ENERGY BALANCE AROUND THE REACTOR

 Energy At Input:

We know that

Q                                 = m C_P 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

Q_IN                               = (n _H2O C_{P H2O} + n _EO C_{P EO}) (195-25)

                        Q_IN                               = {(5966.7)(75.276) + (497.6)(96.5)}(170)

                        Q_IN                               = 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                   +          H2O                    =        MEG

                       

∆Hr1                            = [(-386)] – [(-129)+1(-242)]

                                                            = -15 kJ/mole

                                                           = -15,000 kJ/Kmole

Reaction 2:-

                        2EO                 +          H2O              =              DEG

∆Hr2                            = [(-571)] – [2(-129)+(-242)]

                                                            = -71 KJ/mole

                                                            = -71,000 kJ/Kmole

Reaction 3:-

           

3EO                 +          H2O              =            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

Q_OUT                = 129286 * 4.24 * (468-298)

                                    Q_OUT                = 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