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 25% salt? Find ṁ2.
ṁ1 = 1000 kg/hx1 = 40 %x2 = 15 %x3 = 25 %
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 = 1500 kg/hxF = 15 %xP = 50 %
W = kg/h
3.Energy balanceMedium
Hot ethanol cools water in a counter-flow heat exchanger with no heat loss. Find the water outlet temperature tout.
ṁh = 2 kg/scp,h = 2.44 kJ/(kg·K)Th: 120 → 80 °Cṁc = 2 kg/scp,c = 4.18 kJ/(kg·K)tin = 30 °C
tout = °C
4.Energy balanceHard
Hot glycol mix 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 = 1.5 kg/scp,h = 3.6 kJ/(kg·K)Th: 160 → 80 °Ct: 30 → 50 °Ccp,c = 4.18 kJ/(kg·K)loss = 10 %
ṁc = kg/s
5.Ideal gas lawMedium
A tank of 20 L contains 400 g of CH₄ at 45 °C. Treat it as an ideal gas and find the pressure P.
m = 400 gM = 16.04 g/molV = 20 LT = 45 °CR = 8.314 J/(mol·K)
P = kPa
6.Ideal gas lawHard
A fixed amount of Ar goes from 400 kPa, 2 L, 35 °C to 350 kPa and 170 °C. Find the new volume V2.
P1 = 400 kPaV1 = 2 LT1 = 35 °CP2 = 350 kPaT2 = 170 °C
V2 = L
7.StoichiometryMedium
4 g of Al and 80 g of Cl2 react according to 2Al + 3Cl2 → 2AlCl3. What mass m of AlCl3 can form?
M(Al) = 26.98M(Cl2) = 70.9M(AlCl3) = 133.33 g/mol
m = g
8.StoichiometryHard
In the reaction 2Na + Cl2 → 2NaCl, 12 g of Na reacts completely (the other reactants are in excess) and 18.91 g of NaCl is collected. Find the percent yield Y.
M(Na) = 22.99 g/molM(NaCl) = 58.44 g/molactual: 18.91 g
Y = %
3Solutions
1.Mass balance
Answer: ṁ2 = 1500 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) = 1000·(40 − 25)/(25 − 15) = 1500 kg/h
Answer
ṁ2 = 1500 kg/h
2.Mass balance
Answer: W = 1050 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 = 1500·15/50 = 450 kg/h
Total balance
W = F − P = 1500 − 450 = 1050 kg/h
Answer
W = 1050 kg/h
3.Energy balance
Answer: tout = 53.35 °C
Exchanger energy balance: with no losses, ṁhcp,h(Th,in − Th,out) = ṁccp,c(tout − tin).
Heat released by the hot stream
Q = 2·2.44·(120 − 80) = 195.2 kW
Heat picked up by the water
tout = tin + Q/(ṁccp,c) = 30 + 195.2/(2·4.18) = 53.35 °C
Answer
tout = 53.35 °C
4.Energy balance
Answer: ṁc = 4.651 kg/s
Energy balance with losses: heat released = heat absorbed + heat lost.
Heat released
Qh = 1.5·3.6·(160 − 80) = 432 kW
Heat that reaches the water
Qc = (1 − 0.1)·432 = 388.8 kW
Water flow rate
ṁc = Qc/(cp,c·Δt) = 388.8/(4.18·20) = 4.651 kg/s
Answer
ṁc = 4.651 kg/s
5.Ideal gas law
Answer: P = 3298 kPa
Ideal gas law: PV = nRT, with n = m/M.
Moles
n = m/M = 400/16.04 = 24.94 mol
T = 45 + 273.15 = 318.1 K
Pressure
P = nRT/V = 24.94·8.314·318.1/0.02 = 3298 kPa
Answer
P = 3298 kPa
6.Ideal gas law
Answer: V2 = 3.287 L
Combined gas law: for a fixed amount of ideal gas, P1V1/T1 = P2V2/T2.
Absolute temperatures
T1 = 308.1 K, T2 = 443.1 K
Combined gas law
V2 = V1·(P1/P2)·(T2/T1) = 2·(400/350)·(443.1/308.1) = 3.287 L
Answer
V2 = 3.287 L
7.Stoichiometry
Answer: m = 19.77 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(Al) = 4/26.98 = 0.1483 mol
n(Cl2) = 80/70.9 = 1.128 mol
Limiting reactant
0.1483/2 = 0.07413, 1.128/3 = 0.3761 → Al
Mass of product
n(AlCl3) = 0.07413·2 = 0.1483 mol
m = 0.1483·133.33 = 19.77 g
Answer
m = 19.77 g
8.Stoichiometry
Answer: Y = 61.99 %
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(Na) = 12/22.99 = 0.522 mol
m = 0.522·2/2·58.44 = 30.5 g
Percent yield
Y = 18.91/30.5·100 = 61.99%
Answer
Y = 61.99%
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