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 1550 kg/h · 10% sugarStream 2ṁ₂ = ? · 45% sugarProduct35% sugar

How much of stream 2 must be added each hour to stream 1 so that the blended product contains 35% sugar? Find ṁ2.

ṁ1 = 550 kg/hx1 = 10 %x2 = 45 %x3 = 35 %
ṁ2 = kg/h
2.Mass balanceHard
EvaporatorFeed F2100 kg/h · 5% sugarProduct PP = ? · 40%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 = 2100 kg/hxF = 5 %xP = 40 %
P = kg/h
3.Energy balanceMedium
hot sidecold sideethanol · 150 °C70 °Cwater · 20 °Ct = ? °C

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

ṁh = 0.5 kg/scp,h = 2.44 kJ/(kg·K)Th: 150 → 70 °Cṁc = 3 kg/scp,c = 4.18 kJ/(kg·K)tin = 20 °C
tout = °C
4.Energy balanceHard
hot sidecold sideethanol · 120 °C80 °Cwater · 10 °C75 °C · ṁ = ?5% lost

Hot ethanol heats water in an exchanger that loses 5% of the heat released by the hot stream to the surroundings. Find the water flow rate ṁc needed.

ṁh = 1.5 kg/scp,h = 2.44 kJ/(kg·K)Th: 120 → 80 °Ct: 10 → 75 °Ccp,c = 4.18 kJ/(kg·K)loss = 5 %
ṁc = kg/s
5.Ideal gas lawMedium
330 g of O₂P = 400 kPaT = 45 °CV = ?

What volume V does 330 g of O₂ occupy at 400 kPa and 45 °C? Treat it as an ideal gas.

m = 330 gM = 32 g/molP = 400 kPaT = 45 °CR = 8.314 J/(mol·K)
V = L
6.Ideal gas lawHard
State 1State 2

A fixed amount of CH₄ goes from 150 kPa, 10 L, 25 °C to 700 kPa and 160 °C. Find the new volume V2.

P1 = 150 kPaV1 = 10 LT1 = 25 °CP2 = 700 kPaT2 = 160 °C
V2 = L
7.StoichiometryMedium
m(X)n(X)n(H2​O)m(H2​O)÷ M(X)× 2/ν× M(H2​O)

8 g of CH4 and 20 g of O2 react according to CH4 + 2O2 → CO2 + 2H2O. What mass m of H2O can form?

M(CH4) = 16.04M(O2) = 32M(H2O) = 18.02 g/mol
m = g
8.StoichiometryHard
m(CH4​)n(CH4​)n(CO2​)m(CO2​)÷ M(CH4​)× 1/1× M(CO2​)

In the reaction CH4 + 2O2 → CO2 + 2H2O, 20 g of CH4 reacts completely (the other reactants are in excess) and 30.18 g of CO2 is collected. Find the percent yield Y.

M(CH4) = 16.04 g/molM(CO2) = 44.01 g/molactual: 30.18 g
Y = %

3Solutions

1.Mass balance

Answer: ṁ2 = 1375 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) = 550·(10 − 35)/(35 − 45) = 1375 kg/h

Answer

ṁ2 = 1375 kg/h
2.Mass balance

Answer: P = 262.5 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 = 2100·5/40 = 262.5 kg/h

Answer

P = 262.5 kg/h
3.Energy balance

Answer: tout = 27.78 °C

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

Heat released by the hot stream

Q = 0.5·2.44·(150 − 70) = 97.6 kW

Heat picked up by the water

tout = tin + Q/(ṁccp,c) = 20 + 97.6/(3·4.18) = 27.78 °C

Answer

tout = 27.78 °C
4.Energy balance

Answer: ṁc = 0.5119 kg/s

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

Heat released

Qh = 1.5·2.44·(120 − 80) = 146.4 kW

Heat that reaches the water

Qc = (1 − 0.05)·146.4 = 139.1 kW

Water flow rate

ṁc = Qc/(cp,c·Δt) = 139.1/(4.18·65) = 0.5119 kg/s

Answer

ṁc = 0.5119 kg/s
5.Ideal gas law

Answer: V = 68.19 L

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

Moles

n = 330/32 = 10.31 mol
T = 45 + 273.15 = 318.1 K

Volume

V = nRT/P = 10.31·8.314·318.1/400 = 68.19 L

Answer

V = 68.19 L
6.Ideal gas law

Answer: V2 = 3.113 L

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

Absolute temperatures

T1 = 298.1 K, T2 = 433.1 K

Combined gas law

V2 = V1·(P1/P2)·(T2/T1) = 10·(150/700)·(433.1/298.1) = 3.113 L

Answer

V2 = 3.113 L
7.Stoichiometry

Answer: m = 11.26 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(CH4) = 8/16.04 = 0.4988 mol
n(O2) = 20/32 = 0.625 mol

Limiting reactant

0.4988/1 = 0.4988, 0.625/2 = 0.3125 → O2

Mass of product

n(H2O) = 0.3125·2 = 0.625 mol
m = 0.625·18.02 = 11.26 g

Answer

m = 11.26 g
8.Stoichiometry

Answer: Y = 55 %

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) = 20/16.04 = 1.247 mol
m = 1.247·1/1·44.01 = 54.88 g

Percent yield

Y = 30.18/54.88·100 = 55%

Answer

Y = 55%

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