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 1700 kg/h · 30% NaOHStream 2ṁ₂ = ? · 10% NaOHProduct25% NaOH

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

ṁ1 = 700 kg/hx1 = 30 %x2 = 10 %x3 = 25 %
ṁ2 = kg/h
2.Mass balanceHard
EvaporatorFeed F2000 kg/h · 16% NaOHProduct PP = ? · 60%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 = 2000 kg/hxF = 16 %xP = 60 %
P = kg/h
3.Energy balanceMedium
hot sidecold sideethanol · 130 °C60 °Cwater · 10 °Ct = ? °C

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

ṁh = 1 kg/scp,h = 2.44 kJ/(kg·K)Th: 130 → 60 °Cṁc = 1.5 kg/scp,c = 4.18 kJ/(kg·K)tin = 10 °C
tout = °C
4.Energy balanceHard
hot sidecold sideglycol mix · 140 °C100 °Cwater · 20 °C70 °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 = 2.5 kg/scp,h = 3.6 kJ/(kg·K)Th: 140 → 100 °Ct: 20 → 70 °Ccp,c = 4.18 kJ/(kg·K)loss = 10 %
ṁc = kg/s
5.Ideal gas lawMedium
320 g of O₂P = 200 kPaT = 50 °CV = ?

What volume V does 320 g of O₂ occupy at 200 kPa and 50 °C? Treat it as an ideal gas.

m = 320 gM = 32 g/molP = 200 kPaT = 50 °CR = 8.314 J/(mol·K)
V = L
6.Ideal gas lawHard
State 1State 2

A fixed amount of He goes from 400 kPa, 10 L, 35 °C to 800 kPa and 190 °C. Find the new volume V2.

P1 = 400 kPaV1 = 10 LT1 = 35 °CP2 = 800 kPaT2 = 190 °C
V2 = L
7.StoichiometryMedium
m(X)n(X)n(H2​O)m(H2​O)÷ M(X)× 2/ν× M(H2​O)

16 g of H2 and 64 g of O2 react according to 2H2 + O2 → 2H2O. What mass m of H2O can form?

M(H2) = 2.016M(O2) = 32M(H2O) = 18.02 g/mol
m = g
8.StoichiometryHard
m(Fe2​O3​)n(Fe2​O3​)n(CO2​)m(CO2​)÷ M(Fe2​O3​)× 3/1× M(CO2​)

In the reaction Fe2O3 + 3CO → 2Fe + 3CO2, 16 g of Fe2O3 reacts completely (the other reactants are in excess) and 9.26 g of CO2 is collected. Find the percent yield Y.

M(Fe2O3) = 159.69 g/molM(CO2) = 44.01 g/molactual: 9.26 g
Y = %

3Solutions

1.Mass balance

Answer: ṁ2 = 233.3 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) = 700·(30 − 25)/(25 − 10) = 233.3 kg/h

Answer

ṁ2 = 233.3 kg/h
2.Mass balance

Answer: P = 533.3 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 = 2000·16/60 = 533.3 kg/h

Answer

P = 533.3 kg/h
3.Energy balance

Answer: tout = 37.24 °C

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

Heat released by the hot stream

Q = 1·2.44·(130 − 60) = 170.8 kW

Heat picked up by the water

tout = tin + Q/(ṁccp,c) = 10 + 170.8/(1.5·4.18) = 37.24 °C

Answer

tout = 37.24 °C
4.Energy balance

Answer: ṁc = 1.55 kg/s

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

Heat released

Qh = 2.5·3.6·(140 − 100) = 360 kW

Heat that reaches the water

Qc = (1 − 0.1)·360 = 324 kW

Water flow rate

ṁc = Qc/(cp,c·Δt) = 324/(4.18·50) = 1.55 kg/s

Answer

ṁc = 1.55 kg/s
5.Ideal gas law

Answer: V = 134.3 L

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

Moles

n = 320/32 = 10 mol
T = 50 + 273.15 = 323.1 K

Volume

V = nRT/P = 10·8.314·323.1/200 = 134.3 L

Answer

V = 134.3 L
6.Ideal gas law

Answer: V2 = 7.515 L

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

Absolute temperatures

T1 = 308.1 K, T2 = 463.1 K

Combined gas law

V2 = V1·(P1/P2)·(T2/T1) = 10·(400/800)·(463.1/308.1) = 7.515 L

Answer

V2 = 7.515 L
7.Stoichiometry

Answer: m = 72.08 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(H2) = 16/2.016 = 7.937 mol
n(O2) = 64/32 = 2 mol

Limiting reactant

7.937/2 = 3.968, 2/1 = 2 → O2

Mass of product

n(H2O) = 2·2 = 4 mol
m = 4·18.02 = 72.08 g

Answer

m = 72.08 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(Fe2O3) = 16/159.69 = 0.1002 mol
m = 0.1002·3/1·44.01 = 13.23 g

Percent yield

Y = 9.26/13.23·100 = 70%

Answer

Y = 70%

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