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 20% sugar? Find ṁ2.
ṁ1 = 900 kg/hx1 = 15 %x2 = 35 %x3 = 20 %
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 = 500 kg/hxF = 17 %xP = 55 %
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 = 0.5 kg/scp,h = 2.1 kJ/(kg·K)Th: 160 → 70 °Cṁc = 1 kg/scp,c = 4.18 kJ/(kg·K)tin = 15 °C
tout = °C
4.Energy balanceHard
Hot ethanol heats water in an exchanger that loses 15% of the heat released by the hot stream to the surroundings. Find the water flow rate ṁc needed.
ṁh = 2 kg/scp,h = 2.44 kJ/(kg·K)Th: 160 → 130 °Ct: 15 → 90 °Ccp,c = 4.18 kJ/(kg·K)loss = 15 %
ṁc = kg/s
5.Ideal gas lawMedium
A tank of 30 L contains 310 g of N₂ at 30 °C. Treat it as an ideal gas and find the pressure P.
m = 310 gM = 28.01 g/molV = 30 LT = 30 °CR = 8.314 J/(mol·K)
P = kPa
6.Ideal gas lawHard
A fixed amount of Ar goes from 150 kPa, 20 L, 40 °C to 150 kPa and 90 °C. Find the new volume V2.
P1 = 150 kPaV1 = 20 LT1 = 40 °CP2 = 150 kPaT2 = 90 °C
V2 = L
7.StoichiometryMedium
4 g of Al and 48 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 2Al + 3Cl2 → 2AlCl3, 30 g of Al reacts completely (the other reactants are in excess) and 139.4 g of AlCl3 is collected. Find the percent yield Y.
M(Al) = 26.98 g/molM(AlCl3) = 133.33 g/molactual: 139.4 g
Y = %
3Solutions
1.Mass balance
Answer: ṁ2 = 300 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) = 900·(15 − 20)/(20 − 35) = 300 kg/h
Answer
ṁ2 = 300 kg/h
2.Mass balance
Answer: W = 345.5 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 = 500·17/55 = 154.5 kg/h
Total balance
W = F − P = 500 − 154.5 = 345.5 kg/h
Answer
W = 345.5 kg/h
3.Energy balance
Answer: tout = 37.61 °C
Exchanger energy balance: with no losses, ṁhcp,h(Th,in − Th,out) = ṁccp,c(tout − tin).
Heat released by the hot stream
Q = 0.5·2.1·(160 − 70) = 94.5 kW
Heat picked up by the water
tout = tin + Q/(ṁccp,c) = 15 + 94.5/(1·4.18) = 37.61 °C
Answer
tout = 37.61 °C
4.Energy balance
Answer: ṁc = 0.3969 kg/s
Energy balance with losses: heat released = heat absorbed + heat lost.
Heat released
Qh = 2·2.44·(160 − 130) = 146.4 kW
Heat that reaches the water
Qc = (1 − 0.15)·146.4 = 124.4 kW
Water flow rate
ṁc = Qc/(cp,c·Δt) = 124.4/(4.18·75) = 0.3969 kg/s
Answer
ṁc = 0.3969 kg/s
5.Ideal gas law
Answer: P = 929.8 kPa
Ideal gas law: PV = nRT, with n = m/M.
Moles
n = m/M = 310/28.01 = 11.07 mol
T = 30 + 273.15 = 303.1 K
Pressure
P = nRT/V = 11.07·8.314·303.1/0.03 = 929.8 kPa
Answer
P = 929.8 kPa
6.Ideal gas law
Answer: V2 = 23.19 L
Combined gas law: for a fixed amount of ideal gas, P1V1/T1 = P2V2/T2.
Absolute temperatures
T1 = 313.1 K, T2 = 363.1 K
Combined gas law
V2 = V1·(P1/P2)·(T2/T1) = 20·(150/150)·(363.1/313.1) = 23.19 L
Answer
V2 = 23.19 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) = 48/70.9 = 0.677 mol
Limiting reactant
0.1483/2 = 0.07413, 0.677/3 = 0.2257 → 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 = 94 %
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(Al) = 30/26.98 = 1.112 mol
m = 1.112·2/2·133.33 = 148.3 g
Percent yield
Y = 139.4/148.3·100 = 94%
Answer
Y = 94%
ExamProblems · examproblems.com/courses/material-and-energy-balances