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 11000 kg/h · 45% sugarStream 2ṁ₂ = ? · 5% sugarProduct30% sugar

How much of stream 2 must be added each hour to stream 1 so that the blended product contains 30% sugar? Find ṁ2.

ṁ1 = 1000 kg/hx1 = 45 %x2 = 5 %x3 = 30 %
ṁ2 = kg/h
2.Mass balanceHard
EvaporatorFeed F2200 kg/h · 7% NaOHProduct P50% 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 = 2200 kg/hxF = 7 %xP = 50 %
W = kg/h
3.Energy balanceMedium
hot sidecold sideethanol · 180 °C60 °Cwater · 10 °Ct = ? °C

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

ṁh = 2.5 kg/scp,h = 2.44 kJ/(kg·K)Th: 180 → 60 °Cṁc = 1.5 kg/scp,c = 4.18 kJ/(kg·K)tin = 10 °C
tout = °C
4.Energy balanceHard
hot sidecold sideglycol mix · 130 °C70 °Cwater · 25 °C45 °C · ṁ = ?5% lost

Hot glycol mix heats water in an exchanger that loses 5% 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: 130 → 70 °Ct: 25 → 45 °Ccp,c = 4.18 kJ/(kg·K)loss = 5 %
ṁc = kg/s
5.Ideal gas lawMedium
320 g of N₂V = 10 LT = 55 °CP = ?

A tank of 10 L contains 320 g of N₂ at 55 °C. Treat it as an ideal gas and find the pressure P.

m = 320 gM = 28.01 g/molV = 10 LT = 55 °CR = 8.314 J/(mol·K)
P = kPa
6.Ideal gas lawHard
Ar, M = 39.95 g/molP = 600 kPaT = 40 °Cρ = ?

Find the density ρ of Ar at 600 kPa and 40 °C, treating it as an ideal gas.

M = 39.95 g/molP = 600 kPaT = 40 °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)

8 g of Na and 16 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(H2​)n(H2​)n(H2​O)m(H2​O)÷ M(H2​)× 2/2× M(H2​O)

In the reaction 2H2 + O2 → 2H2O, 10 g of H2 reacts completely (the other reactants are in excess) and 69.72 g of H2O is collected. Find the percent yield Y.

M(H2) = 2.016 g/molM(H2O) = 18.02 g/molactual: 69.72 g
Y = %

3Solutions

1.Mass balance

Answer: ṁ2 = 600 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) = 1000·(45 − 30)/(30 − 5) = 600 kg/h

Answer

ṁ2 = 600 kg/h
2.Mass balance

Answer: W = 1892 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 = 2200·7/50 = 308 kg/h

Total balance

W = F − P = 2200 − 308 = 1892 kg/h

Answer

W = 1892 kg/h
3.Energy balance

Answer: tout = 126.7 °C

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

Heat released by the hot stream

Q = 2.5·2.44·(180 − 60) = 732 kW

Heat picked up by the water

tout = tin + Q/(ṁccp,c) = 10 + 732/(1.5·4.18) = 126.7 °C

Answer

tout = 126.7 °C
4.Energy balance

Answer: ṁc = 4.909 kg/s

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

Heat released

Qh = 2·3.6·(130 − 70) = 432 kW

Heat that reaches the water

Qc = (1 − 0.05)·432 = 410.4 kW

Water flow rate

ṁc = Qc/(cp,c·Δt) = 410.4/(4.18·20) = 4.909 kg/s

Answer

ṁc = 4.909 kg/s
5.Ideal gas law

Answer: P = 3117 kPa

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

Moles

n = m/M = 320/28.01 = 11.42 mol
T = 55 + 273.15 = 328.1 K

Pressure

P = nRT/V = 11.42·8.314·328.1/0.01 = 3117 kPa

Answer

P = 3117 kPa
6.Ideal gas law

Answer: ρ = 9.207 kg/m³

Ideal gas density: ρ = PM/(RT).

Rearrange the gas law

PV = (m/M)RT → ρ = m/V = PM/(RT)
T = 40 + 273.15 = 313.1 K

Density

ρ = 600·39.95/(8.314·313.1) = 9.207 kg/m³

Answer

ρ = 9.207 kg/m³
7.Stoichiometry

Answer: m = 20.34 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) = 8/22.99 = 0.348 mol
n(Cl2) = 16/70.9 = 0.2257 mol

Limiting reactant

0.348/2 = 0.174, 0.2257/1 = 0.2257 → Na

Mass of product

n(NaCl) = 0.174·2 = 0.348 mol
m = 0.348·58.44 = 20.34 g

Answer

m = 20.34 g
8.Stoichiometry

Answer: Y = 78 %

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(H2) = 10/2.016 = 4.96 mol
m = 4.96·2/2·18.02 = 89.38 g

Percent yield

Y = 69.72/89.38·100 = 78%

Answer

Y = 78%

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