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 1400 kg/h · 20% saltStream 2ṁ₂ = ? · 35% saltProduct30% salt

How much of stream 2 must be added each hour to stream 1 so that the blended product contains 30% salt? Find ṁ2.

ṁ1 = 400 kg/hx1 = 20 %x2 = 35 %x3 = 30 %
ṁ2 = kg/h
2.Mass balanceHard
EvaporatorFeed F1500 kg/h · 5% 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 = 1500 kg/hxF = 5 %xP = 50 %
W = kg/h
3.Energy balanceMedium
hot sidecold sideethanol · 130 °C70 °Cwater · 10 °Ct = ? °C

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

ṁh = 1.5 kg/scp,h = 2.44 kJ/(kg·K)Th: 130 → 70 °Cṁc = 3 kg/scp,c = 4.18 kJ/(kg·K)tin = 10 °C
tout = °C
4.Energy balanceHard
hot sidecold sidelight oil · 130 °C70 °Cwater · 10 °C75 °C · ṁ = ?15% lost

Hot light oil heats water in an exchanger that loses 15% 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: 130 → 70 °Ct: 10 → 75 °Ccp,c = 4.18 kJ/(kg·K)loss = 15 %
ṁc = kg/s
5.Ideal gas lawMedium
320 g of CH₄V = 30 LT = 95 °CP = ?

A tank of 30 L contains 320 g of CH₄ at 95 °C. Treat it as an ideal gas and find the pressure P.

m = 320 gM = 16.04 g/molV = 30 LT = 95 °CR = 8.314 J/(mol·K)
P = kPa
6.Ideal gas lawHard
State 1State 2

A fixed amount of CH₄ goes from 350 kPa, 5 L, 20 °C to 500 kPa and 150 °C. Find the new volume V2.

P1 = 350 kPaV1 = 5 LT1 = 20 °CP2 = 500 kPaT2 = 150 °C
V2 = L
7.StoichiometryMedium
m(X)n(X)n(CO2​)m(CO2​)÷ M(X)× 3/ν× M(CO2​)

10 g of C3H8 and 8 g of O2 react according to C3H8 + 5O2 → 3CO2 + 4H2O. What mass m of CO2 can form?

M(C3H8) = 44.1M(O2) = 32M(CO2) = 44.01 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, 10 g of N2 reacts completely (the other reactants are in excess) and 11.06 g of NH3 is collected. Find the percent yield Y.

M(N2) = 28.02 g/molM(NH3) = 17.03 g/molactual: 11.06 g
Y = %

3Solutions

1.Mass balance

Answer: ṁ2 = 800 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) = 400·(20 − 30)/(30 − 35) = 800 kg/h

Answer

ṁ2 = 800 kg/h
2.Mass balance

Answer: W = 1350 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 = 1500·5/50 = 150 kg/h

Total balance

W = F − P = 1500 − 150 = 1350 kg/h

Answer

W = 1350 kg/h
3.Energy balance

Answer: tout = 27.51 °C

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

Heat released by the hot stream

Q = 1.5·2.44·(130 − 70) = 219.6 kW

Heat picked up by the water

tout = tin + Q/(ṁccp,c) = 10 + 219.6/(3·4.18) = 27.51 °C

Answer

tout = 27.51 °C
4.Energy balance

Answer: ṁc = 0.1971 kg/s

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

Heat released

Qh = 0.5·2.1·(130 − 70) = 63 kW

Heat that reaches the water

Qc = (1 − 0.15)·63 = 53.55 kW

Water flow rate

ṁc = Qc/(cp,c·Δt) = 53.55/(4.18·65) = 0.1971 kg/s

Answer

ṁc = 0.1971 kg/s
5.Ideal gas law

Answer: P = 2035 kPa

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

Moles

n = m/M = 320/16.04 = 19.95 mol
T = 95 + 273.15 = 368.1 K

Pressure

P = nRT/V = 19.95·8.314·368.1/0.03 = 2035 kPa

Answer

P = 2035 kPa
6.Ideal gas law

Answer: V2 = 5.052 L

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

Absolute temperatures

T1 = 293.1 K, T2 = 423.1 K

Combined gas law

V2 = V1·(P1/P2)·(T2/T1) = 5·(350/500)·(423.1/293.1) = 5.052 L

Answer

V2 = 5.052 L
7.Stoichiometry

Answer: m = 6.602 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) = 10/44.1 = 0.2268 mol
n(O2) = 8/32 = 0.25 mol

Limiting reactant

0.2268/1 = 0.2268, 0.25/5 = 0.05 → O2

Mass of product

n(CO2) = 0.05·3 = 0.15 mol
m = 0.15·44.01 = 6.602 g

Answer

m = 6.602 g
8.Stoichiometry

Answer: Y = 90.99 %

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) = 10/28.02 = 0.3569 mol
m = 0.3569·2/1·17.03 = 12.16 g

Percent yield

Y = 11.06/12.16·100 = 90.99%

Answer

Y = 90.99%

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