Page 2 of 6 PROBLEM 1 (20 points) A 8.5x10 gallon reactor is used to reat wastew
ID: 562307 • Letter: P
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Page 2 of 6 PROBLEM 1 (20 points) A 8.5x10 gallon reactor is used to reat wastewater from an industrial source. Wastewate enters the reactor at a rate of 445 gallons per minute with a pollutant concentration of 750 mg/ Pollutant decay occurs in the reactor with a first-order reaction rate constant of 0.25 hour (a) [15 pts] Assume that the reactor can be modeled as an ideal completely-mixed continuou flow reactor (CMFR). What would be the expected effluent pollutant concentration (in mg exiting the reactor at steady-state? mia-mout-kvc )(C 4452 (150 )-445(cout)-9,25 ( 8.5xlo5) (e) = G avt 1,567 ml (1 b) [5 pts) What would be the reactor's hydraulic residence time (in hours)?Explanation / Answer
for any order reaction, loss of moles due to chemical reacton= KCAnV, n= order of reaction and V= volume, K= rate cosntant, CA= concentration of A. for 1st order n=1 and for a Completely mixed stirred tank reactor (CFSTR) at steady state for 1st order reaction
FAO( molar flow rate in )= FA( molar flow rate out )+ KCA( loss due to chemical reaction)
FA= FAO*(1-XA)
FAO= FAO*(1-XA)+ KCAV
FAOXA= KCAV (1)
but T= space time = V/VO, V= volume of reactor and Vo= volumetric flow rate, T= V/VO= VCAO/FAO
V/FAO= T/CAO= XA/KCA
T= CAOXA/KCA
CAO= initial concentration and XA= conversion = 1-CA/CAO
T= space time = volume of reactor/ volumetric flow rate= 8.5*105/445 gallons per minute =1910 min= 1910/60 hrs =31.83 hrs
hence 31.83= CAO*(1-CA/CAO)/ KCA = (CAO-CA)/KCA
CA= CAO/(1+K*31.83)
CAO= 750 mg/L and K= rate constant = 0.25/hr
CA= 750/(1+31.83*0.25)=83.73 mg/L
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