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 1500 kg/h · 50% sugarStream 2ṁ₂ = ? · 5% sugarProduct45% sugar

How much of stream 2 must be added each hour to stream 1 so that the blended product contains 45% sugar? Find ṁ2.

ṁ1 = 500 kg/hx1 = 50 %x2 = 5 %x3 = 45 %
ṁ2 = kg/h
2.Mass balanceHard
EvaporatorFeed F600 kg/h · 20% NaOHProduct PP = ? · 60%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 = 600 kg/hxF = 20 %xP = 60 %
P = kg/h
3.Energy balanceMedium
hot sidecold sidelight oil · 130 °C100 °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 = 1 kg/scp,h = 2.1 kJ/(kg·K)Th: 130 → 100 °Cṁc = 3 kg/scp,c = 4.18 kJ/(kg·K)tin = 25 °C
tout = °C
4.Energy balanceHard
hot sidecold sideglycol mix · 160 °C130 °Cwater · 10 °C45 °C · ṁ = ?15% lost

Hot glycol mix 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 = 3.6 kJ/(kg·K)Th: 160 → 130 °Ct: 10 → 45 °Ccp,c = 4.18 kJ/(kg·K)loss = 15 %
ṁc = kg/s
5.Ideal gas lawMedium
420 g of CH₄P = 750 kPaT = 40 °CV = ?

What volume V does 420 g of CH₄ occupy at 750 kPa and 40 °C? Treat it as an ideal gas.

m = 420 gM = 16.04 g/molP = 750 kPaT = 40 °CR = 8.314 J/(mol·K)
V = L
6.Ideal gas lawHard
State 1State 2

A fixed amount of CO₂ goes from 150 kPa, 2 L, 35 °C to 400 kPa and 90 °C. Find the new volume V2.

P1 = 150 kPaV1 = 2 LT1 = 35 °CP2 = 400 kPaT2 = 90 °C
V2 = L
7.StoichiometryMedium
m(X)n(X)n(H2​O)m(H2​O)÷ M(X)× 2/ν× M(H2​O)

12 g of H2 and 32 g of O2 react according to 2H2 + O2 → 2H2O. What mass m of H2O can form?

M(H2) = 2.016M(O2) = 32M(H2O) = 18.02 g/mol
m = g
8.StoichiometryHard
m(Fe2​O3​)n(Fe2​O3​)n(CO2​)m(CO2​)÷ M(Fe2​O3​)× 3/1× M(CO2​)

In the reaction Fe2O3 + 3CO → 2Fe + 3CO2, 6 g of Fe2O3 reacts completely (the other reactants are in excess) and 4.51 g of CO2 is collected. Find the percent yield Y.

M(Fe2O3) = 159.69 g/molM(CO2) = 44.01 g/molactual: 4.51 g
Y = %

3Solutions

1.Mass balance

Answer: ṁ2 = 62.5 kg/h

Steady-state mass balance: in = out for the total and for each component. Blending problems combine both.

Balances

Total: ṁ3 = ṁ1 + ṁ2
sugar: ṁ1x1 + ṁ2x2 = (ṁ1 + ṁ2)x3

Solve for ṁ2

ṁ2 = ṁ1(x1 − x3)/(x3 − x2) = 500·(50 − 45)/(45 − 5) = 62.5 kg/h

Answer

ṁ2 = 62.5 kg/h
2.Mass balance

Answer: P = 200 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 = 600·20/60 = 200 kg/h

Answer

P = 200 kg/h
3.Energy balance

Answer: tout = 30.02 °C

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

Heat released by the hot stream

Q = 1·2.1·(130 − 100) = 63 kW

Heat picked up by the water

tout = tin + Q/(ṁccp,c) = 25 + 63/(3·4.18) = 30.02 °C

Answer

tout = 30.02 °C
4.Energy balance

Answer: ṁc = 1.255 kg/s

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

Heat released

Qh = 2·3.6·(160 − 130) = 216 kW

Heat that reaches the water

Qc = (1 − 0.15)·216 = 183.6 kW

Water flow rate

ṁc = Qc/(cp,c·Δt) = 183.6/(4.18·35) = 1.255 kg/s

Answer

ṁc = 1.255 kg/s
5.Ideal gas law

Answer: V = 90.9 L

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

Moles

n = 420/16.04 = 26.18 mol
T = 40 + 273.15 = 313.1 K

Volume

V = nRT/P = 26.18·8.314·313.1/750 = 90.9 L

Answer

V = 90.9 L
6.Ideal gas law

Answer: V2 = 0.8839 L

Combined gas law: for a fixed amount of ideal gas, P1V1/T1 = P2V2/T2.

Absolute temperatures

T1 = 308.1 K, T2 = 363.1 K

Combined gas law

V2 = V1·(P1/P2)·(T2/T1) = 2·(150/400)·(363.1/308.1) = 0.8839 L

Answer

V2 = 0.8839 L
7.Stoichiometry

Answer: m = 36.04 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(H2) = 12/2.016 = 5.952 mol
n(O2) = 32/32 = 1 mol

Limiting reactant

5.952/2 = 2.976, 1/1 = 1 → O2

Mass of product

n(H2O) = 1·2 = 2 mol
m = 2·18.02 = 36.04 g

Answer

m = 36.04 g
8.Stoichiometry

Answer: Y = 90.91 %

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(Fe2O3) = 6/159.69 = 0.03757 mol
m = 0.03757·3/1·44.01 = 4.961 g

Percent yield

Y = 4.51/4.961·100 = 90.91%

Answer

Y = 90.91%

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