fl.OW IRRIGATION 610

PROBLEMS

. 1. Explain various types C'f canals, according to various classification systems. .

2. Explain. various consideration for alignment of a canal. 3: Distinguish between a contqur canalElnd a ridge canal. 3. Write a note on inundation canal. 5. DeM:ribe the Bandhara irrigation system.

G Irrigation Channels :

Silt "Fheories

14.1. GENERAL

, ,

.I

I \

The canal which .takes off from a river has to draw a fair share of silt flowing in the river. This silt is carried either in suspension or along the bed of the canal. The silt load carried by the canal imposes a difficult problem in a channel design oh alluvial soils. The veiocity to be allOWed in a channel design should be such that the silt flowing in the channel is not dropped on the bed. In case a channel silts up, its capacity redu~ and so it will irrigate less area. Also the, velocity should not be large enough 10 erode away the bed and sides of the channel. If the sides and bed of a channel are eroded away, the cross-section increases and besides other damages b~use of scour, 'its full supply depth :d~reases ; it can, therefore, command much less area, A velocity whicbwill just keep the silt in suspension, with,out sa"Juring the channel is known as non-silting and non-scouring velocity. . I

Many inyestigators have worked un varioUs eXisting channels towards .the design ofnon-silting-non-,scouring channel section. Notable amongst \hem are works done by Mr. R.G.KennedyofPunjab Irrigation Department and Mr. Gerald Lacey, Chief Engineer Irrigation Depart-ment V.P. The ideas p{bpagated J:>y them have rome to be known as, kennedy's theory and Lacey's, theory respectively. '.

. For the design of an irrigation channel, . the design discharge Q, and the surface and soil properties such as the rugosity coefficient N and silt factor f are ' known. The problem consists in determinatiqn of the four unknowrts : (i) Area ,of cross- section (.4), (ii) hydraulic

(611)'

, 612 IRRIGATION CHANNELS: SILTTI-lEORIES

mean depth (R), (iii) velocity of flow, an~ (iv) the bed slope (S). To start with, following two equations are available :

Q = A x V (Continuity equation) ... (1) and V = f(N, R, S) (Flow equation) ... (2)

The flow Eq. (2) may be Manning's equation, or Kutter's equation or any other similar equation. However, since there are four unknowns, two more equations must be available for the complete

~nd unique solution. The additional two eqatlons may be obtained from the following criteria :

(i) Providing channel of b.est discharging section thus gettinf another equation between A and R.

(ii) Limiting equation ofv.elocity from considerations of scouring and sitting.

(iii) Governing the slope by ' the availagle ground slope. (iv) Fixing a !iuitable BID ratio on experience. Criterion (i) is not good (or alluvial soil where a non-silting

non-scouring velocity is a must. Criterion (ii) and (iii) are utilised by Kennedy's method of design of channels. Wood's table gives a table of suitable BID ratio for various discharges. However, Lacey's

theory furnishes four equations for the complete determination of the four unknowns, without depending upon the earlier flow equation by Manning or Kutter. .

14.2. KENNEDY'S THEORY Kennedy selected.a number of sites on Upper Bari Doab Canal

'System, one of the oldest in Punjab (Pakistan) for carrying out in-vestigations about velocity and depth of the channel. The sites selected by him did not require any silt cleaance for more than thirty years and were thus supposed to be flowing with non- silting non-scouring velocity.

Kennedy's study revealed the following : 1. The flowing water has to counteract some amount of friction

against the bed of the' canal. This gives rise to vertical eddies rising up gently to the surface. These eddies are responsible for keeping most of . the silt in suspension. Some eddies may start from sides but these are for most of its part horizontal and so d~ not have any silt supporting power. The silt supporting power is, therefore, proportional to the bed width of the stream and not to its wetted

41 perimeter. . 2. He also defined critical velocity as non-silting-non-scouring

velocity and gave a relation between critical velocity to the depth of flowing water. The relation is '

IRRIGATION CHANNELS: SILT THEORIES 613

where

[or In general

Vo = 0.55 DO.64 . .. (14.J)

Vo = O.84Do.64 in F.P.S~ units] ... [14.1(0)] Vo =CD" ...,;

V ='Critical velocity, in metres/second D = Depth of water over ' bed portion" of

a channel in metres' n = any index number.

Since the equation has been derived on the basis of observations on one canal only, it is applicable to only those channels which are- flowing in sandy silt of the same quality or grade as that of Upper Ban Doab Canal system. .

~ennedy later realised the importance of silt grade on critical velocity and .introduced a factor m known as critical velocity ralio (C.V.R) in his equation. The equation is then written as

V = 0.55 m DO.64 V . where

... (14.2) m = C.V.R. = PO ... (14.2(a)l

Sand coarser than the standard were assigned value of m from 1.1 to 1.2 and tbose finer than the standard from 0.9 to 0.8. Generally, in a system of canal, higher C. V.R. is assumed in head reaches and lower value of C.V.R. is assumed towards its tail end.

The value of constant C in Eq. (14.1) for various grades of material may be assumed as folloM :

T1PUt1{~

14ht Sandy ailt

Coarser light silt Sandy loaql

Coarse silt

TABLE 14.l , VALVE OF C

.

ValMeojC

0.53

0.59

0.65 -

0.70

The value of m in Eq. 14.2 for differeD{ types of silts are given in table 14.2.

614

T,pu/sir

IRRIGATION yHANNELS: SILTnlEORlES .' .

TABLE 14.2 VALUE OF ..

Yalueo/'"

1. Light sandy silt in the riven of Norther1l India 1.00

1.10 2. Somewhat coarser light sandy sill 1.20 3. Sandy. loamy silt 4. Rather coarser silt or debris of hard soil

1.30 .

0.70 S. SUt of rivel)ndus in Sind

Kennedy made use of Kutter's equation for finding the mean velocity of flow in the channel:

23 t ~ + 0.~155 .. V = . 0.00155 N RS ". , ... (14.3)

1+(23: .. s )TR / " (in metric units)

Thus the limitations of Kutter's equatiOn beCOme incorporated in Kennedy's design procedure. Kennedy. did npt give any equation for any -slope of the canat The stope is decided according to the slope of ihe ground available. From the ploUed longitudinal slOpe of the alignment, there is always a certain range of slOpe, that can be given to a channel at the site. For every such slope; a separate channel can be designed to suit a given discharge. For example, when Q = 2 cumecs, N = 0.0225 and !1J = 1, we get the following channel dimensions, for various slOpes :

' ..

Slope B D

. ; BID

("') ("') .

./

. 1 jn SOOO 7 0.68 10.8

1 in 4000 3.2 0.85 3.8

1 in 2000 l.S 1.4 1.07

,

Unfortunately, Kennedy's theory provides no clue to which of ;hannels wilr be best suited for a particular discharge and type of soil available. Hence some guidance is necessary to find out as to which channel would suit the need beSt. To provide this guidance, suitable .BID ratios were fixed in various state departments on' the basis of experience. Wood'S normal table is one such a;iterion which. was prepared in Punjab and is very much in us~. An extract of Wood's table, alongwith its metric conversiqn. is given in Table .14.3.

IRRIGA nON CHANNELS: SILT TIlEORIES

~ ~ ~ 'I) .,.. 8 .... .,.. --.. ..,. .." '"

....

~ E Q cO 00 00 00

fi ~ c; M ~ 'I) ~ ~ ~ .... "1 --. ...; ...; .... .... M o.t!.,

. ~ ~ ... . ~ ~ -0 C! ~ 00 ...; .... - ....

~.~/ . ~ .... ~ ~ ~ i i

'.

0- ~ ~ C!! - .,.. .... ~ .

s

.,.. ~ '" '" ~~ VI l- v S; &; .... 0-00 cO cO 00 _0 0 ..,. 0- Q ~ ~~ ~ .,.. ~ --M ~ ~f'i M

~ 1"'\

616 IRRIGATION CHANNELS: SILTnlEORIES ' ....

Thus, if BID ratio is ~f::lected from the recommended valud, one more equation will be available. Hence unique design of the channel can be ~one, including tbe bed slope using Kennedy's theory. This has been illustrated in case- 2 of 14.3 and in example 14.2.

14.3. KENNED\Y1S MEtHODOF CHANNEL DESIGN PROCEDURE For the design of channel, the fonowing equations are used:

Q=AXV 23 + 1. + 0.0015:-N 'S .-

V = 000155 N vRS ... (2) _ 1 + (.23 + . s JTR

~M ( V = 0.55 mD ... 3) . The value of rugosity coefficient N for earthen channels of N~rth India, as given ~y Buckley are as follows : -

Channel of conditions Value of N 1. Very good 0.02~ 2. Good 3:- Indiffernt 4. Poor

0.025 0.0275 0.03

Table 14.4-gives the Central Water and Power Commission (CWPC) practice 0' selecting Kutter's roefficient N fQr the canals taking off from reserviors carrying silt free water.

TABLE 14.4 CWPC ,PRACTICE FOR N

T,,,. of Ccu..t DiscItGrI. VGIu gfN Soil (CfUIIS)

1. SOU other than rock tlpto 0.14 OJJ3 0.14 to 1.4 0.025 14 to 14 ~ 0.0225 Above 14 0';020

2. Rocky cuts - (I) When rock portipn at least 1S em above the excavatl'd bed level is left out

'. f iD workiDI Oilt - cross-

. ' IeClioDaI area 0.035';' O.OS

-

,

(iI) When no portioD above bed level is left out 0.05 - 0.080

\

,

i

----~

IRRIGATION CHANNELS : SILTTIlEORIES 617

The Irrigation Department of Uttar Pradesh adopts the foll liwing value of N :

Discharge (cumecs) upto 14 .

N 0.0225

above 14 0.020 In should be noted that the value of N also depends upon

the nature or 'COndition of the channel surface, as is clear from Table 14.5.

I

NaIJ4n of clullfrul srufau

1. Alluvial Regime

2. Loose gravel

TABLE 14.5 VALUE CF N

-

ValiI'ofN ---r--~

P~M .Good Fair

0.077 0.02 0.0225 0.OC7S 0.030 0.033

bad ,

0.02.5 I

0.035 - -3. Rock cut, smooth and regUlar 0.025 0.030 0.033 0.03.';

4. Rock cut, irregular 0.03S 0.040 .0.045 0.050 5. With cobble bed and eanhen 0.025 0.030 0.035 0.040 banks

-----6. With earthen bed and rubble 0.015 0.030 0.033 0.035 sides

DESIGN PROCEDURE

Case 1. Gillen Q, N, m and S ({'rom L-section) 1. A'isume a trial value of D in metres. 2. Calculate the velocity V from Eq. (3),

V = 0.55Do.64

3. Get area of section A from Eq. (1).

A=~ 4. Knowing D and A, calculate the bed wi jth B.

be The side slope of the canal in alluvial soil is assumed to ~. : 1 when the canal has run for some time.

A =8D+D2 2

.... /

618 IRRIGATION CHANNELS: SILTTI-lEORIES

From which B can calculated. 5. Calculate the perimeter and the hydraulic mean depth from

the following re.lations : P=B+DVS

A BD +D2/2 ' R = P = -B-+-D-"-Y5""-

6. Calculate the actual mean velocity of flow from Kutter's equation (Eq. 3.). If this value of velocity is the same as that found in step 2, the assumed depth is correct. If not,. repeat the calculations wilh ... a changed ~alue of D till the ~o velocities are the same.

o I

I-- _& -4

FIG. 14.1. CHANNEL SECTION.

Thus we get the solution for a given value of S. Case 1 is the most common one and its solution is normally done with the help of Garret's diagram. Garret's diagrams [Plate 6(a), (b)], provide solution of Kennedy's regime equation and Kut.ter's flow equation for various values of slopes and have been explained in Chapter 15.

Case ~. Given Q, N, m And BID ratio from Wood's table (1) Calculate A in terms

B Let D =X

B = Dx

of D.

, D2' 2 D2 A=BD+T = ... D +T

or A = D2(x + O.S). (2) The value of velocity V is known in t{:;rms of D by

Kennedy's equation. V = 0.55 m DO.64

/rRRIGATION CHANNELS : SILT THEORIES 619

Substitute the values of V and A in the continuity equation and solve for D. Thus

Q = A x V = D2(X + 0.5) x 0.55 m DO.64 or Q = 0.55m (x + O.S)D2 , 64

1 Hence D - [ Q ]2.64 0.55~(x + 0.5) In the above relation Q, m and x are known. Hence D is

determined. 3. Knowing D, calculate B

relations and R from the following

B =xD

and R == BD + D2/2 B+DVS

4. Calculate the velocity V from Kennedy's equation V;" 0.55 m DOM

5. Knowing: V and R, determine the slope S from Kutter's flow equation. The equation can be solved by trial and error.

14.4. SILT SUPPORTING CAPACITY ACCORDING;, TO KENNEDY'S mEORY According to Kennedy's theory. the amount of silt held in

suspension is proportional to the upward force of vertical eddies. and varies as the bed 'width (B) and some power of the velocity of flow in the channel. Let Vo = critical velocity in the channel

or

where

Qt = quantity of silt transported by the channel ,Q = discharge in the channel. Qt ex BV8

Qt =aBVO' a = some constant.

Let p = % of silt in water = ~ . . Qt=pQ.

... (i)

620

or

or

IRRIGATION CHANNELS: SILT THEORIES

Assuming a wide channel with vertical slopes,

Q = BD.vo Q/ = p.BD.vo

Equating (i) and (ii), we get aBV 8 = pB.D.vo v, ,,-1_1.D o --p.

a

I 1

Vo = (; ) /I - I Dn:T 1

(approx.) ... (ii)

T" CD ;;-::-r ( ... ) '-0 = ... tit

This equation is similar to Kennedy's equation

Vo = CDY = CD0 64 (iv) EqJating the indices of both these equations, we get

1 --=0.64 n-1

n = 2.56:::: 5/2. Hence. the quantity of silt transported is given by

Q/=aBV S/ 2 o ... (14.4) Kennedy did not determine the value of the constant Il. Example J 4.1. Design an irrigation channel to carry a discharge

"f 45 cumecs. Assume N = 0.0225 amj m = 1. The channel has 1 bed slope of 0.16 metre per kiloml tre.

SolutioI' : 1. Assume

2.

3.

4.

a trial depth D equal to 1.8 m V = 0.55 mL 0.64

= 0.55 x 1 x 1.8.64 = 0.8 m/sec Q 45 2

A = V = 0.8 = 56.2 m V=BD = D2/2

56.2 = B(1.8) + (1~)2 From which B .= 30.3 m.

IRRIGATION CHANNELS : SlU TIlEORlES

where

"

5. Perimeter P=B+DVs

= 30.3 + 1.8Vs = 34.32 A 56.2

,R = P = 34.32 = 1.64 m 6. V=CYRS

S = 0.16 1000

23-+ -L + 0.00155 x 1000 C = 0.0225 0.16

1 + (23 + 0.00155 1000) 0.0225 0.16 x ,/t.64

= 23 + 44.4 + 9.7 = 49 1 + (23 + 9.7)~

r" = 49 vi 1.64 x O.i6 1000

= 0.793. 7. Ratio of velocity found in steps (6) aM (2)

::::: 0.793 = 0.991::::: 1. 0.8

Hence, the assumed D is satisfactory.

621

Example 14.2. Design an irrigation canal to cany a discharge of 1.4 cumecs. Assume N = 0.0225 , m = 1 and BID = 5.7.

Solution 1.

.. k.

B D =5.7

A :: BD + ~1 = D2 [~ + 0.5 J = D2(5.7 + 0.5) = 6.2 D2

V = 0.55 m DO 64

= 0.55 x 1 x Dl!.64

Q = A x V = (6.2 D2 ) ( 0.55 D O.64 )

622 IRRIGATION CHANNELS: SILTlliEORiES

/ 1

D = [0.55 ~ 6.2]264 _ ( 14 )2.164 - 0.55 x 6.2

= 1.705 :::: 1.71 m.

3. B = 5.7 D = 5.7 x 1.71 = 9.73 m

and

or

which

or

4.

5.

or

BD + D2/2 (9.73 x 1.71) + ~(1.71)2 R = B + DVS = --9-.-73-+-1-.7-'1 {5=-=5--

= 1.37 m.

V = 0.55m DO.64 = 0.55 X (1.71)64 = 0.775 misec-.

23 + 1.. + 0.00155 V= N S YRS

1 + (23 + Q.(0155) N S TR

23 + --L + 0.00155 0.775 = 0.0225 S

1 + (23 + 0.00155 ,0.0225 ..fl.37 x S S {!ITf

0.775 = 23 + 44.5 + 0.OOi55

1 + 0.44 + ' 2.98 x 1.17 x S1I2 -r x 10-5

reduces to

0i~:; ( 1.44 + 2;8 x 10-5 ) = ( 67.5 + 1.~S x 10-3 .) SlI,2

67.5 S312 + 1.55 X 10-3 S1I2 - 0.954 S = 1.98 X 10-5

S3/2 + 2.29 x 10-S S1/2 - 1.41 x 10-2 S = 2.93 X 10-7

Solving this by trial and error, we get

IRRIGATlON CHANNELS: S!LTTHEORIES 623

1 S = 5100'

]4. S. DRAWBACKS IN KENNEDY'S THEORY (1) Ken'nedy did not notice the importance of BID ratio:' (2) He aimed to find out only the average regime conditions

for the design of a channel. (3) No account was laken of silt concentr:ation and bed load,

and the complex silt carry.ing phenomenon was inCorporated in a single factor m. .

(4) Silt grade and silt charge were not defined. (5) Kennedy did nqt give any slope equation. (6) Kennedy used Kutter'S equation for the determination of

the mean velocity and, therefore, the limitations of Kutter's equation got incorporated in Kennedy's theory of channel design.

Further Work on Kennedy's Theory Kennedy's equation could not be made universally applicable

even after the introduction of critical velocity ratio. Regional formulae were thereafter developed to suit the requirements of different canal system by different investigators. In this formula both the coefficient of C as will as power of D varied. A..general feature 'of these formulae indi~ted a lower value of C for finer soils. The value of C and n in Eq. 14.1 (b) for different canal systems are as follows:

SeriolNo.

1

2

3

4

TABLE 14.6 VALVES OF C AND 11

Canal system

Godavari Della

Krishna Western Delta

Lower Chenab Canal

gyptlan Canals

Lindley's jTonnulae

c "

0.391 0.55

0.530 0.52

0.567 0.57

0.283 0.73

It has already been emphasised that depen

624 IRRIGATION CHANNELS: SrLT TI-lEORlES

section. As to which of these. channels will suit the requirement can not be found out from Kennedy's theory. In order to make channel design more definite, Lindley of Punjab irrigation proposed his formulae in 1919. He proposed the following two formulae one relatmg V and D; similar to Kennedy and the other relating V and B.

V = 0.567 D057 ... [14.5(a)] and V = O.274If~5 ... [14.5(b)]

By eliminating V from Eq; [14.5 (a),(b)J, another relation as originally proposed by Lindley can be obtained. This is given by

B = 7.80 D1.61 ... (14.6)

14.6. LACEY'S REGIME THEORY "Dimensions, width, depth and slope of a regime channel to

carry a given discharge loaded with a given silt charge are all fixed by nature. This idea was first put forward by Lacey. Lindley's thclJf)' is also based on the same con~pt. Lacey succeeded in evolving more generally applicable equatiOns based on his own experiments and the experiments of the past . investigators.

Regime channel. Lacey defined regime channel as a st,,:ble channel transporting a regime silt charge. A channel will be in regime if it flows in coherent unlimited alluvium of the same character as that transported and the silt grade and silt charge are all constant

Incoherent alluvium. It is a soil composed of loose granular graded material which can be scoured with the same ease with which it is deposited.

Regime silt charge. It is the minimum transported load consistant with ful!y active bed:

Regime silt grade. This indicates the gradation between the small and the big particles. It should not be taken to mean the average mean diameter of a particle.

Regime conditions. A channel is said to be in regime when the following conditions are satisfied.

1. The channel is flowing in unlimited incoherent alluvium of the same character as that .transported.

2. Silt grade and silt charge is constant. . 3. Discharge is constant:

IRRIGATION CHANNELS: SILTTIfEORIES 625

If the above three conditions are met with fully, then the channel is said to be in true regime. However, it is seldom that the above conditions are realized in field. Hence, Lacey g:J\'l,! the idea of initial and final regime for actual channel.

Initial regime. One of t~e conditions of attaining regime of a channel is that there should be freedom for the channel to form its own section. Initial regime is the state of channel that has formed its section only and yet not secured the longitudinal slope.

Final regime. When a channel is constructed with defective slope, it tries to throw off the incoherent silt on the bed to increase their slopes. To altain the final regime the channel forms its section first before the final slope. The channel after attaining its section and longitudinal slope, will be said to be in final regime.

Permanent regime. When a channel is protected on the bed and side with some kind of protecting material the channel section cannot be Scoured up and so there is no possibility of change of section or longitudinal slope: ; the channel will then be said to be in permanent regime. Regime theory is not applicable to such channels.

There is only one section and only one longitudinal slope at which the channel will carry a particular discharge with a particular silt grade. Natural silt transporting channels have a tendency to assume a semi-elliptical section. The coarser the silt; greater is the waterway of such a channel and narrower the depth. The finer the silt, greater is the depth and the channel closely approximates a semi-circle.

, I

I I

, , ,

, ,

,

FIG. 14.2. CHANNEL SECfION ACCORDING TO LACEY'S TI-lEORY.

62.6 IRRIGATION CHANNELS: SILT THEORIES

vi If a channel is constructed with too small a cross-section for ~ particular discharge and the slope steeper th.an required, scour will occur till final regime is attained. Similarly silting will occur in a channel till final regime is attained in a channel with a wider cross-section and flatter slope than requi!'ed.

L,~ey also states that the silt is kept in suspension due to the force of vertical eddies. According to him, the eddies are generated from bed and sides, both normal to surface of generation. Hence, vertical component of eddies generated from sides will also support the silt. Lacey, therrore, assumed hydraulic mean depth (R) as variable, unlike Kennedy who assumed depth D as variable. Since Lacey assumed a semi-ellipse as the cross-section of ~ regime channel assumption R as a variable seem to more logical.

Lacey's Regime Equations On the basis of arguments mentioned above, Lacey plotted

a large mass of data to obtain a relationship between (i) V vs. R and (ii) A vs. V. Lacey recognised the importance of silt grade in the problem and introduced a function f known as silt factor i '1 the regime relationship. The first two equations, originally sug-gested by Lacey in F.P.S, units , converted in M.K.S., units, are given below :

V = Y~fR Af2 = 140.0V 5

... (14.7)

... ( 14.8)

where A = area of the channel section and V = velocity of flow in it.

Regime Flow Equation By plotting a large mass of data from a number of different

sources, Lacey obtained the relationship V = lo.8R213 5 1/ 3 .. . (4.9)

where 5 = slope of water ~urface This equation is very 'useful for finding velocities during high

flood&. A beginner has not to worry himseIr to find out the exact values of N in order to apply Manning's and Kutter'S formulae. Equation 14.9 contains no sill factor and rugosity coefficient. Such variables were considered to be implicit in the actual depth and slope adopted by the channel in loose alluvium.

IRRIGATION CHANNELS : SILTTHEOR!ES 627

Eqs. 14.7, 14.8 and 14.9 are known as Lacey's fundamental equations from which other flow equations can be derived as shown below.

1. Perimeter Discharge (P-Q) Relation By combining Eq. 14.7 and Eq.14.8 a rcldllonship between

P and Q can be obtained. Raising both sides of Eq. 14.7 to the fourth power. we get

V4 = ~f2R2 25 .

Eliminating [2 between this equation and equation 14.8, we get fr.om equation 14.8

or

or

or

A [2:;4 J = t4U.O V 5 A [ __ 25~] = 140 V

4R-Multiplying both sides of this equation by A, we get

But

25A 2 -y = 140VA = 140 Q 4R

A2 2 R2 = r

P 2 = 4 x 140 Q 25

P = 4.75 Ri ... (14.10) Equation 14.10 is a very imporlant equation in as much as

it is universally true. Power of Q docs not vary, though the coefficient 4.75 varies between certain range.

Or

2. JI-Q1 Relation A relation between V-(J-f can be obtained from Eq. Multiplying Eq. 14.8 by V on both sides, we get

A VJ~ = -140.0 V 6 Qf'!. = 140.0 V 6

14.8.

v= [~1~/6 _ 140.0 J -'. . .. (14.11)

628 IRRIGA nON CHANNELS : SILT TI-lEORrES ~,

This relation is utilized in channel design by Lacey's theory. The value of Q of [ in a channel design are known. The value of V can, .therefore, be. readily found from Eq.14.11.

3. Rt>gime.Slope Equations (5 - Q - f.) I (R - 5 - f) and (5 - [- q) relationships. From Eq.14.9, cubing both sides of the equation

V 3 = 1260 R2 S ... (14.12.) From Eq. 14.7 'cubing both sides of the equation

,. V 3 = (215)3/2[3/2 R3/2

Hence, (2/5)3/2[312 R 312 = 1260 R2 S [ 3/2 s- '

- 4980R1/2 ... (14.13)

or ..

Equation 14.12 can also be written as

(V 2 )5/3 1

5 = R 1260 (RJI)1/3 ... (14.14)

From Eq. 14.7, V 21R = ~[ RV=q

where q = discharge per unit width.

S = (~) 5/3 t /3 or . [5/3

... (14.15) = 0.000178 173 q

The above equation can also be represented in terms of total discharge.

Rewritting equation 14.14,

( V2 )5/3 1

S = R 1260 (RV)1t3 Putting R = AlP, we get

(V 2 \ 5/3 1 .

S = If" J '[A ]1/3 1260 P V 2,5/3 pI/3

IV --= \ If ; 126(l (Q)1 / 3

...

IRRIGATION CHANNELS: SILT THEORIES 6~

But P =4.75VQ S = (2/5)5/3[5/3 (4.75) 113 QI /6

1260 Q1/3 ... (14.16) (5/3

S = ---.::.J_-r= 334OQI/6

4. Regime Scour Depth Relation From Eq. 14.7 as alrea4y derived

2 25 y2 R = '4 /2

5 y2 R="2/

From Eq. 14.11, we get ,

Hence

2= [~]lt3 V 140 5[Q'21 1/31 R=- W_ _ 9 140 J /

= 0.47 (q/f)l/l Since q =RV Substituting for R from Eq. 14.7,

, 5 v3 1="2/

Substituting y from Eq. 14.11, we get

=~ .!.[~]3/6 2' f 140

= 0.21 Qlt2

... (14.17)

. .,

Substituting for Q = (n!r r in equation 14.17, we get ... (14.18) R = 0.47 fll/j [O.~I]213

= 1.35lllfJll3 ... (14.19) 5. fAcey's Non.Rejime Flow Equatio~ Eq. 14.9 is applicable only for regiine flow. The equation to

be derived now will be applicable fQ.r bot,h regime and non regime channels. I. '

From Eq. 14.12, ,V 3 1=: 1260 R2S

V = 35.5 (R/V)J/2VRS ' or ... {14.20(a)I

630

,

IRRIGATION CHANNELS: SlLTTIlEORIES

Equation 14.20 is similar to Chezy's equation.

Now putting

Chezy's

v = KVjR we get C _ 35.5 R1I4

- Klt2f 1l4 . Again putting

35.5 K' Kl/2/1/4 = NtJ

.. l14.7.0(b)]

where Na = Lacey's absolute rugosity coefficient. Lacey called N .. as absolute rugosity coefficient to emphasize that this No depends only upon grade and density of boundary material and'is independent upon channel condilion.

632

~ . ~

~

~

~ I

~ l,a j ~ ~ ~

~ ~ ;J til

..

..

...

~ ~

'~ ...;

~ ~

.!;!

~

.~

...; ~ ~

~ '0'

.~ ~

t

IRRIGATION CHANNELS : SILT THEORIES

-c I- 0.. ~ I 1'\ .,... ... ... ... ~ :q ... ~ ... .... - ~ ..;j ..;j ~ ... I- ~ -.i ..;j ..;j ~ ... ..... ..... ..;j .., .., ~ .... .... .... ..... - ..... - ... ,.... , II

...

;:; " ... ..

" I - ."

.., ." '"" ~ ~ .. .. .. ~ ~, .... .... .

....

.... ... ....

Ys y~ ~~. -~ ;:;c:: ;:; ..... _, ~ f'\ -~ ~ ~ ;;;--E' ~ .~ ~L" ~L .. ~..,,, ~ ~ ~f ....;f ~~ " ~ r:( ~ -~ ~ CI ~ N' ~ ~ .... ~'" $"'!~;;;~

001

63~ IRRIGATION CHANNELS; SILT THEORIES

I; 5. Find out bed width B and depth D of the channel section since A and P are known. The side slope of an irrigation channel

D2 is usually i: 1. Hence, Area A = BD + T

Hence

and

P=B+DVs P - "p2 -6.944A

D = 3.472 B = P- 2.236D

5 V2 6. Calclllate R = 2: 7

BD + D2/2 Also cal :ulate R = -=-:..--=---=

B + 2.23D Both the values of R should be the same; this will provide

a numerical ch ~ck from steps 1 to 5. f5/3

7. Find the slope S = 1 6 3340Q / Example 14.3. A channel section has 10 be designed [or the

[ollowing data : Discharge Q= 30 cumecs Silt factor [= 1.00 Side slope = ~ : 1 Find also the longitudinal slope. Solution 1. The value

2. Velocity

3. Area

4.

5:

of [ is given as 1.00

[30 x 1 ]1/6 V = -w> = 0.773 m/sec

A = 0.;~3 = 38.S sq.m p = 4.75.fQ

=: 4.75 V30 = 26 m p':" "p2 -6.994A

D = 3.472 = 26 - Y'-6-76---6-.94-4-x-3-S.-g = 1.67 m.

3.472 B = P - 2.236 D = 26 ~ 2;236 x 1.67 = 22.6 nt.

6 .. Hydraulic mean radius.

IRRIGATION CHANNELS; SILTTIiEORIES

Also

Hence, checked.

7. Slope

5 V2 R=--2 f 51 2

= "2 x 1 (0.773) = 1.49 m. R _ BD+D2/2

- B+DVS _ 1.67 (22.26) + 1.672/2 - 22.26 + 1.67 VS

38.8 149 = 26.0 =. m.

[5/3 1 S= = __ _ 3340 (Q)I~t; 3340(30)1/6

1 1

635

= 3340(1.764) = 5900 . Hence, the channel has a bed width B = 22.26 m and a depth

of 1.67 m. The longitudinal slope S = 1/5900.

14.8. COMPARISON OF KENNED\"'S AND LACEY'S THEORY 1. Kennedy introduced the term C. V.R. (m) in his equations

to make it applicable for channels of different grades of silt, but he did not give any idea to measure the value of m. Lacey introduced the concept of silt factor f in equations and suggested a method of determining the value of [ by relating it with particle size.

2. Kennedy assumed that silt is kept in sUlopension because of eddies generated from the bed only, and so he proposed a relation between V and D. Lacey assumed that silt is kept in silspension because of the normal components of eddies generated from the entire perimeter and so he proposed a rel~tion between V and R.

3. Kenned} assumed Kutter's formula for finding the value o"mean velocity where in the value of N is to be assumed arbitrarily. Lacey gave his own formula for the velocity and thus a designer has not to choose anything arbitranly.

4. Kennedy gave no formula for determination of longitudinal slope of the canal. The slope to be given to the canal is based on experience or on Wood's table. Lacey gave a formula for the longitudinal slope of a regime channel.

5. Lacey pro'posed that .he shape of a regime channel should be a semi ellips~ince the channel section is l trapezoidal in shape,

..

636 IRRIGATION CHANNELS: SILTTI-IEORIES

it can never attain final regime. Kennedy simply gave the idea 'that a non-silting and non-scouring channel will be a regime channel.

6. Lacey's theory as applied to channel design does not involve any trial and error proCed!.ue whereas Kennedy's theory i!1volves ' a trial procedure for design of channel.

7. Lacey made a distinction between two types of resistance in alluvial channels, one determined by grain size and the other due to irtegularities of the channel. Kennedy did not make any such distinction.

8. Basic concept of the theorie!, is the same that the silt remainS in suspension due to the force of vertical eddies.

14.9. DEFECfS IN lACEY'S mEORY 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 singli!

factor f and dependence on single factor f is not adequate. 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.

4. Lacey's equations do not include a concentration of silt as variable.

5. 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 t~e total discharge of channel.

6. The effect of silt attrition was also ignored. The silt size does actually go on decreasing by the process of attrition among the rolling silt particles dragged along the ~ed.

7: Lacey did f10t properly define the silt grade and silt charge.

8. Lacey. however, considers that a regime channel is inherently free from external restraint and shock and has, therfore, a constant Na for a given size material. In so far as regime channel is a sediment transporting channel and will normally have a changing pattern of bed ripple formation. this statement is unliJ{ely , to be correct.

9. Lacey introduced semi ellipse as ideal shape of a regime channel which is not correct.

10. Strictly speaking an artificial channel is not a regime channel, and regime theory is not applicable to it.

IRRIGATION CHANNELS: SILTTIiEORlES 6TI

14.10. SEDIMENT TRANSPORT The quantity of solids enteriog the channel is an important

factor which controls the cross- section and shape of a true regime channel. Regime theory does not consider this important factor in channel design. However, it ,has now been realised that the channel design will not be successful unless a detailed study of sediment transport has been made.

two

Bed

The sediment parts :

1. Bed load 2. Suspended

load

mOving in a fluid can be broadly divided in

load.

Bed load may be defined as the load of bed mat~rial in the layer where suspension is impossible for fluid dynamic reasons. Sedi-ment grains in the bed layer are not vertically supported by the flow, but rest on the bed almost continuously, while sliding, rolling and jumping along. Their weight is supported by the static grain particles in the non-mOving bed. They move regularly exchanging places with the similar particles in non-moving bed. Suspended Load

In case of suspended load the weight of the particles is suported by the surrounding fluid. Since the particles..Ontinuously settle with its settling velocity Vr, in relation to the surrounding fluid , it is kept in suspension soley because the flow provides an upward motion due to turbulent exchange due to which a fluid is continuously ex-changed over definite distances beJween horizontal layers. The rising flUId originates frorti Jower layers of higher concentration and des~ cending fluid originates from higher layer of lower concentration so that a surplus of upwdrd moving sediment particles over downward moving sediment particle occurs. This surplus provides an upward motion of particles which counterbalances the general settling of sediment. Suspended load cause an additional hydrosatic pressure on the bed.

The. concentration of suspended load C at a height y abov( , the bottom can be determined in terms of known concentratior Ca at a reference point at a hight a above the bottom by the (ollowin! equation :

w

.= [D-y_a_]KV; CD y D-a w =:= fall velocity of grain

... (14.27) where in still water

638 !RRIGATION CHANNELS: SILT rnEORIES

D = Depth of water K = Van Karman's conl'tant = 0.4

V. = Shear' velocity = ~ P 'ro = Intensity of shear stress at bottom.

Rate of bed load transport The initiation of bed load movement is caused by the drag

force exerted on it by moving water. Many investigators have worked o.n the correct estimation for the rate of bed transport but none of the ideas extended so far could be wiedly accepted.

We shall di$cUSS herein two most widely used equations, that are due to Meyer peter and Einstein.

MEYER PETER'S EQUATION . Meyer Peter gave a dimensionless equation based, for the first

time, on rational laws. His equation can be reduced in a simple form as given below :

. (b (n') 3/2 (Y )!/3 " 3 Q n y ... S D = 0.047 (y - y..,) d + 0.25; (gS)'"1 ... (14.28)

where

and . as

So bed share

and and

D = depth of water S = slope of channel d = grain diameter gs = rate of bed load transport per unit

width of channel n I = Manning's n for plane bed n = actual value of Manning's n on rippled be Q = discharge if sides were frictionless Q.. = actual discharge

y = specific weight of sediment particle Yw = specific weight of water

~ = take into account correction for sides Yw R S = shear stress T = Yw DS

R z D for a wi.

where A. , B. and ~ are universal constants with a values Tjo

of 44.5, 0.143 and 2.0 respectively.

where

. . [ 1/2 1 1/2 ] -!P.. -!P. 9t ~ . _ f

4>. -. -. ( ) (3) ... ~14.31) lb Ib as a - P gd gp =rate al which bed load moves through

the unit width of cross-section d = diameter of sediment particle ip = fraction qp which is of diameter d a = mass density of sediment particle P = mass density of water ib = proportion of grains of diameter d

For uni[oml soils in the bed.

640 IRRIGATION CHANNELS: Sn..TlliEORIES

Ill. - W. relationship bas been plotted by Einstein in the form of a graph shown in Fig. 14.3. If the value of '1'. is known, Ill. can be found oul and so q~ can be calculated from Eq. 14.14. The Ill, 'I' relation could be expressed for uniform bed material by

= _1_ e - 0.39111' 0.465

40 30 11. I 1 1 11 \ \ 1 I 1 -, 20 I 1 0'00\1 0-0001

O!ol T 0-01 ;- 7 c>-OO?1 1 H' . '0 I I I ~.,II Ll ~, 0;001 8.3 I :07' ,7'._ .1 I I ... 1 " -1 ~,' ~ ~ 1 I'

,I \1 2'0 I I ill 1.....-1 ~ 11 111 1/ I L

vY 1 ,1'0 I I III U .0'8 I I

to., -"A' 0 ~ 0.:5 L..Ll-0.4 rir~L I 0 " -;/ --. I .A III I I I I I 0'1-' lXV I III II I '--j o,LI.ld" II I I 1I1111 ' 0 '0 ~1111 T I I I J

~ .. -FIG. 14.3. EINSTEIN'S BED LOAD FUNCTION.

Example 14.4. Design a channel section by Kennedy's theoT), given the following data

Discharge Q = 2828 cumecs Kutter's N = 0.0225 Critical velocity ratio m = 1 Side slope = ~ : 1

BID = 7.6 Find also the bed slope of the channel.

..

IAAIGATION CHANNELS: SILTnlEORIES

SoluUon Let the depth of the channel be D Let width B of the channel be 7.6 D . Velocity in the cbann~l

V = 0.55 m DO. 64

641

... (14.2) Keeping m= I, V = 0:55 d64., . . Area A of . the channel .. section having a side slope i : 1

Since

=BD +D2/2 ..; 7."6 D2 + D2/2 = 8.1 D2

Q =;A V . 28 ~ (8.1 Ir)(0.55 Do.64 ) = 4.46 D 264

DiM ' Z,8 '_ 628 . = 4'.46 - .

. D = 2.01 m . B = 15.3 m

Hydraulic mean radius R = AlP BD +D2/2 B+V'5D

= 8.1.02

Keeping we get Keeping

7.6D + 2.23D D = 2.01 m R = 1.66 m m = 1,

where

or

v = 0.55 (2.01)64 = 9.86 By applying Kutter'S formula

V=CYRS

23 + .1 + 0.00155 C= N S

1 + (23 + 0.00155) N S TR 23 + 1.. + 0.00155

N S _ r.:;;:;

( 000155) N vRS = 0.86 1+ 23+ s TR

... (14.6)

or (23+_1_+ 0.00155)SII2 = 0.674 [1 + (23 + 0.00155) 0.0225J 0.0225 S S 1.659

64~ , . IRRIGATION CHANNELS: SILT THEORIES

or ' 67.4 S3/2 _ 0.885 S + 0.00155 SI/2 - 1.42 x 10 - s = 0 __ ,--' - 1

By trial and error S = 6250 . Check

( 1 ) 3,,2 ' (1)., , ( 1 ) \/2 -5

,_ ,~?.4 6250 -~5 6250 +tf.QQl5S 6250 -1.42 X 10 =0 13.65 X 10- 5 - 14.18 X 10- 5 -+ 1.96)( '10- 5 - 1.42 X 10- 5=0

15.61 - 15.60:::: 0- . 1

Hence, S = 6250 B:::: 15.3 m D = 2.01 m.

Example 14.5. Find the channel section and discharge Q that can be aI/OWed to flow in i~ if BID = 5.7 , Bed slope = 5~ and N = 0.0225. Use Kennedy's theory.

Solution BID =5.7

Hence, B = 5.7 0 . For a channel with side slope ~: 1

R _~ _ B D + D2/2 _ 5.7 D2 + 0.5 d - P - B + V5 D - 5.7 D + 2.23D

0.2D2 = 7.93 D = 0.781 D.

Assume critical veloctiy ratio m = 1. Then velocity V = 0.55Do . 64 by Kennedy's formula

= c..fiiS by Kutter'S formula 23 + 1.. + 0.00155

Hence, 0.55 DO.64 = ,N 0001;5 N VRs ~+ (23+ . S )V1f Putting N = 0.0225

23 + 1 R.H.S. = - o:o22s + 0.00l~5 x 5000 1 + [23 + 0.00155 x 5000] 00225 Yo.781 DS

75 1 V'o'.781 15

. 15 ,.---1 + 30.75 x 0.0225 vO.78WS

0.885 :.;t5 =

IRRIGATION CHANNELS: SILT THEORIES 643

_ 75.15 X 0.885 DI/2 - (1 + 0.781 D."I/~)70.7

Equating L.H.S. and R.M.S. we get 70.7 x 0.55 DO.64 + 0.55 X 0.781 x 70.7 D.l4 = 66.5 Dli2

38.88Do,64 - 66.5 D.s + 30.31 VO. 14 = ,0. Solving the above equation by ' trial and error, we get .

D = 1.7 m ,Hence, B = 5.7 x 1.7 = 9.69 m

V = 0.55 x 1 x (1.7),64 = 0.773 mlsec

or A =BD+D212 '2 2

= 6.2 D = 6.2 (1.7) = 17.91 m2

Discharge Q =A V= 17.91 x 0.773 = 13.85 cumecs.

EJample 14.6. Desig7: an irrigation channel in alluvial so~1 according to Lacey's silt theory, given Jhe following data :

Full supply discharge = 15 m J I sec Lacey's silt factor = 1.0 Channel side slopes =! : 1.

Solution 1. Velocity V in the channel

_ r ~ ],116 _ 15 x 1.1 i 16 ' - l 140 - [ 140 J = 0.69 m/sec.

2. Area A of the channel section

=~ = Q~~9 = 21.75 m2.

3. Hydraulic 'mean radius, R

4 .

... 5 V 2 _ 5.(0.69)2 -27- ' 2,' =: 1.19 m.

A . R = 1.19~, B + D..fS

~ '

644 IRRIGATION (l{ANNIiu : SILT ntEORrnS

21.75 ... B+D{S _~ 21.75

or ' B + D v 5 = 1.19 = 18.25 I ... (i) ... (U)

and A = BD + D2/ 2 = 21.75 Solving equation (1) and (it), We get

B = 15.1 m D = 1.38 m P=4.75VQ

.. 4.75{f5 = 18.3 m 5. Check

Also, ' P = B+ D..rs == 18.25. Note. In example 14.3, first A and P were foun~ out from

which 'B and D were determined. A check was made 6y finding out the value of R by Lacey's formul~ as well as from the values of Band D. In this example, first A and R were found out' in the order to determine Band D. A numerical check has been made in step (5) by finding values of p ' from the calculated values of Band D and from that of Lacey's formula. Both the methods are correct and students can follow any procedure.

1 1 4. Slope 5 = 3340 (Q)1/6 = 3340 (15)1 /'"

1 =,5'260' .

646 lSRIGATION CHANNELS: SILT ntEORIES

= 0.681 kglm2

fer = 0.047 (y - y ... ) d = 0.047 (1.67) x 103 x 3 x ' 10- 4 kg/m2

= 0.0235.

Hence qs = 4700 [0.681 X 0.668 - 0.0235)3/2

= 1330 kg/m/hour. \ Quantity of bed load moved according to Einstein equation

[Ref. Eq. 14.29 and 14.3O} : For uniform soils .

a-D d '1'. = 'I' = :::--L--

. ,p R'S

= (Q=.!)L 1 R'S Since G = alp.

But R' is given by the formula

(n' )3/2

R'=R -n where n' , and n are ;IS denned in Eq. 14.28 and R' is as

defi-,ed in Eq.14.32. For wide channel R = D = 3 m

and ('" , 3/2

. n) =0.668 previously.

, 312 R' = R (~) =: 3 x 0.668

as . calculated

=,2.004 ' m 'I' = 2.65 - 1 x 3 x 4400 x 10 - 4

. 1 2.004 x 1 = 1.085.

From Einstein curves. (Fig. 14.3) For 'P. = 'I" = 1.085, we get

'P. = IP = 8.0

AlsO. _q~ ( U )1/2 (' 1 )1/2 IP-- - -0, 0- P gd1 = q~ (~)1I2 (_1 )1/2

0, G - 1 gd3

IRRIGATION CHANNELS: SILTrnEORIES "";.": ....

where

-=--

:: -.!lL' (_1 ) 1/2 (_1 ) 112 Gr... G - 1 gd1

q, = 8.0 x 2.65 x 103"; 1.65 x [9.81 x ,32t 2 x 10 - 12/7 = 1595 ~g/m/hour.

Hence. rate of bed load transport : By Meyer-Peter equation = 1330 kg/m/hour. By ' Einstein equation = 1595 qlJnlhour . Concentration of suspended load .

C. = 400 ppm at 0.3. m above bed = 400 X 10- 6 X 103 .= 0.4kg/m2

W = Fall velocity = 0.03 m/sec.

V. = Shear velocity

=~ =";gR'S P

R' = 2.004 m

V. - V 9.81x2.004 - 4000 . - 0.0668 mlsec K = 0.4::0 Von Karman's COnstant

W 0.03 V.I( = Q.0668 x 0.4

= 1.121. ,From q. 14.27. .. .

C - fQ::..f. _a ] V:K C; . ' y d-a = 3 - Y 0.3] 1 .12~.

.y 0.7 C = 0.0342 [3/y - 1)1.121

.: 0.0342 x [3 - 1)1.121 '. = 0.0342 x 2.175 = 0.0745 = 74.5 ,ppm.:

.I

641

+48 IRRIOATION CHANNELS~ SlLTnlEQIU1!S

PROBLEMS

.. ' . 1. Describe t~s .5I1t iheory. 'A cbannel is siltin8 badly in the ~cb. How. woukl you proceed to determine its cause and wbat remedics would you &ugest.

2. Explain the procedure .of dcsigninl a channel with ICCnocdy's theory. 3. For a cbannel, the discharge (Q), rugo&ity (N), critiCal velocity ratiO.

(m) and the bed widtb~eptb ratio (BID) are JiYCD. &plain boW WO\lld you design the chanoel usin& \ Kennedy's theory 1 '

, . ~. Derive an expression for tbc aiIt ~pporting c:apIdty of a cbaDnCl , ~,to Kennedy's theOry.

~ Yo &plain Lacey's silt tbeory. ' 6. Deacribe the mCtbod of desipliDl a canal ~ 011 Llrey's theorY.

, 7. Compare ~s aod LICey's lilt tbCOicI. why .. II Lacey's conceptio(l superior to that of Keaacdy'l ? '

8. Wbatdo you U1MSentIDd by (eI) rqime c:hIDDCIs, (b) initial and permanent regime of cbaDncts ? .

9. U&ing LIccy'I theory, delip u , irfi&IdOO ch8,ane1 (or the foIkJWing data : .

ImdJarIC Q - SO c:umeca Sift r.;tor f- 1 Side IIopeS - !: 1 [AM _ 29:6 m ; D - 1.98 , S 1/6400) 10 UW, I..IiCey'I t.ic reJime cquatklDS. derie' .. expession for

Lacey'a ICOUf depth. . . . , 11. WIth tbe belp of a.ie rqimc cquatiODl JM5h by La'cey, ~rive

tile ~-dilcbarF rdatiCJIIIbip. . 12. 1bc IIope of a c:banDel1a ... d II 1/5900. F~ the cbannct .

ICCtiOO and the maximum, dIIcbIrF "'** caD be aDoMd to' aow In it. Tate . lJM#'I lilt rIctor ,-~. 1be dIIIf!l~ ,~~'~ .... aide IIope& !: 1. l--- b %Z.3 m, ~ - 1.97 m, Q - 30 c:umec!I)

13. uliDl Kamedy'l u.i" deIinI i c:bIImd ~. (or tbe f0Q0wi08 data :

DiIoI:M'F .. Q 14 cumccs Kutten N O.021S CrticII' wb:Ily raiio~ 1 ' Side ~.!:1 Bed tIIDpe (AM. b.~ 9.7 m ; II. 1.711D) ' 1 ' -~

-

G Design Procedure for a~

Irrigation Channel '

~

15.1. LONGInJDINAL SECfION OF CANAL After baving fixed tbe alfgament of canal on sbajra sheet u

iDdicated in chapter '13, the longitudinal section of the alignment is taken on the field. The lOngitud~ settlon should therafter be, plotted to a borizontal scale of 1 an = 160 m ' and to vertical scale of ~ em = 1 m. Vertical scale sbould . be changed if desired accordin, to tbe magnitude of faU available. .

For drawing 'longitudinal section, following steps are generaUy . followed : .

. (1) Plot the ground level along the alignment of tbe channel with the ~ference to convenient datum.

(2) Mark the fuU supply level (F.s.L.) and the bed level of the parent channel just on the upstream of head regulator of the oft'tating channel for reference purposes.

(3) Draw tbe F.s.L. of : ~e otr-taking channel keeping foUowing i-