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 1800 kg/h · 30% sugarStream 2ṁ₂ = ? · 10% 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 = 800 kg/hx1 = 30 %x2 = 10 %x3 = 20 %
ṁ2 = kg/h
2.Mass balanceHard
EvaporatorFeed F1600 kg/h · 19% NaOHProduct PP = ? · 50%Vapour W (pure water)0% NaOH

An evaporator concentrates a NaOH solution by boiling off pure water. Find the product flow P in kg/h.

F = 1600 kg/hxF = 19 %xP = 50 %
P = kg/h
3.Energy balanceMedium
hot sidecold sideethanol · 140 °C60 °Cwater · 30 °Ct = ? °C

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

ṁh = 0.5 kg/scp,h = 2.44 kJ/(kg·K)Th: 140 → 60 °Cṁc = 0.5 kg/scp,c = 4.18 kJ/(kg·K)tin = 30 °C
tout = °C
4.Energy balanceHard
hot sidecold sidelight oil · 150 °C70 °Cwater · 30 °C65 °C · ṁ = ?15% lost

Hot light oil 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 = 2.1 kJ/(kg·K)Th: 150 → 70 °Ct: 30 → 65 °Ccp,c = 4.18 kJ/(kg·K)loss = 15 %
ṁc = kg/s
5.Ideal gas lawMedium
410 g of CO₂V = 10 LT = 30 °CP = ?

A tank of 10 L contains 410 g of CO₂ at 30 °C. Treat it as an ideal gas and find the pressure P.

m = 410 gM = 44.01 g/molV = 10 LT = 30 °CR = 8.314 J/(mol·K)
P = kPa
6.Ideal gas lawHard
N₂, M = 28.01 g/molP = 350 kPaT = 120 °Cρ = ?

Find the density ρ of N₂ at 350 kPa and 120 °C, treating it as an ideal gas.

M = 28.01 g/molP = 350 kPaT = 120 °CR = 8.314 J/(mol·K)
ρ = kg/m³
7.StoichiometryMedium
m(X)n(X)n(CO2​)m(CO2​)÷ M(X)× 1/ν× M(CO2​)

12 g of CH4 and 10 g of O2 react according to CH4 + 2O2 → CO2 + 2H2O. What mass m of CO2 can form?

M(CH4) = 16.04M(O2) = 32M(CO2) = 44.01 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, 16 g of CaCO3 reacts completely (the other reactants are in excess) and 3.87 g of CO2 is collected. Find the percent yield Y.

M(CaCO3) = 100.09 g/molM(CO2) = 44.01 g/molactual: 3.87 g
Y = %

3Solutions

1.Mass balance

Answer: ṁ2 = 800 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) = 800·(30 − 20)/(20 − 10) = 800 kg/h

Answer

ṁ2 = 800 kg/h
2.Mass balance

Answer: P = 608 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 = 1600·19/50 = 608 kg/h

Answer

P = 608 kg/h
3.Energy balance

Answer: tout = 76.7 °C

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

Heat released by the hot stream

Q = 0.5·2.44·(140 − 60) = 97.6 kW

Heat picked up by the water

tout = tin + Q/(ṁccp,c) = 30 + 97.6/(0.5·4.18) = 76.7 °C

Answer

tout = 76.7 °C
4.Energy balance

Answer: ṁc = 1.952 kg/s

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

Heat released

Qh = 2·2.1·(150 − 70) = 336 kW

Heat that reaches the water

Qc = (1 − 0.15)·336 = 285.6 kW

Water flow rate

ṁc = Qc/(cp,c·Δt) = 285.6/(4.18·35) = 1.952 kg/s

Answer

ṁc = 1.952 kg/s
5.Ideal gas law

Answer: P = 2348 kPa

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

Moles

n = m/M = 410/44.01 = 9.316 mol
T = 30 + 273.15 = 303.1 K

Pressure

P = nRT/V = 9.316·8.314·303.1/0.01 = 2348 kPa

Answer

P = 2348 kPa
6.Ideal gas law

Answer: ρ = 2.999 kg/m³

Ideal gas density: ρ = PM/(RT).

Rearrange the gas law

PV = (m/M)RT → ρ = m/V = PM/(RT)
T = 120 + 273.15 = 393.1 K

Density

ρ = 350·28.01/(8.314·393.1) = 2.999 kg/m³

Answer

ρ = 2.999 kg/m³
7.Stoichiometry

Answer: m = 6.877 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(CH4) = 12/16.04 = 0.7481 mol
n(O2) = 10/32 = 0.3125 mol

Limiting reactant

0.7481/1 = 0.7481, 0.3125/2 = 0.1563 → O2

Mass of product

n(CO2) = 0.1563·1 = 0.1563 mol
m = 0.1563·44.01 = 6.877 g

Answer

m = 6.877 g
8.Stoichiometry

Answer: Y = 55.01 %

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) = 16/100.09 = 0.1599 mol
m = 0.1599·1/1·44.01 = 7.035 g

Percent yield

Y = 3.87/7.035·100 = 55.01%

Answer

Y = 55.01%

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