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 1900 kg/h · 30% sugarStream 2ṁ₂ = ? · 5% sugarProduct10% sugar

How much of stream 2 must be added each hour to stream 1 so that the blended product contains 10% sugar? Find ṁ2.

ṁ1 = 900 kg/hx1 = 30 %x2 = 5 %x3 = 10 %
ṁ2 = kg/h
2.Mass balanceHard
EvaporatorFeed F1400 kg/h · 16% sugarProduct PP = ? · 55%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 = 1400 kg/hxF = 16 %xP = 55 %
P = kg/h
3.Energy balanceMedium
hot sidecold sidelight oil · 170 °C70 °Cwater · 30 °Ct = ? °C

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

ṁh = 1.5 kg/scp,h = 2.1 kJ/(kg·K)Th: 170 → 70 °Cṁc = 1 kg/scp,c = 4.18 kJ/(kg·K)tin = 30 °C
tout = °C
4.Energy balanceHard
hot sidecold sidelight oil · 170 °C130 °Cwater · 15 °C40 °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 = 1.5 kg/scp,h = 2.1 kJ/(kg·K)Th: 170 → 130 °Ct: 15 → 40 °Ccp,c = 4.18 kJ/(kg·K)loss = 15 %
ṁc = kg/s
5.Ideal gas lawMedium
50 g of ArV = 10 LT = 25 °CP = ?

A tank of 10 L contains 50 g of Ar at 25 °C. Treat it as an ideal gas and find the pressure P.

m = 50 gM = 39.95 g/molV = 10 LT = 25 °CR = 8.314 J/(mol·K)
P = kPa
6.Ideal gas lawHard
O₂, M = 32 g/molP = 900 kPaT = 150 °Cρ = ?

Find the density ρ of O₂ at 900 kPa and 150 °C, treating it as an ideal gas.

M = 32 g/molP = 900 kPaT = 150 °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)

24 g of Mg and 48 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(CH4​)n(CH4​)n(H2​O)m(H2​O)÷ M(CH4​)× 2/1× M(H2​O)

In the reaction CH4 + 2O2 → CO2 + 2H2O, 12 g of CH4 reacts completely (the other reactants are in excess) and 21.03 g of H2O is collected. Find the percent yield Y.

M(CH4) = 16.04 g/molM(H2O) = 18.02 g/molactual: 21.03 g
Y = %

3Solutions

1.Mass balance

Answer: ṁ2 = 3600 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) = 900·(30 − 10)/(10 − 5) = 3600 kg/h

Answer

ṁ2 = 3600 kg/h
2.Mass balance

Answer: P = 407.3 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 = 1400·16/55 = 407.3 kg/h

Answer

P = 407.3 kg/h
3.Energy balance

Answer: tout = 105.4 °C

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

Heat released by the hot stream

Q = 1.5·2.1·(170 − 70) = 315 kW

Heat picked up by the water

tout = tin + Q/(ṁccp,c) = 30 + 315/(1·4.18) = 105.4 °C

Answer

tout = 105.4 °C
4.Energy balance

Answer: ṁc = 1.025 kg/s

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

Heat released

Qh = 1.5·2.1·(170 − 130) = 126 kW

Heat that reaches the water

Qc = (1 − 0.15)·126 = 107.1 kW

Water flow rate

ṁc = Qc/(cp,c·Δt) = 107.1/(4.18·25) = 1.025 kg/s

Answer

ṁc = 1.025 kg/s
5.Ideal gas law

Answer: P = 310.2 kPa

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

Moles

n = m/M = 50/39.95 = 1.252 mol
T = 25 + 273.15 = 298.1 K

Pressure

P = nRT/V = 1.252·8.314·298.1/0.01 = 310.2 kPa

Answer

P = 310.2 kPa
6.Ideal gas law

Answer: ρ = 8.186 kg/m³

Ideal gas density: ρ = PM/(RT).

Rearrange the gas law

PV = (m/M)RT → ρ = m/V = PM/(RT)
T = 150 + 273.15 = 423.1 K

Density

ρ = 900·32/(8.314·423.1) = 8.186 kg/m³

Answer

ρ = 8.186 kg/m³
7.Stoichiometry

Answer: m = 39.79 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) = 24/24.31 = 0.9872 mol
n(O2) = 48/32 = 1.5 mol

Limiting reactant

0.9872/2 = 0.4936, 1.5/1 = 1.5 → Mg

Mass of product

n(MgO) = 0.4936·2 = 0.9872 mol
m = 0.9872·40.3 = 39.79 g

Answer

m = 39.79 g
8.Stoichiometry

Answer: Y = 78 %

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(CH4) = 12/16.04 = 0.7481 mol
m = 0.7481·2/1·18.02 = 26.96 g

Percent yield

Y = 21.03/26.96·100 = 78%

Answer

Y = 78%

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