Design Procedure in Türkiye - An Example...

"Network of the Sanitary Sewer System"...

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 : Qwaste = 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 : qwaste = Q waste / T T = 391.20 / 20,000 = 0.01956 L / sec . m

"Explanatory Cross - Section"...


Design table is shown below :

From To L k Lxk qwastexLxk Qupper Qadditional Qdesign 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%)
5 4 135 1 135 2.64 0.00 0.00 2.64 518.70 516.70 1.30 1.50 517.40 515.20 2.20 16.3
4 6 155 1 155 3.03 9.30 0.00 12.33 516.70 515.50 2.70 2.30 514.00 513.20 0.80 5.2
7 6 135 1 135 2.64 0.00 0.00 2.64 519.20 515.50 1.30 1.50 517.90 514.00 3.90 28.9
6 8 110 1 110 2.15 14.97 0.00 17.12 515.50 515.30 2.30 2.50 513.20 512.80 0.40 3.6
9 8 135 1 135 2.64 0.00 0.00 2.64 520.00 515.30 1.30 1.50 518.70 513.80 4.90 36.3
8 X - - - - 29.76 0.00 - 515.30 - 3.00 - 512.30 - - -

From To D Qfill Vfill Qdesign / Qfill Vdesign / Vfill h / D Vdesign hdesign No.drop Drop Total drop Z b - from Z b - to Hydraulic regime
5 4 20 36.44 1.16 0.07 0.59 0.18 0.68 3.6 - - - 517.20 515.00 Turbulence
4 6 20 20.42 0.65 0.60 1.04 0.56 0.68 11.2 - - - 513.80 513.00 Turbulence
7 6 20 48.38 1.54 0.05 0.54 0.15 0.83 3.0 - - - 517.70 513.80 Turbulence
6 8 30 50.19 0.71 0.34 0.91 0.40 0.65 12.0 - - - 512.90 512.50 Turbulence
9 8 20 54.04 1.72 0.05 0.54 0.15 0.93 3.0 - - - 518.50 513.60 Turbulence
8 X 30 - - - - - - - - - - 512.00 - -

"Hydraulic Elements"...

Table for determination of the hydraulic regime is given below. If hdesign < hcritical, 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
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 : Vfull and Qfull
Solution : Note that R is the hydraulic radius and not the geometric radius of the pipe
Vfull = (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
Vfull = (1.49 / n) (Dia / 4) 2 / 3 (S) 1 / 2
Vfull = (1.49 / 0.013) (18 / 4) 2 / 3 (0.01) 1 / 2 = 5.96 ft / sec
Qfull = (Vfull) (Area)
Qfull = (Vfull) [ (p) (Dia) 2 / 4 ]
Qfull = (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
Qfull = (1.49 / n) (Dia / 4) 2 / 3 (S) 1 / 2 (Area)
Qfull = (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
Qfull = (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) [ (Qfull) (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 : Dfull = 18 in, Q = 7 cfs (the design flow - i.e., what the actual flow will be under design conditions), Vfull = 5.96 fps (what the velocity would be at full flow), Qfull = 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 / Qfull = 7 / 10.5 = 0.67
From figure related to hydraulic elements, d / dfull = 0.61
d = (dfull) (0.61) = 11 in
From figure related to hydraulic elements, V / Vfull = 1.08
V = (Vfull) (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 / dfull 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 : Vfull = (1.49 / n) (Dia / 4) 2 / 3 (S) 1 / 2 = 2.46 ft / sec
Flow in a full pipe : Qfull = (Vfull) (Area) = 9.79 ft 3 / sec
Q / Qfull = 0.77
d / dfull = 0.66 (read this from figure related to hydraulic elements - see note 1 below)
V / Vfull = 1.12 (V / Vfull is read from figure related to hydraulic elements - see note 2 below)
V = (1.12) (Vfull) = 2.76 ft / sec
Note 1: Start with the value of Q / Qfull 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 / dfull from the vertical axis.
Note 2: From the d / dfull 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 / Vfull from the horizontal axis.