L2. Sewer Hydraulics

Based on Dr. Fahid Rabah lecture notes (Sanitary Eng. Course)

The Islamic University of Gaza- Environmental Engineering Department

Networks Design and Pumping Stations

EENV 5315

hL

Gravity flow: full flow Gravity flow: Partial flow Pressure flow: full flow

S= HGL = slope of the

sewer

R = Af/Pf = D/4

S= HGL = slope of the

sewer

R =Ap/Pp

S= h/L = slope of the HGL

R=D/4

Sewer Hydraulics

Many formulas are used to solve the flow parameters in sewers were discussed in the

hydraulic course. The most used formula for sanitary sewers is Manning equation:

2

1

3

21

SRn

V …………………………………………………. …...(1)

2

1

3

8312.0

SDn

Q ………………………………………………...(2)

Note: Equation 1 is used for calculating

the velocity in pipes either flowing full or

partially full. Equation 2 same for flow

rate.

R = Hydraulic radius (Area/ wetted parameter)

S = slope.

n = manning coefficient.

D = pipe diameter.

Q = flow rate.

2

1

3

21S

fR

nfV …………………………………………………………..…..(3)

2

1

3

21S

pR

npV ………………………………………………………..…….(4)

Vƒ = velocity flowing full

VP = velocity flowing partially

Note: Equations 3 and 4 are the same as Equation 1, but they are written using the

subscript (ƒ) and (P), to indicate flowing full and partially full, respectively:

Manning equation

In sanitary sewers the flow is not constant; consequently the depth of flow is varying

as mentioned above. In this case it is difficult to find the hydraulic radius to apply

Manning’s equation. For partially full pipe the following relations are applied:

2cos1

2

1

D

d……………………… (5)

2360

Sin

Af

Ap…………………… (6)

2

3601

Sin

Rf

Rp……………………..(7)

3

2

f

p

f

p

R

R

V

V…………………………… (8)

ff

pp

f

p

VA

VA

Q

Q………………………….(9)

d = partial flow depth.

R = Hydraulic radius (P = partial, ƒ = full)

Ө = flow angle in degrees.

Maximum capacity of the pipe when d/D = 0.95

A = Flow area.

Maximum velocity in the pipe occurs at d/D=0.81

Ө

d

D

d

DӨ

Example 1Find the diameter of the pipe required to carry a design flow of 0.186 m3/s when

flowing partially, d/D = 0.67, slope = 0.4%, n = 0.013 use the relations of partial

flow. Solution

1. Find the flow angle : ?

2cos1

2

1

D

d= 0.67

From this relation ? = 219.75o

2. Find Qp/Qƒ :

2

3601

Sin

Rf

Rp=

75.2192

75.2193601

Sin= 1.167

2360

Sin

Af

Ap=

2

75.219

360

75.219 Sin= 0.712

3

2

f

p

f

p

R

R

V

V= (1.167)

2/3 = 1.1088

ff

pp

f

p

VA

VA

Q

Q= 0.712*1.1088 = 0.789

d

DӨ

4. Find the diameter of the pipe (D):

2

1

3

8312.0

SDnf

Q = 21

004.0

8

013.0

312.0 3D = 0.2355

D = (0.15517)3/8

= 0.497 ? 0.50m ? 20" (design pipe diameter)

5. Find the partial flow velocity (VP):

2

1

3

21S

fR

nfV =

2

1

3

2

004.04

497.0

013.0

1

fV = 1.203 m/s

fV

fQ

fA =

203.1

2355.0 = 0.196 m

2 (or

4

2Df

A

)

Ap = Aƒ *0.712 = 0.196* 0.712 = 0.139 m2

Vp = Vƒ *1.109 = 1.203*1.109= 1.33 m/s

3. Calculate Qƒ:

Qƒ = 789.0

PQ

= 789.0

186.0= 0.2355 m

3/s

It is noticed that it is

quite long procedure to

go through the above

calculations for each

pipe in the system of

large numbers of pipes.

The alternative

procedure is to use the

nomographs of

Mannings equation and

the partial flow curves.

Example 2Find the diameter of the pipe required to carry a design flow of 0.186 m3/s when

flowing partially, d/D = 0.67, slope = 0.4%, n = 0.013 using the nomographs and

partial flow curves .Solution

From partial flow curves:

Start from the Y-axis with 67.0D

d, and draw a

horizontal line until you intersect the Q curve (for n = constant,

the dashed line), then draw a vertical line to intersect the X-

axis at .78.0

fQ

PQ

Extend the horizontal line until it intersects the velocity curve,

then draw a vertical line to intersect the X-axis (for n =

constant, the dashed line) at 12.1

fV

PV

.

Calculate Qƒ : 78.0P

Q

fQ = 238.0

78.0

186.0 m

3/s

Use the nomographs for pipes flowing full to find D:

Locate the slope ( 0.004) on the “S” axis.

Locate the manning coefficient “n” ( 0.013) on the “n” axis.

Draw a line connecting “S” and “n” and extended it until it intersects

the Turning Line.

Locate the Qƒ ( 0.238 m3/s) on the “Q” axis.

Draw a line connecting “Qƒ” and the point of intersection on the

Turning Line and find the diameter “D” by reading the value that this

line intersect the D axis at. D = 500 mm = 20”.

find the Velocity “Vƒ” by reading the value that this line intersect the V

axis at. Vƒ = 1.2 m/s.

Calculate Vp = Vƒ * 1.12 = 1.2*1.12= 1.34 m/s.

Partial Flow Curves

d/Dd/D

d

D

Vp/VfQp/Qf

21.1

Partial Flow Curves Q &V

d / D