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Lecture 2.-sewer-hydraulics1
1. L2. Sewer Hydraulics
The Islamic University of Gaza- Civil Engineering Department
Sanitary Engineering- ECIV 4325
Based on Dr. Fahid Rabah lecture notes
2. HGL
HGL
HGL
h
L
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
3. 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
2
1
SR
n
V = …………………………………………………. …...(1)
2
1
3
8
312.0
SD
n
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
21
S
f
R
nf
V = … … … … … … … … … … … … … … … … … … … … … … ..… ..(3)
2
1
3
21
SpR
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
4. 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:
−=
2
cos1
2
1 θ
D
d
……………………… (5)
−=
π
θθ
2360
Sin
Af
Ap
…………………… (6)
−=
πθ
θ
2
360
1
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
D
Ө
5. 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
1. Find the flow angle : ?
−=
2
cos1
2
1 θ
D
d
= 0.67
From this relation ? = 219.75o
2. Find Qp/Qƒ :
−=
πθ
θ
2
360
1
Sin
Rf
Rp
=
( )
−
75.2192
75.219360
1
π
Sin
= 1.167
−=
π
θθ
2360
Sin
Af
Ap
=
−
π2
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
D
Ө
6. 4. Find the diameter of the pipe (D):
2
1
3
8
312.0
SD
nf
Q = = ( )
2
1
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
21
S
f
R
nf
V = = ( )
2
1
3
2
004.0
4
497.0
013.0
1
=
f
V = 1.203 m/s
f
V
f
Q
f
A = =
203.1
2355.0
= 0.196 m2
(or
4
2D
f
A
π
= )
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
P
Q
=
789.0
186.0
= 0.2355 m3
/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.
7. 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
From partial flow curves:
• Start from the Y-axis with 67.0=⇒
D
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=⇒
f
Q
P
Q
• 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=⇒
f
V
P
V
.
•Calculate Qƒ :
78.0
P
Q
f
Q =⇒ = 238.0
78.0
186.0
= m3
/s
8. 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.
