Water Harvesting Agronomic Aspects

7/29/2019 Water Harvesting Agronomic Aspects

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Hydrology

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Storage Capacity & Yield

What storage capacity is required?

Water demand for irrigation

Water demand for livestock

Water demand for domestic use

Assessment of annual safe yield in seasonal streams

Determination of peak discharge

Spillway design


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What Storage Capacity Is Required?

Storage Capacity Required =

Water Demand + Losses in the reservoir

Livestock Domesticuse

Irrigation


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Irrigation water demand

What Storage Capacity is Required?

Ir (m3

) = 10 x ETcrop (mm) x Ca (ha)

Eff

Ir : Irrigation water requirements in cubic metersfor the whole dry period

Etcrop : Crop water requirement in mm during the dry period

Ca : Area irrigated with water from the reservoir in ha

Eff : Overall water application efficiency


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Crop water requirement, ETcrop


ETcrop = Kc x ETo

ETcrop : Crop water requirement in mm per unit of time

Kc : Crop factor (crop coefficient)

ETo : Reference crop evapotranspiration in mm per unit of time


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ETo

Cool Humid

Cloudy

Little or no wind

Hot Dry

Sunny

Windy

ETo


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Computation of ETo

Method Temp. Humid. Wind Sunsh. R ad. Evap. Env.

Blaney-Criddle M E E E E

Radiation M E E M (M) E

Penman M M M M (M) E

Pan E E M M

Minimum data requirement for various ETo computation methods

M: measures data; E: estimated data; (M): if available, but not essential


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Class A evaporation pan


Class a Evaporation Pan: measurement of Epan

the pan is installed in the field 15 cm above

the ground

the pan is filled with water 5 cm below the rim

the water is allowed to evaporate during certain

period of time (usually 24 hours)

measurement is usually taken at 7:00 hours

rainfall, if any, is measured simultaneously

the difference between the two measured

water depths yields the pan evaporation rate:

Epan (mm / 24 hours)

Source: Critchley & Siegert 1991

Pan dimension:

Diameter = 1.21 m

Depth = 25 cm


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Blaney-Criddle method

ETo = p (0.46 Tmean + 8)Tmean : Mean daily temperature (

0C)

P : mean daily percentage of annual day time hours

Pan Evaporation method

ETo = Epan x Kpan

Epan : Evaporation from pan (standard pan is Class A pan)

Kpan: Pan factor. Varies between 0.35 to 0.85.Average value is 0.7


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Indicative values of ETo

Adapted fromCritchley & Siegert1991

Climatic Zone

Arid

Semi-Arid

Low(


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Crop factor, Kc

Varies for different crops

Varies with growing stages

Varies in climateVariation of Kc in growing stages

Initial Dev.

Mid

season

Late

season

Cropfactor,

Kc


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Crop factor, Kc

Crop factors for the most commonly grown crops under WH;Adapted from Critchley & Siegert 1991

C rop I nitia l

stage

da y s C rop

dev.

stage

da y s M id-

season

stage

da y s Late

season

da y s S e ason

av erage

C otton 0.45 30 0.75 50 1.15 55 0.75 45 0.82

M aize 0.40 20 0.80 35 1.15 40 0.70 30 0.82

M ille t 0.35 15 0.70 25 1.10 40 0.65 25 0.79

Sorghum 0.35 20 0.75 30 1.10 40 0.65 30 0.78

Grain/small 0.35 20 0.75 30 1.10 60 0.65 40 0.78

Legumes 0.45 15 0.75 25 1.10 35 0.50 15 0.79

Groundnuts 0.45 25 0.75 35 1.05 45 0.70 25 0.79


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Water demand for livestock

WL = NL x Ac x T

1000

WL: Water needed for livestock during thecritical period in cubic meters

NL : Number of animals to be watered fromthe reservoir

Ac : Average rate of animal waterconsumption in liters per day per animal

25 - 60 liters/animal/day

T : Duration of the critical period in days

Average daily water consumptionof selected animals

Animal Consumption

(l/d)

Camel 50

Cattle 25 - 35

Sheep 5 - 15

Goat 5 - 15

Donkey 16 - 20

Chicken 15 - 20/100 heads


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Water demand for domestic use

Wd = Po x Dc x T

1000

Wd : Domestic water supply during a critical period in cubic meters

Po : Users of the reservoir

Dc : Average rate of water consumption in liters

per day per person

40 liters / person / day

T : Duration of the critical period in days


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Losses: Evaporation and Seepage

Evaporation losses:

Can be calculated or measured

using Pan A

Seepage losses:

Difficult to assess as it depends onpermeability of the prevailing soil

As a rule of thumb can beassumed equal to ETo losses

Evaporation losses

Seepage losses

Reservoir

Total losses = Evaporation + Seepagelosses losses


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Example : Computation of required water storage capacity

Crop data: Selected crop is Sorghum, Average Kc is 0.7

Livestock data: Total livestock number(using the reservoir) 500, water consumption is50l/day/animal

Population data: Number of users is 400, water consumption 40l/day/person

General data: Average reference crop evapotranspiration during the dry period is 6mm/d,Irrigated area is 2ha, overall water application efficiency is 40%, dry period is 90 days, seepageloss is assumed to be equal to ETo losses. Reservoir: Surface area is 1000 m , bottom andside walls area is 1500 m

Irrigation water demand

Etcrop = Kc x Eto= 0.7 x 6mm/d = 4.2mm/d

Etcrop(dry period) = Etcrop(day) x dry period

= 4.2mm/d x 90d = 378mm

Ir = 10 x Etcrop x Ca = 10 x 378 x 2 = 18900m3

Ef 0.4

Livestock water demand

WL = NL x Ac x T = 500 x 50 x 90 = 2250 m3

1000 1000

Domestic water demand

Wd = Po x Dc x T = 400 x 40 x 90 = 1440 m3

1000 1000


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Example:Computa

tionofrequiredwaterstoragecapacity

Total water demand = Ir + WL + Wd

= 18,900 + 2,250 + 1,440 = 22,590m3

Losses = Evaporation losses + Seepage losses

Evaporation losses = ETo X Surface area of reservoir =

6mm/d x 1000 m = 6 m /d

Seepage losses assumed equal to 6mm/d

Total seepage losses = Seepage loss X Bottom and side wall areas =

6mm/d X 1500 m = 9 m /d

Total losses = (Evaporation losses + Seepage losses ) X Dry period =

(6 + 9 m /d) x 90d = 1350 m

Required Storage Capacity = Total water demand + Losses =

22,590 + 1,350 = 23,940 m3


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Assessment of Annual Safe Yield in Seasonal Streams

Assessment of annual safe yield:

Get annual runoff or rainfall amount

Make frequency analysis

Probability in %

99.8 67 1

Annua

lrunoffinm3


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Determination of Peak Discharge

Peak discharge: the maximum amount of runoff (discharge) forwhich the structures are designed.

Computation Methods

Adapted fromBaban 1996

-Flood Frequency analysis

< 5000 km2Unit Hydrograph

-Empirical

< 50km2Rational Method

Applicable

catchment areaMethods


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Rational Formula (method)


Qp = Kr x I x C

3.6

Qp : peak flood discharge (m3/s)Kr : rational runoff coefficient (different from K,

annual or seasonal runoff coefficient)I : rainfall intensity (mm/hr)C : area of the catchment (km2)


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Rational Method: Steps

Step 1:

Compute time of concentration, tc, with the following formula.tc = 0.019471 x L0.77

G0.385


tc : time of concentration (minutes)

L : maximum length of travel of water (m)G : slope of the catchment, H/L, H is the difference

in elevation between the outlet & remote point.


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RemotePoint

Outletpoint

A

B

Catchment map

L

H

Section A-B

A B

Step 1: (Continued)

Determination of H and L

from the catchment map.

Rational Method : Steps



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Step 2: Find the corresponding intensity (I) for duration equal totc for a certain return period, from the Intensity - duration -frequency curve prepared for the area.



AverageInten

sity(mm/hr)

Return period -years

15

50

100

Duration (h) tc

I

Intensity -duration -frequency curve(sample)


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Step 3: Select suitable rational runoff coefficient Kr, having knowledgeof the surface cover of the catchment area. Use the following table.

Values ofrational runoffcoefficient kr;adapted fromBaban 1996

0.70

0.60

0.40

0.30

Tight claycultivated

woodland

Sandy loam

cultivated

woodland

Hilly Area

0.50

0.40

0.20

0.10

Tight claycultivated

woodland

Sandy loam

cultivated

woodland

Flat Area

Value of KrLand coverGeneral Slope


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Step 4: Compute Qp using the rational formula.

Qp = Kr x I x C

3.6


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Example: Determination of peak discharge

Runoff from a catchment, which is 0.9 km2 large, is required to be harvestedand stored in a small earth reservoir. Determine the 15 year(probability of 6.7%)peak discharge for designing the dam spillway.

From the topographic map of the area & field observation:

length of the catchment along the main drain (L) is 800m

slope (G) of the land is 0.7%

catchment is woodland on sandy loam

From a meteorological station nearby, the following data was obtained.

Maximum depth of rainfall with 15 year return period

Duration (minutes) 5 10 20 30 40 60

Depth of rainfall (mm) 17 26 40 50 57 62


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Example: Determination of Peak discharge

Step 1: Computation of tc

tc = 0.019471 x L0.77 = 0.019471 x 8000.77 = 22.6 minutes

G0.385 0.0070.385

Step 2: Look for the corresponding intensity for duration of 22.6 minutes .

From the given table, interpolating for 22.6 minutes of duration yields 42.6mm ofrainfall i.e. I = (42.6mm/22.6minutes) x 60 minutes/hour = 113.1 mm/hr

Step 3: Select suitable value for rational runoff coefficient Kr For sandy loam, woodland and hilly area, Kr = 0.3

Step 4: Use rational formula to compute Qp

Qp = (Kr x I x c)/3.6 = (0.3 x 113.1 x 0.9)/3.6 = 8.48m3/s


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Spillway Design

Spillway is a structure constructed at a dam site, for effectivelydisposal of the dam surplus water from upstream to downstream.

B) Lined spillway

DamReservoir

A) Natural spillway

Dam

Reservoir

Safety valve of a dam!


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Spillway Design

Design Criteria

Enough capacity to dispose peak discharge

Maximum permissible velocity not to erode parts of structures

Qp

Cost

Risk


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Spillway Design

Design Steps - Fixing flow depth & bottom width

Step 1: Compute the peak discharge Qp,(use rational formula)

Step 3: Compute Flow area,A.

A =Q pv

Step 2: Compute hydraulic radius R,(use Mannings formula)

v : mean velocityn : mannings roughness coefficients : bed slope

v ns

R = ( )3/2


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Spillway Design


Step 4: Compute wetted

perimeter, P.

P

A

R

yP P A H H

H H1

4 2 1

2 2 1

2 2

2

yP P A H H

H H2

4 2 1

2 2 1

2 2

2

Step 5: Compute flow depth, y.The solution gives two values.

P

yb 1

H

Side slope: H: V ( H: 1)


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Spillway Design


Step 6: Compute bottom width, b. Two valuesare computed corresponding to yvalues.

From (y1, b1) & (y2, b2) select the reasonablepairs, i.e. non-negative one.

b P y H 1 1 22 1

b P y H 2 22

2 1


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Spillway Design

Example: Spillway design

A spillway is to be designed using the following input values.Peak flow , Qp = 50m3/s

Spillway channel slope, s = 0.02

Roughness coefficient , n = 0.03

Maximum permissible velocity = 4.0m/sSide slope of spillway channel 3: 1 (H:V)


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Spillway Design

Example:S

pillwaydesign

Step 1: Qp= 50 m3/ (given)

Step 2: Compute R, = 0.78 m

Step 3: ComputeA, A = Qp/V= 50/4 = 12.5 m 2

Step 4: Compute P, P = A/R = 12.5/0.78 = 16.03 m

Step 5: Compute y (y1 & y2),

4 0 03

0 02

3 2

.

.

/

Rvn

s

3 2/

yP P A H H

H H1

4 2 1

2 2 1

2 2

2

16 03 16 03 4 12 5 2 3 1 3

2 2 3 1 3

2 2

2

. . .

= 3.84 m

yP P A H H

H H2

2 2

2

4 2 1

2 2 1

16 03 16 03 4 12 5 2 3 1 3

2 2 3 1 3

2 2

2

. . .

= 0.98 m


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Spillway Design

Example:Spillwaydesign

Step 6: Compute b (b1 & b2)

The flow depth value of 3.83m yields a negative bottom width

and hence rejected. Therefore, the correct values are:

Flow depth (y ) = 0.98 m &

Bottom width (b) = 9.83 m

b P y H 1 1

22 1

16 03 2 384 3 1

2. . = - 8.26

b P y H 2 22

2 1 16 03 2 0 98 3 12. . = 9.83


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The required storage capacity is fixed by water demand

(Irrigation, livestock and domestic) and losses in the reservoir

Evapotranspiration varies over growing stages, in climate and variety

of crops

Average rate of animal water consumption is 25 - 60 liters/animal/day

Human average rate of water consumption is 40 liters/person/day

Spillway is safety valve of a dam

Summary
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Water Harvesting Agronomic Aspects

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