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 1700 kg/h · 10% saltStream 2ṁ₂ = ? · 45% saltProduct20% salt

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

ṁ1 = 700 kg/hx1 = 10 %x2 = 45 %x3 = 20 %
ṁ2 = kg/h
2.Mass balanceHard
EvaporatorFeed F700 kg/h · 6% saltProduct PP = ? · 30%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 = 700 kg/hxF = 6 %xP = 30 %
P = kg/h
3.Energy balanceMedium
hot sidecold sidelight oil · 180 °C110 °Cwater · 20 °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 → 110 °Cṁc = 1.5 kg/scp,c = 4.18 kJ/(kg·K)tin = 20 °C
tout = °C
4.Energy balanceHard
hot sidecold sideglycol mix · 170 °C100 °Cwater · 15 °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 = 1 kg/scp,h = 3.6 kJ/(kg·K)Th: 170 → 100 °Ct: 15 → 90 °Ccp,c = 4.18 kJ/(kg·K)loss = 15 %
ṁc = kg/s
5.Ideal gas lawMedium
20 g of N₂P = 400 kPaT = 85 °CV = ?

What volume V does 20 g of N₂ occupy at 400 kPa and 85 °C? Treat it as an ideal gas.

m = 20 gM = 28.01 g/molP = 400 kPaT = 85 °CR = 8.314 J/(mol·K)
V = L
6.Ideal gas lawHard
State 1State 2

A fixed amount of O₂ goes from 250 kPa, 20 L, 15 °C to 300 kPa and 180 °C. Find the new volume V2.

P1 = 250 kPaV1 = 20 LT1 = 15 °CP2 = 300 kPaT2 = 180 °C
V2 = L
7.StoichiometryMedium
m(X)n(X)n(AlCl3​)m(AlCl3​)÷ M(X)× 2/ν× M(AlCl3​)

6 g of Al and 24 g of Cl2 react according to 2Al + 3Cl2 → 2AlCl3. What mass m of AlCl3 can form?

M(Al) = 26.98M(Cl2) = 70.9M(AlCl3) = 133.33 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, 30 g of N2 reacts completely (the other reactants are in excess) and 20.06 g of NH3 is collected. Find the percent yield Y.

M(N2) = 28.02 g/molM(NH3) = 17.03 g/molactual: 20.06 g
Y = %

3Solutions

1.Mass balance

Answer: ṁ2 = 280 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) = 700·(10 − 20)/(20 − 45) = 280 kg/h

Answer

ṁ2 = 280 kg/h
2.Mass balance

Answer: P = 140 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 = 700·6/30 = 140 kg/h

Answer

P = 140 kg/h
3.Energy balance

Answer: tout = 90.33 °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 − 110) = 441 kW

Heat picked up by the water

tout = tin + Q/(ṁccp,c) = 20 + 441/(1.5·4.18) = 90.33 °C

Answer

tout = 90.33 °C
4.Energy balance

Answer: ṁc = 0.6833 kg/s

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

Heat released

Qh = 1·3.6·(170 − 100) = 252 kW

Heat that reaches the water

Qc = (1 − 0.15)·252 = 214.2 kW

Water flow rate

ṁc = Qc/(cp,c·Δt) = 214.2/(4.18·75) = 0.6833 kg/s

Answer

ṁc = 0.6833 kg/s
5.Ideal gas law

Answer: V = 5.315 L

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

Moles

n = 20/28.01 = 0.714 mol
T = 85 + 273.15 = 358.1 K

Volume

V = nRT/P = 0.714·8.314·358.1/400 = 5.315 L

Answer

V = 5.315 L
6.Ideal gas law

Answer: V2 = 26.21 L

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

Absolute temperatures

T1 = 288.1 K, T2 = 453.1 K

Combined gas law

V2 = V1·(P1/P2)·(T2/T1) = 20·(250/300)·(453.1/288.1) = 26.21 L

Answer

V2 = 26.21 L
7.Stoichiometry

Answer: m = 29.65 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(Al) = 6/26.98 = 0.2224 mol
n(Cl2) = 24/70.9 = 0.3385 mol

Limiting reactant

0.2224/2 = 0.1112, 0.3385/3 = 0.1128 → Al

Mass of product

n(AlCl3) = 0.1112·2 = 0.2224 mol
m = 0.2224·133.33 = 29.65 g

Answer

m = 29.65 g
8.Stoichiometry

Answer: Y = 55.01 %

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(N2) = 30/28.02 = 1.071 mol
m = 1.071·2/1·17.03 = 36.47 g

Percent yield

Y = 20.06/36.47·100 = 55.01%

Answer

Y = 55.01%

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