watershed development ppt

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suveing,levelling,hydrology,check dam,waste weir,contour bund, nalla bund, agronomy,economics,extension work

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Central soil and water conservation training and research institute, Bellary presentation on major topics surveying & leveling agricultural economics watershed & structures forestry soil science conservational agronomy :

Central soil and water conservation training and research institute, Bellary presentation on major topics surveying & leveling agricultural economics watershed & s tructures forestry soil science conservational agronomy

introduction:

introduction Soil and water conservation is - in its outlines-  a worldwide strategy in the context of a sustainable natural resource management. Soil and water conservation are those activities at the local level which maintain or enhance the productive capacity of the land including soil . prevention of soil erosion, compaction, salinity; conservation or drainage of water and. maintenance or improvement of soil fertility.

SURVEYING:

SURVEYING It a branch of science in which lines, angles, distance are established and measured on or near the earth surface. Types of survey: Chain Survey Compass Surve y

Survey Lines:

Survey Lines Base Line Check Line Tie Line

PowerPoint Presentation:

TYPES OF CHAIN Types of chain Gunter Chain: 66’ , No. of Links= 100 Revenue Chain: 33’ , No. of Links=16 Engineer’s Chain: 100’ , No. of Links=100 Metric Chain: 30m, No. of Links=100

Chain Survey:

Chain Survey Chain surveying the simplest method of surveying in which only linear measurements are made and no angular measurements are taken. Methods Of Chain Survey Offset Method Triangulation Method

Offset method :

Offset method ∆AGD Area=16.65m 2 □ GDEF Area=26.19m 2 ∆FEB Area=26.52m 2 ∆BHJ Area=124.7m 2 □IKJH Area=349.8m 2 ∆AKI Area=64.87m 2 Total Area=608.73m 2 B Scale 1cm=5m

Triangulation method :

Triangulation method Formula used Heron’s formula Area Of ∆ S= Where S= Semi perimeter a , b, and c are the sides of triangle

Practical work:

Practical work Area of ∆ABC=276.45m 2 Area of ∆ABD=36.96m 2 Area of ∆ ACE=227.89m 2 Total Area=541.3m 2 Scale 1cm=5m

Compass Survey:

Compass Survey

Principle of compass surveying:

Principle of compass surveying The principle of compass surveying is traversing, which involves a series of connected lines. The magnetic bearing of the lines are measured by prismatic compass and the distances of the lines are measured by the chain.

Terms Used In Compass Survey:

Terms Used In Compass Survey Fore and Back Bearing

Calculation:

Calculation BB=FB+/-180° Use the + ve sign when FB<180° Use the – ve sign when FB> 180°

PowerPoint Presentation:

Traversing : Scheme of control points consisting of a series of connected lines. Methods Of Traversing *Open Traversing *Closed Traversing

Open Traversing:

Open Traversing An open, or free traverse (link traverse), consists of a series of linked traverse lines which do not return to the starting point to form a polygon Open survey is utilized in plotting a strip of land which can then be used to plan a route in road construction

Open Traversing(Practical Work):

Open Traversing(Practical Work) Line Length in (m) Fore Bearing Back Bearing Remark AB 21.7 60 o 30’ 237 o 30 Starting point BC 47.7 30 o 211 o 30’ CD 41.9 257 o 76 o 45’ DE 39.0 345 o 164 o 30’ EF 13.66 63 o 45’ 244 o FG 21.4 325 o 30’ 144 o GH 17.77 335 o 154 o 30’ HI 30.5 307 o 70’ 127 o 30’ IJ 128.3 246 o 67 o 45’ JK 90.7 124 o 302 o 15’ End point at field boundry

Graph(open traversing):

Graph(open traversing) Scale: 1cm=10m N

Closed Survey:

Closed Survey A closed traverse (polygonal, or loop traverse) is a series of linked traverse lines where the terminal point closes at the starting point

Closed Traversing(practical work):

Closed Traversing(practical work) Line Length in (m) Fore bearing Back bearing AB 25 339 o 15’ 162 o 30’ BC 51.2 349 o 166 o 30’ CD 24 73 o 45’ 252 o 30’ DE 15 143 o 15’ 324 o 45’ EF 31.2 151 o 45’ 332 o FA 48.5 213 o 33 o

Graph closed traversing:

Graph closed traversing

LEVELLING:

LEVELLING Leveling is a branch of surveying, the object of which is to, Find the elevation of a given point with respect to the given or assumed Datum. Establish a point at a given elevation with respect to the given or assumed Datum.

Instrument Used in leveling:

Instrument Used in leveling Dumpy Level Level Staff Cross staff Chain(30m) Tape(30m) Ranging Rod

Classification of leveling:

Classification of leveling Simple Leveling: The operation of leveling for determining the difference in elevation, if not too great between two points visible from single position of the level is Known as simple Leveling Differential Leveling: this method is used in order to find the difference in elevation between two points , by changing the position of level.

PowerPoint Presentation:

Continue Fly Leveling Profile Leveling Contour Survey Direct Method Block/ Grid Method

Terms used in Leveling:

Terms used in Leveling Reduced Level(RL): Height or depth of a point above or below the assumed datum is called Reduced level Benchmark: Benchmark is a fixed reference point of known elevation. It may be of the following types

PowerPoint Presentation:

1. GTS Benchmark( Grand Trigonometric Survey) 2. Permanent Benchmark 3. Arbitrary Benchmark Mean Sea Level(M.S.L): M.S.L. is obtained by making hourly observations of the tides at any place over a period of 19 years. MSL adopted by Survey of India is now Bombay which was Karachi earlier

PowerPoint Presentation:

Height Of Instrument: The elevation of the line of sight with respect to assumed datum is known as HI. Back sight: The first sight taken on a leveling staff held at a point of known elevation. B.S enables the surveyor to obtain HI + sight. Fore sight: It is the last staff reading taken from a setting of the level.

PowerPoint Presentation:

Change Point: The point on which both the foresight and back sight are taken during the operation of leveling is called change point Intermediate Sight: The foresight taken on a leveling staff held at a point b/w two turning points, to determine the elevation of that point, is known as intermediate sight.

Adjustment Of level:

Adjustment Of level Permanent Adjustment: Required if some error is there in instrument. Temporarily Adjustment: Adjustments which are made for every setting of a level before commencing the leveling. It includes Setting up the level Leveling up Head Adjustment of foot screw Elimination of parallax

Profile leveling:

Profile leveling Profile leveling also called longitudinal leveling or sectioning, is the process of determining the elevation of a series of points measured intervals along a selected or predetermined point.

Practical work (Graphs):

Practical work (Graphs) Cross sectional View of N ala Fig 1 0 Chainage Fig 2 8.75 Chainage Fig 3 18.75 Chainage Scale : y axis 1cm=2cm X axis 1cm=1m

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Fig 4 13.38 Chainage Fig 5 18.70 Chainage Fig 6 13.1 Chainage

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Fig 7 16.9 Chainage Fig 8 13.3 Chainage Fig 9 11.4 Chainage

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Fig 10 29.10 Chainage Fig 11 30.2 Chainage

Longitudinal View of nala:

Longitudinal View of nala Scale: x axis 1cm=20m y axis 1cm=0.1m Distance Waste weir Gabion str.

Structure’s location:

Structure’s location Waste weir or loose boulder at 18.70 chainage Gabion Structure: At 13.30 chainage of 16m length because the width of nala is 16m Total length=18m, width=1m, Height of foundation=0.3m, height above the ground=0.7m In rest 16m length, put 2-2m box and join them, Fill the stones into the box according to the size that means big stones at bottom.

Contour surveying:

Contour surveying Direct Method : In this method the contours to be plotted are actually located on aground with level or hand level by making a various point on each contour. Block/Grid Method : In this method field is divided into a series of squares. The size of squares varies according to the nature of ground.

Block/grid survey(Practical work):

Block/grid survey(Practical work)

economics:

economics It is a branch of social sciences which deals with the human behaviour in relation to the optimum utilization of resources. Total Economic Value(TEV) Non-use value Use value Direct use value Consumptive Non-consumptive Indirect use value Option value Option Bequest

Identification of watershed benefits and costs.:

Identification of watershed benefits and costs. Level/tier Benefits Costs To the farmers Financial benefits Overall soil improvement Societal benefits Higher yield land portion for structures Contribution (10% of total cost) To the watershed development authority Implementation and experimenting of tech. Revenue collection Wasteland reclamation Involvement/manpower Human resources expenditure To the nation food security betterment Additional employment Environmental effects Opportunity cost of capital Depletion of non-renewable resources

Economic evaluation of watershed programs:

Economic evaluation of watershed programs It can be done with the help of following parameters. 1 Benefit Cost Ratio (BCR), 2 payback period, 3 N et P resent Worth/Value, 4 Internal Rate of Return (IRR).

PowerPoint Presentation:

Exercise 1 :-One farmer is getting R s 1000/acre /year as profit from his 5 acre land. He spent Rs20,000 for land improvement .Quality of land improved and productivity also increased. Now he is getting R s 1,500/acre/year as profit. Calculate the payback period for his investment. Sol.: Before land improvement profit (5 acre) =5*1,000 = 5000 After land improvement profit became (5 acre) =5*1500 =7,500 So , net profit generated for 5 acre land =>7500-5000= 2,500 total amount invested = 20,0000 So, payback period = total invested amount net profit generated = 20,000 2,500 = 8 years

PowerPoint Presentation:

So ,from the above problem we can deduce that after investing a sum of 20,000 on land the farmer is getting 1500/acre/year which was earlier 1000/acre/year. so he is getting R s 500/acre/year ,he is having 5 acre of land so every year he is getting extra 2500 ,so by the end of 8 years his investment money will be accrued and after that the extra money will be profits and soil condition is also improved it has intangible benefits , which cannot be calculated.

PowerPoint Presentation:

Exercise 2:- priyanka K.T. (1998) observed in her study that benefit cost ratio of the fuel wood plantation in the moderate to low rainfall areas are very low . The data regarding the total expenditure for different works carried out during different years and the revenue from the plantation ,where as follows. EXPENDITURES REVENUE Calculate the BCR considering 8% discount rate to support the result at 1996-97 period. year activities Expenditure(in Rs) 1987-88 Digging of cattle proof boundary trench 25,800.00 1987-88 Ripping of the area 22,435.00 1988-89 Plantation, watch and ward 88,985.00 1989-90 Maintenance ,watch and ward 29,500.00 1990-91 Maintenance , watch and ward 20,907.00 1996-97 Sale of plantation 3,66,618.00

PowerPoint Presentation:

Solution:- appreciated value of expenditures made in various years at the base year 1996-67. Sale price of plantation in 1996-97=3,66,618.00 BCR = Benefit/cost. Year Expenditure in respective year Appreciated value of investment in 1996-97 1987-88 48,235.00 96,421.98 1988-89 88,985.00 1,64,705.20 1989-90 29,500.00 50,557.80 1990-91 20,907.00 33,176.00 Total 3,44,860

PowerPoint Presentation:

B.C.R= = 366618/344860 =1.064 From above solution we got the value of BCR as 1.064 ,which is slightly greater than 1,this makes the project feasible and profitable. if the BCR value of any project is higher than 1 it means that the project is feasible and promises profit in future.

PowerPoint Presentation:

Exercise 3:- Following details are available for a water harvesting structure. Catchment area-1000ha Cost on structure- Rs 35 lakhs Area protected-200ha Area reclaimed-50ha Command area-400ha Reclamation cost- Rs 5,000/ha Expenses on field channels and other structures- Rs 2,000/ha Analyze the project with appropriate assumptions. Solution: Costs Structure cost = Rs 35 lakhs Reclamation cost =50*5000 =2,50,000 =2.5 lakhs Expenses on field channel=2000*400 = 8,00,000 =8 lakhs Total expenses =35+2.5+8= R s 45.5 lakhs.

PowerPoint Presentation:

Benefits Productive benefits: Increased yield=5 quintal/ha Rate =1000/quintal, area =400ha Total productive benefit=400*5*1000 =20,00,000 Protective benefits: extra returns per hectare =2,000 Area =200ha Total protective benefit=200*2,000 =4,00,000 Reclamation benefits: extra return per hectare =2,000 Area=50 ha Total reclamation benefit=50*2000 =1,00,000 Land value appreciation: Increase in command area land rate/ha=25,000

PowerPoint Presentation:

Command area =400ha Total appreciation in land value= 400*25,000 =1,00,00,000 Land appreciation for protected area=2000*1000 =20,00,000. Land appreciation for reclaimed area=50*20,000 =10,00,000. Total benefits from watershed=20+4+1+100+20+10=155 lakhs B.C.R=

PowerPoint Presentation:

BCR=155/45.5 =3.40 NPV=155-45.5=109.5 lakhs So, BCR value for the above watershed activity is 3.4,which is very high than the required value(1) so this project is feasible . T he Net Present Value (NPV) of the project is also very high so it is lucrative in terms of returns , in the above solution opportunity cost is not included , considering the opportunity cost of money the profit generated will be much higher than what is obtained now.

Watershed development :

Watershed development

Watershed development:

Watershed development Watershed area mainly has three types of land use 1.Forest area Nonarable land 2.Pasture land 3.Arable land

PowerPoint Presentation:

In arable land soil and water conservation structures: 1.Bunding 2.Water ways 3.Farm pond 4.Loose boulders 5.Waste weir

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Soil and water conservation structures between nonarable land: 1. D iversion drain 2. Nallah bund 3. Check dam

Peak run off rate estimation by Rational formula:

Peak run off rate estimation by Rational formula Q= CIA/360 Q =Peak runoff ( Cu.m/sec) C=Runoff Coefficient (Weighted mean ) I = Design Intensity of rainfall (mm/hr) for the design frequency and for duration equal to the time of concentration. A = Catchment area (Hectares)

Table of Runoff coefficient value(C) :

Table of Runoff coefficient value(C)

PowerPoint Presentation:

Runoff Coefficient (C): weighted value of “C” =(A1C1 + A2 C2 + A3C3+ ……+AnCn) / (A1+A2+A3+….+An) To calculate I (design rainfall intensity) I = 2xIo / 1+Tc where, T c = time of concentration Kirpitch formula , Tc = 0.01947x( k)⁰ · ⁷⁷ where, k = √(Lᶟ/H) L= length of nallah H= level difference Iₒ = rainfall intensity in mm/hr

Map for calculation of Iₒ:

Map for calculation of Iₒ

PowerPoint Presentation:

Type of structure Recurrence interval 1. Earthen structure like bunds, terraces , waterways, diversion drains, and dry stone works 10 years 2. Semi permanent masonary structures like small check dams , waste weir etc. 25 years 3. Permanent structures made of cement concrete and RCC and other large structures 50 years

PROBLEM: calculate the discharge of a watershed having area 120ha out of that 20ha is forest area having 11% slope and sandy loam soil condition ,10ha pasture land with 7% slope having silty loam soil and the remaining 90ha is under cultivated land with only 3%slope under clay soil the major nala length is 800m &level diff. from farthest point to end point is 50m the watershed is located an 16 ° latitude &76 ° longitude type of structure is a small check dam. :

PROBLEM: calculate the discharge of a watershed having area 120ha out of that 20ha is forest area having 11% slope and sandy loam soil condition ,10ha pasture land with 7% slope having silty loam soil and the remaining 90ha is under cultivated land with only 3%slope under clay soil the major nala length is 800m &level diff. from farthest point to end point is 50m the watershed is located an 16 ° latitude &76 ° longitude type of structure is a small check dam. Solution: Runoff coefficient(C) from table : c for forest area =0.30 c for pasture area =0.36 c for argil. Area =0.60 weighted value of C=0.3x20+0.36x10+0.6x90/120 =0.53

PowerPoint Presentation:

Design intensity(I) = 2Iₒ/1+Tc Where Tc = 0.01947× {√(Lᶟ/H)} ⁰ · ⁷⁷ = 0.01947 × {√(800ᶟ/50)} ⁰ · ⁷⁷ = 9.73min = 0.162hrs Iₒ = 60 mm/hr from rainfall intensity map for 16° latitude &76 ° longitude for 25 years R.I. So, I= 2 ×60/(1+0.162) = 103.44 mm/hr From Rational formula , Q = CIA/360 =0.53×103.44×120/360 =18.27 cum/s

Check dam:

Check dam Check dams mainly classified as: Temporary check dams, Ex.:- loose boulders Permanent check dams, Ex.:- drop spillway Check dams are used for controlling the soil erosion and runoff in small and medium sized gullies. Components of check dam:- Head wall, head wall extension, side wall, apron, wing wall, weir, end sill, cut off wall, toe wall

Check dam in netranahalli watershed:

Check dam in netranahalli watershed Length of weir

Dam height and weir height in check dam:

Dam height and weir height in check dam Dam height Weir height Head wall extension

Head wall extension of check dam:

Head wall extension of check dam Head wall extension

Length of apron of check dam:

Length of apron of check dam Length of apron Side wall

End wall of check dam:

End wall of check dam End wall

Design of check dam:

Design of check dam Peak rate of runoff, Q= CIA/360 Q for rectangular weir = 1.71LH ³′² where, L= length of weir = width of nala, m H= height of weir , m From this we can find H because L and Q is known . total height of weir = H+ free board Free board is 0.15 to 0.30m

PowerPoint Presentation:

3. Height of dam,D = nala height-weir height 4. Head wall extension = 2H+0.3 5. Length of apron = 2D 6. Height of wing wall and side wall = 2H 7. Wall thickness, head wall = 0.45m side wall= 0.3m wing wall=0.3m

PowerPoint Presentation:

Problem (cont.) : design a rectangular weir from the data given in previous problem and following data: Catchment area = 120 hac, Nallah width = 15m . Calculate head wall extension, length of apron, dam height for 3.5 m nala depth. Solution: from above problem Q= 18027 cumec Q for rectangular weir = 1.71LH ³′² 18.27 = 1.71 * 15 * H ³′² H = 0.71 m add free board(.29) , H = .71+.29 = 1.0 m Head wall extension = 2H+0.3 = 2.3m Height of dam, D = nala height – weir height = 3.5 – 1 =2.5m Length of apron = 2D = 2* 2.5 = 5 m

Diversion drain :

Diversion drain Diversion drain is excavated to intercept the runoff from the area situated above (nonarable land) for protecting arable lands down below and to conduct it safely to natural nalas. Design of diversion drain calculate total area (nonarable) in hectares . Use rational formula, discharge Q=CIA/360 for 10 years frequency .

Diversion drain :

Diversion drain Upstream side downstream berm Top width Side slope

Diversion drain in Ramsagara watershed:

Diversion drain in Ramsagara watershed

PowerPoint Presentation:

3. Q =VA area of cross-section , A= (b+zd)d where, b=bottom width , m d=depth of drain, m z=side slope top width of drain , T = b+2dz velocity of flow , V by manning’s formula = C R 2/3 S 1/2 where, C= 1/n , n= manning’s roughness coefficient R= hydraulic radius = A/P ,m

PowerPoint Presentation:

P= wetted perimeter , m = b+ 2d {√(z²+1)} S= grade of diversion drain (0.2 to 0.3 %) V should be in between 2 – 6 m/s 4. Length of drain = perimeter of hillock , m 5. Depth of diversion drain is assumed as a) d = 0.5 to 1 m in rough terrains . b) d = 1.5 to 2 m in marginal terrains. 6. Construct stabilizers(local stones) to reduce velocity when fall of bed slope is >30 cm. 7. Excavated earth is put on D/S with leaving a berm of 0.6m and vegetative barriers on U/S side

PowerPoint Presentation:

Problem :- Calculate the peak discharge and design the diversion drain in forest land having catchment area 30 ha and sandy soil . I =90mm/hr ,slope of land 20% using modified ‘C’ value. Solution: Q=CIA/360 Q=0.2x90x30/360 1 1.5 1 1 Q= 1.5 cumec assume V = 0.6m/sec non erosive velocity Q =VA 1:1 1.5=0.6A A=2.5 sqm A=( b+zd )d where, b=bottom width d=depth 1.5 2.5=( b+zd )d (assume d=1) 2.5=(b+1)1 2.5-1=b b=1.5m T = b+2dz = 1.5+2*1*1 = 3.5m

PowerPoint Presentation:

Problem : calculate the discharge and Design diversion drain from above problem data by using ‘C’ value from the table. Solution: C = 0.3 from table Q=CIA/360 Q=0.3x90x30/360 = 2.25 cumec assume V = 0.6 m/sec non erosive velocity Q = AV A = Q/V = 2.25/0.6 = 3.75 m² Now , A = (b+zd)d assume d = 1m 3.75 = (b+1)1 b = 2.75 m T = 2.75 + 2*1*1 = 4.75m

Farm pond:

Farm pond it is a water harvesting/storage structure in arable land. Types:- Embankment type. 2. Dug out type. Embankment type pond is built across the stream in areas of gentle to moderately slope. Dug out type pond are constructed by excavating the soil , relatively in level areas.

Farm pond:

Farm pond Depth gauging scale

Inlet of farm pond:

Inlet of farm pond INLET

Outlet of farm pond :

Outlet of farm pond OUTLET Stop dam

Design of dug out type farm pond:

Design of dug out type farm pond Calculate the runoff volume(V1) from catchment area(A). V1=A × d where, d= runoff depth i.e. some % of rainfall. 2. Calculate design runoff volume(V) i.e. some % of total runoff volume(V1). 3. Side slope(z:1) of farm pond:- (A) for red soil= 1.5:1 (B) for black soil= 2:1

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4. Depth(d) of farm pond can be assumed according to farm pond capacity, it should not more than 3m. 5. Bottom width(b)= b= √( 3V - d 3 Z 2 ) - dz √ 3d 6. Top width (T) = b+2dz 7. Capacity of farm pond can be determined by trapezoidal rule V = (A₁+A₂)×H/2 where , A₁ and A₂ are areas b/w 2 successive contours H = vertical interval of contours 8. Volume of excavation for construction of pond by prismodial formula V = (A+4B+C)*D/6 9. Design of inlet and outlet such as mechanical and emergency spillway.

Problem : design a farm pond in red soil region from the following information : catchment area = 5ha, mean annual rainfall = 450mm runoff = 10% of total mean annual rainfall, assume 50% of runoff collection for design , side slope can be assured 1.5:1  solution: 10% of annual rainfall = 450 x 0.10 = 45 mm total runoff volume for 45 mm from 5ha = 45/1000 x 5 x10000 =2250 cubic m design runoff volume (v) = 50% of total runoff volume = 0.50 x 2250 = 1125 cubic m then , b= √(3v - d3z2) - dz √3d :

Problem : design a farm pond in red soil region from the following information : catchment area = 5ha, mean annual rainfall = 450mm runoff = 10% of total mean annual rainfall, assume 50% of runoff collection for design , side slope can be assured 1.5:1 solution: 10% of annual rainfall = 450 x 0.10 = 45 mm total runoff volume for 45 mm from 5ha = 45/1000 x 5 x10000 =2250 cubic m design runoff volume (v) = 50% of total runoff volume = 0.50 x 2250 = 1125 cubic m then , b= √( 3v - d 3 z 2 ) - dz √3d

PowerPoint Presentation:

b = Bottom width V = Volume = 1125 mᶟ Z = Side slope = 1.5 assume d= Depth = 2.5m b= √ ( 3 x1125 – 2.5 ᶟ x 1.5²) – 2.5x1.5 √ 3 x2.5 = 21.10–3.75 = 17.35m Top width = T = b +2dz = 17.35 + 2 x 2.5 x 1.5 = 24.85m

Bunding :

Bunding It is a soil conservation measure , used for retaining the water , creating obstruction and thus to control erosion. Bunds are embankment type structures, constructed across the slope. By bunding practice entire area is divided into several small parts, there by effective slope length, thus reducing soil erosion.

Types of Bund::

Types of Bund: CONTOUR BUND: constructed on contour of area. used in relatively low rainfall (<600mm/year) area for the purpose of controlling soil erosion and to store rain water. Suitable for land having slope of 2 to 6%. Black soil is not suitable for contour bund.

Contour bund:

Contour bund Contour bund

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Bund with borrow pits, waste weir and revetment

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2. GRADED BUND: When a grade is provided to bund is called GB. Constructed in relatively medium to high RF(>700mm/year). Suitable for black soil. Purpose of controlling soil erosion and to store rain water. Suitable for land having slope of 2 to 6%.

Specification of contour bund:

Specification of contour bund Soil Type Land Slope (%) VI (m) Common Cross section (Sqm) Side slope Surplussing arrangement Deep black Upto 3 0.9 1.61 1.5:1 Waste weir Shallow black Upto 3 1.0 1.0 to 1.5 1.5:1 Waste weir Red and Lateritic Upto 3 0.5 1.0 to 1.5 1.3:1or 1.5:1 Open ends with vegetative checks

Typical spacing of contour bund:

Typical spacing of contour bund Slope % VI (m) (S/3 + 2 ) 0.3 HI (m) VI/Slope % X 100 Length of Bunds (m) (10,000)/HI 1.0% 0.70 70 145 1.5% 0.75 50 200 2.0% 0.80 40 250 2.5% 0.85 35 205 3.0% 0.90 30 335

Design of contour bund:

Design of contour bund Spacing of bund by formula Ramser’s formula VI = (S/3+2)0.3 b) USDA formula VI = (S/4+2)0.3 c) Cox formula VI = (XS+Y)0.3 where, X= rainfall factors Y= infiltration and crop cover factor

Values of X and Y for Cox formula:

Values of X and Y for Cox formula Rainfall Annual rainfall (cm) Value of X Intake Crop cover during erosive period of rains Y values Scanty 64 0.8 Below average Low coverage 1.0 Moderate 64-90 0.6 Average or above Good coverage 2.0 Heavy >90 0.4 One of above favorable & Other unfavourable Good coverage 1.5 Value of X Y values

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2. Horizontal interval (HI) = (VI/slope) x 100 3. Rainfall excess (Re) = Rainfall x % runoff 100 (Rainfall of 24 hrs, 10 yr. recurrence interval) 4. Depth of impounding (h)= (VI x Re)/50 5. Depth of temporary storage = 0.3 m 6. Free board (25% inclusive of settlement allowance) = 0. 25 (h+0.3)m 7. Total height of the bund = h + 0.25 (h +0. 3) m.

Select top width and slope of bund depending on soil type:

Select top width and slope of bund depending on soil type Type of the soil Top width (m) side slope Sandy 0.5 2:1 Loamy 0.4 1.5:1 Clayey 0.3 1:1 or 1.5:1

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9. Computation of the bottom width and cross section area ‘A’ 10. Total length of bund/ha L = 10,000 x 1. 3 HI 30% extra length of soil bunds. 11. Earthwork in bunding/ ha V = L X A V= Volume of bund, cum per ha L= Length of bund, m per ha A= C/S area of bund, sqm

Specification of graded bund:

Specification of graded bund Soil Type Slope (%) VI (m) Cross Section (Sqm) Side slope grade Black Upto 5 0.75 to 1.0 0.6 to 0.87 1.5:1 0.1 to 0.3 Red Upto 5 0.75 to 1.0 0.6 to 0.87 1.3:1 0.2 to 0.4 Lateritic 5 to 6 0.75 to 1.5 0.34 to 0.56 1.3:1 0.2 to 0.4

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Steps in design of graded bund are similar to that of contour bund.

Waste weir (WW):

Waste weir (WW) WW(surplus weirs)or rubble/grass outlets are normally provided in valley points by using loose stones properly embedded in soil to avoid scouring and to drain the excess water accumulated against bund. WW are constructed when catchment area is <40 hac. For larger catchment areas, water diversion is necessary. constructed in series from ridge to valley.

Waste weir:

Waste weir Upstream side Downstream side Length

Specification of waste weir:

Specification of waste weir width is equal to width of waterway. crest height in black soils= 15 to 20 cm in red soils = 30 to 40 cm upstream slope = 1.5:1 down stream slope =3:1 whenever open ends are used for draining excess water; the ends are to be vegetated to prevent cutting and scouring. 2m long murram or hard soil packing may be given to either ends of WW in continuation of bund.

Gabion structure:

Gabion structure gabion is a ‘Italic’ word in which small-small stones combined with G.I. wire mesh, to form a large stone and placed across the nala to control heavy flow there by silt. this structure is comparatively strong under both compression and tensile strength.

Gabion structure in Netranahalli watershed:

Gabion structure in Netranahalli watershed Length of gabion width

Gabion structure with vegetative barriers to reduce runoff velocity:

Gabion structure with vegetative barriers to reduce runoff velocity Vegetative barriers gabion

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Technical specification 1. G.I wire 10-14 gauge 2. Foundation = 0.3m 3. Height above ground = 0.70 m 4. Length inside nala = 1 m at both sides 5. Total Length of gabion = width of nala+2m 6. Wire mesh size = 3 inch so, stones should be >3 inch size 7. Spacing for 1-3% slope = 50 m for 3-5% slope = 30 m

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Gabion should be constructed maximum in a 2 m box and join them, filled with stones and tie them together. Bigger stones should be in the bottom and smaller stones (not < 3inch) at the top. Binding should be proper.

Contour trenching:

Contour trenching Contour Trench/’V’ ditches are trenches dug on contour in non-arable lands of more than 3% slope to hold run off for conservation and reducing erosion. They are established for development of trees and grass species and are adoptable in areas with annual rainfall of up to 950 mm. contour trenches have been used on all slopes, trenching on slopes exceeding 20% is not advisable either technically or economically.

Trenches are categorized in 3 types:

Trenches are categorized in 3 types 1.Continuous trenches: Continuous contour trenches are recommended for storage of water in low rainfall relatively flat areas receiving storms of mild intensity. 2. Graded trenches These are drainage type ditches for intercepting and safe disposal of surface flow in very high rainfall areas and impermeable black soils. The grade is given so that the intercepted runoff from the above will be carried safely at non-erosive velocity to the vertical drain without overflow.

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3. Staggered or interrupted trenches In high rainfall areas with highly dissected topography staggered trenches are usually adopted. Staggered trenches are of shorter lengths in a row and are arranged along the contour with inter space between them.

Earthen dam for water harvesting (NALABUND) :

Earthen dam for water harvesting (NALABUND) Nala bund is an earthen structure constructed across the nala/gully in order to store the runoff water flowing through the nala during rainy season. Objectives: reducing the velocity of flow, storing the runoff and thereby allowing it to percolate into the soil profile which in turn helps to enhance the water table of the downstream area. This structure also prevents the silt flowing down and causing the siltation of reservoirs in the downstream side, which can affect the storage capacity of the reservoir.

Site selection of nala bund:

Site selection of nala bund 1.First and foremost requirement is that it should have sufficient catchment to fetch the runoff required for storage. 2. The upstream side of the location there should have enough area for water storage. 3. The nala site selected for the structure should have a relatively narrow cross section. 4. Should be located on the straight stretch of nala. 5. There should be provision for locating surplussing weir on one of the banks. 6. The nala bed should have good hard soil for proper bondage between the structure and natural soil profile. A hard rock foundation may have less bondage with the proposed structure, hence discouraged.

Nala bund:

Nala bund

Design of nala bund:

Design of nala bund 1. Top width, W = Z/5 +3 Where: W= width of crest (m), Z=Height of embankment above the stream bed(m). 2.EMBANKMENT SIDE SLOPES: The side slope of the nala bund depends primarily on stability of the material used for embankment.

Recommended side slopes for earthen embankment:

Recommended side slopes for earthen embankment

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3.CORE WALL: The core wall is a centrally provided fairly impervious wall in the dam. 4. KEY TRENCH: This is a bondage/foundation component of the structure to ensure the stability for the embankment almost like foundation of a structure. 5. Spillway : This is a vent /channel provided at the full tank level in order to dispose of the excess runoff coming in. 6.FREE BOARD: Free board is the additional height of the bund provided to avoid water overtopping the embankment during unexpected flow of runoff 7.REVETMENT (wave protection): Since this an earthen structure and it will be coming in contact with the water in the upstream side of the dam, in order to with stand against the wave action of storage .

forestry:

forestry

Forest for conservation of natural resources:

Forest for conservation of natural resources Forest: An area set aside for the production of timber and other produce or maintain under woody vegetation. Theory and practice and creation Conservation Scientific management of forest Utilization of their resources

Conservation forestry :

Conservation forestry Production of forest product Restoration & maintenances of resource base

Conservation forestry:

Conservation forestry Its need & scope At the global level 15% of the earth’s forest & woodland disappeared during the last one and half century as a result of human activities. Its aim is to prevent erosion from the fertile agricultural land as well as production as socially acceptable uses.

The role of forest in functioning of watershed:

The role of forest in functioning of watershed Conserves soil moisture Maintain soil temperature Infiltration increases Root binding capacity increases Prevent soil erosion

Objective of agroforestry :

Objective of agroforestry To utilize available farm resource. Production of fuel, fodder, food, wood etc. Integration of trees with agricultural land and animal production. To maintain ecological balance. To check erosion hazard. To improve employment potential and rural economy.

Extension:

Extension

Pra (participatory rural apprisal:

Pra (participatory rural apprisal The PRA technique is an useful technique for use in analysis of any situation STEPS Social Mapping Resource Mapping Seasonal analysis Transect walk Preference ranking Historical time line ITK

CONSERVATION AGRONAMY:

CONSERVATION AGRONAMY

In situ soil &moisture conservation measures:

In situ soil &moisture conservation measures Tillage a)Conservation tillage b)Conventional tillage Graded furrow Vegetative barriers Repeated inter culturing Graded border strips Zing terrace Contour cultivation Compartmental bunding Tied ridges & furrows Broad furrow & ridges Scooping Border planting method

Effect of in situ moisture conservation on soil physical properties :

Effect of in situ moisture conservation on soil physical properties Soil temperature Bulk density Penetration resistance S oil compaction Soil aggregation & pore space Runoff & soil loss Nutrient losses Crop growth & yield

Definition & concept of Water shed management:

Definition & concept of Water shed management Watershed is the integration of technologies with in the natural boundaries of drainage area for optimum development of land , water, & plant resources to meet the basic needs of people & animal in sustainable manner.

Components of watershed management:

Components of watershed management Treatment of arable & non arable land for effective in situ & ex situ moisture conservation Identification of sound crop production system & its implementation through development & input agencies Developing suitable infra structure facilities & people organizations to maintain developed resources

Soil science:

Soil science

LAND CAPABILITY CLASSIFICATION :

The systematic arrangement of land into various categories according to its capability to sustain particular land use without land degradation. LAND CAPABILITY CLASSIFICATION

OBJECTIVES OF LCC:

OBJECTIVES OF LCC It makes available the technical data contained in a soil survey map in a simple & practical language Indicates the hazards of soil erosion Indicates the most intensive , profitable & safe use of any piece of land

Land capability groups:

Land capability groups Land suitable for cultivation and other uses (Class I to IV lands) Land not suitable for agriculture but well suited for forestry, grass land and wild life (Class V to VIII)

Influence of effective soil depth on LCC:

Influence of effective soil depth on LCC

Influence of soil texture on LCC:

Influence of soil texture on LCC

Influence of slope on LCC:

Influence of slope on LCC

Influence of erosion on LCC:

Influence of erosion on LCC

Influence of climate on LCC:

Influence of climate on LCC

PowerPoint Presentation:

Determination of bulk density, particle density and pore space Readings taken from soil samples in lab: Sl no. Weight of soil taken , W(gm) Volume of soil taken, V1 (ml) Volume of water added, V2(ml) Volume of soil+water Volume of soil+water at end of exp V3(ml) 1. 30 23.5 50 72.5 59.5 2. 30 20.5 50 70.5 62.5

calculation:

calculation Pore space volume(V4) = (V1+V2)-V3 V4 = 23.5+50-59.5= 14ml % pore space = V4/V1*100 = 14/23.5*100 = 59.57% bulk density = weight of soil/volume of soil = 30/23.5 = 1.27 gm/cc Particle density = weight of soil/(V1-V4) = 30/(23.5- 14 ) = 3.157 gm/cc

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2. Pore space volume(V4) = (V1+V2)-V3 = 20.5+50-62.5 = 8ml %pore space = = V4/V1*100 = 8/20.5*100 = 39.02% bulk density = weight of soil/volume of soil = 30/20.5= 1.46 gm/cc Particle density = weight of soil/(V1-V4) = 30/(20.5-8) = 2.4 gm/cc

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