| Name: | Oneeq20b - Equilibrium conversion in a tubular reactor - revised form | ||
| Source: | N. Brauner, M. Shacham and M. Cutlip, Chem. Eng. Educ., 30 (1), 20-25 (1996). | ||
| Reference/s | |||
| Model: | 1 implicit equation, indep. variable name: xa | ||
| Average difficulty level | |||
| Constraints: 0<=xa<=1 | |||
| Discontinuities: none in the range of interest | |||
| Initial range: xamin=0.75, xamax=1.2 | |||
| Solved by | Shacham, M., POLYMATH 5.1, build 19, April 1, 2001 | ||
| Model Eqs. | Equilibrium conversion in a tubular reactor - revised form |POLVER05_1 | ||
| f(xa)=-ra/FA0 # | |||
| T=313 # | |||
| P0=10 # | |||
| FA0=20*P0/(0.082*450) # | |||
| k1=1.277*1.e9*exp(-90000/(8.31*T)) # | |||
| k2=1.29*1.e11*exp(-135000/(8.31*T)) # | |||
| xa1=1+xa # | |||
| ra = (-k1*P0*(1-xa)/xa1+k2*P0*P0*xa*xa/(xa1^2)) # | |||
| xa(min)=.75, xa(max)=1.2 | |||
| Variable/function values | Variable | Value | f(x) |
| xa | 0.85 | 1.79372E-07 | |
| T | 313 | ||
| P0 | 10 | ||
| FA0 | 5.420054201 | ||
| k1 | 1.19915E-06 | ||
| k2 | 3.71E-12 | ||
| xa1 | 1.85 | ||
| ra | -9.72205E-07 | ||
| Root | Variable | Value | f(x) |
| xa | 0.999984253901100 | 2.9717E-13 | |
| T | 313 | ||
| P0 | 10 | ||
| FA0 | 5.42005420054200 | ||
| k1 | 1.1991500E-06 | ||
| k2 | 3.7120483E-12 | ||
| xa1 | 1.99998425390100 | ||
| ra | -1.6106679E-12 | ||
| Additional information | Elimination of the denominator from the expression | ||
| of ra eliminates the discontinuity and | |||
| alleviates considerably the solution. | |||
| The scaling problem remains. Function value is very small | |||
| throughout the entire range of interest (order of E-8). | |||
| Function plot | |||