Your firm is designing a reactor to treat chemical waste containing ethylene gly
ID: 479843 • Letter: Y
Question
Your firm is designing a reactor to treat chemical waste containing ethylene glycol. The initial concentration of the chemical is 500 mg/L. With treatment the waste will degrade at 0.4/hr. Q=8000 gpd (gallons per day) a. Your first preliminary design is a CSTR 5000 gallon volume reactor. What is the SS diluents concentration? b. Your second preliminary design is a PFR, again a 5000 gallon volume reactor. What is the outlet concentration? c. Your third preliminary design is a series of 4 (1250 gallon each reactor) CSTRs. What is the outlet concentration from the final, 4^th in series, reactor ? (Q same, 8000 gpd).Explanation / Answer
From the rate constant of the reaction, the reaction is 1st order. For 1st order reaction in a CFSTR,
T= V/VO= CAOXA/-rA, -rA= rate of reaction = KCA= KCAO*(1-XA), CAO= initial concentration of A=500 mg/L XA= conversion = 1-CA/CAO, CA= concentration at any time and K= rate constant =0.4/hr, V= Volume of reactor=5000 gallons, Vo= Volumetric flow rate =8000 gallons per day
Hence T= space time = (5000/8000) days=0.625 days= 0.625*24hrs =15 hrs
KT= XA/(1-XA)= 15*0.4 = XA/(1-XA), 6= XA/(1-XA), 6-6XA= XA, XA= 6/7 = 0.857
But XA= 1-CA/CAO, CA/CAO= 1-XA= 1-0.857= 0.143, CA= 800*0.143 mg/L
For PFR, KT= -ln(CA/CAO), 6 = -ln(CA/CAO), CA/CAO= 0.0025, CA=800*0.0025=2 mg/L
3. for the case of 4 CFSTRS in series, CA/CAO= 1/(1+KT)4,
T= V/VO= 1250/8000 days = 0.15625 days= 3.75 hr
CA= concentration at the outlet of 4th reactor
KT= 0.4*3.75= 1.5 , CA/CAO= 1/(1+1.5)4 = 0.0256, CA= 800*0.0256 mg/L=20.48 mg/L
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