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 1900 kg/h · 20% NaOHStream 2ṁ₂ = ? · 50% 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 = 900 kg/hx1 = 20 %x2 = 50 %x3 = 30 %
ṁ2 = kg/h
2.Mass balanceHard
EvaporatorFeed F2600 kg/h · 6% NaOHProduct PP = ? · 30%Vapour W (pure water)0% NaOH

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

F = 2600 kg/hxF = 6 %xP = 30 %
P = kg/h
3.Energy balanceMedium
hot sidecold sideethanol · 180 °C130 °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 = 1.5 kg/scp,h = 2.44 kJ/(kg·K)Th: 180 → 130 °Cṁc = 0.5 kg/scp,c = 4.18 kJ/(kg·K)tin = 30 °C
tout = °C
4.Energy balanceHard
hot sidecold sidelight oil · 170 °C100 °Cwater · 25 °C50 °C · ṁ = ?15% lost

Hot light oil 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 = 2.1 kJ/(kg·K)Th: 170 → 100 °Ct: 25 → 50 °Ccp,c = 4.18 kJ/(kg·K)loss = 15 %
ṁc = kg/s
5.Ideal gas lawMedium
430 g of CH₄V = 50 LT = 20 °CP = ?

A tank of 50 L contains 430 g of CH₄ at 20 °C. Treat it as an ideal gas and find the pressure P.

m = 430 gM = 16.04 g/molV = 50 LT = 20 °CR = 8.314 J/(mol·K)
P = kPa
6.Ideal gas lawHard
Ar, M = 39.95 g/molP = 750 kPaT = 60 °Cρ = ?

Find the density ρ of Ar at 750 kPa and 60 °C, treating it as an ideal gas.

M = 39.95 g/molP = 750 kPaT = 60 °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​)

30 g of N2 and 16 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(Al)n(Al)n(AlCl3​)m(AlCl3​)÷ M(Al)× 2/2× M(AlCl3​)

In the reaction 2Al + 3Cl2 → 2AlCl3, 24 g of Al reacts completely (the other reactants are in excess) and 100.8 g of AlCl3 is collected. Find the percent yield Y.

M(Al) = 26.98 g/molM(AlCl3) = 133.33 g/molactual: 100.8 g
Y = %

3Solutions

1.Mass balance

Answer: ṁ2 = 450 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) = 900·(20 − 30)/(30 − 50) = 450 kg/h

Answer

ṁ2 = 450 kg/h
2.Mass balance

Answer: P = 520 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 = 2600·6/30 = 520 kg/h

Answer

P = 520 kg/h
3.Energy balance

Answer: tout = 117.6 °C

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

Heat released by the hot stream

Q = 1.5·2.44·(180 − 130) = 183 kW

Heat picked up by the water

tout = tin + Q/(ṁccp,c) = 30 + 183/(0.5·4.18) = 117.6 °C

Answer

tout = 117.6 °C
4.Energy balance

Answer: ṁc = 2.989 kg/s

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

Heat released

Qh = 2.5·2.1·(170 − 100) = 367.5 kW

Heat that reaches the water

Qc = (1 − 0.15)·367.5 = 312.4 kW

Water flow rate

ṁc = Qc/(cp,c·Δt) = 312.4/(4.18·25) = 2.989 kg/s

Answer

ṁc = 2.989 kg/s
5.Ideal gas law

Answer: P = 1307 kPa

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

Moles

n = m/M = 430/16.04 = 26.81 mol
T = 20 + 273.15 = 293.1 K

Pressure

P = nRT/V = 26.81·8.314·293.1/0.05 = 1307 kPa

Answer

P = 1307 kPa
6.Ideal gas law

Answer: ρ = 10.82 kg/m³

Ideal gas density: ρ = PM/(RT).

Rearrange the gas law

PV = (m/M)RT → ρ = m/V = PM/(RT)
T = 60 + 273.15 = 333.1 K

Density

ρ = 750·39.95/(8.314·333.1) = 10.82 kg/m³

Answer

ρ = 10.82 kg/m³
7.Stoichiometry

Answer: m = 36.47 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) = 30/28.02 = 1.071 mol
n(H2) = 16/2.016 = 7.937 mol

Limiting reactant

1.071/1 = 1.071, 7.937/3 = 2.646 → N2

Mass of product

n(NH3) = 1.071·2 = 2.141 mol
m = 2.141·17.03 = 36.47 g

Answer

m = 36.47 g
8.Stoichiometry

Answer: Y = 85 %

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) = 24/26.98 = 0.8895 mol
m = 0.8895·2/2·133.33 = 118.6 g

Percent yield

Y = 100.8/118.6·100 = 85%

Answer

Y = 85%

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