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 1900 kg/h · 5% saltStream 2ṁ₂ = ? · 35% 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 = 900 kg/hx1 = 5 %x2 = 35 %x3 = 15 %
ṁ2 = kg/h
2.Mass balanceHard
EvaporatorFeed F2100 kg/h · 11% NaOHProduct P30% 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 = 2100 kg/hxF = 11 %xP = 30 %
W = kg/h
3.Energy balanceMedium
hot sidecold sidelight oil · 180 °C110 °Cwater · 15 °Ct = ? °C

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

ṁh = 2.5 kg/scp,h = 2.1 kJ/(kg·K)Th: 180 → 110 °Cṁc = 3 kg/scp,c = 4.18 kJ/(kg·K)tin = 15 °C
tout = °C
4.Energy balanceHard
hot sidecold sidelight oil · 120 °C80 °Cwater · 15 °C55 °C · ṁ = ?10% lost

Hot light oil 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 = 0.5 kg/scp,h = 2.1 kJ/(kg·K)Th: 120 → 80 °Ct: 15 → 55 °Ccp,c = 4.18 kJ/(kg·K)loss = 10 %
ṁc = kg/s
5.Ideal gas lawMedium
240 g of HeP = 700 kPaT = 15 °CV = ?

What volume V does 240 g of He occupy at 700 kPa and 15 °C? Treat it as an ideal gas.

m = 240 gM = 4.003 g/molP = 700 kPaT = 15 °CR = 8.314 J/(mol·K)
V = L
6.Ideal gas lawHard
State 1State 2

A fixed amount of He goes from 100 kPa, 20 L, 25 °C to 550 kPa and 230 °C. Find the new volume V2.

P1 = 100 kPaV1 = 20 LT1 = 25 °CP2 = 550 kPaT2 = 230 °C
V2 = L
7.StoichiometryMedium
m(X)n(X)n(H2​O)m(H2​O)÷ M(X)× 4/ν× M(H2​O)

12 g of C3H8 and 48 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, 24 g of Na reacts completely (the other reactants are in excess) and 42.71 g of NaCl is collected. Find the percent yield Y.

M(Na) = 22.99 g/molM(NaCl) = 58.44 g/molactual: 42.71 g
Y = %

3Solutions

1.Mass balance

Answer: ṁ2 = 450 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) = 900·(5 − 15)/(15 − 35) = 450 kg/h

Answer

ṁ2 = 450 kg/h
2.Mass balance

Answer: W = 1330 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 = 2100·11/30 = 770 kg/h

Total balance

W = F − P = 2100 − 770 = 1330 kg/h

Answer

W = 1330 kg/h
3.Energy balance

Answer: tout = 44.31 °C

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

Heat released by the hot stream

Q = 2.5·2.1·(180 − 110) = 367.5 kW

Heat picked up by the water

tout = tin + Q/(ṁccp,c) = 15 + 367.5/(3·4.18) = 44.31 °C

Answer

tout = 44.31 °C
4.Energy balance

Answer: ṁc = 0.2261 kg/s

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

Heat released

Qh = 0.5·2.1·(120 − 80) = 42 kW

Heat that reaches the water

Qc = (1 − 0.1)·42 = 37.8 kW

Water flow rate

ṁc = Qc/(cp,c·Δt) = 37.8/(4.18·40) = 0.2261 kg/s

Answer

ṁc = 0.2261 kg/s
5.Ideal gas law

Answer: V = 205.2 L

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

Moles

n = 240/4.003 = 59.96 mol
T = 15 + 273.15 = 288.1 K

Volume

V = nRT/P = 59.96·8.314·288.1/700 = 205.2 L

Answer

V = 205.2 L
6.Ideal gas law

Answer: V2 = 6.137 L

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

Absolute temperatures

T1 = 298.1 K, T2 = 503.1 K

Combined gas law

V2 = V1·(P1/P2)·(T2/T1) = 20·(100/550)·(503.1/298.1) = 6.137 L

Answer

V2 = 6.137 L
7.Stoichiometry

Answer: m = 19.61 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) = 12/44.1 = 0.2721 mol
n(O2) = 48/32 = 1.5 mol

Limiting reactant

0.2721/1 = 0.2721, 1.5/5 = 0.3 → C3H8

Mass of product

n(H2O) = 0.2721·4 = 1.088 mol
m = 1.088·18.02 = 19.61 g

Answer

m = 19.61 g
8.Stoichiometry

Answer: Y = 70.01 %

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) = 24/22.99 = 1.044 mol
m = 1.044·2/2·58.44 = 61.01 g

Percent yield

Y = 42.71/61.01·100 = 70.01%

Answer

Y = 70.01%

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