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IWM I

APPLICATION VOLUME CALCULATIONS

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BORDER AREA

AREA = 1360’ x 80’ = 108,800 SQ. FEET

HOW MANY ACRES ARE IN THIS BORDER ?

1360 Feet

80 Feet

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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) ?

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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

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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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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

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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.

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EFFICIENCYWhat 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= 45,333cu.ft.

54,400+ 10152 ft3= 64,552cu.ft

=70.2%

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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%

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WHEEL ROLL SPRINKLERS

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 ?

80 FT.

1360 FT

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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

FEEL AND APPEARANCE

25% - 50%

50% -75%

CLAY LOAM

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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

Coverage @ 10 psi (.7 bar)

Throw Distance Data

PerformancePerformance

DescriptionPressure

Range Nozzle Range

ConcaveMediumGroove

6 - 40 psi.41 - 2.8 bar

3TN#9 - #50

3TN NOZ. SIZE #24 #36 #44DIAMETER 26’ 36’ 32’

7.9 M 11.0 M 9.8 M

3 ft. Mounting Height(.9 m )

UNDERSTAND PRODUCT PERFORMANCE• Throw Distance• Droplet Size• Uniformity

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’ ?

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WHY A ROTATOR vs. A SPRAY? SPRINKLER PERFORMANCE

Adequate soil moisture monitoring

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