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 · 45% saltStream 2ṁ₂ = ? · 10% saltProduct25% salt

How much of stream 2 must be added each hour to stream 1 so that the blended product contains 25% salt? Find ṁ2.

ṁ1 = 400 kg/hx1 = 45 %x2 = 10 %x3 = 25 %
ṁ2 = kg/h
2.Mass balanceHard
EvaporatorFeed F900 kg/h · 20% saltProduct P50% 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 = 900 kg/hxF = 20 %xP = 50 %
W = kg/h
3.Energy balanceMedium
hot sidecold sideglycol mix · 150 °C120 °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: 150 → 120 °Cṁc = 0.5 kg/scp,c = 4.18 kJ/(kg·K)tin = 10 °C
tout = °C
4.Energy balanceHard
hot sidecold sideethanol · 170 °C70 °Cwater · 25 °C60 °C · ṁ = ?10% lost

Hot ethanol 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 = 2.5 kg/scp,h = 2.44 kJ/(kg·K)Th: 170 → 70 °Ct: 25 → 60 °Ccp,c = 4.18 kJ/(kg·K)loss = 10 %
ṁc = kg/s
5.Ideal gas lawMedium
410 g of HeV = 50 LT = 85 °CP = ?

A tank of 50 L contains 410 g of He at 85 °C. Treat it as an ideal gas and find the pressure P.

m = 410 gM = 4.003 g/molV = 50 LT = 85 °CR = 8.314 J/(mol·K)
P = kPa
6.Ideal gas lawHard
State 1State 2

A fixed amount of He goes from 100 kPa, 5 L, 35 °C to 400 kPa and 70 °C. Find the new volume V2.

P1 = 100 kPaV1 = 5 LT1 = 35 °CP2 = 400 kPaT2 = 70 °C
V2 = L
7.StoichiometryMedium
m(X)n(X)n(NH3​)m(NH3​)÷ M(X)× 2/ν× M(NH3​)

16 g of N2 and 24 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(N2​)n(N2​)n(NH3​)m(NH3​)÷ M(N2​)× 2/1× M(NH3​)

In the reaction N2 + 3H2 → 2NH3, 12 g of N2 reacts completely (the other reactants are in excess) and 9.04 g of NH3 is collected. Find the percent yield Y.

M(N2) = 28.02 g/molM(NH3) = 17.03 g/molactual: 9.04 g
Y = %

3Solutions

1.Mass balance

Answer: ṁ2 = 533.3 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·(45 − 25)/(25 − 10) = 533.3 kg/h

Answer

ṁ2 = 533.3 kg/h
2.Mass balance

Answer: W = 540 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 = 900·20/50 = 360 kg/h

Total balance

W = F − P = 900 − 360 = 540 kg/h

Answer

W = 540 kg/h
3.Energy balance

Answer: tout = 61.67 °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·(150 − 120) = 108 kW

Heat picked up by the water

tout = tin + Q/(ṁccp,c) = 10 + 108/(0.5·4.18) = 61.67 °C

Answer

tout = 61.67 °C
4.Energy balance

Answer: ṁc = 3.753 kg/s

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

Heat released

Qh = 2.5·2.44·(170 − 70) = 610 kW

Heat that reaches the water

Qc = (1 − 0.1)·610 = 549 kW

Water flow rate

ṁc = Qc/(cp,c·Δt) = 549/(4.18·35) = 3.753 kg/s

Answer

ṁc = 3.753 kg/s
5.Ideal gas law

Answer: P = 6100 kPa

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

Moles

n = m/M = 410/4.003 = 102.4 mol
T = 85 + 273.15 = 358.1 K

Pressure

P = nRT/V = 102.4·8.314·358.1/0.05 = 6100 kPa

Answer

P = 6100 kPa
6.Ideal gas law

Answer: V2 = 1.392 L

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

Absolute temperatures

T1 = 308.1 K, T2 = 343.1 K

Combined gas law

V2 = V1·(P1/P2)·(T2/T1) = 5·(100/400)·(343.1/308.1) = 1.392 L

Answer

V2 = 1.392 L
7.Stoichiometry

Answer: m = 19.45 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) = 16/28.02 = 0.571 mol
n(H2) = 24/2.016 = 11.9 mol

Limiting reactant

0.571/1 = 0.571, 11.9/3 = 3.968 → N2

Mass of product

n(NH3) = 0.571·2 = 1.142 mol
m = 1.142·17.03 = 19.45 g

Answer

m = 19.45 g
8.Stoichiometry

Answer: Y = 61.97 %

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) = 12/28.02 = 0.4283 mol
m = 0.4283·2/1·17.03 = 14.59 g

Percent yield

Y = 9.04/14.59·100 = 61.97%

Answer

Y = 61.97%

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