Proble 1: Adiabatic PFR Operation You are running the following gas-phase reacti

Question

Proble 1: Adiabatic PFR Operation You are running the following gas-phase reaction in an adiabatic and isobaric PFR: A+2B C The reaction rate is given by-r,-k CB2. The rate constant, k, is known to obey the Arrhenius equation with E 500]/mol. At 300K, k is measured to have a value of 0.0055 L/(mol's) a. Given the feed temperature is 500K, find the outlet conversion and outlet temperature. [T = 500K, X = 0.0032] Parameters: V = 150L CP,A = CI,B = 150 J/ (mol*K) vo = 50 L/s Yao = 0.2 yB0 = 0.6 Fo=5mol/s IR =-7000 J / mol b. If the reactor temperature must never exceed Tmax 600K and you wish to have a final conversion of 0.95, what feed temperature will allow you to use the smallest reactor volume possible? What is this volume? To 593K, V-70000L

Explanation / Answer

A)Using Arrhenius equation

AT 300 K the value of the fraction is

        e -EA/RT    = e- 500/8.31*300

                     =8.18*10-1

At 500 K the value of the fraction is

    e-EA/RT= e-500/8.31*500

              =8.86*10-1

You can see that the fraction of the molecules able to react has (8.18 to 8.86*10-1) 8% increased by increasing the temperature by 200K.

B)Determinge Xe for a PFR with no pressure drop p=p0

The reactor volume when X=0.8Xe

Given yao=0.2

First Calculate Xe

kc= Xe(1+xe)/2YAO(1-Xe)2

Xe0.89

X=0.8Xe=0.711

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