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:
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AWHC (in/ft) |
AWHC (in) |
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ANSWER:
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AWHC (in/ft) |
AWHC (in) |
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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