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 1700 kg/h · 15% sugarStream 2ṁ₂ = ? · 30% 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 = 700 kg/hx1 = 15 %x2 = 30 %x3 = 20 %
ṁ2 = kg/h
2.Mass balanceHard
EvaporatorFeed F1300 kg/h · 18% NaOHProduct P55% NaOHVapour W (pure water)W = ? kg/h

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

F = 1300 kg/hxF = 18 %xP = 55 %
W = kg/h
3.Energy balanceMedium
hot sidecold sidelight oil · 180 °C80 °Cwater · 20 °Ct = ? °C

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

ṁh = 2.5 kg/scp,h = 2.1 kJ/(kg·K)Th: 180 → 80 °Cṁc = 2 kg/scp,c = 4.18 kJ/(kg·K)tin = 20 °C
tout = °C
4.Energy balanceHard
hot sidecold sidelight oil · 130 °C70 °Cwater · 10 °C35 °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 = 2.5 kg/scp,h = 2.1 kJ/(kg·K)Th: 130 → 70 °Ct: 10 → 35 °Ccp,c = 4.18 kJ/(kg·K)loss = 10 %
ṁc = kg/s
5.Ideal gas lawMedium
310 g of N₂V = 50 LT = 85 °CP = ?

A tank of 50 L contains 310 g of N₂ at 85 °C. Treat it as an ideal gas and find the pressure P.

m = 310 gM = 28.01 g/molV = 50 LT = 85 °CR = 8.314 J/(mol·K)
P = kPa
6.Ideal gas lawHard
State 1State 2

A fixed amount of He goes from 300 kPa, 10 L, 40 °C to 300 kPa and 60 °C. Find the new volume V2.

P1 = 300 kPaV1 = 10 LT1 = 40 °CP2 = 300 kPaT2 = 60 °C
V2 = L
7.StoichiometryMedium
m(X)n(X)n(H2​O)m(H2​O)÷ M(X)× 2/ν× M(H2​O)

24 g of H2 and 8 g of O2 react according to 2H2 + O2 → 2H2O. What mass m of H2O can form?

M(H2) = 2.016M(O2) = 32M(H2O) = 18.02 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, 8 g of Al reacts completely (the other reactants are in excess) and 32.42 g of AlCl3 is collected. Find the percent yield Y.

M(Al) = 26.98 g/molM(AlCl3) = 133.33 g/molactual: 32.42 g
Y = %

3Solutions

1.Mass balance

Answer: ṁ2 = 350 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) = 700·(15 − 20)/(20 − 30) = 350 kg/h

Answer

ṁ2 = 350 kg/h
2.Mass balance

Answer: W = 874.5 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 = 1300·18/55 = 425.5 kg/h

Total balance

W = F − P = 1300 − 425.5 = 874.5 kg/h

Answer

W = 874.5 kg/h
3.Energy balance

Answer: tout = 82.8 °C

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

Heat released by the hot stream

Q = 2.5·2.1·(180 − 80) = 525 kW

Heat picked up by the water

tout = tin + Q/(ṁccp,c) = 20 + 525/(2·4.18) = 82.8 °C

Answer

tout = 82.8 °C
4.Energy balance

Answer: ṁc = 2.713 kg/s

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

Heat released

Qh = 2.5·2.1·(130 − 70) = 315 kW

Heat that reaches the water

Qc = (1 − 0.1)·315 = 283.5 kW

Water flow rate

ṁc = Qc/(cp,c·Δt) = 283.5/(4.18·25) = 2.713 kg/s

Answer

ṁc = 2.713 kg/s
5.Ideal gas law

Answer: P = 659.1 kPa

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

Moles

n = m/M = 310/28.01 = 11.07 mol
T = 85 + 273.15 = 358.1 K

Pressure

P = nRT/V = 11.07·8.314·358.1/0.05 = 659.1 kPa

Answer

P = 659.1 kPa
6.Ideal gas law

Answer: V2 = 10.64 L

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

Absolute temperatures

T1 = 313.1 K, T2 = 333.1 K

Combined gas law

V2 = V1·(P1/P2)·(T2/T1) = 10·(300/300)·(333.1/313.1) = 10.64 L

Answer

V2 = 10.64 L
7.Stoichiometry

Answer: m = 9.01 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(H2) = 24/2.016 = 11.9 mol
n(O2) = 8/32 = 0.25 mol

Limiting reactant

11.9/2 = 5.952, 0.25/1 = 0.25 → O2

Mass of product

n(H2O) = 0.25·2 = 0.5 mol
m = 0.5·18.02 = 9.01 g

Answer

m = 9.01 g
8.Stoichiometry

Answer: Y = 82 %

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) = 8/26.98 = 0.2965 mol
m = 0.2965·2/2·133.33 = 39.53 g

Percent yield

Y = 32.42/39.53·100 = 82%

Answer

Y = 82%

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