1. APPLICATION VOLUME CALCULATIONS
IWM I
APPLICATION VOLUME CALCULATIONS

2. BORDER AREA AREA = 1360’ x 80’ = 108,800 SQ. FEET
HOW MANY ACRES ARE IN THIS BORDER ?

3. 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) ?

4. 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

5. APPLICATION DEPTHS CONSIDERATIONS: Rooting depth AWHC MAD
LEACHING NEEDS
WATER SUPPLY
IWR or TR - 21

6. EFFECTIVE ROOT ZONE

7. Soil and Rooting Conditions

9. 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

10. 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.

11. 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= ,333cu.ft.
54, ft3= 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%

13. WHEEL ROLL SPRINKLERS
1360 FT
80 FT.
40’x60’ sprinkler lateral having 34 heads, could cover the same area in hour sets.
What would be the flow rate for the lateral ?
What would be the output for each head ?

14. 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 = cfs = 308 gpm or
9.1 gpm per sprinkler

15. FEEL AND APPEARANCE
25% - 50%
50% -75%
CLAY LOAM

16. AREAS OF A CIRCLE FULL CIRCLE AREA = π R2 A = 3.14 X (1320’)2
= ft 2
( acres)
SO A HALF CIRCLE = ½ OF
THE FULL CIRCLE
= 63 acres ΠΠ

17. UNDERSTAND PRODUCT PERFORMANCE Throw Distance Droplet Size Uniformity
Throw Distance Data
3 ft. Mounting Height
(.9 m )
3TN NOZ. SIZE #24 #36 #44
DIAMETER 26’ 36’ 32’
7.9 M 11.0 M 9.8 M
10 psi (.7 bar)

18. 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’ ?

19. SPRINKLER PERFORMANCE
WHY A ROTATOR vs. A SPRAY?

21. Adequate soil moisture monitoring