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 1550 kg/h · 15% sugarStream 2ṁ₂ = ? · 40% 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 = 550 kg/hx1 = 15 %x2 = 40 %x3 = 25 %
ṁ2 = kg/h
2.Mass balanceHard
EvaporatorFeed F3300 kg/h · 18% sugarProduct PP = ? · 45%Vapour W (pure water)0% sugar

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

F = 3300 kg/hxF = 18 %xP = 45 %
P = kg/h
3.Energy balanceMedium
hot sidecold sideglycol mix · 180 °C60 °Cwater · 20 °Ct = ? °C

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

ṁh = 1 kg/scp,h = 3.6 kJ/(kg·K)Th: 180 → 60 °Cṁc = 1.5 kg/scp,c = 4.18 kJ/(kg·K)tin = 20 °C
tout = °C
4.Energy balanceHard
hot sidecold sideglycol mix · 160 °C60 °Cwater · 20 °C55 °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 = 3 kg/scp,h = 3.6 kJ/(kg·K)Th: 160 → 60 °Ct: 20 → 55 °Ccp,c = 4.18 kJ/(kg·K)loss = 5 %
ṁc = kg/s
5.Ideal gas lawMedium
480 g of O₂P = 350 kPaT = 100 °CV = ?

What volume V does 480 g of O₂ occupy at 350 kPa and 100 °C? Treat it as an ideal gas.

m = 480 gM = 32 g/molP = 350 kPaT = 100 °CR = 8.314 J/(mol·K)
V = L
6.Ideal gas lawHard
State 1State 2

A fixed amount of CH₄ goes from 250 kPa, 2 L, 35 °C to 150 kPa and 100 °C. Find the new volume V2.

P1 = 250 kPaV1 = 2 LT1 = 35 °CP2 = 150 kPaT2 = 100 °C
V2 = L
7.StoichiometryMedium
m(X)n(X)n(CO2​)m(CO2​)÷ M(X)× 3/ν× M(CO2​)

16 g of Fe2O3 and 48 g of CO react according to Fe2O3 + 3CO → 2Fe + 3CO2. What mass m of CO2 can form?

M(Fe2O3) = 159.69M(CO) = 28.01M(CO2) = 44.01 g/mol
m = g
8.StoichiometryHard
m(N2​)n(N2​)n(NH3​)m(NH3​)÷ M(N2​)× 2/1× M(NH3​)

In the reaction N2 + 3H2 → 2NH3, 30 g of N2 reacts completely (the other reactants are in excess) and 32.09 g of NH3 is collected. Find the percent yield Y.

M(N2) = 28.02 g/molM(NH3) = 17.03 g/molactual: 32.09 g
Y = %

3Solutions

1.Mass balance

Answer: ṁ2 = 366.7 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) = 550·(15 − 25)/(25 − 40) = 366.7 kg/h

Answer

ṁ2 = 366.7 kg/h
2.Mass balance

Answer: P = 1320 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 = 3300·18/45 = 1320 kg/h

Answer

P = 1320 kg/h
3.Energy balance

Answer: tout = 88.9 °C

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

Heat released by the hot stream

Q = 1·3.6·(180 − 60) = 432 kW

Heat picked up by the water

tout = tin + Q/(ṁccp,c) = 20 + 432/(1.5·4.18) = 88.9 °C

Answer

tout = 88.9 °C
4.Energy balance

Answer: ṁc = 7.013 kg/s

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

Heat released

Qh = 3·3.6·(160 − 60) = 1080 kW

Heat that reaches the water

Qc = (1 − 0.05)·1080 = 1026 kW

Water flow rate

ṁc = Qc/(cp,c·Δt) = 1026/(4.18·35) = 7.013 kg/s

Answer

ṁc = 7.013 kg/s
5.Ideal gas law

Answer: V = 133 L

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

Moles

n = 480/32 = 15 mol
T = 100 + 273.15 = 373.1 K

Volume

V = nRT/P = 15·8.314·373.1/350 = 133 L

Answer

V = 133 L
6.Ideal gas law

Answer: V2 = 4.036 L

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

Absolute temperatures

T1 = 308.1 K, T2 = 373.1 K

Combined gas law

V2 = V1·(P1/P2)·(T2/T1) = 2·(250/150)·(373.1/308.1) = 4.036 L

Answer

V2 = 4.036 L
7.Stoichiometry

Answer: m = 13.23 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) = 16/159.69 = 0.1002 mol
n(CO) = 48/28.01 = 1.714 mol

Limiting reactant

0.1002/1 = 0.1002, 1.714/3 = 0.5712 → Fe2O3

Mass of product

n(CO2) = 0.1002·3 = 0.3006 mol
m = 0.3006·44.01 = 13.23 g

Answer

m = 13.23 g
8.Stoichiometry

Answer: Y = 88 %

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(N2) = 30/28.02 = 1.071 mol
m = 1.071·2/1·17.03 = 36.47 g

Percent yield

Y = 32.09/36.47·100 = 88%

Answer

Y = 88%

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