INTERCEPTION DRAINS, WATERWAYS AND INTERCEPTION DRAINS, WATERWAYS AND WEIRSWEIRS

Interception drainsInterception drains• also - cut-off drains, stormwater drainsalso - cut-off drains, stormwater drains• "standard" design is dangerous"standard" design is dangerous• estimation of peak runoff, especially where estimation of peak runoff, especially where

there are unqualified field staff, is most there are unqualified field staff, is most difficult aspect of designdifficult aspect of design

• Can use Hudson's (if anything, it errs on side Can use Hudson's (if anything, it errs on side of over-design) or similar simple methodof over-design) or similar simple method

Simplified approach to design of ditches where Simplified approach to design of ditches where bottom width equals depthbottom width equals depth

Determine cross section area.Determine cross section area.

velocity safe .Maxdischarge Peakarea tionseccross Minimum

Determine side slope and maximum velocityDetermine side slope and maximum velocity

Use following table or data from handoutUse following table or data from handout

Depth = h = bottom widthDepth = h = bottom width

hh

hh

side slopeside slope(horizontal / vertical)(horizontal / vertical)= z= z

Decide on Manning's n valueDecide on Manning's n value

A more detailed list of maximum velocities and A more detailed list of maximum velocities and Manning’s n can be printed out from the Excel Manning’s n can be printed out from the Excel module on Manning's equationmodule on Manning's equation

Determine minimum depthDetermine minimum depthCross-sectional area, A is given byCross-sectional area, A is given by

)1(2 zhA

where h is the depth and where h is the depth and side slope (horizontal : vertical) is z:1side slope (horizontal : vertical) is z:1This is shown for a range of side slopes in next This is shown for a range of side slopes in next 2 graphs. Use them to estimate the depth for a 2 graphs. Use them to estimate the depth for a required area.required area.

Graph of Area v. depthGraph of Area v. depth

Figure 1A

Figure 1B

Determining the minimum hydraulic radiusDetermining the minimum hydraulic radius

For the given situation, R is given by:For the given situation, R is given by:

))2h/Sin(+(h)1(2

=R zh

where where q q is the angle of the side slope. is the angle of the side slope. R can be estimated from Figure 2.R can be estimated from Figure 2.

The only unknown now is the required slope of The only unknown now is the required slope of the channel.the channel.To do this we first have to find the hydraulic To do this we first have to find the hydraulic radius from the depth and side slope.radius from the depth and side slope.

Figure 2. Hydraulic radius v. depth

Determine the channel slopeDetermine the channel slope

By rearranging Manning's equation, the slope By rearranging Manning's equation, the slope can be calculated from :can be calculated from :

2

32R

vnS

If the velocity is at the maximum, the slope is If the velocity is at the maximum, the slope is the only unknown.the only unknown.If the velocity is allowed to be less than the If the velocity is allowed to be less than the maximum, the required cross-sectional area maximum, the required cross-sectional area will increase, and so will the required R value. will increase, and so will the required R value. Thus, the maximum slope (and minimum cross-Thus, the maximum slope (and minimum cross-sectional area) is given when the velocity is the sectional area) is given when the velocity is the maximum allowed.maximum allowed.

Maximum grade for channel terracesMaximum grade for channel terraces

Thomas (1997) suggests the following Thomas (1997) suggests the following maximummaximum

slopes for diversion ditches:-slopes for diversion ditches:-

light subsoil (sand, silt, sandy loam)light subsoil (sand, silt, sandy loam) 0.1 to 0.2 0.1 to 0.2 %%

heavy subsoil (clay and clay loam)heavy subsoil (clay and clay loam) 0.4 0.4 to 0.5%to 0.5%

ExampleExamplePeak flow = 2 mPeak flow = 2 m33secsec-1-1

Maximum velocity = 1.5 m secMaximum velocity = 1.5 m sec-1-1, , side slope is 1:1 side slope is 1:1 n = 0.03n = 0.03Thus, minimum cross-sectional area Thus, minimum cross-sectional area = = 2/1.5 2/1.5

= 1.33 m= 1.33 m22

From Figure 1, the minimum cross-sectional From Figure 1, the minimum cross-sectional area corresponds to a depth of 0.8 m.area corresponds to a depth of 0.8 m.From Figure 2, this corresponds to a hydraulic From Figure 2, this corresponds to a hydraulic radius of about 0.43.radius of about 0.43.

2

0.430.03 x 5.10.67S

From the equation for the slopeFrom the equation for the slope

i.e. 0.63%i.e. 0.63%

In practice, slopes are limited to 0.5% for small In practice, slopes are limited to 0.5% for small ditches like this. ditches like this.

If we took a velocity of 1.0, If we took a velocity of 1.0, the required cross-sectional areas would have the required cross-sectional areas would have been 2 mbeen 2 m22; ; the depth would have been 1 m, the depth would have been 1 m, the hydraulic radius would have been 0.54 and the hydraulic radius would have been 0.54 and the slope would be 0.2 %. the slope would be 0.2 %.

A grade of 0.5% and a depth of 1 m would A grade of 0.5% and a depth of 1 m would therefore work OK because as the slope therefore work OK because as the slope increases the depth decreases. increases the depth decreases.

The relationship between the allowable slopes The relationship between the allowable slopes and allowable depths for this example is given and allowable depths for this example is given in Figure 3.in Figure 3.

Figure 3. Relationship between slope and depth for the given example

A triangular cross-section - common for channel A triangular cross-section - common for channel terraces has a cross-sectional area, A ofterraces has a cross-sectional area, A of

zh2

A where h is the centre depth and z:1 where h is the centre depth and z:1 (horizontal/vertical) is the side slope of the (horizontal/vertical) is the side slope of the channel (given for different soil types).channel (given for different soil types).Again, the minimum cross-sectional area occurs Again, the minimum cross-sectional area occurs when the velocity is a maximum.when the velocity is a maximum.

Another special case - triangular cross-sectionAnother special case - triangular cross-section

21

)zA(h minmin Thus the minimum depth: Thus the minimum depth:

The hydraulic radius (csa/perimeter) for the The hydraulic radius (csa/perimeter) for the triangular section can be shown to be:triangular section can be shown to be:

212 )z1(2

hr

Note - when h is a minimum, r is a minimum and Note - when h is a minimum, r is a minimum and v is a maximum. So hv is a maximum. So hminmin can also be written as: can also be written as:

min2

min r)z1(2h 21

From From

so we can replace this in the equation for hso we can replace this in the equation for hminmin::

21

23

21

max )zA())(z1(2h mins

nv2min

So :So :

32

min3

2

22max

342

Az

nv)z1(52.2s

So:So:

General case for trapeziodal cross-section channelsGeneral case for trapeziodal cross-section channels

Referring to Figure, the cross-section area, A is Referring to Figure, the cross-section area, A is given by:given by:

A d w d z 2

R w d d z

d w d z

2 2 2 0 5

2

( ( ) ) .

We need to solve this for s, but there are an We need to solve this for s, but there are an infinity of combinations of depth and width. infinity of combinations of depth and width. Trying to solve this analytically soon becomes Trying to solve this analytically soon becomes impossible and the text books recommend a impossible and the text books recommend a trial and error approach. trial and error approach.

As before, Manning’s equation is:As before, Manning’s equation is:

Fortunately computers allow us to do lots of Fortunately computers allow us to do lots of calculations rapidly and there is an Excel calculations rapidly and there is an Excel module that can be downloaded that enables module that can be downloaded that enables various solutions to be selected.various solutions to be selected.

Demonstration

WaterwaysWaterwaysIf interception drain cannot be dischargedIf interception drain cannot be dischargeddirectly, construct waterway and take waterdirectly, construct waterway and take waterdirectly downhilldirectly downhillBecause some soil borrowed from side of bund,Because some soil borrowed from side of bund,construct short bunds at right anglesconstruct short bunds at right anglesShould be established with grass (or at least Should be established with grass (or at least

somesomeform of vegetation) before discharge of waterform of vegetation) before discharge of waterfrom drain.from drain.

Velocity of water flowing in a channel is related Velocity of water flowing in a channel is related to:to:

• slopeslope• roughnessroughness• shapeshape

Manning's equation for shallow flowsManning's equation for shallow flowsUsed for calculating flow in any channel, but Used for calculating flow in any channel, but here specifically adapted for calculation of here specifically adapted for calculation of waterway dimensionwaterway dimension

nRs 3

25.0

v

s = slope (s in 1)s = slope (s in 1)n = roughness factorn = roughness factorR = hydraulic radius of channel R = hydraulic radius of channel

(1)(1)

perimeter wettedoflength areasection -cross

Rearranging (1):Rearranging (1):

21

32

S

vnR

and soand so

4

3

23

S

vnR

Must find solution to this where s is slope of Must find solution to this where s is slope of land, v is maximum safe velocity or lessland, v is maximum safe velocity or lessWe can simplifiy this because the width of We can simplifiy this because the width of waterways will normally be large compared waterways will normally be large compared with depth.with depth.

hWWhR 2

Ignoring h compared to W on bottom, R is Ignoring h compared to W on bottom, R is approximately = happroximately = h

(This is why for overland flow, the hydraulic (This is why for overland flow, the hydraulic radius is simply the depth of water.)radius is simply the depth of water.)

Discharge per unit width of waterway will then Discharge per unit width of waterway will then be given by:be given by:

43

23

25

s

nvvhq

Obtain n & v from tables as before.Obtain n & v from tables as before.

ExampleExampleFind depth and width of waterway to discharge Find depth and width of waterway to discharge 500 l/sec down slope of 3% over silt loam with 500 l/sec down slope of 3% over silt loam with moderate vegetation cover.moderate vegetation cover.Maximum permissible velocity is 0.75 m/sec; n Maximum permissible velocity is 0.75 m/sec; n is 0.033. Slope as decimal is 0.03, so depth of is 0.033. Slope as decimal is 0.03, so depth of flow is given by: flow is given by:

75.0

5.1

)03.0(

)033.0x75.0(h

= = 0.054 m0.054 m

Thus q Thus q = = 0.054 x 0.75 cu/m/sec/m0.054 x 0.75 cu/m/sec/m

== 0.0405 cu. m/sec/m = 40.5 l/sec/m0.0405 cu. m/sec/m = 40.5 l/sec/m

Thus waterway should be 500/40.5 = 12.3 m andThus waterway should be 500/40.5 = 12.3 m anddepth should be 5.4 cm. depth should be 5.4 cm.

A safety margin of at least another 10 cm should A safety margin of at least another 10 cm should always be added to the depth of a waterway always be added to the depth of a waterway and so in this case the bund height should be and so in this case the bund height should be 19 cm. 19 cm.

Often the calculated width will be quite wide. If Often the calculated width will be quite wide. If there is a lot of land pressure, the width can there is a lot of land pressure, the width can be reduced by constructing check dams across be reduced by constructing check dams across the channel to reduce the slope or using lock the channel to reduce the slope or using lock and spill drains instead.and spill drains instead.

• Or split into 2 - one on either edge of farmOr split into 2 - one on either edge of farm

• By using thick vegetation such as napier By using thick vegetation such as napier grass with a higher n, the width could have grass with a higher n, the width could have been reduced considerably (with a been reduced considerably (with a corresponding increase in the allowable corresponding increase in the allowable depth of water)depth of water)

• If check dams are used, it would be possible If check dams are used, it would be possible to cultivate the waterway, creating a kind of to cultivate the waterway, creating a kind of basin irrigation, e.g. for high value crop, basin irrigation, e.g. for high value crop, preferably with good ground cover.preferably with good ground cover.

A more normal cross-section of waterway…A more normal cross-section of waterway…

Nalah plugs can function as waterwaysNalah plugs can function as waterways

Broad crested weirsBroad crested weirs

Flow over a broad crested weir such as one you mayFlow over a broad crested weir such as one you mayhave in a field bund to discharge surplus flow ishave in a field bund to discharge surplus flow isgiven by:given by:

Q = 1.7 B hQ = 1.7 B h1.51.5

where where B is the width of the weir (m);B is the width of the weir (m);h is the depth of flow over the weir (m); h is the depth of flow over the weir (m); Q is the discharge (mQ is the discharge (m33/sec)/sec)

Lock and spill drainsLock and spill drains