h

Answer

ṁ2 = 125 kg/h
2.Mass balance

Answer: W = 256.7 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 = 700·19/30 = 443.3 kg/h

Total balance

W = F − P = 700 − 443.3 = 256.7 kg/h

Answer

W = 256.7 kg/h
3.Energy balance

Answer: tout = 48.84 °C

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

Heat released by the hot stream

Q = 2.5·2.1·(140 − 110) = 157.5 kW

Heat picked up by the water

tout = tin + Q/(ṁccp,c) = 30 + 157.5/(2·4.18) = 48.84 °C

Answer

tout = 48.84 °C
4.Energy balance

Answer: ṁc = 2.662 kg/s

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

Heat released

Qh = 1.5·2.44·(160 − 80) = 292.8 kW

Heat that reaches the water

Qc = (1 − 0.05)·292.8 = 278.2 kW

Water flow rate

ṁc = Qc/(cp,c·Δt) = 278.2/(4.18·25) = 2.662 kg/s

Answer

ṁc = 2.662 kg/s
5.Ideal gas law

Answer: V = 39.39 L

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

Moles

n = 410/39.95 = 10.26 mol
T = 50 + 273.15 = 323.1 K

Volume

V = nRT/P = 10.26·8.314·323.1/700 = 39.39 L

Answer

V = 39.39 L
6.Ideal gas law

Answer: ρ = 4.396 kg/m³

Ideal gas density: ρ = PM/(RT).

Rearrange the gas law

PV = (m/M)RT → ρ = m/V = PM/(RT)
T = 110 + 273.15 = 383.1 K

Density

ρ = 500·28.01/(8.314·383.1) = 4.396 kg/m³

Answer

ρ = 4.396 kg/m³
7.Stoichiometry

Answer: m = 61.01 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(Na) = 24/22.99 = 1.044 mol
n(Cl2) = 40/70.9 = 0.5642 mol

Limiting reactant

1.044/2 = 0.522, 0.5642/1 = 0.5642 → Na

Mass of product

n(NaCl) = 0.522·2 = 1.044 mol
m = 1.044·58.44 = 61.01 g

Answer

m = 61.01 g
8.Stoichiometry

Answer: Y = 90.99 %

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) = 20/22.99 = 0.8699 mol
m = 0.8699·2/2·58.44 = 50.84 g

Percent yield

Y = 46.26/50.84·100 = 90.99%

Answer

Y = 90.99%

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