A process is used to make product Z by a 1st order reaction (k=0.186 min-1) in 4
ID: 1070103 • Letter: A
Question
A process is used to make product Z by a 1st order reaction (k=0.186 min-1) in 4 ideal mixed flow reactors in series. The required overall conversion is 96.0% using an inlet reactant concentration = 5.7 mol/L (both fixed values) Original process operating conditions: Flowrate = 500 L/hr A. Calculate the size (L) of an individual reactor. ANSWER:55.379L B. What is the new flowrate (L/hr)? How does this change the production rate of Z (calculate the new production rate as % of original production rate)? If only use 2 MFR what is q? (Use same equation but with n =2 and V=5 5.4 L) ANSWER: 154 L/hr, 30.8 percent of original production C. If you were to shift production over to using only the PFR, what would the flowrate be (L/hr)? How does this change the production rate of Z (calculate the new production rate as % of original production rate)? ANSWER: 346.705L/hr. Production Rate: 1.897 times 10^3 mol/hr Production Rate percent: .693 Can someone please work out how to get the answers? thank you.Explanation / Answer
For a 1st order reaction when carried out in equal volumes of CSTRS
C4/CO= 1/(1+KT)4
T= V/VO and C4/CO= 1-0.96= 0.04
0.04 = 1/(1+0.186*T)4
1+0.186T= 2.24
0.186T =1.24
T = 1.24/0.1=86 =6.7min
T =V/VO, VO= volumetric flow rate = 500/60 L/min = 8.33L/min
V =volume of reactor = 6.7*8.33 L= 55.8 L
For two CSTRS
C4/CO = 1/(1+KT)2 =
0.04 = 1/(1+KT)2
1+KT= 5, KT= 4
T= 4/0.186= 21.5 min
V/VO= 21.5
VO = 55.8/21.5 =2.6 L/min =2.6*60 L/hr= 156 L/hr
Change i n production = 100*(156/500)= 31.2%
For a PFR using single reactor for 1st order reaction
KT= -ln(1-XA)
XA= converion =0.96
-ln(0.04)= 0.186*T
T= 17.3058 min
T= V/VO= 17.3058
V= 2*55.8 = 111.6
Vo=111.6/17.3058 =6.44 L/min
Vo = 6.44*60 L/hr =387 L/hr
Change in production =100* 387/500 =77.4%
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