Question

Please answer #5

4. What is the theoretical oxygen demand for a wastewater flow of 25 MGD (million gallons per day) containing 100 mg/L ethanol (C2H5OH)? 5. If this wastewater was discharged into a stream flowing at 10 m/s with an ultimate BOD of 5 mg/L, dissolved oxygen (DO) of 5mg/L, and deoxygenation rate coefficient ka of 0.2/day: (a) Find ultimate BOD of mixture just downstream of wastewater discharge. State all assumptions used. (b) In your opinion, is the wastewater discharge permissible without treatment and why? (c) What ultimate BOD can you expect 10 km downstream, assuming uniform cross sectional area of 20 m??

Answer 1

We will have to take the flow rate of wastewater as 25 million gallons per day. 1 gallon =3.785 L 1 million = 10^6 Thus 25 million gallons per day = 25*3.785*10^61/day =35*3.785*10^6/3600*24V/ =1095.2 Us 1000 L = 1m13 Thus flow rate=1.09 m^3/s Parta- Stream is discharged in to a stream flowing at a rate of 10 m3/s After mixing these two steams we will have flow rate of 10-1.09 =1109 m^3/s We know that BOD IS just because of the microorganisms and thus when we mix the two streams the concentration will be changed and thus the resulting concentration can be given - C1V1+C2V2/(V1+V2) Here C1 and C2 are the concentration of the species and V1 and V2 are their volumetric flow rates. Here C1 = O since the flow water stream is pure, also V1 - 10 m 3 /s C2 =5 mg/L and V2 = 1.09 Keeping these values in the equation BOD just after nixing tusse.wo1.22:5/1102=0.5.budd,

Part b We know that in the wastewater steam we have the owlnisrds that will consume the oxygen and thus they should not be allowed to mix with the fresh water. But in the case we have a usable facility located at a far point that will consume the water after some days we can say that it can be mixed since with time BOD will decrcase due to the decy. Partc Since we have 5 mg/l. of oxycen available as well so the microorganisms will keep consuming the oxygen and BOD will decrease with time. We can write-rate of change of BOD with time as d(BOD)/d1=KBOD Let us denote BOD by A for simplicity. Thus-da/dt = ka. And thus In (N/A)=-Kt, equation 1 Here an is the initial BOD and A is BOD after t days. Volumetric flow rate = 11.09 m^3/s Volumetric flow rate = cross sectional area "velocity We are given with the cross sectional area of 20 m^2. Velocity = 11.09/20 = 0.55 m/s Given we have a point located 10 km away. Time taken for the stream to reach that point = distance /velocity = 10000/0.55 = 18034.2 sec

=5 hours =5/24=0.208 days Keeping this value of time in the equation 1 and keeping the value of K as 0.2/day In A/ALE-KT In(A/0.5 -0.2*0.208 A=0.48 mg/l Thus the BOD after reaching 10 km = 0.48 mg/L Hope this helps Thanks