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).
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