Lacey’s Regime Lacey’s Regime TheoryTheoryGerald Lacey -- 1930

Lacey followed Lindley’s hypothesis:“dimensions and slope of a channel to carry a given discharge and silt load in easily erodable soil are uniquily determined by nature”.

According to Lacey:“Silt is kept in suspension by the vertical component of eddies generated at all points of forces normal to the wetted perimeter”.

Regime Channel“A channel is said to in regime, if there is neither silting nor scouring”. According to Lacey there may be three regime conditions: (i) True regime; (ii) Initial regime; and (iii) Final regime.

(1)True regime

A channel shall be in 'true regime' if the following conditions are satisfied:

(i) Discharge is constant;(ii) Flow is uniform;(iii) Silt charge is constant; i.e. the amount of silt is constant;(iv) Silt grade is constant; i.e., the type and size of silt is always the same; and (v) Channel is flowing through a material which can be scoured as easily as it can be deposited (such soil is known as incoherent alluvium), and is of the same grade as is transported.

But in practice, all these conditions can never be satisfied. And, therefore, artificial channels can never be in 'true regime’; they can either be in initial regime or final regime.

(ii) Initial regime bed slope of a channel varies cross-section or wetted perimeter remains unaffected

(iii) Final regime all the variables such as perimeter, depth, slope, etc. are

equally free to vary and achieve permanent stability, called Final Regime.

In such a channel,The coarser the silt, the flatter is the semi-ellipse.The finer the silt, the more nearly the section attains a semi-circle.

Lacey’s Equations:Lacey’s Equations:Fundamental Equations:

Derived Equations:

(Lacey’s Non-regime flow equation)

RVffRV

2

25or

52

52 140VAf

21

32

8.10 SRV

QP 75.4

61

2

140

QfV

21

23

4980R

fS

61

35

3340Q

fS

21

431 SR

NV

a

Lacey’s Channel Design Lacey’s Channel Design ProcedureProcedure

Problem:Design an irrigation channel in alluvial soil from following data using Lacey’s theory:Discharge = 15.0 cumec; Lacey’s silt factor = 1.0; Side slope = ½ : 1Solution:

sec/ 689.0)140

115()140

( 61

612

mQfV

2 77.21689.015 m

VQA

m 18.4 1575.475.4 QP

m 36.1742.3

)77.21(944.6)4.18(4.18742.3

944.6 22

APPD

m 36.15)36.1(54.185 DPB

m 185.11

)689.0(25

25 22

fVR

52451

)15(3340

)1(

3340 61

35

61

35

Q

fS

Problem:The slope of an irrigation channel is 0.2 per thousand. Lacey’s silt factor = 1.0, channel side slope = ½ : 1. Find the full supply discharge and dimensions of the channel.Data:S = 0.2/1000 = (0.2 x 5) / (1000 x 5) = 1/5000Solution:

cumecS

fQQ

fS 25.115000

133401

33403340

63

5

61

35

mS

fRR

fS 008.15000

149801)

4980(

4980

2

22

3

21

23

mQP 93.1525.1175.475.4

206.16008.193.15 mPRA

m 153.1742.3

)06.16(944.6)93.15(93.15742.3

944.6 22

APPD

m 35.13)153.1(593.155 DPB

Problem:Design an earthen channel of 10 cumec capacity. The value of Lacey’s silt factor in the neighboring canal system is 0.9. General grade of the country is 1 in 8000.Data:Q = 10 cumec; f = 0.9; Sn=1/8000; B = ?; D = ?; Sreq= ?.Solution:

Which is steeper than the natural grade of the country (i.e. 1 in 8000), therefore not feasible.

m/sec 622.0140

9.010140

61

261

2

QfV

2m 08.16622.010

VQA

m 02.151075.475.4 QP

m 25.1742.3

)08.16(944.6)02.15(02.15742.3

944.6 22

APPD

m 22.12)25.1(502.155 DPB

5844

1103340

9.0

3340 61

35

61

35

Q

fSreq

Now putting S = 1/8000 in the relationship

Hence silt factor will be reduced to 0.7454 by not allowing coarser silt to enter the canal system by providing silt ejectors and silt excluders.

i.e. silt having mean diameter > 0.179 mm will not be allowed to enter the canal system.

7454.0108000133403340

3340

53

615

36

1

61

35

SQf

Q

fS

mm 179.076.176.12

5050

fDDf

Lacey's Shock Lacey's Shock TheoryTheoryLacey considered absolute rugosity coefficient Na as;

Constant andIndependent of channel dimensions.

In practice Na varies because;V-S and y-f relationships are logarithmic,Due to irregularities or mounds in the sides and bed of

the channel (ripples), pressure on front is more than the pressure on the rear.

The resistance to flow due to this difference of pressure on the two sides of the mound is called form resistance.

Lacey termed this loss as shock loss, which is different from frictional resistance or tangential drag.

Shock loss = f (size, shape and spacing)

Total resistance = frictional resistance + shock loss (due to bed) (due to irregularities)

Lacey suggested:Na should remain constantSlope should be splited

to overcome friction andto meet shock loss

i.e.

where, s = slope required to withstand shock losses.

According to LaceyNa = 0.025 with shock lossNa = 0.0225 without shock loss

Therefore, s = 0.19 S

i.e. for a channel in good condition19 % slope for shock loss

and 81 % slope for friction

214

31 sSRN

Va

214321430225.01

025.01

sSRSR