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 · 10% NaOHStream 2ṁ₂ = ? · 50% NaOHProduct35% NaOH

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

ṁ1 = 650 kg/hx1 = 10 %x2 = 50 %x3 = 35 %
ṁ2 = kg/h
2.Mass balanceHard
EvaporatorFeed F3600 kg/h · 16% saltProduct P35% 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 = 3600 kg/hxF = 16 %xP = 35 %
W = kg/h
3.Energy balanceMedium
hot sidecold sidelight oil · 140 °C100 °Cwater · 10 °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: 140 → 100 °Cṁc = 1.5 kg/scp,c = 4.18 kJ/(kg·K)tin = 10 °C
tout = °C
4.Energy balanceHard
hot sidecold sidelight oil · 150 °C120 °Cwater · 25 °C60 °C · ṁ = ?10% lost

Hot light oil 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 = 1.5 kg/scp,h = 2.1 kJ/(kg·K)Th: 150 → 120 °Ct: 25 → 60 °Ccp,c = 4.18 kJ/(kg·K)loss = 10 %
ṁc = kg/s
5.Ideal gas lawMedium
250 g of CH₄V = 20 LT = 80 °CP = ?

A tank of 20 L contains 250 g of CH₄ at 80 °C. Treat it as an ideal gas and find the pressure P.

m = 250 gM = 16.04 g/molV = 20 LT = 80 °CR = 8.314 J/(mol·K)
P = kPa
6.Ideal gas lawHard
State 1State 2

A fixed amount of He goes from 150 kPa, 20 L, 30 °C to 550 kPa and 130 °C. Find the new volume V2.

P1 = 150 kPaV1 = 20 LT1 = 30 °CP2 = 550 kPaT2 = 130 °C
V2 = L
7.StoichiometryMedium
m(X)n(X)n(NaCl)m(NaCl)÷ M(X)× 2/ν× M(NaCl)

10 g of Na and 24 g of Cl2 react according to 2Na + Cl2 → 2NaCl. What mass m of NaCl can form?

M(Na) = 22.99M(Cl2) = 70.9M(NaCl) = 58.44 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, 12 g of CH4 reacts completely (the other reactants are in excess) and 27.99 g of CO2 is collected. Find the percent yield Y.

M(CH4) = 16.04 g/molM(CO2) = 44.01 g/molactual: 27.99 g
Y = %

3Solutions

1.Mass balance

Answer: ṁ2 = 1083 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·(10 − 35)/(35 − 50) = 1083 kg/h

Answer

ṁ2 = 1083 kg/h
2.Mass balance

Answer: W = 1954 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 = 3600·16/35 = 1646 kg/h

Total balance

W = F − P = 3600 − 1646 = 1954 kg/h

Answer

W = 1954 kg/h
3.Energy balance

Answer: tout = 30.1 °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·(140 − 100) = 126 kW

Heat picked up by the water

tout = tin + Q/(ṁccp,c) = 10 + 126/(1.5·4.18) = 30.1 °C

Answer

tout = 30.1 °C
4.Energy balance

Answer: ṁc = 0.5813 kg/s

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

Heat released

Qh = 1.5·2.1·(150 − 120) = 94.5 kW

Heat that reaches the water

Qc = (1 − 0.1)·94.5 = 85.05 kW

Water flow rate

ṁc = Qc/(cp,c·Δt) = 85.05/(4.18·35) = 0.5813 kg/s

Answer

ṁc = 0.5813 kg/s
5.Ideal gas law

Answer: P = 2288 kPa

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

Moles

n = m/M = 250/16.04 = 15.59 mol
T = 80 + 273.15 = 353.1 K

Pressure

P = nRT/V = 15.59·8.314·353.1/0.02 = 2288 kPa

Answer

P = 2288 kPa
6.Ideal gas law

Answer: V2 = 7.254 L

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

Absolute temperatures

T1 = 303.1 K, T2 = 403.1 K

Combined gas law

V2 = V1·(P1/P2)·(T2/T1) = 20·(150/550)·(403.1/303.1) = 7.254 L

Answer

V2 = 7.254 L
7.Stoichiometry

Answer: m = 25.42 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(Na) = 10/22.99 = 0.435 mol
n(Cl2) = 24/70.9 = 0.3385 mol

Limiting reactant

0.435/2 = 0.2175, 0.3385/1 = 0.3385 → Na

Mass of product

n(NaCl) = 0.2175·2 = 0.435 mol
m = 0.435·58.44 = 25.42 g

Answer

m = 25.42 g
8.Stoichiometry

Answer: Y = 85.01 %

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 = 0.7481·1/1·44.01 = 32.93 g

Percent yield

Y = 27.99/32.93·100 = 85.01%

Answer

Y = 85.01%

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