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 1950 kg/h · 50% saltStream 2ṁ₂ = ? · 20% saltProduct40% salt

How much of stream 2 must be added each hour to stream 1 so that the blended product contains 40% salt? Find ṁ2.

ṁ1 = 950 kg/hx1 = 50 %x2 = 20 %x3 = 40 %
ṁ2 = kg/h
2.Mass balanceHard
EvaporatorFeed F2800 kg/h · 18% sugarProduct P35% 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 = 2800 kg/hxF = 18 %xP = 35 %
W = kg/h
3.Energy balanceMedium
hot sidecold sideglycol mix · 140 °C70 °Cwater · 15 °Ct = ? °C

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

ṁh = 3 kg/scp,h = 3.6 kJ/(kg·K)Th: 140 → 70 °Cṁc = 1.5 kg/scp,c = 4.18 kJ/(kg·K)tin = 15 °C
tout = °C
4.Energy balanceHard
hot sidecold sideglycol mix · 130 °C100 °Cwater · 25 °C90 °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 = 0.5 kg/scp,h = 3.6 kJ/(kg·K)Th: 130 → 100 °Ct: 25 → 90 °Ccp,c = 4.18 kJ/(kg·K)loss = 15 %
ṁc = kg/s
5.Ideal gas lawMedium
430 g of HeV = 20 LT = 0 °CP = ?

A tank of 20 L contains 430 g of He at 0 °C. Treat it as an ideal gas and find the pressure P.

m = 430 gM = 4.003 g/molV = 20 LT = 0 °CR = 8.314 J/(mol·K)
P = kPa
6.Ideal gas lawHard
He, M = 4.003 g/molP = 300 kPaT = 200 °Cρ = ?

Find the density ρ of He at 300 kPa and 200 °C, treating it as an ideal gas.

M = 4.003 g/molP = 300 kPaT = 200 °CR = 8.314 J/(mol·K)
ρ = kg/m³
7.StoichiometryMedium
m(X)n(X)n(MgO)m(MgO)÷ M(X)× 2/ν× M(MgO)

24 g of Mg and 24 g of O2 react according to 2Mg + O2 → 2MgO. What mass m of MgO can form?

M(Mg) = 24.31M(O2) = 32M(MgO) = 40.3 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, 8 g of Na reacts completely (the other reactants are in excess) and 14.24 g of NaCl is collected. Find the percent yield Y.

M(Na) = 22.99 g/molM(NaCl) = 58.44 g/molactual: 14.24 g
Y = %

3Solutions

1.Mass balance

Answer: ṁ2 = 475 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) = 950·(50 − 40)/(40 − 20) = 475 kg/h

Answer

ṁ2 = 475 kg/h
2.Mass balance

Answer: W = 1360 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 = 2800·18/35 = 1440 kg/h

Total balance

W = F − P = 2800 − 1440 = 1360 kg/h

Answer

W = 1360 kg/h
3.Energy balance

Answer: tout = 135.6 °C

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

Heat released by the hot stream

Q = 3·3.6·(140 − 70) = 756 kW

Heat picked up by the water

tout = tin + Q/(ṁccp,c) = 15 + 756/(1.5·4.18) = 135.6 °C

Answer

tout = 135.6 °C
4.Energy balance

Answer: ṁc = 0.1689 kg/s

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

Heat released

Qh = 0.5·3.6·(130 − 100) = 54 kW

Heat that reaches the water

Qc = (1 − 0.15)·54 = 45.9 kW

Water flow rate

ṁc = Qc/(cp,c·Δt) = 45.9/(4.18·65) = 0.1689 kg/s

Answer

ṁc = 0.1689 kg/s
5.Ideal gas law

Answer: P = 12197 kPa

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

Moles

n = m/M = 430/4.003 = 107.4 mol
T = 0 + 273.15 = 273.1 K

Pressure

P = nRT/V = 107.4·8.314·273.1/0.02 = 12197 kPa

Answer

P = 12197 kPa
6.Ideal gas law

Answer: ρ = 0.3053 kg/m³

Ideal gas density: ρ = PM/(RT).

Rearrange the gas law

PV = (m/M)RT → ρ = m/V = PM/(RT)
T = 200 + 273.15 = 473.1 K

Density

ρ = 300·4.003/(8.314·473.1) = 0.3053 kg/m³

Answer

ρ = 0.3053 kg/m³
7.Stoichiometry

Answer: m = 39.79 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(Mg) = 24/24.31 = 0.9872 mol
n(O2) = 24/32 = 0.75 mol

Limiting reactant

0.9872/2 = 0.4936, 0.75/1 = 0.75 → Mg

Mass of product

n(MgO) = 0.4936·2 = 0.9872 mol
m = 0.9872·40.3 = 39.79 g

Answer

m = 39.79 g
8.Stoichiometry

Answer: Y = 70.02 %

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) = 8/22.99 = 0.348 mol
m = 0.348·2/2·58.44 = 20.34 g

Percent yield

Y = 14.24/20.34·100 = 70.02%

Answer

Y = 70.02%

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