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 1500 kg/h · 10% sugarStream 2ṁ₂ = ? · 35% sugarProduct20% sugar

How much of stream 2 must be added each hour to stream 1 so that the blended product contains 20% sugar? Find ṁ2.

ṁ1 = 500 kg/hx1 = 10 %x2 = 35 %x3 = 20 %
ṁ2 = kg/h
2.Mass balanceHard
EvaporatorFeed F800 kg/h · 13% NaOHProduct PP = ? · 55%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 = 800 kg/hxF = 13 %xP = 55 %
P = kg/h
3.Energy balanceMedium
hot sidecold sidelight oil · 170 °C100 °Cwater · 25 °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: 170 → 100 °Cṁc = 2 kg/scp,c = 4.18 kJ/(kg·K)tin = 25 °C
tout = °C
4.Energy balanceHard
hot sidecold sideglycol mix · 160 °C80 °Cwater · 30 °C80 °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 = 3 kg/scp,h = 3.6 kJ/(kg·K)Th: 160 → 80 °Ct: 30 → 80 °Ccp,c = 4.18 kJ/(kg·K)loss = 10 %
ṁc = kg/s
5.Ideal gas lawMedium
170 g of HeP = 550 kPaT = 95 °CV = ?

What volume V does 170 g of He occupy at 550 kPa and 95 °C? Treat it as an ideal gas.

m = 170 gM = 4.003 g/molP = 550 kPaT = 95 °CR = 8.314 J/(mol·K)
V = L
6.Ideal gas lawHard
State 1State 2

A fixed amount of N₂ goes from 200 kPa, 5 L, 15 °C to 650 kPa and 150 °C. Find the new volume V2.

P1 = 200 kPaV1 = 5 LT1 = 15 °CP2 = 650 kPaT2 = 150 °C
V2 = L
7.StoichiometryMedium
m(X)n(X)n(AlCl3​)m(AlCl3​)÷ M(X)× 2/ν× M(AlCl3​)

4 g of Al and 8 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(Al)n(Al)n(AlCl3​)m(AlCl3​)÷ M(Al)× 2/2× M(AlCl3​)

In the reaction 2Al + 3Cl2 → 2AlCl3, 16 g of Al reacts completely (the other reactants are in excess) and 55.35 g of AlCl3 is collected. Find the percent yield Y.

M(Al) = 26.98 g/molM(AlCl3) = 133.33 g/molactual: 55.35 g
Y = %

3Solutions

1.Mass balance

Answer: ṁ2 = 333.3 kg/h

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

Balances

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

Solve for ṁ2

ṁ2 = ṁ1(x1 − x3)/(x3 − x2) = 500·(10 − 20)/(20 − 35) = 333.3 kg/h

Answer

ṁ2 = 333.3 kg/h
2.Mass balance

Answer: P = 189.1 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 = 800·13/55 = 189.1 kg/h

Answer

P = 189.1 kg/h
3.Energy balance

Answer: tout = 77.75 °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·(170 − 100) = 441 kW

Heat picked up by the water

tout = tin + Q/(ṁccp,c) = 25 + 441/(2·4.18) = 77.75 °C

Answer

tout = 77.75 °C
4.Energy balance

Answer: ṁc = 3.721 kg/s

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

Heat released

Qh = 3·3.6·(160 − 80) = 864 kW

Heat that reaches the water

Qc = (1 − 0.1)·864 = 777.6 kW

Water flow rate

ṁc = Qc/(cp,c·Δt) = 777.6/(4.18·50) = 3.721 kg/s

Answer

ṁc = 3.721 kg/s
5.Ideal gas law

Answer: V = 236.3 L

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

Moles

n = 170/4.003 = 42.47 mol
T = 95 + 273.15 = 368.1 K

Volume

V = nRT/P = 42.47·8.314·368.1/550 = 236.3 L

Answer

V = 236.3 L
6.Ideal gas law

Answer: V2 = 2.259 L

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

Absolute temperatures

T1 = 288.1 K, T2 = 423.1 K

Combined gas law

V2 = V1·(P1/P2)·(T2/T1) = 5·(200/650)·(423.1/288.1) = 2.259 L

Answer

V2 = 2.259 L
7.Stoichiometry

Answer: m = 10.03 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) = 4/26.98 = 0.1483 mol
n(Cl2) = 8/70.9 = 0.1128 mol

Limiting reactant

0.1483/2 = 0.07413, 0.1128/3 = 0.03761 → Cl2

Mass of product

n(AlCl3) = 0.03761·2 = 0.07522 mol
m = 0.07522·133.33 = 10.03 g

Answer

m = 10.03 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(Al) = 16/26.98 = 0.593 mol
m = 0.593·2/2·133.33 = 79.07 g

Percent yield

Y = 55.35/79.07·100 = 70%

Answer

Y = 70%

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