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 1200 kg/h · 30% sugarStream 2ṁ₂ = ? · 10% 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 = 200 kg/hx1 = 30 %x2 = 10 %x3 = 25 %
ṁ2 = kg/h
2.Mass balanceHard
EvaporatorFeed F4700 kg/h · 6% sugarProduct P50% sugarVapour W (pure water)W = ? kg/h

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

F = 4700 kg/hxF = 6 %xP = 50 %
W = kg/h
3.Energy balanceMedium
hot sidecold sideglycol mix · 170 °C130 °Cwater · 30 °Ct = ? °C

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

ṁh = 1.5 kg/scp,h = 3.6 kJ/(kg·K)Th: 170 → 130 °Cṁc = 3 kg/scp,c = 4.18 kJ/(kg·K)tin = 30 °C
tout = °C
4.Energy balanceHard
hot sidecold sidelight oil · 140 °C100 °Cwater · 25 °C50 °C · ṁ = ?10% lost

Hot light oil 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 = 2.1 kJ/(kg·K)Th: 140 → 100 °Ct: 25 → 50 °Ccp,c = 4.18 kJ/(kg·K)loss = 10 %
ṁc = kg/s
5.Ideal gas lawMedium
100 g of N₂P = 700 kPaT = 25 °CV = ?

What volume V does 100 g of N₂ occupy at 700 kPa and 25 °C? Treat it as an ideal gas.

m = 100 gM = 28.01 g/molP = 700 kPaT = 25 °CR = 8.314 J/(mol·K)
V = L
6.Ideal gas lawHard
N₂, M = 28.01 g/molP = 850 kPaT = 140 °Cρ = ?

Find the density ρ of N₂ at 850 kPa and 140 °C, treating it as an ideal gas.

M = 28.01 g/molP = 850 kPaT = 140 °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)

20 g of Mg and 24 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(N2​)n(N2​)n(NH3​)m(NH3​)÷ M(N2​)× 2/1× M(NH3​)

In the reaction N2 + 3H2 → 2NH3, 8 g of N2 reacts completely (the other reactants are in excess) and 8.85 g of NH3 is collected. Find the percent yield Y.

M(N2) = 28.02 g/molM(NH3) = 17.03 g/molactual: 8.85 g
Y = %

3Solutions

1.Mass balance

Answer: ṁ2 = 66.67 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) = 200·(30 − 25)/(25 − 10) = 66.67 kg/h

Answer

ṁ2 = 66.67 kg/h
2.Mass balance

Answer: W = 4136 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 = 4700·6/50 = 564 kg/h

Total balance

W = F − P = 4700 − 564 = 4136 kg/h

Answer

W = 4136 kg/h
3.Energy balance

Answer: tout = 47.22 °C

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

Heat released by the hot stream

Q = 1.5·3.6·(170 − 130) = 216 kW

Heat picked up by the water

tout = tin + Q/(ṁccp,c) = 30 + 216/(3·4.18) = 47.22 °C

Answer

tout = 47.22 °C
4.Energy balance

Answer: ṁc = 0.7234 kg/s

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

Heat released

Qh = 1·2.1·(140 − 100) = 84 kW

Heat that reaches the water

Qc = (1 − 0.1)·84 = 75.6 kW

Water flow rate

ṁc = Qc/(cp,c·Δt) = 75.6/(4.18·25) = 0.7234 kg/s

Answer

ṁc = 0.7234 kg/s
5.Ideal gas law

Answer: V = 12.64 L

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

Moles

n = 100/28.01 = 3.57 mol
T = 25 + 273.15 = 298.1 K

Volume

V = nRT/P = 3.57·8.314·298.1/700 = 12.64 L

Answer

V = 12.64 L
6.Ideal gas law

Answer: ρ = 6.931 kg/m³

Ideal gas density: ρ = PM/(RT).

Rearrange the gas law

PV = (m/M)RT → ρ = m/V = PM/(RT)
T = 140 + 273.15 = 413.1 K

Density

ρ = 850·28.01/(8.314·413.1) = 6.931 kg/m³

Answer

ρ = 6.931 kg/m³
7.Stoichiometry

Answer: m = 33.16 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) = 20/24.31 = 0.8227 mol
n(O2) = 24/32 = 0.75 mol

Limiting reactant

0.8227/2 = 0.4114, 0.75/1 = 0.75 → Mg

Mass of product

n(MgO) = 0.4114·2 = 0.8227 mol
m = 0.8227·40.3 = 33.16 g

Answer

m = 33.16 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(N2) = 8/28.02 = 0.2855 mol
m = 0.2855·2/1·17.03 = 9.724 g

Percent yield

Y = 8.85/9.724·100 = 91.01%

Answer

Y = 91.01%

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