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 1600 kg/h · 5% NaOHStream 2ṁ₂ = ? · 35% NaOHProduct25% NaOH

How much of stream 2 must be added each hour to stream 1 so that the blended product contains 25% NaOH? Find ṁ2.

ṁ1 = 600 kg/hx1 = 5 %x2 = 35 %x3 = 25 %
ṁ2 = kg/h
2.Mass balanceHard
EvaporatorFeed F1300 kg/h · 13% saltProduct PP = ? · 45%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 = 1300 kg/hxF = 13 %xP = 45 %
P = kg/h
3.Energy balanceMedium
hot sidecold sideethanol · 150 °C80 °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: 150 → 80 °Cṁc = 2 kg/scp,c = 4.18 kJ/(kg·K)tin = 10 °C
tout = °C
4.Energy balanceHard
hot sidecold sideethanol · 160 °C110 °Cwater · 20 °C50 °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 = 2 kg/scp,h = 2.44 kJ/(kg·K)Th: 160 → 110 °Ct: 20 → 50 °Ccp,c = 4.18 kJ/(kg·K)loss = 5 %
ṁc = kg/s
5.Ideal gas lawMedium
270 g of ArV = 10 LT = 90 °CP = ?

A tank of 10 L contains 270 g of Ar at 90 °C. Treat it as an ideal gas and find the pressure P.

m = 270 gM = 39.95 g/molV = 10 LT = 90 °CR = 8.314 J/(mol·K)
P = kPa
6.Ideal gas lawHard
State 1State 2

A fixed amount of He goes from 200 kPa, 20 L, 20 °C to 700 kPa and 100 °C. Find the new volume V2.

P1 = 200 kPaV1 = 20 LT1 = 20 °CP2 = 700 kPaT2 = 100 °C
V2 = L
7.StoichiometryMedium
m(X)n(X)n(NH3​)m(NH3​)÷ M(X)× 2/ν× M(NH3​)

20 g of N2 and 10 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(Fe2​O3​)n(Fe2​O3​)n(CO2​)m(CO2​)÷ M(Fe2​O3​)× 3/1× M(CO2​)

In the reaction Fe2O3 + 3CO → 2Fe + 3CO2, 16 g of Fe2O3 reacts completely (the other reactants are in excess) and 9.26 g of CO2 is collected. Find the percent yield Y.

M(Fe2O3) = 159.69 g/molM(CO2) = 44.01 g/molactual: 9.26 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
NaOH: ṁ1x1 + ṁ2x2 = (ṁ1 + ṁ2)x3

Solve for ṁ2

ṁ2 = ṁ1(x1 − x3)/(x3 − x2) = 600·(5 − 25)/(25 − 35) = 1200 kg/h

Answer

ṁ2 = 1200 kg/h
2.Mass balance

Answer: P = 375.6 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 = 1300·13/45 = 375.6 kg/h

Answer

P = 375.6 kg/h
3.Energy balance

Answer: tout = 40.65 °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·(150 − 80) = 256.2 kW

Heat picked up by the water

tout = tin + Q/(ṁccp,c) = 10 + 256.2/(2·4.18) = 40.65 °C

Answer

tout = 40.65 °C
4.Energy balance

Answer: ṁc = 1.848 kg/s

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

Heat released

Qh = 2·2.44·(160 − 110) = 244 kW

Heat that reaches the water

Qc = (1 − 0.05)·244 = 231.8 kW

Water flow rate

ṁc = Qc/(cp,c·Δt) = 231.8/(4.18·30) = 1.848 kg/s

Answer

ṁc = 1.848 kg/s
5.Ideal gas law

Answer: P = 2041 kPa

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

Moles

n = m/M = 270/39.95 = 6.758 mol
T = 90 + 273.15 = 363.1 K

Pressure

P = nRT/V = 6.758·8.314·363.1/0.01 = 2041 kPa

Answer

P = 2041 kPa
6.Ideal gas law

Answer: V2 = 7.274 L

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

Absolute temperatures

T1 = 293.1 K, T2 = 373.1 K

Combined gas law

V2 = V1·(P1/P2)·(T2/T1) = 20·(200/700)·(373.1/293.1) = 7.274 L

Answer

V2 = 7.274 L
7.Stoichiometry

Answer: m = 24.31 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) = 20/28.02 = 0.7138 mol
n(H2) = 10/2.016 = 4.96 mol

Limiting reactant

0.7138/1 = 0.7138, 4.96/3 = 1.653 → N2

Mass of product

n(NH3) = 0.7138·2 = 1.428 mol
m = 1.428·17.03 = 24.31 g

Answer

m = 24.31 g
8.Stoichiometry

Answer: Y = 70 %

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(Fe2O3) = 16/159.69 = 0.1002 mol
m = 0.1002·3/1·44.01 = 13.23 g

Percent yield

Y = 9.26/13.23·100 = 70%

Answer

Y = 70%

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