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 · 50% NaOHStream 2ṁ₂ = ? · 5% NaOHProduct20% NaOH

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

ṁ1 = 400 kg/hx1 = 50 %x2 = 5 %x3 = 20 %
ṁ2 = kg/h
2.Mass balanceHard
EvaporatorFeed F900 kg/h · 18% sugarProduct P65% sugarVapour W (pure water)W = ? kg/h

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

F = 900 kg/hxF = 18 %xP = 65 %
W = kg/h
3.Energy balanceMedium
hot sidecold sideglycol mix · 140 °C60 °Cwater · 10 °Ct = ? °C

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

ṁh = 2.5 kg/scp,h = 3.6 kJ/(kg·K)Th: 140 → 60 °Cṁc = 1.5 kg/scp,c = 4.18 kJ/(kg·K)tin = 10 °C
tout = °C
4.Energy balanceHard
hot sidecold sideglycol mix · 130 °C60 °Cwater · 15 °C40 °C · ṁ = ?15% lost

Hot glycol mix 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 = 2 kg/scp,h = 3.6 kJ/(kg·K)Th: 130 → 60 °Ct: 15 → 40 °Ccp,c = 4.18 kJ/(kg·K)loss = 15 %
ṁc = kg/s
5.Ideal gas lawMedium
400 g of CO₂V = 30 LT = 75 °CP = ?

A tank of 30 L contains 400 g of CO₂ at 75 °C. Treat it as an ideal gas and find the pressure P.

m = 400 gM = 44.01 g/molV = 30 LT = 75 °CR = 8.314 J/(mol·K)
P = kPa
6.Ideal gas lawHard
State 1State 2

A fixed amount of He goes from 350 kPa, 10 L, 10 °C to 350 kPa and 120 °C. Find the new volume V2.

P1 = 350 kPaV1 = 10 LT1 = 10 °CP2 = 350 kPaT2 = 120 °C
V2 = L
7.StoichiometryMedium
m(X)n(X)n(NH3​)m(NH3​)÷ M(X)× 2/ν× M(NH3​)

6 g of N2 and 80 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(Al)n(Al)n(AlCl3​)m(AlCl3​)÷ M(Al)× 2/2× M(AlCl3​)

In the reaction 2Al + 3Cl2 → 2AlCl3, 4 g of Al reacts completely (the other reactants are in excess) and 13.84 g of AlCl3 is collected. Find the percent yield Y.

M(Al) = 26.98 g/molM(AlCl3) = 133.33 g/molactual: 13.84 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
NaOH: ṁ1x1 + ṁ2x2 = (ṁ1 + ṁ2)x3

Solve for ṁ2

ṁ2 = ṁ1(x1 − x3)/(x3 − x2) = 400·(50 − 20)/(20 − 5) = 800 kg/h

Answer

ṁ2 = 800 kg/h
2.Mass balance

Answer: W = 650.8 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 = 900·18/65 = 249.2 kg/h

Total balance

W = F − P = 900 − 249.2 = 650.8 kg/h

Answer

W = 650.8 kg/h
3.Energy balance

Answer: tout = 124.8 °C

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

Heat released by the hot stream

Q = 2.5·3.6·(140 − 60) = 720 kW

Heat picked up by the water

tout = tin + Q/(ṁccp,c) = 10 + 720/(1.5·4.18) = 124.8 °C

Answer

tout = 124.8 °C
4.Energy balance

Answer: ṁc = 4.1 kg/s

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

Heat released

Qh = 2·3.6·(130 − 60) = 504 kW

Heat that reaches the water

Qc = (1 − 0.15)·504 = 428.4 kW

Water flow rate

ṁc = Qc/(cp,c·Δt) = 428.4/(4.18·25) = 4.1 kg/s

Answer

ṁc = 4.1 kg/s
5.Ideal gas law

Answer: P = 876.9 kPa

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

Moles

n = m/M = 400/44.01 = 9.089 mol
T = 75 + 273.15 = 348.1 K

Pressure

P = nRT/V = 9.089·8.314·348.1/0.03 = 876.9 kPa

Answer

P = 876.9 kPa
6.Ideal gas law

Answer: V2 = 13.88 L

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

Absolute temperatures

T1 = 283.1 K, T2 = 393.1 K

Combined gas law

V2 = V1·(P1/P2)·(T2/T1) = 10·(350/350)·(393.1/283.1) = 13.88 L

Answer

V2 = 13.88 L
7.Stoichiometry

Answer: m = 7.293 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) = 6/28.02 = 0.2141 mol
n(H2) = 80/2.016 = 39.68 mol

Limiting reactant

0.2141/1 = 0.2141, 39.68/3 = 13.23 → N2

Mass of product

n(NH3) = 0.2141·2 = 0.4283 mol
m = 0.4283·17.03 = 7.293 g

Answer

m = 7.293 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(Al) = 4/26.98 = 0.1483 mol
m = 0.1483·2/2·133.33 = 19.77 g

Percent yield

Y = 13.84/19.77·100 = 70.01%

Answer

Y = 70.01%

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