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 11000 kg/h · 20% sugarStream 2ṁ₂ = ? · 35% sugarProduct25% sugar

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

ṁ1 = 1000 kg/hx1 = 20 %x2 = 35 %x3 = 25 %
ṁ2 = kg/h
2.Mass balanceHard
EvaporatorFeed F800 kg/h · 7% saltProduct PP = ? · 65%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 = 800 kg/hxF = 7 %xP = 65 %
P = kg/h
3.Energy balanceMedium
hot sidecold sideethanol · 120 °C90 °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 = 2.5 kg/scp,h = 2.44 kJ/(kg·K)Th: 120 → 90 °Cṁc = 1.5 kg/scp,c = 4.18 kJ/(kg·K)tin = 25 °C
tout = °C
4.Energy balanceHard
hot sidecold sideethanol · 180 °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 = 2.5 kg/scp,h = 2.44 kJ/(kg·K)Th: 180 → 110 °Ct: 20 → 50 °Ccp,c = 4.18 kJ/(kg·K)loss = 5 %
ṁc = kg/s
5.Ideal gas lawMedium
410 g of O₂P = 450 kPaT = 65 °CV = ?

What volume V does 410 g of O₂ occupy at 450 kPa and 65 °C? Treat it as an ideal gas.

m = 410 gM = 32 g/molP = 450 kPaT = 65 °CR = 8.314 J/(mol·K)
V = L
6.Ideal gas lawHard
State 1State 2

A fixed amount of He goes from 250 kPa, 2 L, 25 °C to 750 kPa and 160 °C. Find the new volume V2.

P1 = 250 kPaV1 = 2 LT1 = 25 °CP2 = 750 kPaT2 = 160 °C
V2 = L
7.StoichiometryMedium
m(X)n(X)n(NH3​)m(NH3​)÷ M(X)× 2/ν× M(NH3​)

10 g of N2 and 40 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(C6​H12​O6​)n(C6​H12​O6​)n(C2​H5​OH)m(C2​H5​OH)÷ M(C6​H12​O6​)× 2/1× M(C2​H5​OH)

In the reaction C6H12O6 → 2C2H5OH + 2CO2, 20 g of C6H12O6 reacts completely (the other reactants are in excess) and 9.61 g of C2H5OH is collected. Find the percent yield Y.

M(C6H12O6) = 180.16 g/molM(C2H5OH) = 46.07 g/molactual: 9.61 g
Y = %

3Solutions

1.Mass balance

Answer: ṁ2 = 500 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) = 1000·(20 − 25)/(25 − 35) = 500 kg/h

Answer

ṁ2 = 500 kg/h
2.Mass balance

Answer: P = 86.15 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 = 800·7/65 = 86.15 kg/h

Answer

P = 86.15 kg/h
3.Energy balance

Answer: tout = 54.19 °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 − 90) = 183 kW

Heat picked up by the water

tout = tin + Q/(ṁccp,c) = 25 + 183/(1.5·4.18) = 54.19 °C

Answer

tout = 54.19 °C
4.Energy balance

Answer: ṁc = 3.235 kg/s

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

Heat released

Qh = 2.5·2.44·(180 − 110) = 427 kW

Heat that reaches the water

Qc = (1 − 0.05)·427 = 405.6 kW

Water flow rate

ṁc = Qc/(cp,c·Δt) = 405.6/(4.18·30) = 3.235 kg/s

Answer

ṁc = 3.235 kg/s
5.Ideal gas law

Answer: V = 80.05 L

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

Moles

n = 410/32 = 12.81 mol
T = 65 + 273.15 = 338.1 K

Volume

V = nRT/P = 12.81·8.314·338.1/450 = 80.05 L

Answer

V = 80.05 L
6.Ideal gas law

Answer: V2 = 0.9685 L

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

Absolute temperatures

T1 = 298.1 K, T2 = 433.1 K

Combined gas law

V2 = V1·(P1/P2)·(T2/T1) = 2·(250/750)·(433.1/298.1) = 0.9685 L

Answer

V2 = 0.9685 L
7.Stoichiometry

Answer: m = 12.16 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) = 10/28.02 = 0.3569 mol
n(H2) = 40/2.016 = 19.84 mol

Limiting reactant

0.3569/1 = 0.3569, 19.84/3 = 6.614 → N2

Mass of product

n(NH3) = 0.3569·2 = 0.7138 mol
m = 0.7138·17.03 = 12.16 g

Answer

m = 12.16 g
8.Stoichiometry

Answer: Y = 93.95 %

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(C6H12O6) = 20/180.16 = 0.111 mol
m = 0.111·2/1·46.07 = 10.23 g

Percent yield

Y = 9.61/10.23·100 = 93.95%

Answer

Y = 93.95%

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