chap5-2 - lacey
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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
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