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 · 50% NaOHStream 2ṁ₂ = ? · 15% NaOHProduct30% NaOH

How much of stream 2 must be added each hour to stream 1 so that the blended product contains 30% NaOH? Find ṁ2.

ṁ1 = 650 kg/hx1 = 50 %x2 = 15 %x3 = 30 %
ṁ2 = kg/h
2.Mass balanceHard
EvaporatorFeed F3100 kg/h · 19% sugarProduct PP = ? · 45%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 = 3100 kg/hxF = 19 %xP = 45 %
P = kg/h
3.Energy balanceMedium
hot sidecold sidelight oil · 180 °C130 °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 = 0.5 kg/scp,h = 2.1 kJ/(kg·K)Th: 180 → 130 °Cṁc = 1 kg/scp,c = 4.18 kJ/(kg·K)tin = 25 °C
tout = °C
4.Energy balanceHard
hot sidecold sideethanol · 140 °C60 °Cwater · 30 °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 = 3 kg/scp,h = 2.44 kJ/(kg·K)Th: 140 → 60 °Ct: 30 → 75 °Ccp,c = 4.18 kJ/(kg·K)loss = 5 %
ṁc = kg/s
5.Ideal gas lawMedium
10 g of HeV = 30 LT = 45 °CP = ?

A tank of 30 L contains 10 g of He at 45 °C. Treat it as an ideal gas and find the pressure P.

m = 10 gM = 4.003 g/molV = 30 LT = 45 °CR = 8.314 J/(mol·K)
P = kPa
6.Ideal gas lawHard
O₂, M = 32 g/molP = 350 kPaT = 50 °Cρ = ?

Find the density ρ of O₂ at 350 kPa and 50 °C, treating it as an ideal gas.

M = 32 g/molP = 350 kPaT = 50 °CR = 8.314 J/(mol·K)
ρ = kg/m³
7.StoichiometryMedium
m(X)n(X)n(MgO)m(MgO)÷ M(X)× 2/ν× M(MgO)

8 g of Mg and 32 g of O2 react according to 2Mg + O2 → 2MgO. What mass m of MgO can form?

M(Mg) = 24.31M(O2) = 32M(MgO) = 40.3 g/mol
m = g
8.StoichiometryHard
m(Na)n(Na)n(NaCl)m(NaCl)÷ M(Na)× 2/2× M(NaCl)

In the reaction 2Na + Cl2 → 2NaCl, 10 g of Na reacts completely (the other reactants are in excess) and 23.13 g of NaCl is collected. Find the percent yield Y.

M(Na) = 22.99 g/molM(NaCl) = 58.44 g/molactual: 23.13 g
Y = %

3Solutions

1.Mass balance

Answer: ṁ2 = 866.7 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·(50 − 30)/(30 − 15) = 866.7 kg/h

Answer

ṁ2 = 866.7 kg/h
2.Mass balance

Answer: P = 1309 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 = 3100·19/45 = 1309 kg/h

Answer

P = 1309 kg/h
3.Energy balance

Answer: tout = 37.56 °C

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

Heat released by the hot stream

Q = 0.5·2.1·(180 − 130) = 52.5 kW

Heat picked up by the water

tout = tin + Q/(ṁccp,c) = 25 + 52.5/(1·4.18) = 37.56 °C

Answer

tout = 37.56 °C
4.Energy balance

Answer: ṁc = 2.958 kg/s

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

Heat released

Qh = 3·2.44·(140 − 60) = 585.6 kW

Heat that reaches the water

Qc = (1 − 0.05)·585.6 = 556.3 kW

Water flow rate

ṁc = Qc/(cp,c·Δt) = 556.3/(4.18·45) = 2.958 kg/s

Answer

ṁc = 2.958 kg/s
5.Ideal gas law

Answer: P = 220.3 kPa

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

Moles

n = m/M = 10/4.003 = 2.498 mol
T = 45 + 273.15 = 318.1 K

Pressure

P = nRT/V = 2.498·8.314·318.1/0.03 = 220.3 kPa

Answer

P = 220.3 kPa
6.Ideal gas law

Answer: ρ = 4.169 kg/m³

Ideal gas density: ρ = PM/(RT).

Rearrange the gas law

PV = (m/M)RT → ρ = m/V = PM/(RT)
T = 50 + 273.15 = 323.1 K

Density

ρ = 350·32/(8.314·323.1) = 4.169 kg/m³

Answer

ρ = 4.169 kg/m³
7.Stoichiometry

Answer: m = 13.26 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(Mg) = 8/24.31 = 0.3291 mol
n(O2) = 32/32 = 1 mol

Limiting reactant

0.3291/2 = 0.1645, 1/1 = 1 → Mg

Mass of product

n(MgO) = 0.1645·2 = 0.3291 mol
m = 0.3291·40.3 = 13.26 g

Answer

m = 13.26 g
8.Stoichiometry

Answer: Y = 90.99 %

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(Na) = 10/22.99 = 0.435 mol
m = 0.435·2/2·58.44 = 25.42 g

Percent yield

Y = 23.13/25.42·100 = 90.99%

Answer

Y = 90.99%

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