Water Harvesting Agronomic Aspects
-
Upload
haithamelramlawi7503 -
Category
Documents
-
view
225 -
download
0
Transcript of Water Harvesting Agronomic Aspects
-
7/29/2019 Water Harvesting Agronomic Aspects
1/35
Hydrology
Unit6+7
Storage
Capacity
and Yield
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
-
7/29/2019 Water Harvesting Agronomic Aspects
2/35
Hydrology
Unit6+7
Storage
Capacity
and Yield
What Storage Capacity Is Required?
Storage Capacity Required =
Water Demand + Losses in the reservoir
Livestock Domesticuse
Irrigation
-
7/29/2019 Water Harvesting Agronomic Aspects
3/35
Hydrology
Unit6+7
Storage
Capacity
and Yield
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
-
7/29/2019 Water Harvesting Agronomic Aspects
4/35
Hydrology
Unit6+7
Storage
Capacity
and Yield
Crop water requirement, ETcrop
What Storage Capacity is Required?
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
-
7/29/2019 Water Harvesting Agronomic Aspects
5/35
Hydrology
Unit6+7
Storage
Capacity
and Yield
What Storage Capacity is Required?
ETo
Cool Humid
Cloudy
Little or no wind
Hot Dry
Sunny
Windy
ETo
-
7/29/2019 Water Harvesting Agronomic Aspects
6/35
Hydrology
Unit6+7
Storage
Capacity
and Yield
What Storage Capacity is Required?
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
-
7/29/2019 Water Harvesting Agronomic Aspects
7/35
Hydrology
Unit6+7
Storage
Capacity
and Yield
Class A evaporation pan
What Storage Capacity is Required?
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
-
7/29/2019 Water Harvesting Agronomic Aspects
8/35
Hydrology
Unit6+7
Storage
Capacity
and Yield
What Storage Capacity is Required?
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
-
7/29/2019 Water Harvesting Agronomic Aspects
9/35
Hydrology
Unit6+7
Storage
Capacity
and Yield
What Storage Capacity is Required?
Indicative values of ETo
Adapted fromCritchley & Siegert1991
Climatic Zone
Arid
Semi-Arid
Low(
-
7/29/2019 Water Harvesting Agronomic Aspects
10/35
Hydrology
Unit6+7
Storage
Capacity
and Yield
What Storage Capacity is Required?
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
-
7/29/2019 Water Harvesting Agronomic Aspects
11/35
Hydrology
Unit6+7
Storage
Capacity
and Yield
What Storage Capacity is Required?
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
-
7/29/2019 Water Harvesting Agronomic Aspects
12/35
Hydrology
Unit6+7
Storage
Capacity
and Yield
What Storage Capacity is Required?
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
-
7/29/2019 Water Harvesting Agronomic Aspects
13/35
Hydrology
Unit6+7
Storage
Capacity
and Yield
What Storage Capacity is Required?
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
-
7/29/2019 Water Harvesting Agronomic Aspects
14/35
Hydrology
Unit6+7
Storage
Capacity
and Yield
What Storage Capacity is Required?
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
-
7/29/2019 Water Harvesting Agronomic Aspects
15/35
Hydrology
Unit6+7
Storage
Capacity
and Yield
What Storage Capacity is Required?
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
-
7/29/2019 Water Harvesting Agronomic Aspects
16/35
Hydrology
Unit6+7
Storage
Capacity
and Yield
What Storage Capacity is Required?
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
-
7/29/2019 Water Harvesting Agronomic Aspects
17/35
Hydrology
Unit6+7
Storage
Capacity
and Yield
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
-
7/29/2019 Water Harvesting Agronomic Aspects
18/35
Hydrology
Unit6+7
Storage
Capacity
and Yield
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
-
7/29/2019 Water Harvesting Agronomic Aspects
19/35
Hydrology
Unit6+7
Storage
Capacity
and Yield
Rational Formula (method)
Determination of Peak Discharge
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)
-
7/29/2019 Water Harvesting Agronomic Aspects
20/35
Hydrology
Unit6+7
Storage
Capacity
and Yield
Rational Method: Steps
Step 1:
Compute time of concentration, tc, with the following formula.tc = 0.019471 x L0.77
G0.385
Determination of Peak Discharge
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.
-
7/29/2019 Water Harvesting Agronomic Aspects
21/35
Hydrology
Unit6+7
Storage
Capacity
and Yield
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
Determination of Peak Discharge
-
7/29/2019 Water Harvesting Agronomic Aspects
22/35
Hydrology
Unit6+7
Storage
Capacity
and Yield
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.
Determination of Peak Discharge
Rational Method : Steps
AverageInten
sity(mm/hr)
Return period -years
15
50
100
Duration (h) tc
I
Intensity -duration -frequency curve(sample)
-
7/29/2019 Water Harvesting Agronomic Aspects
23/35
Hydrology
Unit6+7
Storage
Capacity
and Yield
Determination of Peak Discharge
Rational Method : Steps
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
-
7/29/2019 Water Harvesting Agronomic Aspects
24/35
Hydrology
Unit6+7
Storage
Capacity
and Yield
Determination of Peak Discharge
Rational Method : Steps
Step 4: Compute Qp using the rational formula.
Qp = Kr x I x C
3.6
-
7/29/2019 Water Harvesting Agronomic Aspects
25/35
Hydrology
Unit6+7
Storage
Capacity
and Yield
Determination of Peak Discharge
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
-
7/29/2019 Water Harvesting Agronomic Aspects
26/35
Hydrology
Unit6+7
Storage
Capacity
and Yield
Determination of Peak Discharge
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
-
7/29/2019 Water Harvesting Agronomic Aspects
27/35
Hydrology
Unit6+7
Storage
Capacity
and Yield
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!
-
7/29/2019 Water Harvesting Agronomic Aspects
28/35
Hydrology
Unit6+7
Storage
Capacity
and Yield
Spillway Design
Design Criteria
Enough capacity to dispose peak discharge
Maximum permissible velocity not to erode parts of structures
Qp
Cost
Risk
-
7/29/2019 Water Harvesting Agronomic Aspects
29/35
Hydrology
Unit6+7
Storage
Capacity
and Yield
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
-
7/29/2019 Water Harvesting Agronomic Aspects
30/35
Hydrology
Unit6+7
Storage
Capacity
and Yield
Spillway Design
Design Steps - Fixing flow depth & bottom width
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)
-
7/29/2019 Water Harvesting Agronomic Aspects
31/35
Hydrology
Unit6+7
Storage
Capacity
and Yield
Spillway Design
Design Steps - Fixing flow depth & bottom width
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
-
7/29/2019 Water Harvesting Agronomic Aspects
32/35
Hydrology
Unit6+7
Storage
Capacity
and Yield
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)
-
7/29/2019 Water Harvesting Agronomic Aspects
33/35
Hydrology
Unit6+7
Storage
Capacity
and Yield
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
-
7/29/2019 Water Harvesting Agronomic Aspects
34/35
Hydrology
Unit6+7
Storage
Capacity
and Yield
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
-
7/29/2019 Water Harvesting Agronomic Aspects
35/35
Hydrology
Unit6+7
Storage
Capacity
and Yield
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
http://localhost/var/www/apps/conversion/tmp/scratch_10/hydro-unit08e.ppt