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 · 45% saltStream 2ṁ₂ = ? · 5% saltProduct30% salt

How much of stream 2 must be added each hour to stream 1 so that the blended product contains 30% salt? Find ṁ2.

ṁ1 = 500 kg/hx1 = 45 %x2 = 5 %x3 = 30 %
ṁ2 = kg/h
2.Mass balanceHard
EvaporatorFeed F1500 kg/h · 12% saltProduct P40% saltVapour W (pure water)W = ? kg/h

An evaporator concentrates a salt solution by boiling off pure water. Find the water evaporated W in kg/h.

F = 1500 kg/hxF = 12 %xP = 40 %
W = 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 = 2.5 kg/scp,h = 2.1 kJ/(kg·K)Th: 130 → 100 °Cṁc = 0.5 kg/scp,c = 4.18 kJ/(kg·K)tin = 25 °C
tout = °C
4.Energy balanceHard
hot sidecold sideglycol mix · 170 °C110 °Cwater · 30 °C45 °C · ṁ = ?10% lost

Hot glycol mix heats water in an exchanger that loses 10% 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: 170 → 110 °Ct: 30 → 45 °Ccp,c = 4.18 kJ/(kg·K)loss = 10 %
ṁc = kg/s
5.Ideal gas lawMedium
210 g of CO₂V = 30 LT = 55 °CP = ?

A tank of 30 L contains 210 g of CO₂ at 55 °C. Treat it as an ideal gas and find the pressure P.

m = 210 gM = 44.01 g/molV = 30 LT = 55 °CR = 8.314 J/(mol·K)
P = kPa
6.Ideal gas lawHard
N₂, M = 28.01 g/molP = 1000 kPaT = 0 °Cρ = ?

Find the density ρ of N₂ at 1000 kPa and 0 °C, treating it as an ideal gas.

M = 28.01 g/molP = 1000 kPaT = 0 °CR = 8.314 J/(mol·K)
ρ = kg/m³
7.StoichiometryMedium
m(X)n(X)n(NH3​)m(NH3​)÷ M(X)× 2/ν× M(NH3​)

20 g of N2 and 10 g of H2 react according to N2 + 3H2 → 2NH3. What mass m of NH3 can form?

M(N2) = 28.02M(H2) = 2.016M(NH3) = 17.03 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, 10 g of CH4 reacts completely (the other reactants are in excess) and 24.97 g of CO2 is collected. Find the percent yield Y.

M(CH4) = 16.04 g/molM(CO2) = 44.01 g/molactual: 24.97 g
Y = %

3Solutions

1.Mass balance

Answer: ṁ2 = 300 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) = 500·(45 − 30)/(30 − 5) = 300 kg/h

Answer

ṁ2 = 300 kg/h
2.Mass balance

Answer: W = 1050 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 = 1500·12/40 = 450 kg/h

Total balance

W = F − P = 1500 − 450 = 1050 kg/h

Answer

W = 1050 kg/h
3.Energy balance

Answer: tout = 100.4 °C

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

Heat released by the hot stream

Q = 2.5·2.1·(130 − 100) = 157.5 kW

Heat picked up by the water

tout = tin + Q/(ṁccp,c) = 25 + 157.5/(0.5·4.18) = 100.4 °C

Answer

tout = 100.4 °C
4.Energy balance

Answer: ṁc = 3.1 kg/s

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

Heat released

Qh = 1·3.6·(170 − 110) = 216 kW

Heat that reaches the water

Qc = (1 − 0.1)·216 = 194.4 kW

Water flow rate

ṁc = Qc/(cp,c·Δt) = 194.4/(4.18·15) = 3.1 kg/s

Answer

ṁc = 3.1 kg/s
5.Ideal gas law

Answer: P = 433.9 kPa

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

Moles

n = m/M = 210/44.01 = 4.772 mol
T = 55 + 273.15 = 328.1 K

Pressure

P = nRT/V = 4.772·8.314·328.1/0.03 = 433.9 kPa

Answer

P = 433.9 kPa
6.Ideal gas law

Answer: ρ = 12.33 kg/m³

Ideal gas density: ρ = PM/(RT).

Rearrange the gas law

PV = (m/M)RT → ρ = m/V = PM/(RT)
T = 0 + 273.15 = 273.1 K

Density

ρ = 1000·28.01/(8.314·273.1) = 12.33 kg/m³

Answer

ρ = 12.33 kg/m³
7.Stoichiometry

Answer: m = 24.31 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(N2) = 20/28.02 = 0.7138 mol
n(H2) = 10/2.016 = 4.96 mol

Limiting reactant

0.7138/1 = 0.7138, 4.96/3 = 1.653 → N2

Mass of product

n(NH3) = 0.7138·2 = 1.428 mol
m = 1.428·17.03 = 24.31 g

Answer

m = 24.31 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) = 10/16.04 = 0.6234 mol
m = 0.6234·1/1·44.01 = 27.44 g

Percent yield

Y = 24.97/27.44·100 = 91.01%

Answer

Y = 91.01%

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