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 1700 kg/h · 40% NaOHStream 2ṁ₂ = ? · 20% NaOHProduct30% NaOH

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

ṁ1 = 700 kg/hx1 = 40 %x2 = 20 %x3 = 30 %
ṁ2 = kg/h
2.Mass balanceHard
EvaporatorFeed F4800 kg/h · 17% saltProduct PP = ? · 30%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 = 4800 kg/hxF = 17 %xP = 30 %
P = kg/h
3.Energy balanceMedium
hot sidecold sideglycol mix · 140 °C60 °Cwater · 25 °Ct = ? °C

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

ṁh = 2.5 kg/scp,h = 3.6 kJ/(kg·K)Th: 140 → 60 °Cṁc = 1.5 kg/scp,c = 4.18 kJ/(kg·K)tin = 25 °C
tout = °C
4.Energy balanceHard
hot sidecold sideethanol · 150 °C80 °Cwater · 20 °C60 °C · ṁ = ?15% lost

Hot ethanol 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 = 2.5 kg/scp,h = 2.44 kJ/(kg·K)Th: 150 → 80 °Ct: 20 → 60 °Ccp,c = 4.18 kJ/(kg·K)loss = 15 %
ṁc = kg/s
5.Ideal gas lawMedium
20 g of CH₄V = 10 LT = 20 °CP = ?

A tank of 10 L contains 20 g of CH₄ at 20 °C. Treat it as an ideal gas and find the pressure P.

m = 20 gM = 16.04 g/molV = 10 LT = 20 °CR = 8.314 J/(mol·K)
P = kPa
6.Ideal gas lawHard
He, M = 4.003 g/molP = 600 kPaT = 190 °Cρ = ?

Find the density ρ of He at 600 kPa and 190 °C, treating it as an ideal gas.

M = 4.003 g/molP = 600 kPaT = 190 °CR = 8.314 J/(mol·K)
ρ = kg/m³
7.StoichiometryMedium
m(X)n(X)n(MgO)m(MgO)÷ M(X)× 2/ν× M(MgO)

24 g of Mg and 80 g of O2 react according to 2Mg + O2 → 2MgO. What mass m of MgO can form?

M(Mg) = 24.31M(O2) = 32M(MgO) = 40.3 g/mol
m = g
8.StoichiometryHard
m(Mg)n(Mg)n(MgO)m(MgO)÷ M(Mg)× 2/2× M(MgO)

In the reaction 2Mg + O2 → 2MgO, 12 g of Mg reacts completely (the other reactants are in excess) and 15.52 g of MgO is collected. Find the percent yield Y.

M(Mg) = 24.31 g/molM(MgO) = 40.3 g/molactual: 15.52 g
Y = %

3Solutions

1.Mass balance

Answer: ṁ2 = 700 kg/h

Steady-state mass balance: in = out for the total and for each component. Blending problems combine both.

Balances

Total: ṁ3 = ṁ1 + ṁ2
NaOH: ṁ1x1 + ṁ2x2 = (ṁ1 + ṁ2)x3

Solve for ṁ2

ṁ2 = ṁ1(x1 − x3)/(x3 − x2) = 700·(40 − 30)/(30 − 20) = 700 kg/h

Answer

ṁ2 = 700 kg/h
2.Mass balance

Answer: P = 2720 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 = 4800·17/30 = 2720 kg/h

Answer

P = 2720 kg/h
3.Energy balance

Answer: tout = 139.8 °C

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

Heat released by the hot stream

Q = 2.5·3.6·(140 − 60) = 720 kW

Heat picked up by the water

tout = tin + Q/(ṁccp,c) = 25 + 720/(1.5·4.18) = 139.8 °C

Answer

tout = 139.8 °C
4.Energy balance

Answer: ṁc = 2.171 kg/s

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

Heat released

Qh = 2.5·2.44·(150 − 80) = 427 kW

Heat that reaches the water

Qc = (1 − 0.15)·427 = 362.9 kW

Water flow rate

ṁc = Qc/(cp,c·Δt) = 362.9/(4.18·40) = 2.171 kg/s

Answer

ṁc = 2.171 kg/s
5.Ideal gas law

Answer: P = 303.9 kPa

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

Moles

n = m/M = 20/16.04 = 1.247 mol
T = 20 + 273.15 = 293.1 K

Pressure

P = nRT/V = 1.247·8.314·293.1/0.01 = 303.9 kPa

Answer

P = 303.9 kPa
6.Ideal gas law

Answer: ρ = 0.6237 kg/m³

Ideal gas density: ρ = PM/(RT).

Rearrange the gas law

PV = (m/M)RT → ρ = m/V = PM/(RT)
T = 190 + 273.15 = 463.1 K

Density

ρ = 600·4.003/(8.314·463.1) = 0.6237 kg/m³

Answer

ρ = 0.6237 kg/m³
7.Stoichiometry

Answer: m = 39.79 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(Mg) = 24/24.31 = 0.9872 mol
n(O2) = 80/32 = 2.5 mol

Limiting reactant

0.9872/2 = 0.4936, 2.5/1 = 2.5 → Mg

Mass of product

n(MgO) = 0.4936·2 = 0.9872 mol
m = 0.9872·40.3 = 39.79 g

Answer

m = 39.79 g
8.Stoichiometry

Answer: Y = 78.02 %

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(Mg) = 12/24.31 = 0.4936 mol
m = 0.4936·2/2·40.3 = 19.89 g

Percent yield

Y = 15.52/19.89·100 = 78.02%

Answer

Y = 78.02%

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