Irrigation Water Management

Soil-Water-Plant Relationships

5-Practice Irrigation Calculations


1. For a clay soil, Field Capacity (FC) = 48%, Permanent Wilting Point (PWP) = 28%, what is Available Water Holding Capcity (AWHC) (%)?

AWHC (%) = FC (%) - PWP (%) = 48% - 28% = 20%


2. For a clay soil, Field Capacity = 48%, Permanent Wilting Point = 28%. What are Field Capacity, Permenent Wilting Point and Available Water Holding Capacity in in/ft?

FC: 48% x 12 in/ft = 5.76 in/ft

PWP: 28% x 12 in/ft = 3.36 in/ft

AWHC: 20 % x 12 in/ft = 2.40 in/ft


3. If PWP = 1.8 in/ft and AWHC = 1.6 in/ft, what is FC (in/ft)?

FC = PWP + AWHC = 1.8 in/ft + 1.6 in/ft = 3.4 in/ft


4. If PWP = 1.8 in/ft and AWHC = 1.6 in/ft. What are FC, PWP and AWHC in %?

FC: 100 x 3.4 in/ft ÷ 12 in/ft = 28%

PWP: 100 x 1.8 in/ft ÷ 12 in/ft = 15%

AWHC: 100 x 1.6 in/ft ÷ 12 in/ft = 13%


5. In a particular sandy loam soil, FC = 14% and AWHC = 6%. What is PWP measured in %?

PWP = FC - AWHC = 14% - 6% = 8%


6. In a particular sandy loam soil, FC = 14% and AWHC = 6%. What are FC, PWP and AWHC in in/ft?

FC: 14% x 12 in/ft = 1.68 in/ft

PWP: 8% x 12 in/ft = 0.96 in/ft

AWHC: 6 % x 12 in/ft = 0.72 in/ft


7. Complete the following table:

Layer

Depth

Thickness

Layer
AWHC
(in/ft)

Layer
AWHC
(in)

  [ans:]

  1

  0-12"

  

  0.75

  

  [0.75]

  2

  12"-36"

  

  1.60

  

  [3.20]

  3

  36"-48"

  

  2.20

  

  [2.20]

  all

  0-48"

  

  Total

  

  [6.15]

ANSWER:

Layer

Depth

Thickness

Layer
AWHC
(in/ft)

Layer
AWHC
(in)

  1

  0-12"

1 ft

  0.75

0.75

  2

  12"-36"

2 ft

  1.60

3.20

  3

  36"-48"

1 ft

  2.20

2.20

  all

  0-48"

4 ft

  Total

6.15


8. In a sandy loam soil, with AWHC = 1.25 in/ft, barley with a 4 ft root zone is grown. How much water is available to the crop (in)?

AWHC (in) = AWHC (in/ft) x RZ (ft) = 1.25 in/ft x 4 ft = 5 in


9. In a sandy loam soil, with AWHC = 1.25 in/ft, barley with a 4 ft root zone is grown. If the irrigation manager chooses the Management Allowable Depletion (MAD) to be 60%, what is the maximum allowed soil moisture depletion (Max SMD) (in)?

max SMD = AWHC (in) x MAD (%) = 5 in x 60% = 3 in


10. In a sandy loam soil, with AWHC = 1.25 in/ft, barley with a 4 ft root zone is grown. If the barley uses water at the rate of ETc (Crop Evapotranspiration) = 0.25 in/day, what is the soil moisture depletion 6 days after an irrigation?

amount = rate x time
SMD (in) = ETc (in/day) x time (days) = 0.25 in/day x 6 days = 1.5 in


11. In a sandy loam soil, with AWHC = 1.25 in/ft, barley with a 4 ft root zone is grown. How long between irrigations?

time = amount ÷ rate
time = max SMD ÷ ETc = 3 in ÷ 0.25 in/day = 12 days


12. Carrots with a 2 foot root zone are grown on a silt loam soil with AWHC = 1.25 in/ft. The irrigation manager chooses MAD = 40%. What is the maximum soil depletion allowed?

max SMD = AWHC (in/ft) x RZ (ft) x MAD (%)
max SMD = 1.25 in/ft x 2 ft x 40% = 1 in


13. Carrots with a 2 foot root zone are grown on a silt loam soil with AWHC = 1.25 in/ft. The irrigation manager chooses MAD = 40%. The carrots use water at the rate of 0.2 in/day. How many days can be allowed between irrigations?

time = amount ÷ rate
time = max SMD ÷ ETc = 1 in ÷ 0.2 in/day = 5 days


14. It has been 8 days since the last full irrigation of a loam soil. During that time, ETc has averaged 0.35 in/day. What is the soil moisture depletion (in)?

amount = rate x time
SMD (in) = ETc (in/day) x time (days) = 0.35 in/day x 8 days = 2.8 in


15. Corn with a 6 ft root zone is grown on a soil with AWHC = 1.5 in/ft. ETc = 0.35 in/day. The irrigation manager chooses MAD = 65%. How many days can you schedule between irrigations?

max SMD = AWHC (in/ft) x RZ (ft) x MAD (%)
max SMD = 1.5 in/ft x 6 ft x 65% = 5.85 in
time = amount ÷ rate
time = max SMD ÷ ETc = 5.85 in ÷ 0.35 in/day = 16.7 days - use 16 days

Note:
actual SMD just prior to irrigation = 16 days x 0.35 in/day = 5.60 in


16. Corn with a 6 ft root zone is grown on a soil with AWHC = 1.5 in/ft. ETc = 0.35 in/day. The irrigation manager chooses MAD = 65%. When you do irrigate, what net irrigation should be applied?

actual SMD just prior to irrigation = 16 days x 0.35 in/day = 5.60 in


17. Corn with a 6 ft root zone is grown on a soil with AWHC = 1.5 in/ft. ETc = 0.35 in/day. The irrigation manager chooses MAD = 65%. If the field size is 40 acres, what is the net volume of water needed (ac-ft).

volume = area x depth = 40 ac x 5.6 in x 1 ft/12 in = 18.67 ac-ft


18. Corn with a 6 ft root zone is grown on a soil with AWHC = 1.5 in/ft. ETc = 0.35 in/day. The irrigation manager chooses MAD = 65%. If the irrigation efficiency (IE) is 75%, what volume of water must the pump deliver (ac-ft)?

gross = net ÷ IE/100 = 18.67 ac-ft/0.75 = 24.89 ac-ft


19. Corn with a 6 ft root zone is grown on a soil with AWHC = 1.5 in/ft. ETc = 0.35 in/day. The irrigation manager chooses MAD = 65%. In order to complete this irrigation in 22 hours, the pump must deliver water at what flow rate (cfs and gpm)?

24.89 ac-ft x 12 in/ft = 299 ac-in
rate = amount ÷ time = 299 ac-in ÷ 22 hr
rate = 13.59 cfs
13.59 cfs x 448.8 gpm/cfs = 6100 gpm


20. A crop for which ETc = 0.28 in/day is irrigated with a sprinkler system. The system applies 3.6 inches and achieves 70% IE during an irrigation. How long before the irrigation is needed again (days)?

net = gross x IE/100 = 3.6 in x 0.70 = 2.52 in
time = amount ÷ rate = 2.52 in ÷ 0.28 in/day = 9 days


21. A crop with a 4 ft root zone is grown on a soil with AWHC = 2 in/ft. For MAD = 50%, what is the net application amount at each irrigation (in)?

net application = max SMD = AWHC (in/ft) x RZ (ft) x MAD (%)
net application = 2 in/ft x 4 ft x 50% = 4 in


22. A crop with a 4 ft root zone is grown on a soil with AWHC = 2 in/ft. If IE = 65%, how much water must the irrigation system deliver (in)?

gross = net ÷ IE/100 = 4 in ÷ 0.65 = 6.15 in


23. A crop with a 4 ft root zone is grown on a soil with AWHC = 2 in/ft. It is desired to deliver this irrigation with a sprinkler system operating for a 22 hour period. What should the application rate be (in/hr)?

rate = amount ÷ time = 6.15 in ÷ 22 hr = 0.28 in/hr


24. A crop with a 4 ft root zone is grown on a soil with AWHC = 2 in/ft. If the sprinkler system uses a 40' x 60' spacing, what should the sprinkler flow rate be (gpm)?

gpm = rate (in/hr) x area (sq-ft) ÷ 96.3
gpm = 0.28 in/hr x 40' x 60' ÷ 96.3 = 6.97 gpm