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 1200 kg/h · 10% saltStream 2ṁ₂ = ? · 30% saltProduct15% salt

How much of stream 2 must be added each hour to stream 1 so that the blended product contains 15% salt? Find ṁ2.

ṁ1 = 200 kg/hx1 = 10 %x2 = 30 %x3 = 15 %
ṁ2 = kg/h
2.Mass balanceHard
EvaporatorFeed F3800 kg/h · 20% 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 = 3800 kg/hxF = 20 %xP = 30 %
P = kg/h
3.Energy balanceMedium
hot sidecold sideglycol mix · 150 °C100 °Cwater · 10 °Ct = ? °C

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

ṁh = 2 kg/scp,h = 3.6 kJ/(kg·K)Th: 150 → 100 °Cṁc = 1 kg/scp,c = 4.18 kJ/(kg·K)tin = 10 °C
tout = °C
4.Energy balanceHard
hot sidecold sidelight oil · 180 °C130 °Cwater · 10 °C60 °C · ṁ = ?10% lost

Hot light oil 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 = 1 kg/scp,h = 2.1 kJ/(kg·K)Th: 180 → 130 °Ct: 10 → 60 °Ccp,c = 4.18 kJ/(kg·K)loss = 10 %
ṁc = kg/s
5.Ideal gas lawMedium
270 g of ArV = 50 LT = 65 °CP = ?

A tank of 50 L contains 270 g of Ar at 65 °C. Treat it as an ideal gas and find the pressure P.

m = 270 gM = 39.95 g/molV = 50 LT = 65 °CR = 8.314 J/(mol·K)
P = kPa
6.Ideal gas lawHard
State 1State 2

A fixed amount of O₂ goes from 100 kPa, 2 L, 10 °C to 200 kPa and 210 °C. Find the new volume V2.

P1 = 100 kPaV1 = 2 LT1 = 10 °CP2 = 200 kPaT2 = 210 °C
V2 = L
7.StoichiometryMedium
m(X)n(X)n(CO2​)m(CO2​)÷ M(X)× 1/ν× M(CO2​)

6 g of CH4 and 48 g of O2 react according to CH4 + 2O2 → CO2 + 2H2O. What mass m of CO2 can form?

M(CH4) = 16.04M(O2) = 32M(CO2) = 44.01 g/mol
m = g
8.StoichiometryHard
m(CaCO3​)n(CaCO3​)n(CaO)m(CaO)÷ M(CaCO3​)× 1/1× M(CaO)

In the reaction CaCO3 → CaO + CO2, 10 g of CaCO3 reacts completely (the other reactants are in excess) and 4.76 g of CaO is collected. Find the percent yield Y.

M(CaCO3) = 100.09 g/molM(CaO) = 56.08 g/molactual: 4.76 g
Y = %

3Solutions

1.Mass balance

Answer: ṁ2 = 66.67 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) = 200·(10 − 15)/(15 − 30) = 66.67 kg/h

Answer

ṁ2 = 66.67 kg/h
2.Mass balance

Answer: P = 2533 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 = 3800·20/30 = 2533 kg/h

Answer

P = 2533 kg/h
3.Energy balance

Answer: tout = 96.12 °C

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

Heat released by the hot stream

Q = 2·3.6·(150 − 100) = 360 kW

Heat picked up by the water

tout = tin + Q/(ṁccp,c) = 10 + 360/(1·4.18) = 96.12 °C

Answer

tout = 96.12 °C
4.Energy balance

Answer: ṁc = 0.4522 kg/s

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

Heat released

Qh = 1·2.1·(180 − 130) = 105 kW

Heat that reaches the water

Qc = (1 − 0.1)·105 = 94.5 kW

Water flow rate

ṁc = Qc/(cp,c·Δt) = 94.5/(4.18·50) = 0.4522 kg/s

Answer

ṁc = 0.4522 kg/s
5.Ideal gas law

Answer: P = 380 kPa

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

Moles

n = m/M = 270/39.95 = 6.758 mol
T = 65 + 273.15 = 338.1 K

Pressure

P = nRT/V = 6.758·8.314·338.1/0.05 = 380 kPa

Answer

P = 380 kPa
6.Ideal gas law

Answer: V2 = 1.706 L

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

Absolute temperatures

T1 = 283.1 K, T2 = 483.1 K

Combined gas law

V2 = V1·(P1/P2)·(T2/T1) = 2·(100/200)·(483.1/283.1) = 1.706 L

Answer

V2 = 1.706 L
7.Stoichiometry

Answer: m = 16.46 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(CH4) = 6/16.04 = 0.3741 mol
n(O2) = 48/32 = 1.5 mol

Limiting reactant

0.3741/1 = 0.3741, 1.5/2 = 0.75 → CH4

Mass of product

n(CO2) = 0.3741·1 = 0.3741 mol
m = 0.3741·44.01 = 16.46 g

Answer

m = 16.46 g
8.Stoichiometry

Answer: Y = 84.96 %

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(CaCO3) = 10/100.09 = 0.09991 mol
m = 0.09991·1/1·56.08 = 5.603 g

Percent yield

Y = 4.76/5.603·100 = 84.96%

Answer

Y = 84.96%

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