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 · 15% saltStream 2ṁ₂ = ? · 35% saltProduct20% salt

How much of stream 2 must be added each hour to stream 1 so that the blended product contains 20% salt? Find ṁ2.

ṁ1 = 600 kg/hx1 = 15 %x2 = 35 %x3 = 20 %
ṁ2 = kg/h
2.Mass balanceHard
EvaporatorFeed F1900 kg/h · 10% sugarProduct PP = ? · 35%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 = 1900 kg/hxF = 10 %xP = 35 %
P = kg/h
3.Energy balanceMedium
hot sidecold sideethanol · 120 °C60 °Cwater · 30 °Ct = ? °C

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

ṁh = 2.5 kg/scp,h = 2.44 kJ/(kg·K)Th: 120 → 60 °Cṁc = 1.5 kg/scp,c = 4.18 kJ/(kg·K)tin = 30 °C
tout = °C
4.Energy balanceHard
hot sidecold sideglycol mix · 140 °C60 °Cwater · 10 °C40 °C · ṁ = ?15% lost

Hot glycol mix 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 = 1 kg/scp,h = 3.6 kJ/(kg·K)Th: 140 → 60 °Ct: 10 → 40 °Ccp,c = 4.18 kJ/(kg·K)loss = 15 %
ṁc = kg/s
5.Ideal gas lawMedium
330 g of ArP = 100 kPaT = 30 °CV = ?

What volume V does 330 g of Ar occupy at 100 kPa and 30 °C? Treat it as an ideal gas.

m = 330 gM = 39.95 g/molP = 100 kPaT = 30 °CR = 8.314 J/(mol·K)
V = L
6.Ideal gas lawHard
State 1State 2

A fixed amount of CH₄ goes from 100 kPa, 5 L, 40 °C to 450 kPa and 160 °C. Find the new volume V2.

P1 = 100 kPaV1 = 5 LT1 = 40 °CP2 = 450 kPaT2 = 160 °C
V2 = L
7.StoichiometryMedium
m(X)n(X)n(CO2​)m(CO2​)÷ M(X)× 3/ν× M(CO2​)

20 g of C3H8 and 24 g of O2 react according to C3H8 + 5O2 → 3CO2 + 4H2O. What mass m of CO2 can form?

M(C3H8) = 44.1M(O2) = 32M(CO2) = 44.01 g/mol
m = g
8.StoichiometryHard
m(C3​H8​)n(C3​H8​)n(H2​O)m(H2​O)÷ M(C3​H8​)× 4/1× M(H2​O)

In the reaction C3H8 + 5O2 → 3CO2 + 4H2O, 20 g of C3H8 reacts completely (the other reactants are in excess) and 17.98 g of H2O is collected. Find the percent yield Y.

M(C3H8) = 44.1 g/molM(H2O) = 18.02 g/molactual: 17.98 g
Y = %

3Solutions

1.Mass balance

Answer: ṁ2 = 200 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) = 600·(15 − 20)/(20 − 35) = 200 kg/h

Answer

ṁ2 = 200 kg/h
2.Mass balance

Answer: P = 542.9 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 = 1900·10/35 = 542.9 kg/h

Answer

P = 542.9 kg/h
3.Energy balance

Answer: tout = 88.37 °C

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

Heat released by the hot stream

Q = 2.5·2.44·(120 − 60) = 366 kW

Heat picked up by the water

tout = tin + Q/(ṁccp,c) = 30 + 366/(1.5·4.18) = 88.37 °C

Answer

tout = 88.37 °C
4.Energy balance

Answer: ṁc = 1.952 kg/s

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

Heat released

Qh = 1·3.6·(140 − 60) = 288 kW

Heat that reaches the water

Qc = (1 − 0.15)·288 = 244.8 kW

Water flow rate

ṁc = Qc/(cp,c·Δt) = 244.8/(4.18·30) = 1.952 kg/s

Answer

ṁc = 1.952 kg/s
5.Ideal gas law

Answer: V = 208.2 L

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

Moles

n = 330/39.95 = 8.26 mol
T = 30 + 273.15 = 303.1 K

Volume

V = nRT/P = 8.26·8.314·303.1/100 = 208.2 L

Answer

V = 208.2 L
6.Ideal gas law

Answer: V2 = 1.537 L

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

Absolute temperatures

T1 = 313.1 K, T2 = 433.1 K

Combined gas law

V2 = V1·(P1/P2)·(T2/T1) = 5·(100/450)·(433.1/313.1) = 1.537 L

Answer

V2 = 1.537 L
7.Stoichiometry

Answer: m = 19.8 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(C3H8) = 20/44.1 = 0.4535 mol
n(O2) = 24/32 = 0.75 mol

Limiting reactant

0.4535/1 = 0.4535, 0.75/5 = 0.15 → O2

Mass of product

n(CO2) = 0.15·3 = 0.45 mol
m = 0.45·44.01 = 19.8 g

Answer

m = 19.8 g
8.Stoichiometry

Answer: Y = 55 %

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(C3H8) = 20/44.1 = 0.4535 mol
m = 0.4535·4/1·18.02 = 32.69 g

Percent yield

Y = 17.98/32.69·100 = 55%

Answer

Y = 55%

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