0
6.0kviews
Drawbacks in Kennedy Theory
1 Answer
0
44views

Drawbacks in Kennedy Theory

Kennedy’s theory suffers from the drawbacks (i) It does not take cognisance of the width or the shape of the channel which have to be assumed, i.e., importance of bed width and depth ratio is ignored, (ii) It involves use of Chezy’s formula and Kutter’s N for working out mean velocity and as such incorporates limitations of those relations in this theory, (iii) Adoption of arbitary value of N as 0.0225 is not correct, (iv) Design of only average regime channel was aimed at, (v) Did not specify regime water surface slope realtion. His diagram, however, show that steeper slopes are required for small channel and flatter ones for large channels, (vi) Silt concentration and bed load were not considered, (vii) Silt grade and silt charge were not defined,(viii) Design of channel involves trial and error as the velocity worked out with assumed depth should give the required discharge for the section and at the same time satisfy Kenndey’s equation, and (ix) Simply stated that CVR varies according to the silt charge and silt grade no method to measure value of n or CVR applicable for channels of different silt grades.

Design of Canals by Lacey's Theory

Discharge and mean diameter if silt (mm) are given.

Design Procedure: (i) Fix the value of silt factor f, i.e., $f=1.76m^0{}^.{}^5$, (ii)Calculate velocity, $V=(Qf^2/140)^1{}^/{}^6$,(iii)Calculate hydraulic mean depth, $R=2.5V^2/f$,(iv) Calculate area, A=Q/V, (v) Calculate wetted perimeter, $P=4.75Q^1{}^/{}^2$, (vi) Assuming side slope 1/2:1, calculate bed width b and depth D, A and P being known:

A=$BD+0.5D^2$,$P=B+\sqrt{5}D$, and $A=R \times P$

or alternatively, $D=\frac{P-\sqrt{P^2-6.944A}}{3.472}$

and, $B=P-\sqrt{5}D$

(vii) Calculate $R=\frac{BD+0.5D^2}{B+\sqrt{5}D}$

It should tally with the value obtained in step (iii) above.

(viii) Calculate slope, $S=0.0003f^5{}^/{}^3/Q^1{}^/{}^6$.

Alternatively, discharge and water surface slope ( within the range of Lacey's f=1.0 to 0.8) are given. The Lacey's f is first determined from the known discharge and slope by the relation, $S=0.0003F^5{}^/{}^3/Q^1{}^/{}^6$. The remaining procedure is as above.

Drawbacks in Kennedy And Lacey Theories

Both the well known theories are deficient in the respects (i) Both aimed to fined out average regime conditions but none took cognisance of the effect of varied silt conductive powers of outlets and offtakes on the regime of the channel, (ii) Both ignored the effect of silt attrition, (iii) Both did not consider scale effect. River and a minor or a distributary may have Lacey's silt factor unity or Kennedy's CVR unity but in fact they carry silt charge and silt grade many times different, and (v) Both did not precisely define the silt charge and the channel size for unity silt factor or unit CVR.

Solved Examples

Example 5.1 Design a channel by Kennedy Formula, discharge =15cumecs, slope =0.0002, n=0.0225 and m=1.

Solution Assume FS depth =1.75m

Kennedy $V_0 =0.55mD^0{}^.{}^6{}^4$

= $0.55\times 1 \times (1.75)^0{}^.{}^6{}^4=0.786m/s$.

A$=Q/V = 15/0.786 = 19.06m^2$

Also A=$(B+0.5D)D$, assuming 1/2:1 side slope

19.06$= (B+0.5 \times 1.75)1.75$

R=$A/P = 19.06/13.91 = 1.37m$

Kutter's $C=\frac{23+(\frac{0.00155}{0.0002}+\frac{2}{0.0225})}{1+(23+\frac{0.00155}{0.0002})\frac{0.0225}{\sqrt{1.37}}} =47.3$

Also $V=C\sqrt{RS} =47.3\sqrt{137 \times 0.0002} =0.783m/s$

which is almost the same as worked out earlier, hence acceptable. Therefore, Bed Width =10m, FS depth =1.75m and Side slope =1/2:1.

Altenative method:

For 15 cumecs discharge, b/d ratio read from Fig. 5.4=5.75

Bed Width $=5.75 \times 1.75 = 10m$

Area $=(10+2.236 \times 1.75)1.75=19.03m^2$

P$=(10+2.236 \times 1.75)=13.9m$

R$=A/P=19.03/13.9=1.37$ C$=\frac{23+(\frac{1}{0.0225}+\frac{0.00155}{0.0002})}{1+(23+\frac{0.00155}{0.0002})\frac{0.0225}{\sqrt{1.37}}} =47.26$

$V=C\sqrt{RS} =47.26\sqrt{1.37 \times 0.0002} =0.78m/s$

Q=A.V=$19.03 \times 0.78 =156 cumecs$

C.V.R. =$V/V_0 = 0.78/0.786=0.99$

Example 5.2 An irrigation canal having side slope 1:1 has bottom width of 3 m. It runs at a depth of 1m with a bed slope of 1 in 2500. Manning's value of n=0.025. Determine whether the canal will be sitting or scouring or remain stable. Use Manning's Kennedy's Equations.

Solution $A=(3+1)1=4m^2$

P$=3+2\sqrt{2}=5.83m$

R$=A/P=4/5.83=0.69m$

By Mannig's equation $V=\frac{1}{n}R^2{}^/{}^3S^1{}^/{}^2$

V$=\frac{1}{0.028}(0.69)^2{}^/{}^3(1/2500)^1{}^/{}^2=0.558m/s$.

Kenndey's non-silting, non-scouring velocity, assuming m=1.0

$V_0=0.55D^0{}^.{}^6{}^4$,$=0.55(1.0)^0{}^.{}^6{}^4=0.55m/s$.

Since the velocity as per Manning's formula is approximately the same as Kennedy's non-silting, non-scouring velocity, the channel is stable.

Example 5.3 Design an irrigation canal in clayey alluvial soil for full supply discharge =35 cumecs, coefficient of roughness 0.025, canal side slope 1:1, longitudinal slope 1 m in 5000. Also check for critical velocity ratio, allowable CVR is 0.9 to 1.1.

Solution Assume depth =2.0m

As per Kennedy's formula $V_0=0.55mD^0{}^.{}^6{}^4$, assuming m=1.0

$V_0=0.55(2.0)^0{}^.{}^6{}^4=0.86m/s$.

Now, A=40.70=(B+2)2

or B=18.35, say 18.5m

P=B+2.828D

P$=18.5+2.828 \times 2 =24.16m$

R=A/P =40.70/24.16=1.68

As per Kutter's formula,C$=\frac{23+(\frac{1}{0.0225}+\frac{0.00155}{0.0002})}{1+(23+\frac{0.00155}{0.0002})\frac{0.025}{\sqrt{1.68}}} =44.50$

$V=C\sqrt{RS} =44.50\sqrt{1.68 \times 0.0002} =0.82m/s$

CVR=$V/V_0=0.82/0.86=0.95$

which is within the allowable value of 0.9 to 1.1, hence O.K.

Example 5.4 Design the canal in example 5.3 by Lacey theory.

Solution $S=f^5{}^/{}^3/3340Q^1{}^/{}^6$

$1/5000=f^5{}^/{}^3/3340(35)^1{}^/{}^6$

or f=1.12

Also $N=0.0225f^1{}^/{}^4$

0.025$=0.0225f^1{}^/{}^4$

or f=1.52

Thus, we have got two different values of f. Further it is stipulated that CVR Allowable is 0.9 to 1.1.

But $f=(V/V_0)^2) or (0.9)^2 to (1.1)^2$, which implies that the value of f as 1.12 is acceptable.

Now, $V=(\frac{Q_f^2}{140})^1{}^/{}^6=[\frac{35(1.12)^2}{140}]^1{}^/{}^6 =0.82m/s$

A$=Q/V=35/0.82=42.68m^2$

R$=2.5V^2/f=2.5(0.82)^2/1.12=1.5m$

Also, A=PR

or 42.68$=P \times 1.5$ or P=28.45m

ALso 42.68$=BD+D^2$ with 1:1 side slope ---(i)

And 28.45=B+2.828D ---(ii)

From (i) and (ii) D=1.68m

From (i) with D=1.68, B=23.75m

N$=0.0225f^1{}^/{}^4=0.0225(1.12)^1{}^/{}^4 = 0.02315$

Channel by Lacey is shallow, and wide 1.68 x 23.75m against 2.0 x 18.5 m by Kennedy formula in example 5.3.

Example 5.5 Design a canal by Lacey's theory for 40 cumecs discharge, and f=0.9.

Solution R=0.47(40/0.9))^1{}^/{}^3=1.66m

P$=4.75(40)^0{}^.{}^5=30m$

D=$\frac{P-\sqrt{P^2-(6.944 \times R \times P)}}{3.472}$

D=$\frac{30.0-\sqrt{30.0^2-(6.944 \times 1.66 \times 30)}}{3.472}$

D=1.86m

B=P-2.236D with 1/2:1 side slope

=30.0-2.236 x 1.86

=25.84 Say 26m

V$=[40(0.9)^2/140]^1{}^/{}^6$

=0.78m/s

$S=(0.9)^5{}^/{}^3/3340(40)^1{}^/{}^6=1/7356$

Example 5.6 Design an irrigation channel in non-alluvial soil by Chezy-Kutter and Kennedy formula for a discharge of 30 cumecs, n =0.02, slope 0.000125 and side slope 1/2:1.

Solution Assume bed width 16m and depth 2.15m

Cross sectional area, A=(16+1.075)2.15=36.71$m^2$

Wetted perimeter, P=(16+1.118 x 2.15 x 2) =20.81m

Hydraulic mean depth, R=A/P=36.71/20.81=1.764m

Kutter's C$=\frac{23+(\frac{1}{N}+\frac{0.00155}{S})}{1+(23+\frac{0.00155}{S})\frac{N}{\sqrt{R}}}$

C$=\frac{23+(\frac{1}{0.02}+\frac{0.00155}{0.000125})}{1+(23+\frac{0.00155}{0.000125})\frac{0.02}{\sqrt{1.764}}} =55.78$ Chezy, $V=C\sqrt{RS} =55.78\sqrt{17.64 \times 0.000125}$

V=0.83m/s.

Now, Q=A.V=36.71 x 0.83=30 cumecs

Critical velocity $V_0=0.55mD^0{}^.{}^6{}^4=0.55(2.15)^0{}^.{}^6{}^4=0.90m/s$

Critical velocity ratio, CVR = 0.83/0.90=0.92

Please log in to add an answer.

#### Continue reading...

The best way to discover useful content is by searching it.