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 1400 kg/h · 15% saltStream 2ṁ₂ = ? · 35% saltProduct30% salt

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

ṁ1 = 400 kg/hx1 = 15 %x2 = 35 %x3 = 30 %
ṁ2 = kg/h
2.Mass balanceHard
EvaporatorFeed F800 kg/h · 11% saltProduct PP = ? · 35%Vapour W (pure water)0% salt

An evaporator concentrates a salt solution by boiling off pure water. Find the product flow P in kg/h.

F = 800 kg/hxF = 11 %xP = 35 %
P = kg/h
3.Energy balanceMedium
hot sidecold sideglycol mix · 120 °C90 °Cwater · 20 °Ct = ? °C

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

ṁh = 2 kg/scp,h = 3.6 kJ/(kg·K)Th: 120 → 90 °Cṁc = 1 kg/scp,c = 4.18 kJ/(kg·K)tin = 20 °C
tout = °C
4.Energy balanceHard
hot sidecold sideethanol · 140 °C80 °Cwater · 20 °C45 °C · ṁ = ?10% lost

Hot ethanol 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 = 2.44 kJ/(kg·K)Th: 140 → 80 °Ct: 20 → 45 °Ccp,c = 4.18 kJ/(kg·K)loss = 10 %
ṁc = kg/s
5.Ideal gas lawMedium
150 g of N₂V = 50 LT = 45 °CP = ?

A tank of 50 L contains 150 g of N₂ at 45 °C. Treat it as an ideal gas and find the pressure P.

m = 150 gM = 28.01 g/molV = 50 LT = 45 °CR = 8.314 J/(mol·K)
P = kPa
6.Ideal gas lawHard
N₂, M = 28.01 g/molP = 200 kPaT = 80 °Cρ = ?

Find the density ρ of N₂ at 200 kPa and 80 °C, treating it as an ideal gas.

M = 28.01 g/molP = 200 kPaT = 80 °CR = 8.314 J/(mol·K)
ρ = kg/m³
7.StoichiometryMedium
m(X)n(X)n(H2​O)m(H2​O)÷ M(X)× 2/ν× M(H2​O)

20 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(CaCO3​)n(CaCO3​)n(CaO)m(CaO)÷ M(CaCO3​)× 1/1× M(CaO)

In the reaction CaCO3 → CaO + CO2, 24 g of CaCO3 reacts completely (the other reactants are in excess) and 11.83 g of CaO is collected. Find the percent yield Y.

M(CaCO3) = 100.09 g/molM(CaO) = 56.08 g/molactual: 11.83 g
Y = %

3Solutions

1.Mass balance

Answer: ṁ2 = 1200 kg/h

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

Balances

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

Solve for ṁ2

ṁ2 = ṁ1(x1 − x3)/(x3 − x2) = 400·(15 − 30)/(30 − 35) = 1200 kg/h

Answer

ṁ2 = 1200 kg/h
2.Mass balance

Answer: P = 251.4 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 = 800·11/35 = 251.4 kg/h

Answer

P = 251.4 kg/h
3.Energy balance

Answer: tout = 71.67 °C

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

Heat released by the hot stream

Q = 2·3.6·(120 − 90) = 216 kW

Heat picked up by the water

tout = tin + Q/(ṁccp,c) = 20 + 216/(1·4.18) = 71.67 °C

Answer

tout = 71.67 °C
4.Energy balance

Answer: ṁc = 1.891 kg/s

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

Heat released

Qh = 1.5·2.44·(140 − 80) = 219.6 kW

Heat that reaches the water

Qc = (1 − 0.1)·219.6 = 197.6 kW

Water flow rate

ṁc = Qc/(cp,c·Δt) = 197.6/(4.18·25) = 1.891 kg/s

Answer

ṁc = 1.891 kg/s
5.Ideal gas law

Answer: P = 283.3 kPa

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

Moles

n = m/M = 150/28.01 = 5.355 mol
T = 45 + 273.15 = 318.1 K

Pressure

P = nRT/V = 5.355·8.314·318.1/0.05 = 283.3 kPa

Answer

P = 283.3 kPa
6.Ideal gas law

Answer: ρ = 1.908 kg/m³

Ideal gas density: ρ = PM/(RT).

Rearrange the gas law

PV = (m/M)RT → ρ = m/V = PM/(RT)
T = 80 + 273.15 = 353.1 K

Density

ρ = 200·28.01/(8.314·353.1) = 1.908 kg/m³

Answer

ρ = 1.908 kg/m³
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) = 20/16.04 = 1.247 mol
n(O2) = 20/32 = 0.625 mol

Limiting reactant

1.247/1 = 1.247, 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 = 87.97 %

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(CaCO3) = 24/100.09 = 0.2398 mol
m = 0.2398·1/1·56.08 = 13.45 g

Percent yield

Y = 11.83/13.45·100 = 87.97%

Answer

Y = 87.97%

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