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 1600 kg/h · 15% sugarStream 2ṁ₂ = ? · 40% sugarProduct30% sugar

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

ṁ1 = 600 kg/hx1 = 15 %x2 = 40 %x3 = 30 %
ṁ2 = kg/h
2.Mass balanceHard
EvaporatorFeed F900 kg/h · 9% 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 = 900 kg/hxF = 9 %xP = 50 %
W = kg/h
3.Energy balanceMedium
hot sidecold sidelight oil · 160 °C100 °Cwater · 30 °Ct = ? °C

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

ṁh = 3 kg/scp,h = 2.1 kJ/(kg·K)Th: 160 → 100 °Cṁc = 1 kg/scp,c = 4.18 kJ/(kg·K)tin = 30 °C
tout = °C
4.Energy balanceHard
hot sidecold sidelight oil · 140 °C70 °Cwater · 30 °C45 °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 = 2.5 kg/scp,h = 2.1 kJ/(kg·K)Th: 140 → 70 °Ct: 30 → 45 °Ccp,c = 4.18 kJ/(kg·K)loss = 5 %
ṁc = kg/s
5.Ideal gas lawMedium
70 g of N₂V = 20 LT = 65 °CP = ?

A tank of 20 L contains 70 g of N₂ at 65 °C. Treat it as an ideal gas and find the pressure P.

m = 70 gM = 28.01 g/molV = 20 LT = 65 °CR = 8.314 J/(mol·K)
P = kPa
6.Ideal gas lawHard
N₂, M = 28.01 g/molP = 750 kPaT = 60 °Cρ = ?

Find the density ρ of N₂ at 750 kPa and 60 °C, treating it as an ideal gas.

M = 28.01 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​)

6 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(CH4​)n(CH4​)n(H2​O)m(H2​O)÷ M(CH4​)× 2/1× M(H2​O)

In the reaction CH4 + 2O2 → CO2 + 2H2O, 4 g of CH4 reacts completely (the other reactants are in excess) and 4.94 g of H2O is collected. Find the percent yield Y.

M(CH4) = 16.04 g/molM(H2O) = 18.02 g/molactual: 4.94 g
Y = %

3Solutions

1.Mass balance

Answer: ṁ2 = 900 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) = 600·(15 − 30)/(30 − 40) = 900 kg/h

Answer

ṁ2 = 900 kg/h
2.Mass balance

Answer: W = 738 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 = 900·9/50 = 162 kg/h

Total balance

W = F − P = 900 − 162 = 738 kg/h

Answer

W = 738 kg/h
3.Energy balance

Answer: tout = 120.4 °C

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

Heat released by the hot stream

Q = 3·2.1·(160 − 100) = 378 kW

Heat picked up by the water

tout = tin + Q/(ṁccp,c) = 30 + 378/(1·4.18) = 120.4 °C

Answer

tout = 120.4 °C
4.Energy balance

Answer: ṁc = 5.568 kg/s

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

Heat released

Qh = 2.5·2.1·(140 − 70) = 367.5 kW

Heat that reaches the water

Qc = (1 − 0.05)·367.5 = 349.1 kW

Water flow rate

ṁc = Qc/(cp,c·Δt) = 349.1/(4.18·15) = 5.568 kg/s

Answer

ṁc = 5.568 kg/s
5.Ideal gas law

Answer: P = 351.3 kPa

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

Moles

n = m/M = 70/28.01 = 2.499 mol
T = 65 + 273.15 = 338.1 K

Pressure

P = nRT/V = 2.499·8.314·338.1/0.02 = 351.3 kPa

Answer

P = 351.3 kPa
6.Ideal gas law

Answer: ρ = 7.584 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 = 750·28.01/(8.314·333.1) = 7.584 kg/m³

Answer

ρ = 7.584 kg/m³
7.Stoichiometry

Answer: m = 7.293 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) = 6/28.02 = 0.2141 mol
n(H2) = 16/2.016 = 7.937 mol

Limiting reactant

0.2141/1 = 0.2141, 7.937/3 = 2.646 → N2

Mass of product

n(NH3) = 0.2141·2 = 0.4283 mol
m = 0.4283·17.03 = 7.293 g

Answer

m = 7.293 g
8.Stoichiometry

Answer: Y = 54.97 %

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(CH4) = 4/16.04 = 0.2494 mol
m = 0.2494·2/1·18.02 = 8.988 g

Percent yield

Y = 4.94/8.988·100 = 54.97%

Answer

Y = 54.97%

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