NRCS -IWM II 1 IWM I APPLICATION VOLUME CALCULATIONS

NRCS -IWM II 2 BORDER AREA u AREA = 1360’ x 80’ = 108,800 SQ. FEET HOW MANY ACRES ARE IN THIS BORDER ? 1360 Feet 80 Feet

NRCS -IWM II 3 APPLICATION VOLUME u AREA =108,800 SQ. FT. = 2.5 ACRES APPLY 6 INCHES (GROSS) u VOLUME = 2.5 AC. x 0.5’ =1.25 AC -FT OR u 108,800 Sq. Ft. x 0.5 Ft. =54,400 Cu.Ft. u For an 8 hour set time what minimum flow is needed in CFS (Cu. Ft./ sec) ?

NRCS -IWM II 4 FLOW NEEDS u 8Hrs. x 3600 Sec/Hr. u =28,800 Sec. u 54,400 cu.ft. 28,800 sec. u = 1.88 CFS minimum flow needed for an eight hour set

NRCS -IWM II 5 APPLICATION DEPTHS CONSIDERATIONS: u Rooting depth u AWHC u MAD u LEACHING NEEDS u WATER SUPPLY u IWR or TR - 21

NRCS -IWM II 6 EFFECTIVE ROOT ZONE

NRCS -IWM II 7 Soil and Rooting Conditions

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9 DEEP PERCOLATION u 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. u 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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NRCS -IWM II 11 TRAVEL TIME u TRAVEL TIME IS THE TIME NEEDED FOR THE WETTED FRONT TO TRAVEL FROM THE ENTRANCE POINT TO THE FAR REACHES OF THE SET. u THE INITIAL RATE OF TRAVEL SLOWS DOWN AS MORE AREA IN THE FIELD ABSORBS THE WATER AND REDUCES THE HEAD THAT DRIVES THE FLOW. u 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 12 EFFICIENCY What is the application efficiency for a 5 inch net ? (50% MAD of 10 in. AWC) (assume no runoff water) Total H 2 O applied = 54,400 ft ,152 ft 3 = 64,552 ft 3. 5”/12” x 108,800 ft 2 = 45,333cu.ft. 54, ft 3 =64,552cu.ft =70.2%

NRCS -IWM II 13 RUNOFF WATER u If this was a graded border instead of a level border, there would be water running off the low end of the field. u 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 14 WHEEL ROLL SPRINKLERS u 40’x60’ sprinkler lateral having 34 heads, could cover the same area in hour sets. u What would be the flow rate for the lateral ? u What would be the output for each head ? 80 FT FT

NRCS -IWM II 15 SPRINKLER APPLICATION u 1360 ft.x40 ft. = 54,400 sq. ft. a 6 inch application uses 27,200 cu. ft. u 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

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

NRCS -IWM II 17 AREAS OF A CIRCLE FULL CIRCLE AREA = π R 2 A= 3.14 X (1320’) 2 = ft 2 ( acres) SO A HALF CIRCLE = ½ OF THE FULL CIRCLE = 63 acres

10 psi (.7 bar) Throw Distance Data Performance 3TN NOZ. SIZE #24 #36#44 DIAMETER 26’ 36’ 32’ 7.9 M11.0 M9.8 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’ ?

NRCS -IWM II 20 WHY A ROTATOR vs. A SPRAY? SPRINKLER PERFORMANCE

Adequate soil moisture monitoring

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NRCS -IWM II 24 Apply water when the crop needs it most