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 · 30% sugarStream 2ṁ₂ = ? · 15% 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 = 30 %x2 = 15 %x3 = 20 %
ṁ2 = kg/h
2.Mass balanceHard
EvaporatorFeed F1000 kg/h · 8% saltProduct PP = ? · 30%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 = 1000 kg/hxF = 8 %xP = 30 %
P = kg/h
3.Energy balanceMedium
hot sidecold sideethanol · 170 °C60 °Cwater · 20 °Ct = ? °C

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

ṁh = 3 kg/scp,h = 2.44 kJ/(kg·K)Th: 170 → 60 °Cṁc = 1.5 kg/scp,c = 4.18 kJ/(kg·K)tin = 20 °C
tout = °C
4.Energy balanceHard
hot sidecold sideethanol · 140 °C60 °Cwater · 15 °C80 °C · ṁ = ?15% lost

Hot ethanol 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 = 2.44 kJ/(kg·K)Th: 140 → 60 °Ct: 15 → 80 °Ccp,c = 4.18 kJ/(kg·K)loss = 15 %
ṁc = kg/s
5.Ideal gas lawMedium
40 g of HeP = 150 kPaT = 45 °CV = ?

What volume V does 40 g of He occupy at 150 kPa and 45 °C? Treat it as an ideal gas.

m = 40 gM = 4.003 g/molP = 150 kPaT = 45 °CR = 8.314 J/(mol·K)
V = L
6.Ideal gas lawHard
CH₄, M = 16.04 g/molP = 1000 kPaT = 70 °Cρ = ?

Find the density ρ of CH₄ at 1000 kPa and 70 °C, treating it as an ideal gas.

M = 16.04 g/molP = 1000 kPaT = 70 °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)

6 g of Na and 10 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(CaCO3​)n(CaCO3​)n(CO2​)m(CO2​)÷ M(CaCO3​)× 1/1× M(CO2​)

In the reaction CaCO3 → CaO + CO2, 24 g of CaCO3 reacts completely (the other reactants are in excess) and 9.29 g of CO2 is collected. Find the percent yield Y.

M(CaCO3) = 100.09 g/molM(CO2) = 44.01 g/molactual: 9.29 g
Y = %

3Solutions

1.Mass balance

Answer: ṁ2 = 1300 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·(30 − 20)/(20 − 15) = 1300 kg/h

Answer

ṁ2 = 1300 kg/h
2.Mass balance

Answer: P = 266.7 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 = 1000·8/30 = 266.7 kg/h

Answer

P = 266.7 kg/h
3.Energy balance

Answer: tout = 148.4 °C

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

Heat released by the hot stream

Q = 3·2.44·(170 − 60) = 805.2 kW

Heat picked up by the water

tout = tin + Q/(ṁccp,c) = 20 + 805.2/(1.5·4.18) = 148.4 °C

Answer

tout = 148.4 °C
4.Energy balance

Answer: ṁc = 1.221 kg/s

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

Heat released

Qh = 2·2.44·(140 − 60) = 390.4 kW

Heat that reaches the water

Qc = (1 − 0.15)·390.4 = 331.8 kW

Water flow rate

ṁc = Qc/(cp,c·Δt) = 331.8/(4.18·65) = 1.221 kg/s

Answer

ṁc = 1.221 kg/s
5.Ideal gas law

Answer: V = 176.2 L

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

Moles

n = 40/4.003 = 9.993 mol
T = 45 + 273.15 = 318.1 K

Volume

V = nRT/P = 9.993·8.314·318.1/150 = 176.2 L

Answer

V = 176.2 L
6.Ideal gas law

Answer: ρ = 5.622 kg/m³

Ideal gas density: ρ = PM/(RT).

Rearrange the gas law

PV = (m/M)RT → ρ = m/V = PM/(RT)
T = 70 + 273.15 = 343.1 K

Density

ρ = 1000·16.04/(8.314·343.1) = 5.622 kg/m³

Answer

ρ = 5.622 kg/m³
7.Stoichiometry

Answer: m = 15.25 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) = 6/22.99 = 0.261 mol
n(Cl2) = 10/70.9 = 0.141 mol

Limiting reactant

0.261/2 = 0.1305, 0.141/1 = 0.141 → Na

Mass of product

n(NaCl) = 0.1305·2 = 0.261 mol
m = 0.261·58.44 = 15.25 g

Answer

m = 15.25 g
8.Stoichiometry

Answer: Y = 88.03 %

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(CaCO3) = 24/100.09 = 0.2398 mol
m = 0.2398·1/1·44.01 = 10.55 g

Percent yield

Y = 9.29/10.55·100 = 88.03%

Answer

Y = 88.03%

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