Drain Pipe Calculations
Drain Pipe Calculations
Drain Pipe Calculations
Bimlesh Kumar
E-mail: bimk@iitg.ernet.in
Introduction
Sewer System
Fundamental Hydrology for Sewer Design
Fundamental Hydraulics for Sewer Design
Sewer System
Sewer
What is sewer?
Sewer is an artificial conduit or system of
conduits used to remove sewage and to
provide drainage.
Sewage
Sewage is the mainly liquid waste containing
some solids produced by humans which
t i ll consists
typically
i t off
-washing water
-faeces
-urine
-laundry
y waste
-other material from household and
industry
History
In the 20th century developed world,
Sewers are usually pipelines that begin with
connecting pipes from buildings to one or
more levels of larger underground horizontal
mains, which terminate at sewage treatment
f iliti
facilities.
Vertical pipes, called manhole, connect the
mains to the surface.
Sewers are generally gravity powered, though
pump may be used if necessary.
Sewer Systems
Q = 100pA1/2
Q = max flow (cfs)
p = Myers rating
A = area (sq.mi.)
Rational Method
Developed in 1800s in England as the first
dimensionall correct eq
dimensionally
equation.
ation
Used by 90% of engineers still today.
Equation assumes that Q is a function of rainfall
intensity applied uniformly over the watershed for
a duration D.
Equation also assumes that frequency of Q is
equal to frequency of rainfall intensity.
The proper rainfall duration is equal to the time of
concentration.
Rational Method
The equation is
Q = 1.008CIA
Where
Rational Method
What is needed?
1) Time of concentration
2) A set of rainfall intensity-duration-frequency
curve (IDF curve))
3) Drainage area size
4)) An estimate of the coefficient C
IDF Curve
Frequency
Return Period
Frequency
Rational Method
C is know as runoff coefficient and can be found
for the different land uses
If land
l d use iis mixed,
i d you can calculate
l l t a
composite C value as follows:
C = (CAAA+CBAB)/(AA+AB) or
C = (CiAi)/(Ai)
Where
Example 1
A storm drain system
y
consisting of two inlets and
pipe is to be designed using
rational method. A schematic
of the system is shown.
Determine the peak flow
rates to be used in sizing the
two pipes and inlets.
Rainfall intensity (in/hr) as a
function of t is:
Example 1
Size Inlet 1 and pipe 1:
Area A and B contribute
T k largest
Take
l
t tc
t = 12 min
i
A = 5+3 = 8 acre
C = (5*0.2+3*0.3)/8
(5*0 2 3*0 3)/8 = 0.24
0 24
I = 30/(12+5)0.7 = 4.13 in/hr
Q = CIA = 0.24*4.13*8 = 7.9 cfs
Example 1
Size Inlet 2:
Flow from area C contributes
T k tc
Take
t = 8 min
i
A = 4 acre
C = 0.4
04
I = 30/(8+5)0.7 = 4.98 in/hr
Q = CIA = 0.4*4.98*4 = 8.0 cfs
Example 1
Size p
pipe
p 2:
Flow from all areas
Take tc = 12+1 = 13 min
A = 5+4+3 = 12 acre
C = (5*0.2+4*0.4+3*0.3)/12
(5 0.2 4 0.4 3 0.3)/12
= 0.29
I = 30/(13+5)0.7 = 3.97 in/hr
Q = CIA = 0.29
0 29*3
3.97
97*12
12 = 13.8
13 8 cfs
Open Channel
TYPE OF OPEN CHANNEL
Most open channel flow occur in drainage
structures and facilities.
Various forms of open channel types such as
man-made ditch, natural stream, sewer etc.
In case flow sewer system, open channel flow
conditions exist,
exist even though the flow takes
place in a pipe.
Open Channel
Open Channel
v=
2/3
1/ 2
xS
n
AxR2 / 3 xS1/ 2
Q=
n
s = KQ2
Values of Mannings
Roughness Coefficient, n
Example 2
A concrete channel (n=0.013),
(n=0 013) rectangular in shape
and 1.25 m wide, must carry water at a uniform rate of
flow of 2000 L/s and a depth of 0.75 n.
Determine the required channel bottom slope for this
channel.
AxR2 / 3 xS1/ 2
Q=
n
Example 2
Solution
A = 1.25x0.75
1 25x0 75 = 0.938
0 938 m2
P = 0.75+1.25+0.75 = 2.75 m
R = A/P = 0.938/2.75 = 0.341 m
Therefore,
S = [(nQ)/(AR)2/3)]2
= [(
[(0.013x2.0)/(0.938x0.341)
) (
)2/3]
= 0.003
So
So,
So = 0.003
0 003
Example 3
A 500 mm asbestos cement sewer pipe (n=0.012)
(n=0 012) has
been installed with an invert slopes of 0.008.
Determine
D
t
i
the
th capacity
it off flow
fl
when
h
thi
this pipe
i
is
i
flowing half full. Assume the flow is uniform.
Solution
Example 4
For the trapezoidal channel shown in figure,
figure determine
the slope of the channel if the capacity of flow has to
be 4500 L/s. Assume uniform flow and n=0.025
Example 4
Solution
Pipe Flow
H
Hazen-Williams
Willi
FFormula
l
Q = 0.849CAR0.63 s0.54
Values of C in the Hazen-William Formula and of n in
the Manning Formula.
Example 5
A cast
cast-iron
iron water pipe,
pipe 400 mm in diameter
diameter, carries water at a
rate of 0.125 cms. Determine, by means of the Hazen-Williams
formula. The slope of the hydraulic gradient of this pipe and the
velocity
y of flow.
Solution
1. Graphical solution
Use the nomograph, line up the known value,
d=381, the actual diameter of a nominal 400-m
pipe and Q=0
pipe,
Q 0.125,
125 and find
S = 0.0045 m/m
v = 1.09 m/s
Example 5
Solution
2 Mathematical solution
2.
From table, for d = 400 and C = 100, find K = 0.232
and A = 0.114.
0 114 Hence
s = 0.232x0.1251.85 = 0.005 m/m
and v = Q/A = 0.125/0.114 = 1.09 m/s
Example 6
An asbestos cement water pipe (C=140) with a diameter of 300
mm has a slope of the hydraulic gradient of 0.0025
0 0025 m/m.
m/m
Determine, using the Hazen-Williams formula, the capacity of the
pipe and the velocity of flow.
Solution
1. Graphical solution
Example 6
Solution
2 Mathematical solution
2.
Example 7
A concrete pipe, 400 mm in diameter, carries water a rate of 125
L/s Determine by means of the Manning formula the slope of the
L/s.
hydraulic gradient of this pipe and the velocity of flow. Assume
n=0.013.
Solution
1. Graphical solution
Example 7
Solution
2 Mathematical solution
2.
Summary
Hazen Williams formula
Hazen-Williams
Manning formula
In fact
fact, these formulas:
are developed for use with the flow of water.
are applicable to incompressible fluids with a relative
density of 1.0.
Common practice,
p
,
Hazen-Williams formula exclusively for applications to
pipe flow.
Manning formula for gravity flow.
flow