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 1300 kg/h · 30% saltStream 2ṁ₂ = ? · 10% saltProduct15% salt

How much of stream 2 must be added each hour to stream 1 so that the blended product contains 15% salt? Find ṁ2.

ṁ1 = 300 kg/hx1 = 30 %x2 = 10 %x3 = 15 %
ṁ2 = kg/h
2.Mass balanceHard
EvaporatorFeed F2000 kg/h · 13% saltProduct P50% 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 = 13 %xP = 50 %
W = kg/h
3.Energy balanceMedium
hot sidecold sideethanol · 170 °C130 °Cwater · 10 °Ct = ? °C

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

ṁh = 1.5 kg/scp,h = 2.44 kJ/(kg·K)Th: 170 → 130 °Cṁc = 0.5 kg/scp,c = 4.18 kJ/(kg·K)tin = 10 °C
tout = °C
4.Energy balanceHard
hot sidecold sideethanol · 120 °C90 °Cwater · 20 °C35 °C · ṁ = ?15% lost

Hot ethanol 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 = 0.5 kg/scp,h = 2.44 kJ/(kg·K)Th: 120 → 90 °Ct: 20 → 35 °Ccp,c = 4.18 kJ/(kg·K)loss = 15 %
ṁc = kg/s
5.Ideal gas lawMedium
340 g of HeP = 200 kPaT = 45 °CV = ?

What volume V does 340 g of He occupy at 200 kPa and 45 °C? Treat it as an ideal gas.

m = 340 gM = 4.003 g/molP = 200 kPaT = 45 °CR = 8.314 J/(mol·K)
V = L
6.Ideal gas lawHard
N₂, M = 28.01 g/molP = 450 kPaT = 190 °Cρ = ?

Find the density ρ of N₂ at 450 kPa and 190 °C, treating it as an ideal gas.

M = 28.01 g/molP = 450 kPaT = 190 °CR = 8.314 J/(mol·K)
ρ = kg/m³
7.StoichiometryMedium
m(X)n(X)n(NH3​)m(NH3​)÷ M(X)× 2/ν× M(NH3​)

24 g of N2 and 80 g of H2 react according to N2 + 3H2 → 2NH3. What mass m of NH3 can form?

M(N2) = 28.02M(H2) = 2.016M(NH3) = 17.03 g/mol
m = g
8.StoichiometryHard
m(CH4​)n(CH4​)n(H2​O)m(H2​O)÷ M(CH4​)× 2/1× M(H2​O)

In the reaction CH4 + 2O2 → CO2 + 2H2O, 30 g of CH4 reacts completely (the other reactants are in excess) and 57.3 g of H2O is collected. Find the percent yield Y.

M(CH4) = 16.04 g/molM(H2O) = 18.02 g/molactual: 57.3 g
Y = %

3Solutions

1.Mass balance

Answer: ṁ2 = 900 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) = 300·(30 − 15)/(15 − 10) = 900 kg/h

Answer

ṁ2 = 900 kg/h
2.Mass balance

Answer: W = 1480 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 = 2000·13/50 = 520 kg/h

Total balance

W = F − P = 2000 − 520 = 1480 kg/h

Answer

W = 1480 kg/h
3.Energy balance

Answer: tout = 80.05 °C

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

Heat released by the hot stream

Q = 1.5·2.44·(170 − 130) = 146.4 kW

Heat picked up by the water

tout = tin + Q/(ṁccp,c) = 10 + 146.4/(0.5·4.18) = 80.05 °C

Answer

tout = 80.05 °C
4.Energy balance

Answer: ṁc = 0.4962 kg/s

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

Heat released

Qh = 0.5·2.44·(120 − 90) = 36.6 kW

Heat that reaches the water

Qc = (1 − 0.15)·36.6 = 31.11 kW

Water flow rate

ṁc = Qc/(cp,c·Δt) = 31.11/(4.18·15) = 0.4962 kg/s

Answer

ṁc = 0.4962 kg/s
5.Ideal gas law

Answer: V = 1123 L

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

Moles

n = 340/4.003 = 84.94 mol
T = 45 + 273.15 = 318.1 K

Volume

V = nRT/P = 84.94·8.314·318.1/200 = 1123 L

Answer

V = 1123 L
6.Ideal gas law

Answer: ρ = 3.273 kg/m³

Ideal gas density: ρ = PM/(RT).

Rearrange the gas law

PV = (m/M)RT → ρ = m/V = PM/(RT)
T = 190 + 273.15 = 463.1 K

Density

ρ = 450·28.01/(8.314·463.1) = 3.273 kg/m³

Answer

ρ = 3.273 kg/m³
7.Stoichiometry

Answer: m = 29.17 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(N2) = 24/28.02 = 0.8565 mol
n(H2) = 80/2.016 = 39.68 mol

Limiting reactant

0.8565/1 = 0.8565, 39.68/3 = 13.23 → N2

Mass of product

n(NH3) = 0.8565·2 = 1.713 mol
m = 1.713·17.03 = 29.17 g

Answer

m = 29.17 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) = 30/16.04 = 1.87 mol
m = 1.87·2/1·18.02 = 67.41 g

Percent yield

Y = 57.3/67.41·100 = 85.01%

Answer

Y = 85.01%

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