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 1800 kg/h · 35% sugarStream 2ṁ₂ = ? · 20% sugarProduct25% sugar

How much of stream 2 must be added each hour to stream 1 so that the blended product contains 25% sugar? Find ṁ2.

ṁ1 = 800 kg/hx1 = 35 %x2 = 20 %x3 = 25 %
ṁ2 = kg/h
2.Mass balanceHard
EvaporatorFeed F2200 kg/h · 20% sugarProduct P40% 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 = 2200 kg/hxF = 20 %xP = 40 %
W = kg/h
3.Energy balanceMedium
hot sidecold sideethanol · 150 °C60 °Cwater · 30 °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: 150 → 60 °Cṁc = 1 kg/scp,c = 4.18 kJ/(kg·K)tin = 30 °C
tout = °C
4.Energy balanceHard
hot sidecold sideglycol mix · 140 °C70 °Cwater · 30 °C90 °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 = 0.5 kg/scp,h = 3.6 kJ/(kg·K)Th: 140 → 70 °Ct: 30 → 90 °Ccp,c = 4.18 kJ/(kg·K)loss = 15 %
ṁc = kg/s
5.Ideal gas lawMedium
150 g of O₂V = 30 LT = 95 °CP = ?

A tank of 30 L contains 150 g of O₂ at 95 °C. Treat it as an ideal gas and find the pressure P.

m = 150 gM = 32 g/molV = 30 LT = 95 °CR = 8.314 J/(mol·K)
P = kPa
6.Ideal gas lawHard
State 1State 2

A fixed amount of Ar goes from 250 kPa, 5 L, 20 °C to 600 kPa and 170 °C. Find the new volume V2.

P1 = 250 kPaV1 = 5 LT1 = 20 °CP2 = 600 kPaT2 = 170 °C
V2 = L
7.StoichiometryMedium
m(X)n(X)n(Fe)m(Fe)÷ M(X)× 2/ν× M(Fe)

12 g of Fe2O3 and 10 g of CO react according to Fe2O3 + 3CO → 2Fe + 3CO2. What mass m of Fe can form?

M(Fe2O3) = 159.69M(CO) = 28.01M(Fe) = 55.85 g/mol
m = g
8.StoichiometryHard
m(CH4​)n(CH4​)n(CO2​)m(CO2​)÷ M(CH4​)× 1/1× M(CO2​)

In the reaction CH4 + 2O2 → CO2 + 2H2O, 20 g of CH4 reacts completely (the other reactants are in excess) and 49.94 g of CO2 is collected. Find the percent yield Y.

M(CH4) = 16.04 g/molM(CO2) = 44.01 g/molactual: 49.94 g
Y = %

3Solutions

1.Mass balance

Answer: ṁ2 = 1600 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) = 800·(35 − 25)/(25 − 20) = 1600 kg/h

Answer

ṁ2 = 1600 kg/h
2.Mass balance

Answer: W = 1100 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 = 2200·20/40 = 1100 kg/h

Total balance

W = F − P = 2200 − 1100 = 1100 kg/h

Answer

W = 1100 kg/h
3.Energy balance

Answer: tout = 135.1 °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·(150 − 60) = 439.2 kW

Heat picked up by the water

tout = tin + Q/(ṁccp,c) = 30 + 439.2/(1·4.18) = 135.1 °C

Answer

tout = 135.1 °C
4.Energy balance

Answer: ṁc = 0.427 kg/s

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

Heat released

Qh = 0.5·3.6·(140 − 70) = 126 kW

Heat that reaches the water

Qc = (1 − 0.15)·126 = 107.1 kW

Water flow rate

ṁc = Qc/(cp,c·Δt) = 107.1/(4.18·60) = 0.427 kg/s

Answer

ṁc = 0.427 kg/s
5.Ideal gas law

Answer: P = 478.2 kPa

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

Moles

n = m/M = 150/32 = 4.688 mol
T = 95 + 273.15 = 368.1 K

Pressure

P = nRT/V = 4.688·8.314·368.1/0.03 = 478.2 kPa

Answer

P = 478.2 kPa
6.Ideal gas law

Answer: V2 = 3.149 L

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

Absolute temperatures

T1 = 293.1 K, T2 = 443.1 K

Combined gas law

V2 = V1·(P1/P2)·(T2/T1) = 5·(250/600)·(443.1/293.1) = 3.149 L

Answer

V2 = 3.149 L
7.Stoichiometry

Answer: m = 8.394 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(Fe2O3) = 12/159.69 = 0.07515 mol
n(CO) = 10/28.01 = 0.357 mol

Limiting reactant

0.07515/1 = 0.07515, 0.357/3 = 0.119 → Fe2O3

Mass of product

n(Fe) = 0.07515·2 = 0.1503 mol
m = 0.1503·55.85 = 8.394 g

Answer

m = 8.394 g
8.Stoichiometry

Answer: Y = 91.01 %

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(CH4) = 20/16.04 = 1.247 mol
m = 1.247·1/1·44.01 = 54.88 g

Percent yield

Y = 49.94/54.88·100 = 91.01%

Answer

Y = 91.01%

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