dymanics of open channel flow
TRANSCRIPT
Dynamics of Fluid Flow in Open Channel
Flow: Concept of open channel, Chezy’s and
Manning’s equations for open channel and their
application.
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Fluid Mechanics
Dr. Mohsin Siddique
Assistant Professor
Outcome of Today’s Lecture
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� After completing this lecture…
� The students should be able to:
� Understand the concepts and basic equations used in open channel flow
� Determine the velocity and discharge using Chezy’s and Manning’s equation
� Understand the concept of most economical sections
Open Channel Flow
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� An open channel is the one in which stream is not complete enclosed by solid boundaries and therefore has a free surface subjected only to atmosphere pressure.
� The flow in such channels is not caused by some external head, but rather only by gravitational component along the slope of channel. Thus open channel flow is also referred to as free surface flow or gravity flow.
� Examples of open channel are
� Rivers, canals, streams, & sewerage system etc
Open Channel Flow
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Thal Canal Indus river
Comparison between open channel flow and
pipe flow
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Aspect Open Channel Pipe flow
Cause of flow Gravity force (provided by sloping bottom)
Pipes run full and flow takes place under hydraulic pressure.
Cross-sectional shape
Open channels may have any shape, e.g., triangular, rectangular, trapezoidal, parabolic or circular etc
Pipes are generally round in cross-section which is uniform along length
Surface roughness
Varies with depth of flow Varies with type of pipe material
Piezometrichead
(z+h), where h is depth of channel
(z+P/γ) where P is the pressure in pipe
Velocity distribution
Maximum velocity occurs at a little distance below the water surface. The shape of the velocity profile is dependent on the channel roughness.
The velocity distribution is symmetrical about the pipe axis. Maximum velocity occurs at the pipe center and velocity at pipe walls reduced to zero.
Types of Channels
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� Natural Channels: It is one with irregular sections of varying shapes, developing in natural way. .e.g., rivers, streams etc
� Artificial Channels: It is the one built artificially for carrying water for various purposes. e.g., canals,
� Open Channel: A channel without any cover at the top. e.g., canals, rivers streams etc
� Covered Channels: A channel having cover at the top. e.g., partially filled conduits carrying water
� Prismatic Channels: A channel with constant bed slope and cross-section along its length.
Types of flow in open channels
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� Steady and unsteady flow
� Uniform and non-uniform flow
� Laminar and Turbulent flow
� Subcritical, critical and supercritical flow
Same definition with pipe flows
� Laminar and Turbulent flow: For open channels, it is defined with Reynolds No. as;
νh
e
VRR =
ννh
e
VRVDR
4==
Remember in pipe flows
Therefore,For laminar flow: Re <= 500For transitional flow: 500 <Re< 1000For Turbulent flow: Re >= 1000
For laminar flow: Re <= 2000
Types of flow in open channels
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� Subcritical, Critical and Supercritical Flow. These are classified with Froude number.
� Froude No. (Fr). It is ratio of inertial force to gravitational force of flowing fluid. Mathematically, Froude no. is
If Fr. < 1, Flow is subcritical flowFr. = 1, Flow is critical flowFr. > 1, Flow is supercritical flow
gh
VFr =
Where, V is average velocity of flow, h is depth of flow and g is gravitational acceleration
Definitions
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� Depth of Flow: It is the vertical distance of the lowest point of a channel section(bed of the channel) from the free surface.
� Depth of Flow Section: It is depth of flow normal to bed of the channel.
� Top Width: It is the width of channel section at the free surface.
� Wetted Area: It is the cross-sectional area of the flow section of channel.
� Wetted Perimeter: It is the length of channel boundary in contact with the flowing water at any section.
� Hydraulic Radius: It is ratio of cross-sectional area of flow to wetted perimeter.
Open channel formulae for uniform flow
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� For uniform flow in open channels, following formulae are widely used
oRSCV =
2/13/21oSR
nV =
Here, V=Average flow velocityR=Hydraulic radiusSo=Channel bed slope
C= Chezy’s constant
n= Manning’s Roughness coefficient
1. Chezy’s Formula: Antoine de Chezy (1718-1798), a French bridge and hydraulic expert, proposed his formula in 1775.
2. Manning’s Formula: Rober Manning (An Irish engineer) proposed the following relation for Chezy’s coefficient C
( ) 6/1/1 RnC =
According to which Chezy’s equation can be written as
Derivation of Chezy’s formula
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� In uniform flow the cross-sectional through which flow occurs is constant along the channel and so also is the velocity. Thus y1=y2=yo and V1=V2 =V and the channel bed, water surface and energy line are parallel to one another.
According to force balance along the direction of flow; we can write,
PLALFF oτθγ =+− sin21
F1= Pressure force at section 1
F2= Pressure force at section 2
W= Weight of fluid between section 1 and 2=So= slope of channel
θ= Inclination of channel with horizontal line
τo= shearing stress
P= Wetted perimeter
L= length between sections
V= Avg. Flow velocityyo= depth of flow
ALγ
Derivation of Chezy’s formula
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θγθγθγ
τ sinsinsin
RP
A
PL
ALo ===
x
zzSo
∆
−= 21
( ) ( )x
yzyzSw
∆
+−+= 2211
( ) ( )
x
hS
x
gvyzgvyzS
L
∆=
∆
++−++=
2/2/ 222111
θsin=≈= SSS wo
For channels with So<0.1, we can safely assume that
oo RSγτ =
Therefore;
Derivation of Chezy’s formula
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τo (shearing stress) can also be expressed as (already discussed)
2
2V
C fo ρτ =
Comparing both equations of τo we get;
o
foo
f
of
RSCV
fCRSf
gRS
C
gV
RSV
C
=
===
=
4/82
2
2
Q
γρ
Where C is Chezy’s Constant whose value depend upon the type of channel surface
f
gC
8=Q
Relation b/w f and C
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� As f and C are related, the same consideration that are present for determination of friction factor, f, for pipe flows also applies here.
4/82
fCf
g
C
gC f
f
=== Q
Empirical Relations for Chezy’s Constant, C
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� Although Chezy’s equation is quite simple, the selection of a correct value of C is rather difficult. Some of the important formulae developed for Chezy’s Constant C are;
� 1. Bazin Formula: A French hydraulic engineer H. Bazin (1897) proposed the following empirical formula for C
RKC
/181
6.157
+=
R= Hydraulic RadiusK=Bazin Constant
The value of K depends upon the type of channel surface
Empirical Relations for Chezy’s Constant, C
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� 2. Kutter’s Formula: Two Swiss engineers Ganguillet and Kutterproposed following formula for determination of C
R= Hydraulic Radiusn=Manning’s roughness coefficient
� 3. Manning’s Formula:Rober Manning (An Irish engineer) proposed the following relation for Chezy’s coefficient C
2/13/21oSR
nV =
n= Manning’s Roughness coefficient
( ) 6/1/1 RnC =
The values of n depends upon nature of channel surface
BG units SI units
Empirical Relations for Chezy’s Constant, C
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Relation b/w f and n
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� Since
� Also
� It mean n and f can also be related with each other.
� Hence
4/82
fCf
g
C
gC f
f
=== Q
( ) 6/1/1 RnC =
g
fRn
8486.1 6/1=
g
fRn
8
6/1= SI
BG
Chezy’s and Manning’s Equations in SI and BG System
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� Chezy’s Equation � Manning’s Equation
2/13/21oSR
nV =
( ) 2/13/23 1/ oSAR
nsmQ =
( ) 2/13/2486.1oSAR
ncfsQ =
SI
BG
oRSCV =
oRSCAQ =
Value of C is determine from respective BG or SI Kutter’sformula.
C= Chezy’s Constant
A= Cross-sectional area of flow A= Cross-sectional area of flow
Problem-1
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� Water is flowing in a 2-m-wide rectangular, brick channel (n=0.016) at a depth of 120 cm. The bed slope is 0.0012. Estimate the flow rate using the Manning’s equation.
� Solution: First, calculate the hydraulic radius
� Manning’s equation (for SI units) provides
Problem-2
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� Compute the flow rate for a depth of 2, 4, 6 and 8ft.
( ) 2/13/2486.1oSAR
ncfsQ =
For BG units
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� Let’s consider a trapezoidal channel having bottom width, b, depth of flow, d, and side slope, S.
Trapezoidal section
h
b
s
Sh Sh
1
( ) 2222
2
22
sec
hShShA/hhShbPimeterWetted Per
ShbhAa of flowtional areCross-
++−=++==
+==
Shh
Ab
ShbhA
−=
+= 2
b+2Sh
1Sh 2 +θ
Problem-3
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� Solution
( ) 2/13/23 1/ oSAR
nsmQ =
For SI units 4
y
Problem-4
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2/13/2486.1oSR
nV =
For BG units
6
3
Problem-5
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( ) 2/13/23 1/ oSAR
nsmQ =
For SI units
321 QQQQ ++=
Problem-7
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( ) 2/13/23 1/ oSAR
nsmQ =
For SI units
Solution: (a)
Problem-7
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� (b) ( ) 2/13/23 1/ oSAR
nsmQ =For SI units
Most Economical Section
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� From Manning’s formula, we can write that
� For a given channel of slope, So, area of cross-section, A, and roughness, n, we can simplify above equation as
� It emphasis that discharge will be maximum, when Rh is maximum and for a given cross-section, Rh will be maximum if perimeter is minimum.
� Therefore, the most economical section (also called best section or most efficient section) is the one which gives maximum discharge for a given area of cross-section (say excavation for channel shape).
ohSARn
Q1
∝
PQ
P
AQRQ h
1∝⇒∝⇒∝
Most economical rectangular section
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� Let’s consider a rectangular channel as shown in figure in which width of channel is b and depth of flow is h.
b
h
2h/A2hbPPerimeter Wetted
bhAflow of area sectional-Cross
+=+==
==
h
� For most economical section, perimeter should be minimum. i.e.,
( ) ( ) 02h/A2hbP/dh
0P/dh
=+=+=
=
hdh
d
dh
dd
d
2/2
2
202
2
2
2
bhorhb
hbh
hAh
A
==
=
=⇒=+−
� Hence for most economical rectangular section, width is twice the depth of channel
Problem
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� Let’s consider a trapezoidal channel having bottom width, b, depth of flow is d, and side slope, S, as shown in figure
Most economical trapezoidal section
h
b
s
Sh Sh
1
( ) 2222
2
hSh2/AhSh2bPPerimeter Wetted
ShbhAflow of area sectional-Cross
++−=++==
+==
Shh
� For most economical section, perimeter should be minimum. i.e.,
( ) 0hSh2/A0 22 =++−⇒= Shhdh
d
dh
dP
Shh
b −=
+=
A
ShbhA 2
b+2Sh
1Sh 2 +θ
Most economical trapezoidal section
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� Hence for most economical trapezoidal section, top width is twice the length of one sloping side or half of
top width is equal to length of one sloping side
( ) 01S201S2h/A 2
2
2 =++−−⇒=++− Sh
AShh
dh
d
1Sh2Sh2b1Sh2
Sh2b
1S2Sh2b
1S2Shb
1S2Shb
1S2Shbh
1S2
22
2
22
2
2
22
2
+=+⇒+=+
+=+
+=++
⇒+=++
+=++
⇒+=+
h
h
Sh
hS
h
Sh
Sh
A
Most economical trapezoidal section
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� For given width, b, and depth, h, perimeter becomes only the function of side slope, S,. So if we estimate value of S that provide minimum P then we have;
( ) ( ) 01S2h/A0hhS2/A0 2222 =++−⇒=++−⇒= ShhdS
dShh
dS
d
dS
dP
( ) ( )
( ) ( ) SSh
ShSh
21Sh21S
01Sh2021S2
12h
22
2/1212/12
=+−⇒=+−
=++−⇒=
×++−
−−
Squaring both sides of equation, we get
3
1
3
141S 222 =⇒=⇒=+− SSS
If sloping sides make an angle θ with the horizontal than S=tanθ
oS 60
3
1tan =⇒== θθ
Thank you
� Questions….
� Feel free to contact:
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