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 1650 kg/h · 30% NaOHStream 2ṁ₂ = ? · 20% NaOHProduct25% NaOH

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

ṁ1 = 650 kg/hx1 = 30 %x2 = 20 %x3 = 25 %
ṁ2 = kg/h
2.Mass balanceHard
EvaporatorFeed F2000 kg/h · 14% saltProduct P65% saltVapour W (pure water)W = ? kg/h

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

F = 2000 kg/hxF = 14 %xP = 65 %
W = kg/h
3.Energy balanceMedium
hot sidecold sideethanol · 140 °C80 °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: 140 → 80 °Cṁc = 3 kg/scp,c = 4.18 kJ/(kg·K)tin = 20 °C
tout = °C
4.Energy balanceHard
hot sidecold sideglycol mix · 140 °C90 °Cwater · 25 °C90 °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 = 1.5 kg/scp,h = 3.6 kJ/(kg·K)Th: 140 → 90 °Ct: 25 → 90 °Ccp,c = 4.18 kJ/(kg·K)loss = 15 %
ṁc = kg/s
5.Ideal gas lawMedium
480 g of CH₄V = 50 LT = 20 °CP = ?

A tank of 50 L contains 480 g of CH₄ at 20 °C. Treat it as an ideal gas and find the pressure P.

m = 480 gM = 16.04 g/molV = 50 LT = 20 °CR = 8.314 J/(mol·K)
P = kPa
6.Ideal gas lawHard
State 1State 2

A fixed amount of CH₄ goes from 400 kPa, 10 L, 30 °C to 550 kPa and 100 °C. Find the new volume V2.

P1 = 400 kPaV1 = 10 LT1 = 30 °CP2 = 550 kPaT2 = 100 °C
V2 = L
7.StoichiometryMedium
m(X)n(X)n(H2​O)m(H2​O)÷ M(X)× 4/ν× M(H2​O)

16 g of C3H8 and 64 g of O2 react according to C3H8 + 5O2 → 3CO2 + 4H2O. What mass m of H2O can form?

M(C3H8) = 44.1M(O2) = 32M(H2O) = 18.02 g/mol
m = g
8.StoichiometryHard
m(H2​)n(H2​)n(H2​O)m(H2​O)÷ M(H2​)× 2/2× M(H2​O)

In the reaction 2H2 + O2 → 2H2O, 6 g of H2 reacts completely (the other reactants are in excess) and 41.83 g of H2O is collected. Find the percent yield Y.

M(H2) = 2.016 g/molM(H2O) = 18.02 g/molactual: 41.83 g
Y = %

3Solutions

1.Mass balance

Answer: ṁ2 = 650 kg/h

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

Balances

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

Solve for ṁ2

ṁ2 = ṁ1(x1 − x3)/(x3 − x2) = 650·(30 − 25)/(25 − 20) = 650 kg/h

Answer

ṁ2 = 650 kg/h
2.Mass balance

Answer: W = 1569 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 = 2000·14/65 = 430.8 kg/h

Total balance

W = F − P = 2000 − 430.8 = 1569 kg/h

Answer

W = 1569 kg/h
3.Energy balance

Answer: tout = 25.84 °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·(140 − 80) = 73.2 kW

Heat picked up by the water

tout = tin + Q/(ṁccp,c) = 20 + 73.2/(3·4.18) = 25.84 °C

Answer

tout = 25.84 °C
4.Energy balance

Answer: ṁc = 0.8447 kg/s

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

Heat released

Qh = 1.5·3.6·(140 − 90) = 270 kW

Heat that reaches the water

Qc = (1 − 0.15)·270 = 229.5 kW

Water flow rate

ṁc = Qc/(cp,c·Δt) = 229.5/(4.18·65) = 0.8447 kg/s

Answer

ṁc = 0.8447 kg/s
5.Ideal gas law

Answer: P = 1459 kPa

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

Moles

n = m/M = 480/16.04 = 29.93 mol
T = 20 + 273.15 = 293.1 K

Pressure

P = nRT/V = 29.93·8.314·293.1/0.05 = 1459 kPa

Answer

P = 1459 kPa
6.Ideal gas law

Answer: V2 = 8.952 L

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

Absolute temperatures

T1 = 303.1 K, T2 = 373.1 K

Combined gas law

V2 = V1·(P1/P2)·(T2/T1) = 10·(400/550)·(373.1/303.1) = 8.952 L

Answer

V2 = 8.952 L
7.Stoichiometry

Answer: m = 26.15 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(C3H8) = 16/44.1 = 0.3628 mol
n(O2) = 64/32 = 2 mol

Limiting reactant

0.3628/1 = 0.3628, 2/5 = 0.4 → C3H8

Mass of product

n(H2O) = 0.3628·4 = 1.451 mol
m = 1.451·18.02 = 26.15 g

Answer

m = 26.15 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(H2) = 6/2.016 = 2.976 mol
m = 2.976·2/2·18.02 = 53.63 g

Percent yield

Y = 41.83/53.63·100 = 78%

Answer

Y = 78%

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