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 1550 kg/h · 40% NaOHStream 2ṁ₂ = ? · 10% NaOHProduct30% NaOH

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

ṁ1 = 550 kg/hx1 = 40 %x2 = 10 %x3 = 30 %
ṁ2 = kg/h
2.Mass balanceHard
EvaporatorFeed F4600 kg/h · 7% saltProduct PP = ? · 65%Vapour W (pure water)0% salt

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

F = 4600 kg/hxF = 7 %xP = 65 %
P = kg/h
3.Energy balanceMedium
hot sidecold sideethanol · 170 °C90 °Cwater · 15 °Ct = ? °C

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

ṁh = 2.5 kg/scp,h = 2.44 kJ/(kg·K)Th: 170 → 90 °Cṁc = 1.5 kg/scp,c = 4.18 kJ/(kg·K)tin = 15 °C
tout = °C
4.Energy balanceHard
hot sidecold sidelight oil · 150 °C120 °Cwater · 15 °C70 °C · ṁ = ?5% lost

Hot light oil 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 = 1 kg/scp,h = 2.1 kJ/(kg·K)Th: 150 → 120 °Ct: 15 → 70 °Ccp,c = 4.18 kJ/(kg·K)loss = 5 %
ṁc = kg/s
5.Ideal gas lawMedium
170 g of CH₄V = 10 LT = 50 °CP = ?

A tank of 10 L contains 170 g of CH₄ at 50 °C. Treat it as an ideal gas and find the pressure P.

m = 170 gM = 16.04 g/molV = 10 LT = 50 °CR = 8.314 J/(mol·K)
P = kPa
6.Ideal gas lawHard
State 1State 2

A fixed amount of CH₄ goes from 150 kPa, 10 L, 25 °C to 300 kPa and 240 °C. Find the new volume V2.

P1 = 150 kPaV1 = 10 LT1 = 25 °CP2 = 300 kPaT2 = 240 °C
V2 = L
7.StoichiometryMedium
m(X)n(X)n(NH3​)m(NH3​)÷ M(X)× 2/ν× M(NH3​)

8 g of N2 and 16 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(C6​H12​O6​)n(C6​H12​O6​)n(C2​H5​OH)m(C2​H5​OH)÷ M(C6​H12​O6​)× 2/1× M(C2​H5​OH)

In the reaction C6H12O6 → 2C2H5OH + 2CO2, 6 g of C6H12O6 reacts completely (the other reactants are in excess) and 2.52 g of C2H5OH is collected. Find the percent yield Y.

M(C6H12O6) = 180.16 g/molM(C2H5OH) = 46.07 g/molactual: 2.52 g
Y = %

3Solutions

1.Mass balance

Answer: ṁ2 = 275 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) = 550·(40 − 30)/(30 − 10) = 275 kg/h

Answer

ṁ2 = 275 kg/h
2.Mass balance

Answer: P = 495.4 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 = 4600·7/65 = 495.4 kg/h

Answer

P = 495.4 kg/h
3.Energy balance

Answer: tout = 92.83 °C

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

Heat released by the hot stream

Q = 2.5·2.44·(170 − 90) = 488 kW

Heat picked up by the water

tout = tin + Q/(ṁccp,c) = 15 + 488/(1.5·4.18) = 92.83 °C

Answer

tout = 92.83 °C
4.Energy balance

Answer: ṁc = 0.2603 kg/s

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

Heat released

Qh = 1·2.1·(150 − 120) = 63 kW

Heat that reaches the water

Qc = (1 − 0.05)·63 = 59.85 kW

Water flow rate

ṁc = Qc/(cp,c·Δt) = 59.85/(4.18·55) = 0.2603 kg/s

Answer

ṁc = 0.2603 kg/s
5.Ideal gas law

Answer: P = 2847 kPa

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

Moles

n = m/M = 170/16.04 = 10.6 mol
T = 50 + 273.15 = 323.1 K

Pressure

P = nRT/V = 10.6·8.314·323.1/0.01 = 2847 kPa

Answer

P = 2847 kPa
6.Ideal gas law

Answer: V2 = 8.606 L

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

Absolute temperatures

T1 = 298.1 K, T2 = 513.1 K

Combined gas law

V2 = V1·(P1/P2)·(T2/T1) = 10·(150/300)·(513.1/298.1) = 8.606 L

Answer

V2 = 8.606 L
7.Stoichiometry

Answer: m = 9.724 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) = 8/28.02 = 0.2855 mol
n(H2) = 16/2.016 = 7.937 mol

Limiting reactant

0.2855/1 = 0.2855, 7.937/3 = 2.646 → N2

Mass of product

n(NH3) = 0.2855·2 = 0.571 mol
m = 0.571·17.03 = 9.724 g

Answer

m = 9.724 g
8.Stoichiometry

Answer: Y = 82.12 %

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(C6H12O6) = 6/180.16 = 0.0333 mol
m = 0.0333·2/1·46.07 = 3.069 g

Percent yield

Y = 2.52/3.069·100 = 82.12%

Answer

Y = 82.12%

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