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.

New problemsBack to the course

In the print window, choose “Save as PDF” to keep a copy on your device.

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 11000 kg/h · 35% sugarStream 2ṁ₂ = ? · 20% sugarProduct30% sugar

How much of stream 2 must be added each hour to stream 1 so that the blended product contains 30% sugar? Find ṁ2.

ṁ1 = 1000 kg/hx1 = 35 %x2 = 20 %x3 = 30 %
ṁ2 = kg/h
2.Mass balanceHard
EvaporatorFeed F700 kg/h · 8% NaOHProduct PP = ? · 30%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 = 700 kg/hxF = 8 %xP = 30 %
P = kg/h
3.Energy balanceMedium
hot sidecold sideglycol mix · 180 °C110 °Cwater · 20 °Ct = ? °C

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

ṁh = 0.5 kg/scp,h = 3.6 kJ/(kg·K)Th: 180 → 110 °Cṁc = 2 kg/scp,c = 4.18 kJ/(kg·K)tin = 20 °C
tout = °C
4.Energy balanceHard
hot sidecold sideglycol mix · 140 °C70 °Cwater · 15 °C75 °C · ṁ = ?5% lost

Hot glycol mix heats water in an exchanger that loses 5% 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: 140 → 70 °Ct: 15 → 75 °Ccp,c = 4.18 kJ/(kg·K)loss = 5 %
ṁc = kg/s
5.Ideal gas lawMedium
30 g of HeV = 10 LT = 60 °CP = ?

A tank of 10 L contains 30 g of He at 60 °C. Treat it as an ideal gas and find the pressure P.

m = 30 gM = 4.003 g/molV = 10 LT = 60 °CR = 8.314 J/(mol·K)
P = kPa
6.Ideal gas lawHard
He, M = 4.003 g/molP = 500 kPaT = 140 °Cρ = ?

Find the density ρ of He at 500 kPa and 140 °C, treating it as an ideal gas.

M = 4.003 g/molP = 500 kPaT = 140 °CR = 8.314 J/(mol·K)
ρ = kg/m³
7.StoichiometryMedium
m(X)n(X)n(H2​O)m(H2​O)÷ M(X)× 4/ν× M(H2​O)

16 g of C3H8 and 10 g of O2 react according to C3H8 + 5O2 → 3CO2 + 4H2O. What mass m of H2O can form?

M(C3H8) = 44.1M(O2) = 32M(H2O) = 18.02 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, 10 g of N2 reacts completely (the other reactants are in excess) and 11.06 g of NH3 is collected. Find the percent yield Y.

M(N2) = 28.02 g/molM(NH3) = 17.03 g/molactual: 11.06 g
Y = %

3Solutions

1.Mass balance

Answer: ṁ2 = 500 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) = 1000·(35 − 30)/(30 − 20) = 500 kg/h

Answer

ṁ2 = 500 kg/h
2.Mass balance

Answer: P = 186.7 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 = 700·8/30 = 186.7 kg/h

Answer

P = 186.7 kg/h
3.Energy balance

Answer: tout = 35.07 °C

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

Heat released by the hot stream

Q = 0.5·3.6·(180 − 110) = 126 kW

Heat picked up by the water

tout = tin + Q/(ṁccp,c) = 20 + 126/(2·4.18) = 35.07 °C

Answer

tout = 35.07 °C
4.Energy balance

Answer: ṁc = 0.4773 kg/s

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

Heat released

Qh = 0.5·3.6·(140 − 70) = 126 kW

Heat that reaches the water

Qc = (1 − 0.05)·126 = 119.7 kW

Water flow rate

ṁc = Qc/(cp,c·Δt) = 119.7/(4.18·60) = 0.4773 kg/s

Answer

ṁc = 0.4773 kg/s
5.Ideal gas law

Answer: P = 2076 kPa

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

Moles

n = m/M = 30/4.003 = 7.494 mol
T = 60 + 273.15 = 333.1 K

Pressure

P = nRT/V = 7.494·8.314·333.1/0.01 = 2076 kPa

Answer

P = 2076 kPa
6.Ideal gas law

Answer: ρ = 0.5827 kg/m³

Ideal gas density: ρ = PM/(RT).

Rearrange the gas law

PV = (m/M)RT → ρ = m/V = PM/(RT)
T = 140 + 273.15 = 413.1 K

Density

ρ = 500·4.003/(8.314·413.1) = 0.5827 kg/m³

Answer

ρ = 0.5827 kg/m³
7.Stoichiometry

Answer: m = 4.505 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(C3H8) = 16/44.1 = 0.3628 mol
n(O2) = 10/32 = 0.3125 mol

Limiting reactant

0.3628/1 = 0.3628, 0.3125/5 = 0.0625 → O2

Mass of product

n(H2O) = 0.0625·4 = 0.25 mol
m = 0.25·18.02 = 4.505 g

Answer

m = 4.505 g
8.Stoichiometry

Answer: Y = 90.99 %

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(N2) = 10/28.02 = 0.3569 mol
m = 0.3569·2/1·17.03 = 12.16 g

Percent yield

Y = 11.06/12.16·100 = 90.99%

Answer

Y = 90.99%

ExamProblems · examproblems.com/courses/material-and-energy-balances