17 design of earthen canals PDF

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Lecture 17

Design of Earthen Canals I. General • • • •

• • • • • • •

• •

•

Much of this information applies in general to both earthen and lined canals Attempt to balance cuts and fills to avoid waste material and or the need for “borrow pits” along the canal It is expensive to move earth long distances, and or to move it in large volumes Many large canals zigzag across the terrain to accommodate natural slopes; this makes A borrow pit the canal longer than it may need to be, but earthwork is less Canals may also follow the contours along hilly or mountainous terrain Of course, canal routing must also consider the location of water delivery points In hilly and mountainous terrain, canals generally follow contour gradients equal to the design bed slope of the canal Adjustments can be made by applying geometrical equations, but usually a lot of hand calculations and trial-and-error are required As previously discussed, it is generally best to follow the natural contour of the land such that the longitudinal bed slope is acceptable Most large- and medium-size irrigation canals have longitudinal slopes from 0.00005 to 0.001 m/m A typical design value is 0.000125 m/m, but in mountainous areas the slope may be as high as 0.001 m/m: elevation change is more than enough With larger bed slopes the problems of sedimentation can be lessened An earthen channel In the technical literature, it is possible to find many papers and articles on canal design, including application of mathematical optimization techniques (e.g. FAO Irrig & Drain Paper #44), some of which are many years old The design of new canals is not as predominant as it once was

II. Earthen Canal Design Criteria • •

Design cross sections are usually trapezoidal Field measurements of many older canals will also show that this is the range of averaged side slopes, even though they don’t appear to be trapezoidal in shape

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•

When canals are built on hillsides, a berm on the uphill side should be constructed to help prevent sloughing and landslides, which could block the canal and cause considerable damage if the canal is breached

III. Earth Canal Design: Velocity Limitations •

In designing earthen canals it is necessary to consider erodibility of the banks and bed -- this is an “empirical” exercise, and experience by the designer is valuable Below are four methods applied to the design of earthen channels The first three of these are entirely empirical All of these methods apply to open channels with erodible boundaries in alluvial soils carrying sediment in the water

• • •

1. 2. 3. 4.

Kennedy Formula Lacey Method Maximum Velocity Method Tractive-Force Method

1. Kennedy Formula • •

Originally developed by British on a canal system in Pakistan Previously in wide use, but not used very much today

(

Vo = C1 havg

)

C2

(1)

where Vo is the velocity (fps); and havg is the mean water depth (ft) • • • • •

The resulting velocity is supposed to be “just right”, so that neither erosion nor sediment deposition will occur in the channel The coefficient (C1) and exponent (C2) can be adjusted for specific conditions, preferably based on field measurements C1 is mostly a function of the characteristics of the earthen material in the channel C2 is dependent on the silt load of the water Below are values for the coefficient and exponent of the Kennedy formula:

Table 1. Calibration values for the Kennedy formula. C1 0.56 0.84 0.92 1.01 1.09 Gary P. Merkley

Material extremely fine soil fine, light sandy soil coarse, light sandy soil sandy, loamy silt coarse silt or hard silt debris 192

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C2 0.64 0.50

Sediment Load water containing very fine silt clear water Kennedy Formula (clear water: C2 = 0.50)

1.2 C1 C1 C1 C1 C1

Velocity, V o (m/s)

1.0

0.8

= = = = =

0.56 0.84 0.92 1.01 1.09

0.6

0.4

0.2

0.0 0.0

0.5

1.0

1.5

2.0

2.5

3.0

Depth, db (m)

Figure 1. Velocity values versus water depth for the Kennedy formula with clear water. 2. Lacey Method • •

Developed by G. Lacey in the early part of the 20th century based on data from India, Pakistan, Egypt and elsewhere Supports the “Lindley Regime Concept”, in which Lindley wrote: “when an artificial channel is used to convey silty water, both bed and banks scour or fill, changing depth, gradient and width, until a state of balance is attained at which the channel is said to be in regime”

• •

There are four relationships in the Lacey method All four must be satisfied to achieve “regime” conditions

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1. Velocity

V = 1.17 fR

(2)

Wp = 2.67 Q

(3)

R = 0.47 3 Q / f

(4)

2. Wetted Perimeter 3. Hydraulic Radius 4. Bed Slope

S = 0.000547

f2 / 3 Q1/ 6

(5)

where,

f = 1.76 dm

(6)

and, dm is the mean diameter of the bed and side slope materials (mm); V is the mean velocity over the cross-section (fps); Wp is the wetted perimeter (ft); R is the hydraulic radius (ft); S is the longitudinal bed slope (ft/ft); and Q is discharge (cfs) • • • •

The above relationships can be algebraically manipulated to derive other dependent relationships that may be convenient for some applications For example, solve for S in terms of discharge Or, solve for dm as a function of R and V Here are two variations of the equations: 12 V = 0.00124 d11/ / S (7) m

and, 1/12 (8) V = 0.881 Q1/ 6dm

•

A weakness in the above method is that it considers particle size, dm, but not cohesion & adhesion

Lacey General Slope Formula:

V=

1.346 0.75 R S Na

(9)

where Na is a roughness factor, defined as:

Na = 0.0225f 0.25 ≅ 0.9nR0.083 (10) where n is the Manning roughness factor Gary P. Merkley

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• • •

This is for uniform flow conditions Applies to both regime and non-regime conditions Appears similar to the Manning equation, but according to Lacey it is more representative of flow in alluvial channels

3. Maximum Velocity Method • • •

This method gives the maximum permissible mean velocity based on the type of bed material and silt load of the water It is basically a compilation of field data, experience, and judgment Does not consider the depth of flow, which is generally regarded as an important factor in determining velocity limits

Table 2. Maximum permissible velocities recommended by Fortier and Scobey Velocity (fps) Clear Water with water colloidal silt 1.5 2.5 1.75 2.5 2 3 2 3.5 2.5 3.5 2.5 3.5 3.75 5 3.75 5 6 6 2.5 5 4 6 5 5.5

Material Fine sand, colloidal Sandy loam, non-colloidal Silt loam, non-colloidal Alluvial silt, non-colloidal Firm loam soil Volcanic ash Stiff clay, highly colloidal Alluvial silt, colloidal Shales and hard "pans" Fine gravel Coarse gravel Cobble and shingle

Table 3. USBR data on permissible velocities for non-cohesive soils

Material Silt Fine sand Medium sand Coarse sand Fine gravel Medium gravel Coarse gravel Fine pebbles Medium pebbles Coarse pebbles Large pebbles

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Particle Mean velocity diameter (mm) (fps) 0.005-0.05 0.49 0.05-0.25 0.66 0.25 0.98 1.00-2.50 1.80 2.50-5.00 2.13 5.00 2.62 10.00-15.00 3.28 15.00-20.00 3.94 25.00 4.59 40.00-75.00 5.91 75.00-200.00 7.87-12.80

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IV. Introduction to the Tractive Force Method • • • • • • • • • • • • •

This method is to prevent scouring, not sediment deposition This is another design methodology for earthen channels, but it is not 100% empirical, unlike the previously discussed methods It is most applicable to the design of earthen channels with erodible boundaries (wetted perimeter) carrying clear water, and earthen channels in which the material forming the boundaries is much coarser than the transported sediment The tractive force is that which is exerted on soil particles on the wetted perimeter of an earthen channel by the water flowing in the channel The “tractive force” is actually a shear stress multiplied by an area upon which the stress acts A component of the force of gravity on the side slope material is added to the analysis, whereby gravity will tend to cause soil particles to roll or slide down toward the channel invert (bed, or bottom) The design methodology treats the bed of the channel separately from the side slopes The key criterion is whether the tractive + gravity forces are less than the “critical” tractive force of the materials along the wetted perimeter of the channel If this is true, the channel should not experience scouring (erosion) from the flow of water within Thus, the critical tractive force is the threshold value at which scouring would be expected to begin This earthen canal design approach is for the prevention of scouring, but not for the prevention of sediment deposition The design methodology is for trapezoidal or rectangular cross sections This methodology was developed by the USBR

V. Forces on Bed Particles •

The friction force (resisting particle movement) is:

Ws tanθ

(11)

where θ is the angle of repose of the bed material and Ws is the weight of a soil particle

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• • •

Use the angle of repose for wet (not dry) material θ will be larger for most wet materials Note that “tan θ” is the angle of repose represented as a slope Angle of Repose for Non-Cohesive Earthen Material

Angle of repose (degrees)

42 40

Very angular

38

Moderately angular Slightly angular

36

Slightly rounded Moderately rounded

34

Very rounded

32 30 28 26 24 22 20 1

10

100

Soil particle size (mm)

Figure 2. Angle of repose (degrees from horizontal), θ, for non-cohesive earthen materials (adapted from USBR Hyd Lab Report Hyd-366). •

The shear force on a bed particle is:

aTbed (12) where “a” is the effective particle area and Tbed (lbs/ft2 or N/m2) is the shear stress exerted on the particle by the flow of water in the channel •

When particle movement is impending on the channel bed, expressions 1 and 2 are equal, and:

W s tan θ = aTbed

(13)

or,

Tbed = BIE 5300/6300 Lectures

Ws tan θ a 197

(14)

Gary P. Merkley

VI. Forces on Side-Slope Particles •

The component of gravity down the side slope is:

Ws sin φ

(15)

where φ is the angle of the side slope, as defined in the figure below

Figure 3. Force components on a soil particle along the side slope of an earthen channel. •

If the inverse side slope is m, then:

⎛ 1⎞ φ = tan− 1 ⎜ ⎟ ⎝m⎠ •

(16)

The force on the side slope particles in the direction of water flow is:

aTside (17) where Tside is the shear stress (lbs/ft2 or N/m2) exerted on the side slope particle by the flow of water in the channel • •

Note: multiply lbs/ft2 by 47.9 to convert to N/m2 Combining Eqs. 15 & 17, the resultant force on the side slope particles is downward and toward the direction of water flow, with the following magnitude:

Ws2 sin2 φ + a2 Ts2ide •

•

(18)

The resistance to particle movement on the side slopes is due to the orthogonal component of Eq. 15, Wscosφ, as shown in the above figure, multiplied by the coefficient of friction, tan θ Thus, when particle movement is impending on the side slopes:

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2

2

2 2

Ws cos φ tan θ = Ws sin φ + a Tside •

Solving Eq. 19 for Tside:

Tside = •

(19)

Ws cos2 φ tan2 θ − sin2 φ a

(20)

Applying trigonometric identities and simplifying:

Tside =

Ws t an2 φ cos φ tan θ 1 − a tan2 θ

(21)

or,

Ws sin2 φ Tside = tan θ 1 − a sin2 θ

(22)

VII. Tractive Force Ratio • • •

As defined in Eq. 14, Tbed is the critical shear on bed particles As defined in Eqs. 20-22, Tside is the critical shear on side slope particles The tractive force ratio, K, is defined as:

K=

Tside Tbed

(23)

where Tside and Tbed are the critical (threshold) values defined in Eqs. 4 & 9-11 •

Then:

K = 1−

sin2 φ sin2 θ

= cos φ 1 −

tan2 φ

(24)

tan2 θ

VIII. Design Procedure • •

The design procedure is based on calculations of maximum depth of flow, h Separate values are calculated for the channel bed and the side slopes, respectively

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• • • •

It is necessary to choose values for inverse side slope, m, and bed width, b to calculate maximum allowable depth in this procedure Limits on side slope will be found according to the angle of repose and the maximum allowable channel width Limits on bed width can be set by specifying allowable ranges on the ratio of b/h, where b is the channel base width and h is the flow depth Thus, the procedure involves some trial and error Step 0 • • • •

Specify the desired maximum discharge in the channel Identify the soil characteristics (particle size gradation, cohesion) Determine the angle of repose of the soil material, θ Determine the longitudinal bed slope, So, of the channel

Step 1 • • •

Determine the critical shear stress, Tc (N/m2 or lbs/ft2), based on the type of material and particle size from Fig. 3 or 4 (note: 47.90 N/m2 per lbs/ft2) Fig. 3 is for cohesive material; Fig. 4 is for non-cohesive material Limit φ according to θ (let φ ≤ θ)

Step 2 • •

Choose a value for b Choose a value for m

Step 3 • • • •

Gary P. Merkley

Calculate φ from Eq. 16 Calculate K from Eq. 24 Determine the max shear stress fraction (dimensionless), Kbed, for the channel bed, based on the b/h ratio and Fig. 6 Determine the max shear stress fraction (dimensionless), Kside, for the channel side slopes, based on the b/h ratio and Fig. 7

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Permissible Tc for Cohesive Material

Tc (N/m2)

100

Lean clay Clay Heavy clay Sandy clay

10

1 0.1

1.0

10.0

Void Ratio

Figure 4. Permissible value of critical shear stress, Tc, in N/m2, for cohesive earthen material (adapted from USBR Hyd Lab Report Hyd-352). About Figure 4: The “void ratio” is the ratio of volume of pores to volume of solids. Note that it is greater than 1.0 when there is more void space than that occupied by solids. The void ratio for soils is usually between 0.3 and 2.0.

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Permissible T c for Non-Cohesive Material 100

2

Tc (N/m )

Clear water Low content of fine sediment High content of fine sediment Coarse, non-cohesive material

10

size for which gradation gives 25% of the material being larger in size

1 0.1

1.0

10.0

100.0

Average particle diameter (mm)

Figure 5. Permissible value of critical shear stress, Tc, in N/m2, for noncohesive earthen material (adapted from USBR Hyd Lab Report Hyd-352). • • •

The three curves at the left side of Fig. 5 are for the average particle diameter The straight line at the upper right of Fig. 5 is not for the “average particle diameter,” but for the particle size at which 25% of the material is larger in size This implies that a gradation (sieve) analysis has been performed on the earthen material

particle gradation 75%

smallest

Gary P. Merkley

25%

largest

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•

The three curves at the left side of Fig. 5 (d ≤ 5 mm) can be approximated as follows: Clear water:

Tc = 0.0759d 3 − 0.269 d2 + 0.947 d + 1.08 (25) Low sediment:

Tc = 0.0756 d3 − 0.241d2 + 0.872d + 2.26 (26) High sediment:

Tc = −0.0321d3 + 0.458d2 + 0.190d + 3.83 (27) where Tc is in N/m2; and d is in mm •

The portion of Fig 5. corresponding to “coarse material” (d > 5 mm) is approximated as: Coarse material:

Tc = 2.17 d0.75 (28) • • •

Equations 25-28 are for diameter, d, in mm; and Tc in N/m2 Equations 25-28 give Tc values within ±1% of the USBR-published data Note that Eq. 28 is exponential, which is required for a straight-line plot with loglog scales

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Figure 6. Kbed values as a function of the b/h ratio. Notes: This figure was made using data from USBR Hydraulic Lab Report Hyd-366. The ordinate values are for maximum shear stress divided by γ hSo, where γ = ρ g, h is water depth, and So is longitudinal bed slope. Both the ordinate & abscissa values are dimensionless.

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Figure 7. Kside values as a function of the b/h ratio. Notes: This figure was made using data from USBR Hydraulic Lab Report Hyd-366. The ordinate values are for maximum shear stress divided by γ hSo, where γ = ρg, h is water depth, and So is longitudinal bed slope. Both the ordinate and abscissa values are dimensionless. BIE 5300/6300 Lectures

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• • •

Regression analysis can be performed on the plotted data for Kbed & Kside This is useful to allow interpolations that can be programmed, instead of reading values off the curves by eye The following regression results give sufficient accuracy for the max shear stress fractions: 0.153

K bed K bed

⎛b ⎞ ≅ 0.792 ⎜ ⎟ ⎝h ⎠ ⎛b ⎞ ≅ 0.00543 ⎜ ⎟ + 0.947 ⎝h ⎠

for 1 ≤ b / h ≤ 4 (29)

for 4 ≤ b / h ≤ 10

for trapezoidal cross sections; and,

K side ≅

AB + C ( b / h) B + ( b / h)

D

(30)

D

where,

A = −0.0592( m) + 0.347( m) + 0.193 2

− 0.000311( m) B = 2.30 − 1.56e

7.23

− 0.00143( m) C = 1.14 − 0.395e

D = 1.58 − 3.06e −35.2 (m )

(31)

(32) 5.63

(33)

− 3.29

(34)

for 1 ≤ m ≤ 3, and where e is the base of natural logarithms • •

• • •

Equations 29 give Kbed to within ±1% of the values from the USBR data for 1 ≤ b/h ≤ 10 Equations 30-34 give Kside to within ±2% of the values from the USBR data for 1...