Silt Factor f

Lacey’s Regime Theory: Lacey carried out a detailed study in designing suitable channels in alluvial soils. He developed the regime theory and formulated a number of expressions based on his observations. The salient features of Lacey‘s theory are stated as follows. 1. In a channel constructed in alluvial soil to carry a certain discharge, the bottom width, depth and bed slope of the channel will undergo modifications by silting and scouring till equilibrium is attained. The channel is now said to be a regime channel. (A regime channel is defined as a stable channel whose bed width, depth and side slopes have undergone modifications by silting and scouring and are so adjusted that equilibrium is attained.) A channel is said to be in regime when the following conditions are satisfied. a. The Channel is flowing in unlimited incoherent alluvium of the same character as that of the transported sediment. (Incoherent alluvium is a soil composed of loose granular material which can be scoured and deposited with the same ease.) b. Silt grade (silt size) and silt charge (silt concentration) is the same throughout the channel. c. Discharge in the canal is constant. 2. The silt carried by the flowing water in the canal is kept in suspension by vertical eddies generated from the bed as well as from the sides of the canal. 3. The silt grade also plays an important role in controlling the regime conditions of the channel. The silt factor is given by the relationship 𝑓 = 1.76√𝑑 where 𝑑 represents diameter of silt particle in mm.

Lacey’s procedures for designing unlined canals: In Lacey‘s method for designing unlined canals in alluvial soils for a known discharge Q and a mean diameter of silt particle ‗d‘, the required quantities are calculated as follows. 1. The mean velocity of flow is computed from the relation.

V=(Qf ÷140) 1/6

Where Q is discharge in m3/s, f is silt factor given by 𝑓 = 1.76√𝑑,

d is diameter of silt

particle in mm. 2. Calculate the cross sectional area of flow

A=

Q V

3. Knowing the side slope express ‘A‘ in terms of B and D A = B D + K D2 4. Determine the required wetted perimeter from the relationship P= 4.75 Q -------(1) 5. Express the wetted perimeter in terms of B and D P= B + 2 D√1 + 𝐾 2 -------------(2) 6. From equations 1 and 2 solve for values of B and D 7. Calculate ‗Hydraulic mean radius‘ (R) from the relationship 

R

5 V 2     -----(3) 2  f 

8. Also calculate hydraulic mean radius ‗R‘ from the relationship R=

A P

2

=

BD + 𝐾 D B+2D√1+𝐾2

----- (4)

9. If the values of hydraulic mean radius R worked out from equations 3 and 4 are the same then the design is OK.   10. Calculated the Bed slope from the equationS  

5

f

3 1

  

 3340Q 6 

 

Draw backs in Lacey‘s theory

1. The theory does not give a clear description of physical aspects of the problem. 2. It does not define what actually governs the characteristics of an alluvial channel. 3. The derivation of various formulae depends upon a single factor f and dependence on single factor f is not adequate.

4. There are different phases of flow on bed and sides and hence different values of silt factor for bed and side should have been used. 5. Lacey‘s equations do not include a concentration of silt as variable. 6. Lacey did not take into account the silt left in channel by water that is lost in absorption which is as much as 12 to 15% of the total discharge of channel. 7. The effect of silt accumulation was also ignored. The silt size does actually go on decreasing by the process attrition among the rolling silt particles dragged along the bed. 8. Lacey did not properly define the silt grade and silt charge. 9. Lacey introduced semiellipse as ideal shape of a regime channel which is not correct. Problems 1. Design a irrigation channel in alluvial soil according to Lace’s silt theory folloeing data. Full supply discharge= 10cumecs Lacey’s silt factor= 0.9 Side slopes of channel= 1/2H: 1V Solution: 1. Velocity = 2 V  Qf 140 

1

 6   m/s

V= (10x 0.92/140)1/6 = 0.62m/s 2. Area= A= Q / V= 10 / 0.62= 16.13m2 3. Hydraulic mean radius= R

5 V 2     2  f 

= 5/2(0.622/0.9) = 1.07m 4. Perimeter=A/R= 16.13/1.07= 15.07m = B+ 2 D√1 + 𝐾 2

for the

5.



5    3 Bed slope=S   f  1   6  3340Q 



 S   0.9 1    6  3340(10)  5 3





S= 1/ 5844= 1/5850 2. The slope of a channel in alluvium is 1/4000, Lacey’s silt factor is 0.9 and side slopes are 1/2H : 1V. Find the channel section and maximum discharge which can be allowed to flow in it. Solution :

From the equation 



  S 1    3340Q 6  



5

 f

3

  Q 1    3340S 6  5

f

3

  f  Q   1   3340S 6 

 = (4000/ 3340)6x (0.9)10 = 1.03m/s From equation  f 3 / 2  S   1/ 2  4980R  





 f 3  R   2   4980S  = (0.9)3 x 4000x 4000/ (4980x4980) = 0.47m

From the equation P= 4.75 Q P= 4.75√1.03 =4.82m A= PR= 4.82x 0.47= 2.27m2 A= BD+ D2/2 = 2.27

---------equ 1

P= B+ D√5 = 4.82 ------------------ equ 2

Solving equation 1 and 2 we get B= 3.48m D= 0.6m