 Design of simple portable sprinkler and drip irrigation system
Required data :
crop to be planted and climate type
soil type and peak consumptive use
wind speed( average)
Area of plot
elevation of collection chamber above irrigated land
supply line length . etc

peak consumptive use calculation for different crops :
from cropwat 8.0 ,
for Asarghat hydrological station kamalbajar achham
station: Asara-ghat achham
altitude=1388 m
latitude= 28.95 deg.
Month
January
February
March
April
May
June
July
August
September
October
November
December
Average
Min Temp
°C
4.2
5
8.1
12.7
15.2
17
17.2
17.1
15.4
12.1
8.6
5.9
11.5
longitude=81.45
deg.
Max
Temp
Humidity Wind
Sun
Rad
Eto
°C
%
km/day
Hours
MJ/m²/day mm/day
11.8
57
86
6.3
12
1.63
13.3
55
104
6.9
14.7
2.13
17.4
52
121
7
17.4
2.97
22.5
40
147
8.1
20.9
4.4
24.7
44
147
7.9
21.7
4.88
24.9
73
130
4.3
16.7
3.68
23
89
121
2.9
14.4
2.91
22.8
87
104
3.1
14
2.83
21.7
86
95
4.4
14.4
2.73
20.1
60
86
6.6
15
2.75
16.7
51
78
7.3
13.5
2.11
13.9
46
78
6.7
11.8
1.72
19.4
62
108
6
15.5
2.9
thus , mean monthly air temperature (Tmean) =( Tmax mean + Tmin mean )/2 = ( 11.5 + 19.4)/2= 15.45
0
C
table (a.) Lengths of crop development stages for various planting periods and climatic
regions (days)
Crop
Init.
(Lini)
Dev.
(Ldev)
Mid
(Lmid)
Late
(Llate)
Total
Plant
Date
Region
a. Small Vegetables
Broccoli
35
45
40
15
Cabbage
30
35
30
15
20
30
30
40
30
Carrots
Cauliflower
Onion (seed)
Sweet peppers
(bell)
Tomato
110
Sept
Calif. Desert,
USA
Sept
Calif. Desert,
USA
20
100
Oct/Jan
Arid climate
60
20
150
Feb/Mar
Mediterranean
50
90
30
200
Oct
Calif. Desert,
USA
35
50
40
15
140
Sept
Calif. Desert,
USA
35
50
45
10
140
Feb
Mediterranean
20
45
165
45
275
Sept
Calif. Desert,
USA
35
40
20
125
April/June
Europe and
Medit.
30
40
110
30
210
October
Arid Region
30
40
40
25
135
January
Arid Region
35
40
50
30
155
Apr/May
Calif., USA
25
40
60
30
155
Jan
Calif. Desert,
USA
35
45
70
30
180
Oct/Nov
Arid Region
30
40
45
30
145
April/May
Mediterranean
25/30
50/30
135
Table (b) FAO Crop Coefficients (Kc) values :
CROP
Crop Development Stages
Total
Initial
Crop
MidLate
At
growing
development season season harvest period
Cabbage
0.40.5
0.7-0.8
0.951.1
0.91.0
0.80.95
0.7-0.8
Cotton
0.40.4
0.7-0.8
1.05125
0.80.9
0.650.7
0.8-0.9
Grape
0.350.55
0.6-0.8
0.70.9
0.60.8
0.550.7
0.550.75
Groundout
0.40.5
0.7-0.8
0.951.1
0.750.85
0.550.6
0.75-0.8
Dry
0.40.6
0.7-0.8
0.951.1
0.850.9
0.750.85
0.8-0.9
Green
0.40.6
0.6-0.75
0.951.05
0.951.05
0.951.05
0.65-0.8
Pea, fresh
0.40.5
0.7-0.85
1.051.2
1.01.15
0.951.1
0.8-0.95
Pepper,
fresh
0.30.4
0.6-0.75
0.951.1
0.851.0
0.8-0.9
0.7-0.8
Potato
0.40.5
0.7-0.8
1.051.2
0.850.95
0.70.75
0.75-0.9
Tomato
0.40.5
0.7-0.8
1.051.25
0.80.95
0.60.65
0.75-0.9
Onion
Using BLANNEY-CRIDDLE method to find ET0 ( Reference evapotranspiration) and then ET crop
= ET0 * Kc
from BLANNEY-CRIDDLE method, we have ET0 = P (.457 * Tmean +8.128)
where, P= mean percentage of annual daytime hours and
mm/day
Tmean = mean daily temperature in 0C
the value of P depends on latitudes
Table 1 MEAN DAILY PERCENTAGE (p) OF ANNUAL DAYTIME HOURS FOR DIFFERENT
LATITUDES
Latitude North Jan Feb Mar Apr
South July Aug Sept Oct
60°
.15 .20 .26 .32
55
.17 .21 .26 .32
50
.19 .23 .27 .31
45
.20 .23 .27 .30
40
.22 .24 .27 .30
35
.23 .25 .27 .29
30
.24 .25 .27 .29
25
.24 .26 .27 .29
20
.25 .26 .27 .28
15
.26 .26 .27 .28
10
.26 .27 .27 .28
5
.27 .27 .27 .28
0
.27 .27 .27 .27
May
Nov
.38
.36
.34
.34
.32
.31
.31
.30
.29
.29
.28
.28
.27
June July Aug
Dec Jan Feb
.41 .40 .34
.39 .38 .33
.36 .35 .32
.35 .34 .32
.34 .33 .31
.32 .32 .30
.32 .31 .30
.31 .31 .29
.30 .30 .29
.29 .29 .28
.29 .29 .28
.28 .28 .28
.27 .27 .27
Sept Oct
Mar Apr
.28 .22
.28 .23
.28 .24
.28 .24
.28 .25
.28 .25
.28 .26
.28 .26
.28 .26
.28 .27
.28 .27
.28 .27
.27 .27
Nov
May
.17
.18
.20
.21
.22
.23
.24
.25
.25
.26
.26
.27
.27
Dec
June
.13
.16
.18
.20
.21
.22
.23
.24
.25
.25
.26
.27
.27
for Asaaraghat station , lalitude = 28.950 N
assuming Crops( cabbage , potato, pepper, tomato ) are transplanted in October 1st
thus, from table 1 , P= .29
also Tmean = 15.45 0C
therefore, ET0 = P (.457 * Tmean +8.128)
mm/day
= .29 ( .457 * 15.45+ 8.128)
= 4.40 mm/day
now using value of crop coefficient for different growth stages of crops and finding the crop
water requirements as follow :
Table (i) crop water requirements of CROP : CABBAGE
Growth stages
( days)
Kc
ET0 (mm/day)
ETcrop
(mm/day)
Initial
(20-30 )
.45
4.4
1.98
Development
(30-35)
.75
4.4
3.3
Mid season
(20-30)
1.025
4.4
4.5
Late season
(10-20)
.95
4.4
4.2
At harvest
(5-10)
.875
4.4
3.85
here, maximum ET crop = 4.5 mm/ day
Thus, peak consumptive use of cabbage = 4.5 mm/day
Table (ii) crop water requirements of CROP : TOMATO
Growth stages
( days)
Kc
ET0 (mm/day)
ETcrop
(mm/day)
Initial
(30 )
.45
4.4
1.98
Development
(40)
.75
4.4
3.3
Mid season
(45)
1.15
4.4
5.06
Late season
(30)
.875
4.4
3.85
At harvest
(15)
.625
4.4
2.75
here, maximum ET crop = 5.06 mm/ day
Thus, peak consumptive use of tomato = 5.06 mm/day
Table (iii) crop water requirements of CROP : POTATO
Growth stages
( days)
Kc
ET0 (mm/day)
ETcrop
(mm/day)
Initial
(20 )
.45
4.4
1.98
Development
(40)
.75
4.4
3.3
Mid season
(45)
1.125
4.4
4.9
Late season
(30)
.9
4.4
3.96
At harvest
(20)
.725
4.4
3.19
here, maximum ET crop = 4.95 mm/ day
Thus, peak consumptive use of tomato = 4.9 mm/day
Table (iv) crop water requirements of CROP : PEPPER
Growth stages
( days)
Kc
ET0 (mm/day)
ETcrop
(mm/day)
Initial
(25 )
.35
4.4
1.54
Development
(35)
.675
4.4
2.97
Mid season
(40)
1.025
4.4
4.5
Late season
(20)
.925
4.4
4.07
At harvest
(10)
.85
4.4
3.74
here, maximum ET crop = 4.5 mm/ day
Thus, peak consumptive use of tomato = 4.5 mm/day
Q.NO. 1) Design of portable sprinkler system @ kamalbajar Achham .
Required datas :
crop : potatoes
soil type : clay loam
climate : moderately hot and humid
peak consumptive use for potatoes : 4.9 mm/day (from table(iii) above)
wind speed( average) : 108 km/day =1.25 m/s (from cropwatt 8.0 )
Area of plot : 120 m * 40 m
elevation of collection chamber above irrigated land : 45 m
supply line : 400 m
f= darcy friction factor = 0.023 ( S.P.P ) & 0.057( H.P.P)
solutions:
Step(1) : Computation of irrigation interval
from table 8.2 ( H.I.S. book)


available water for clay loam soil = 90-210 mm per meter of soil depth
extractable water for clay loam soil= 45-105 mm per meter soil (say 75 mm/ m
soil depth)
from table 8.3 ( H.I.S. book)

crop rooting depth for potatoes =0.3– 0 .5 m ( say 0.4 m)
thus, water available at root zone = 0.4 * 75 =30 mm
and irrigation interval = (water available at root zone)/(peak consumptive use )= 30/4.9
= 6.12
say 6 days ( # for safer side)
Assuming sprinkler application efficiency = 70 % ( for moderate climate)
now, Required application = ( peak consumptive use * irrigation interval) /application efficiency
= 4.9 * 6/ 0.7
= 42 mm every 7 days
Assuming 8 hours irrigation per day
rate of application = 42/8 = 5.2 mm/hr.
check intake rate for clay loam :
from table 8.5 ( H.I.S. book), intake rate of clay loam soil =( 6 – 8) mm /hr i.e. estimated
application rate of sprinkler( 5.2 mm/hr) < intake rate of clay loam . Hence ok
Step(2) : Sprinkler selection
from sprinkler performance table as shown below select sprinkler to deliver 5.2
mm/hr .
Spacing ( along lateral * of lateral ) / precipitation
rate(mm/hr)
P
Q
D
nozzle size (mm)
(atm.)
(m3/hr)
(m)
6m*5m 6m*6m 6m*7m 6m*8m 6m*10m
2.2
1
0.16
9
5.3
4.4
3.8
3.3
2.5
1.5
0.195
10
6.5
5.4
4.6
4.1
3
2
0.22
11
7.3
6.1
5.2
4.6
3.4
2.5
0.25
12
8.3
6.9
6
5.2
3.9
3
0.275
12
9.2
7.6
6.6
5.7
4.1
where , Q= Discharge per sprinkler , P= working pressure , D= coverage diameter
since there is prevailing wind along the direction of lateral , spacing along the lateral must be
reduced by 60 % of coverage diameter and spacing in the other direction ( no wind) is 65 % of
diameter
(for square and rectangular spacing with wind speed = 1.25 m/s < 2 m/s along lateral and no
wind in the other direction )
from sprinkler performance table above :
Q= Discharge per sprinkler = .22 m3/s
P= working pressure = 2 atm = 2 * 10.34 =20.68 m 0f water
D= coverage diameter= 11 m
thus, spacing of sprinkler along lateral = .6 * 11= 6.6 m ( say 6 m) and spacing of the laterals (
spacing in other direction)= .65* 11= 7.15m (say 7 m)
Step(3) : Design of laterals (BC)
maximum allowable pressure variation in the lateral = 20 % of 20.68 m= 4.14 m
length of each lateral = 60 m
leaving 3 m side clearance on each side ( left and right ) along lateral and 2.5 on each side of the
lateral we have,
main supply line
distribution line
A
lateral L1
2.5 m
day 6
16
day 1
7m
day 5
day 4
B
BBB
22
fs
day 2
40444434y6544
6m
day 3
C
B
lateral L2 ( 10 sprinkler per lateral)
irrigated plot area ( 120 m * 40 m )
no. of sprinkler along lateral = [{( length of lateral – 2* clearance )/ (spacing of sprinkler along
lateral)}+1]
= [{(60-2 *30)/6} +1]
= 10 (as shown in figure above )
C
for 10 no. of sprinkler, total discharge at head of lateral ( discharge per lateral ) = 10* 0.22
= 2.2 m3/hr
assume 2 no. of laterals operating per day.
from table 8.6 ( H.I.S BOOK ) correction factor for 10 no. of sprinklers ( C ) = 0.402
using maximum allowable pressure variation( head loss ) in lateral ( hL= 4.14 m ) we find the
required diameter of lateral pipe as follow :
using soft polythene pipe ( S.P.P ) with f = 0.023 as lateral pipe
we have , hL =[C* ( f *L * V 2 ) / 2gd ] ………………..(1)
but v= QL /A =4QL /𝜋𝑑 2 where d= required diameter of lateral pipe and QL = flow per lateral
using ( hL= 4.14 m , f= 0.023,C= 0.402 , QL= 2.2 m3/hr = 6.11 * 10-4 m3/s , g= 9.81 m/ s2 , v=
4Q/𝜋𝑑 2 and L = length of lateral =60 m in eq. 1 ) by back calculation we get dreq.= 0.026m =26
mm ( minimum req. diameter )
as increased size of pipe will reduce velocity of jet and hence head loss so choose
32 mm ( > dreq. and market available ) diameter ( safer side ) soft polythene pipe for the laterals .
check headloss for 32 mm pipe :
for 32 mm pipe , hL =[ ( f *L * V 2 ) / 2gd ]
(with v= ( 4* 6.11 * 10-4 / 𝜋 ∗ 0.0322 ) =.76 m/s
thus, hL = 1.3 m < 4.14 (hL max. ) . ok
hence adopt 32 mm dia. S.P.P pipe for the laterals .
assuming valve loss = 0.5 m ,
required pressure at head of lateral = 20.68 + 0.5 +1.3
= 22.48 m ( Say 22.5 m)
Step(4) : Design of distribution pipe (AB)
As the head loss in distribution pipe will vary according to position of operating laterals at one
time. When laterals L1 & L2 are in operations the head loss in distribution pipe will be
maximum , thus diameter of distribution line must be selected such that distribution head loss
is within acceptable limit when laterals L1 & L2 in operations.
here, discharge per lateral = 2.2 m3/hr
no. of laterals operating at one time = 2
thus , maximum flow in distribution pipe= 2 * 2.2 = 4.4 m3/hr
length of distribution pipe line = width of plot = 40 m
friction loss of all fittings in the distribution pipe = about 20 % of length of distribution line
= 0.2 * 40 = 8 m
tus, total equivalent length of distribution pipe L= 48 m
recommended ( allowable ) head loss in distribution pipe = 30 % 0f pressure at head of lateral
= 0.3 * 22.5 = 6.75 m
using hard polythene pipe ( H.P.P ) with f = 0.057 as distribution pipe
using head loss euation , hL =[ ( f *L * V 2 ) / 2gd ……………(2)
where , V = QD /A =4QD /𝜋𝑑 2 (QD= maximum flow in distribution pipe=
4.4 m3/hr=1.22 * 10-3 m3/s , f= 0.057 , L= 48 m & g= 9.81 m/ s2
from eqn. 2 by back calculation we get d req. = 0.035 m = 35 mm
thus adopt 40 mm( safer side ) diameter hard polythene pipe as distribution line .
check headloss for 40 mm pipe :
for 40 mm pipe , hL =[ ( f *L * V 2 ) / 2gd ]
(with [ (V= 4* 1.222 * 10-3/ 𝜋 ∗ 0.042 )=.97 m/s , f=0.057 , L = 48 m ]
thus, hL = 3.28 m < 6.75 m (hL allowable . ) . ok
Step(5) : Design of supply line
here, length of supply line = 400 m = 0 .4 km
maximum flow in supply line = 4.4 m3/hr= 1.22 * 10-3 m3/s
available head = 45 m ( as collection chamber is 45 m above the irrigated land)
thus, required pressure at A = pressure at head of lateral + actual head loss in distribution line
= 22.5 + 3.28
= 25.78 m
Therefore, Allowable loss in the supply line = 45- 25.78 =19.22 m
assuming 50 mm dia. Hard polythene pipe as supply line , we have
actual head loss through the supply line is :
for 50 mm pipe , hL =[ ( f *L * V 2 ) / 2gd ]………….(3)
(with [ V= ( 4* 1.222 * 10-3/ 𝜋 ∗ 0.052 )=.62 m/s , f= 0.057 , L=400 m ]
from eqn. 3,
hL = 9.01 m < 19.22 m ( allowable loss )
hence ok .
hence Adopt 50 mm H.P.P as the supply line
Q.NO. 2) Design of simple farmer's portable drip irrigation system @ kamalbajar Achham .
Required datas :
crop : pepper
soil type : silty loam
Dripper efficiency = 90 %
Ground slope = -1 %
peak consumptive use for pepper : 4.5 mm/day (from table(iv) above)
wind speed( average) : 108 km/day = 1.25 m/s (from cropwatt 8.0 )
Area of plot : 60 m * 45 m
elevation of collection chamber above irrigated land : 45 m
supply line : 400 m
f= darcy friction factor = 0.057 ( hard Polythene Pipe)
solutions:
Step(1) : Computation of irrigation interval
from table 8.2 ( H.I.S. book)


available water for clay loam soil = 60-120 mm per meter of soil depth
extractable water for clay loam soil= 30-60 mm per meter soil (say 45 mm/ m
soil depth)
from table 8.3 ( H.I.S. book)

crop rooting depth for peppers =0.5– 1.0 m ( say 0.75 m)
thus, water available at root zone = 0.75 * 45 =33.75 mm
and irrigation interval = (water available at root zone)/(peak consumptive use )=33.75/4.5
= 7.5
say 7 days ( # for safer side)
Assuming dripper application efficiency = 90 %
now, Required application = ( peak consumptive use * irrigation interval) /application efficiency
= 4.5* 7/ 0.9
= 35 mm every 7 days
Assuming 12 hours irrigation per day
rate of application = 35/12 = 2.92 mm/hr.
check intake rate for silty loam :
from table 8.5 ( H.I.S. book), intake rate of clay loam soil =( 7– 10) mm /hr i.e. estimated
application rate of dripper( 2.92 mm/hr) < intake rate of clay loam . Hence ok
Step(2) : Drip line selection
Lateral spacing = 1.5 m ( determined by row spacing of crop )
no. of laterals = width of plot / lateral spacing = 45/1.5=30
Dripper spacing (spacing dripper along laterals ) = 0.5 m ( determined by type of crop and
manufacturers dripline specification )
Assuming dripper discharge (discharge per dripper) of 2 litrs/ hr. = 2 * 10-3 m3/hr
application rate = (discharge per dripper)/( lateral spacing * dripper spacing )
= ( 2 * 10-3 m3/hr )/ (1.5 *0.5 m2)= 2.66 * 10-3 m/hr
= 2.66 mm/hr
Required no. of hours of irrigation per day = water available to root zone / application rate
= 33.75 mm/ 2.66 mm/hr=12.68 hr
thus, Required no. of hours of irrigation per day =12.7 hrs per day
Step(3) : Design of lateral ( BC)
Assume 16 mm diameter of dripline with average dripper flow rate 2 l/hr and working pressure
of 1 atm.( 10.34 m water) per dripper would be used
maximum allowable pressure variation in the lateral = 20 % of 1 atm.
= 0.2 * 10.34 m of water
= 2.06 m
length of lateral = 60 m ( length of plot )
no. of dripper per lateral = length pf lateral /dripper spacing
= 60 / 0.5
= 120
since discharge per dripper is 2 L/hr
thus , for 120 dripper per lateral ( total discharge at head of lateral ) = 120 * 2 = 240 L/hr
from table 8.6 ( H.I.S BOOK ) correction factor for 120 no. of drippers ( C ) = 0.351
friction loss in 16 mm dia. low flow thin walled soft polythene pipe , hL = C *( f *L * V 2 ) / 2gd
….(1)
where , f= 0.057 , L= 60 m
V = QL /A =4QL /𝜋𝑑 2 = ( 0.24 *4 )/( 3600 * 𝜋 ∗ 0.0162)
= 0.332 m/s
thus, from eqn. (1) , hL = 0.42 m < 2.06 m ( allowable loss) ok
hence adopt 16 mm dia. soft thin walled polythene pipe as lateral
assume valve loss = 0.5 m
Required pressure at the head of lateral = 10.34 m + 0.5 m+ 0.42 m= 11.26 m
from table 8.7 ( H.I.S book ), maximum length of 16 mm dia. dripper lateral for 1 atm pressure ,
dripper discharge 2 l/hr , dripper spacing = 0.5 m ground slope -1 % and maximum flow
variation = 10 % is 95 m but for maximum allowable head loss of 2.06 m ,
L allowable = 102 m > 95 hence ok .
main supply line
irrigation land ( 60 m * 45 m )
distribution line
DAY 7
DAY 6
DAY 5
DAY 4
DAY 3
DAY 2
DAY 1
laterals (total 30 laterals with120 dripers per laterals and 4 laterals operating per day )
Step(4) : Design of distribution pipe (AB)
Discharge per lateral = 240 L/hr = 0.24 m3/hr
no. of laterals operating at one time = 4 (assumed )
maximum flow in distribution pipe = 4 * 0.24 = 0.96 m3/hr
length of distribution line = no. of lateral * lateral spacing = 30 * 1.5 = 45 m
friction loss of all fitiings in distribution line = 20 % of length of pipe = 0.2 * 45 = 9 m
thus, total equivalent length of distribution line = 45+9 =54 m
recommended ( allowable ) head loss in distribution pipe = 20 % of pressure at head of lateral
= 0.2 *11.26 m = 2.25 m
using hard polythene pipe ( H.P.P ) with f = 0.057 as distribution pipe
using head loss euation , hL =[ ( f *L * V 2 ) / 2gd ……………(2)
where , V = QD /A =4QD /𝜋𝑑 2 (QD= maximum flow in distribution pipe=
0.96 m3/hr=2.667 * 10-4 m3/s , f= 0.057 , L= 54 m & g= 9.81 m/ s2
from eqn. 2 by back calculation we get d req. = 0.021 m = 21 mm
thus adopt 25 mm dia. ( safer side ) hard polythene pipe as distribution line
check headloss for 25 mm pipe :
for 25 mm pipe , hL =[ ( f *L * V 2 ) / 2gd ]
(with [ (V= 4* 2.667 * 10-4/ 𝜋 ∗ 0.0252 )=.54 m/s , f=0.057 , L = 54 m ]
thus, hL = 1.83 m < 2.25 m (hL allowable . ) . ok
Step(5) : Design of supply line
here, length of supply line = 400 m = 0 .4 km
maximum flow in supply line = 0.96 m3/hr= 2.667 * 10-4 m3/s
available head = 45 m ( as collection chamber is 45 m above the irrigated land)
thus, required pressure at A = pressure at head of lateral + actual head loss in distribution line
= 11.26 + 1.83
= 13.09 m
Therefore, Allowable loss in the supply line = 45-13.09 =31.91 m
assuming 32 mm dia. Hard polythene pipe as supply line , we have
actual head loss through the supply line is :
for 32 mm pipe , hL =[ ( f *L * V 2 ) / 2gd ]………….(3)
(with [ V= ( 4* 2.667 * 10-4/ 𝜋 ∗ 0.0322 )=.332m/s , f= 0.057 , L=400 m ]
from eqn. 3,
hL = 4.0 m < 33.2 m ( allowable loss )
hence ok .
hence Adopt 32 mm H.P.P as the supply line