IWM I
APPLICATION VOLUME
CALCULATIONS
NRCS -IWM II
1
BORDER AREA
1360 Feet
80 Feet
 AREA
= 1360’ x 80’ = 108,800 SQ. FEET
HOW MANY ACRES ARE IN THIS BORDER ?
NRCS -IWM II
2
APPLICATION VOLUME
 AREA
=108,800 SQ. FT. = 2.5 ACRES
APPLY 6 INCHES (GROSS)
 VOLUME = 2.5 AC. x 0.5’ =1.25 AC -FT
OR
 108,800
Sq. Ft. x 0.5 Ft. =54,400 Cu.Ft.
 For
an 8 hour set time what minimum
flow is needed in CFS (Cu. Ft./ sec) ?
NRCS -IWM II
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FLOW NEEDS
 8Hrs.
x 3600 Sec/Hr.
 =28,800 Sec.

54,400 cu.ft.
28,800 sec.

= 1.88 CFS
minimum flow needed
for an eight hour set
NRCS -IWM II
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APPLICATION DEPTHS
CONSIDERATIONS:
 Rooting depth
 AWHC
 MAD
 LEACHING NEEDS
 WATER SUPPLY
 IWR or TR - 21
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EFFECTIVE ROOT ZONE
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Soil and Rooting Conditions
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NRCS -IWM II
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DEEP PERCOLATION

THE UPPER END OF THE FIELD CONTINUES TO
TAKE IN WATER DURING THE ENTIRE TIME IT
TAKES FOR THE WETTED FRONT TO REACH
THE END OF THE FIELD.

IF THE TRAVEL TIME TAKES 1.5 HOURS, AT OUR
PREVIOUS FLOW RATE OF 1.88 CFS WHAT
VOLUME OF WATER IS ADDED ?
1.5 HRS. x 3600 SEC./ HR. x 1.88 CFS =10,152 CU FT
NRCS -IWM II
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TRAVEL TIME

TRAVEL TIME IS THE TIME
NEEDED FOR THE WETTED
FRONT TO TRAVEL FROM THE
ENTRANCE POINT TO THE FAR
REACHES OF THE SET.

THE INITIAL RATE OF TRAVEL
SLOWS DOWN AS MORE AREA
IN THE FIELD ABSORBS THE
WATER AND REDUCES THE
HEAD THAT DRIVES THE FLOW.

IN A GRADED BORDER IT MAY
TAKE MORE THAN AN HOUR
DEPENDING ON SOILS, BUT A
WHEEL ROLL SPRINKLER TAKES
ONLY MINUTES.
NRCS -IWM II
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EFFICIENCY
What is the application efficiency for a 5
inch net ? (50% MAD of 10 in. AWC)
(assume no runoff water)
Total H2O applied = 54,400 ft3 + 10,152 ft3 = 64,552 ft3.
5”/12” x 108,800 ft2=
54,400+ 10152 ft3=
NRCS -IWM II
45,333cu.ft.
64,552cu.ft
=70.2%
12
RUNOFF WATER
 If
this was a graded border instead of a
level border, there would be water
running off the low end of the field.
 If there were 10,000 cu.ft. water loss,
what is the application efficiency ?
45,333 cu.ft.
74,552 cu.ft
=60.8%
NRCS -IWM II
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WHEEL ROLL SPRINKLERS
1360 FT
80 FT.
 40’x60’
sprinkler lateral having 34 heads,
could cover the same area in 2 - 11 hour sets.
 What would be the flow rate for the lateral ?
 What would be the output for each head ?
NRCS -IWM II
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SPRINKLER APPLICATION
 1360

ft.x40 ft. = 54,400 sq. ft.
a 6 inch application uses 27,200 cu. ft.
11 hrs x 3600 sec. per hr. =39,600 sec.
27,200 cu.ft
39,600 sec
flow rate = 0.6868 cfs = 308 gpm or
9.1 gpm per sprinkler
NRCS -IWM II
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FEEL AND
APPEARANCE
25% - 50%
50% -75%
CLAY LOAM
AREAS OF A CIRCLE
FULL CIRCLE AREA
= π R2
A = 3.14 X (1320’)2
= 5473911 ft 2
(125.66 acres)
SO A HALF CIRCLE = ½ OF
THE FULL CIRCLE
= 63 acres
NRCS -IWM II
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UNDERSTAND PRODUCT PERFORMANCE
• Throw Distance
Performance
• Droplet Size
Pressure
Description
Range
Nozzle Range
• Uniformity
Concave
Medium
Groove
6 - 40 psi
.41 - 2.8 bar
3TN
#9 - #50
Throw Distance Data
3 ft. Mounting Height
(.9 m )
3TN NOZ. SIZE
DIAMETER
#24
26’
#36
36’
#44
32’
7.9 M
11.0 M
9.8 M
Coverage @ 10 psi (.7 bar)
BUT, A SPRAY HEAD SHOOTS ALL THE
WATER OUT ABOUT THE SAME DISTANCE
IN A “DONUT” A
ROUND THE HEAD
SO WHAT’S THE GREEN AREA IF THE INSIDE CIRCLE
HAS A 10’ RADIUS AND THE OUTSIDE CIRCLE IS 15’ ?
SPRINKLER
WHY A ROTATOR vs.PERFORMANCE
A SPRAY?
NRCS -IWM II
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Adequate soil moisture monitoring
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