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 = 750 kg/hx1 = 35 %x2 = 20 %x3 = 30 %
ṁ2 = kg/h
2.Mass balanceHard
An evaporator concentrates a salt solution by boiling off pure water. Find the water evaporated W in kg/h.
F = 3500 kg/hxF = 12 %xP = 40 %
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 = 3 kg/scp,h = 2.1 kJ/(kg·K)Th: 120 → 70 °Cṁc = 1.5 kg/scp,c = 4.18 kJ/(kg·K)tin = 30 °C
tout = °C
4.Energy balanceHard
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: 160 → 80 °Ct: 25 → 75 °Ccp,c = 4.18 kJ/(kg·K)loss = 10 %
ṁc = kg/s
5.Ideal gas lawMedium
A tank of 30 L contains 330 g of CO₂ at 85 °C. Treat it as an ideal gas and find the pressure P.
m = 330 gM = 44.01 g/molV = 30 LT = 85 °CR = 8.314 J/(mol·K)
P = kPa
6.Ideal gas lawHard
A fixed amount of O₂ goes from 350 kPa, 5 L, 20 °C to 700 kPa and 50 °C. Find the new volume V2.
P1 = 350 kPaV1 = 5 LT1 = 20 °CP2 = 700 kPaT2 = 50 °C
V2 = L
7.StoichiometryMedium
12 g of Na and 20 g of Cl2 react according to 2Na + Cl2 → 2NaCl. What mass m of NaCl can form?
M(Na) = 22.99M(Cl2) = 70.9M(NaCl) = 58.44 g/mol
m = g
8.StoichiometryHard
In the reaction CaCO3 → CaO + CO2, 10 g of CaCO3 reacts completely (the other reactants are in excess) and 3.92 g of CaO is collected. Find the percent yield Y.
M(CaCO3) = 100.09 g/molM(CaO) = 56.08 g/molactual: 3.92 g
Y = %
3Solutions
1.Mass balance
Answer: ṁ2 = 375 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) = 750·(35 − 30)/(30 − 20) = 375 kg/h
Answer
ṁ2 = 375 kg/h
2.Mass balance
Answer: W = 2450 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 = 3500·12/40 = 1050 kg/h
Total balance
W = F − P = 3500 − 1050 = 2450 kg/h
Answer
W = 2450 kg/h
3.Energy balance
Answer: tout = 80.24 °C
Exchanger energy balance: with no losses, ṁhcp,h(Th,in − Th,out) = ṁccp,c(tout − tin).
Heat released by the hot stream
Q = 3·2.1·(120 − 70) = 315 kW
Heat picked up by the water
tout = tin + Q/(ṁccp,c) = 30 + 315/(1.5·4.18) = 80.24 °C
Answer
tout = 80.24 °C
4.Energy balance
Answer: ṁc = 2.101 kg/s
Energy balance with losses: heat released = heat absorbed + heat lost.
Heat released
Qh = 2.5·2.44·(160 − 80) = 488 kW
Heat that reaches the water
Qc = (1 − 0.1)·488 = 439.2 kW
Water flow rate
ṁc = Qc/(cp,c·Δt) = 439.2/(4.18·50) = 2.101 kg/s
Answer
ṁc = 2.101 kg/s
5.Ideal gas law
Answer: P = 744.2 kPa
Ideal gas law: PV = nRT, with n = m/M.
Moles
n = m/M = 330/44.01 = 7.498 mol
T = 85 + 273.15 = 358.1 K
Pressure
P = nRT/V = 7.498·8.314·358.1/0.03 = 744.2 kPa
Answer
P = 744.2 kPa
6.Ideal gas law
Answer: V2 = 2.756 L
Combined gas law: for a fixed amount of ideal gas, P1V1/T1 = P2V2/T2.
Absolute temperatures
T1 = 293.1 K, T2 = 323.1 K
Combined gas law
V2 = V1·(P1/P2)·(T2/T1) = 5·(350/700)·(323.1/293.1) = 2.756 L
Answer
V2 = 2.756 L
7.Stoichiometry
Answer: m = 30.5 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(Na) = 12/22.99 = 0.522 mol
n(Cl2) = 20/70.9 = 0.2821 mol
Limiting reactant
0.522/2 = 0.261, 0.2821/1 = 0.2821 → Na
Mass of product
n(NaCl) = 0.261·2 = 0.522 mol
m = 0.522·58.44 = 30.5 g
Answer
m = 30.5 g
8.Stoichiometry
Answer: Y = 69.96 %
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(CaCO3) = 10/100.09 = 0.09991 mol
m = 0.09991·1/1·56.08 = 5.603 g
Percent yield
Y = 3.92/5.603·100 = 69.96%
Answer
Y = 69.96%
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