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 1550 kg/h · 35% saltStream 2ṁ₂ = ? · 15% 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 = 550 kg/hx1 = 35 %x2 = 15 %x3 = 25 %
ṁ2 = kg/h
2.Mass balanceHard
EvaporatorFeed F4500 kg/h · 12% saltProduct P45% 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 = 4500 kg/hxF = 12 %xP = 45 %
W = kg/h
3.Energy balanceMedium
hot sidecold sideglycol mix · 170 °C130 °Cwater · 20 °Ct = ? °C

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

ṁh = 0.5 kg/scp,h = 3.6 kJ/(kg·K)Th: 170 → 130 °Cṁc = 0.5 kg/scp,c = 4.18 kJ/(kg·K)tin = 20 °C
tout = °C
4.Energy balanceHard
hot sidecold sideglycol mix · 120 °C70 °Cwater · 20 °C45 °C · ṁ = ?10% lost

Hot glycol mix 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 = 2 kg/scp,h = 3.6 kJ/(kg·K)Th: 120 → 70 °Ct: 20 → 45 °Ccp,c = 4.18 kJ/(kg·K)loss = 10 %
ṁc = kg/s
5.Ideal gas lawMedium
240 g of O₂P = 550 kPaT = 90 °CV = ?

What volume V does 240 g of O₂ occupy at 550 kPa and 90 °C? Treat it as an ideal gas.

m = 240 gM = 32 g/molP = 550 kPaT = 90 °CR = 8.314 J/(mol·K)
V = L
6.Ideal gas lawHard
Ar, M = 39.95 g/molP = 650 kPaT = 130 °Cρ = ?

Find the density ρ of Ar at 650 kPa and 130 °C, treating it as an ideal gas.

M = 39.95 g/molP = 650 kPaT = 130 °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)

4 g of Na and 64 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(C3​H8​)n(C3​H8​)n(H2​O)m(H2​O)÷ M(C3​H8​)× 4/1× M(H2​O)

In the reaction C3H8 + 5O2 → 3CO2 + 4H2O, 8 g of C3H8 reacts completely (the other reactants are in excess) and 12.29 g of H2O is collected. Find the percent yield Y.

M(C3H8) = 44.1 g/molM(H2O) = 18.02 g/molactual: 12.29 g
Y = %

3Solutions

1.Mass balance

Answer: ṁ2 = 550 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) = 550·(35 − 25)/(25 − 15) = 550 kg/h

Answer

ṁ2 = 550 kg/h
2.Mass balance

Answer: W = 3300 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 = 4500·12/45 = 1200 kg/h

Total balance

W = F − P = 4500 − 1200 = 3300 kg/h

Answer

W = 3300 kg/h
3.Energy balance

Answer: tout = 54.45 °C

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

Heat released by the hot stream

Q = 0.5·3.6·(170 − 130) = 72 kW

Heat picked up by the water

tout = tin + Q/(ṁccp,c) = 20 + 72/(0.5·4.18) = 54.45 °C

Answer

tout = 54.45 °C
4.Energy balance

Answer: ṁc = 3.1 kg/s

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

Heat released

Qh = 2·3.6·(120 − 70) = 360 kW

Heat that reaches the water

Qc = (1 − 0.1)·360 = 324 kW

Water flow rate

ṁc = Qc/(cp,c·Δt) = 324/(4.18·25) = 3.1 kg/s

Answer

ṁc = 3.1 kg/s
5.Ideal gas law

Answer: V = 41.17 L

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

Moles

n = 240/32 = 7.5 mol
T = 90 + 273.15 = 363.1 K

Volume

V = nRT/P = 7.5·8.314·363.1/550 = 41.17 L

Answer

V = 41.17 L
6.Ideal gas law

Answer: ρ = 7.747 kg/m³

Ideal gas density: ρ = PM/(RT).

Rearrange the gas law

PV = (m/M)RT → ρ = m/V = PM/(RT)
T = 130 + 273.15 = 403.1 K

Density

ρ = 650·39.95/(8.314·403.1) = 7.747 kg/m³

Answer

ρ = 7.747 kg/m³
7.Stoichiometry

Answer: m = 10.17 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) = 4/22.99 = 0.174 mol
n(Cl2) = 64/70.9 = 0.9027 mol

Limiting reactant

0.174/2 = 0.08699, 0.9027/1 = 0.9027 → Na

Mass of product

n(NaCl) = 0.08699·2 = 0.174 mol
m = 0.174·58.44 = 10.17 g

Answer

m = 10.17 g
8.Stoichiometry

Answer: Y = 93.99 %

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(C3H8) = 8/44.1 = 0.1814 mol
m = 0.1814·4/1·18.02 = 13.08 g

Percent yield

Y = 12.29/13.08·100 = 93.99%

Answer

Y = 93.99%

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