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 11000 kg/h · 5% sugarStream 2ṁ₂ = ? · 45% sugarProduct10% sugar

How much of stream 2 must be added each hour to stream 1 so that the blended product contains 10% sugar? Find ṁ2.

ṁ1 = 1000 kg/hx1 = 5 %x2 = 45 %x3 = 10 %
ṁ2 = kg/h
2.Mass balanceHard
EvaporatorFeed F2700 kg/h · 8% NaOHProduct P40% NaOHVapour W (pure water)W = ? kg/h

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

F = 2700 kg/hxF = 8 %xP = 40 %
W = kg/h
3.Energy balanceMedium
hot sidecold sideethanol · 140 °C60 °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 kg/scp,h = 2.44 kJ/(kg·K)Th: 140 → 60 °Cṁc = 1 kg/scp,c = 4.18 kJ/(kg·K)tin = 30 °C
tout = °C
4.Energy balanceHard
hot sidecold sideethanol · 140 °C70 °Cwater · 20 °C65 °C · ṁ = ?10% lost

Hot ethanol 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 = 2 kg/scp,h = 2.44 kJ/(kg·K)Th: 140 → 70 °Ct: 20 → 65 °Ccp,c = 4.18 kJ/(kg·K)loss = 10 %
ṁc = kg/s
5.Ideal gas lawMedium
70 g of CO₂V = 30 LT = 85 °CP = ?

A tank of 30 L contains 70 g of CO₂ at 85 °C. Treat it as an ideal gas and find the pressure P.

m = 70 gM = 44.01 g/molV = 30 LT = 85 °CR = 8.314 J/(mol·K)
P = kPa
6.Ideal gas lawHard
CO₂, M = 44.01 g/molP = 200 kPaT = 50 °Cρ = ?

Find the density ρ of CO₂ at 200 kPa and 50 °C, treating it as an ideal gas.

M = 44.01 g/molP = 200 kPaT = 50 °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​)

8 g of N2 and 32 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(Mg)n(Mg)n(MgO)m(MgO)÷ M(Mg)× 2/2× M(MgO)

In the reaction 2Mg + O2 → 2MgO, 30 g of Mg reacts completely (the other reactants are in excess) and 30.83 g of MgO is collected. Find the percent yield Y.

M(Mg) = 24.31 g/molM(MgO) = 40.3 g/molactual: 30.83 g
Y = %

3Solutions

1.Mass balance

Answer: ṁ2 = 142.9 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) = 1000·(5 − 10)/(10 − 45) = 142.9 kg/h

Answer

ṁ2 = 142.9 kg/h
2.Mass balance

Answer: W = 2160 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 = 2700·8/40 = 540 kg/h

Total balance

W = F − P = 2700 − 540 = 2160 kg/h

Answer

W = 2160 kg/h
3.Energy balance

Answer: tout = 76.7 °C

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

Heat released by the hot stream

Q = 1·2.44·(140 − 60) = 195.2 kW

Heat picked up by the water

tout = tin + Q/(ṁccp,c) = 30 + 195.2/(1·4.18) = 76.7 °C

Answer

tout = 76.7 °C
4.Energy balance

Answer: ṁc = 1.634 kg/s

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

Heat released

Qh = 2·2.44·(140 − 70) = 341.6 kW

Heat that reaches the water

Qc = (1 − 0.1)·341.6 = 307.4 kW

Water flow rate

ṁc = Qc/(cp,c·Δt) = 307.4/(4.18·45) = 1.634 kg/s

Answer

ṁc = 1.634 kg/s
5.Ideal gas law

Answer: P = 157.9 kPa

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

Moles

n = m/M = 70/44.01 = 1.591 mol
T = 85 + 273.15 = 358.1 K

Pressure

P = nRT/V = 1.591·8.314·358.1/0.03 = 157.9 kPa

Answer

P = 157.9 kPa
6.Ideal gas law

Answer: ρ = 3.276 kg/m³

Ideal gas density: ρ = PM/(RT).

Rearrange the gas law

PV = (m/M)RT → ρ = m/V = PM/(RT)
T = 50 + 273.15 = 323.1 K

Density

ρ = 200·44.01/(8.314·323.1) = 3.276 kg/m³

Answer

ρ = 3.276 kg/m³
7.Stoichiometry

Answer: m = 9.724 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) = 8/28.02 = 0.2855 mol
n(H2) = 32/2.016 = 15.87 mol

Limiting reactant

0.2855/1 = 0.2855, 15.87/3 = 5.291 → N2

Mass of product

n(NH3) = 0.2855·2 = 0.571 mol
m = 0.571·17.03 = 9.724 g

Answer

m = 9.724 g
8.Stoichiometry

Answer: Y = 61.99 %

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) = 30/24.31 = 1.234 mol
m = 1.234·2/2·40.3 = 49.73 g

Percent yield

Y = 30.83/49.73·100 = 61.99%

Answer

Y = 61.99%

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