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 1250 kg/h · 5% saltStream 2ṁ₂ = ? · 30% saltProduct20% salt

How much of stream 2 must be added each hour to stream 1 so that the blended product contains 20% salt? Find ṁ2.

ṁ1 = 250 kg/hx1 = 5 %x2 = 30 %x3 = 20 %
ṁ2 = kg/h
2.Mass balanceHard
EvaporatorFeed F500 kg/h · 5% saltProduct PP = ? · 35%Vapour W (pure water)0% salt

An evaporator concentrates a salt solution by boiling off pure water. Find the product flow P in kg/h.

F = 500 kg/hxF = 5 %xP = 35 %
P = kg/h
3.Energy balanceMedium
hot sidecold sideethanol · 130 °C100 °Cwater · 15 °Ct = ? °C

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

ṁh = 2 kg/scp,h = 2.44 kJ/(kg·K)Th: 130 → 100 °Cṁc = 1 kg/scp,c = 4.18 kJ/(kg·K)tin = 15 °C
tout = °C
4.Energy balanceHard
hot sidecold sideglycol mix · 130 °C90 °Cwater · 15 °C50 °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 = 1.5 kg/scp,h = 3.6 kJ/(kg·K)Th: 130 → 90 °Ct: 15 → 50 °Ccp,c = 4.18 kJ/(kg·K)loss = 5 %
ṁc = kg/s
5.Ideal gas lawMedium
200 g of O₂V = 10 LT = 5 °CP = ?

A tank of 10 L contains 200 g of O₂ at 5 °C. Treat it as an ideal gas and find the pressure P.

m = 200 gM = 32 g/molV = 10 LT = 5 °CR = 8.314 J/(mol·K)
P = kPa
6.Ideal gas lawHard
State 1State 2

A fixed amount of CH₄ goes from 250 kPa, 20 L, 15 °C to 500 kPa and 120 °C. Find the new volume V2.

P1 = 250 kPaV1 = 20 LT1 = 15 °CP2 = 500 kPaT2 = 120 °C
V2 = L
7.StoichiometryMedium
m(X)n(X)n(H2​O)m(H2​O)÷ M(X)× 2/ν× M(H2​O)

20 g of CH4 and 80 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(Fe2​O3​)n(Fe2​O3​)n(Fe)m(Fe)÷ M(Fe2​O3​)× 2/1× M(Fe)

In the reaction Fe2O3 + 3CO → 2Fe + 3CO2, 8 g of Fe2O3 reacts completely (the other reactants are in excess) and 4.76 g of Fe is collected. Find the percent yield Y.

M(Fe2O3) = 159.69 g/molM(Fe) = 55.85 g/molactual: 4.76 g
Y = %

3Solutions

1.Mass balance

Answer: ṁ2 = 375 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) = 250·(5 − 20)/(20 − 30) = 375 kg/h

Answer

ṁ2 = 375 kg/h
2.Mass balance

Answer: P = 71.43 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 = 500·5/35 = 71.43 kg/h

Answer

P = 71.43 kg/h
3.Energy balance

Answer: tout = 50.02 °C

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

Heat released by the hot stream

Q = 2·2.44·(130 − 100) = 146.4 kW

Heat picked up by the water

tout = tin + Q/(ṁccp,c) = 15 + 146.4/(1·4.18) = 50.02 °C

Answer

tout = 50.02 °C
4.Energy balance

Answer: ṁc = 1.403 kg/s

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

Heat released

Qh = 1.5·3.6·(130 − 90) = 216 kW

Heat that reaches the water

Qc = (1 − 0.05)·216 = 205.2 kW

Water flow rate

ṁc = Qc/(cp,c·Δt) = 205.2/(4.18·35) = 1.403 kg/s

Answer

ṁc = 1.403 kg/s
5.Ideal gas law

Answer: P = 1445 kPa

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

Moles

n = m/M = 200/32 = 6.25 mol
T = 5 + 273.15 = 278.1 K

Pressure

P = nRT/V = 6.25·8.314·278.1/0.01 = 1445 kPa

Answer

P = 1445 kPa
6.Ideal gas law

Answer: V2 = 13.64 L

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

Absolute temperatures

T1 = 288.1 K, T2 = 393.1 K

Combined gas law

V2 = V1·(P1/P2)·(T2/T1) = 20·(250/500)·(393.1/288.1) = 13.64 L

Answer

V2 = 13.64 L
7.Stoichiometry

Answer: m = 44.94 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) = 80/32 = 2.5 mol

Limiting reactant

1.247/1 = 1.247, 2.5/2 = 1.25 → CH4

Mass of product

n(H2O) = 1.247·2 = 2.494 mol
m = 2.494·18.02 = 44.94 g

Answer

m = 44.94 g
8.Stoichiometry

Answer: Y = 85.06 %

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) = 8/159.69 = 0.0501 mol
m = 0.0501·2/1·55.85 = 5.596 g

Percent yield

Y = 4.76/5.596·100 = 85.06%

Answer

Y = 85.06%

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