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 1350 kg/h · 5% sugarStream 2ṁ₂ = ? · 40% sugarProduct25% sugar

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

ṁ1 = 350 kg/hx1 = 5 %x2 = 40 %x3 = 25 %
ṁ2 = kg/h
2.Mass balanceHard
EvaporatorFeed F4000 kg/h · 17% NaOHProduct P50% NaOHVapour W (pure water)W = ? kg/h

An evaporator concentrates a NaOH solution by boiling off pure water. Find the water evaporated W in kg/h.

F = 4000 kg/hxF = 17 %xP = 50 %
W = kg/h
3.Energy balanceMedium
hot sidecold sideglycol mix · 150 °C110 °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: 150 → 110 °Cṁc = 3 kg/scp,c = 4.18 kJ/(kg·K)tin = 15 °C
tout = °C
4.Energy balanceHard
hot sidecold sidelight oil · 130 °C70 °Cwater · 25 °C60 °C · ṁ = ?15% lost

Hot light oil 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 = 2.1 kJ/(kg·K)Th: 130 → 70 °Ct: 25 → 60 °Ccp,c = 4.18 kJ/(kg·K)loss = 15 %
ṁc = kg/s
5.Ideal gas lawMedium
480 g of CO₂V = 50 LT = 85 °CP = ?

A tank of 50 L contains 480 g of CO₂ at 85 °C. Treat it as an ideal gas and find the pressure P.

m = 480 gM = 44.01 g/molV = 50 LT = 85 °CR = 8.314 J/(mol·K)
P = kPa
6.Ideal gas lawHard
N₂, M = 28.01 g/molP = 1000 kPaT = 130 °Cρ = ?

Find the density ρ of N₂ at 1000 kPa and 130 °C, treating it as an ideal gas.

M = 28.01 g/molP = 1000 kPaT = 130 °CR = 8.314 J/(mol·K)
ρ = kg/m³
7.StoichiometryMedium
m(X)n(X)n(MgO)m(MgO)÷ M(X)× 2/ν× M(MgO)

10 g of Mg and 64 g of O2 react according to 2Mg + O2 → 2MgO. What mass m of MgO can form?

M(Mg) = 24.31M(O2) = 32M(MgO) = 40.3 g/mol
m = g
8.StoichiometryHard
m(CaCO3​)n(CaCO3​)n(CaO)m(CaO)÷ M(CaCO3​)× 1/1× M(CaO)

In the reaction CaCO3 → CaO + CO2, 12 g of CaCO3 reacts completely (the other reactants are in excess) and 5.92 g of CaO is collected. Find the percent yield Y.

M(CaCO3) = 100.09 g/molM(CaO) = 56.08 g/molactual: 5.92 g
Y = %

3Solutions

1.Mass balance

Answer: ṁ2 = 466.7 kg/h

Steady-state mass balance: in = out for the total and for each component. Blending problems combine both.

Balances

Total: ṁ3 = ṁ1 + ṁ2
sugar: ṁ1x1 + ṁ2x2 = (ṁ1 + ṁ2)x3

Solve for ṁ2

ṁ2 = ṁ1(x1 − x3)/(x3 − x2) = 350·(5 − 25)/(25 − 40) = 466.7 kg/h

Answer

ṁ2 = 466.7 kg/h
2.Mass balance

Answer: W = 2640 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 = 4000·17/50 = 1360 kg/h

Total balance

W = F − P = 4000 − 1360 = 2640 kg/h

Answer

W = 2640 kg/h
3.Energy balance

Answer: tout = 20.74 °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·(150 − 110) = 72 kW

Heat picked up by the water

tout = tin + Q/(ṁccp,c) = 15 + 72/(3·4.18) = 20.74 °C

Answer

tout = 20.74 °C
4.Energy balance

Answer: ṁc = 1.83 kg/s

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

Heat released

Qh = 2.5·2.1·(130 − 70) = 315 kW

Heat that reaches the water

Qc = (1 − 0.15)·315 = 267.8 kW

Water flow rate

ṁc = Qc/(cp,c·Δt) = 267.8/(4.18·35) = 1.83 kg/s

Answer

ṁc = 1.83 kg/s
5.Ideal gas law

Answer: P = 649.5 kPa

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

Moles

n = m/M = 480/44.01 = 10.91 mol
T = 85 + 273.15 = 358.1 K

Pressure

P = nRT/V = 10.91·8.314·358.1/0.05 = 649.5 kPa

Answer

P = 649.5 kPa
6.Ideal gas law

Answer: ρ = 8.357 kg/m³

Ideal gas density: ρ = PM/(RT).

Rearrange the gas law

PV = (m/M)RT → ρ = m/V = PM/(RT)
T = 130 + 273.15 = 403.1 K

Density

ρ = 1000·28.01/(8.314·403.1) = 8.357 kg/m³

Answer

ρ = 8.357 kg/m³
7.Stoichiometry

Answer: m = 16.58 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(Mg) = 10/24.31 = 0.4114 mol
n(O2) = 64/32 = 2 mol

Limiting reactant

0.4114/2 = 0.2057, 2/1 = 2 → Mg

Mass of product

n(MgO) = 0.2057·2 = 0.4114 mol
m = 0.4114·40.3 = 16.58 g

Answer

m = 16.58 g
8.Stoichiometry

Answer: Y = 88.05 %

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(CaCO3) = 12/100.09 = 0.1199 mol
m = 0.1199·1/1·56.08 = 6.724 g

Percent yield

Y = 5.92/6.724·100 = 88.05%

Answer

Y = 88.05%

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