Monday, 11 April 2011

ENERGY BALANCE AROUND THE REACTOR

 Energy At Input:


We know that
Q                                 = m CP DT
Datum Temperature                              = 25 0C
Input Temperature                                = 195 0C
Mean Temperature                               = 110 0C
∆T= 195 -25 = 170 0C
Cp of Ethylene Oxide at 110 0C           = 96.5 KJ/Kmole oK
Cp of Water at 110 0C                        = 75.275 KJ/Kmole oK
QIN                               = (n H2O CP H2O + n EO CP EO) (195-25)
                        QIN                               = {(5966.7)(75.276) + (497.6)(96.5)}(170)
                        QIN                               = 8.4 * 10 7 KJ

Energy Generation Due To Reaction:


Component
Heat of Formation
EG
-386 KJ/mole
EO
-129 KJ/mole
Water
-242 KJ/mole
DEG
-571 KJ/mole
TEG


-654 KJ/mole




Reaction 1:-
           

EO                   +          H
2O                    =        MEG

                       
∆Hr1                            = [(-386)] – [(-129)+1(-242)]
                                                            = -15 kJ/mole
                                                           = -15,000 kJ/Kmole


Reaction 2:-


                        2EO                 +          H
2O              =              DEG


∆Hr2                            = [(-571)] – [2(-129)+(-242)]
                                                            = -71 KJ/mole
                                                            = -71,000 kJ/Kmole

Reaction 3:-
           

3EO                 +          H
2O              =            TEG

                       
∆Hr3                            = [(-654)] – [3(-129)+(-242)]
                                                            = -25 kJ/mole
                                                            = -25,000 kJ/Kmole


Total Heat of Reaction = ∆Hro
      = [(-15000)*(411.4)] + [(-71,000)*(40)] + [(-25,000)*(2)]
                                                            = - 9.0 *106 KJ
Energy At Output:

Mean CP of product mixture                 = 4.24 KJ/Kg K
Mass of mixture leaving the Reactor      = 129286 Kg

We know that
                                     Q                    = m CP DT
QOUT                = 129286 * 4.24 * (468-298)
                                    QOUT                = 9.3 *10 7 KJ

According to law of conservation of energy:

 { Transfer  }    =    { Transfer }    +    {Energy }
   of  energy               of  energy            generation   
   out of system           into system             within          
through system       through system           system
   boundary                   boundary

   9.3 *10 7 KJ    =       8.4 *10 7 KJ + 9.0 *10 6 KJ
   9.3 *10 7 KJ    =       9.3 *10 7 KJ

1 comment:

  1. why you multiply to total heat of reaction by 411.4 , 40 and 2.
    what is the reason

    ReplyDelete