1. Find: DOB mg L A B C chloride=0 2.3 3.7 7.5 9.7 Stream A C Q [m3/s]
6.2
T [C]
19.4
21.0
D [mg/L]
4.49
3.58 o o mg L B A C
flow
chloride=0
2.3
3.7
7.5
9.7
Find the dissolved oxygen in Stream B, in milligrams per liter. [pause] In this problem, ---

2. Find: DOB mg L A B C chloride=0 2.3 3.7 7.5 9.7 Stream A C Q [m3/s]
6.2
T [C]
19.4
21.0
D [mg/L]
4.49
3.58 o o mg L B A C
flow
chloride=0
2.3
3.7
7.5
9.7
Stream A, and Stream B, converge, to create Stream C.

3. Find: DOB mg L A B C chloride=0 2.3 3.7 7.5 9.7 Stream A C Q [m3/s]
6.2
T [C]
19.4
21.0
D [mg/L]
4.49
3.58 o o mg L B A C
flow
chloride=0
2.3
3.7
7.5
9.7
The flowrates, temperatures, and oxygen deficits, in Streams A and C, are provided. [pause] For sake of convenience, ----

4. Find: DOB mg L A B C Stream Q [m3/s] T [C] D [mg/L] A 3.7 21.0 4.49 B
6.2
19.4
3.58 A C
flow
we’ll add an additional row, for stream B. The concentrations of dissolved oxygen --- B

5. Find: DOB mg L A B C DOA * QA + DOB * QB = DOC * QC Stream Q [m3/s]
T [C]
D [mg/L] o A
3.7
21.0
4.49 B o C
6.2
19.4
3.58
DOA * QA + DOB * QB = DOC * QC B A C
flow
in the three streams are related to each other by their flowrates.

6. Find: DOB mg L A B C DOA * QA + DOB * QB = DOC * QC Stream Q [m3/s]
T [C]
D [mg/L] o A
3.7
21.0
4.49 B o C
6.2
19.4
3.58
DOA * QA + DOB * QB = DOC * QC B A C
flow
The dissolved oxygen concentration in Stream B is a function of the dissolved oxygen concentrations in ---

7. Find: DOB mg L A B C DOA * QA + DOB * QB = DOC * QC DOB = Stream
Q [m3/s]
T [C]
D [mg/L] o A
3.7
21.0
4.49 B o C
6.2
19.4
3.58
DOA * QA + DOB * QB = DOC * QC B A C
flow
Streams A and C, as well as the flowrates, in all 3 streams. [pause] The problem statement provides the flowrates ---
DOC * QC - DOA * QA
DOB = QB

8. Find: DOB mg L A B C DOA * QA + DOB * QB = DOC * QC DOB = Stream
Q [m3/s]
T [C]
D [mg/L] o A
3.7
21.0
4.49 B o C
6.2
19.4
3.58
DOA * QA + DOB * QB = DOC * QC B A C
flow
in streams A and C. And the flowrate in Stream B is equal to ---
DOC * QC - DOA * QA
DOB = QB

9. Find: DOB mg L A B C DOA * QA + DOB * QB = DOC * QC DOB = Stream
Q [m3/s]
T [C]
D [mg/L] o A
3.7
21.0
4.49 B o C
6.2
19.4
3.58
DOA * QA + DOB * QB = DOC * QC B A C
flow
the flowrate in stream C, minus the flowrate in stream A, which equals, ----
DOC * QC - DOA * QA
DOB = QB
QB = QC - QA

10. Find: DOB mg L A B C DOA * QA + DOB * QB = DOC * QC DOB = Stream
Q [m3/s]
T [C]
D [mg/L] o A
3.7
21.0
4.49 B o C
6.2
19.4
3.58
DOA * QA + DOB * QB = DOC * QC B A C
flow
2.5 meters cubed per second. We’ll add this value to the table.
DOC * QC - DOA * QA
DOB = QB
QB = QC - QA
QB = 2.5 [m3/s]

11. Find: DOB mg L A B C DOA * QA + DOB * QB = DOC * QC DOB = Stream
Q [m3/s]
T [C]
D [mg/L] o A
3.7
21.0
4.49 B
2.5 o C
6.2
19.4
3.58
DOA * QA + DOB * QB = DOC * QC B A C
flow
[pause] The last two unknown variables, ---
DOC * QC - DOA * QA
DOB = QB
QB = QC - QA
QB = 2.5 [m3/s]

12. Find: DOB mg L A B C DOA * QA + DOB * QB = DOC * QC DOB = Stream
Q [m3/s]
T [C]
D [mg/L] o A
3.7
21.0
4.49 B
2.5 o C
6.2
19.4
3.58
DOA * QA + DOB * QB = DOC * QC B A C
flow
are the dissolved oxygen concentrations, in Stream A and Stream C. [pause]
DOC * QC - DOA * QA
DOB = QB

13. Find: DOB mg L A B C DO = DOSAT - D Stream Q [m3/s] T [C] D [mg/L] A
3.7
21.0
4.49 B
2.5 o C
6.2
19.4
3.58
DO = DOSAT - D B A C
flow
The concentration of dissolved oxygen ---

14. Find: DOB mg L A B C DO = DOSAT - D Stream Q [m3/s] T [C] D [mg/L] A
3.7
21.0
4.49 B
2.5 o C
6.2
19.4
3.58
DO = DOSAT - D B A C
flow
in a stream, equals, the saturated --- mg
dissolved oxygen L

15. Find: DOB mg L A B C DO = DOSAT - D Stream Q [m3/s] T [C] D [mg/L] A
3.7
21.0
4.49 B
2.5 o C
6.2
19.4
3.58
DO = DOSAT - D B A C
flow
concentration of dissolved oxygen, minus, ---
saturated mg
dissolved oxygen L mg
dissolved oxygen L

16. Find: DOB mg L A B C DO = DOSAT - D Stream Q [m3/s] T [C] D [mg/L] A
3.7
21.0
4.49 B
2.5 o C
6.2
19.4
3.58
oxygen mg
DO = DOSAT - D
deficit L B A C
flow
the oxygen deficit. For this equation, we’ll be working in units of ---
saturated mg
dissolved oxygen L mg
dissolved oxygen L

17. Find: DOB mg L A B C DO = DOSAT - D Stream Q [m3/s] T [C] D [mg/L] A
3.7
21.0
4.49 B
2.5 o C
6.2
19.4
3.58
oxygen mg
DO = DOSAT - D
deficit L B A C
flow
milligrams per liter, and we’re out to calculate the dissolved oxygen for ---
saturated mg
dissolved oxygen L mg
dissolved oxygen L

18. Find: DOB mg L A B C DOA= DOA, SAT - DA Stream Q [m3/s] T [C] D [mg/L]
3.7
21.0
4.49 B
2.5 o C
6.2
19.4
3.58
DOA= DOA, SAT - DA B A C
flow
Stream A, [pause] and for Stream C.

19. Find: DOB mg L A B C DOA= DOA, SAT - DA DOC= DOC, SAT - DC Stream
Q [m3/s]
T [C]
D [mg/L] o A
3.7
21.0
4.49 B
2.5 o C
6.2
19.4
3.58
DOA= DOA, SAT - DA B A C
flow
DOC= DOC, SAT - DC
For each of these equation we’ve been provided the oxygen deficits, D, ----

20. Find: DOB mg L A B C DOA= DOA, SAT - DA DOC= DOC, SAT - DC Stream
Q [m3/s]
T [C]
D [mg/L] o A
3.7
21.0
4.49 B
2.5 o C
6.2
19.4
3.58
DOA= DOA, SAT - DA B A C
flow
DOC= DOC, SAT - DC
of 4.49 milligrams per liter, and 3.58 milligrams per liter. The saturated concentration of dissolved oxygen ---

21. Find: DOB mg L A B C DOA= DOA, SAT - DA DOC= DOC, SAT - DC ƒ(T,CCl-)
Stream
Q [m3/s]
T [C]
D [mg/L] o A
3.7
21.0
4.49 B
2.5 o C
6.2
19.4
3.58
DOA= DOA, SAT - DA B A C
flow
DOC= DOC, SAT - DC
is a function of the temperature and chloride concentration, of the water. [pause] Since the problem statement provides the ----
ƒ(T,CCl-)

22. Find: DOB mg L A B C DOA= DOA, SAT - DA DOC= DOC, SAT - DC ƒ(T,CCl-)
Stream
Q [m3/s]
T [C]
D [mg/L] o A
3.7
21.0
4.49 B
2.5 o C
6.2
19.4
3.58
DOA= DOA, SAT - DA B A C
flow
DOC= DOC, SAT - DC
temperatures and chloride concentrations, the corresponding saturated DO concentrations are looked up in a table of values, and equal ---
ƒ(T,CCl-) mg L
chloride=0

23. Find: DOB mg L DOA= DOA, SAT - DA = 8.99 -4.49 DOC= DOC, SAT - DC
Stream
Q [m3/s]
T [C]
D [mg/L] o A
3.7
21.0
4.49 B
2.5 o C
6.2
19.4
3.58
DOA= DOA, SAT - DA
8.99 milligrams per liter in Stream A, and 9.28 milligrams per liter in Stream C. The resulting dissolved oxygen concentrations are --- mg L mg
= 8.99
-4.49 L
DOC= DOC, SAT - DC mg L mg
= 9.28
-3.58 L

24. Find: DOB mg L DOA= DOA, SAT - DA = 8.99 -4.49 = 4.50
Stream
Q [m3/s]
T [C]
D [mg/L] o A
3.7
21.0
4.49 B
2.5 o C
6.2
19.4
3.58
DOA= DOA, SAT - DA
4.50 milligrams per liter, and 5.70 milligrams per liter, for Streams A and C, respectively. --- mg mg mg L
= 8.99
-4.49
= 4.50 L L
DOC= DOC, SAT - DC mg mg mg L
= 9.28
-3.58
= 5.70 L L

25. Find: DOB mg L DOA= DOA, SAT - DA = 8.99 -4.49 = 4.50
Stream
Q [m3/s]
T [C]
D [mg/L]
DO [mg/L] o A
3.7
21.0
4.49
4.50 B
2.5 o C
6.2
19.4
3.58
5.70
DOA= DOA, SAT - DA
[pause] Returning to our original equation for the concentration of --- mg mg mg
= 8.99
-4.49
= 4.50 L L L
DOC= DOC, SAT - DC mg mg mg
= 9.28
-3.58
= 5.70 L L L

26. Find: DOB mg L A B C DOB = Stream Q [m3/s] T [C] D [mg/L] DO [mg/L] A
3.7
21.0
4.49
4.50 B
2.5 o C
6.2
19.4
3.58
5.70
DOC*QC - DOA*QA
DOB = B A C
flow
dissolved oxygen in Stream B, the known values are plugged into the equation, --- QB

27. Find: DOB mg L A B C DOB = Stream Q [m3/s] T [C] D [mg/L] DO [mg/L] A
3.7
21.0
4.49
4.50 B
2.5 o C
6.2
19.4
3.58
5.70
DOC*QC - DOA*QA
DOB = B A C
flow
and the concentration of dissolved oxygen, in Stream B equals, --- QB

28. Find: DOB mg L A B C DOB = DOB =7.48 Stream Q [m3/s] T [C] D [mg/L]
DO [mg/L] o A
3.7
21.0
4.49
4.50 B
2.5 o C
6.2
19.4
3.58
5.70
DOC*QC - DOA*QA
DOB = B A C
flow
7.48 milligrams per liter. QB mg L
DOB =7.48

29. Find: DOB mg L A B C DOB = 2.3 3.7 7.5 9.7 DOB =7.48 Stream Q [m3/s]
T [C]
D [mg/L]
DO [mg/L] o A
3.7
21.0
4.49
4.50 B
2.5 o C
6.2
19.4
3.58
5.70
DOC*QC - DOA*QA
DOB = B A C
flow
2.3
3.7
7.5
9.7
When reviewing the possible solutions, --- QB mg L
DOB =7.48

30. Find: DOB mg L AnswerC A B C DOB = 2.3 3.7 7.5 9.7 DOB =7.48 Stream
Q [m3/s]
T [C]
D [mg/L]
DO [mg/L] o A
3.7
21.0
4.49
4.50 B
2.5 o C
6.2
19.4
3.58
5.70
DOC*QC - DOA*QA
DOB = B A C
flow
2.3
3.7
7.5
9.7
the answer is C. QB mg L
DOB =7.48
AnswerC

31. ? 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