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 1750 kg/h · 30% sugarStream 2ṁ₂ = ? · 20% 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 = 750 kg/hx1 = 30 %x2 = 20 %x3 = 25 %
ṁ2 = kg/h
2.Mass balanceHard
EvaporatorFeed F4500 kg/h · 15% saltProduct P35% saltVapour W (pure water)W = ? kg/h

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

F = 4500 kg/hxF = 15 %xP = 35 %
W = kg/h
3.Energy balanceMedium
hot sidecold sideethanol · 180 °C120 °Cwater · 20 °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: 180 → 120 °Cṁc = 0.5 kg/scp,c = 4.18 kJ/(kg·K)tin = 20 °C
tout = °C
4.Energy balanceHard
hot sidecold sideethanol · 170 °C110 °Cwater · 15 °C30 °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 = 3 kg/scp,h = 2.44 kJ/(kg·K)Th: 170 → 110 °Ct: 15 → 30 °Ccp,c = 4.18 kJ/(kg·K)loss = 5 %
ṁc = kg/s
5.Ideal gas lawMedium
410 g of HeP = 200 kPaT = 95 °CV = ?

What volume V does 410 g of He occupy at 200 kPa and 95 °C? Treat it as an ideal gas.

m = 410 gM = 4.003 g/molP = 200 kPaT = 95 °CR = 8.314 J/(mol·K)
V = L
6.Ideal gas lawHard
State 1State 2

A fixed amount of CH₄ goes from 250 kPa, 2 L, 15 °C to 800 kPa and 240 °C. Find the new volume V2.

P1 = 250 kPaV1 = 2 LT1 = 15 °CP2 = 800 kPaT2 = 240 °C
V2 = L
7.StoichiometryMedium
m(X)n(X)n(H2​O)m(H2​O)÷ M(X)× 2/ν× M(H2​O)

20 g of H2 and 80 g of O2 react according to 2H2 + O2 → 2H2O. What mass m of H2O can form?

M(H2) = 2.016M(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, 6 g of Na reacts completely (the other reactants are in excess) and 13.88 g of NaCl is collected. Find the percent yield Y.

M(Na) = 22.99 g/molM(NaCl) = 58.44 g/molactual: 13.88 g
Y = %

3Solutions

1.Mass balance

Answer: ṁ2 = 750 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) = 750·(30 − 25)/(25 − 20) = 750 kg/h

Answer

ṁ2 = 750 kg/h
2.Mass balance

Answer: W = 2571 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 = 4500·15/35 = 1929 kg/h

Total balance

W = F − P = 4500 − 1929 = 2571 kg/h

Answer

W = 2571 kg/h
3.Energy balance

Answer: tout = 90.05 °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·(180 − 120) = 146.4 kW

Heat picked up by the water

tout = tin + Q/(ṁccp,c) = 20 + 146.4/(0.5·4.18) = 90.05 °C

Answer

tout = 90.05 °C
4.Energy balance

Answer: ṁc = 6.655 kg/s

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

Heat released

Qh = 3·2.44·(170 − 110) = 439.2 kW

Heat that reaches the water

Qc = (1 − 0.05)·439.2 = 417.2 kW

Water flow rate

ṁc = Qc/(cp,c·Δt) = 417.2/(4.18·15) = 6.655 kg/s

Answer

ṁc = 6.655 kg/s
5.Ideal gas law

Answer: V = 1567 L

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

Moles

n = 410/4.003 = 102.4 mol
T = 95 + 273.15 = 368.1 K

Volume

V = nRT/P = 102.4·8.314·368.1/200 = 1567 L

Answer

V = 1567 L
6.Ideal gas law

Answer: V2 = 1.113 L

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

Absolute temperatures

T1 = 288.1 K, T2 = 513.1 K

Combined gas law

V2 = V1·(P1/P2)·(T2/T1) = 2·(250/800)·(513.1/288.1) = 1.113 L

Answer

V2 = 1.113 L
7.Stoichiometry

Answer: m = 90.1 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(H2) = 20/2.016 = 9.921 mol
n(O2) = 80/32 = 2.5 mol

Limiting reactant

9.921/2 = 4.96, 2.5/1 = 2.5 → O2

Mass of product

n(H2O) = 2.5·2 = 5 mol
m = 5·18.02 = 90.1 g

Answer

m = 90.1 g
8.Stoichiometry

Answer: Y = 91.01 %

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(Na) = 6/22.99 = 0.261 mol
m = 0.261·2/2·58.44 = 15.25 g

Percent yield

Y = 13.88/15.25·100 = 91.01%

Answer

Y = 91.01%

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