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 1250 kg/h · 50% saltStream 2ṁ₂ = ? · 10% saltProduct35% salt

How much of stream 2 must be added each hour to stream 1 so that the blended product contains 35% salt? Find ṁ2.

ṁ1 = 250 kg/hx1 = 50 %x2 = 10 %x3 = 35 %
ṁ2 = kg/h
2.Mass balanceHard
EvaporatorFeed F2900 kg/h · 9% saltProduct P60% 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 = 2900 kg/hxF = 9 %xP = 60 %
W = kg/h
3.Energy balanceMedium
hot sidecold sidelight oil · 170 °C60 °Cwater · 10 °Ct = ? °C

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

ṁh = 2.5 kg/scp,h = 2.1 kJ/(kg·K)Th: 170 → 60 °Cṁc = 3 kg/scp,c = 4.18 kJ/(kg·K)tin = 10 °C
tout = °C
4.Energy balanceHard
hot sidecold sideglycol mix · 170 °C80 °Cwater · 15 °C55 °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 = 1.5 kg/scp,h = 3.6 kJ/(kg·K)Th: 170 → 80 °Ct: 15 → 55 °Ccp,c = 4.18 kJ/(kg·K)loss = 5 %
ṁc = kg/s
5.Ideal gas lawMedium
80 g of ArP = 750 kPaT = 45 °CV = ?

What volume V does 80 g of Ar occupy at 750 kPa and 45 °C? Treat it as an ideal gas.

m = 80 gM = 39.95 g/molP = 750 kPaT = 45 °CR = 8.314 J/(mol·K)
V = L
6.Ideal gas lawHard
State 1State 2

A fixed amount of He goes from 200 kPa, 2 L, 10 °C to 500 kPa and 140 °C. Find the new volume V2.

P1 = 200 kPaV1 = 2 LT1 = 10 °CP2 = 500 kPaT2 = 140 °C
V2 = L
7.StoichiometryMedium
m(X)n(X)n(NH3​)m(NH3​)÷ M(X)× 2/ν× M(NH3​)

24 g of N2 and 16 g of H2 react according to N2 + 3H2 → 2NH3. What mass m of NH3 can form?

M(N2) = 28.02M(H2) = 2.016M(NH3) = 17.03 g/mol
m = g
8.StoichiometryHard
m(C3​H8​)n(C3​H8​)n(CO2​)m(CO2​)÷ M(C3​H8​)× 3/1× M(CO2​)

In the reaction C3H8 + 5O2 → 3CO2 + 4H2O, 12 g of C3H8 reacts completely (the other reactants are in excess) and 28.02 g of CO2 is collected. Find the percent yield Y.

M(C3H8) = 44.1 g/molM(CO2) = 44.01 g/molactual: 28.02 g
Y = %

3Solutions

1.Mass balance

Answer: ṁ2 = 150 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) = 250·(50 − 35)/(35 − 10) = 150 kg/h

Answer

ṁ2 = 150 kg/h
2.Mass balance

Answer: W = 2465 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 = 2900·9/60 = 435 kg/h

Total balance

W = F − P = 2900 − 435 = 2465 kg/h

Answer

W = 2465 kg/h
3.Energy balance

Answer: tout = 56.05 °C

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

Heat released by the hot stream

Q = 2.5·2.1·(170 − 60) = 577.5 kW

Heat picked up by the water

tout = tin + Q/(ṁccp,c) = 10 + 577.5/(3·4.18) = 56.05 °C

Answer

tout = 56.05 °C
4.Energy balance

Answer: ṁc = 2.761 kg/s

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

Heat released

Qh = 1.5·3.6·(170 − 80) = 486 kW

Heat that reaches the water

Qc = (1 − 0.05)·486 = 461.7 kW

Water flow rate

ṁc = Qc/(cp,c·Δt) = 461.7/(4.18·40) = 2.761 kg/s

Answer

ṁc = 2.761 kg/s
5.Ideal gas law

Answer: V = 7.062 L

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

Moles

n = 80/39.95 = 2.003 mol
T = 45 + 273.15 = 318.1 K

Volume

V = nRT/P = 2.003·8.314·318.1/750 = 7.062 L

Answer

V = 7.062 L
6.Ideal gas law

Answer: V2 = 1.167 L

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

Absolute temperatures

T1 = 283.1 K, T2 = 413.1 K

Combined gas law

V2 = V1·(P1/P2)·(T2/T1) = 2·(200/500)·(413.1/283.1) = 1.167 L

Answer

V2 = 1.167 L
7.Stoichiometry

Answer: m = 29.17 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(N2) = 24/28.02 = 0.8565 mol
n(H2) = 16/2.016 = 7.937 mol

Limiting reactant

0.8565/1 = 0.8565, 7.937/3 = 2.646 → N2

Mass of product

n(NH3) = 0.8565·2 = 1.713 mol
m = 1.713·17.03 = 29.17 g

Answer

m = 29.17 g
8.Stoichiometry

Answer: Y = 77.99 %

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(C3H8) = 12/44.1 = 0.2721 mol
m = 0.2721·3/1·44.01 = 35.93 g

Percent yield

Y = 28.02/35.93·100 = 77.99%

Answer

Y = 77.99%

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