A wastewater is discharged into a river traveling at a velocity of 5 km/day. Dir
ID: 882485 • Letter: A
Question
A wastewater is discharged into a river traveling at a velocity of 5 km/day. Directly after the discharge point, the river has a dissolved-oxygen content of 5 mg/L and an ultimate BOD of 30 mg/L. The waste has a BOD decay coefficient k1 of 0.25/day. The stream has a rcaeration rate coefficient k2 of 0.6/day. The temperature is 25C and Henry?s law constant for oxygen at 25C is 1.29x 1O^-3 mol/(L.atm). a) What is the initial dissolved-oxygen deficit, D0? b) What is the location of the critical point, in time and distance? c) What is the dissolved-oxygen deficit at the critical point? d) What is the dissolved-oxygen concentration at the critical point? e) Suppose that the dissolved oxygen (DO) concentration in the river must be at least 4.0 mg/L. What would the ultimate BOD directly after the discharge point (L0) have to be to assure that this minimum is met throughout the river?Explanation / Answer
Solubility of water at 25 Deg.C =8.11 mg/liter
Dissolved oxygen contents of the river = 5 mg/liter
a) Initial dissolved oxygen deficit= 8.11-5 =3.11 mg/liter
b) Critcal time tc= (1/(R-K) log(R/K)[1-D0(R-K)/(KLo)]
R= Reoxygenation rate constant = 0.6/day
K= Deoxygenation rate constnat =0.25/day
D0 = initial oxygen deficit = 3.11 mg/liter
L0= BOD at time t=0 =30 mg/liter
R-K =0.6-0.25=0.35/day
R/K= 0.6/0.25 =2.4
D0*(R-K)/ KL0= 3.11*0.35/(0.25*30) =0.14153
Y=1- D0*(R-K)/ KL0 =1-0.14153 = 0.854867
log(R/K *Y)= log(2.4*0.854867)=log(2.05168) =0.31211
tc= (1/0.35)*0.31211 =2.857143*0.31211 = 0.891 days
critcial distnace= critical time * velocity= 0.891* 5km/day= 4.445 km
critcial deficit =(K/R)* L0 10-Ktc = (0.25/0.6)*30 10-0.25*0.891 = 12.5*0.5988= 7.485 mg/liter
critcial deficit= saturation concentration- oxygen concentration
7.485= 9.11-oxygen concentration
oxygen concentration = 9.11-7.485=1.625 mg/liter
e) DC= (K/R)* L0 10-Ktc= (0.25/0.6)*L0* 10-0.25*0.89 =0.41667*LO*0.5987
4 =0.41667*0.5987L0= 0.249LO=
L0= 4/0.249=16 mg/liter
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