Sponsored Links
This presentation is the property of its rightful owner.
1 / 12

# HGL PowerPoint PPT Presentation

Sewer Hydraulics. Gravity flow: full flow. Gravity flow: Partial flow. Pressure flow: full flow. L. h. HGL. HGL. HGL. S= HGL = slope of the sewer R =A p /P p. S= h/L = slope of the HGL R=D/4. S= HGL = slope of the sewer R = A f /P f = D/4. Manning equation.

### Download Presentation

HGL

An Image/Link below is provided (as is) to download presentation

Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author.While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server.

- - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - -

#### Presentation Transcript

Sewer Hydraulics

Gravity flow: full flow

Gravity flow: Partial flow

Pressure flow: full flow

L

h

HGL

HGL

HGL

S= HGL = slope of the sewer

R =Ap/Pp

S= h/L = slope of the HGL

R=D/4

S= HGL = slope of the sewer

R = Af/Pf = D/4

Manning equation

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:

R = Hydraulic radius (Area/ wetted parameter)

S = slope.

n = manning coefficient.

D = pipe diameter.

Q = flow rate.

Note:Equation 1 is used for calculating the velocity in pipes either flowing full or partially full. Equation 2 same for flow rate.

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:

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:

D

Ө

d

D

• 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

Ө

d

Example 1

Find 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

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 2

Find 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

D

d

Partial Flow Curves

d/D

d/D

Qp/Qf

Vp/Vf

Partial Flow Curves Q &V

d / D

1.1

2