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 1400 kg/h · 50% saltStream 2ṁ₂ = ? · 5% saltProduct45% salt

How much of stream 2 must be added each hour to stream 1 so that the blended product contains 45% salt? Find ṁ2.

ṁ1 = 400 kg/hx1 = 50 %x2 = 5 %x3 = 45 %
ṁ2 = kg/h
2.Mass balanceHard
EvaporatorFeed F3800 kg/h · 11% saltProduct P30% saltVapour W (pure water)W = ? kg/h

An evaporator concentrates a salt solution by boiling off pure water. Find the water evaporated W in kg/h.

F = 3800 kg/hxF = 11 %xP = 30 %
W = kg/h
3.Energy balanceMedium
hot sidecold sideglycol mix · 130 °C80 °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 = 1 kg/scp,h = 3.6 kJ/(kg·K)Th: 130 → 80 °Cṁc = 3 kg/scp,c = 4.18 kJ/(kg·K)tin = 10 °C
tout = °C
4.Energy balanceHard
hot sidecold sideglycol mix · 150 °C100 °Cwater · 25 °C50 °C · ṁ = ?10% lost

Hot glycol mix 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 = 1.5 kg/scp,h = 3.6 kJ/(kg·K)Th: 150 → 100 °Ct: 25 → 50 °Ccp,c = 4.18 kJ/(kg·K)loss = 10 %
ṁc = kg/s
5.Ideal gas lawMedium
480 g of N₂V = 20 LT = 85 °CP = ?

A tank of 20 L contains 480 g of N₂ at 85 °C. Treat it as an ideal gas and find the pressure P.

m = 480 gM = 28.01 g/molV = 20 LT = 85 °CR = 8.314 J/(mol·K)
P = kPa
6.Ideal gas lawHard
State 1State 2

A fixed amount of CH₄ goes from 300 kPa, 2 L, 20 °C to 500 kPa and 210 °C. Find the new volume V2.

P1 = 300 kPaV1 = 2 LT1 = 20 °CP2 = 500 kPaT2 = 210 °C
V2 = L
7.StoichiometryMedium
m(X)n(X)n(MgO)m(MgO)÷ M(X)× 2/ν× M(MgO)

20 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(Fe2​O3​)n(Fe2​O3​)n(CO2​)m(CO2​)÷ M(Fe2​O3​)× 3/1× M(CO2​)

In the reaction Fe2O3 + 3CO → 2Fe + 3CO2, 16 g of Fe2O3 reacts completely (the other reactants are in excess) and 11.64 g of CO2 is collected. Find the percent yield Y.

M(Fe2O3) = 159.69 g/molM(CO2) = 44.01 g/molactual: 11.64 g
Y = %

3Solutions

1.Mass balance

Answer: ṁ2 = 50 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) = 400·(50 − 45)/(45 − 5) = 50 kg/h

Answer

ṁ2 = 50 kg/h
2.Mass balance

Answer: W = 2407 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 = 3800·11/30 = 1393 kg/h

Total balance

W = F − P = 3800 − 1393 = 2407 kg/h

Answer

W = 2407 kg/h
3.Energy balance

Answer: tout = 24.35 °C

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

Heat released by the hot stream

Q = 1·3.6·(130 − 80) = 180 kW

Heat picked up by the water

tout = tin + Q/(ṁccp,c) = 10 + 180/(3·4.18) = 24.35 °C

Answer

tout = 24.35 °C
4.Energy balance

Answer: ṁc = 2.325 kg/s

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

Heat released

Qh = 1.5·3.6·(150 − 100) = 270 kW

Heat that reaches the water

Qc = (1 − 0.1)·270 = 243 kW

Water flow rate

ṁc = Qc/(cp,c·Δt) = 243/(4.18·25) = 2.325 kg/s

Answer

ṁc = 2.325 kg/s
5.Ideal gas law

Answer: P = 2551 kPa

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

Moles

n = m/M = 480/28.01 = 17.14 mol
T = 85 + 273.15 = 358.1 K

Pressure

P = nRT/V = 17.14·8.314·358.1/0.02 = 2551 kPa

Answer

P = 2551 kPa
6.Ideal gas law

Answer: V2 = 1.978 L

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

Absolute temperatures

T1 = 293.1 K, T2 = 483.1 K

Combined gas law

V2 = V1·(P1/P2)·(T2/T1) = 2·(300/500)·(483.1/293.1) = 1.978 L

Answer

V2 = 1.978 L
7.Stoichiometry

Answer: m = 33.16 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) = 20/24.31 = 0.8227 mol
n(O2) = 24/32 = 0.75 mol

Limiting reactant

0.8227/2 = 0.4114, 0.75/1 = 0.75 → Mg

Mass of product

n(MgO) = 0.4114·2 = 0.8227 mol
m = 0.8227·40.3 = 33.16 g

Answer

m = 33.16 g
8.Stoichiometry

Answer: Y = 87.99 %

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(Fe2O3) = 16/159.69 = 0.1002 mol
m = 0.1002·3/1·44.01 = 13.23 g

Percent yield

Y = 11.64/13.23·100 = 87.99%

Answer

Y = 87.99%

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