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.

New problemsBack to the course

In the print window, choose “Save as PDF” to keep a copy on your device.

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 1950 kg/h · 40% NaOHStream 2ṁ₂ = ? · 15% NaOHProduct35% NaOH

How much of stream 2 must be added each hour to stream 1 so that the blended product contains 35% NaOH? Find ṁ2.

ṁ1 = 950 kg/hx1 = 40 %x2 = 15 %x3 = 35 %
ṁ2 = kg/h
2.Mass balanceHard
EvaporatorFeed F2100 kg/h · 17% NaOHProduct P30% NaOHVapour W (pure water)W = ? kg/h

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

F = 2100 kg/hxF = 17 %xP = 30 %
W = kg/h
3.Energy balanceMedium
hot sidecold sidelight oil · 140 °C100 °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 kg/scp,h = 2.1 kJ/(kg·K)Th: 140 → 100 °Cṁc = 1 kg/scp,c = 4.18 kJ/(kg·K)tin = 30 °C
tout = °C
4.Energy balanceHard
hot sidecold sidelight oil · 150 °C120 °Cwater · 25 °C50 °C · ṁ = ?10% lost

Hot light oil heats water in an exchanger that loses 10% 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: 150 → 120 °Ct: 25 → 50 °Ccp,c = 4.18 kJ/(kg·K)loss = 10 %
ṁc = kg/s
5.Ideal gas lawMedium
430 g of O₂P = 450 kPaT = 65 °CV = ?

What volume V does 430 g of O₂ occupy at 450 kPa and 65 °C? Treat it as an ideal gas.

m = 430 gM = 32 g/molP = 450 kPaT = 65 °CR = 8.314 J/(mol·K)
V = L
6.Ideal gas lawHard
O₂, M = 32 g/molP = 700 kPaT = 160 °Cρ = ?

Find the density ρ of O₂ at 700 kPa and 160 °C, treating it as an ideal gas.

M = 32 g/molP = 700 kPaT = 160 °CR = 8.314 J/(mol·K)
ρ = kg/m³
7.StoichiometryMedium
m(X)n(X)n(H2​O)m(H2​O)÷ M(X)× 2/ν× M(H2​O)

20 g of CH4 and 20 g of O2 react according to CH4 + 2O2 → CO2 + 2H2O. What mass m of H2O can form?

M(CH4) = 16.04M(O2) = 32M(H2O) = 18.02 g/mol
m = g
8.StoichiometryHard
m(CH4​)n(CH4​)n(H2​O)m(H2​O)÷ M(CH4​)× 2/1× M(H2​O)

In the reaction CH4 + 2O2 → CO2 + 2H2O, 6 g of CH4 reacts completely (the other reactants are in excess) and 9.44 g of H2O is collected. Find the percent yield Y.

M(CH4) = 16.04 g/molM(H2O) = 18.02 g/molactual: 9.44 g
Y = %

3Solutions

1.Mass balance

Answer: ṁ2 = 237.5 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) = 950·(40 − 35)/(35 − 15) = 237.5 kg/h

Answer

ṁ2 = 237.5 kg/h
2.Mass balance

Answer: W = 910 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 = 2100·17/30 = 1190 kg/h

Total balance

W = F − P = 2100 − 1190 = 910 kg/h

Answer

W = 910 kg/h
3.Energy balance

Answer: tout = 70.19 °C

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

Heat released by the hot stream

Q = 2·2.1·(140 − 100) = 168 kW

Heat picked up by the water

tout = tin + Q/(ṁccp,c) = 30 + 168/(1·4.18) = 70.19 °C

Answer

tout = 70.19 °C
4.Energy balance

Answer: ṁc = 0.5426 kg/s

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

Heat released

Qh = 1·2.1·(150 − 120) = 63 kW

Heat that reaches the water

Qc = (1 − 0.1)·63 = 56.7 kW

Water flow rate

ṁc = Qc/(cp,c·Δt) = 56.7/(4.18·25) = 0.5426 kg/s

Answer

ṁc = 0.5426 kg/s
5.Ideal gas law

Answer: V = 83.95 L

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

Moles

n = 430/32 = 13.44 mol
T = 65 + 273.15 = 338.1 K

Volume

V = nRT/P = 13.44·8.314·338.1/450 = 83.95 L

Answer

V = 83.95 L
6.Ideal gas law

Answer: ρ = 6.22 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 = 700·32/(8.314·433.1) = 6.22 kg/m³

Answer

ρ = 6.22 kg/m³
7.Stoichiometry

Answer: m = 11.26 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) = 20/16.04 = 1.247 mol
n(O2) = 20/32 = 0.625 mol

Limiting reactant

1.247/1 = 1.247, 0.625/2 = 0.3125 → O2

Mass of product

n(H2O) = 0.3125·2 = 0.625 mol
m = 0.625·18.02 = 11.26 g

Answer

m = 11.26 g
8.Stoichiometry

Answer: Y = 70.02 %

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(CH4) = 6/16.04 = 0.3741 mol
m = 0.3741·2/1·18.02 = 13.48 g

Percent yield

Y = 9.44/13.48·100 = 70.02%

Answer

Y = 70.02%

ExamProblems · examproblems.com/courses/material-and-energy-balances