Mass balance
- Steady state
- in = out
- Total balance
- ṁ1 + ṁ2 = ṁ3
- Component balance
- ṁ1x1 + ṁ2x2 = ṁ3x3
- Evaporator
- F·xF = P·xP, W = F − P
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.
1Formulas and methods
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
- Write and balance the chemical equation.
- Convert the given mass to moles: n = m/M.
- Use the coefficients as the mole ratio to find the moles of the substance you want.
- Convert back to mass with m = n·M. With two reactants, find the limiting reactant first.
2Practice problems
1.Mass balanceMedium
How much of stream 2 must be added each hour to stream 1 so that the blended product contains 30% sugar? Find ṁ2.
ṁ1 = 800 kg/hx1 = 50 %x2 = 10 %x3 = 30 %
ṁ2 = kg/h
2.Mass balanceHard
An evaporator concentrates a sugar solution by boiling off pure water. Find the water evaporated W in kg/h.
F = 2100 kg/hxF = 20 %xP = 65 %
W = kg/h
3.Energy balanceMedium
Hot light oil cools water in a counter-flow heat exchanger with no heat loss. Find the water outlet temperature tout.
ṁh = 1.5 kg/scp,h = 2.1 kJ/(kg·K)Th: 160 → 110 °Cṁc = 1 kg/scp,c = 4.18 kJ/(kg·K)tin = 25 °C
tout = °C
4.Energy balanceHard
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: 150 → 70 °Ct: 25 → 70 °Ccp,c = 4.18 kJ/(kg·K)loss = 5 %
ṁc = kg/s
5.Ideal gas lawMedium
A tank of 5 L contains 80 g of Ar at 80 °C. Treat it as an ideal gas and find the pressure P.
m = 80 gM = 39.95 g/molV = 5 LT = 80 °CR = 8.314 J/(mol·K)
P = kPa
6.Ideal gas lawHard
A fixed amount of CH₄ goes from 250 kPa, 20 L, 35 °C to 750 kPa and 160 °C. Find the new volume V2.
P1 = 250 kPaV1 = 20 LT1 = 35 °CP2 = 750 kPaT2 = 160 °C
V2 = L
7.StoichiometryMedium
20 g of Fe2O3 and 40 g of CO react according to Fe2O3 + 3CO → 2Fe + 3CO2. What mass m of Fe can form?
M(Fe2O3) = 159.69M(CO) = 28.01M(Fe) = 55.85 g/mol
m = g
8.StoichiometryHard
In the reaction 2Na + Cl2 → 2NaCl, 30 g of Na reacts completely (the other reactants are in excess) and 69.4 g of NaCl is collected. Find the percent yield Y.
M(Na) = 22.99 g/molM(NaCl) = 58.44 g/molactual: 69.4 g
Y = %
3Solutions
1.Mass balance
Answer: ṁ2 = 800 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) = 800·(50 − 30)/(30 − 10) = 800 kg/h
Answer
ṁ2 = 800 kg/h
2.Mass balance
Answer: W = 1454 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 = 2100·20/65 = 646.2 kg/h
Total balance
W = F − P = 2100 − 646.2 = 1454 kg/h
Answer
W = 1454 kg/h
3.Energy balance
Answer: tout = 62.68 °C
Exchanger energy balance: with no losses, ṁhcp,h(Th,in − Th,out) = ṁccp,c(tout − tin).
Heat released by the hot stream
Q = 1.5·2.1·(160 − 110) = 157.5 kW
Heat picked up by the water
tout = tin + Q/(ṁccp,c) = 25 + 157.5/(1·4.18) = 62.68 °C
Answer
tout = 62.68 °C
4.Energy balance
Answer: ṁc = 0.7273 kg/s
Energy balance with losses: heat released = heat absorbed + heat lost.
Heat released
Qh = 0.5·3.6·(150 − 70) = 144 kW
Heat that reaches the water
Qc = (1 − 0.05)·144 = 136.8 kW
Water flow rate
ṁc = Qc/(cp,c·Δt) = 136.8/(4.18·45) = 0.7273 kg/s
Answer
ṁc = 0.7273 kg/s
5.Ideal gas law
Answer: P = 1176 kPa
Ideal gas law: PV = nRT, with n = m/M.
Moles
n = m/M = 80/39.95 = 2.003 mol
T = 80 + 273.15 = 353.1 K
Pressure
P = nRT/V = 2.003·8.314·353.1/0.005 = 1176 kPa
Answer
P = 1176 kPa
6.Ideal gas law
Answer: V2 = 9.371 L
Combined gas law: for a fixed amount of ideal gas, P1V1/T1 = P2V2/T2.
Absolute temperatures
T1 = 308.1 K, T2 = 433.1 K
Combined gas law
V2 = V1·(P1/P2)·(T2/T1) = 20·(250/750)·(433.1/308.1) = 9.371 L
Answer
V2 = 9.371 L
7.Stoichiometry
Answer: m = 13.99 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(Fe2O3) = 20/159.69 = 0.1252 mol
n(CO) = 40/28.01 = 1.428 mol
Limiting reactant
0.1252/1 = 0.1252, 1.428/3 = 0.476 → Fe2O3
Mass of product
n(Fe) = 0.1252·2 = 0.2505 mol
m = 0.2505·55.85 = 13.99 g
Answer
m = 13.99 g
8.Stoichiometry
Answer: Y = 91.01 %
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(Na) = 30/22.99 = 1.305 mol
m = 1.305·2/2·58.44 = 76.26 g
Percent yield
Y = 69.4/76.26·100 = 91.01%
Answer
Y = 91.01%
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