Transcript Design Irrigation System

Design Irrigation System II
Asher Azenkot
1
Local Head Losses
The local head loss due to a local
disturbance in water flow is proportional to
the head velocity.
V2
h  K
2g
K - Coefficient
2
Hydraulic Valve Local Head Loss
3
Filter Local Head Loss
4
Metzer drip line
Water flow velocity reduced gradually along
the lateral pipe
5
Local head loss in “head connector”
m3/h
6
Local head loss
Flow rate
Local head loss m
7
Example:
A 12" valve (K = 2.5) is installed in 1,250
meters long pipe (12” and C = 130). What is
the total head loss due to the valve and the
pipe when the water flow rate is = 100, 200
and 400 m3/h.
The pipe cross section area is:
A12" 
  0.32
4
 0.07m 2
8
Continue:
Q m3/h
100
200
400
0.39
0.78
1.57
V /2g
0.01
0.03
0.13
Local head loss
0.02
0.08
0.31
J ‰ 12” pipe
0.6
2.2
7.8
Head loss in pipe
0.8
2.7
9.8
Total head loss
0.82
2.78
10.11
% local head loss
2.5
2.96
3.1
Velocity
2
9
Continue:
If an 8" valve is replaced the 12", what will
be the new total head loss?
Q m3/h
100
200
400
Local head loss (8")
0.1
0.4
1.55
Head loss (12” Pipe)
0.62
2.38
8.75
Total head loss (m)
0.72
2.78
10.2
% local head loss
16.1
16.8
17.7
10
Lateral Pipes
A lateral pipe is characterized by a continuous
decline in water discharge along the pipe. The flow
rate starts at Qu (m3/h) at the upstream end and
ends up with a q1 (m3/h) downstream. (Lateral pipe
is abide by: 1. A same size of pipe, 2. even distance
between outlets, 3. a same outlet (sprinkler or
emitter)
rate. of the head loss is done in two step
The flow
calculation
1.
The head loss is calculated by assuming the pipe is
2.
The outcome is multiplied by the coefficient F
Qu = n*q
D
3q
2q
q
Sl
95 m
11
Coefficient F
Plastic later
Al lateral
n
F1
F2
F3
F1
F2
F3
5
10
12
15
20
25
30
40
50
100
0.469
0.415
0.406
0.398
0.389
0.384
0.381
0.376
0.374
0.369
0.337
0.35
0.352
0.355
0.357
0.358
0.359
0.36
0.361
0.362
0.41
0.384
0.381
0.377
0.373
0.371
0.37
0.368
0.367
0.366
0.457
0.402
0.393
0.385
0.376
0.371
0.368
0.363
0.361
0.356
0.321
0.336
0.338
0.341
0.343
0.345
0.346
0.347
0.348
0.349
0.396
0.371
0.367
0.363
0.36
0.358
0.357
0.355
0.354
0.352
1. F1 to be used when the distance from the lateral inlet to the first outlet is
meters.
2. F2 to be used when the first outlet is near the lateral inlet.
3. F3 to be used when the distance from the lateral inlet to the first outlet is
meters.
Sl
Sl/2
12
Characteristics of a Lateral pipe

The sprinkler pressure along the lateral pipe
decline faster along the first 40% of the length
than afterwards (figure 2).

The sprinkler flow rate along the lateral pipe
declines faster along the first 40% of the
length (figure 1).

The location of the sprinkler (or emitter) with the
average pressure and flow rate is 40% away from the
lateral’s inlet.

Three quarter of the lateral head loss takes
place along the first two fifth sections (40%).
13
Fig. 1: Flow rate reduction in a plain pipe
and in a lateral with sprinklers.
2
25
1.9
20
1.8
15
1.7
10
1.6
5
1.5
0
0
12
24
36
48
60
72
84
96 108 120 132 144 156
sprinkler flow
1.9 1.88 1.86 1.84 1.83 1.82 1.81 1.8 1.8 1.8 1.79 1.79 1.79
% reduction
100 98.7 97.6 96.8 96.1 95.6 95.2 94.9 94.6 94.5 94.5 94.5 94.5
line flow23.75 23.75 21.84 19.96 18.11 16.26 14.4312.61 10.8 8.99 7.19 5.39 3.59 1.79
sprinkler flow
line flow
14
120
120
100
100
% of head loss
Sprinkler pressure m
Fig. 2: Head loss and percent of head loss
80
80
60
60
40
40
20
20
0
0
12 24 36 48 60 72 84 96 108 120 132 144 156
pressure (m) 40 38.5 37.2 36.1 35.2 34.4 33.9 33.4 33 32.8 32.6 32.5 32.5 32.5
Pres Red 100 96 93 90 88 86 85 84 83
82 82 81 81 81
% of reduction
0 20 37 52 64 75 81 88 93
96 99 100 100 100
% of length
0
8 15 23 31 38 46 54 62
69 77 85 92 100
Plain line 100 96 92 88 84 80 76 72 68
64 60 56 52 48
pressure (m)
Pres Red
% of reduction
Plain line
15
16
17
18
Head loss Calculation Along lateral

Select a suitable sprinkler or emitter with a
required Hs, qs and sl from a catalogue (figure 3).

The number of sprinklers (n) along the lateral is
determined by (L/sl).

The discharge rate at the
determined by (Qu = n x qs ).

The lateral diameter (D) should comply with
maximum head loss of 20%.

The head loss along a lateral (Qu, q, D and L) is
computed by:
o
Assuming the lateral pipe is plain and.
o
The outcome is multiplied by F factor.
lateral
inlet
is
19
Fig. 3 - Naan 233
20
Head loss in drip lateral pipe
 A modified Hazen-Williams head loss equation:
L
N  q 1.85
H L  2.78  10  F  4.87  (
)
D
C
6
HL = head loss along a lateral drip line
L = lateral length (m)
D = internal diameter (m)
N = number of emitters
q = average emitter flow rate (m3/h)
C = Hazen-Williams coefficient (130 - 120 for
F
polyethylene
pipe with ID < 16 mm)
= 0.37 for more than 20 emitters
21
Hydro P.C. & Hydro P.C.N.D - 1.2* L/H MAXIMUM
RECOMMENDED DRIPLINE LENGTH (m)
PIPE DIAMETER -16/13.8 (OD/ID)
Emitter Spacing (m)
Pressure
Slope
(m)
%
0.2
0.3
0.4
0.5
0.6
0.7
0.8
1
1.25
1.5
0
65
92
117
140
161
181
201
236
278
315
2
71
104
135
166
195
223
251
305
371
434
4
76
113
150
186
221
256
290
357
439
519
0
75
107
135
162
186
209
231
273
321
365
2
81
117
152
185
217
248
278
337
408
476
4
85
126
165
204
241
279
314
386
473
558
0
83
118
149
179
206
232
256
302
355
405
2
88
128
165
201
235
268
301
363
438
510
4
92
136
178
219
258
298
335
410
501
590
0
90
127
161
193
223
251
278
327
385
438
2
94
137
176
214
250
286
319
385
464
539
4
98
144
188
232
273
314
354
431
525
617
0
95
135
172
206
238
267
296
349
411
468
2
100
144
186
226
264
301
337
405
486
566
4
104
152
198
243
286
328
370
450
546
641
0
101
143
181
217
251
282
312
369
434
494
2
105
151
195
237
276
314
352
423
508
588
4
109
159
207
253
298
341
384
467
566
663
0
105
149
190
228
263
296
327
387
455
519
2
109
158
203
246
287
327
366
439
526
611
4
113
165
214
262
308
354
397
482
585
684
0
110
156
198
237
274
309
342
403
475
540
2
114
164
211
256
298
339
378
455
545
630
4
117
171
222
271
319
365
410
497
601
702
0
114
161
205
246
284
320
354
419
493
561
2
118
169
218
264
308
350
390
469
561
650
4
121
176
229
279
328
375
421
510
618
720
0
117
167
212
254
294
331
366
433
510
581
2
121
175
224
272
317
361
402
482
578
668
4
125
181
235
287
337
385
432
523
631
737
0
121
172
219
262
303
342
378
447
526
600
2
125
179
231
279
325
370
413
495
591
684
4
128
186
241
294
345
394
442
535
646
752
10
12
14
16
18
20
22
24
26
28
30
22
Number Of Mamkad spinklers
23
Mamkad mini-sprinkler nozzles
24
Example:
A flat field, 360 x 360 m, is irrigated with a hand
moved aluminum lateral pipe (C = 140). The water
source to the lateral pipe is from a sub-main,
which crosses the center of the field. The
selected sprinklers are Naan 233/92 with a nozzle
size of 4.5 mm, pressure of 25 m (hs) and flow
rate (qs) of 1.44 m3/hr. The space between the
sprinklers is 12 meters apart, and the location of
the first sprinkler is 6 meters away from lateral
inlet. The riser height is 0.8 m and diameter of
3/4".
25
Answer:
lateral 360 m
Submain
The number of sprinklers on the lateral is
180
n
 15sprinklers
12
The length of the lateral (l) is
l = (14 sprinkler x 12 m apart) + 6 m = 174 meters
26
Continue
The inlet flow rate of the lateral is
Qu = 15 (sprinklers) x 1.44 m3/h = 21.6 m3/h
The maximum allowed head loss (20%) throughout the field
20
h 
 25  5meters
100
For a plain 2" aluminum pipe - the hydraulic
gradient out of Hazen Williams is: J = 188.9
‰
27
Continue:
The head loss in a 2" (plain) aluminum pipe is as follow
J
h
J  L 188.9 174
1,000  h 

 32.9m
L
1,000
1,000
The F factor for 15 sprinklers is
11.9m  5m
F15 = 0.363
h f  h  F15  32.9  0.363  11.9m
For a 3" aluminum pipe - the hydraulic gradient
out of a table or ruler is: J = 26.2‰
28
Continue:
The head loss in a 3" (plain) aluminum pipe is as follo
J  L 26.2 174
h 

 4.6m eters
1,000
1,000
The F factor for 15 sprinklers is
F15 = 0.363
h f  h  F15  4.6  0.363  1.66m eters
1.66  5m
The difference 5 - 1.66 = 3.34 meters head loss
which will be used as the head loss for the submain pipe.
29
A Lateral Inlet Pressure
P=?
D
Qu = n*q
3q
2q
Sl
30
A Lateral Inlet Pressure
The pressure head at the lateral inlet (hu) is
determined by:
3
hu  hs   h f  riser  lh
4
hu - lateral inlet pressure head
hs - pressure head of selected sprinkler
hf - head loss along lateral
riser – the length (height) of the riser
l h - local head loss (incurred between laleral
pipe and
sprinkler)
31
Local head loss
Flow rate
Local head loss m
32
Example:
Following the previous example, what is the inlet pressu
hf = 1.66 meters
riser height = 0.8 meters
hs = 25 meters
3
hu  hs   h f  riser
4
3
hu  25   1.66  0.8  27m  2.7 atm osphere
4
33
Inlet pressure in case of a Lateral pipe
Laid out on a Slop
The inlet pressure of a lateral pipe which laid
out along a slope is as follows:
3
z
hu  hs   h f  riser 
4
2
hu - the lateral inlet pressure
hs - pressure head of selected
sprinkler
hf - head loss along lateral
riser - riser height
Z

- adjustment for an upward
2
Z
slope

2
- adjustment for an downward
slope
34
Example:
Following the previous example, but this time
with: a. 2% downward slope, or b. 2% upward
slope.
The difference elevation
2 between the two ends
 Z  174 
 3.48meters
is as follows:
100
a. 2% downward slope
3
z
hu  hs   h f  riser 
4
2
3
3.48
hu  25   1.66  0.8 
 25.3m
4
2
35
Cont.
The pressure by the last sprinkler is as follows:
hu  h f  Z
25.3  1.66  3.48  27.12m
The head loss between lateral inlet and last sprinkler is:
25.3  27.12  1.82 m
36
Cont.
27.12m
360 m
25.3m
P=30.3m
P=28.5m
Sub-main
28.5 – 25.3 m = 3.2 m is taken place along the sub-main pipe.
Therefore, the pressure at the head of the field is 28.5 m.
37
Continue:
b. 2% upward slope
3
z
hu  hs   h f  riser 
4
2
3
3.48
hu  25   1.66  0.8 
 28.78m
4
2
The total head loss throughout the lateral pipe is:
hd  1.66  3.48  5.14meters
5.14 meters are just the permitted 20% head loss. Therefore,
nothing is left for the sub-main. In this case, pressure
regulators should be installed in every lateral inlets or
selecting a wider pipe.
38
Maximum Permitted Head loss
39
Distribution of water and pressure
s
40
“The 20% rule”
In order to maintain up to 10% difference in flow
rate between sprinklers or emitter within a sub-plot,
then the pressure difference inside the plot should
be less than 20%.
Q  C  A 2 g  H
or
Q K H

Q - flow rate
C – coefficient, which depends on a nozzle type
A - cross section area of a nozzle
H - pressure head
X - exponent which depends on the flow pattern.
41
Pressure m
Pressure Vs. Flow
Liter/hr
42
O-tif Flow rate Vs. Pressure
2lph
4lph
8lph
16lph
Recommended
pressure range
(
lph
)
Flow rate
Brown
Black
Green
Purple
.
.
.
.
.
.
.
.
Inlet pressure (kg/cm )
43
3/8" Rondo Nozzles - Flow rate Vs. Pressure
)
(gph
Flow rate
Plastro Red
Plastro Green
Plastro Dark blue
Plastro Blue
Plastro Black
Pressure (psi)
44
Supertif - flow rate Vs. Pressure
Flow rate (lph)
l/h
l/h
l/h
l/h
l/h
l/h
Inlet pressure (m)
45
Ram & Agriplas Flow rate Vs Pressure
46
Flow rate Vs Pressure
47
Temperature Vs CV (tapes drip-line)
48
Example:
What is the expected difference discharge
between the two ends of a lateral sprinkler?
When the hydraulic gradient along a lateral pipe
is 20%.
The flow rate of a sprinkler is as follows:
QK H
49
Continue:
Two identical sprinklers have a same coefficient:
K  H2
Q2

Q1
K  H1
Q2

Q1
H2

H1
H2
H1
H 2  0.8  H1 (20%difference)
Q2

Q1
0.8 H1
 0.8  0.89  90%
H1
The difference in flow rate between the two ends is
10% (within 20% rule), once the exponent is 0.5.
50
TUFFTIF Dripper Flow Rate Table
TYPE
Q=AxH^B *
COLOR
(l/h)
A
B
BLACK
BLACK
GREEN
(2)
(4)
(8)
0.615
1.3643
2.7257
0.539
0.478
0.465
RED
(11)
4.3329
0.414
%DH
FLOW RATE (LPH) AT PRESSURE
2m
4m
6m
8m
10m
12m
14m
16m
18m
19.3
22
22.7
0.9
1.9
3.8
1.3
2.6
5.2
1.6
3.2
6.3
1.9
3.7
7.2
2.1
4.1
8
2.3
4.5
8.7
2.6
4.8
9.3
2.7
5.1
9.9
2.9
5.4
10.5
25.8
5.8
7.7
9.1
10.3
11.3
12.1
12.9
13.7
14.4
Q=LPH H=m.
%DH - MAXIMUM ALLOWABLE HEAD LOSS DIFFERENCE.
ALONG A LATERAL (AS PERCENT OF THE WORKING PRESSURE.
TO KEEP FLOW RATE DIFFERENCEC ≤ 10%
OF WORKING FLOW RATE.
51
Tufftif performance chart
20
18
16
BLACK (2)
12
BLACK (4)
10
GREEN (8)
6
RED
(11)
)
8
(lph
FLOW RATE
14
4
2
0
0
2m
4m
6m
8m
10m
12m
14m
16m
18m
20m
25m
30m
PRESSURE (m)
52
Example for a micro sprinklers
A polyethylene lateral pipe, grade 4, has 10
micro-sprinklers at 10 meters apart, while the
first is only one half way. The flow rate of the
selected sprinkler is qs = 120 l/h at hs = 20
meters. The riser’s height is 0.15 meter (can be
ignored). What is the required pipe for the
qs=120l/h
lateral pipe?
h =20m
s
D = ? mm
0.15 m
Q=1.2m3/h
10 m
95 m
53
Continue:
n = 10 micro-sprinklers
length (L) = (9 sprinkler x 10 m) + 5 m = 95 meters
F10 = 0.384
10 120
3
Qu 
 1.2m / h
1,000
54
Continue:
The maximum allowable in the field is as
follows:
20
 20  4meters
100
For a 20 mm polyethylene pipe grade 4 (ID 16.6
mm), the hydraulic gradient found out of a slide
ruler or monograph Q = 1.2 m3/h is J = 18.5%.
95
h  18.5 
 17.5m
100
h f  h  F10
h f  17.5  0.384  6.74m
6.74 meters is exceeding the allowable 4 meters (20%)
55
Continue:
The hydraulic gradient for a 25 mm P.E. pipe (ID 21.2
mm) and Q = 1.2 m3/h is J = 5.8%.
95
h  5.8 
 5.5m
100
h f  5.5  0.384  2.1m
The head loss of 2.1 meters is less than the
allowable 4 m (20%). The maximum allowed head
loss along the manifold is 4 m - 2.1 m = 1.9 meters.
56
Continue:
The required pressure by the lateral inlet pipe is as follow
3
hu  20   2.1  21.5meters
4
57
Design an Irrigation System

Option 1 - The rule of 20% is applied to all the
outlets (either sprinklers or drips) on the same
subplot. Any excess pressure over 20% between
the subplots is controlled by flow pressure
regulators.
58
Lay out of drip line without pressure
regulator
59
Drip line inlet without pressure
regulator
60
Design an Irrigation System
Option 2 - The rule of 20% is applied to a
single lateral, and pressure regulators control
the pressure difference between the laterals.
61
Inlet with pressure regulator
62
Design an Irrigation System

Option 3
- The difference pressure along a
lateral pipe exceeds the 20% head loss by any
desired amount, and the excess pressure should be
reduced by pressure or flow regulators in each
emitters or sprinklers.
63
Example:
Ten micro-sprinklers are installed along a plastic
lateral pipe (grade 4) at 10 m (32.8 ft) apart (the
first sprinkler is 5 meters). The flow rate of the
selected sprinkler is qs = 120 l/h (0.5 GPM), at a
pressure of hs = 20 meters. The riser height is 0.15
m (which can be ignored). What is the appropriate
lateral pipe diameter, if the field is designed and
abided by options 1, 2 and 3?
n = 10,
L = 95 m
F10 =0.384
10 120
Q
 1.2m3 / hr
1,000
64
Continue:
Option 1:
For 20 mm - The hydraulic gradient For a 20
mm P.E. pipe and Q = 1.2 m3/h is J = 18.5%.
95
h  18.5 
 17.5m
100
h  h  F10  17.5  0.384  6.72m
6.72 meters exceed the allowable 4 meters (20%)
65
Continue:
For 25 mm - The hydraulic gradient For a 25
mm P.E. pipe and Q = 1.2 m3/h is J = 5.8%.
95
 5.5m
100
h  h  F10  5.5  0.384  2.11m
h  5.8 
The head loss difference 4 - 2.11 = 1.89 m, which
is available for the manifold head loss.
The inlet lateral pressure is:
3
hu  20   2.11  21.58meters  2.15atmosphere
4
66
Continue:
Option 2:
If the allowable pressure variation along the
lateral pipe is 4 meters, then 25 mm P.E. pipe is
too much and 20 mm pipe too small. Therefore, a
combination of the two pipes can be used.
The design procedure for the combined lateral pipe
is:
• Try first D = 25 mm along 35 meters (n = 4) and
D = 20 mm along 60 meters (n = 6)
67
Continue:
Compute the head loss for a pipe D = 25, L = 95
m, n = 10 and Q = 1.2 m3/h (from previous
calculation which it was found 2.11 m)
Compute the head loss for D = 25, L = 60 m, n
= 6 and F6 = 0.458
6  120
Q
 0.72m3 / hr
1,000
68
Continue:
From a table or a slide ruler - the hydraulic
gradient for D = 25 mm and Q = 0.72 is J = 2.4%.
60
h  2.4 
 1.4m
100
h f  h  F6  1.4  0.458  0.64
The head loss for D = 25 mm and 35 meter long
with four sprinklers is
2.11 m - 0.64 = 1.47 meter
L = 35 m
h = h95 – h60
L = 60
m
=
L = 95 m h
h = 2.11 0.64
mm
69
Continue:
Compute the head loss in D = 20 mm, L = 60
meters, n = 6 and F6 = 0.458 and
6  120
Q
 0.72m3 / hr
1,000
From tables or a slide ruler the hydraulic
gradient for D = 20 mm and Q = 0.72 is J =
7.6%.
60
h  7.6 
 4.5m
100
h f  h  F6  4.5  0.458  2.06m
70
Continue:
The total head loss along the combined 25
and 20 mm lateral is as follows:
hf  h25  h20  1.47  2.06  3.53m
Since 3.5 m is too less than 4.0 meters.
Therefore, it is possible to try a shorter 25 mm
pipe with a length of 25 m and n = 3 and a longer
20 mm diameter pipe along 70 m and n = 7. The
previous procedure should be repeated. The new
head loss is 4.5 meters, which exceeds the limit
of 4 meters - (20% rule).
71
Continue:
The lateral inlet pressure requirement is as follows:
3
hu  20   3.5  22.7m  23m  2.3atmosphere
4
72
Continue:
Option 3:
The lateral pipe is design either with flow or
pressure regulators in every - sprinkler. The
laterals diameter can be reduced to 20 mm or
even further to 16 mm.
In case of 20 mm diameter pipe, the head loss is
6.72 meters (see Option 1). Therefore, the
pressure requirement at the last lateral inlet is:
hu  20  6.72  26.72m  27m  2.7at.
The lateral inlet pressure - the entire head loss is added
to the required sprinkler pressure.
73
Example:
A manifold was installed along the center of a
rectangular field (100 x 100 m). The lateral pipes
were hooked up to the two sides of the manifold
pipe. The difference in elevation between the
center and the end of the field is 2 meters (either
positive or negative). Each lateral pipe has eight
120 l/hr micro-sprinklers at 6 meters apart and the
pressure (hs) is 25 meters. What is the required
diameter of the lateral pipess, if the system is
designed and abides by option 1?
74
Answer:
The lateral's head loss along the two sides of the
manifold should be close enough (in away that
the total head loss due to the difference in
elevation and friction on both sides of the
manifold should be almost the same).
6m
75
Continue:
The maximum head loss between the sprinklers
throughout the field is:
20
 25  5meters
100
For a 20 mm (ID = 16.6 mm) lateral pipe on the two
sides:
n=8
F8 = 0.394
L = (7 sprinklers x 6 m) + 3 m = 45 m
120
Q  8
 0.96m3 / hr
1,000
76
Continue:
The hydraulic gradient for Q = 0.96 m3/hr
and D = 20 mm (ID = 16.6 mm) is J =
12.5%
45
h  12.5 
 5.62m
100
h f  5.62  0.394  2.21
The inlet pressure on the downward slope lateral is:
3
2
hu  25   2.21   25.6m  2.56 at .
4
2
77
Continue:
The pressure by the last sprinkler is
h8  25.6  2.21 2  25.4m  2.5atmosphere
The pressure difference between the two ends is
hd  hu  h8  25 .6  25 .4  0.2 m  0.02 at .
78
Continue:
The inlet pressure by the lateral upward is:
3
2
hu  25   2.21   27.65m  2.76 at .
4
2
The pressure by the last sprinkler is
h8  27.65  2.21 2  23.4m  2.3at.
The head loss along the upward lateral is
27.65 m - 23.4 m = 4.25 m, which is less than 5 m - 20%
rule
79
Continue:
The pressure requirement for the upward laterals inlet is
27.65 m and for downward laterals inlet is only 25.6 m.
The head loss along the upward laterals is 4.25 m, almost
all the permitted 20% (5 m). Therefore either:
•the upward lateral will be increased to 25 mm or more,
•the manifold can be reallocated to a higher position.
• (or pressure regulators should be installed by the lateral
inlets,)
80
Continue:
When 20 mm lateral pipes are in used, the values of
hu for both sides of the manifold vary by 27.65 - 25.6
= 2.05 m .
To avoid this difference (hu) in the pressure, the
upward 20 mm laterals can be replaced by 25 mm.
The inlet pressure (hu) for 25 mm is 26.5 m. Therefore,
the the difference inlet pressure for both sides of the
manifold will be less, only 26.5 - 25.7 = 0.8 m.
Less expensive alternative is by reallocating the
manifold away from the center of the field to a higher
point. That way, 6 sprinklers will be on the upward
side and 10 sprinklers on the downward laterals.
81
Continue:
Downward laterals:
D = 20 mm
n = 10 L = (9 sprinkler x 6 m) + 3 m = 57 m
Q = 1.2 m3/hr
F10 = 0.384

J = 18.5%
57
h  18.5 
 10.53m
100
h f  h  F10  10.53  0.384  4.15m  4.2m
82
Continue:
The difference elevation is as follows:
2
57
 4%  4 
 2.28m
50
100
The pressure at the lateral inlet is as follows:
3
2.28
hu  25   4.2 
 27.1m
4
2
83
Continue:
The pressure head by the last lateral sprinkler is
hd  hu  hf  Z  27.1  4.2  2.28  25.2m
The head loss along the downward lateral is
h  hu  hd  27.1  25.2  1.9m
84
Continue:
Upward laterals:
n=6
F6 = 0.405 L = (5 sprinklers x 6) + 3 = 33
m Q = 0.72 m3/hr D=20 mm (ID=16.6mm) 
J
= 7.6%
33
h  7.6 
 2.5m
100
h f  h  F6  2.5  0.405  1.01m  1m
The elevation difference is for a slope of 4% is:
33
4
 1.32 m
100
85
Continue:
The pressure head at the lateral inlet is:
3
Z
3
1.32
hu  hs   h f 
 25   1.0 
 26.51m
4
2
4
2
The pressure head at the last lateral sprinkler is
hd  hu  hf  Z  26.511.0 1.32  24.19m
86
Continue:
The head loss along the upward lateral is
h  hu  hd  26.51 24.19  2.32m
The values of hu for both sides are 27.1 m and
26.51 m which is practically the same. The
maximum head loss is 2.3 m, so 2.7 meters are
available as a head loss for the manifold.
87
Design of a manifold pipe
The manifold is a pipe with multiple outlets with
the same space between the outlets, therefore
the manifold is designed the same way as a
lateral.
88
Example:
A fruit tree plot (96 x 96 m) is designed for
irrigation with a solid set system. A manifold is
laid throughout the center of the field. The whole
plot is irrigated simultaneously. The flow rate of
the selected micro-sprinkler is qs = 0.11 m3/hr at a
pressure (hs) of 2.0 atmosphere. The space
between the micro-sprinklers along the lateral is 8
meters (26.24 ft.) and between the laterals is 6
meters (19.68 ft.). What is the required diameter of
the pipes? (The local head loss is 10% of the total
head loss and is taken in account).
89
Answer:
lateral
qs=.11m3/h
Ps=20m
6m
8
m
manifold
90
Continue:
The maximum allowable pressure head variation
20
20 
 4meters .
100
The number of micro-sprinkler on every lateral i
48
6
8
91
Continue:
F6= 0.405
L = (5 sprinklers x 8 m) + 4 m = 44 m
Q = 0.11 m3/hr x 6 sprinklers = 0.66 m3/hr
For 16 mm P.E. lateral pipe (ID = 12.8 mm)
The hydraulic gradient for 16 mm P.E. pipe
(ID=12.8mm) and Q = 0.66 m3/hr is J = 22.3%
44
h  22.3 
 9.8m
100
92
Continue:
The head loss along 16 mm lateral pipe
(including 10% local head loss) is as follows:
hf  (10%local head  loss)  h  F6  1.1 9.8  0.405 4.36m
4.36 m head loss exceeds the allowable 4 meter
(20%). So we have to try the head loss for 20 mm
P.E. lateral pipe.
93
Continue:
The hydraulic gradient for 20 mm P.E. pipe (ID =
16mm) and Q = 0.66 m3/hr is J = 6.5%
44
h  6.5 
 2.86m
100
The head loss in 20-mm lateral pipe (including
10 local head loss) is
h f  1.1  2.8  0.405  1.3m
94
Continue:
1.3 m head loss is less than 4 m (20%) and can
be selected as a lateral.
The lateral inlet pressure is:
3
3
hu  hs   h f  20   1.3  21m
4
4
The water pressure at the last micro-sprinkler
on the lateral pipe is, as follows:
h6  hu  h f  21  1.3  19.7m
95
Continue:
Manifold Design:
The number of laterals is
96
(N) 
 2  (two sides)  32
6
F32 = 0.376
L = (31 laterals x 6 m) + 3 = 93 m
Q = 32 x 0.66 = 21.1 m3/hr
96
Continue:
The hydraulic gradient for 63 mm P.E. pipe (ID = 58.2
mm) and Q = 21.1 m3/hr is
J = 7.2%
93
h  7.2 
 6.69
100
The head loss in 63 mm P.E. pipe (including 10%
local head loss) is as follows:
hf  (10% local head loss)  h  F32  1.1 6.69 0.376  2.76m
97
Continue:
The pressure by the manifold inlet is as follows:
3
Z
3
hum  hul   h f 
 21   2.76  23.07 m
4
2
4
The pressure by the last lateral inlet is as follows:
hd  hum  h f  23.07  2.76  20.31m
The maximum pressure in the entire plot is at
the first lateral inlet 23.07 meters (2.3
atmosphere)
98
Continue:
The minimum pressure throughout the system
is at the last sprinkler on the last lateral, which
is as follows:
20.31 - 1.3 = 19.01 m
The pressure difference between the first and last
sprinkler is as follows:
23.07 - 19.01 = 4.06 m (i.e. just above 4 meters
(20%))
99
Distribution of water and pressure
s
100
Distribution of water and pressure
p max.
haverage
qaverage
q min
<10%
101
Designing of Irrigation System

Considerations: soil, topography, water supply and
quality, kind of crops, climate.
o
o
o
o
o
o
o
o
Soil – infiltration rate, field capacity, (the lighter the soil
is - a higher advantage to drip system). ‫המטרה‬
Topography – the steeper the terrain - a higher
advantage to drip system.
Water supply – availability (time), pressure and quantity
Water quality – salinity (chlorine, SAR, B, heavy metal
or any other toxic), hardness, Fe, Mn, total suspended
material and type.
Crop – as the root system shallower a higher advantage
to micro irrigation system (closing spacing).
Price – as the expected income is relative higher - a
better water distribution system is an advantage.
Crop – Layout of the crop and type.
Climate – evaporation, wind pattern, crop protection
(high or low temperature)
102
Continue:


Farm schedule.
o
Working time.
o
Crop related activity – such as chemical application,
harvesting, weeds control and so on.
Water application:
o
Estimate water application depth at each irrigation cycle.
o
Determine the peak period of daily water consumption.
o
Determine the frequency of water supply.
103
Continue:

Irrigation system:
o
Consider several alternative types of irrigation
systems.
o
Determine the sprinklers or emitters spacing,
discharge, nozzle sizes, water pressure. ‫טפטוף המטרה‬
o
Determine the minimum number of sprinklers or
emitters (or a size of subplot) which must be operated
simultaneously.
104
Continue:

Irrigation layout:
o
Divide the field into sub-plots according to the crops,
availability of water and number of shifts (in one
complete irrigation cycle).
o
Determine the best layout of main and laterals.
o
Determine the required lateral size.
o
Determine the size of a main pipe.
o
Select a pump.
105
Continue

Prepare plans, schedules, and instructions for a
proper layout and operation.

Prepare a schematic diagram for each set of submains
or
manifolds
which
can
operate
simultaneously.

Prepare a diagram to show the discharge, pressure
requirement, elevation and pipe length.

Select appropriate pipes, starting at the downstream
end and ending up by the water source.
106
Combination of pipes
The total head loss along 300 meters PVC (grade 6) pipe is
15 m, with a flow rate of 180 m3/h. Which size of pipes are
required?
Q=180 m3/h
300 m
h=15m
107
Answer
The hydraulic gradient for 160 mm (ID 150.2 mm) is: 3.4%
Therefore, the head loss for 300 m long pipe is: 10.2 m (too
big pipe)
The hydraulic gradient for 140 mm (ID 131.4 mm) is: 6.4%
Therefore, the head loss for 300 m long pipe is: 19.2 m (too
small pipe).
Therefore, a combination of the two can make it.
108
Cont.
Q=150 m3/h
300 m
L
300-L
3.4  L 6.4  (300 L)
15 

100
100
3L  1920 1500
L  140m
140 m 160 mm PVC pipe + 160 m 140 mm PVC pipe
109
Example:
A flat field with two plots, each plot is divided into
six subplots. The selected system for this field is
drip irrigation. The flow rate in each subplot is 21
m3/hr and the pressure requirement to the submain inlets is 25 meters. The interval of water
supply is every three days and only one shift a
day. Therefore, two subplots in each plot must be
irrigated simultaneously. The main pipes are
made of PVC (C = 150) and are buried 0.6 meters
deep. The local head losses is up to 10% of the
longitudinal head losses. The pump pressure is
50 meters and with a flow rate of 84m3/h (local
head loss due primary filter and others pump
110
attached accessories is 10m).
Continue:
1
D
3
50m
96m
C
2
50m
96m
B
4
1’
5’
E
5
A
6
3’
C’
B’
A’
2’
4’
6
96m
96m
Q=84m3/h F Pump
111
Pressure requirement
Qu=21m3
Hu=25 m
Qu=84m3
Hu=40m
112
Continue:
The sequence of water application is as follow
First day
-
1, 5, 1' and 5' plots
Second day -
2, 6, 2' and 6' plots
Third day
3, 4, 3' and 4' plots
-
113
Continue:
The diagram for the first and second day of water
supply is:
1
2Q=42m3/h
C
2
L=50m
E
2Q=42m3/h
1’
5’
C’
2’
L=50m
5
Q=21m3/h
250m
D
3
B
4
L=192m
hu=25.6m
A
6
3’
Q=21m3/h
B’
4’
L=192
A’
6
F Pump (Q=84m3/H= 50m)
114
Continue:
The diagram for the third day of water supply is:
1
D
3
5
hu=25.6m
Q=0m3/h
2Q=42m3/h
C
2
B
4
A
6
L=146m
1’
5’
E
2Q=42m3/h C’
2’
L=146m
3’
Q=0m3/h
B’
A’
4’
6
F Pump (Q=84m3/h, 40m)
115
Main pipes’ diagram for first and
second day
D
E
F
120 m
84 m3/h
300 m
42 m3/h
A
C
192m
21m3/h
116
Main pipes’ diagram for third day
(Case 2)
D
E
F
120 m
84 m3/h
B
396 m
42 m3/h
117
Continue:
The following table presents the the head loss
(including 10%) for local head loss for selected
pipe:
Discharge
(m3/h)
pipe
length
(m)
2"
3"
4"
5"
21
A-C
192
31.7
4.5
1.1
0.4
50
146
250
8.3
42
C-D
B-D
D-E
1.2
12.1
20.7
0.3
3
5.2
0.1
1
1.8
84
E-F
120
8.8
3
6"
1.2
118
Continue:
Case 1: Design system for the first and second day.
hu - 25.6 m (including the depth of the main pipe 0.6 m)
pump pressure - 40 m
Total head loss = 40 - 25.6 = 14.4 m
Head loss for the selected pipe:
A-C 192m Q = 21 m3/hr 3" pipe = 4.5 m
Q = 42 m3/hr 4" pipe
C-D
50m
= 1m
D-E
250m Q = 42 m3/hr 4" pipe = 5.2 m
The pressure by E is : 25.6 + 4.5 + 1.0 + 5.2 = 36.3 m.
The head loss available for E-F = 40 – 36.3 = 3.7 m
119
Pressure diagram for case 1
30.1m
31.1m
D
4” 50m
C
25.6m
A
3” 192m
36.3m E
40m
F
120
Continue:
For E - F (pump) section:
4" pipe is too small (8.8 m head loss, 6.7%), on the
other hand 5" pipe is too much (3 m head loss,
2.3%). Therefore, a combination of the two is
selected for E-F section.
1.1(6.7 
L( 4")
100
 2.3 
120 L
100
)  3.7
L (4") = 13.7 m
L (5“) = 120 – 13.7 = 106.3 m
121
Case 2
Hu-25.6
D
146m3
B
42 m
42 m3
E
F
Hu-40m
122
Continue
Case 2: Design system for the third day:
hu – 25.6 m (including the depth of a main pipe 0.6 m)
After pump pressure - 40 m
Maximum head loss = 40 – 25.6 = 14.4 m
Selected pipes –
B-D Q = 42 m3/hr 4" pipe
=3m
D-E Q = 42 m3/hr 4" pipe
= 5.2 m
The pressure head by E tee is 26.6+3+5.2=33.8 m.
The head loss available for E-F = 40 – 33.8 = 6.2 m
123
Case 2
H-28.6m
D
H-33.8m
42 m3
4” 146m
Hu-25.6
B
E
F
Hu-40m
124
Continue
The pressure in case 2 at point E is 33.8 m which is lower than
in case 2 which was 36.3 m (pressure different of 2.5 m). .
Therefore, the pipe for DB should be reduced, in a way that the
pressure at a point D should be a same as case 1 which is
3
42
m
31.2m.
Hu-25.6
H-31.1m
D
146m
B
E
A combination of pipes 3” and 4” for DB section with
5.6 m head loss should be selected.
125
Continue
Section DB: the hydraulic gradient for 3” pipe with 42
m3 is 7.5% and for 4” is 1.9%
146 L
L3"
1.1(7.5 
 1.9 
)  5.5
100
100
3” pipe is = 39.75 m and for 4” is 106.25 m
(The actual length should take in account the
commercial pipe length)
126
Case 2
H-31.2m
D
4”
106.25m
3”
39.75 m
Hu-25.6
B
42 m3
E
In this situation the size of pipes DE and EF are a same as in case
1.
In case 2, the required pipes are smaller than case 1 (DB). (The
selected pipe in case 1 are too small to maintain the pressure
requirement in case 2)
127
Pipe diagram for case 1 & 2
Hu-25.6m
D
4” 106.75m
C
3” 135.25m
B
A
E
F
128
Head loss in case 1
For section A – C (192m 21m3/h):
3” 137.4 m
4” 54.6m
For section C – D (42m3/h)
4” 50 m
For section D – E (42 m3/h):
4” 250 m
For section E – F (120m 84m3/h):
4” 13.7 m
5: 106.3 m
The total head loss from A to F is
- 3.2 m
- 0.3 m
-1m
- 5.2 m
-1m
- 2.7 m
-13.4 m
129
Pressure diagram for case 1
H-30.1
D
H-29.1m
4” 104.6m
C
H-35.3m
H-39m
25.6m
3” 137.4m
B
A
E
F
130
Head loss in case 2
For section D – B (1146m 42m3/h):
3” 39.75 m
4” 106.25 m
For section D – E (42 m3/h):
4” 250 m
For section E – F (120m 84m3/h):
4” 13.7 m
5”: 106.3 m
The total head loss from A to F is
- 3.3 m
- 2.2 m
- 5.2 m
-1m
- 2.7 m
14.4 m
131
Pressure diagram for case 2
H-31.1
D
H-28.9m
4” 106.25m
C
25.6m
3” 39.75m
B
H-36.3m
E
H-40m
F
Case 2 -‫צריך להוסיף מפה דומה ל‬
132
Diagram for E – A’ case 2
H-31.1
D
H-28.9m
4” 104.6m
C
25.6m
3” 137.4m
B
A
H-36.3m
25.6m
E
B’
H-40m
F
133
The pipe for E – A’ for case 1
E 42m3/h C’
H-35.3m
50 m
Q – 21m3/h
A’
H-25.6m
192 m
The head loss E – A’:
35.3 – 25.6 = 9.7 m
The length of E – C’ is 50 m 42 m3/h  for 3” pipe the
head loss is 4.1 m
The length of C’ A’ is 192 m 21m3/h  the required head
loss for this section is: 9.7 – 4.1 = 5.6m
The hydraulic gradient for 2” is 15% and for 3” is 2.1%
and combination of the two will make it.
134
Continue
The hydraulic gradient for 2” is 15% and for 3” is 2.1% and
combination of the two will make it.
L2"
192  L
1.1(15 
 2.1 
)  5.6
100
100
8.2 m 2” pipe, and 183.8 m 3” pipe.
The head loss for E-C’ 3” pipe (42 m3/h is 4.1 m.
The total head loss for E-A’ is 9.7 m.
A’
C’
E
42m3/h
3”
233.8 m
Q – 21m3/h
2”
8.2m
135
The pipe for E – B’ for case 2
E
42m3/h
H-36.3m
B’
H-25.6m
146 m
The head loss E – B’:
36.3 – 25.6 = 10.7 m
The length of E – B’ is 146 m 42 m3/h
The head loss for 3” is too much 12.1 m
and for 4” is too small 3 m
A combination of the two will make it, the hydraulic
gradient for 3” is 7.5% and for 4” is 1.9%.
136
The combination pipes for E – B’
L3"
146  L
1.1(7.5 
 1.9 
)  10.7
100
100
3” pipe
4” pipe
124.1 m
21.9 m
137
Diagram for E – A’ for case 2
H-30.1
D
H-35.3m
E
H-29.1m
4” 104.6m
C
4”
21.9m
25.6m
3” 137.4m
B
A
B’
25.6m
3”
211.9m
2” A’
8.2m
H-39m
F
Since the size of the pipe E-A’ increased, therefore we have to
reconsider the pipe size for EA’ in case 1.
138
Section E-A’ case 1
25.6m
C’
E
4”
21.9m
3”
211.9m
2”
8.2m
A’
The head loss E – B’:
35.3 – 25.6 = 9.7 m
The length of E – C’ is 50 m 42 m3/h
The head loss for 4” 21.9 m is 0.5 m
The head loss for 3” 28. 1 m is 2.3 m
The length of C’ – A’ is 192 m 21 m3/h
The head loss for 3” 183.8 m is 4.3 m
The head loss for 2” 8.2 m is 1.4 m
The total head loss is 0.5 + 2.3 + 4.3 + 1.4 = 8.5 m
Therefore, the length of 2” pipe can be extended a little bit by (to
increase the head loss): 9.7 – 8.5 = 1.2 m:
139
Continue
The length of B – A is 96 m with 21 m3/h
The current head loss is:
for 3” 183.8 m is
-4.3 m
for 2” 8.2 m is
-1.4 m
The total head loss should be increased by 1.2 m
4.3 + 1.4 + 1.2 = 6.9 m
The hydraulic gradient for 2” and 21m3/h is 15% and
for 3” pipe is 2.1%
L2"
192  L
1.1(15 
 2.1 
)  6.9
100
100
The length of 2” pipe is 17.4 m
and 3” pipe is 174.6 m
140
Final Diagram for E – A’
H-30.1
D
H-29.1m
4” 104.6m
C
H-35.3m
E
4”
C’
4”
21.9m
25.6m
3” 137.4m
B
B’
3”
202.7m
A
25.6m
2” A’
17.4m
H-39m
5”
F
To overcome the pressure differences, pressure regulator
should be installed by every risers.
141
142
143
To overcome the pressure differences, pressure regulator
should be installed by every risers.
144
‫‪145‬‬
‫צריך להוסיף את בחינה של גודל צינורות ’‪ E-A‬עבור ‪case2‬‬
‫אשר נדרש לקטרים אחרים‬
Case 2
H-31.2m
D
4”
104.6m
3”
41.4 m
Hu-25.6
B
42 m3
E
146
Continue
The pipe selection for F - A is as follows:
A-D 3" 242 meters
D-E 4" 250 meters
E-F 6" 120 meters
147
Continue
The pipe selection for the section E-A' for case 1 is as fo
Hu=38m
C’
50 m
E
42m3/h
Hu=25.6
B’
A’
192m
21m3/h
148
Continue
The head loss is
38.0 - 25.6 = 12.4 m
Selection of pipes A'-C' Q = 21 m3/hr 2" pipe = 7.7 m
The head loss available for E-C'
38 - 7.7 - 25.6 = 4.7 m
149
Continue
2" pipe is too small (6.5 m), on the other hand
3" pipe is too much (2.4 m). Therefore, a
combination of the two pipes is required for EC' section. The two pipes which required for EC section are:
L
50  L
1.1(43 
 130 
)  4.7
1,000
1,000
L = 33 m
L(3") = 33 meters and 2" pipe 27 meters
150
Continue
The design for the section E-A' for case 2 is as
follows:
A’
H=38m
B’
C’
E
146m
42m3/h
QEB' = 42 m3/hr 2" pipe = 7.7 m
The hydraulic gradient for E-B’ section is 38.0 - 25.6 = 12
151
Continue
2" pipe is too small (18.9 m), on the other hand 3"
pipe is too much (6.9m) Therefore, a mix of the
two is required for F-B', section.
1.1(43 
L
146  L
 130 
)  12.4
1,000
1,000
L = 95
L (3") = 95 m and 2" pipe 51 m
152
Continue:
The selected pipe for E-B' section will be as in
case 2, 95 meters 3" pipe and 51 meters 2"
pipe.
Pressure regulators on some of the risers
should be considered to compensate the total
head loss over 20% (including the head loss
inside each plot).
153