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 · 50% saltStream 2ṁ₂ = ? · 5% saltProduct25% salt

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

ṁ1 = 300 kg/hx1 = 50 %x2 = 5 %x3 = 25 %
ṁ2 = kg/h
2.Mass balanceHard
EvaporatorFeed F2600 kg/h · 11% sugarProduct P40% sugarVapour W (pure water)W = ? kg/h

An evaporator concentrates a sugar solution by boiling off pure water. Find the water evaporated W in kg/h.

F = 2600 kg/hxF = 11 %xP = 40 %
W = kg/h
3.Energy balanceMedium
hot sidecold sidelight oil · 160 °C70 °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 = 2 kg/scp,h = 2.1 kJ/(kg·K)Th: 160 → 70 °Cṁc = 1 kg/scp,c = 4.18 kJ/(kg·K)tin = 25 °C
tout = °C
4.Energy balanceHard
hot sidecold sideethanol · 130 °C90 °Cwater · 20 °C70 °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 = 1.5 kg/scp,h = 2.44 kJ/(kg·K)Th: 130 → 90 °Ct: 20 → 70 °Ccp,c = 4.18 kJ/(kg·K)loss = 5 %
ṁc = kg/s
5.Ideal gas lawMedium
460 g of ArP = 250 kPaT = 20 °CV = ?

What volume V does 460 g of Ar occupy at 250 kPa and 20 °C? Treat it as an ideal gas.

m = 460 gM = 39.95 g/molP = 250 kPaT = 20 °CR = 8.314 J/(mol·K)
V = L
6.Ideal gas lawHard
State 1State 2

A fixed amount of Ar goes from 350 kPa, 10 L, 15 °C to 450 kPa and 120 °C. Find the new volume V2.

P1 = 350 kPaV1 = 10 LT1 = 15 °CP2 = 450 kPaT2 = 120 °C
V2 = L
7.StoichiometryMedium
m(X)n(X)n(H2​O)m(H2​O)÷ M(X)× 4/ν× M(H2​O)

6 g of C3H8 and 40 g of O2 react according to C3H8 + 5O2 → 3CO2 + 4H2O. What mass m of H2O can form?

M(C3H8) = 44.1M(O2) = 32M(H2O) = 18.02 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, 20 g of Na reacts completely (the other reactants are in excess) and 31.52 g of NaCl is collected. Find the percent yield Y.

M(Na) = 22.99 g/molM(NaCl) = 58.44 g/molactual: 31.52 g
Y = %

3Solutions

1.Mass balance

Answer: ṁ2 = 375 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·(50 − 25)/(25 − 5) = 375 kg/h

Answer

ṁ2 = 375 kg/h
2.Mass balance

Answer: W = 1885 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 = 2600·11/40 = 715 kg/h

Total balance

W = F − P = 2600 − 715 = 1885 kg/h

Answer

W = 1885 kg/h
3.Energy balance

Answer: tout = 115.4 °C

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

Heat released by the hot stream

Q = 2·2.1·(160 − 70) = 378 kW

Heat picked up by the water

tout = tin + Q/(ṁccp,c) = 25 + 378/(1·4.18) = 115.4 °C

Answer

tout = 115.4 °C
4.Energy balance

Answer: ṁc = 0.6655 kg/s

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

Heat released

Qh = 1.5·2.44·(130 − 90) = 146.4 kW

Heat that reaches the water

Qc = (1 − 0.05)·146.4 = 139.1 kW

Water flow rate

ṁc = Qc/(cp,c·Δt) = 139.1/(4.18·50) = 0.6655 kg/s

Answer

ṁc = 0.6655 kg/s
5.Ideal gas law

Answer: V = 112.3 L

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

Moles

n = 460/39.95 = 11.51 mol
T = 20 + 273.15 = 293.1 K

Volume

V = nRT/P = 11.51·8.314·293.1/250 = 112.3 L

Answer

V = 112.3 L
6.Ideal gas law

Answer: V2 = 10.61 L

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

Absolute temperatures

T1 = 288.1 K, T2 = 393.1 K

Combined gas law

V2 = V1·(P1/P2)·(T2/T1) = 10·(350/450)·(393.1/288.1) = 10.61 L

Answer

V2 = 10.61 L
7.Stoichiometry

Answer: m = 9.807 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) = 6/44.1 = 0.1361 mol
n(O2) = 40/32 = 1.25 mol

Limiting reactant

0.1361/1 = 0.1361, 1.25/5 = 0.25 → C3H8

Mass of product

n(H2O) = 0.1361·4 = 0.5442 mol
m = 0.5442·18.02 = 9.807 g

Answer

m = 9.807 g
8.Stoichiometry

Answer: Y = 62 %

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) = 20/22.99 = 0.8699 mol
m = 0.8699·2/2·58.44 = 50.84 g

Percent yield

Y = 31.52/50.84·100 = 62%

Answer

Y = 62%

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