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 1950 kg/h · 20% saltStream 2ṁ₂ = ? · 50% saltProduct40% salt

How much of stream 2 must be added each hour to stream 1 so that the blended product contains 40% salt? Find ṁ2.

ṁ1 = 950 kg/hx1 = 20 %x2 = 50 %x3 = 40 %
ṁ2 = kg/h
2.Mass balanceHard
EvaporatorFeed F3200 kg/h · 5% sugarProduct P55% sugarVapour W (pure water)W = ? kg/h

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

F = 3200 kg/hxF = 5 %xP = 55 %
W = kg/h
3.Energy balanceMedium
hot sidecold sidelight oil · 180 °C80 °Cwater · 25 °Ct = ? °C

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

ṁh = 3 kg/scp,h = 2.1 kJ/(kg·K)Th: 180 → 80 °Cṁc = 2 kg/scp,c = 4.18 kJ/(kg·K)tin = 25 °C
tout = °C
4.Energy balanceHard
hot sidecold sideethanol · 150 °C80 °Cwater · 15 °C55 °C · ṁ = ?10% lost

Hot ethanol 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 = 2.44 kJ/(kg·K)Th: 150 → 80 °Ct: 15 → 55 °Ccp,c = 4.18 kJ/(kg·K)loss = 10 %
ṁc = kg/s
5.Ideal gas lawMedium
40 g of HeV = 5 LT = 55 °CP = ?

A tank of 5 L contains 40 g of He at 55 °C. Treat it as an ideal gas and find the pressure P.

m = 40 gM = 4.003 g/molV = 5 LT = 55 °CR = 8.314 J/(mol·K)
P = kPa
6.Ideal gas lawHard
O₂, M = 32 g/molP = 550 kPaT = 200 °Cρ = ?

Find the density ρ of O₂ at 550 kPa and 200 °C, treating it as an ideal gas.

M = 32 g/molP = 550 kPaT = 200 °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 20 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(C6​H12​O6​)n(C6​H12​O6​)n(C2​H5​OH)m(C2​H5​OH)÷ M(C6​H12​O6​)× 2/1× M(C2​H5​OH)

In the reaction C6H12O6 → 2C2H5OH + 2CO2, 6 g of C6H12O6 reacts completely (the other reactants are in excess) and 2.61 g of C2H5OH is collected. Find the percent yield Y.

M(C6H12O6) = 180.16 g/molM(C2H5OH) = 46.07 g/molactual: 2.61 g
Y = %

3Solutions

1.Mass balance

Answer: ṁ2 = 1900 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) = 950·(20 − 40)/(40 − 50) = 1900 kg/h

Answer

ṁ2 = 1900 kg/h
2.Mass balance

Answer: W = 2909 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 = 3200·5/55 = 290.9 kg/h

Total balance

W = F − P = 3200 − 290.9 = 2909 kg/h

Answer

W = 2909 kg/h
3.Energy balance

Answer: tout = 100.4 °C

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

Heat released by the hot stream

Q = 3·2.1·(180 − 80) = 630 kW

Heat picked up by the water

tout = tin + Q/(ṁccp,c) = 25 + 630/(2·4.18) = 100.4 °C

Answer

tout = 100.4 °C
4.Energy balance

Answer: ṁc = 0.9194 kg/s

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

Heat released

Qh = 1·2.44·(150 − 80) = 170.8 kW

Heat that reaches the water

Qc = (1 − 0.1)·170.8 = 153.7 kW

Water flow rate

ṁc = Qc/(cp,c·Δt) = 153.7/(4.18·40) = 0.9194 kg/s

Answer

ṁc = 0.9194 kg/s
5.Ideal gas law

Answer: P = 5452 kPa

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

Moles

n = m/M = 40/4.003 = 9.993 mol
T = 55 + 273.15 = 328.1 K

Pressure

P = nRT/V = 9.993·8.314·328.1/0.005 = 5452 kPa

Answer

P = 5452 kPa
6.Ideal gas law

Answer: ρ = 4.474 kg/m³

Ideal gas density: ρ = PM/(RT).

Rearrange the gas law

PV = (m/M)RT → ρ = m/V = PM/(RT)
T = 200 + 273.15 = 473.1 K

Density

ρ = 550·32/(8.314·473.1) = 4.474 kg/m³

Answer

ρ = 4.474 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) = 20/70.9 = 0.2821 mol

Limiting reactant

0.261/2 = 0.1305, 0.2821/1 = 0.2821 → 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 = 85.05 %

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(C6H12O6) = 6/180.16 = 0.0333 mol
m = 0.0333·2/1·46.07 = 3.069 g

Percent yield

Y = 2.61/3.069·100 = 85.05%

Answer

Y = 85.05%

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