Monday, 11 April 2011

DESIGN OF REACTOR


In the designing of a reactor, following things should be in mind

  • Type of Reactor
  • Method of Reactor
  • Size of Reactor

5.4.1: TYPE OF REACTOR:

                        There are three basic types of reactors.
  • Batch Reactor
  • Plug Flow Reactor
  • Continuous Stirred Tank Reactor

In most of industries and for the commercial scale production plug flow reactors and continuous stirred tank reactors are used. On the small scale batch reactor are used. Because our process is continuous process therefore, we can use plug flow reactor or continuous stirred tank reactor to achieve our goal.
Reactor is selected on the basis of following parameter:
  • Conversion                                                                               
  • Selectivity
  • Productivity
  • Safety                                                                       
  • Economics
  • Availability
  • Flexibility
  • Compatibility with processing
  • Energy utilization
  • Feasibility
  • Investment operating cost
By taking these above points in mind, we select plug flow reactor for our process of production of  Mono- Ethylene Glycol. Because
a.       Mostly used for MEG production
b.      High conversion
c.       No back mixing
d.      Less volume than CSTR
e.       Economical

5.4.2: TYPE OF REACTOR:
According to our process conditions, our reaction takes place at 15 atm pressure and 195 0C temperature. For this purpose, we needed isothermal reactor, which operate at constant temperature. Therefore we use isothermal plug flow reactor.

5.4.3: SIZE OF REACTOR:
For isothermal PFR reactor, the performance equation is given as



Where
            V         = Volume of reactor
            FAo         = Flow rate of reactant in
            XA          = Conversion
            -rA          = Rate of reaction
As we are using 1200 % excess of water, so reaction will be Pseudo First Order reaction
So



Reaction Temperature               = 195 0C
Reaction Pressure                                 = 15 atm
k (reaction constant)                             = 0.23/min
Conversion                                           = 85%
Volume of the reactor =?
We need 
Total Molar Flow Rate In=         = 497.4 Kmoles/hr
         = Kmoles of Ethylene Oxide/Volume of the mixture
Kmoles of Ethylene Oxide                    = 497.4 Kmoles
Now we need Volume of the mixture;
Volume of the mixture               = Total weight of the mixture/Density of the mixture                         
Total weight of the mixture                    = 129324 Kg/hr
Now Density of the mixture;
Density of Water                                  =868.5 Kg/m3
Density of EO                           =405.9 Kg/m3
Density of the mixture               =  (0.169*405.9) + (0.83*868.5)
                                                            = 68.59 + 720.85
                                                            = 789.44 kg/ m3
By using Equation for Volume of the mixture, so
Volume of the mixture                   = 163.817 m3
And
                                                     = 3.036 kmole/m3
Now
Volume of the Reactor
Thus
            Volume of PFR                        = 22.5 m3
Residence Time:

                                                                          = 22.5/163.817
                                                                          = 8.23 min.
Length and Diameter of Reactor:
Volume of reactor = V              = π/4 (D2L)                                        
L/D                  = 200 (P.F.R)
L                      = 200 * D                                                                    
V                     = 50 π D3
D                     = (V/50 π)1/3
D                     = (22.5/50* π)1/3
D                     = 0.52 m                                                                           
L                      = 200 * 0.52 = 104 m

Selection of Pipe:    
Nominal Pipe Size                               = 2 inch            Schedule No. 80         
            Di                                 =1.939 inch      = 0.04925 m
            Do                               =2.38 inch        =0.0605 m
Flow Area of one pipe                          = 1.905 * 10-3 m2
Required Flow Area for Plug Flow        = π/4 (0.52) 2                                       
                                                            = 0.212 m2
No. of pipes required                           = 0.212/1.905 * 10-3
                                                            = 112 pipes
Assume 8 passes
No. of  pipe in each pass                      =112/8             =14pipes
Length of each pipe                              =104/8
                                                            = 13 m
Pitch:
Triangular Pitch
                        PT                                           =1.25 Do
                                                            =1.25(0.0605)
                                                            =0.08m
Pipe Spacing:
                                                            =0.08-.0605    =0.02m
Bundle Diameter:
Selection: U tube and Fixed.
Db        = Do(nT/K1)1/n1
For 8 passes
n1 = 2.675
K1 = 0.0365
Do = 2.38 in = 60.452 mm
nT = 112
Db        = 60.452(112/0.0365) 1/2.675
Db        = 1216.00 mm 
            = 1.22m
Form fig12.10 (C.R vol. 6)

Inside Shell Diameter- Bundle Diameter = 18mm
Inside Shell Diameter                            =18 + 1216
                                                            =1234mm
                                                            =1.234 m

No comments:

Post a Comment