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 1800 kg/h · 50% saltStream 2ṁ₂ = ? · 15% saltProduct40% salt

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

ṁ1 = 800 kg/hx1 = 50 %x2 = 15 %x3 = 40 %
ṁ2 = kg/h
2.Mass balanceHard
EvaporatorFeed F3100 kg/h · 18% sugarProduct PP = ? · 30%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 = 3100 kg/hxF = 18 %xP = 30 %
P = kg/h
3.Energy balanceMedium
hot sidecold sideglycol mix · 160 °C130 °Cwater · 15 °Ct = ? °C

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

ṁh = 0.5 kg/scp,h = 3.6 kJ/(kg·K)Th: 160 → 130 °Cṁc = 3 kg/scp,c = 4.18 kJ/(kg·K)tin = 15 °C
tout = °C
4.Energy balanceHard
hot sidecold sideglycol mix · 160 °C100 °Cwater · 30 °C60 °C · ṁ = ?15% lost

Hot glycol mix heats water in an exchanger that loses 15% 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: 160 → 100 °Ct: 30 → 60 °Ccp,c = 4.18 kJ/(kg·K)loss = 15 %
ṁc = kg/s
5.Ideal gas lawMedium
400 g of CH₄V = 5 LT = 10 °CP = ?

A tank of 5 L contains 400 g of CH₄ at 10 °C. Treat it as an ideal gas and find the pressure P.

m = 400 gM = 16.04 g/molV = 5 LT = 10 °CR = 8.314 J/(mol·K)
P = kPa
6.Ideal gas lawHard
He, M = 4.003 g/molP = 150 kPaT = 190 °Cρ = ?

Find the density ρ of He at 150 kPa and 190 °C, treating it as an ideal gas.

M = 4.003 g/molP = 150 kPaT = 190 °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 Fe2O3 and 8 g of CO react according to Fe2O3 + 3CO → 2Fe + 3CO2. What mass m of CO2 can form?

M(Fe2O3) = 159.69M(CO) = 28.01M(CO2) = 44.01 g/mol
m = g
8.StoichiometryHard
m(Mg)n(Mg)n(MgO)m(MgO)÷ M(Mg)× 2/2× M(MgO)

In the reaction 2Mg + O2 → 2MgO, 10 g of Mg reacts completely (the other reactants are in excess) and 15.58 g of MgO is collected. Find the percent yield Y.

M(Mg) = 24.31 g/molM(MgO) = 40.3 g/molactual: 15.58 g
Y = %

3Solutions

1.Mass balance

Answer: ṁ2 = 320 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) = 800·(50 − 40)/(40 − 15) = 320 kg/h

Answer

ṁ2 = 320 kg/h
2.Mass balance

Answer: P = 1860 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 = 3100·18/30 = 1860 kg/h

Answer

P = 1860 kg/h
3.Energy balance

Answer: tout = 19.31 °C

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

Heat released by the hot stream

Q = 0.5·3.6·(160 − 130) = 54 kW

Heat picked up by the water

tout = tin + Q/(ṁccp,c) = 15 + 54/(3·4.18) = 19.31 °C

Answer

tout = 19.31 °C
4.Energy balance

Answer: ṁc = 3.66 kg/s

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

Heat released

Qh = 2.5·3.6·(160 − 100) = 540 kW

Heat that reaches the water

Qc = (1 − 0.15)·540 = 459 kW

Water flow rate

ṁc = Qc/(cp,c·Δt) = 459/(4.18·30) = 3.66 kg/s

Answer

ṁc = 3.66 kg/s
5.Ideal gas law

Answer: P = 11741 kPa

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

Moles

n = m/M = 400/16.04 = 24.94 mol
T = 10 + 273.15 = 283.1 K

Pressure

P = nRT/V = 24.94·8.314·283.1/0.005 = 11741 kPa

Answer

P = 11741 kPa
6.Ideal gas law

Answer: ρ = 0.1559 kg/m³

Ideal gas density: ρ = PM/(RT).

Rearrange the gas law

PV = (m/M)RT → ρ = m/V = PM/(RT)
T = 190 + 273.15 = 463.1 K

Density

ρ = 150·4.003/(8.314·463.1) = 0.1559 kg/m³

Answer

ρ = 0.1559 kg/m³
7.Stoichiometry

Answer: m = 12.57 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(Fe2O3) = 16/159.69 = 0.1002 mol
n(CO) = 8/28.01 = 0.2856 mol

Limiting reactant

0.1002/1 = 0.1002, 0.2856/3 = 0.0952 → CO

Mass of product

n(CO2) = 0.0952·3 = 0.2856 mol
m = 0.2856·44.01 = 12.57 g

Answer

m = 12.57 g
8.Stoichiometry

Answer: Y = 93.98 %

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(Mg) = 10/24.31 = 0.4114 mol
m = 0.4114·2/2·40.3 = 16.58 g

Percent yield

Y = 15.58/16.58·100 = 93.98%

Answer

Y = 93.98%

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