An activated sludge reactor is designed to remove the BOD from the wastewater. E

Question

An activated sludge reactor is designed to remove the BOD from the wastewater. Experiments on the rate of biodegradation provided the data below. Using the data below, determine if the rate of BOD consumption is a zero, 1^st or 2^nd order reaction rate. Show all 3 graphs. Determine the value of the kinetic constant for the rate model you choose. If the wastewater spends 6 hours in the activated sludge tank, what would you expect the effluent BOD to be? How big (gallons capacity) must the tank be (minimum size) if we use a design value of 15 mg/l BOD in the effluent and the unit is treating 25 MGD of wastewater ? The influent concentration was measured at time 0.

Explanation / Answer

for zero order -dCA/dt= K, CA= CAO-Kt

where CA =concentration of BOD at any time, CAO= Concentration of BOD at zero time, K is rate constant

if the reaction is 1st order, the plot of BOD vs t has to be linear.

for 1st order -dCA/dt= KCA or lnCA = lnCAO-Kt

So a plot CA vs t will be expontential.

where K is 1st order rate constant

for 2nd order, -dCA/dt= KCA2, when integrated 1/CA =1/CAO + Kt

the plots for zero order and 1st order can be combine into a single plot of CA vs t while for 2nd order,

the plot 1/CA vs t is drawn which shows it to be strainght line with less R2 value than 1st order.

the plots are shown below

the reaction is 1st order CA= CAO*e(-Kt), K= rate constant = 0.048/min

for 6 hrs= 6*60min= 360 min

CA= 170*e(-0.048*360) = 3.12*10-8 mg/L

4 Assuming the reactor to Be CSTR for 1st order reaction

T= V/VO= (CO-C1)/0.048C1

25 Mgd=25Mgd/day= (25/24*60 ) Mgd/min=0.0173 Mgd/min

V/0.0173 = (170-15)/(0.048*15)= 215

Vo= 215*0.0173=3.72 Mgd

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