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 1450 kg/h · 50% sugarStream 2ṁ₂ = ? · 5% sugarProduct10% sugar

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

ṁ1 = 450 kg/hx1 = 50 %x2 = 5 %x3 = 10 %
ṁ2 = kg/h
2.Mass balanceHard
EvaporatorFeed F4600 kg/h · 7% NaOHProduct PP = ? · 55%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 = 4600 kg/hxF = 7 %xP = 55 %
P = kg/h
3.Energy balanceMedium
hot sidecold sidelight oil · 180 °C140 °Cwater · 30 °Ct = ? °C

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

ṁh = 3 kg/scp,h = 2.1 kJ/(kg·K)Th: 180 → 140 °Cṁc = 0.5 kg/scp,c = 4.18 kJ/(kg·K)tin = 30 °C
tout = °C
4.Energy balanceHard
hot sidecold sidelight oil · 120 °C70 °Cwater · 30 °C90 °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 = 2 kg/scp,h = 2.1 kJ/(kg·K)Th: 120 → 70 °Ct: 30 → 90 °Ccp,c = 4.18 kJ/(kg·K)loss = 15 %
ṁc = kg/s
5.Ideal gas lawMedium
210 g of N₂P = 700 kPaT = 70 °CV = ?

What volume V does 210 g of N₂ occupy at 700 kPa and 70 °C? Treat it as an ideal gas.

m = 210 gM = 28.01 g/molP = 700 kPaT = 70 °CR = 8.314 J/(mol·K)
V = L
6.Ideal gas lawHard
State 1State 2

A fixed amount of He goes from 250 kPa, 20 L, 30 °C to 200 kPa and 190 °C. Find the new volume V2.

P1 = 250 kPaV1 = 20 LT1 = 30 °CP2 = 200 kPaT2 = 190 °C
V2 = L
7.StoichiometryMedium
m(X)n(X)n(NaCl)m(NaCl)÷ M(X)× 2/ν× M(NaCl)

30 g of Na and 20 g of Cl2 react according to 2Na + Cl2 → 2NaCl. What mass m of NaCl can form?

M(Na) = 22.99M(Cl2) = 70.9M(NaCl) = 58.44 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, 4 g of N2 reacts completely (the other reactants are in excess) and 4.13 g of NH3 is collected. Find the percent yield Y.

M(N2) = 28.02 g/molM(NH3) = 17.03 g/molactual: 4.13 g
Y = %

3Solutions

1.Mass balance

Answer: ṁ2 = 3600 kg/h

Steady-state mass balance: in = out for the total and for each component. Blending problems combine both.

Balances

Total: ṁ3 = ṁ1 + ṁ2
sugar: ṁ1x1 + ṁ2x2 = (ṁ1 + ṁ2)x3

Solve for ṁ2

ṁ2 = ṁ1(x1 − x3)/(x3 − x2) = 450·(50 − 10)/(10 − 5) = 3600 kg/h

Answer

ṁ2 = 3600 kg/h
2.Mass balance

Answer: P = 585.5 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 = 4600·7/55 = 585.5 kg/h

Answer

P = 585.5 kg/h
3.Energy balance

Answer: tout = 150.6 °C

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

Heat released by the hot stream

Q = 3·2.1·(180 − 140) = 252 kW

Heat picked up by the water

tout = tin + Q/(ṁccp,c) = 30 + 252/(0.5·4.18) = 150.6 °C

Answer

tout = 150.6 °C
4.Energy balance

Answer: ṁc = 0.7117 kg/s

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

Heat released

Qh = 2·2.1·(120 − 70) = 210 kW

Heat that reaches the water

Qc = (1 − 0.15)·210 = 178.5 kW

Water flow rate

ṁc = Qc/(cp,c·Δt) = 178.5/(4.18·60) = 0.7117 kg/s

Answer

ṁc = 0.7117 kg/s
5.Ideal gas law

Answer: V = 30.56 L

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

Moles

n = 210/28.01 = 7.497 mol
T = 70 + 273.15 = 343.1 K

Volume

V = nRT/P = 7.497·8.314·343.1/700 = 30.56 L

Answer

V = 30.56 L
6.Ideal gas law

Answer: V2 = 38.19 L

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

Absolute temperatures

T1 = 303.1 K, T2 = 463.1 K

Combined gas law

V2 = V1·(P1/P2)·(T2/T1) = 20·(250/200)·(463.1/303.1) = 38.19 L

Answer

V2 = 38.19 L
7.Stoichiometry

Answer: m = 32.97 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(Na) = 30/22.99 = 1.305 mol
n(Cl2) = 20/70.9 = 0.2821 mol

Limiting reactant

1.305/2 = 0.6525, 0.2821/1 = 0.2821 → Cl2

Mass of product

n(NaCl) = 0.2821·2 = 0.5642 mol
m = 0.5642·58.44 = 32.97 g

Answer

m = 32.97 g
8.Stoichiometry

Answer: Y = 84.94 %

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) = 4/28.02 = 0.1428 mol
m = 0.1428·2/1·17.03 = 4.862 g

Percent yield

Y = 4.13/4.862·100 = 84.94%

Answer

Y = 84.94%

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