Question

The elementary irreversible gas phase reaction A --> B + C is carried out in a PFR packed with catalyst. Pure A enters the reactor at a volumetric flowrate of 20 dm³ /s at a pressure of 10 atm and 450K. Consider that the heat is removed by a heat exchanger jacketing the reactor. The flowrate of coolant through the jacket is sufficiently high so that the ambient exchanger temperature is constant at 50^oC.

C_pA=40 J/mol.K HfA=-70 kJ/mol

C_pB=25 J/mol.K H_fB=-50 kJ/mol   (T_R=273 K)

C_pC=15 J/mol.K H_fC=-40 kJ/mol

k = 0.133 exp (E/R [ 1/450 ? 1/T ]) dm³ kg cat. s with E = 31.4 kJ/mol

U= 100 W/m2 .K

D_t=10 cm

p_b=5 g/cm3

a) Plot the conversion and temperature down the plug flow reactor until 80% conversion is achieved. Discuss the effect of feed temperature on the temperature profile and conversion.

b) Calculate the amount of catalyst required to achieve 80% conversion

Please do not copy the previous solutions. This question is different

Answer 1

The given gas phase reaction

A --> B + C

volumetric flowrate v = 20 dm3 /s

pressure P0 = 10 atm

Temperature T0 = 450K

Design equation for catalytic PFR

dX/dW = - rA'/FA0

Rate constant

k = 0.133 exp ( E/R*[ 1/450 ? 1/T ] )

E = 31.4 kJ/mol x 1000J/kJ = 31400 J/mol

Heat of reaction = sum of heat of formation of products - sum of heat of formation of reactants

H rxn = HfC + HfB - HfA

= - 40 - 50 + 70

= - 20 kJ/mol

= - 20000 J/mol

Cp = CpC + CpB - CpA

= 15 + 25 - 40

= 0

Energy balance equation

From the polymath

For the differential and explicit equations

conversion (X) along with the plug flow reactor until 80%

Temperature (T) along with the plug flow reactor until 80%

Part b

T (at X = 0.8) = 450 + 500*0.8

T = 850 K

k = 0.133 exp ( 31400/8.314*[ 1/450 ? 1/850] )

k = 6.9 dm3 kg cat. s

amount of catalyst

W = FA0*X/(-rA')

= CA0*v*X/(kCA)

CA0 = P0/RT0 = 10 atm / 0.0821 dm3-atm/mol-K x 450 K

= 0.2707 mol/dm3

CA = CA0 ( 1 - X)T /(1+X)T0

= 0.2707 (1-0.8)*850/(1+0.8)*450

= 0.0568 mol/dm3

W = 0.2707*20*0.8/(6.9*0.0568)

= 39.42 Kg