CHAPTER 2: REACTOR DESIGN A – Phenol From journal, we set P A0 = 0.42 atm For vapor phase reaction at initial temperature, T 0 = 150 0 !atio " 2#Phenol = 4#1 at P = 2 atm $ % = & %0 #& A0 = 4#1 = 4 From journal, we use' PF! that act as stan'ar' shell an' tu(e e)chan*er reactor. Tu(e =+ 15 inches + -uter iameter, - = 1./ inch 4 .2 mm3 + nner iameter, = 1. inch 44./5mm3 6all thic7ness = 0.0 5 inch 1. 5mm3 esi*n e8uation9FA0 dx dW = r A , dW dx = FA 0 − rA ' !ate :aw9 r A ; = k 1 k A P A ( k B P B ) 2 ( 1 +k A P A +k B P B +k C P C +k D P D ) 3 <toichiometr&9 P A = P A0 1 3 P% = P A0 $ % – (#a 3 = P A0 4 5#2 3 P = P A0 P= P A0 om(ine >8uation9 ∫ 0 w dW = ∫ 0 0.75 F A 0 ( 1 +k A P A + k B P B +k C P C +k D P D ) 3 dX nsert all 'ata in pol&math, there?ore we *et9 6 = @12. 14 7* re?er Ta(le 2.13 P ( = P c 1 – 3
CHAPTER 2: REACTOR DESIGN A PhenolFrom journal, we set PA0 =
0.42 atmFor vapor phase reaction at initial temperature, T0 =
1500CRatio H2/Phenol = 4/1 at P = 2 atmB = yB0/yA0 = 4/1 = 4From
journal, we used PFR that act as standard shell and tube heat
exchanger reactor.Tube => 15 inches -> Outer Diameter, OD =
1.9 inch (48.26mm) -> Inner Diameter, ID = 1.77 inch
(44.95mm)Wall thickness = 0.065 inch (1.65mm)
Design equation:FA0 = -rA, = Rate Law:-rA = Stoichiometry: PA =
PA0 (1-X)PB = PA0 (B b/a X) = PA0 (4- 5/2 X)PC = PA0XPD =
PA0XCombine Equation: = Insert all data in polymath, therefore we
get: W = 6312.814 kg (refer Table 2.1)Pb = Pc (1 ) = 700 kg/m3From
equation W = L Ac PbL = = = 5707.79mFrom equation FA0 = -rA, = FA0
= -rA,Therefore, V = WSo, V = 6312.814 m3From equation, n = = = 700
tubesTherefore, L of reactor:L = = 8.15 m with 700 tubes.
