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Exercises reactors and heterogeneous reactors
| Content Provider | Scilit |
|---|---|
| Author | Schmal, Martin |
| Copyright Year | 2014 |
| Description | Given the average residence time, dXA (1 + εAXA) (−rA) (20.2) t=−1 k ln(1 −XA) (20.3) The value of the rate constant can be calculated at 150◦C (reaction temperature), which is: ln k=−5400 T + 12.5 where: k= 0.649 min−1 As t= 3 min XA = 0.857 To calculate the volume, one has the PFR equation: τ = V v0 dXA ( − rA) Substituting (−rA), we obtain the following solution: τ =−1 k [(1 + εA) ln (1 −XA) + εAXA] (20.4) Calculating εA: εA =−0.25 Substituting these values into Equation 20.4, we get: τ = 2.58 min But F0 = 100mol/min However: FA0 + FB0 = 100 FA0 + 3FA0 = 100 FA0 = 25 mol/min C0 = PRT = 2.88 × 10 −1mol/L For v0: v0 = F0C0 = 346 L/min Therefore, the volume of PFR reactor is: V = 0.89 m3 In the CSTR reactor, assuming the same conversion, we use the equation: V v0 = τ =CA0 XA( − rA) Replacing the rate and substituting these values for the same conversion, we obtain: τ = 4.71 min With the same input flow, we get: V = 1.63 m3 This gives us a ratio of: VCSTR VPFR = 1.8 SE.2 The reaction A k−→R+ S is irreversible and first order. It is conducted in a PFR with 50 tubes, each with “½'' in diameter and 1.0 m of height. 200 kg/h of reactant A (MW = 80 g/gmol) with 30% inert is introduced at a pressure of 50 bar at 500◦C. The output conversion is 80%. Calculate the average residence time. Book Name: Chemical Reaction Engineering |
| Related Links | https://content.taylorfrancis.com/books/download?dac=C2011-0-08524-8&isbn=9780429217326&doi=10.1201/b16743-24&format=pdf |
| Ending Page | 572 |
| Page Count | 90 |
| Starting Page | 483 |
| DOI | 10.1201/b16743-24 |
| Language | English |
| Publisher | Informa UK Limited |
| Publisher Date | 2014-04-04 |
| Access Restriction | Open |
| Subject Keyword | Book Name: Chemical Reaction Engineering Chemical Engineering Residence Time Reactor Average Residence Fa0 K Ln |
| Content Type | Text |
| Resource Type | Chapter |
