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 1950 kg/h · 15% sugarStream 2ṁ₂ = ? · 40% sugarProduct20% sugar

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

ṁ1 = 950 kg/hx1 = 15 %x2 = 40 %x3 = 20 %
ṁ2 = kg/h
2.Mass balanceHard
EvaporatorFeed F4900 kg/h · 12% NaOHProduct P50% NaOHVapour W (pure water)W = ? kg/h

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

F = 4900 kg/hxF = 12 %xP = 50 %
W = kg/h
3.Energy balanceMedium
hot sidecold sidelight oil · 150 °C80 °Cwater · 20 °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: 150 → 80 °Cṁc = 2 kg/scp,c = 4.18 kJ/(kg·K)tin = 20 °C
tout = °C
4.Energy balanceHard
hot sidecold sideglycol mix · 120 °C70 °Cwater · 30 °C80 °C · ṁ = ?10% lost

Hot glycol mix 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 = 2.5 kg/scp,h = 3.6 kJ/(kg·K)Th: 120 → 70 °Ct: 30 → 80 °Ccp,c = 4.18 kJ/(kg·K)loss = 10 %
ṁc = kg/s
5.Ideal gas lawMedium
260 g of HeV = 10 LT = 75 °CP = ?

A tank of 10 L contains 260 g of He at 75 °C. Treat it as an ideal gas and find the pressure P.

m = 260 gM = 4.003 g/molV = 10 LT = 75 °CR = 8.314 J/(mol·K)
P = kPa
6.Ideal gas lawHard
State 1State 2

A fixed amount of CO₂ goes from 150 kPa, 20 L, 35 °C to 200 kPa and 170 °C. Find the new volume V2.

P1 = 150 kPaV1 = 20 LT1 = 35 °CP2 = 200 kPaT2 = 170 °C
V2 = L
7.StoichiometryMedium
m(X)n(X)n(CO2​)m(CO2​)÷ M(X)× 1/ν× M(CO2​)

6 g of CH4 and 40 g of O2 react according to CH4 + 2O2 → CO2 + 2H2O. What mass m of CO2 can form?

M(CH4) = 16.04M(O2) = 32M(CO2) = 44.01 g/mol
m = g
8.StoichiometryHard
m(CaCO3​)n(CaCO3​)n(CO2​)m(CO2​)÷ M(CaCO3​)× 1/1× M(CO2​)

In the reaction CaCO3 → CaO + CO2, 12 g of CaCO3 reacts completely (the other reactants are in excess) and 3.27 g of CO2 is collected. Find the percent yield Y.

M(CaCO3) = 100.09 g/molM(CO2) = 44.01 g/molactual: 3.27 g
Y = %

3Solutions

1.Mass balance

Answer: ṁ2 = 237.5 kg/h

Steady-state mass balance: in = out for the total and for each component. Blending problems combine both.

Balances

Total: ṁ3 = ṁ1 + ṁ2
sugar: ṁ1x1 + ṁ2x2 = (ṁ1 + ṁ2)x3

Solve for ṁ2

ṁ2 = ṁ1(x1 − x3)/(x3 − x2) = 950·(15 − 20)/(20 − 40) = 237.5 kg/h

Answer

ṁ2 = 237.5 kg/h
2.Mass balance

Answer: W = 3724 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 = 4900·12/50 = 1176 kg/h

Total balance

W = F − P = 4900 − 1176 = 3724 kg/h

Answer

W = 3724 kg/h
3.Energy balance

Answer: tout = 55.17 °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·(150 − 80) = 294 kW

Heat picked up by the water

tout = tin + Q/(ṁccp,c) = 20 + 294/(2·4.18) = 55.17 °C

Answer

tout = 55.17 °C
4.Energy balance

Answer: ṁc = 1.938 kg/s

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

Heat released

Qh = 2.5·3.6·(120 − 70) = 450 kW

Heat that reaches the water

Qc = (1 − 0.1)·450 = 405 kW

Water flow rate

ṁc = Qc/(cp,c·Δt) = 405/(4.18·50) = 1.938 kg/s

Answer

ṁc = 1.938 kg/s
5.Ideal gas law

Answer: P = 18800 kPa

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

Moles

n = m/M = 260/4.003 = 64.95 mol
T = 75 + 273.15 = 348.1 K

Pressure

P = nRT/V = 64.95·8.314·348.1/0.01 = 18800 kPa

Answer

P = 18800 kPa
6.Ideal gas law

Answer: V2 = 21.57 L

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

Absolute temperatures

T1 = 308.1 K, T2 = 443.1 K

Combined gas law

V2 = V1·(P1/P2)·(T2/T1) = 20·(150/200)·(443.1/308.1) = 21.57 L

Answer

V2 = 21.57 L
7.Stoichiometry

Answer: m = 16.46 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(CH4) = 6/16.04 = 0.3741 mol
n(O2) = 40/32 = 1.25 mol

Limiting reactant

0.3741/1 = 0.3741, 1.25/2 = 0.625 → CH4

Mass of product

n(CO2) = 0.3741·1 = 0.3741 mol
m = 0.3741·44.01 = 16.46 g

Answer

m = 16.46 g
8.Stoichiometry

Answer: Y = 61.97 %

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(CaCO3) = 12/100.09 = 0.1199 mol
m = 0.1199·1/1·44.01 = 5.276 g

Percent yield

Y = 3.27/5.276·100 = 61.97%

Answer

Y = 61.97%

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