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 1200 kg/h · 50% 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 = 200 kg/hx1 = 50 %x2 = 10 %x3 = 15 %
ṁ2 = kg/h
2.Mass balanceHard
EvaporatorFeed F3400 kg/h · 13% sugarProduct P60% 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 = 3400 kg/hxF = 13 %xP = 60 %
W = kg/h
3.Energy balanceMedium
hot sidecold sideethanol · 150 °C60 °Cwater · 25 °Ct = ? °C

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

ṁh = 1 kg/scp,h = 2.44 kJ/(kg·K)Th: 150 → 60 °Cṁc = 1.5 kg/scp,c = 4.18 kJ/(kg·K)tin = 25 °C
tout = °C
4.Energy balanceHard
hot sidecold sideethanol · 180 °C150 °Cwater · 20 °C80 °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 kg/scp,h = 2.44 kJ/(kg·K)Th: 180 → 150 °Ct: 20 → 80 °Ccp,c = 4.18 kJ/(kg·K)loss = 10 %
ṁc = kg/s
5.Ideal gas lawMedium
20 g of CH₄V = 5 LT = 80 °CP = ?

A tank of 5 L contains 20 g of CH₄ at 80 °C. Treat it as an ideal gas and find the pressure P.

m = 20 gM = 16.04 g/molV = 5 LT = 80 °CR = 8.314 J/(mol·K)
P = kPa
6.Ideal gas lawHard
State 1State 2

A fixed amount of CH₄ goes from 250 kPa, 5 L, 10 °C to 650 kPa and 220 °C. Find the new volume V2.

P1 = 250 kPaV1 = 5 LT1 = 10 °CP2 = 650 kPaT2 = 220 °C
V2 = L
7.StoichiometryMedium
m(X)n(X)n(H2​O)m(H2​O)÷ M(X)× 4/ν× M(H2​O)

20 g of C3H8 and 24 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(N2​)n(N2​)n(NH3​)m(NH3​)÷ M(N2​)× 2/1× M(NH3​)

In the reaction N2 + 3H2 → 2NH3, 16 g of N2 reacts completely (the other reactants are in excess) and 15.95 g of NH3 is collected. Find the percent yield Y.

M(N2) = 28.02 g/molM(NH3) = 17.03 g/molactual: 15.95 g
Y = %

3Solutions

1.Mass balance

Answer: ṁ2 = 1400 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) = 200·(50 − 15)/(15 − 10) = 1400 kg/h

Answer

ṁ2 = 1400 kg/h
2.Mass balance

Answer: W = 2663 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 = 3400·13/60 = 736.7 kg/h

Total balance

W = F − P = 3400 − 736.7 = 2663 kg/h

Answer

W = 2663 kg/h
3.Energy balance

Answer: tout = 60.02 °C

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

Heat released by the hot stream

Q = 1·2.44·(150 − 60) = 219.6 kW

Heat picked up by the water

tout = tin + Q/(ṁccp,c) = 25 + 219.6/(1.5·4.18) = 60.02 °C

Answer

tout = 60.02 °C
4.Energy balance

Answer: ṁc = 0.2627 kg/s

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

Heat released

Qh = 1·2.44·(180 − 150) = 73.2 kW

Heat that reaches the water

Qc = (1 − 0.1)·73.2 = 65.88 kW

Water flow rate

ṁc = Qc/(cp,c·Δt) = 65.88/(4.18·60) = 0.2627 kg/s

Answer

ṁc = 0.2627 kg/s
5.Ideal gas law

Answer: P = 732.2 kPa

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

Moles

n = m/M = 20/16.04 = 1.247 mol
T = 80 + 273.15 = 353.1 K

Pressure

P = nRT/V = 1.247·8.314·353.1/0.005 = 732.2 kPa

Answer

P = 732.2 kPa
6.Ideal gas law

Answer: V2 = 3.349 L

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

Absolute temperatures

T1 = 283.1 K, T2 = 493.1 K

Combined gas law

V2 = V1·(P1/P2)·(T2/T1) = 5·(250/650)·(493.1/283.1) = 3.349 L

Answer

V2 = 3.349 L
7.Stoichiometry

Answer: m = 10.81 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(H2O) = 0.15·4 = 0.6 mol
m = 0.6·18.02 = 10.81 g

Answer

m = 10.81 g
8.Stoichiometry

Answer: Y = 82.01 %

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(N2) = 16/28.02 = 0.571 mol
m = 0.571·2/1·17.03 = 19.45 g

Percent yield

Y = 15.95/19.45·100 = 82.01%

Answer

Y = 82.01%

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