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 · 40% 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 = 500 kg/hx1 = 40 %x2 = 20 %x3 = 25 %
ṁ2 = kg/h
2.Mass balanceHard
EvaporatorFeed F500 kg/h · 5% saltProduct PP = ? · 55%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 = 55 %
P = kg/h
3.Energy balanceMedium
hot sidecold sideethanol · 160 °C120 °Cwater · 25 °Ct = ? °C

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

ṁh = 3 kg/scp,h = 2.44 kJ/(kg·K)Th: 160 → 120 °Cṁc = 1 kg/scp,c = 4.18 kJ/(kg·K)tin = 25 °C
tout = °C
4.Energy balanceHard
hot sidecold sideglycol mix · 130 °C60 °Cwater · 30 °C60 °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 kg/scp,h = 3.6 kJ/(kg·K)Th: 130 → 60 °Ct: 30 → 60 °Ccp,c = 4.18 kJ/(kg·K)loss = 5 %
ṁc = kg/s
5.Ideal gas lawMedium
440 g of O₂V = 30 LT = 15 °CP = ?

A tank of 30 L contains 440 g of O₂ at 15 °C. Treat it as an ideal gas and find the pressure P.

m = 440 gM = 32 g/molV = 30 LT = 15 °CR = 8.314 J/(mol·K)
P = kPa
6.Ideal gas lawHard
State 1State 2

A fixed amount of He goes from 200 kPa, 5 L, 15 °C to 700 kPa and 250 °C. Find the new volume V2.

P1 = 200 kPaV1 = 5 LT1 = 15 °CP2 = 700 kPaT2 = 250 °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 32 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(Al)n(Al)n(AlCl3​)m(AlCl3​)÷ M(Al)× 2/2× M(AlCl3​)

In the reaction 2Al + 3Cl2 → 2AlCl3, 30 g of Al reacts completely (the other reactants are in excess) and 81.54 g of AlCl3 is collected. Find the percent yield Y.

M(Al) = 26.98 g/molM(AlCl3) = 133.33 g/molactual: 81.54 g
Y = %

3Solutions

1.Mass balance

Answer: ṁ2 = 1500 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·(40 − 25)/(25 − 20) = 1500 kg/h

Answer

ṁ2 = 1500 kg/h
2.Mass balance

Answer: P = 45.45 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 = 500·5/55 = 45.45 kg/h

Answer

P = 45.45 kg/h
3.Energy balance

Answer: tout = 95.05 °C

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

Heat released by the hot stream

Q = 3·2.44·(160 − 120) = 292.8 kW

Heat picked up by the water

tout = tin + Q/(ṁccp,c) = 25 + 292.8/(1·4.18) = 95.05 °C

Answer

tout = 95.05 °C
4.Energy balance

Answer: ṁc = 1.909 kg/s

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

Heat released

Qh = 1·3.6·(130 − 60) = 252 kW

Heat that reaches the water

Qc = (1 − 0.05)·252 = 239.4 kW

Water flow rate

ṁc = Qc/(cp,c·Δt) = 239.4/(4.18·30) = 1.909 kg/s

Answer

ṁc = 1.909 kg/s
5.Ideal gas law

Answer: P = 1098 kPa

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

Moles

n = m/M = 440/32 = 13.75 mol
T = 15 + 273.15 = 288.1 K

Pressure

P = nRT/V = 13.75·8.314·288.1/0.03 = 1098 kPa

Answer

P = 1098 kPa
6.Ideal gas law

Answer: V2 = 2.594 L

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

Absolute temperatures

T1 = 288.1 K, T2 = 523.1 K

Combined gas law

V2 = V1·(P1/P2)·(T2/T1) = 5·(200/700)·(523.1/288.1) = 2.594 L

Answer

V2 = 2.594 L
7.Stoichiometry

Answer: m = 18.02 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) = 32/32 = 1 mol

Limiting reactant

1.247/1 = 1.247, 1/2 = 0.5 → O2

Mass of product

n(H2O) = 0.5·2 = 1 mol
m = 1·18.02 = 18.02 g

Answer

m = 18.02 g
8.Stoichiometry

Answer: Y = 55 %

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(Al) = 30/26.98 = 1.112 mol
m = 1.112·2/2·133.33 = 148.3 g

Percent yield

Y = 81.54/148.3·100 = 55%

Answer

Y = 55%

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