SR

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- Example of Batch and Continuous reactive process
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- Group 12 Syumayyah Binti Rasis (191141282)
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- Example 1 - Combustion of Methane
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- Methane is burned with oxygen to yield carbon dioxide and water. The feed contains 20 mole% CH4, 60% O2, and 20% CO2 and a 90% conversion of the limiting reactant is achieved. Calculate the molar composition of the product stream using (1) balance on molecular species, (2) atomic balances and (3) the extent of reaction.
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- Solution:
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- Basis 100 mol Feed
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- 100 mole
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- 0.200 CH4 0.600 O2 0.200 CO2
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- Since a 2:1 ratio of O2 to CH4 would be stoichiometric and the actual ratio is 3:1, CH4 is the limiting reactant and O2 is in excess. Before the balances are written, the given process information should be used to determine the unknown variables or relation between them. In this case, the methane conversion of 90% tells us that 10% of the methane fed to the reactor emerges in the product, or
- 00:25
- n1 = 0.100(20.0 mol CH4 fed) = 2.0 mol CH4
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- Now all the remains are the balances. We will proceed by each of the indicated methods.
- 00:31
- 1. Molecular Balances
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- CH4 reacted = 20.0 mol CH4 fed 0.900 mol react = 18 mol CH4 react mol fed
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- CO2 balance (Input + generation = output)
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- H2O balance (generation = input)
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- O2 balance (Input = output + consumption)
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- 100 mol 0.600 mol O2 = n2 + 18 mol CH4 react 2 mol O2 react mol 1 mol CH4 react
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- n2 = (60-36) mol O2 = 24 mol O2
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- In summary, the output quatities are 2 mol CH4, 24 mol O2, 38 mol CO2 and 36 mol H2O, for a total of 100 mol. (Since 3 moles of product are produced for every 3 moles of reactants consumed, it should come as no suprised that total moles in is equal to total moles out.) The mole fraction of the product gas components are thus: 0.02 mol CH4/mol 0.24 mol O2/mol 0.38 mol CO2/mol 0.36 mol H2O/mol
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- 2. Atomic Balances
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- Referring to the flowchart, we see that a balance on atomic carbon involves only one unknown (n3), and a balance on atomic hydrogen also involves one unknown (n4), but a balance on oxygen involves three unknowns. We should therefore write C and H balances first, and then O balance to determine the remaining unknown variables. All atomic balance have the form input equal to output. (We will just determine the component amounts; calculation of mole fractions then follows as in part 1.
- 01:07
- C balance
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- 20.0 mol CH4 1 mol C + 20.0 mol CO2 1 mol C = 2.0 mol CH4 1 mol C 1 mol CH4 1 mol CO2 1 mol CH4 + n3 mol CO2 1 mol C 1 mol CO2
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- n3 = 38 mol CO2
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- H balance
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- 20.0 mol CH4 4 mol H = 2.0 mol CH4 4 mol H + n4 mol H2O 2 mol H 1 mol CH4 1 mol CH4 1 mol H2O
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- n4 = 36 mol H2O
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- O balance
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- 60.0 mol O2 2 mol C + 20.0 mol CO2 2 mol O = (n2 mol O2)(2) 1 mol O2 1 mol CO2 + (38 mol CO2)(2) + (36 mol H2O)(1)
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- n2 = 24 mol O2
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- Confirming the results obtained using molecular balances.
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- 3. Extent of Reaction
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- As discussed earlier, for all reactive species
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- n (out) = n (in) + β(i) ξ
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- For species in this problem, we may write
- 01:36
- n1 = 20.0 mol - ξ n2 = 60.0 mol - 2ξ n3 = 20.0 mol + ξ n4 = 0 + 2ξ
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- ξ = 18.0 mol n2 = 60.0 mol - 2(18.0 mol) = 24.0 mol O2 n3 = 20.0 mol + 18.0 mol = 38.0 mol CO2 n4 = 2(18.0 mol) = 36.0 mol H2O
- 01:38
- Again, the previous solutions obtained.
- 01:39
- Thankyou!