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بسم الله الرحمن الرحيم
Open Channel Flow)3- )الجريان في القنوات المفتوحةrd
Class Dr. Sataa A. Al-Bayati(09-10)
Introduction:
What is the difference between the flow in open-channel & closed pipe?
Open channel flow is flow of a liquid in a conduit in which upper surface of
liquid (free surface) is in contact with atmosphere.
Natural channels: rivers, & streams.
Manmade channels: irrigation canals, sewer lines, storm drains, & street
gutters.
What are the important parameters to be calculated in open-channel?
Fig.(1) Flow in an open-channel
2
Fig.(2) Uniform flow in an open-channel
Calculation of Q using Moody diagram:
The discharge can be calculated as follows,
oRSf
gAAVQ
8 ------- (1)
Where:
A = cross section area
ƒ = friction factor
R = hydraulic radius = A/P
P = wetted perimeter
So= channel slope
What is Moody Diagram Fig.(3)?
V2/2g
3
Fig.(3) Moody diagram: friction factors for commercial pipe (Featherstone,
R. E. & Nalluri, C., 1995)
4
Example (1):
Determine the discharge in a long rectangular concrete channel that is 5ft
wide, that has a slope of 0.002, & in which the water depth is 2ft. Use
K=5×10-3
ft, ν = 1.22 × 10-5
ft2/s.
Solution:
What are the relation between D & R?
Given: rectangular ch., b = 5ft, So = 0.002, y = 2ft, K = 5x10-3
ft, &
ν = 1.22×10-5
ft2/s.
Q =?
522
52
P
AR
= 1.11ft
as D = 4R
11.14
105
4
3
R
K s
= 1.13 x 10-3
Assume ƒ = 0.020
Eq. (1)
oRSf
gAAVQ
8
002.011.102.0
2.32852
= 53.5ft3/s
V = Q/A = 53.5 / 10
= 5.35ft/s
RVRe
4
51022.1
11.1435.5
eR
= 1.95 × 106
From Moody diagram, Fig. (3), ƒ = 0.02 OK
5
Stop & take Q = 53.5cfs.
Note: if ƒ is not equal repeat the calculation with the new ƒ.
*************
Calculation of Q using Manning Equation:
Manning Equation
What is Manning equation? Used for what?
2/13/249.1oSAR
nQ (EI) ----------- (2-a)
2/13/21oSAR
nQ (SI) ----------- (2-b)
Where:
n = Manning coefficient
Dividing the actual parameters of Eqs.(2) by the flowing full parameters we
get the following ( o or f = flowing full):
2/13/2
2/13/2
oo
o
o SnR
SnR
V
V
2/13/2
2/13/2
ooo
o
o SRnA
SnAR
Q
Q
From these two forms Fig.(4) is plotted.
Fig. (4) Flow characteristics of a circular section, Q/Qf vs. z=y/df.
V/Vf vs. z=y/df
6
Fig. (4) Flow characteristics of a circular section
For quick solution Eqs.(2) is arranged in Fig. (5).
Fig. (5) Chart for flow of water in pipes flowing full, (Q, D, n, V, & S).
Need to know 3-parameters → find the other 2-parameters.
8
Example (2):
Determine the discharge in a 3ft sewer pipe if the depth of flow is 1ft, &
pipe slope is 0.0019. Assume n = 0.012.
Solution:
Given: pipe ch., d = 3ft, y = 1ft, S = 0.0019, & n = 0.012.
Q =?
Use Fig. (5).
Draw a straight line through the points for n = 0.012 & S = 0.0019.
This line intersects match line.
Draw a line through intersection point & the 3ft diameter point.
Full discharge, Qf = 20mgd.
y/df = 1/3 = 0.333 → Fig. (4) → Q/Qf = 0.2
Q = 0.2Qf
= 0.2 × 20
= 4mgd.
*********************
Flow in Channel of Trapezoidal Cross Section)الجريان في القنوات شبه منحرف(
Fig. (6) Determining the normal depth.
A R2/3
/ b8/3
vs. y/b (trapezoidal)
A R2/3
/ do8/3
vs. y/do (circular)
Where:
b = bottom channel width,
do = pipe diameter.
10
Example (3):
Determine the normal depth for a trapezoidal channel with side slopes of 1
vertical to 2 horizontal, a bottom width of 8ft, discharge of 200cfs, channel
bottom slope of 1ft in 1000ft, & n = 0.012.
Solution:
Given: trapezoidal ch., 1:z = 1:2, b =8ft,Q =200cfs, So= 1/1000, &n = 0.012.
y =?
We need (A R2/3
)/ b8/3
, so we do the following steps:
2/13/249.1oSAR
nQ
2/1
3/2
49.1 oS
nQAR
Divide the two sides by b8/3
,
3/82/13/8
3/2
49.1 bS
nQ
b
AR
o
3/82/1 )8()001.0(49.1
012.0200
= 0.199
Go to Fig. (6) with (A R2/3
)/ b8/3
= 0.199 → y/b = 0.33
y = 0.33 × b
= 0.33 × 8
= 2.64ft.
********************
Design of Erodible Channels)القنوات القابلة للتآكل(
They are channels constructed in erodible material e.g. soil or fine gravel.
These channels may be eroded. Methods for designing erodible channels are:
1. Permissible velocity method.
2. Tractive force method.
Permissible velocity method:
11
Table (1) Permissible side slopes for different materials
Material Side slopes
Rock Nearly vertical
Stiff clay or earth with concrete lining 1: ½ to 1:1
Firm soil 1:1
Loam sandy soil 1:2
Sandy loam 1:3
Table (2) Maximum permissible velocities for different materials, with n
values
Material V, ft/s n
Fine sand 1.50 0.020
Sandy loam 1.75 0.020
Silt loam 2.00 0.020
Firm loam 2.50 0.020
Stiff clay 3.75 0.025
Fine gravel 2.50 0.020
Coarse gravel 4.00 0.025
Example (4):
An unlined irrigation canal is to be constructed in a firm loam soil. The slope
is to be 0.0006, & it is to carry a water flow of 100cfs. Determine an
appropriate cross section for this canal.
Solution:
Given: Firm loam soil, So = 0.0006, & Q = 100cfs
y& b =?
From Table (1), firm soil → 1:1
From Table (2): max. permissible V = 2.5ft/s & n = 0.02
2/13/249.1oSR
nV
or
2/3
2/149.1
oS
VnR
12
2/3
2/1)0006.0(49.1
5.202.0
R
= 1.6ft
A = Q/V
= 100/2.5
= 40ft2
R = A/P
P = A/R
= 40/1.6 = 25ft
A = by + zy2 = by + y
2 = 40
_____ _
P = b + 2y√1 + z2 = b + 2√2 y = 25
Solve these two Eqs. simultaneously
b = 19.8ft & y = 1.85ft
**************************
Best Hydraulic Section)أفضل مقطع هايدروليكي(
Section factor = A R2/3
= A (A/P)2/3
For given n & So in Manning eq.
Q = (1/n) A R2/3
So1/2
= (1/n) A (A/P)2/3
So1/2
Q increase with A → Q α A
decrease with P, → Q α 1/P
For given A & shape e.g. rectangular, there will be a certain
y/b → A (A/P)2/3
= max. → Qmax
Such a ratio establishes the best hydraulic section.
Best hydraulic section: is channel proportion that yields a Pmin. & Qmax for a
given A.
Given A: y/b → Pmin.
13
Example (5):
Determine the best hydraulic section for a rectangular channel.
Solution:
A = by --- (1)
P = b + 2y --- (2)
Let A be constant, then from Eq.(1), b = A/y
Take Eq.(2),
P = b + 2y
= A/y + 2y
By differentiation,
022
y
A
dy
dp
A/y2 = 2
But A = by
Then, b/y = 2 → y = b/2
Therefore, best hydraulic section for a rectangular channel occurs when the
depth is one half the widths (½ square).
***********************
The best hydraulic section for;
Trapezoidal channel = 1/2 (hexagon)
Circular = 1/2 (circle)
Triangular = 1/2 (square)
Best hydraulic section → min. excavation → min. cost
Specific Energy)الطاقة النوعية(
Specific Energy = depth + velocity head
E = y + V2/2g
E1 = E2 = ---
2
2
2
22
1
2
122 gA
Qy
gA
Qy ----- (1)
Fig. (7) Relation of y vs. E
14
Fig. (7) Specific energy curve
Characteristics of Critical Flow)خواص الجريان الحرج(
Critical flow occurs at E for a given Q
13
2
c
c
gA
TQ --- (2)
Multiply by Ac & divided by Tc ,
2
2
cc
c
gA
Q
T
A --- (3)
Q = VA,
g
V
T
A c
c
c
2
---- (4)
Where:
T = channel top width
Example (6):
Determine the critical depth in a trapezoidal channel for a discharge of
500cfs. The width of the channel bottom is 20ft, & the sides slope upward at
an angle of 45o.
15
Solution:
Given: Q =500cfs, b = 20ft, & 1:z = 1:1.
yc =?
Eq.(2), 13
2
c
c
gA
TQ
or
g
Q
T
A
c
c
23
= (500)2 / 32.2 = 7764ft
2
But the A & T of trapezoidal channel are,
A = y (b + y) = y (20 + y)
T = b + 2y = 20 + 2y
Therefore,
7764220
20323
y
yy
T
A
c
c
By iteration: choose different y & place in the left side of the equation to
satisfy the right side =7764.
yc = 2.57ft
*****************
For rectangular channel:
c
c
cT
Ay
&
2
2
2
2
cc y
q
A
Q
Where: T
Qq = discharge / unit width
Eq. (3), 2
2
cc
c
gA
Q
T
A
becomes,
16
3
2
g
qyc -----(5)
Eq. (4)
g
V
T
A c
c
c
2
yc = Vc2 /g
1c
c
gy
V Frc (critical Froud number) ----(6)
__
What is Froud number? Fr = V / √g y
Critical flow is unstable character of flow. Therefore, flow in ordinary canals
is subcritical. Critical flow location is useful in discharge measurement,
why?
Eq. (3)
2
2
cc
c
gA
Q
T
A
Rearrangement
2/1
2/3
T
A
g
Q = Zc
_
For each Q /√g & shape → one yc
Fig. (8) curves for determining yc
5.2
oBg
Q
vs. yc/Bo (trapezoidal)
5.2
oDg
Q
vs. yc/Do (circular)
18
Example (7):
Solve example (6) using Fig. (8).
Solution:
5.2bg
Q
=
5.2)20(2.32
500
= 0.0493
Fig. (8) with z = 1 → yc/Bo = 0.13
yc = 0.13 * 20
= 2.6ft
References:
- Chow, V. T., 1959, “ Open-channel Hydraulics” Japan.
- Roberson, J.A., et.al., “ Hydraulic Engineering”, 2nd Ed, John Wiley & Sons. Inc., New York.
- Featherstone, R. E. & Nalluri, C., 1995, "Civil Eng. Hydraulics", 3rd
Ed,
Blackwell Sc. Ltd.
19
H.W: 05-06
M E
Ex.1: wide = 5 + 0.1(no.) = 5.1ft depth = 2 + 0.1(no.) = 2.1ft
Ex.2: Dia. = 3 + 0.1(no.) = 3.1ft D = 6ft & y = 1 + 0.1 = 1.1ft
Ex.3 b = 8 + 0.2(no.) = 8.2ft Q = 200 + 5(no.) = 205cfs
Ex.4 stiff clay sandy loam
Q = 100 + 5(no.) = 105cfs Q = 100 – 2(no.) = 98cfs
Ex.6 Q = 500 + 10(no.) = 510cfs Q = 500 – 10(no.) = 490cfs
Ex.7
H.W: 06-07
M E
Ex.1: wide = 6 + 0.1(no.) = 6.1ft depth = 3 + 0.1(no.) = 3.1ft
Ex.2: D = 4 + 0.1(no.) = 4.1ft D = 6ft & y = 1.5 + 0.1 = 1.6ft
Ex.3 b = 9 + 0.2(no.) = 9.2ft Q = 220 + 5(no.) = 225cfs
Ex.4 Q = 90 - 5(no.) = 85cfs Q = 110 + 5(no.) = 115cfs
Ex.6 Q = 450 - 10(no.) = 440cfs Q = 550 + 10(no.) = 560cfs
Ex.7
H.W: 07-08
E M
Ex.1: width = 6 + 0.1(no.) = 6.1ft depth = 3 + 0.1(no.) = 3.1ft
Ex.2: D = 4 + 0.1(no.) = 4.1ft D = 6ft & y = 1.5 + 0.1 = 1.6ft
Recommended