1. NRCS -IWM II1 IWM II APPLICATION VOLUME CALCULATIONS

2. NRCS -IWM II2 BORDER AREA u AREA = 1360’ x 80’ = 108,800 SQ. FEET HOW MANY ACRES ARE IN THIS BORDER ? 1360 Feet 80 Feet

3. NRCS -IWM II3 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) ?

4. NRCS -IWM II4 APPLICATION DEPTHS CONSIDERATIONS u Rooting depth u AWHC u MAD u LEACHING NEEDS u WATER SUPPLY u IWR or TR - 21

5. NRCS -IWM II5 FLOW NEEDS u 8Hrs. x 3600Sec/Hr. u =28,800 Sec. u 54,400 cu.ft. 28,800 sec. u = 1.88 CFS

6. NRCS -IWM II6 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.

7. NRCS -IWM II7 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

8. NRCS -IWM II8 EFFICIENCY u Total volume = 54,400 + 10,152 cu. ft = 64,552 cu.ft. What is the application efficiency for a 5 inch net ? (50% MAD of 10 in. AWC) ( no waste water) 45,333cu.ft. 64,552cu.ft =70.2%

9. NRCS -IWM II9 WASTEWATER u If this was a graded border instead of a level border, there would be wastewater running off the low end of the field. u If there were 10,000 cu.ft. wastewater, what is the application efficiency ? 45,333 cu.ft. 74,552 cu.ft =60.8%

10. NRCS -IWM II10 WHEEL ROLL SPRINKLERS u 40’x60’ sprinkler lateral having 34 heads, could cover the same area in 2 - 11 hour sets. u What would be the flow rate for the lateral ? u What would be the output for each head ? 80 FT. 1360 FT

11. NRCS -IWM II11 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 = 0.6868 cfs = 308 gpm or 9.1 gpm per sprinkler