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Storm Sewer System Design

Water Resources Engineering (Negros Oriental State University)

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Republic of the Philippines

Negros Oriental State University

College of Engineering and Architecture
Main Campus II, Bajumpandan, Dumaguete City

ENS 438 WATER RESOURCES ENGINEERING

SECTION B / MWF 8:00-9:00 AM

STORM SEWER SYSTEM

DESIGN
SUBMITTED BY:

ENGR. Ph.D.
INSTRUCTOR

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TABLE OF CONTENTS

I. INTRODUCTION ......................................................................................................... 3

II. DRAINAGE DESIGN CRITERIA ............................................................................ 4

1.0 PIPE SIZING .............................................................................................................. 4

2.0 COEFFICIENTS......................................................................................................... 4

3.0 INLET TIME .............................................................................................................. 5

4.0 FORMULA/EQUATIONS ......................................................................................... 6

5.0 MANNING’S EQUATION ........................................................................................ 6

6.0 VELOCITY ................................................................................................................ 7

7.0 SEWER FLOW TIME................................................................................................ 7

8.0 DEPTH OF SOIL COVER ......................................................................................... 8

III. STORM SEWER SYSTEM DESIGN LAYOUT ...................................................... 9

IV. DATA SUMMARY & RESULTS ........................................................................... 10

V. DESIGN COMPUTATIONS ......................................................................................11

1. LINE 1-2 CALCULATION .....................................................................................11

2. LINE 3-4 CALCULATION .....................................................................................14

3. LINE 4-2 CALCULATIONS ...................................................................................17

4. LINE 2–OUTFALL CALCULATIONS .................................................................. 20

5. LINE 5-OUTFALL CALCULATION ..................................................................... 23

6. OUTFALL DISCHARGE CALCULATION........................................................... 26

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I. INTRODUCTION
Design the storm sewer system of the proposed subdivision site development plan below for a
10-year return period.

Figure 1: Subdivision Map

Figure 2: Contour Map

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II. DRAINAGE DESIGN CRITERIA

Storm Sewer system design based on the following criteria for a 10-year return period.
1.0 PIPE SIZING
The calculated pipe must be rounded up to the nearest available commercial pipe size.
1.2 COMMERCIAL PIPE
Table 1: Commercially Available Pipe
Pipe Diameter
12” 300 m
14” 350 m
15” 375 m
16” 400 m
18” 450 m
21” 525 m
24” 600 m
27” 675 m
30” 750 m
36” 900 m
2.0 COEFFICIENTS
2.1 MANNINGS FORMULA, n

Table 2: Manning’s Formula, n

Values of n to be used
n
Nature of surfaces
Min Max
Neat cement surfaces 0.010 0.013
Wood-Stave pipe 0.010 0.013
Plank flumes, planed 0.010 0.014
Vitrified sewer pipe 0.010 0.017
Metal flumes, smooth 0.011 0.015
Concrete, precast 0.011 0.013
Cement mortar surfaces 0.011 0.015
Plank flumes, unplanned 0.011 0.015
Common- clay drainage tile 0.011 0.017
Concrete, monolithic 0.012 0.016
Brick with cement mortar 0.012 0.017
Cast iron-new 0.013 0.017
Cement rubble surfaces 0.017 0.030
Riveted steel 0.017 0.020
Corrugated metal pipe 0.021 0.025
Canals and ditches, smooth earth 0.017 0.025

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2.2 RUNOFF COEFFICIENT FOR RATIONAL METHOD

Runoff Coefficient should be chosen to match the character surface of each area in the
subdivision map.
Table 15.2.3 Runoff Coefficients for Use in the Rational Method
Return period (years)
Character of Surface 2 5 10 25
Developed
Asphaltic 0.73 0.77 0.81 0.86
Concrete/roof 0.75 0.80 0.83 0.88
Grass areas (lawns, parks, etc.)
Poor condition (grass cover less than 50% of the area)
Flat, 0-2% 0.32 0.34 0.37 0.40
Average, 2-7% 0.37 0.40 0.43 0.46
Steep, over 7% 0.40 0.43 0.45 0.49
Fair condition (grass cover 50% to 75% of the area)
Flat, 0-2% 0.25 0.28 0.30 0.34
Average, 2-7% 0.33 0.36 0.38 0.42
Steep, over 7% 0.37 0.40 0.42 0.46
Good condition (grass cover larger than 75% of the area)
Flat, 0-2% 0.21 0.23 0.25 0.29
Average, 2-7% 0.29 0.32 0.35 0.39
Steep, over 7% 0.34 0.37 0.49 0.44
Undeveloped
Cultivated land
Flat, 0-2% 0.31 0.34 0.36 0.40
Average, 2-7% 0.35 0.38 0.41 0.44
Steep, over 7% 0.39 0.42 0.44 0.48
Pasture/range
Flat, 0-2% 0.25 0.28 0.30 0.34
Average, 2-7% 0.33 0.36 0.38 0.42
Steep, over 7% 0.37 0.40 0.42 0.46
Forest/woodlands
Flat, 0-2% 0.20 0.25 0.28 0.31
Average, 2-7% 0.31 0.34 0.36 0.40
Steep, over 7% 0.35 0.39 0.41 0.45

3.0 INLET TIME

Table 3: Inlet Time
Characteristics 끫룂끫뢬 (minutes)
A. Densely developed areas 5-10
B. Flat well-developed districts 10-15
C. Flat residential districts with widely
20-30
spaced street inlets

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4.0 FORMULA/EQUATIONS
4.1 RAINFALL INTENSITY FORMULA
122.6끫뢎
끫뢬 =
(끫룂 + 1)0.83
where,
i = rainfall intensity (mm/hr)
T = return period
t = inlet time (min)
4.2 RUNOFF DISCHARGE FORMULA
끫뢈 = 0.278 끫롬끫뢬 끫롨
where,
끫뢴끫뢴
끫뢬 = ℎ끫뢾

끫롨 = 끫뢰끫뢴2
끫롬 = Runoff Coefficient
끫뢴3
끫뢈 = 끫룀

5.0 MANNING’S EQUATION

2 1
1
끫뢈 = 끫뢶 끫롨끫뢊 3 끫뢌 2

2 1
1
끫뢒 = 끫뢶 끫뢊 3 끫뢌 2

5.1 PIPE DIAMETER EQUATION

5.1.1 ASSUMING THE PIPE IF FLOWING FULL,
Using the A and R, in terms of D,
끫븖끫롮2
끫롨 =
4
끫롮
끫뢊 = 4

Substituting it to Manning’s Equation,

we get,
1 끫븖끫롮2 끫롮 2 1
끫뢈 = ( )( )3 끫뢌 2
끫뢶 4 4
Figure 3: Full Flowing Pipe

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5.1.2 WETTED ANGLE FOR A PARTIALLY FLOWING PIPE,

If the pipe is flowing, more than 50% of the Pipe Area,
끫븖끫뢾 2 끫뷺 1 2
끫롨 = + [끫뢾 sin(360 − 끫뷺)]
360 2
If the pipe is flowing, less than 50% of the Pipe Area,
끫븖끫뢾 2 끫뷺 1 2
끫롨 = − [끫뢾 sin(끫뷺)]
360 2
끫븖끫뢾(끫뷺)
끫뢆 = 180

Note: When beta is greater than

180°, the depth of flow is more
than half full, and when it’s less
than 180°, the depth of flow is
less than half full.
Calculating depth of flow, d,
끫뷺
끫뢢 = 끫뢾 − [끫뢾 cos � 2 �], if beta
(β) is less than 180

Figure 4: Partially Flowing

끫뷺 Pipe
끫뢢 = 끫뢾 + [끫뢾 cos � 2 �], if beta
(β) is greater than 180

6.0 VELOCITY
The calculated velocity must not exceed the allowable maximum velocity to avoid any erosion,
and must not go below the minimum velocity, to avoid anything settling in the pipe.

Minimum Velocity,
3
끫뢒 = 0.75 끫뢴 �끫룀

Maximum Velocity,
3
끫뢒 = 3.0 끫뢴 �끫룀
7.0 SEWER FLOW TIME
The sewer flow time of each line shall be calculated using,
끫롾끫롾끫롾끫롾끫롾끫롾
끫룂 =
끫뢒끫롾끫롾끫뢒끫뢒끫뢒끫롾끫뢒

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8.0 DEPTH OF SOIL COVER

The recommended depth of soil cover must be equal or greater than 0.6 m.

8.1 INVERT ELEVATION

Upper and lower invert elevation shall be solved using this equation,

끫뢐끫뢐끫뢐끫뢐끫뢾 끫롸끫뢶끫롸끫뢐끫뢾끫룂 끫롰끫롰끫뢐끫롸. = 끫뢀끫뢀끫뢶ℎ끫뢸끫롰끫뢐 끫롰끫롰끫뢐끫롸끫뢀끫룂끫뢬끫뢸끫뢶 − 끫롮끫뢐끫뢐끫룂ℎ 끫뢸끫뢸 끫뢌끫뢸끫뢬끫롰 끫롬끫뢸끫롸끫뢐끫뢾 − 끫뢆끫뢬끫뢐끫뢐 끫롮끫뢬끫뢀.
끫롾끫뢸끫롾끫뢐끫뢾 끫롸끫뢶끫롸끫뢐끫뢾끫룂 끫롰끫롰끫뢐끫롸. = 끫뢐끫뢐끫뢐끫뢐끫뢾 끫롸끫뢶끫롸끫뢐끫뢾끫룂 − (끫뢌끫뢸 끫룊 끫롾끫뢐끫뢶끫롾끫룂ℎ)

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III. STORM SEWER SYSTEM DESIGN LAYOUT

Figure 5: Storm Sewer System Design Layout

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IV. DATA SUMMARY & RESULTS

LOCATION BASIC DATA SEWER DESIGN

GROUND
SURFACE CALCULATED DATA
Runoff Rainfall ELEVATION
Length Area Intel Manning's Sewer
Line No. Manhole No. Coefficient Intensity, I MH ELEVATION Invert Elevation (m)
(m) (m2) Time (ti) n SLOPE, Computed Commer Velocity Flowrate, Depth of Flow
,C (mm/hr) cial Pipe flow, d
Upper So Ø (mm) (m/s) Q (m3/s) Time
Lower (m) Ø (mm) (mm) Upper Lower
(m) (min)
Line 1-2 1 2 35 1404 0.83 218.236 7 92.0625 91.7812 0.0080 277.9133 300 1.368 0.0707 0.206 0.426 91.163 90.881 PASSED
Line 3-4 3 4 35 750 0.83 277.093 5 92.1938 91.9375 0.007 244.4899 300 1.214 0.0480 0.164 0.480 91.294 91.038 PASSED
Line 4-2 4 2 28 510 0.83 243.814 6 91.9375 91.7812 0.006 306.7175 350 1.227 0.0766 0.216 0.380 90.988 90.831 PASSED
0.013
Line 2-Outfal 2 Outfall 30 1074 0.83 209.028 7.426 91.7812 91.6438 0.005 455.3653 525 1.448 0.1991 0.319 0.345 90.656 90.519 PASSED
Line 5-Outfal 5 Outfall 28 441 0.83 218.236 7 91.75 91.6438 0.004 207.2343 300 0.782 0.0222 0.127 0.597 90.850 90.744 PASSED
Outfall - 861 0.83 197.911 8 91.6438 0.2607

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V. DESIGN COMPUTATIONS

1. LINE 1-2 CALCULATION

Solve for the time of concentration & Rainfall Intensity :

𝑢끫룀끫뢐 끫룂끫뢬1 = 끫룂𝑐1 = 7 mins

T = 10 years
i = 218.236 mm/hr

Solve for the design runoff, Q, using rational method;

2
C= 0.83 Atotal = 0.0014 km
L= 35 m

끫뢈 = 0.278끫롬끫뢬끫롨

Q = 0.070699 m3 /s

Assuming Pipe is flowing full, for D ;

n = 0.013 So = 0.00804

1 끫븖끫롮 2 끫롮 2 1
끫뢈 = ( )( )3 끫뢌 2
끫뢶 4 4
D = 0.277913 m
D = 277.9133 mm
*Computed Pipe Size
D = 300 mm
*Commercial Pipe Size
r= 0.15 m

Checking if the Pipe is more than or less than half full

Areafull = 0.060661 m

Commercial Pipe Area,

Area = 0.070686 m

Area Percentage = Areafull/Commercial Pipe Area

Area Percentage = 85.82 %

Since it is more than 50% the pipe is flowing more than half full

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Calculating the angle beta (β)

From manning's equation,

5
끫뢈끫뢶 끫롨 3
1 = 2
끫뢌 2 끫뢆 3
5
끫롨 3
0.010252 = 2
끫뢆 3
Since,
끫븖끫뢾2 끫뷺 1 2
끫롨 = + 끫뢾 sin(360 − 끫뷺)
360 2

끫븖끫뢾끫뷺
끫뢆 =
180
5
3
0.000196 (β) + 0.01125 sin (360 - β)
0.010252 =
끫븖 0.15 끫뷺 2
( 3 )
180
Note : For complex equations, the use of calculator is needed to solve beta ( β ).

β = 224.060 *Beta is greater than 180

ϴ = 135.940
𝜃
끫뢢 = 끫뢾 − [끫뢾𝑐끫뢸끫룀
]
2 , if beta (β) is less than 180
𝜃
끫뢢 = 끫뢾 + [끫뢾𝑐끫뢸끫룀
2 , if beta (β) is greater than 180
]
Checking if its equal or not, through substitution.

0.0103 = 0.0103

Solve for the actual velocity.

끫븖끫뢾2 끫뷺 1 2
끫롨 = + 끫뢾 sin(360 − 끫뷺)
360 2
A = 0.051817 m2

끫븖끫뢾끫뷺
끫뢆 =
180
P = 0.586587 m
1 2 1
끫뢒 = 끫뢊 3 끫뢌 2
끫뢶
V = 1.367842 m/s *Velocity is "> V min " and "< V max ".

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Calculate Depth of Flow.

𝜃
끫뢢 = 끫뢾 + [끫뢾𝑐끫뢸끫룀 ]
2
d = 0.206263 m

Solve Sewer Flow Time.

끫롾끫뢐끫뢶끫롾끫룂ℎ
끫룂 =
끫뢒끫뢐끫롰끫뢸𝑐끫뢬끫룂𝑦

35 = 25.5877 seconds
t=
1.367842
= 0.42646 min

Invert Elevation

끫뢐끫뢐끫뢐끫뢐끫뢾 끫롸끫뢶끫롸끫뢐끫뢾끫룂 끫롰끫롰끫뢐끫롸. = 끫뢀끫뢀끫뢶ℎ끫뢸끫롰끫뢐 끫롰끫롰끫뢐끫롸. −끫롮끫뢐끫뢐끫룂ℎ 끫뢸끫뢸 끫롬끫뢸끫롸끫뢐끫뢾 − 끫뢆끫뢬끫뢐끫뢐 끫롮끫뢬끫뢀.

MH 1 Elev. = 92.0625 m

Depth of Cover = 0.6 m

Upper Invert = 91.1625 m

끫롾끫뢸끫롾끫뢐끫뢾 끫롸끫뢶끫롸끫뢐끫뢾끫룂 끫롰끫롰끫뢐끫롸. = 끫뢐끫뢐끫뢐끫뢐끫뢾 끫롸끫뢶끫롸끫뢐끫뢾끫룂 끫롰끫롰끫뢐끫롸. −(끫뢌끫뢸 끫룊 끫롾끫뢐끫뢶끫롾끫룂ℎ)

Lower Invert = 90.8812 m

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2. LINE 3-4 CALCULATION

Solve for the time of concentration & Rainfall Intensity :

𝑢끫룀끫뢐 끫룂끫뢬3 = 끫룂𝑐3 = 5 mins

T = 10 years
i = 277.093 mm/hr

Solve for the design runoff, Q, using rational method;

2
C= 0.83 Atotal = 0.00075 km
L= 35 m

끫뢈 = 0.278끫롬끫뢬끫롨
3
Q = 0.047952 m /s

Assuming Pipe is flowing full, for D ;

n = 0.013 So = 0.00732

1 끫븖끫롮 2 끫롮 2 1
끫뢈 = ( )( )3 끫뢌 2
끫뢶 4 4

D = 0.24449 m
D = 244.4899 mm
*Computed Pipe Size
D = 300 mm
*Commercial Pipe Size
r= 0.15 m

Checking if the Pipe is more than or less than half full

Areafull = 0.046947 m

Commercial Pipe Area,

Area = 0.070686 m

Area Percentage = Areafull/Commercial Pipe Area

Area Percentage = 66.42 %

Since it is more than 50% the pipe is flowing more than half full

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Calculating the angle beta (β)

From manning's equation,

5
끫뢈끫뢶 끫롨 3
1 = 2
끫뢌 2 끫뢆 3
5
끫롨 3
0.007285 = 2
끫뢆 3
Since,
끫븖끫뢾2 끫뷺 1 2
끫롨 = + 끫뢾 sin(360 − 끫뷺)
360 2
끫븖끫뢾끫뷺
끫뢆 =
180 5
3
0.000196 (β) + 0.01125 sin (360 - β)
0.007285 =
끫븖 0.15 끫뷺 2
( )3
180
Note : For complex equations, the use of calculator is needed to solve beta ( β ).

β = 190.8673 *Beta is greater than 180

ϴ = 169.1327
𝜃
끫뢢 = 끫뢾 − [끫뢾𝑐끫뢸끫룀 ] , if beta (β) is less than 180
2
𝜃
끫뢢 = 끫뢾 + [끫뢾𝑐끫뢸끫룀 ] , if beta (β) is greater than 180
2
Checking if its equal or not, through substitution.

0.0073 = 0.0073

Solve for the actual velocity.

끫븖끫뢾2 끫뷺 1 2
끫롨 = + 끫뢾 sin(360 − 끫뷺)
360 2
2
A = 0.039598 m

끫븖끫뢾끫뷺
끫뢆 =
180
P = 0.499689 m
1 2 1
끫뢒 = 끫뢊 3 끫뢌 2
끫뢶
V = 1.214442 m/s *Velocity is "> V min " and "< V max ".

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Calculate Depth of Flow.

𝜃
끫뢢 = 끫뢾 + [끫뢾𝑐끫뢸끫룀 ]
2
d = 0.164204 m

Solve Sewer Flow Time.

끫롾끫뢐끫뢶끫롾끫룂ℎ
끫룂 =
끫뢒끫뢐끫롰끫뢸𝑐끫뢬끫룂𝑦
35 = 28.8198 seconds
t=
1.214442
= 0.48033 min

Invert Elevation

끫뢐끫뢐끫뢐끫뢐끫뢾 끫롸끫뢶끫롸끫뢐끫뢾끫룂 끫롰끫롰끫뢐끫롸. = 끫뢀끫뢀끫뢶ℎ끫뢸끫롰끫뢐 끫롰끫롰끫뢐끫롸. −끫롮끫뢐끫뢐끫룂ℎ 끫뢸끫뢸 끫롬끫뢸끫롸끫뢐끫뢾 − 끫뢆끫뢬끫뢐끫뢐 끫롮끫뢬끫뢀.

MH 3 Elev. = 92.1938 m

Depth of Cover = 0.6 m

Upper Invert = 91.2938 m

끫롾끫뢸끫롾끫뢐끫뢾 끫롸끫뢶끫롸끫뢐끫뢾끫룂 끫롰끫롰끫뢐끫롸. = 끫뢐끫뢐끫뢐끫뢐끫뢾 끫롸끫뢶끫롸끫뢐끫뢾끫룂 끫롰끫롰끫뢐끫롸. −(끫뢌끫뢸 끫룊 끫롾끫뢐끫뢶끫롾끫룂ℎ)

Lower Invert = 91.0375 m

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3. LINE 4-2 CALCULATIONS

Solve for the Time of Concentration & Rainfall Intensity :

𝑢끫룀끫뢐 끫룂끫뢬3 = 5 mins 끫룂끫뢬4 = 6 mins

끫룂𝑐3−4 = 끫룂끫뢬3 + 끫룂𝑓3
끫룂𝑓3 = 0.48033 mins
t c3-4 = 5.480 min
t i4 = 6 min *Use this the highest time

use t c4 = 6 min

T = 10 years
i = 243.814 mm/hr

Solve for the design runoff, Q, using rational method;

2
C= 0.83 Atotal = 0.00051 km
L= 28 m

끫뢈 = 0.278끫롬끫뢬끫롨

Q = 0.0286914 m /s + Q 3-4
3

Q = 0.0286914 m /s + 0.04795 m /s
3 3

Q = 0.0766437 m3 /s

Assuming Pipe is flowing full, for D ;

n = 0.013 So = 0.00558

1 끫븖끫롮 2 끫롮 2 1
끫뢈 = ( )( )3 끫뢌 2
끫뢶 4 4

D = 0.3067175 m
D = 306.71747 mm
*Computed Pipe Size
D = 350 mm
*Commercial Pipe Size
r= 0.175 m

Checking if the Pipe is more than or less than half full

Areafull = 0.0738868 m

Commercial Pipe Area,

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Area = 0.0962113 m

Area Percentage = Areafull/Commercial Pipe Area

Area Percentage = 76.80 %

Since it is more than 50% the pipe is flowing more than half full

Calculating the angle beta (β)

From manning's equation,

5
끫뢈끫뢶 끫롨 3
1 = 2
끫뢌 2 끫뢆 3
5
끫롨 3
0.013336 = 2
끫뢆 3

Since,
끫븖끫뢾 2 끫뷺 1 2
끫롨 = + 끫뢾 sin(360 − 끫뷺)
360 2
끫븖끫뢾끫뷺
끫뢆 =
180 5
3
0.00026725 (β) + 0.01531 sin (360 - β)
0.013336 =
끫븖 0.175 끫뷺 2
( )3
180
Note : For complex equations, the use of calculator is needed to solve beta ( β ).

β= 207.404 *Beta is greater than 180

ϴ= 152.596
𝜃
끫뢢 = 끫뢾 − [끫뢾𝑐끫뢸끫룀 ] , if beta (β) is less than 180
2
𝜃
끫뢢 = 끫뢾 + [끫뢾𝑐끫뢸끫룀 ] , if beta (β) is greater than 180
2

Checking if its equal or not, through substitution.

0.0133 = 0.0133

Solve for the actual velocity.

끫븖끫뢾2 끫뷺 1 2
끫롨 = + 끫뢾 sin(360 − 끫뷺)
360 2
A = 0.0624772 m2

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끫븖끫뢾끫뷺
끫뢆 =
180
P = 0.6334793 m
1 2 1
끫뢒 = 끫뢊 3 끫뢌 2
끫뢶
V = 1.2268264 m/s *Velocity is "> V min " and "< V max ".

Calculate Depth of Flow.

𝜃
끫뢢 = 끫뢾 + [끫뢾𝑐끫뢸끫룀 ]
2
d = 0.2164525 m

Solve Sewer Flow Time.

끫롾끫뢐끫뢶끫롾끫룂ℎ
끫룂 =
끫뢒끫뢐끫롰끫뢸𝑐끫뢬끫룂𝑦
28
t= = 22.8231 seconds
1.2268264
= 0.38039 min

Invert Elevation

끫뢐끫뢐끫뢐끫뢐끫뢾 끫롸끫뢶끫롸끫뢐끫뢾끫룂 끫롰끫롰끫뢐끫롸. = 끫뢀끫뢀끫뢶ℎ끫뢸끫롰끫뢐 끫롰끫롰끫뢐끫롸. −끫롮끫뢐끫뢐끫룂ℎ 끫뢸끫뢸 끫롬끫뢸끫롸끫뢐끫뢾 − 끫뢆끫뢬끫뢐끫뢐 끫롮끫뢬끫뢀.

MH 4 Elev. = 91.9375 m

Depth of Cover = 0.6 m

Upper Invert = 90.9875 m

끫롾끫뢸끫롾끫뢐끫뢾 끫롸끫뢶끫롸끫뢐끫뢾끫룂 끫롰끫롰끫뢐끫롸. = 끫뢐끫뢐끫뢐끫뢐끫뢾 끫롸끫뢶끫롸끫뢐끫뢾끫룂 끫롰끫롰끫뢐끫롸. −(끫뢌끫뢸 끫룊 끫롾끫뢐끫뢶끫롾끫룂ℎ)

Lower Invert = 90.8312 m

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4. LINE 2–OUTFALL CALCULATIONS

Solve for the Time of Concentration & Rainfall Intensity :

𝑢끫룀끫뢐 끫룂끫뢬2 = 7 mins 끫룂끫뢬1 + 끫룂𝑐1−2 = 7.42646 mins *Use the highest
time
끫룂끫뢬4 + 끫룂𝑐4−2 =6.38039 mins
T = 10 years
i = 209.028 mm/hr

Solve for the design runoff, Q, using rational method;

2
C= 0.83 Atotal = 0.00107 km
L= 30 m

끫뢈 = 0.278끫롬끫뢬끫롨

0.0518 m /s + Q 4-2 + Q 1-2

3
Q=
0.0518 m /s + 0.0766 m /s +
3 3
Q= 0.0707 m3 /s
3
Q = 0.19914 m /s

Assuming Pipe is flowing full, for D ;

n = 0.013 So = 0.00458

1 끫븖끫롮 2 끫롮 2 1
끫뢈 = ( )( )3 끫뢌 2
끫뢶 4 4
D = 0.45537 m
D = 455.365 mm
*Computed Pipe Size
D = 525 mm
*Commercial Pipe Size
r = 0.2625 m

Checking if the Pipe is more than or less than half full

Areafull = 0.16286 m

Commercial Pipe Area,

Area = 0.21648 m

Area Percentage = Areafull/Commercial Pipe Area

Area Percentage = 75.23 %

Since it is more than 50% the pipe is flowing more than half full

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Calculating the angle beta (β)

From manning's equation,

5
끫뢈끫뢶 끫롨 3
1 = 2
끫뢌 2 끫뢆 3
5
끫롨 3
0.038254 = 2
끫뢆 3
Since,
끫븖끫뢾 2 끫뷺 1 2
끫롨 = + 끫뢾 sin(360 − 끫뷺)
360 2
끫븖끫뢾끫뷺
끫뢆 =
180 5
3
0.0006 (β) + 0.034 sin (360 - β)
0.038254 =
끫븖 0.26 끫뷺 2
( )3
180
Note : For complex equations, the use of calculator is needed to solve beta ( β ).

β = 204.77 *Beta is greater than 180

ϴ = 155.23
𝜃
끫뢢 = 끫뢾 − [끫뢾𝑐끫뢸끫룀 ] , if beta (β) is less than 180
2
𝜃
끫뢢 = 끫뢾 + [끫뢾𝑐끫뢸끫룀 ] , if beta (β) is greater than 180
2

Checking if its equal or not, through substitution.

0.0383 =0.0383

Solve for the actual velocity.

끫븖끫뢾2 끫뷺 1 2
끫롨 = + 끫뢾 sin(360 − 끫뷺)
360 2
A= 0.14 m2
끫븖끫뢾끫뷺
끫뢆 =
180
P = 0.93815 m
1 2 1
끫뢒 = 끫뢊 3 끫뢌 2
끫뢶
V= 1.448 m/s *Velocity is "> V min " and "< V max ".

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Calculate Depth of Flow.

𝜃
끫뢢 = 끫뢾 + [끫뢾𝑐끫뢸끫룀 ]
2

d= 0.3188 m

Solve Sewer Flow Time.

끫롾끫뢐끫뢶끫롾끫룂ℎ
끫룂 =
끫뢒끫뢐끫롰끫뢸𝑐끫뢬끫룂𝑦
30 = 20.72 seconds
t=
1.45
= 0.3454 min

Invert Elevation

끫뢐끫뢐끫뢐끫뢐끫뢾 끫롸끫뢶끫롸끫뢐끫뢾끫룂 끫롰끫롰끫뢐끫롸. = 끫뢀끫뢀끫뢶ℎ끫뢸끫롰끫뢐 끫롰끫롰끫뢐끫롸. −끫롮끫뢐끫뢐끫룂ℎ 끫뢸끫뢸 끫롬끫뢸끫롸끫뢐끫뢾 − 끫뢆끫뢬끫뢐끫뢐 끫롮끫뢬끫뢀.

MH 2 Elev. = 91.7812 m

Depth of Cover = 0.6 m

Upper Invert = 90.6562 m

끫롾끫뢸끫롾끫뢐끫뢾 끫롸끫뢶끫롸끫뢐끫뢾끫룂 끫롰끫롰끫뢐끫롸. = 끫뢐끫뢐끫뢐끫뢐끫뢾 끫롸끫뢶끫롸끫뢐끫뢾끫룂 끫롰끫롰끫뢐끫롸. −(끫뢌끫뢸 끫룊 끫롾끫뢐끫뢶끫롾끫룂ℎ)

Lower Invert = 90.5188 m

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5. LINE 5-OUTFALL CALCULATION

Solve for the time of concentration & Rainfall Intensity :

𝑢끫룀끫뢐 끫룂끫뢬5 = 끫룂𝑐5 = 7 mins

T = 10 years
i = 218.236 mm/hr

Solve for the design runoff, Q, using rational method;

2
C= 0.83 Atotal = 0.000441 km
L= 28 m
끫뢈 = 0.278끫롬끫뢬끫롨

Q = 0.022207 m3 /s

Assuming Pipe is flowing full, for D ;

n = 0.013 So = 0.00379

1 끫븖끫롮 2 끫롮 2 1
끫뢈 = ( )( )3 끫뢌 2
끫뢶 4 4

D = 0.207234 m
D = 207.2343 mm
*Computed Pipe Size
D = 300 mm
*Commercial Pipe Size
r= 0.15 m

Checking if the Pipe is more than or less than half full

Areafull = 0.03373 m

Commercial Pipe Area,

Area = 0.070686 m

Area Percentage = Areafull/Commercial Pipe Area

Area Percentage = 47.72 %

Since it is less than 50% the pipe is flowing less than half full

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Calculating the angle beta (β)

From manning's equation,

5
끫뢈끫뢶 끫롨 3
1 = 2
끫뢌 2 끫뢆 3
5
끫롨 3
0.004688 = 2
끫뢆 3

Since,
끫븖끫뢾 2 끫뷺 1 2
끫롨 = − 끫뢾 sin(끫뷺)
360 2
끫븖끫뢾끫뷺
끫뢆 =
180 5
3
0.000196 (β) - 0.01125 sin (β)
0.004688 =
끫븖 0.15 β 2
( 3 )
180
Note : For complex equations, the use of calculator is needed to solve beta ( β ).

β = 162.4537 *Beta is less than 180

β
끫뢢 = 끫뢾 − [끫뢾𝑐끫뢸끫룀 ] , if beta (β) is less than 180
2
끫뷺
끫뢢 = 끫뢾 + [끫뢾𝑐끫뢸끫룀 ] , if beta (β) is greater than 180
2

Checking if its equal or not, through substitution.

0.0047 = 0.0047

Solve for the actual velocity.

끫븖끫뢾2 끫뷺 1 2
끫롨 = + 끫뢾 sin(360 − 끫뷺)
360 2
2
A = 0.028506 m

끫븖끫뢾끫뷺
끫뢆 =
180
P = 0.425303 m
1 2 1
끫뢒 = 끫뢊 3 끫뢌 2
끫뢶
V = 0.781687 m/s *Velocity is "> V min " and "< V max ".

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Calculate Depth of Flow.

끫뷺
끫뢢 = 끫뢾 − [끫뢾𝑐끫뢸끫룀 ]
2
d= 0.1271 m

Solve Sewer Flow Time.

끫롾끫뢐끫뢶끫롾끫룂ℎ
끫룂 =
끫뢒끫뢐끫롰끫뢸𝑐끫뢬끫룂𝑦
t= 28 = 35.82 seconds
0.781687
= 0.597 min

Invert Elevation

끫뢐끫뢐끫뢐끫뢐끫뢾 끫롸끫뢶끫롸끫뢐끫뢾끫룂 끫롰끫롰끫뢐끫롸. = 끫뢀끫뢀끫뢶ℎ끫뢸끫롰끫뢐 끫롰끫롰끫뢐끫롸. −끫롮끫뢐끫뢐끫룂ℎ 끫뢸끫뢸 끫롬끫뢸끫롸끫뢐끫뢾 − 끫뢆끫뢬끫뢐끫뢐 끫롮끫뢬끫뢀.

MH 5 Elev. = 91.75 m

Depth of Cover = 0.6 m

Upper Invert = 90.85 m

끫롾끫뢸끫롾끫뢐끫뢾 끫롸끫뢶끫롸끫뢐끫뢾끫룂 끫롰끫롰끫뢐끫롸. = 끫뢐끫뢐끫뢐끫뢐끫뢾 끫롸끫뢶끫롸끫뢐끫뢾끫룂 끫롰끫롰끫뢐끫롸. −(끫뢌끫뢸 끫룊 끫롾끫뢐끫뢶끫롾끫룂ℎ)

Lower Invert = 90.7438 m

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6. OUTFALL DISCHARGE CALCULATION

Solve for the Time of Concentration & Rainfall Intensity :

𝑢끫룀끫뢐 끫룂끫뢬끫뢒𝑢𝑡𝑓𝑎𝑙𝑙 = 8 mins 끫룂끫뢬2 + 끫룂𝑐2−끫뢒 = 7.77186 mins

*Use the highest time
끫룂끫뢬5 + 끫룂𝑐5−끫뢒 = 7.597 mins
T = 10 years
i = 197.911 mm/hr

Solve for the design runoff, Q, using rational method;

2
C= 0.83 Atotal = 0.00086 km

끫뢈 = 0.278끫롬끫뢬끫롨

Q = 0.03932 m3 /s

Calculate Total Discharge for Outfall,

끫뢈끫롾끫뢸𝑡𝑎𝑙 = 끫뢈끫뢒𝑢𝑡𝑓𝑎𝑙𝑙 + 끫뢈2−끫뢒𝑢𝑡𝑓𝑎𝑙𝑙 + 끫뢈5−끫뢒𝑢𝑡𝑓𝑎𝑙𝑙

Q = 0.03932 +0.19914 +0.02221

3
Q = 0.26067 m /s

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