1. Find: Dc mg L at 20 C [mg/L] Water Body Q [m3/s] T [C] BOD5 DO
wastewater
0.2
20.2
100
1.0 o
stream
5.0
15.0
2.0
8.0
Kd=0.2 [day-1]
base e, DO o
at 20 C
DOSAT
Kr=0.3 [day-1] Do Dc
stream
Find the critical oxygen deficit, D sub c, in milligrams per liter. [pause] In this problem, ---
wastewater
DOo
A) C) 2.7
B) D) 7.7 t

2. Find: Dc mg L at 20 C [mg/L] Water Body wastewater stream Q [m3/s] 0.2
15.0
20.2
BOD5
100
2.0 DO
1.0
8.0
5.0 o
Kd=0.2 [day-1]
base e, DO o
DOSAT
Kr=0.3 [day-1]
at 20 C Do Dc
stream
The effluent from a wastewater plant, discharges, into a local stream.
wastewater
DOo
A) C) 2.7
B) D) 7.7 t

3. Find: Dc mg L at 20 C [mg/L] Water Body wastewater stream Q [m3/s] 0.2
15.0
20.2
BOD5
100
2.0 DO
1.0
8.0
5.0 o
Kd=0.2 [day-1]
base e, DO o
at 20 C
DOSAT
Kr=0.3 [day-1] Do Dc
stream
The flowrates, temperatures, 5-day carbonaceous biochemical oxygen demand, at 20 degree Celsius, and the dissolved oxygen values, are provided.
wastewater
DOo
A) C) 2.7
B) D) 7.7 t

4. Find: Dc mg L at 20 C [mg/L] Water Body wastewater stream Q [m3/s] 0.2
15.0
20.2
BOD5
100
2.0 DO
1.0
8.0
5.0 o
Kd=0.2 [day-1]
base e, DO o
at 20 C
DOSAT
Kr=0.3 [day-1] Do Dc
stream
The deoxygenation rate constant, and reoxygenation rate constant, are also provided in base e, at 20 degrees Celsius. [pause]
wastewater
DOo
A) C) 2.7
B) D) 7.7 t

5. Find: Dc mg L at 20 C [mg/L] Water Body wastewater stream Q [m3/s] 0.2
15.0
20.2
BOD5
100
2.0 DO
1.0
8.0
5.0 o
Kd=0.2 [day-1]
base e, DO o
at 20 C
DOSAT
Kr=0.3 [day-1] Do
Kd*BODu Dc
The critical oxygen deficit, in milligrams per liter equals, ---
DC=
* e-Kd * tc Kr
DOo t mg L
critical oxygen deficit

6. Find: Dc mg L at 20 C [mg/L] Water Body wastewater stream Q [m3/s] 0.2
15.0
20.2
BOD5
100
2.0 DO
1.0
8.0
5.0 o
Kd=0.2 [day-1]
base e, o
at 20 C
Kr=0.3 [day-1]
Kd*BODu
the deoxygenation rate constant, K sub d, in days to the minus 1, times, ---
DC=
* e-Kd * tc Kr
deoxygenation rate constant [day-1] mg
critical oxygen deficit L

7. Find: Dc mg L at 20 C [mg/L] Water Body wastewater stream Q [m3/s] 0.2
15.0
20.2
BOD5
100
2.0 DO
1.0
8.0
5.0 o
Kd=0.2 [day-1]
base e, o
at 20 C
Kr=0.3 [day-1]
ultimate biochemical
Kd*BODu
oxygen demand mg L
the ultimate BOD, in milligrams per liter, divided by, ---
DC=
* e-Kd * tc Kr
deoxygenation rate constant [day-1] mg
critical oxygen deficit L

8. Find: Dc mg L at 20 C [mg/L] Water Body wastewater stream Q [m3/s] 0.2
15.0
20.2
BOD5
100
2.0 DO
1.0
8.0
5.0 o
Kd=0.2 [day-1]
base e, o
at 20 C
Kr=0.3 [day-1]
ultimate biochemical
Kd*BODu
oxygen demand mg
the reoxygenation rate constant, K sub r, in days to the minus 1, times e raised to negative 1 times, ---
DC=
* e-Kd * tc Kr L
reoxygenation rate constant [day-1]
deoxygenation rate constant [day-1] mg
critical oxygen deficit L

9. Find: Dc mg L at 20 C [mg/L] Water Body wastewater stream Q [m3/s] 0.2
15.0
20.2
BOD5
100
2.0 DO
1.0
8.0
5.0 o
Kd=0.2 [day-1]
base e, o
at 20 C
Kr=0.3 [day-1]
ultimate biochemical
Kd*BODu
oxygen demand mg
the deoxygenation rate constant, times, ---
DC=
* e-Kd * tc Kr L
reoxygenation rate constant [day-1]
deoxygenation rate constant [day-1] mg
critical oxygen deficit L

10. Find: Dc mg L at 20 C [mg/L] Water Body wastewater stream Q [m3/s] 0.2
15.0
20.2
BOD5
100
2.0 DO
1.0
8.0
5.0 o
Kd=0.2 [day-1]
base e,
critical time [day] o
at 20 C
Kr=0.3 [day-1]
ultimate biochemical
Kd*BODu
oxygen demand mg
the critical time, t sub c, in days. [pause] The problem statement already provides the ---
DC=
* e-Kd * tc Kr L
reoxygenation rate constant [day-1]
deoxygenation rate constant [day-1] mg
critical oxygen deficit L

11. Find: Dc mg L at 20 C [mg/L] Water Body wastewater stream Q [m3/s] 0.2
15.0
20.2
BOD5
100
2.0 DO
1.0
8.0
5.0 o
Kd=0.2 [day-1]
base e,
critical time [day] o
at 20 C
Kr=0.3 [day-1]
ultimate biochemical
Kd*BODu
oxygen demand mg
oxygenation rate constants, K sub d, and K sub r, but we do not know the values of the ---
DC=
* e-Kd * tc Kr L
reoxygenation rate constant [day-1]
deoxygenation rate constant [day-1] mg
critical oxygen deficit L

12. Find: Dc mg L at 20 C [mg/L] Water Body wastewater stream Q [m3/s] 0.2
15.0
20.2
BOD5
100
2.0 DO
1.0
8.0
5.0 o
Kd=0.2 [day-1]
base e,
critical time [day] o
at 20 C
Kr=0.3 [day-1]
ultimate biochemical
Kd*BODu
oxygen demand mg L
ultimate BOD, or, the critical time. [pause] We’ll first solve for the ultimate BOD.
DC=
* e-Kd * tc Kr
reoxygenation rate constant [day-1]
deoxygenation rate constant [day-1] mg
critical oxygen deficit L

13. Find: Dc mg L BODu = at 20 C [mg/L] Water Body wastewater stream
Q [m3/s]
0.2
T [C]
15.0
20.2
BOD5
100
2.0 DO
1.0
8.0
5.0 o
Kd=0.2 [day-1]
base e, o
at 20 C
Kr=0.3 [day-1]
BODt
BODu =
The ultimate BOD, in milligrams per liter equals, ---
(1-e-kd * t) mg L
ultimate biochemical oxygen demand

14. Find: Dc mg L BODu = at 20 C [mg/L] Water Body wastewater stream
Q [m3/s]
0.2
T [C]
15.0
20.2
BOD5
100
2.0 DO
1.0
8.0
5.0 o
Kd=0.2 [day-1]
base e,
biochemical o
at 20 C
Kr=0.3 [day-1]
oxygen
BODt
demand mg L
BODu =
the BOD at a given time period, t, in milligrams per liter, divided by 1 minus e raised to ---
at time t
(1-e-kd * t) mg
ultimate biochemical oxygen demand L

15. Find: Dc mg L BODu = at 20 C [mg/L] Water Body wastewater stream
Q [m3/s]
0.2
T [C]
15.0
20.2
BOD5
100
2.0 DO
1.0
8.0
5.0 o
Kd=0.2 [day-1]
base e,
biochemical o
at 20 C
Kr=0.3 [day-1]
oxygen
BODt
demand mg
BODu =
the minus 1 times K sub d, times the time, ---
at time t
(1-e-kd * t) L
deoxygenation rate constant [day-1] mg
ultimate biochemical oxygen demand L

16. Find: Dc mg L BODu = at 20 C [mg/L] Water Body wastewater stream
Q [m3/s]
0.2
T [C]
15.0
20.2
BOD5
100
2.0 DO
1.0
8.0
5.0 o
Kd=0.2 [day-1]
base e,
biochemical o
at 20 C
Kr=0.3 [day-1]
oxygen
BODt
demand mg
BODu =
t, in days. [pause] Since we’ve been given data for the 5-day BOD, ---
at time t
(1-e-kd * t) L
time [day]
deoxygenation rate constant [day-1] mg
ultimate biochemical oxygen demand L

17. Find: Dc mg L BODu = at 20 C [mg/L] Water Body wastewater stream
Q [m3/s]
0.2
T [C]
15.0
20.2
BOD5
100
2.0 DO
1.0
8.0
5.0 o
Kd=0.2 [day-1]
base e,
biochemical o
at 20 C
Kr=0.3 [day-1]
oxygen
BODt
demand mg
BODu =
we’ll use a time of 5 days. [pause] Next, we’ll calculate the 5-day BOD in the stream, ---
at time t
(1-e-kd * t) L
5 [day]
deoxygenation rate constant [day-1] mg
ultimate biochemical oxygen demand L

18. Find: Dc mg L BODu = at 20 C [mg/L] Water Body wastewater stream
Q [m3/s]
0.2
T [C]
15.0
20.2
BOD5
100
2.0 DO
1.0
8.0
5.0 o
Kd=0.2 [day-1]
base e,
biochemical o
at 20 C
Kr=0.3 [day-1]
oxygen
BOD5
demand mg
BODu =
after it mixes with the wastewater effluent, in milligrams per liter. [pause] The 5-day BOD, after mixing, equals ---
at time t
(1-e-kd * t) L
5 [day]
deoxygenation rate constant [day-1] mg
ultimate biochemical oxygen demand L

19. Find: Dc mg L BODu = at 20 C [mg/L] Water Body wastewater stream
Q [m3/s]
0.2
T [C]
15.0
20.2
BOD5
100
2.0 DO
1.0
8.0
5.0 o
BOD5
biochemical oxygen mg
BODu =
demand at 5 days
(1-e-kd * t) L
BOD5,WW *QWW + BOD5,S*QS
the weighted average of the 5-day BOD values of the 2 water bodies, based on their flowrates.
BOD5=
QWW + QS

20. Find: Dc mg L BODu = at 20 C [mg/L] Water Body wastewater stream
Q [m3/s]
0.2
T [C]
15.0
20.2
BOD5
100
2.0 DO
1.0
8.0
5.0 o
BOD5
biochemical oxygen mg
BODu =
demand at 5 days
(1-e-kd * t) L
BOD5,WW *QWW + BOD5,S*QS
The known values are plugged into the equation, and the 5-day BOD of the two water bodies equals, ---
BOD5=
QWW + QS

21. Find: Dc mg L BODu = BOD5=5.77 at 20 C [mg/L] Water Body wastewater
stream
Q [m3/s]
0.2
T [C]
15.0
20.2
BOD5
100
2.0 DO
1.0
8.0
5.0 o
BOD5
biochemical oxygen mg
BODu =
demand at 5 days
(1-e-kd * t) L
BOD5,WW *QWW + BOD5,S*QS
5.77, milligrams per liter. [pause]
BOD5=
QWW + QS mg L
BOD5=5.77

22. Find: Dc mg L BODu = at 20 C [mg/L] Water Body wastewater stream
Q [m3/s]
0.2
T [C]
15.0
20.2
BOD5
100
2.0 DO
1.0
8.0
5.0 o
BOD5 mg
BODu =
BOD5=5.77 L
(1-e-kd * t)
Kd=0.2 [day-1]
Returning to our equation for the ultimate BOD, ---
t=5 [days]

23. Find: Dc mg L BODu = at 20 C [mg/L] Water Body wastewater stream
Q [m3/s]
0.2
T [C]
15.0
20.2
BOD5
100
2.0 DO
1.0
8.0
5.0 o
BOD5 mg
BODu =
BOD5=5.77 L
(1-e-kd * t)
Kd=0.2 [day-1]
The known variables are plugged in, and the ultimate BOD equals, ---
t=5 [days]

24. Find: Dc mg L BODu = BODu =9.13 at 20 C [mg/L] Water Body wastewater
stream
Q [m3/s]
0.2
T [C]
15.0
20.2
BOD5
100
2.0 DO
1.0
8.0
5.0 o
BOD5 mg
BODu =
BOD5=5.77 L
(1-e-kd * t)
Kd=0.2 [day-1]
9.13, milligrams per liter. [pause] Now that ultimate BOD is calculated, --- mg L
BODu =9.13
t=5 [days]

25. Find: Dc mg L at 20 C [mg/L] Water Body wastewater stream Q [m3/s] 0.2
15.0
20.2
BOD5
100
2.0 DO
1.0
8.0
5.0 o
Kd=0.2 [day-1]
base e,
critical time [day] o
at 20 C
Kr=0.3 [day-1]
ultimate biochemical
Kd*BODu
oxygen demand mg
the last unknown variable is ---
DC=
* e-Kd * tc Kr L
reoxygenation rate constant [day-1]
deoxygenation rate constant [day-1] mg
critical oxygen deficit L

26. Find: Dc mg L at 20 C [mg/L] Water Body wastewater stream Q [m3/s] 0.2
15.0
20.2
BOD5
100
2.0 DO
1.0
8.0
5.0 o
Kd=0.2 [day-1]
base e,
critical time [day] o
at 20 C
Kr=0.3 [day-1]
ultimate biochemical
Kd*BODu
oxygen demand mg
the critical time. [pause] On an oxygen sag curve, ---
DC=
* e-Kd * tc Kr L
reoxygenation rate constant [day-1]
deoxygenation rate constant [day-1] mg
critical oxygen deficit L

27. Find: Dc mg L at 20 C [mg/L] Water Body wastewater stream Q [m3/s] 0.2
15.0
20.2
BOD5
100
2.0 DO
1.0
8.0
5.0 o
Kd*BODu
DC=
* e-Kd * tc DO Kr
DOSAT Do
the concentration of dissolved oxygen initially decreases, and then rebounds back, based on the ---- Dc
DOo t

28. Find: Dc mg L at 20 C [mg/L] Water Body wastewater stream Q [m3/s] 0.2
15.0
20.2
BOD5
100
2.0 DO
1.0
8.0
5.0 o
Kd*BODu
* e-Kd * tc DO
DC= Kr
DOSAT Do
Streeter-Phelps equation which involves the rate constants, ultimate BOD, and initial oxygen deficit. Dc
DOo t

29. Find: Dc mg L at 20 C [mg/L] Water Body wastewater stream Q [m3/s] 0.2
15.0
20.2
BOD5
100
2.0 DO
1.0
8.0
5.0 o
Kd*BODu
* e-Kd * tc DO
DC= Kr
DOSAT Do
D sub c is the critical deficit, when the dissolved oxygen is at its minimum value, which is what were solving for, and which occurs at the --- Dc
DOo t

30. Find: Dc mg L at 20 C [mg/L] Water Body wastewater stream Q [m3/s] 0.2
15.0
20.2
BOD5
100
2.0 DO
1.0
8.0
5.0 o
Kd*BODu
* e-Kd * tc DO
DC= Kr
DOSAT Do
the critical time, t sub c. [pause] The critical time, in days, is a function of ---- Dc
DOo tc t

31. Find: Dc mg L tC= 1-Do at 20 C [mg/L] Water Body wastewater stream
Q [m3/s]
0.2
T [C]
15.0
20.2
BOD5
100
2.0 DO
1.0
8.0
5.0 o
Kd*BODu
DC=
* e-Kd * tc Kr
Kr-Kd 1 Kr
the oxygenation rate constants, the initial oxygen deficit, and the ultimate biochemical oxygen demand.
tC= ln *
1-Do
Kr-Kd * * Kd
Kd*BODu

32. Find: Dc mg L tC= 1-Do at 20 C [mg/L] Water Body wastewater stream
Q [m3/s]
0.2
T [C]
15.0
20.2
BOD5
100
2.0 DO
1.0
8.0
5.0 o
initial
Kd*BODu mg L
DC=
* e-Kd * tc
oxygen Kr
deficit
Kr-Kd 1 Kr
The only unknown variable is, the initial oxygen deficit, D sub o, in milligrams per liter.
tC= ln *
1-Do
Kr-Kd * * Kd
Kd*BODu
Kd=0.2 [day-1]
Kr=0.3 [day-1]
BODu=9.13 [mg/L]

33. Find: Dc mg L tC= 1-Do at 20 C [mg/L] Water Body wastewater stream
Q [m3/s]
0.2
T [C]
15.0
20.2
BOD5
100
2.0 DO
1.0
8.0
5.0 o
initial
Kd*BODu mg L
DC=
* e-Kd * tc
oxygen Kr
deficit
Kr-Kd 1 Kr
To determine the initial oxygen deficit, we’ll subtract the initial concentration of dissolved oxygen, ----
tC= ln *
1-Do
Kr-Kd * * Kd
Kd*BODu
Kd=0.2 [day-1]
Kr=0.3 [day-1]
BODu=9.13 [mg/L]

34. Find: Dc mg L tC= Do=Do,SAT-DOo 1-Do at 20 C [mg/L] Water Body
wastewater
stream
Q [m3/s]
0.2
T [C]
15.0
20.2
BOD5
100
2.0 DO
1.0
8.0
5.0 o
initial
Kd*BODu mg
DC=
* e-Kd * tc
oxygen L Kr
deficit D
Kr-Kd 1 Kr
from the initial saturated concentration, of dissolved oxygen, for the combined flows. The initial concentration of dissolved oxygen is ---
tC= ln *
1-Do
DOSAT
Kr-Kd * * Kd
Kd*BODu DO
Kd=0.2 [day-1]
Do=Do,SAT-DOo
Kr=0.3 [day-1]
BODu=9.13 [mg/L]

35. Find: Dc mg L Do=Do,SAT-DOo DOo= at 20 C [mg/L] Water Body Q [m3/s]
wastewater
0.2
20.2
100
1.0 o
stream
5.0
15.0
2.0
8.0
Do=Do,SAT-DOo D
DOo,WW * QWW + DOo,S * QS
DOo=
the weighted average of the initial concentrations of dissolved oxygen from the two water bodies, weighted by flowrate.
QWW + QS
DOSAT DO

36. Find: Dc mg L Do=Do,SAT-DOo DOo= at 20 C [mg/L] Water Body Q [m3/s]
wastewater
0.2
20.2
100
1.0 o
stream
5.0
15.0
2.0
8.0
Do=Do,SAT-DOo D
DOo,WW * QWW + DOo,S * QS
DOo=
The values are plugged into the equation, and the initial dissolved oxygen ---
QWW + QS
DOSAT DO

37. Find: Dc mg L Do=Do,SAT-DOo DOo= DOo=7.73 [mg/L] at 20 C [mg/L]
Water Body
Q [m3/s]
T [C]
BOD5 DO o
wastewater
0.2
20.2
100
1.0 o
stream
5.0
15.0
2.0
8.0
Do=Do,SAT-DOo D
DOo,WW * QWW + DOo,S * QS
DOo=
concentration equals 7.73 milligrams per liter. [pause] The saturated dissolved oxygen concentration ----
QWW + QS
DOSAT DO
DOo=7.73 [mg/L]

38. Find: Dc mg L Do=Do,SAT-DOo DOo= DOo,SAT= DOo=7.73 [mg/L] at 20 C
Water Body
Q [m3/s]
T [C]
BOD5 DO o
wastewater
0.2
20.2
100
1.0 o
stream
5.0
15.0
2.0
8.0
Do=Do,SAT-DOo D
DOo,WW * QWW + DOo,S * QS
DOo=
is calculated the same way, where the flowrates are plugged in, ---
QWW + QS
DOSAT DO
DOo=7.73 [mg/L]
DOSAT,WW * QWW + DOSAT,S * QS
DOo,SAT=
QWW + QS

39. Find: Dc mg L Do=Do,SAT-DOo DOo= DOo,SAT= DOo=7.73 [mg/L] at 20 C
Water Body
Q [m3/s]
T [C]
BOD5 DO o
wastewater
0.2
20.2
100
1.0 o
stream
5.0
15.0
2.0
8.0
Do=Do,SAT-DOo D
DOo,WW * QWW + DOo,S * QS
DOo=
and the saturated DO concentrations are a function of the
QWW + QS
DOSAT DO
DOo=7.73 [mg/L]
DOSAT,WW * QWW + DOSAT,S * QS
DOo,SAT=
QWW + QS

40. Find: Dc mg L Do=Do,SAT-DOo DOo,SAT= at 20 C [mg/L] Water Body
Q [m3/s]
T [C]
BOD5 DO o
wastewater
0.2
20.2
100
1.0 o
stream
5.0
15.0
2.0
8.0
Do=Do,SAT-DOo D
ƒ(TWW,Cl-WW)
ƒ(TS,Cl-S)
temperature of each body of water.. Assuming no chloride contamination, the saturated DO concentrations are ---
DOSAT DO
DOSAT,WW * QWW + DOSAT,S * QS
DOo,SAT=
QWW + QS

41. Find: Dc mg L Do=Do,SAT-DOo 10.15 9.13 DOo,SAT= at 20 C [mg/L]
Water Body
Q [m3/s]
T [C]
BOD5 DO o
wastewater
0.2
20.2
100
1.0 o
stream
5.0
15.0
2.0
8.0
Do=Do,SAT-DOo D
ƒ(TWW,Cl-WW)
ƒ(TS,Cl-S)
9.13 milligrams per liter, and milligrams per liter, for the wastewater effluent and the stream, respectively. mg L mg L
DOSAT
9.13
10.15 DO
DOSAT,WW * QWW + DOSAT,S * QS
DOo,SAT=
QWW + QS

42. Find: Dc mg L Do=Do,SAT-DOo 10.15 9.13 DOo,SAT= at 20 C [mg/L]
Water Body
Q [m3/s]
T [C]
BOD5 DO o
wastewater
0.2
20.2
100
1.0 o
stream
5.0
15.0
2.0
8.0
Do=Do,SAT-DOo D
ƒ(TWW,Cl-WW)
ƒ(TS,Cl-S)
The initial saturated dissolved oxygen concentration equals, --- mg L mg L
DOSAT
9.13
10.15 DO
DOSAT,WW * QWW + DOSAT,S * QS
DOo,SAT=
QWW + QS

43. Find: Dc mg L Do=Do,SAT-DOo DOo,SAT=10.11 [mg/L] 10.15 9.13 DOo,SAT=
at 20 C
[mg/L]
Water Body
Q [m3/s]
T [C]
BOD5 DO o
wastewater
0.2
20.2
100
1.0 o
stream
5.0
15.0
2.0
8.0
Do=Do,SAT-DOo D
DOo,SAT=10.11 [mg/L]
ƒ(TS,Cl-S)
10.11, milligrams per liter. [pause] mg mg
DOSAT
9.13
10.15 DO L L
DOSAT,WW * QWW + DOSAT,S * QS
DOo,SAT=
QWW + QS

44. Find: Dc mg L Do=Do,SAT-DOo DOo,SAT=10.11 [mg/L] DOo=7.73 [mg/L]
at 20 C
[mg/L]
Water Body
Q [m3/s]
T [C]
BOD5 DO o
wastewater
0.2
20.2
100
1.0 o
stream
5.0
15.0
2.0
8.0
Do=Do,SAT-DOo
DOo=7.73 [mg/L] D
DOo,SAT=10.11 [mg/L]
By knowing the initial DO concentration, and the initial saturated DO concentration, ---
DOSAT DO

45. Find: Dc mg L Do=Do,SAT-DOo DOo,SAT=10.11 [mg/L] DOo=7.73 [mg/L]
at 20 C
[mg/L]
Water Body
Q [m3/s]
T [C]
BOD5 DO o
wastewater
0.2
20.2
100
1.0 o
stream
5.0
15.0
2.0
8.0
Do=Do,SAT-DOo
DOo=7.73 [mg/L] D
DOo,SAT=10.11 [mg/L]
the initial oxygen deficit in the stream equals, ---
DOSAT DO

46. Find: Dc mg L Do=Do,SAT-DOo Do=2.38 DOo,SAT=10.11 [mg/L]
at 20 C
[mg/L]
Water Body
Q [m3/s]
T [C]
BOD5 DO o
wastewater
0.2
20.2
100
1.0 o
stream
5.0
15.0
2.0
8.0
Do=Do,SAT-DOo
DOo=7.73 [mg/L] D
DOo,SAT=10.11 [mg/L]
2.38, milligrams per liter. [pause] At this point, ---
DOSAT mg L DO
Do=2.38

47. Find: Dc mg L tC= 1-Do at 20 C [mg/L] Water Body wastewater stream
Q [m3/s]
0.2
T [C]
15.0
20.2
BOD5
100
2.0 DO
1.0
8.0
5.0 o
initial
Kd*BODu mg L
DC=
* e-Kd * tc
oxygen Kr
deficit
Kr-Kd 1 Kr
we can return to our equation for the critical time, t sub c. Since we know the initial oxygen deficit is ---
tC= ln *
1-Do
Kr-Kd * * Kd
Kd*BODu
Kd=0.2 [day-1]
critical
Kr=0.3 [day-1]
time [days]
BODu=9.13 [mg/L]

48. Find: Dc mg L tC= 2.38 1-Do at 20 C [mg/L] Water Body wastewater
stream
Q [m3/s]
0.2
T [C]
15.0
20.2
BOD5
100
2.0 DO
1.0
8.0
5.0 o mg L
Kd*BODu
2.38
DC=
* e-Kd * tc Kr
Kr-Kd 1 Kr
2.38 milligrams per liter, and we plug in the ---
tC= ln *
1-Do
Kr-Kd * * Kd
Kd*BODu
Kd=0.2 [day-1]
critical
Kr=0.3 [day-1]
time [days]
BODu=9.13 [mg/L]

49. Find: Dc mg L tC= 2.38 1-Do at 20 C [mg/L] Water Body wastewater
stream
Q [m3/s]
0.2
T [C]
15.0
20.2
BOD5
100
2.0 DO
1.0
8.0
5.0 o mg L
Kd*BODu
2.38
DC=
* e-Kd * tc Kr
Kr-Kd 1 Kr
oxygenation rate constants and the ultimate BOD, ---
tC= ln *
1-Do
Kr-Kd * * Kd
Kd*BODu
Kd=0.2 [day-1]
critical
Kr=0.3 [day-1]
time [days]
BODu=9.13 [mg/L]

50. Find: Dc mg L tC= tC=2.66 [days] 1-Do at 20 C [mg/L] Water Body
wastewater
stream
Q [m3/s]
0.2
T [C]
15.0
20.2
BOD5
100
2.0 DO
1.0
8.0
5.0 o
Kd*BODu
DC=
* e-Kd * tc Kr
Kr-Kd 1 Kr
the critical time equals, 2.66 days. The value of 2.66 days is plugged into t sub c, ---
tC= ln *
1-Do
Kr-Kd * * Kd
Kd*BODu
Kd=0.2 [day-1]
Kr=0.3 [day-1]
tC=2.66 [days]
BODu=9.13 [mg/L]

51. Find: Dc mg L tC=2.66 [days] at 20 C [mg/L] Water Body wastewater
stream
Q [m3/s]
0.2
T [C]
15.0
20.2
BOD5
100
2.0 DO
1.0
8.0
5.0 o
Kd*BODu
DC=
* e-Kd * tc Kr
as well as the other variables, and the critical oxygen deficit equals, ---
Kd=0.2 [day-1]
Kr=0.3 [day-1]
tC=2.66 [days]
BODu=9.13 [mg/L]

52. Find: Dc mg L tC=2.66 [days] at 20 C [mg/L] Water Body wastewater
stream
Q [m3/s]
0.2
T [C]
15.0
20.2
BOD5
100
2.0 DO
1.0
8.0
5.0 o
Kd*BODu
DC=
* e-Kd * tc Kr
3.57, milligrams per liter. mg L
DC=3.57
Kd=0.2 [day-1]
Kr=0.3 [day-1]
tC=2.66 [days]
BODu=9.13 [mg/L]

53. Find: Dc mg L tC=2.66 [days] 2.4 2.7 3.6 7.7 at 20 C [mg/L] Water Body
wastewater
stream
Q [m3/s]
0.2
T [C]
15.0
20.2
BOD5
100
2.0 DO
1.0
8.0
5.0 o
Kd*BODu
DC=
* e-Kd * tc
2.4
2.7
3.6
7.7 Kr
When reviewing the possible solutions, --- mg L
DC=3.57
Kd=0.2 [day-1]
Kr=0.3 [day-1]
tC=2.66 [days]
BODu=9.13 [mg/L]

54. Find: Dc mg L AnswerC tC=2.66 [days] 2.4 2.7 3.6 7.7 at 20 C [mg/L]o L
at 20 C
[mg/L]
Water Body
wastewater
stream
Q [m3/s]
0.2
T [C]
15.0
20.2
BOD5
100
2.0 DO
1.0
8.0
5.0 o
Kd*BODu
DC=
* e-Kd * tc
2.4
2.7
3.6
7.7 Kr
the answer is C. mg L
DC=3.57
Kd=0.2 [day-1]
AnswerC
Kr=0.3 [day-1]
tC=2.66 [days]
BODu=9.13 [mg/L]

55. ? Index σ’v = Σ γ d γT=100 [lb/ft3] +γclay dclay 1
Find: σ’v at d = 30 feet
(1+wc)*γw
wc+(1/SG)
σ’v = Σ γ d d
Sand
10 ft
γT=100 [lb/ft3]
100 [lb/ft3]
10 [ft]
20 ft
Clay
= γsand dsand
+γclay dclay A W S
V [ft3]
W [lb]
40 ft
text
wc = 37% ? Δh
20 [ft]
(5 [cm])2 * π/4