1. Find: min D [in] = 12 16 18 24 P=30,000 people 18+P/1000 PF= 4+P/1000
Find the minimum pipe diameter, capital D, in inches. [pause] In this problem, --- D d
S=0.005
max
= 0.70
concrete
gal
average sewage
sewer = 90
generation rate
person*day
pipe

2. Find: min D [in] = 12 16 18 24 P=30,000 people 18+P/1000 PF= 4+P/1000
a concrete sewer pipe, positioned at a half percent slope, conveys the sewage generated by a population of 30,000 people. D d
max
= 0.70
S=0.005
concrete
gal
average sewage
sewer = 90
generation rate
person*day
pipe

3. Find: min D [in] = 12 16 18 24 P=30,000 people 18+P/1000 PF= 4+P/1000
The maximum allowable d over D is 0.70, and the peaking factor and generation rate are provided. [pause] Since the depth of water in the pipe --- D d
max
= 0.70
S=0.005
concrete
gal
average sewage
sewer = 90
generation rate
person*day
pipe

4. Find: min D [in] = d P=30,000 people < 0.70 D 18+P/1000 PF=
is at most 70% of the pipe diameter, --- D d
S=0.005
max
= 0.70
concrete
gal
average sewage
sewer = 90
generation rate
person*day
pipe

5. Find: min D [in] = d P=30,000 people < 0.70 D 18+P/1000 PF=
d<0.7*D
4+P/1000 D d
then the diameter of the pipe is at least, ---- D d
= 0.70
S=0.005
max
concrete
gal
average sewage
sewer = 90
generation rate
person*day
pipe

6. Find: min D [in] = d P=30,000 people < 0.70 D 18+P/1000 d PF=
d<0.7*D
D>
4+P/1000
0.70 D d
the depth of water in the pipe, divided by [pause] Our strategy will be to utilize --- D d
S=0.005
max
= 0.70
concrete
gal
average sewage
sewer = 90
generation rate
person*day
pipe

7. Find: min D [in] = Q= A * R2/3 * S 1/2 d P=30,000 people < 0.70 D
PF=
d<0.7*D
D>
4+P/1000
0.70 k Q=
A * R2/3 * S 1/2 * n D d
Manning’s equation, solve for the area and ---
concrete
gal
average sewage
sewer = 90
generation rate
person*day
pipe

8. Find: min D [in] = Q= A * R2/3 * S 1/2 d P=30,000 people < 0.70 D
PF=
d<0.7*D
D>
4+P/1000
0.70 k Q=
A * R2/3 * S 1/2 * n D d
hydraulic radius, and then, ---
concrete
gal
average sewage
sewer = 90
generation rate
person*day
pipe

9. Find: min D [in] = Q= A * R2/3 * S 1/2 A * R2/3 = d P=30,000 people
< 0.70 D
18+P/1000 d
PF=
d<0.7*D
D>
4+P/1000
0.70 k Q=
A * R2/3 * S 1/2 * n D d
develop an expression for the left hand side of this equation ---
Q * n
A * R2/3 =
k * S 1/2
concrete
gal
average sewage
sewer = 90
generation rate
person*day
pipe

10. Find: min D [in] = Q= A * R2/3 * S 1/2 A * R2/3 = A*R2/3=ƒ(D) d
P=30,000 people
< 0.70 D
18+P/1000 d
PF=
d<0.7*D
D>
4+P/1000
0.70 k Q=
A * R2/3 * S 1/2 * n D d
in terms of the the diameter, D, and then solve for D directly. [pause] So let’s begin by solving ---
Q * n
A * R2/3 =
k * S 1/2
A*R2/3=ƒ(D)
concrete
gal
average sewage
sewer = 90
generation rate
person*day
pipe

11. Find: min D [in] = A * R2/3 = Q * n P=30,000 people k * S 1/2
PF=
4+P/1000 D d
for the right hand side of this equation, which equals, the flowrate times the roughness coefficient, divided by k times the square root of the slope.
S=0.005
concrete
gal
average sewage
sewer = 90
generation rate
person*day
pipe

12. Find: min D [in] = A * R2/3 = Q * n P=30,000 people k * S 1/2
PF=
4+P/1000 D d
We’ve been given the slope, which is 0.005, or 1/2%. [pause] The problem states the ---
S=0.005
concrete
gal
average sewage
sewer = 90
generation rate
person*day
pipe

13. Find: min D [in] ? = A * R2/3 = Q * n P=30,000 people k * S 1/2
PF=
4+P/1000 D d
pipe is concrete, so we can look up it’s roughness coefficient, n, ---
S=0.005
concrete
gal
average sewage
sewer = 90
generation rate
person*day
pipe

14. Find: min D [in] ? = A * R2/3 = Q * n P=30,000 people k * S 1/2
PF=
0.010
plastic
brass
concrete
brick
CMP
0.011
0.013
0.015
0.022
material n
4+P/1000 D d
in a standard table of values. The roughness for concrete is ---
S=0.005
concrete
gal
average sewage
sewer = 90
generation rate
person*day
pipe

15. Find: min D [in] ? = A * R2/3 = Q * n P=30,000 people k * S 1/2
PF=
0.010
plastic
brass
concrete
brick
CMP
0.011
0.013
0.015
0.022
material n
4+P/1000 D d
most always
S=0.005
concrete
gal
average sewage
sewer = 90
generation rate
person*day
pipe

16. Find: min D [in] = 0.013 A * R2/3 = Q * n P=30,000 people k * S 1/2
PF=
0.010
plastic
brass
concrete
brick
CMP
0.011
0.013
0.015
0.022
material n
4+P/1000 D d
[pause] When Manning’s equation is used for English units, ---
S=0.005
concrete
gal
average sewage
sewer = 90
generation rate
person*day
pipe

17. Find: min D [in] = 0.013 A * R2/3 = Q * n P=30,000 people k * S 1/2
PF=
k=1.49
4+P/1000 D d
the value for k equals 1.486, or [pause] Lastly, we turn to our flowrate, Q.
S=0.005
concrete
gal
average sewage
sewer = 90
generation rate
person*day
pipe

18. Find: min D [in] = 0.013 A * R2/3 = Q * n P=30,000 people k * S 1/2
PF=
k=1.49
4+P/1000 D d
The average flowrate of sewage through the pipe equals, ---
S=0.005
concrete
gal
average sewage
sewer = 90
generation rate
person*day
pipe

19. Find: min D [in] = 0.013 A * R2/3 = Qave = P * Q * n P=30,000 people
k * S 1/2
18+P/1000
PF=
k=1.49
4+P/1000
average sewage
Qave = P *
generation rate D d
the total population, P, times the average sewage generation rate. For sake of space, I’ll abbreviate ---
S=0.005
concrete
gal
average sewage
sewer = 90
generation rate
person*day
pipe

20. Find: min D [in] = 0.013 A * R2/3 = Qave = P * Qave = P * Rave Q * n
P=30,000 people
A * R2/3 =
k * S 1/2
18+P/1000
PF=
S=0.005
k=1.49
4+P/1000
average sewage
Qave = P *
generation rate
Qave = P * Rave D d
this generation rate with, R sub a-v-e. [pause] However, in sewer design, we’ll want to account for the peak flowrate through the pipe, ---
gal
average sewage =
Rave=90
generation rate
person*day

21. Find: min D [in] = 0.013 A * R2/3 = Qave = P * Qave = P * Rave
Q * n
0.013
P=30,000 people
A * R2/3 =
k * S 1/2
18+P/1000
PF=
S=0.005
k=1.49
4+P/1000
average sewage
Qave = P *
generation rate
Qave = P * Rave D
Qpeak= Qave* PF d
which equals the average flowrate, times a peaking factor, PF.
gal
average sewage =
Rave=90
generation rate
person*day

22. Find: min D [in] = 0.013 A * R2/3 = Qave = P * Qave = P * Rave
Q * n
0.013
P=30,000 people
A * R2/3 =
k * S 1/2
18+P/1000
PF=
S=0.005
k=1.49
4+P/1000
average sewage
Qave = P *
generation rate
Qave = P * Rave D
Qpeak= Qave* PF(P) d
This peaking factor is a function of the population, P, and assumed to account for both seasonal and daily variation in flowrates.
gal
average sewage =
Rave=90
generation rate
person*day

23. Find: min D [in] = 0.013 A * R2/3 = Qave = P * Qave = P * Rave
Q * n
0.013
P=30,000 people
A * R2/3 =
k * S 1/2
18+P/1000
PF=
S=0.005
k=1.49
4+P/1000
average sewage
Qave = P *
generation rate
Qave = P * Rave D
Qpeak= Qave* PF(P) d
Plugging in our known values for the ---
Qpeak= P * Rave* PF(P)
gal
average sewage =
Rave=90
generation rate
person*day

24. Find: min D [in] = 0.013 A * R2/3 = Qave = P * Qave = P * Rave
Q * n
0.013
P=30,000 people
A * R2/3 =
k * S 1/2
18+P/1000
PF=
S=0.005
k=1.49
4+P/1000
average sewage
Qave = P *
generation rate
Qave = P * Rave D
Qpeak= Qave* PF(P) d
population, ---
Qpeak= P * Rave* PF(P)
gal
average sewage =
Rave=90
generation rate
person*day

25. Find: min D [in] = 0.013 A * R2/3 = Qave = P * Qave = P * Rave
Q * n
0.013
P=30,000 people
A * R2/3 =
k * S 1/2
18+P/1000
PF=
S=0.005
k=1.49
4+P/1000
average sewage
Qave = P *
generation rate
Qave = P * Rave D
Qpeak= Qave* PF(P) d
the generation rate, ---
Qpeak= P * Rave* PF(P)
gal
average sewage =
Rave=90
generation rate
person*day

26. Find: min D [in] = 0.013 A * R2/3 = Qave = P * Qave = P * Rave
Q * n
0.013
P=30,000 people
A * R2/3 =
k * S 1/2
18+P/1000
PF=
S=0.005
k=1.49
4+P/1000
average sewage
Qave = P *
generation rate
Qave = P * Rave D
Qpeak= Qave* PF(P) d
and the peaking factor, after it’s value ---
Qpeak= P * Rave* PF(P)
gal
average sewage =
Rave=90
generation rate
person*day

27. Find: min D [in] = 0.013 A * R2/3 = Qave = P * Qave = P * Rave
Q * n
0.013
P=30,000 people
A * R2/3 =
k * S 1/2
18+P/1000
PF=
S=0.005
k=1.49
4+P/1000
average sewage
Qave = P *
generation rate
Qave = P * Rave D
Qpeak= Qave* PF(P) d
is calculated using the population, P, the peak flowrate computes to, ---
Qpeak= P * Rave* PF(P)
gal
average sewage =
Rave=90
generation rate
person*day

28. Find: min D [in] = 0.013 A * R2/3 = Qave = P * Qave = P * Rave
Q * n
0.013
P=30,000 people
A * R2/3 =
k * S 1/2
18+P/1000
PF=
S=0.005
k=1.49
4+P/1000
average sewage
Qave = P *
generation rate
Qave = P * Rave D
Qpeak= Qave* PF(P) d
4.96 cubic feet per second. [pause] is then plugged into ---
Qpeak= P * Rave* PF(P)
Qpeak= 4.96 [cfs]
gal
average sewage =
Rave=90
generation rate
person*day

29. Find: min D [in] = 0.013 A * R2/3 = Qave = P * Qave = P * Rave
Q * n
0.013
P=30,000 people
A * R2/3 =
k * S 1/2
18+P/1000
PF=
S=0.005
k=1.49
4+P/1000
average sewage
Qave = P *
generation rate
Qave = P * Rave D
Qpeak= Qave* PF(P) d
the variable Q, in addition to the other three values, ---
Qpeak= P * Rave* PF(P)
Qpeak= 4.96 [cfs]
gal
average sewage =
Rave=90
generation rate
person*day

30. Find: min D [in] = 0.013 A * R2/3 = Q= 4.96 [cfs] Q * n
P=30,000 people
A * R2/3 =
k * S 1/2
18+P/1000
PF=
S=0.005
k=1.49
4+P/1000
Q= 4.96 [cfs] D d
and the right hand side of the equation becomes ---
gal
average sewage =
Rave=90
generation rate
person*day

31. Find: min D [in] = A * R2/3 = 0.6137 0.013 A * R2/3 = Q= 4.96 [cfs]
Q * n
0.013
P=30,000 people
A * R2/3 =
k * S 1/2
18+P/1000
PF=
S=0.005
k=1.49
4+P/1000
Q= 4.96 [cfs]
A * R2/3 = D d
[pause] The units have to been dropped from the equation at this step, so we’ll have to keep in mind that the ---
gal
average sewage =
Rave=90
generation rate
person*day

32. Find: min D [in] = A * R2/3 = 0.6137 0.013 A * R2/3 = Q= 4.96 [cfs]
Q * n
0.013
P=30,000 people
A * R2/3 =
k * S 1/2
18+P/1000
PF=
S=0.005
k=1.49
4+P/1000
Q= 4.96 [cfs]
A * R2/3 =
hydraulic D d
area is in units of feet squared, and the hydraulic radius is in units of feet.
radius [ft]
area [ft2]
gal
average sewage =
Rave=90
generation rate
person*day

33. Find: min D [in] = A * R2/3 = 0.6137 0.013 A * R2/3 = Q= 4.96 [cfs]
Q * n
0.013
P=30,000 people
A * R2/3 =
k * S 1/2
18+P/1000
PF=
S=0.005
k=1.49
4+P/1000
Q= 4.96 [cfs]
A * R2/3 =
hydraulic D d
Now it’s time to develop an expression for the left hand side of the equation, in terms of the pipe diameter, capitol D.
radius [ft]
area [ft2]
gal
average sewage =
Rave=90
generation rate
person*day

34. Find: min D [in] = A * R2/3 = 0.6137 R = ƒ (D) A = ƒ (D)
P=30,000 people
A * R2/3 =
18+P/1000
PF=
R = ƒ (D)
4+P/1000
A = ƒ (D) D d
We can divide and conquer this problem by solving for the area and hydraulic radius separately.
gal
average sewage =
Rave=90
generation rate
person*day

35. Find: min D [in] A * R2/3 = 0.6137 R = ƒ (D) A = ƒ (D) A = ƒ (D) D d
If we increase the pipe figure, ---

36. Find: min D [in] θ A * R2/3 = 0.6137 R = ƒ (D) A = ƒ (D) D d
and define the angle, theta, from geometry, we know the cross-sectional area of the water equals, ---

37. Find: min D [in] θ (θ-sin θ) * D2 A * R2/3 = 0.6137 A = 1 8 D d *
1/8 times the quantity, theta minus sin theta, times the diameter squared. [pause] And the hydraulic radius equals, ---

38. Find: min D [in] θ (θ-sin θ) * D2 A * R2/3 = 0.6137 A = sin θ R= 1- θ8 1
sin θ R= 1-
* D * D 4 θ d θ
one quarter times the quantity, 1 minus sin theta divided by theta, time the diameter. [pause] In both of these equations, ---

39. Find: min D [in] θ (θ-sin θ) * D2 A * R2/3 = 0.6137 A = sin θ R= 1- θ8 1
sin θ R= 1-
* D * D 4 θ d θ
radians
theta is measured in radians. [pause] To find theta, ---

40. Find: min D [in] θ (θ-sin θ) * D2 A * R2/3 = 0.6137 A = sin θ R= 1- θ8 1
sin θ R= 1-
* D * D 4 θ d θ
radians
we’ll take a closer look at this triangle, or more specifically, ---

41. Find: min D [in] θ (θ-sin θ) * D2 A * R2/3 = 0.6137 A = sin θ R= 1- θ8 1
sin θ R= 1-
* D * D 4 θ d θ
radians
half the triangle. Since we’re solving for d over D ---

42. Find: min D [in] θ (θ-sin θ) * D2 A * R2/3 = 0.6137 A = sin θ R= 1- θ8 1
sin θ R= 1-
* D * D 4 θ d θ
radians
equal to 0.70, we’ll substitute in --- d
= 0.70 D

43. Find: min D [in] θ (θ-sin θ) * D2 A * R2/3 = 0.6137 A = sin θ R= 1- θ8 1
sin θ R= 1-
* D * D 4 θ θ
0.70 * D
radians
0.70 times the diameter, for the depth. [pause] Since the radius is --- d
= 0.70 D
d= 0.70 * D

44. Find: min D [in] (θ-sin θ) * D2 A * R2/3 = 0.6137 A = sin θ R= 1- θ8 1
sin θ R= 1-
* D * 4 θ
0.7 *D
radians
0.5 * D
half the diameter, we know the small leg of the triangle is --- d
= 0.70 D
d= 0.70 * D

45. Find: min D [in] (θ-sin θ) * D2 A * R2/3 = 0.6137 A = sin θ R= 1- θ8 1
sin θ
0.2 * D R= 1-
* D * 4 θ
0.7 *D
radians
0.5 * D
0.2, times the diameter, and the hypotenuse of the triangle is --- d
= 0.70 D
d= 0.70 * D

46. Find: min D [in] (θ-sin θ) * D2 A * R2/3 = 0.6137 A = sin θ R= 1- θ8 1
sin θ
0.2 * D R= 1-
* D * 4 θ
0.5 * D
0.7 *D
radians
0.5 * D
0.5, times the diameter. [pause] Knowing this is a right triangle, --- d
= 0.70 D
d= 0.70 * D

47. Find: min D [in] α (θ-sin θ) * D2 A * R2/3 = 0.6137 A = sin θ R= 1- θ8 α 1
sin θ
0.2 * D R= 1-
* D * 4 θ
0.5 * D
0.7 *D
radians
0.5 * D
the angle defined as alpha, equals, --- d
= 0.70 D
d= 0.70 * D

48. Find: min D [in] α α = cos-1 (θ-sin θ) * D2 A * R2/3 = 0.6137 A =8 α 1
sin θ
0.2 * D R= 1-
* D * 4 θ
0.5 * D
0.7 *D
radians
0.5 * D
the arc cosine of 0.2 time D divided by 0.5 times D, which equals, --- d
= 0.70 D
0.2 * D
α = cos-1
d= 0.70 * D
0.5 * D

49. Find: min D [in] α α = cos-1 (θ-sin θ) * D2 A * R2/3 = 0.6137 A =8 α 1
sin θ
0.2 * D R= 1-
* D * 4 θ
0.5 * D
0.7 *D
radians
0.5 * D
1.159 radians. [pause] From the diagram, we observe theta equals --- d
= 0.70 D
0.2 * D
α = cos-1
d= 0.70 * D
= rad
0.5 * D

50. Find: min D [in] θ α α α = cos-1 (θ-sin θ) * D2 θ = 2 * π - 2 * α
A * R2/3 = 1
A =
(θ-sin θ) * D2 * 8 α α 1
sin θ R= 1-
* D * 4 θ θ
radians
2 PI radians minus 2 times alpha, or,
θ = 2 * π - 2 * α d
= 0.70 D
0.2 * D
α = cos-1
d= 0.70 * D
= rad
0.5 * D

51. Find: min D [in] θ α α α = cos-1 (θ-sin θ) * D2 θ = 2 * π - 2 * α
A * R2/3 = 1
A =
(θ-sin θ) * D2 * 8 α α 1
sin θ R= 1-
* D * 4 θ θ
radians
theta equals, 3.964, radians. [pause] This value of theta is next plugged into the equations ---
θ = 2 * π - 2 * α d
θ = rad
= 0.70 D
0.2 * D
α = cos-1
d= 0.70 * D
= rad
0.5 * D

52. Find: min D [in] θ α α α = cos-1 (θ-sin θ) * D2 θ = 2 * π - 2 * α
A * R2/3 = 1
A =
(θ-sin θ) * D2 * 8 α α 1
sin θ R= 1-
* D * 4 θ θ
for area and hydraulic radius, and the area equals, ---
θ = 2 * π - 2 * α
θ = 3.964
0.2 * D
α = cos-1
= rad
0.5 * D

53. Find: min D [in] α = cos-1 (θ-sin θ) * D2 θ = 2 * π - 2 * α θ = 3.964
A * R2/3 = 1
A =
(θ-sin θ) * D2
= * D2 * 8 1
sin θ R= 1-
* D * 4 θ
times the diameter squared, and the hydraulic radius equals, ---
θ = 2 * π - 2 * α
θ = 3.964
0.2 * D
α = cos-1
= rad
0.5 * D

54. Find: min D [in] α = cos-1 (θ-sin θ) * D2 θ = 2 * π - 2 * α θ = 3.964
A * R2/3 = 1
A =
(θ-sin θ) * D2
= * D2 * 8 1
sin θ R= 1-
* D
= * D * 4 θ
0.2962, times the diameter. [pause]
θ = 2 * π - 2 * α
θ = 3.964
0.2 * D
α = cos-1
= rad
0.5 * D

55. Find: min D [in] α = cos-1 (θ-sin θ) * D2 θ = 2 * π - 2 * α θ = 3.964
A * R2/3 = 1
A =
(θ-sin θ) * D2
= * D2 * 8 1
sin θ R= 1-
* D
= * D * 4 θ
Returning to the previous equation, these value for area and hydraulic radius are ---
θ = 2 * π - 2 * α
θ = 3.964
0.2 * D
α = cos-1
= rad
0.5 * D

56. Find: min D [in] A * R2/3 = 0.6137 A=0.5873 * D2 R=0.2962 * D D d
substituted in for A, and R, ---

57. Find: min D [in] A * R2/3 = 0.6137 0.2610 * D8/3 = 0.6137
A= * D2
R= * D
* D8/3 = D d
and after a few lines of algebra, the find that the minimum allowable diameter, D, is, ---

58. Find: min D [in] A * R2/3 = 0.6137 0.2610 * D8/3 = 0.6137
A= * D2
R= * D
* D8/3 =
D = [ft] D d
1.378 feet, which equals, ---

59. Find: min D [in] A * R2/3 = 0.6137 0.2610 * D8/3 = 0.6137
A= * D2
R= * D
* D8/3 =
D = [ft]
= [in] D d
16.54 inches [pause] This diameters must be rounded up, ---

60. Find: min D [in] A * R2/3 = 0.6137 0.2610 * D8/3 = 0.6137
A= * D2
R= * D
* D8/3 =
D = [ft] 12 16 18 24
= [in] D d
to the nearest possible solution, ---

61. Find: min D [in] A * R2/3 = 0.6137 0.2610 * D8/3 = 0.6137
A= * D2
R= * D
* D8/3 =
D = [ft] 12 16 18 24
= [in] D d
and the answer is C.
answerC

62. ? 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