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Design Procedure in Türkiye - An Example...

"Network of the Sanitary Sewer System"...
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Basic data for system design are as follows :

Item	Value
Design population (for 30 years)	N ₃₀ = 68,000
Domestic water demand	Q _d = 195.60 L / sec
Domestic water distributed : wastewater collected ratio	P ₁ = 1
Return time ratio	P ₂ = 2
Density coefficient for big streets	k ₁ = 1.5
Density coefficient for small streets	k ₂ = 1
Total length of big streets	L ₁ = 1,700 m
Total length of small streets	L ₂ = 17,450 m
Upper wastewater flow to manhole no . 4 from manhole no . 2	Q _{2 - 4} = 6.66 L / sec
Upper wastewater flow to manhole no . 8 from manhole no . 10	Q _{10 - 8} = 10.00 L / sec
Entering depth to manhole no. 4 from manhole no . 2	h _{u : 2 - 4} = 2.70 m
Entering depth to manhole no. 8 from manhole no . 10	h _{u : 10 - 8} = 3.00 m
Manning friction coefficient	n _M = 0.015

"Design Equations"...

Design procedure is given below :

Wastewater flowrate : Q_waste = P ₁ x P ₂ x Q _d = 1 x 2 x 195.60 = 391.20 L / sec
Total weighted length of the big streets : T₁ = L ₁ x k ₁ = 1,700 x 1.5 = 2,550 m
Total weighted length of the small streets : T₂ = L ₂ x k ₂ = 17,450 x 1.0 = 17,450 m
Total weighted length of the streets : T _T = T ₁ + T ₂ = 2,550 + 17,450 = 20,000 m
Unit wastewater flowrate : q_waste = Q _waste / T _T = 391.20 / 20,000 = 0.01956 L / sec . m

"Explanatory Cross - Section"...

Design table is shown below :

From

Lxk

q_wastexLxk

Q_upper

Q_additional

Q_design

Z_{r - from}

Z_{r - to}

h_{u - from}

h_{u - to}

Z_{u - from}

Z_{u - to}

Z_{u - from} - Z_{u - to}

J (0%)

135

2.64

0.00

2.64

518.70

516.70

1.30

1.50

517.40

515.20

2.20

16.3

155

3.03

9.30

0.00

12.33

516.70

515.50

2.70

2.30

514.00

513.20

0.80

5.2

135

2.64

0.00

2.64

519.20

515.50

1.30

1.50

517.90

514.00

3.90

28.9

110

2.15

14.97

0.00

17.12

515.50

515.30

2.30

2.50

513.20

512.80

0.40

3.6

135

2.64

0.00

2.64

520.00

515.30

1.30

1.50

518.70

513.80

4.90

36.3

29.76

0.00

515.30

3.00

512.30

From

Q_fill

V_fill

Q_design / Q_fill

V_design / V_fill

h / D

V_design

h_design

No.drop

Drop

Total drop

Z_{b - from}

Z_{b - to}

Hydraulic regime

36.44

1.16

0.07

0.59

0.18

0.68

3.6

517.20

515.00

Turbulence

20.42

0.65

0.60

1.04

0.56

0.68

11.2

513.80

513.00

Turbulence

48.38

1.54

0.05

0.54

0.15

0.83

3.0

517.70

513.80

Turbulence

50.19

0.71

0.34

0.91

0.40

0.65

12.0

512.90

512.50

Turbulence

54.04

1.72

0.05

0.54

0.15

0.93

3.0

518.50

513.60

Turbulence

512.00

"Hydraulic Elements"...

Table for determination of the hydraulic regime is given below. If h _design is less than h _critical, the hydrauic regime is on the turbulence section.

Q / (g/d)^1/2(D)^5/2	h _critical / D
0.0001	0.01
0.0010	0.03
0.0027	0.05
0.0107	0.10
0.0238	0.15
0.0418	0.20
0.0646	0.25
0.0921	0.30
0.1241	0.35
0.1603	0.40
0.2011	0.45
0.2459	0.50
0.2949	0.55
0.3484	0.60

"Longitudal Cross - Section of Street 5 - 4"...

Statistical Info About the Sewer System Use in Türkiye...

Population	Population info served by sanitay sewer system ( As of 1995 )
Population	Total	Percentage in Türkiye's population	Percentage in municipalities' population
- 2,001	99,409	0.70	15.93
2,001 - 5,000	1,004,101	18.10	23.84
5,001 - 10,000	1,182,409	37.83	38.25
10,001 - 25,000	2,439,255	55.09	55.26
25,001 - 50,000	2,249,988	57.96	57.96
50,001 - 100,000	3,923,162	76.94	76.94
100,001 - 500,000	7,127,200	79.08	79.08
500,001 - 1,000,000	1,513,948	65.15	65.15
1,000,001 - 5,000,000	6,297,613	89.14	89.14
5,000,000 +	7,327,047	90.00	90.00
Türkiye	33,164,131	53	69

Click here for the "Detailed Information about the Infrastructure Systems of Türkiye ( In Turkish )"...

Click here for the Source : "State Statistics Institute"...

Basic Calculations...

Example - 1...

Given : Circular pipe flowing full, Slope, S = 0.01, n = 0.013, Dia = 18 in.
Required : V_full and Q_full
Solution : Note that R is the hydraulic radius and not the geometric radius of the pipe
V_full = (1.49 / n) (R)^{2 / 3} (S) ^{1 / 2}
At full flow in a circular pipe : R = Area / Wetted perimeter
R = [ (p) (Dia)² / 4 ] / (p) (Dia) = Dia / 4
V_full = (1.49 / n) (Dia / 4)^{2 / 3} (S) ^{1 / 2}
V_full = (1.49 / 0.013) (18 / 4)^{2 / 3} (0.01) ^{1 / 2} = 5.96 ft / sec
Q_full = (V_full) (Area)
Q_full = (V_full) [ (p) (Dia)² / 4 ]
Q_full = (5.96) [ (3.14) (18 / 12)² / 4 ] = 10.5 ft³ / sec

Example - 2...

Given : Q = 7 ft³ / sec, Slope = 0.01, n = 0.013
Required : Select the smallest standard pipe size that will carry the flow at the specified slope. Note: For this class, standard pipe sizes for sewers are: 8", 10", 12", 15", 18", 21", 24", 27", 30", 33", 36", 42", 48", 54", 60", 72", 84", 96"
Solution : Try Dia = 15 inches
Q_full = (1.49 / n) (Dia / 4)^{2 / 3} (S) ^{1 / 2} (Area)
Q_full = (1.49 / 0.013) (15 / 12 / 4)^{2 / 3} (0.01) ^{1 / 2} [ (3.14) (15 / 12)² / 4 ] = 6.5 ft³ / sec
Try Dia = 18 inches
Q_full = (1.49 / 0.013) (18 / 12 / 4)^{2 / 3} (0.01) ^{1 / 2} [ (3.14) (18 / 12)² / 4 ] = 10.5 ft³ / sec
It is also possible to solve for dia in Manning's Equation :
Dia = (1.33) [ (Q_full) (n) ]^{3 / 8} / S^{1 / 2}
Dia = 1.29 ft = 15.4 in
The next largest standard size greater than 15.4 inches is 18 inches.

Example - 3...

This problem illustrates the use of figure related to hydraulic elements in working with partially full sewers.
Given : D_full = 18 in, Q = 7 cfs (the design flow - i.e., what the actual flow will be under design conditions), V_full = 5.96 fps (what the velocity would be at full flow), Q_full = 10.5 cfs (what Q will be when the pipe is flowing full)
Required : (a) depth, d, when Q = 7 cfs (i.e., when the sewer is partially full) and (b) velocity, v, when Q = 7 cfs (i.e., when the sewer is partially full)
Solution : Q / Q_full = 7 / 10.5 = 0.67
From figure related to hydraulic elements, d / d_full = 0.61
d = (d_full) (0.61) = 11 in
From figure related to hydraulic elements, V / V_full = 1.08
V = (V_full) (1.08) = 6.4 ft / sec

Example - 4...

Given : Design flow, Q = 7.5 cfs, n = 0.013
Required : Select a standard pipe size such that : (a) pipe slope is as close as possible to the street slope of 0.001, but the velocity is at least 2 fps and (b) d / d_full is approximately 2/3 (pipe is flowing 2/3 full at design (peak) flow)
Solution : A slope of 0.001 will initially be selected for the slope. If a velocity of 2 fps is not possible, the slope can be modified.
The procedure used in this problem is a "trial and error" type solution using Mathcad. A similar process could be accomplished on a spreadsheet using cell formulas.

Constants	Variables (try different values until the conditions are satisfied)
n = 0.013	Dia = 27 in
7.5 ft³ / sec	S = 0.001

Values computed from constants and variables above :
Hydraulic radius : R = Dia / 4 (true only for a pipe flowing full)
Area of pipe : A = (p) (Dia)² / 4
Velocity in a full pipe : V_full = (1.49 / n) (Dia / 4)^{2 / 3} (S) ^{1 / 2} = 2.46 ft / sec
Flow in a full pipe : Q_full = (V_full) (Area) = 9.79 ft ³ / sec
Q / Q_full = 0.77
d / d_full = 0.66 (read this from figure related to hydraulic elements - see note 1 below)
V / V_full = 1.12 (V / V_full is read from figure related to hydraulic elements - see note 2 below)
V = (1.12) (V_full) = 2.76 ft / sec
Note 1 : Start with the value of Q / Q_full on the horizontal axis of figure related to hydraulic elements. Draw a line vertically upward until it intersects the discharge curve. From the intersection, draw a line horizontally to the left and read d / d_full from the vertical axis.
Note 2 : From the d / d_full value on the vertical axis of figure related to hydraulic elements, draw a line horizontally to the right until it intersects the velocity curve. From the intersection, draw a line vertically downward and read V / V_full from the horizontal axis.

"Hydraulic Elements"...