Babylon University
College of Engineering
Enviromential Engineering Department
Subject:Water Resourses Engineering
Dr: Naba Shakir Hadi
The Best Hydraulic Section:
It is known that the coneyance of achannel section increase with increase in the
hydraulic radius or with decrease in the wetted perimeter.
Section
R=
*For Rectangular Section:
P=B+2y
y
B
A =2
By=2
B=2
R=
R=
For rectangular section
*For Trapezoidal Section:
P=b+2y
+2
+2
+2
y
= 0]
=0
2:1
b
A = by + zy2
Or = ( b + zy ) y
P = b + 2y
2
=0
2
=0
2
2
P = b + 2y
×2
×2
2
+2zy = 0
2
4z2 = 1 + z2
z2 =
4=
Z=
θ=60
2
。
=0
R=
R=
=
R=
Notes:
When the circular section used:
P= 2 π r
A= π r2
R=
=by
Example:
A trapezoidal channel carrying 11.34 m3 /sec is built with non erodible bed
having aslope of 0.0016 and n = 0.025 proportion the section dimension by using best
hydraulic section.
Solution:
Q=
zy
b
zy
R=
A = by + zy2
y
1
z
b
11.34 =
Y = 2m
B = 2.31m
Z=
*Control of Flow:
The control of flow in an open channel is defined as the estiblashment of
adefinitive flow condition in the channel or a definitive relation ship between the
stage and discharge of the flow. When the control of flow is achieved at a certain
section of the channel this section is called a control section.
Solution:
Stage
“rating curve”
Depth
discharge
discharge
alwayes stage above sea surface = 220m or 230m
At the critical state of flow a definitive stage – discharge relation ship as shown in
fiqure:
1)
subcritical
flow(y1 yc)
y1 yc y2
y1
yc
yc
Se Sc
mild
y2
steep
Se Sc
2)
yc
S0
Sc “Steep flow”
3) Free Flow
yn
yc
mild
yn
S0
yc
0.7yc
Sc
4yc
yc
yn
yn yc
S0 Sc “Steep flow”
super critical flow
yc y2
4)
y1
yc
y2
y3
yc = control flow
5)
depth
y1
y
y1
y
y2
yc
y2
z
“hump”
mini
z for yc
Example:
A rectangular channel B= 1.5m , Q= 900L/s, the depth of flow before the hump
is 1m and z=200mm , compute the depth of flow above the hump and calculate the
required z cause the critical flow above it.
Solution:
v2/2g
v2/2g
y1=1m
y
z=200m
“hump”
V1 = Q / A1= 0.9 m3/s / (1.5×1) = 0.6 m/s
V= Q / (1.5 × y) = 0.9 / 1.5 y
z
y=0.765m
=0.3323m
Vc = Q / Ac = 0.9 / ( 1.5 × 0.332 ) = 1.807 m/s
1 = ( 0.6 )2 / 2g = 0.332 + ( 1.807 )2 / 2g + z
z = 0.52 m
*Non uniform flow: (varied flow)
For design of open channel and analysis of their performance the engineer must be
able to predict forms and positions of water surface profiles of varied flow andacquire
some facility in their calculation.
horizontal
v2/2g
Se
hf=Se dx
v2/2g+d(v2/2g)
y
S0
y+dy
S0dx
Horizontal (dx)
surface profile for varied flow
)
in non uniform
hydraulic depth
T = Top surface width
Depth decrease with distance
Aproching
To channel bottom
Depth increace with distance
Direction to up
Driving from channel bottom
“general form”
,
E.L
Se
hf
y1
S0
Z1
y2
constant
Example:
A flow rate 10 m3 /s occure in a rectangular channel n= 0.,013 , 6m in wide ,
slope =0.0001, at point in this channel the depth is 1.5m. how far from this point will
the depth be 1.65m.
Solution:
y
1.50
A
=B×y
9
1.55
1.60
1.65
9.3
9.6
9.9
P
Rave
Vave
V2/2g
(m)
2
9
9.1
9.2
9.3
1
1.011
1.022
1.0325
1.043
1.0538
1.0645
1.111
1.093
1.075
1.0585
1.042
1.026
1.01
0.0629
0.0589
0.0553
0.05199
(E=y+v /2g)
1.5629
0.046
1.6089
0.0464
1.6553
0.04664
1.7099
465
570
709
Also we find
Example:
When the depth of water just upstream from a friction less broad-crested weir of
0.6m height is 0.9m , what is flow rate per meter of crest length can be expected.
/2g
/2g
y1=0.9m
0.6m
Q = 0.28 m3/ s.m