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 1300 kg/h · 45% NaOHStream 2ṁ₂ = ? · 5% NaOHProduct40% NaOH

How much of stream 2 must be added each hour to stream 1 so that the blended product contains 40% NaOH? Find ṁ2.

ṁ1 = 300 kg/hx1 = 45 %x2 = 5 %x3 = 40 %
ṁ2 = kg/h
2.Mass balanceHard
EvaporatorFeed F4800 kg/h · 14% sugarProduct P45% 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 = 4800 kg/hxF = 14 %xP = 45 %
W = kg/h
3.Energy balanceMedium
hot sidecold sidelight oil · 130 °C80 °Cwater · 10 °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: 130 → 80 °Cṁc = 3 kg/scp,c = 4.18 kJ/(kg·K)tin = 10 °C
tout = °C
4.Energy balanceHard
hot sidecold sideethanol · 150 °C100 °Cwater · 30 °C85 °C · ṁ = ?5% lost

Hot ethanol 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 = 3 kg/scp,h = 2.44 kJ/(kg·K)Th: 150 → 100 °Ct: 30 → 85 °Ccp,c = 4.18 kJ/(kg·K)loss = 5 %
ṁc = kg/s
5.Ideal gas lawMedium
40 g of CO₂V = 5 LT = 45 °CP = ?

A tank of 5 L contains 40 g of CO₂ at 45 °C. Treat it as an ideal gas and find the pressure P.

m = 40 gM = 44.01 g/molV = 5 LT = 45 °CR = 8.314 J/(mol·K)
P = kPa
6.Ideal gas lawHard
State 1State 2

A fixed amount of N₂ goes from 100 kPa, 5 L, 10 °C to 700 kPa and 110 °C. Find the new volume V2.

P1 = 100 kPaV1 = 5 LT1 = 10 °CP2 = 700 kPaT2 = 110 °C
V2 = L
7.StoichiometryMedium
m(X)n(X)n(AlCl3​)m(AlCl3​)÷ M(X)× 2/ν× M(AlCl3​)

12 g of Al and 20 g of Cl2 react according to 2Al + 3Cl2 → 2AlCl3. What mass m of AlCl3 can form?

M(Al) = 26.98M(Cl2) = 70.9M(AlCl3) = 133.33 g/mol
m = g
8.StoichiometryHard
m(C3​H8​)n(C3​H8​)n(CO2​)m(CO2​)÷ M(C3​H8​)× 3/1× M(CO2​)

In the reaction C3H8 + 5O2 → 3CO2 + 4H2O, 16 g of C3H8 reacts completely (the other reactants are in excess) and 33.53 g of CO2 is collected. Find the percent yield Y.

M(C3H8) = 44.1 g/molM(CO2) = 44.01 g/molactual: 33.53 g
Y = %

3Solutions

1.Mass balance

Answer: ṁ2 = 42.86 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) = 300·(45 − 40)/(40 − 5) = 42.86 kg/h

Answer

ṁ2 = 42.86 kg/h
2.Mass balance

Answer: W = 3307 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 = 4800·14/45 = 1493 kg/h

Total balance

W = F − P = 4800 − 1493 = 3307 kg/h

Answer

W = 3307 kg/h
3.Energy balance

Answer: tout = 35.12 °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·(130 − 80) = 315 kW

Heat picked up by the water

tout = tin + Q/(ṁccp,c) = 10 + 315/(3·4.18) = 35.12 °C

Answer

tout = 35.12 °C
4.Energy balance

Answer: ṁc = 1.512 kg/s

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

Heat released

Qh = 3·2.44·(150 − 100) = 366 kW

Heat that reaches the water

Qc = (1 − 0.05)·366 = 347.7 kW

Water flow rate

ṁc = Qc/(cp,c·Δt) = 347.7/(4.18·55) = 1.512 kg/s

Answer

ṁc = 1.512 kg/s
5.Ideal gas law

Answer: P = 480.8 kPa

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

Moles

n = m/M = 40/44.01 = 0.9089 mol
T = 45 + 273.15 = 318.1 K

Pressure

P = nRT/V = 0.9089·8.314·318.1/0.005 = 480.8 kPa

Answer

P = 480.8 kPa
6.Ideal gas law

Answer: V2 = 0.9665 L

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

Absolute temperatures

T1 = 283.1 K, T2 = 383.1 K

Combined gas law

V2 = V1·(P1/P2)·(T2/T1) = 5·(100/700)·(383.1/283.1) = 0.9665 L

Answer

V2 = 0.9665 L
7.Stoichiometry

Answer: m = 25.07 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(Al) = 12/26.98 = 0.4448 mol
n(Cl2) = 20/70.9 = 0.2821 mol

Limiting reactant

0.4448/2 = 0.2224, 0.2821/3 = 0.09403 → Cl2

Mass of product

n(AlCl3) = 0.09403·2 = 0.1881 mol
m = 0.1881·133.33 = 25.07 g

Answer

m = 25.07 g
8.Stoichiometry

Answer: Y = 70 %

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(C3H8) = 16/44.1 = 0.3628 mol
m = 0.3628·3/1·44.01 = 47.9 g

Percent yield

Y = 33.53/47.9·100 = 70%

Answer

Y = 70%

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