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 · 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 = 900 kg/hx1 = 45 %x2 = 5 %x3 = 30 %
ṁ2 = kg/h
2.Mass balanceHard
EvaporatorFeed F1800 kg/h · 20% NaOHProduct P35% 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 = 1800 kg/hxF = 20 %xP = 35 %
W = kg/h
3.Energy balanceMedium
hot sidecold sideglycol mix · 150 °C100 °Cwater · 20 °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 kg/scp,h = 3.6 kJ/(kg·K)Th: 150 → 100 °Cṁc = 0.5 kg/scp,c = 4.18 kJ/(kg·K)tin = 20 °C
tout = °C
4.Energy balanceHard
hot sidecold sidelight oil · 130 °C80 °Cwater · 10 °C25 °C · ṁ = ?5% lost

Hot light oil heats water in an exchanger that loses 5% of the heat released by the hot stream to the surroundings. Find the water flow rate ṁc needed.

ṁh = 0.5 kg/scp,h = 2.1 kJ/(kg·K)Th: 130 → 80 °Ct: 10 → 25 °Ccp,c = 4.18 kJ/(kg·K)loss = 5 %
ṁc = kg/s
5.Ideal gas lawMedium
340 g of ArP = 100 kPaT = 90 °CV = ?

What volume V does 340 g of Ar occupy at 100 kPa and 90 °C? Treat it as an ideal gas.

m = 340 gM = 39.95 g/molP = 100 kPaT = 90 °CR = 8.314 J/(mol·K)
V = L
6.Ideal gas lawHard
N₂, M = 28.01 g/molP = 700 kPaT = 150 °Cρ = ?

Find the density ρ of N₂ at 700 kPa and 150 °C, treating it as an ideal gas.

M = 28.01 g/molP = 700 kPaT = 150 °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​)

4 g of N2 and 64 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(H2​)n(H2​)n(H2​O)m(H2​O)÷ M(H2​)× 2/2× M(H2​O)

In the reaction 2H2 + O2 → 2H2O, 8 g of H2 reacts completely (the other reactants are in excess) and 62.93 g of H2O is collected. Find the percent yield Y.

M(H2) = 2.016 g/molM(H2O) = 18.02 g/molactual: 62.93 g
Y = %

3Solutions

1.Mass balance

Answer: ṁ2 = 540 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) = 900·(45 − 30)/(30 − 5) = 540 kg/h

Answer

ṁ2 = 540 kg/h
2.Mass balance

Answer: W = 771.4 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 = 1800·20/35 = 1029 kg/h

Total balance

W = F − P = 1800 − 1029 = 771.4 kg/h

Answer

W = 771.4 kg/h
3.Energy balance

Answer: tout = 106.1 °C

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

Heat released by the hot stream

Q = 1·3.6·(150 − 100) = 180 kW

Heat picked up by the water

tout = tin + Q/(ṁccp,c) = 20 + 180/(0.5·4.18) = 106.1 °C

Answer

tout = 106.1 °C
4.Energy balance

Answer: ṁc = 0.7955 kg/s

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

Heat released

Qh = 0.5·2.1·(130 − 80) = 52.5 kW

Heat that reaches the water

Qc = (1 − 0.05)·52.5 = 49.88 kW

Water flow rate

ṁc = Qc/(cp,c·Δt) = 49.88/(4.18·15) = 0.7955 kg/s

Answer

ṁc = 0.7955 kg/s
5.Ideal gas law

Answer: V = 257 L

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

Moles

n = 340/39.95 = 8.511 mol
T = 90 + 273.15 = 363.1 K

Volume

V = nRT/P = 8.511·8.314·363.1/100 = 257 L

Answer

V = 257 L
6.Ideal gas law

Answer: ρ = 5.573 kg/m³

Ideal gas density: ρ = PM/(RT).

Rearrange the gas law

PV = (m/M)RT → ρ = m/V = PM/(RT)
T = 150 + 273.15 = 423.1 K

Density

ρ = 700·28.01/(8.314·423.1) = 5.573 kg/m³

Answer

ρ = 5.573 kg/m³
7.Stoichiometry

Answer: m = 4.862 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) = 4/28.02 = 0.1428 mol
n(H2) = 64/2.016 = 31.75 mol

Limiting reactant

0.1428/1 = 0.1428, 31.75/3 = 10.58 → N2

Mass of product

n(NH3) = 0.1428·2 = 0.2855 mol
m = 0.2855·17.03 = 4.862 g

Answer

m = 4.862 g
8.Stoichiometry

Answer: Y = 88 %

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(H2) = 8/2.016 = 3.968 mol
m = 3.968·2/2·18.02 = 71.51 g

Percent yield

Y = 62.93/71.51·100 = 88%

Answer

Y = 88%

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