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 1800 kg/h · 30% NaOHStream 2ṁ₂ = ? · 5% NaOHProduct10% NaOH

How much of stream 2 must be added each hour to stream 1 so that the blended product contains 10% NaOH? Find ṁ2.

ṁ1 = 800 kg/hx1 = 30 %x2 = 5 %x3 = 10 %
ṁ2 = kg/h
2.Mass balanceHard
EvaporatorFeed F3800 kg/h · 6% NaOHProduct PP = ? · 40%Vapour W (pure water)0% NaOH

An evaporator concentrates a NaOH solution by boiling off pure water. Find the product flow P in kg/h.

F = 3800 kg/hxF = 6 %xP = 40 %
P = kg/h
3.Energy balanceMedium
hot sidecold sideethanol · 140 °C60 °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 = 0.5 kg/scp,h = 2.44 kJ/(kg·K)Th: 140 → 60 °Cṁc = 3 kg/scp,c = 4.18 kJ/(kg·K)tin = 10 °C
tout = °C
4.Energy balanceHard
hot sidecold sideethanol · 160 °C100 °Cwater · 30 °C70 °C · ṁ = ?10% lost

Hot ethanol 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 = 2.44 kJ/(kg·K)Th: 160 → 100 °Ct: 30 → 70 °Ccp,c = 4.18 kJ/(kg·K)loss = 10 %
ṁc = kg/s
5.Ideal gas lawMedium
230 g of N₂V = 10 LT = 60 °CP = ?

A tank of 10 L contains 230 g of N₂ at 60 °C. Treat it as an ideal gas and find the pressure P.

m = 230 gM = 28.01 g/molV = 10 LT = 60 °CR = 8.314 J/(mol·K)
P = kPa
6.Ideal gas lawHard
State 1State 2

A fixed amount of CO₂ goes from 400 kPa, 20 L, 20 °C to 750 kPa and 220 °C. Find the new volume V2.

P1 = 400 kPaV1 = 20 LT1 = 20 °CP2 = 750 kPaT2 = 220 °C
V2 = L
7.StoichiometryMedium
m(X)n(X)n(NH3​)m(NH3​)÷ M(X)× 2/ν× M(NH3​)

4 g of N2 and 48 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(CH4​)n(CH4​)n(H2​O)m(H2​O)÷ M(CH4​)× 2/1× M(H2​O)

In the reaction CH4 + 2O2 → CO2 + 2H2O, 4 g of CH4 reacts completely (the other reactants are in excess) and 6.74 g of H2O is collected. Find the percent yield Y.

M(CH4) = 16.04 g/molM(H2O) = 18.02 g/molactual: 6.74 g
Y = %

3Solutions

1.Mass balance

Answer: ṁ2 = 3200 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) = 800·(30 − 10)/(10 − 5) = 3200 kg/h

Answer

ṁ2 = 3200 kg/h
2.Mass balance

Answer: P = 570 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 = 3800·6/40 = 570 kg/h

Answer

P = 570 kg/h
3.Energy balance

Answer: tout = 17.78 °C

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

Heat released by the hot stream

Q = 0.5·2.44·(140 − 60) = 97.6 kW

Heat picked up by the water

tout = tin + Q/(ṁccp,c) = 10 + 97.6/(3·4.18) = 17.78 °C

Answer

tout = 17.78 °C
4.Energy balance

Answer: ṁc = 0.394 kg/s

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

Heat released

Qh = 0.5·2.44·(160 − 100) = 73.2 kW

Heat that reaches the water

Qc = (1 − 0.1)·73.2 = 65.88 kW

Water flow rate

ṁc = Qc/(cp,c·Δt) = 65.88/(4.18·40) = 0.394 kg/s

Answer

ṁc = 0.394 kg/s
5.Ideal gas law

Answer: P = 2274 kPa

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

Moles

n = m/M = 230/28.01 = 8.211 mol
T = 60 + 273.15 = 333.1 K

Pressure

P = nRT/V = 8.211·8.314·333.1/0.01 = 2274 kPa

Answer

P = 2274 kPa
6.Ideal gas law

Answer: V2 = 17.94 L

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

Absolute temperatures

T1 = 293.1 K, T2 = 493.1 K

Combined gas law

V2 = V1·(P1/P2)·(T2/T1) = 20·(400/750)·(493.1/293.1) = 17.94 L

Answer

V2 = 17.94 L
7.Stoichiometry

Answer: m = 4.862 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) = 4/28.02 = 0.1428 mol
n(H2) = 48/2.016 = 23.81 mol

Limiting reactant

0.1428/1 = 0.1428, 23.81/3 = 7.937 → N2

Mass of product

n(NH3) = 0.1428·2 = 0.2855 mol
m = 0.2855·17.03 = 4.862 g

Answer

m = 4.862 g
8.Stoichiometry

Answer: Y = 74.99 %

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(CH4) = 4/16.04 = 0.2494 mol
m = 0.2494·2/1·18.02 = 8.988 g

Percent yield

Y = 6.74/8.988·100 = 74.99%

Answer

Y = 74.99%

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