B is by catalytic dehydrogenation of A. A is fed to the reactor at a flow rate o

Question

B is by catalytic dehydrogenation of A. A is fed to the reactor at a flow rate of 200 produced kmol/h at 600 degree C. Two reactions take place: A rightarrow B + C A rightarrow D + E Total conversion of A and selectivity of B to D are 38% and 17, respectively. a) Calculate the product distribution b) If the temperature of outlet stream is 560 degree C, what is the amount of heat should be added? (nonisothermal-nonadiabatic) c) If the reactor operates adiabatically, what will be the exit temperature (adiabatic) d) If the reaction take place isothermally, calculate the heat load (isothermal)

Explanation / Answer

part a

Take basis of 1 hour so the calculation becomes

Moles of A = 200 kmol

conversion = 38 percent = (moles of A reacted/moles of A fed) * 100

this means that moles of A reacted = 76

unreacted moles of A = 124

now 76 moles of a dorm all products combined

but we are gicen the selectivity of products

selectivity of B to D is 17

which means moles of B/Moles of D = 17

that is B=17 D

also B+D= 76

so we get 17 D+D=76

D= 4.22 so E= 4.22

B= 17*4.2 = 71.4 so C= 71.4

part b

now let us equate the energy inflow and energy outflow

let Q be the amount of heat added

Cpa * Ta + Q = Cpa * Ta + Cpb * Tb + Cpc * Tc+ Cpd * Td+ Cpe * Te + Ha A + Hb B + Hc C + Hd D + He E

206*200*873 + Q = ((29.79*124)+(194* 71.4)+(29*71.4)+(165*4.22)+(45*4.22))*833 + (29.79*124)+ (147.56*71.4)+(135.21*4.22)

So Q= 18874333.7898 J = 18874.3 KJ

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