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 1850 kg/h · 30% NaOHStream 2ṁ₂ = ? · 10% NaOHProduct15% NaOH

How much of stream 2 must be added each hour to stream 1 so that the blended product contains 15% NaOH? Find ṁ2.

ṁ1 = 850 kg/hx1 = 30 %x2 = 10 %x3 = 15 %
ṁ2 = kg/h
2.Mass balanceHard
EvaporatorFeed F4000 kg/h · 17% sugarProduct PP = ? · 40%Vapour W (pure water)0% sugar

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

F = 4000 kg/hxF = 17 %xP = 40 %
P = kg/h
3.Energy balanceMedium
hot sidecold sideglycol mix · 170 °C120 °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 kg/scp,h = 3.6 kJ/(kg·K)Th: 170 → 120 °Cṁc = 0.5 kg/scp,c = 4.18 kJ/(kg·K)tin = 30 °C
tout = °C
4.Energy balanceHard
hot sidecold sidelight oil · 140 °C70 °Cwater · 15 °C55 °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 = 0.5 kg/scp,h = 2.1 kJ/(kg·K)Th: 140 → 70 °Ct: 15 → 55 °Ccp,c = 4.18 kJ/(kg·K)loss = 10 %
ṁc = kg/s
5.Ideal gas lawMedium
430 g of ArP = 350 kPaT = 0 °CV = ?

What volume V does 430 g of Ar occupy at 350 kPa and 0 °C? Treat it as an ideal gas.

m = 430 gM = 39.95 g/molP = 350 kPaT = 0 °CR = 8.314 J/(mol·K)
V = L
6.Ideal gas lawHard
CO₂, M = 44.01 g/molP = 250 kPaT = 90 °Cρ = ?

Find the density ρ of CO₂ at 250 kPa and 90 °C, treating it as an ideal gas.

M = 44.01 g/molP = 250 kPaT = 90 °CR = 8.314 J/(mol·K)
ρ = kg/m³
7.StoichiometryMedium
m(X)n(X)n(CO2​)m(CO2​)÷ M(X)× 3/ν× M(CO2​)

16 g of C3H8 and 40 g of O2 react according to C3H8 + 5O2 → 3CO2 + 4H2O. What mass m of CO2 can form?

M(C3H8) = 44.1M(O2) = 32M(CO2) = 44.01 g/mol
m = g
8.StoichiometryHard
m(Na)n(Na)n(NaCl)m(NaCl)÷ M(Na)× 2/2× M(NaCl)

In the reaction 2Na + Cl2 → 2NaCl, 10 g of Na reacts completely (the other reactants are in excess) and 15.76 g of NaCl is collected. Find the percent yield Y.

M(Na) = 22.99 g/molM(NaCl) = 58.44 g/molactual: 15.76 g
Y = %

3Solutions

1.Mass balance

Answer: ṁ2 = 2550 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) = 850·(30 − 15)/(15 − 10) = 2550 kg/h

Answer

ṁ2 = 2550 kg/h
2.Mass balance

Answer: P = 1700 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 = 4000·17/40 = 1700 kg/h

Answer

P = 1700 kg/h
3.Energy balance

Answer: tout = 116.1 °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·(170 − 120) = 180 kW

Heat picked up by the water

tout = tin + Q/(ṁccp,c) = 30 + 180/(0.5·4.18) = 116.1 °C

Answer

tout = 116.1 °C
4.Energy balance

Answer: ṁc = 0.3956 kg/s

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

Heat released

Qh = 0.5·2.1·(140 − 70) = 73.5 kW

Heat that reaches the water

Qc = (1 − 0.1)·73.5 = 66.15 kW

Water flow rate

ṁc = Qc/(cp,c·Δt) = 66.15/(4.18·40) = 0.3956 kg/s

Answer

ṁc = 0.3956 kg/s
5.Ideal gas law

Answer: V = 69.84 L

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

Moles

n = 430/39.95 = 10.76 mol
T = 0 + 273.15 = 273.1 K

Volume

V = nRT/P = 10.76·8.314·273.1/350 = 69.84 L

Answer

V = 69.84 L
6.Ideal gas law

Answer: ρ = 3.644 kg/m³

Ideal gas density: ρ = PM/(RT).

Rearrange the gas law

PV = (m/M)RT → ρ = m/V = PM/(RT)
T = 90 + 273.15 = 363.1 K

Density

ρ = 250·44.01/(8.314·363.1) = 3.644 kg/m³

Answer

ρ = 3.644 kg/m³
7.Stoichiometry

Answer: m = 33.01 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(C3H8) = 16/44.1 = 0.3628 mol
n(O2) = 40/32 = 1.25 mol

Limiting reactant

0.3628/1 = 0.3628, 1.25/5 = 0.25 → O2

Mass of product

n(CO2) = 0.25·3 = 0.75 mol
m = 0.75·44.01 = 33.01 g

Answer

m = 33.01 g
8.Stoichiometry

Answer: Y = 62 %

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(Na) = 10/22.99 = 0.435 mol
m = 0.435·2/2·58.44 = 25.42 g

Percent yield

Y = 15.76/25.42·100 = 62%

Answer

Y = 62%

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