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 1500 kg/h · 35% saltStream 2ṁ₂ = ? · 20% saltProduct25% salt

How much of stream 2 must be added each hour to stream 1 so that the blended product contains 25% salt? Find ṁ2.

ṁ1 = 500 kg/hx1 = 35 %x2 = 20 %x3 = 25 %
ṁ2 = kg/h
2.Mass balanceHard
EvaporatorFeed F4300 kg/h · 10% NaOHProduct PP = ? · 35%Vapour W (pure water)0% NaOH

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

F = 4300 kg/hxF = 10 %xP = 35 %
P = kg/h
3.Energy balanceMedium
hot sidecold sideethanol · 160 °C80 °Cwater · 15 °Ct = ? °C

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

ṁh = 0.5 kg/scp,h = 2.44 kJ/(kg·K)Th: 160 → 80 °Cṁc = 0.5 kg/scp,c = 4.18 kJ/(kg·K)tin = 15 °C
tout = °C
4.Energy balanceHard
hot sidecold sideglycol mix · 170 °C90 °Cwater · 15 °C35 °C · ṁ = ?10% lost

Hot glycol mix 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 = 3.6 kJ/(kg·K)Th: 170 → 90 °Ct: 15 → 35 °Ccp,c = 4.18 kJ/(kg·K)loss = 10 %
ṁc = kg/s
5.Ideal gas lawMedium
390 g of CH₄P = 150 kPaT = 95 °CV = ?

What volume V does 390 g of CH₄ occupy at 150 kPa and 95 °C? Treat it as an ideal gas.

m = 390 gM = 16.04 g/molP = 150 kPaT = 95 °CR = 8.314 J/(mol·K)
V = L
6.Ideal gas lawHard
State 1State 2

A fixed amount of CO₂ goes from 150 kPa, 20 L, 25 °C to 100 kPa and 250 °C. Find the new volume V2.

P1 = 150 kPaV1 = 20 LT1 = 25 °CP2 = 100 kPaT2 = 250 °C
V2 = L
7.StoichiometryMedium
m(X)n(X)n(CO2​)m(CO2​)÷ M(X)× 3/ν× M(CO2​)

16 g of C3H8 and 48 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(Al)n(Al)n(AlCl3​)m(AlCl3​)÷ M(Al)× 2/2× M(AlCl3​)

In the reaction 2Al + 3Cl2 → 2AlCl3, 24 g of Al reacts completely (the other reactants are in excess) and 97.25 g of AlCl3 is collected. Find the percent yield Y.

M(Al) = 26.98 g/molM(AlCl3) = 133.33 g/molactual: 97.25 g
Y = %

3Solutions

1.Mass balance

Answer: ṁ2 = 1000 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) = 500·(35 − 25)/(25 − 20) = 1000 kg/h

Answer

ṁ2 = 1000 kg/h
2.Mass balance

Answer: P = 1229 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 = 4300·10/35 = 1229 kg/h

Answer

P = 1229 kg/h
3.Energy balance

Answer: tout = 61.7 °C

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

Heat released by the hot stream

Q = 0.5·2.44·(160 − 80) = 97.6 kW

Heat picked up by the water

tout = tin + Q/(ṁccp,c) = 15 + 97.6/(0.5·4.18) = 61.7 °C

Answer

tout = 61.7 °C
4.Energy balance

Answer: ṁc = 6.201 kg/s

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

Heat released

Qh = 2·3.6·(170 − 90) = 576 kW

Heat that reaches the water

Qc = (1 − 0.1)·576 = 518.4 kW

Water flow rate

ṁc = Qc/(cp,c·Δt) = 518.4/(4.18·20) = 6.201 kg/s

Answer

ṁc = 6.201 kg/s
5.Ideal gas law

Answer: V = 496.1 L

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

Moles

n = 390/16.04 = 24.31 mol
T = 95 + 273.15 = 368.1 K

Volume

V = nRT/P = 24.31·8.314·368.1/150 = 496.1 L

Answer

V = 496.1 L
6.Ideal gas law

Answer: V2 = 52.64 L

Combined gas law: for a fixed amount of ideal gas, P1V1/T1 = P2V2/T2.

Absolute temperatures

T1 = 298.1 K, T2 = 523.1 K

Combined gas law

V2 = V1·(P1/P2)·(T2/T1) = 20·(150/100)·(523.1/298.1) = 52.64 L

Answer

V2 = 52.64 L
7.Stoichiometry

Answer: m = 39.61 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) = 48/32 = 1.5 mol

Limiting reactant

0.3628/1 = 0.3628, 1.5/5 = 0.3 → O2

Mass of product

n(CO2) = 0.3·3 = 0.9 mol
m = 0.9·44.01 = 39.61 g

Answer

m = 39.61 g
8.Stoichiometry

Answer: Y = 82 %

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(Al) = 24/26.98 = 0.8895 mol
m = 0.8895·2/2·133.33 = 118.6 g

Percent yield

Y = 97.25/118.6·100 = 82%

Answer

Y = 82%

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