OPEN CHANNEL FLOW a c e , • Open-channel flow must have a f r e e s u r f ac whereas pipe flow has none. A f r e e s u r f a c e is s u b j ec e c t t o at a t m o s p h e r ic i c p r e s s u r e .
• The flow therefore always takes place due to the fact that the canal is at a slope and a component of the weight of the liquid causes the foreward motion . • Physical conditions in open-channels vary much more than in pipes. • Flow conditions in open channels are complicated by the position of the free surface which will change with time and space.
Comparison between open channel and pipe flow Pipe flow
Open channel flow
Flow driven by
Pressure work
Gravity (potential energy)
Flow cross section
Known ,fixed
Unknown in advance because flow depth is unknown
Characteristics flow parameters
Velocity deduced from continuity
Flow depth deduced from continuity and momentum equations
Specific boundary conditions
Atmospheric pressure at the free surface
Properties of Open Channels • Artificial channels -These are man made e.g. irrigation canals, navigation canals, spillways ,drainage ditches, culverts . • Usually constructed in regular shape throughout, and has reasonably defined roughness. • Constructed of concrete ,steel or earth. • Natural channelschannels- not regula regularr in shape shape and and usually constructed of earth and the surface roughness coefficient varies. • Affected by erosion erosion and deposition of sediments. sediments.
Flow Classification • The flow in an open channel is classified according to the change in the depth of flow with respect to space and time. • Uniform - if the the depth depth of of flow flow remains remains the same same at every section of the channel. • Non-uniform flow- the depth changes along the length of the channel. • Steady uniform flow -Depth is constant both with time and distance. • Steady non uniform flow -Depth varies with distance, but not with time. • Unsteady flow -Depth varies with both time and distance.
Gradually varied and Rapidly Varied Flow •
•
When the change in the depth occurs abruptly over a short distance, it is a Rapidly Varied Flow (RVF). When the change in depth occurs gradually it is a Gradually Varied Flow (GVF.
Geometric properties of Open Channels • To determine the flow in a canal the shape and size of the canal are important: – Rectangular. – Triangular. – Trapezoidal. – Semi Circular.
• To compare different canal sections, the hydraulic radius and the hydraulic depth are used.
Geometric Properties of Open Channels • Depth (y)-the vertical distance of the lowest point of a channel section from the free surface. • Stage (h)- the vertical vertical distan distance ce of of the the free free surface from an arbitrary datum. • Area (A)- the cross cross section sectional al area of flow flow normal normal to the direction of flow. • Wetted perimeter (P)perimeter (P)- the leng length th of the the wett wetted ed surface measured normal to the direction of flow. • Surface width (B) -the width of the channel section at the free surface. • Hydraulic radius (R)-the ratio of the wetted area to the wetted perimeter (A/P). • Hydraulic mean depth (Dm)-the ratio of the area to the surface width (A/B)
•
•
• •
Hydraulic Sections To minimize the quantity of material required max hydraulic radius should be used. Semi-circle is the most effective but difficult to construct freshly placed concrete tend to slide down the sides Trapezoidal channels are efficient especially for greater discharges. discharges. Rectangular channels can be used where space is not limited.
Rectangular section
Trapezoidal Section
Channel free board • The channel free board is necessary to prevent overtopping due to waves or variations in water level. • Free board variation of 10 -30% of the normal flow depth is acceptable.
Channel depth (m)
Free board height (mm)
<0,25
50
0,25-0,4
75
0,4-0,65
100
0,65-0,9
125
>0,9
150
Fundamental equations • Equations which describe the flow of a fluid are derived from three fundamental laws of physics • Conservation of matter (mass) • Conservation of energy • Conservation of momentum
Laminar and Turbulent flow • For channels the Reynolds number is RV RV Re
• Where:
V-velocity of flow µ-dynamic viscosity ρ-fluid density R-hydraulic radius • Re(channel) <500 flow is laminar • Re(channel)>1000 flow is turbulent
Elements of Channel Section • The channel bottom should have a slope in the direction of flow
The Chezy’s formula • Chezy's formula can be derived by equating the propulsive force due to the weight of the water in the direction of flow with the retarding shear force at the channel boundary.
The Chezy equation • Shear force is proportional to velocity squared
o
2
k V
• Substitute into
V • Then
V
g
C RS RS o
• C-Chezy coefficient
k
RS RS o
The Manning’s formula • In terms of velocity (V) 2
V
1
R 3 S 0 2 n
• In terms of discharge
Q • • • •
1 n
2
1
AR A R 3 S o 2
R-hydraulic radius So- channe channell bed bed slope slope n-Manning’s n-Manning’s roughness coefficient A-Cross sectional sectional area
Typical Values of Manning’s n Channel Type
Surface material & alignment
Manning’s n
River
Earth, straight
0.02-0.025
Earth meandering
0.03-0.05
Gravel(75150mm) straight
0.03-0.04
Earth, straight
0.018-0.025
Rock, straight
0.025-0.045
Lined canals
Concrete
0.012-0.017
Models
Mortar
0.011-0.013
Perspex
0.009
Unlined canals
Examples • A trapezoidal concrete lined lined channel with uniform flow of water has a normal depth of 2 m. The base width is 5 m has equal side slopes at 1:2.The channel bed slope is 0,001 and Manning’s n =0,015. Dynamic viscosity of water is 1,14 x 10 -3 kg/m.s. Calculate the (a) Discharge (b) Mean velocity (c) Reynolds number • If the discharge in the channel given above is 30 m3/s.Find the normal depth of flow.
Example •
Determine the hydraulic radius of a trapezoidal canal with the following dimensions. Bottom width 2,5 m, sloping sides 2,4m at 45 o to the horizontal with a flow depth of 1,5 m.
Example • Determine the flow depth and average flow velocities for a concrete channel with slope 1:2 500 changing to 1: 3 000. Assume a Manning’s n=0,0017.The channel is rectangular with base width of 3 m and must be able to handle a flow rate of 2 m3/s.
Compound Channels Example • If the channel above was to be designed for flooding it might have a trapezoidal channel and then flood plains so that it carries more discharge. • During flooding the water level in the channel given above exceeds the bank full-level of 2,5 m. The flood banks are 10 m wide and are grassed with side slopes of 1:3. The estimated Manning’s n for the flood banks is 0,035. Estimate the discharge for a maximum flood level of 4 m.
Example on Erodible Channels • Determine the floor width (b) and safe flow depth (y) of a trapezoidal spillway with a floor slope of 0.0016 and a flow rate of 7.750 m 3/h.The spillway is built in sandy loam soils. • The n value of a trapezoidal channel in a sandy soil weakens from 0.025 to 0.3 as a result of bad maintenance (no weed control). The channel was initially designed designed to handle handle a flow rate of 2m3/s.Channel slope is 1:2 500. Determine the reduction in flow rate with the new n-value.
Open Channel Flow SPECIFIC ENERGY • If water flows in a canal at a depth y and average velocity v, the specific energy is E y
V
2
2 g
• This is the energy of the liquid in relation to the bottom of a canal
Specific energy • E1= E2= E3 • If the canal width remains constant y1
V 1
2
2 g
y 2
V 2
2
2 g
y 3
V 2
2
2 g
• Q = AV = By1V1= By2V2 and
V 1
Q By B y1
V 2
Q By B y2
Specific energy • If q = Q/B is the flow rate per unit width then V 1
q
y1
V 2
y1
q2 2 gy1
y 3
2
y 2
Ey Ey 2
q y 2
q2 2 gy 2 q
2
2 g
0
Specific Energy • For a certain flow there are two depths at which the water can flow • Consider a canal with a sluice gate • Before the gate water flows slowly with a large depth y 1 (Sub critical) and after sluice flow is fast with a small depth y 2 (supercritical) (supercritical) but the specific energy is the same
Specific Energy • Energy before sluice is stored as potential and after sluice is mainly Kinetic • Critical depth can be determined by differentiating
E y
V
2
2 g
Froude Number • A dimensionless dimensionless ratio ratio of the inertia forces forces to the gravitational gravitational force Fr F r
V gyc
• Fr determine the velocity of the surface wave • Fr>1 supercritical • Fr=1 critical • Fr<1 subcritical.
The Hydraulic Jump • It is the change from shooting flow to tranquil flow which occurs abruptly. • This is due to change in slope slope from being being very steep (high velocity) to gentle which destroys the high velocity and water moves slower at a greater depth. • It occurs when a supercrtical flow meets a subcritical flow. • The resulting flow transition is rapid and involves large energy loss due to turbulence.
The Hydraulic Jump
The Hydraulic Jump • The depth of flow before the jump h1
h2
2
( 1 8Fr 1
2
1)
• Depth of flow after the jump
h2
h1
( 1 8Fr 1
2
1)
2 • Head or Energy loss due to the jump E
(h2
h1 ) 3
4h1 h2
Head losses in a hydraulic Jump • The loss of mechanical energy that takes place in a hydraulic jump may be readily determined from the energy equation.
Velocity after a Hydraulic Jump
Length of a Hydraulic Jump
Example • Water flows in a nearly horizontal canal at a velocity of 17m/s and a depth of 300mm. • (a) Will it be possible for a hydraulic jump to occur? • (b) what will be the length and the velocity be just after the jump? jump? • (c )What percentage of the initial power in the stream remains after the jump?
Solution
Solution
