BRAE Midterm 2
Which one of the following crops or situations most likely takes the largest MAD?
Almonds prior to harvest
Which one of the following irrigation systems is not a pressurized irrigation system?
Contour furrows
Between irrigations, the moisture content of the surface layers of soil can not be drier than PWP?
false
The FDR probe measures % moisture by volume. That is, the difference between FC and ________.
% moisture by volume = FC - SMD The correct answer is: SMD
What is the approximate "x" value of PC emitters (above their minimum pressure)?
0
Where is the majority of all organic matter found in the soil?
0-1 feet below surface
What emitter exponent number is represented by the red curve in the graph?
0.5
How deeply must you probe to determine the wetted soil area per emitter?
1.5 feet
What is the typical water-air-solid proportion of a loam soil at field capacity?
25% air, 25% water, 50% solids
For a typical loam soil, what % water content do irrigators typically have to work with assuming a MAD of 50%?
6%
A soil has a water content at field capacity (% by volume) of 52%. At PWP, the water content is 17%. The root zone is 47 inches and the SMD just prior to irrigation is 6.75 inches. What MAD is the irrigation manager using?
AWHC (%) = FC (%) - PWP (%) AWHC (in/ft) = AWHC (%) * 12 in/ft RZ AWHC (in) = AWHC (in/ft) * RZ (ft) Max allowed SMD (in) = RZ AWHC (in) * MAD (%) MAD (%) = SMD max (in) ÷ RZ AWHC (in) The correct answer is: 41%
A soil has a water content at field capacity (% by volume) of 51%. At PWP, the water content is 20%. The root zone is 49 inches and the SMD just prior to irrigation is 6.80 inches. What MAD is the irrigation manager using?
AWHC (%) = FC (%) - PWP (%) AWHC (in/ft) = AWHC (%) * 12 in/ft RZ AWHC (in) = AWHC (in/ft) * RZ (ft) Max allowed SMD (in) = RZ AWHC (in) * MAD (%) MAD (%) = SMD max (in) ÷ RZ AWHC (in) The correct answer is: 45%
A soil has a water content at field capacity (% by volume) of 53%. At PWP, the water content is 18%. The root zone is 48 inches and the SMD just prior to irrigation is 6.82 inches. What MAD is the irrigation manager using?
AWHC = FC (%) - PWP (%) % * 12 in/ft = in/ft RZ AWHC = in/ft * RZ (ft) = inches Max allowed SMD = MAD * AWHC MAD = Max allowed SMD ÷ AWHC = % The correct answer is: 41%
A soil has a water content at field capacity (% by volume) of 52%. At PWP, the water content is 19%. The root zone is 44 inches and the SMD just prior to irrigation is 6.73 inches. What MAD is the irrigation manager using?
AWHC = FC (%) - PWP (%) % * 12 in/ft = in/ft RZ AWHC = in/ft * RZ (ft) = inches Max allowed SMD = MAD * AWHC MAD = Max allowed SMD ÷ AWHC = % The correct answer is: 46%
A soil has a water content at field capacity (% by volume) of 30%. At PWP, the water content is 18%. The root zone is 42 inches and the SMD just prior to irrigation is 6.81 inches. What is the AWHC (in/ft)?
AWHC = FC (%) - PWP (%) % * 12 in/ft = in/ft The correct answer is: 1.44 in/ft
A soil has a water content at field capacity (% by volume) of 33%. At PWP, the water content is 15%. The root zone is 42 inches and the SMD just prior to irrigation is 6.87 inches. What is the AWHC (in/ft)?
AWHC = FC (%) - PWP (%) % * 12 in/ft = in/ft The correct answer is: 2.16 in/ft
A linear move sprinkler system costing $421 per acre irrigates 108 acres of alfalfa, which requires an annual net application of 70 inches. IE = 80% and labor costs $17 per hour. What is the total annual energy cost if the enrgy use is 18 kwh/ac-in and energy costs 0.15 cents/kwh?
Application = inches * acres = ac-in (net)Gross = net ÷ (IE/100) = ac-in (gross)Energy costs = ac-in (gross) * kw-hr/ac-in * cents/kw-hr = $/yr The correct answer is: 25515 $/yr
A hand move sprinkler system costing $448 per acre irrigates 324 acres of sudan grass, which requires an annual net application of 68 inches. IE = 76% and labor costs $13 per hour. What is the total annual energy cost if the energy use is 16 kw-hr/ac-in and energy costs $0.19/kw-hr?
Application = inches * acres = ac-in (net)Gross = net ÷ (IE/100) = ac-in (gross)Energy costs = ac-in (gross) * kw-hr/ac-in * cents/kw-hr = $/yr The correct answer is: 88128 $/yr
What is the formula for DUlq?
Average of the lowest 1/4 values / Average of the data set
Which one of the following crops or situations most likely takes the largest MAD?
Border strip on sandy soil
What type of MAD do these conditions usually call for?
Border strip → Relatively Large MAD, Shallow root zone → Relatively Small MAD, Corn, during pollination → Relatively Small MAD, Potato or radish → Relatively Small MAD, Strawberries or banana → Relatively Small MAD, Wine grapes prior to harvest → Relatively Large MAD, Any microirrigation system → Relatively Small MAD, Almonds prior to harvest → Relatively Large MAD, Irrigation with saline water → Relatively Small MAD, Root crops like carrot, onion, potato → Relatively Small MAD
Which is the only soil moisture measuring technique to give SMD directly?
By feel/appearance method
Which one of these irrigation systems has the lowest capital cost ($/acre)?
Center pivot without corner system
According to the figure above, where is Available Water Holding Capacity located?
D
SMD totals 7.3 inches for a 36 inch root zone. How deep (inches) into the soil will 3.1 inches of rain water penetrate?
SMD (in/in) = inches depletion / inch root zone Depth (in) = Rain (in) / SMD (in/in) The correct answer is: 15.29
Which one of the following crops or situtations most likely takes the smallest MAD?
Shallow root zone
Given the following flow rate information, what is the DUlq of the field?
DU = Ave LQ / Ave All The LQ is this case is the lowest 3 values.
Given the following flow rate information, what is the DUlq of the field? 1234567891011120.820.941.011.031.161.201.251.301.341.391.431.47
DU = Ave LQ / Ave All The LQ is this case is the lowest 3 values. The correct answer is: 0.77
Grapes with a 3.1 foot root zone are grown in a clay loam soil. AWHC = 1.82 in/ft. ETc = 0.27 in/day and the irrigation manager uses a MAD = 53%. What is the SMD 6 days after irrigation?
Depletion = ETc (in/day) x Time (days) = inches The correct answer is: 1.62 inches
How much rain (inches) will it take to penetrate 3 feet into the soil if the SMD is 0.45 in/ft?
Depth (ft) = Rain (in) / SMD (in/ft) Rain (in) = Depth (ft) * SMD (in/ft) The correct answer is: 1.35
Choose the best match between the limitation and system affected. An answer may be used more than once.
Difficult to use alternate sets to improve CCDU → Solid Set, Can't achieve good DU without runoff → Sloping Furrow, Difficult to use on tall crops like corn → Hand Move, Can't irrigate all of square fields → Center Pivot, Efficiency is poor unless a tailwater return system is used → Sloping Furrow, PC versions don't tend to engage until about 25 psi, and even then don't always work that well → Micro Sprinklers, Typically need sandier soils since the application rate can be quite high to achieve a good DU → Center Pivot, Very large flow rates are required to acheive a good DU with a slope → Border Strip, A compromise between DU and labor cost is typical → Hand Move, Often difficult to keep in alignment → Linear Move
Which is NOT one of the main causes of DUother?
Drainage problems
What is a possible solution for a significant pressure difference between hoses along a manifold?
Either install pressure regulators, or replace hose screen washers with plain washers
If Emitter A and Emitter B have the same flow rate at 15 psi, which would have a higher flow rate at 20 psi? Emitter A - standard turbulent path emitter x = 0.55 Emitter B - pressure compensating emitter x = 0.00 (above 10 psi)
Emitter A would have a high flow rate since it is not pressure compensating - in this case, it would increase by about 17%. Emitter B would remain at the same flow rate. Emitter A
FC is 25%. What is FC in inches per foot?
FC % * 12 in/ft = in/ft The correct answer is: 3.00
FC is 40%. What is FC in inches per foot?
FC % * 12 in/ft = in/ft The correct answer is: 4.80
Clay loam soil, saturation = 4.8 in/ft, AWHC = 1.7 in/ft, PWP = 2.1 in/ft, air dry = 1.4 in/ft. What is the moisture content of the soil (%) if SMD = 0.6 in/ft?
FC = AWHC (in/ft) + PWP (in/ft) Moisture content = FC - SMD = in/ft Moisture content (in/ft) ÷ 12 in/ft * 100 = % The correct answer is: 26.7%
Clay loam soil, saturation = 4.1 in/ft, AWHC = 2.0 in/ft, PWP = 2.1 in/ft, air dry = 1.4 in/ft. What is the moisture content of the soil (%) if SMD = 0.4 in/ft?
FC = AWHC (in/ft) + PWP (in/ft) Moisture content = FC - SMD = in/ft Moisture content (in/ft) ÷ 12 in/ft * 100 = % The correct answer is: 30.8%
Clay loam soil, saturation = 4.2 in/ft, AWHC = 1.9 in/ft, PWP = 1.6 in/ft, air dry = 0.8 in/ft. What is FC (in/ft)?
FC = AWHC (in/ft) + PWP (in/ft) The correct answer is: 3.5
Clay loam soil, PWP = 2.0 in/ft, AWHC = 1.5 in/ft. What is FC (%)?
FC = PWP (in/ft) + AWHC (in/ft) FC = in/ft + in/ft = in/ft in/ft ÷ 12 in/ft * 100 = % The correct answer is: 29.2%
A "torpedo" is a special drill for installing underground pipe.
False
Both moisure conditions and salinity affect plants. Plants respond to a combination of the two. Specifically, plants respond to a DIFFERENCE in moisture tension and osmotic tension (due to salinity).
False
If spatial variability of soil properties is sufficiently high, commercial irrigation schedulers will install and use many sampling sites to characterize the moisture status of the soil.
False
Subirrigation is a type of irrigation system.
False
Weather influences the ET of well watered, full cover crop, as does the type of irrigation system.
False
According to the figure above, where is Field Capacity located?
Field Capacity is dashed line at location B. The correct answer is: B
Find the DU due to differences in pressure with the given information below. Given: x = 0.54, k = 0.468 Emitter Pressure Measurements (in psi)1234567812.313.013.414.715.115.415.816.1
First, determine the flow rate from each emitter using Q = kPx. For example, the first emitter would have a flow rate = 0.468 * 12.30.54 = 1.81 gph. Once you have all of the flow rates, for each emitter, use DU = Ave LQ / Ave All, to determine the DU due to pressure differences. The correct answer is: 0.93
Given the following representative soil log for a field If the effective rooting depth of the crop is 39 inches, and the MAD is 65%, what is the SMD just prior to irrigation?
For each layer, AWHC (in) = AWHC (in/ft) x Thickness (ft) Layer 1 (0 - 18 in): 2.22 in/ft x 18/12 ft = 3.330 in Layer 2a (18 - 39 in): 1.72 in/ft x 21/12 ft = 3.010 in Layer 2b (39 - 42 in): below root zone Layer 3 (42 - 48 in): below root zone Layer 4 (48 - 54 in): below root zone root zone AWHC = 3.330 + 3.010 = 6.34 in SMD just prior to irrigation = max allowed SMD = AWHC for RZ x MAD/100 6.34 inches x 65/100 = 4.121 inches The correct answer is: 4.12
Soil Layer Soil Texture Depth (in) Thickness, ft AWHC, in/ft1 Loamy Sand 0-12 1.0 1.12 2 Clay Loam 12-24 1.0 2.36 3 Silt Loam 24-30 0.5 1.82 4 Clay 30-42 1.0 2.38 If the effective rooting depth of the crop is 27 inches, and the MAD is 50%, what is the SMD just prior to irrigation?
For each layer, AWHC (in) = AWHC (in/ft) x Thickness (ft) Layer 1 (0 - 12 in): 1.12 in/ft x 12/12 ft = 1.120 in Layer 2 (12 - 24 in): 2.36 in/ft x 12/12 ft = 2.360 in Layer 3a (24 - 27 in): 1.82 in/ft x 3/12 ft = 0.455 in Layer 3b (27 - 30 in): below root zone Layer 4 (30 - 42 in): below root zoneroot zone AWHC = 1.120 + 2.360 + 0.455 = 3.935 inSMD just prior to irrigation = max allowed SMD = AWHC for RZ x MAD/1003.935 inches x 50/100 = 1.968 inches The correct answer is: 1.97 inches Question 9 Correct 1.00 points out of 1.00 Flag question Question text Which of these are reduced when a plant undergoes moisture stres
Given the following representative soil log for a field: If the effective rooting depth of the crop is 33 inches, and the MAD is 45%, what is the SMD just prior to irrigation?
For each layer, AWHC (in) = AWHC (in/ft) x Thickness (ft) Layer 1 (0 - 12 in): 1.77 in/ft x 12/12 ft = 1.770 in Layer 2 (12 - 18 in): 1.01 in/ft x 6/12 ft = 0.505 in Layer 3 (18 - 30 in): 2.14 in/ft x 12/12 ft = 2.140 in Layer 4a (30 - 33 in): 1.55 in/ft x 3/12 ft = 0.388 in Layer 4b (33 - 42 in): below root zone root zone AWHC = 1.770 + 0.505 + 2.140 + 0.388 = 4.803 in SMD just prior to irrigation = max allowed SMD = AWHC for RZ x MAD/100 4.803 inches x 45/100 = 2.161 inches The correct answer is: 2.16 inches
Given the following representative soil log for a field: Soil Layer Soil Texture Depth (in) Thickness, ft AWHC, in/ft 1 Sandy Loam 0 - 18 1.5 1.14 2 Clay Loam 18 - 30 1.0 2.44 3 Loamy Sand 30 - 48 1.5 0.93 4 Fine Sand 48 - 60 1.0 0.77 If the effective rooting depth of the crop is 33 inches, and the MAD is 70%, what is the SMD just prior to irrigation?
For each layer, AWHC (in) = AWHC (in/ft) x Thickness (ft) Layer 1 (0 - 18 in): 1.14 in/ft x 18/12 ft = 1.710 in Layer 2 (18 - 30 in): 2.44 in/ft x 12/12 ft = 2.440 in Layer 3a (30 - 33 in): 0.93 in/ft x 3/12 ft = 0.233 in Layer 3b (33 - 48 in): below root zone Layer 4 (48 - 60 in): below root zone root zone AWHC = 1.710 + 2.440 + 0.233 = 4.383 in SMD just prior to irrigation = max allowed SMD = AWHC for RZ x MAD/100 4.383 inches x 70/100 = 3.068 inches The correct answer is: 3.07 inches
Given the following representative soil log for a field: Soil Layer Soil Texture Depth (in) Thickness, ft AWHC, in/ft 1 Silt Loam 0 - 18 1.5 1.91 2 Loam 18 - 30 1.0 1.74 3 Loamy Sand 30 - 36 0.5 0.99 4 Sandy Loam 36 - 42 0.5 1.19 Find the AWHC of the soil root zone for a crop with an effective rooting depth of 27 inches.
For each layer, AWHC (in) = AWHC (in/ft) x Thickness (ft) Layer 1 (0 - 18 in): 1.91 in/ft x 18/12 ft = 2.865 in Layer 2a (18 - 27 in): 1.74 in/ft x 9/12 ft = 1.305 in Layer 2b (27 - 30 in): below root zone Layer 3 (30 - 36 in): below root zone Layer 4 (36 - 42 in): below root zone root zone AWHC = 2.865 + 1.305 = 4.17 in The correct answer is: 4.17 inches
Given the following representative soil log for a field: Soil Layer Soil Texture Depth (in) Thickness, ft AWHC, in/ft Sandy Loam Clay Loam Silt Loam Fine Sand Find the AWHC of the soil root zone for a crop with an effective rooting depth of 45 inches.
For each layer, AWHC (in) = AWHC (in/ft) x Thickness (ft) Layer 1 (0 - 24 in): 1.23 in/ft x 24/12 ft = 2.460 in Layer 2 (24 - 36 in): 2.47 in/ft x 12/12 ft = 2.470 in Layer 3 (36 - 42 in): 1.83 in/ft x 6/12 ft = 0.915 in Layer 4a (42 - 45 in): 0.78 in/ft x 3/12 ft = 0.195 in Layer 4b (45 - 48 in): below root zone root zone AWHC = 2.460 + 2.470 + 0.915 + 0.195 = 6.04 in The correct answer is: 6.04 inches
Given the following representative soil log for a field: Soil Layer Soil Texture Depth (in) Thickness, ft AWHC, in/ft 1 Silt Loam 0 - 36 3.0 1.94 2 Fine Sand 36 - 42 0.5 0.76 3 Clay Loam 42 - 48 0.5 2.49 4 Loamy Sand 48 54 0.5 1.11 Find the AWHC of the entire soil profile (to 54 inches).
For each layer, AWHC (in) = AWHC (in/ft) x Thickness (ft) Layer 1 (0 - 36 in): 1.94 in/ft x 3.0 ft = 5.820 in Layer 2 (36 - 42 in): 0.76 in/ft x 0.5 ft = 0.380 in Layer 3 (42 - 48 in): 2.49 in/ft x 0.5 ft = 1.245 in Layer 4 (48 - 54 in): 1.11 in/ft x 0.5 ft = 0.555 in soil profile AWHC = 5.820 + 0.380 + 1.245 + 0.555 = 8.00 in The correct answer is: 8.00 inches
Given the following representative soil log for a field: Soil Layer Soil Texture Depth (in) Thickness, ft AWHC, in/ft Sandy Loam Fine Sand Silt Loam Clay Loam Find the AWHC of the entire soil profile (to 36 inches).
For each layer, AWHC (in) = AWHC (in/ft) x Thickness (ft) Layer 1 (0 - 6 in): 1.36 in/ft x 0.5 ft = 0.680 in Layer 2 (6 - 18 in): 0.77 in/ft x 1.0 ft = 0.770 in Layer 3 (18 - 30 in): 1.84 in/ft x 1.0 ft = 1.840 in Layer 4 (30 - 36 in): 2.13 in/ft x 0.5 ft = 1.065 in soil profile AWHC = 0.680 + 0.770 + 1.840 + 1.065 = 4.36 in The correct answer is: 4.36 inches
An FDR probe measures the moisture content of a soil. The soil sample is coarse loamy sand from 8 inches deep, it appears moist and makes a weak ball. The moisture content is measured to be 10%. What is the FC for the soil (in/ft)?
From the soil moisture measurement chart, SMD = 0.3 in/ft SMD = FC - soil moisture content ==> FC = SMD + soil moisture content Soil moisture content = % * 12 in/ft = in/ft FC = in/ft + 0.3 in/ft = in/ft The correct answer is: 1.5 in/ft
An FDR probe measures the moisture content of a soil. The soil sample is coarse loamy sand from 16 inches deep, it appears slightly moist and sticks together slightly. The moisture content is measured to be 13%. What is the FC for the soil (in/ft)? Refer to the "Soil Texture Classification" table in the Basic Soil-Plant-Water Relationships section of your book.
From the soil moisture measurement chart, SMD = 0.5 in/ft SMD = FC - soil moisture content ==> FC = SMD + soil moisture content Soil moisture content = % * 12 in/ft = in/ft FC = in/ft + 0.5 in/ft = in/ft The correct answer is: 2.1 in/ft
A hand move sprinkler system costing $358 per acre irrigates 174 acres of alfalfa, which requires an annual net application of 65 inches. What is the total annual cost for labor and maintenance if the IE = 79% and labor costs $15 per hour?
From the text, hand move sprinkler systems require 0.175 hr labor/ac-in applied (Table 4-15), and annual maintenance costs are about 2% (Table 4-18) of the capital cost of the system. $/acre * acres = $ for the irrigation system Maintenance costs = 2% * $ for system = $/yr for maintenance Application = inches * acres = ac-in (net) Gross = net ÷ (IE/100) = ac-in (gross) Labor = 0.175 hr/ac-in * ac-in (gross) * $/hr = $ of labor Total labor + maintenance = annual cost The correct answer is: $38827/yr
A center pivot sprinkler system costing $481 per acre irrigates 165 acres of alfalfa, which requires an annual net application of 58 inches. What is the total annual cost for labor and maintenance if the IE = 80% and labor costs $13 per hour? Hint: Look in the chapter for labor and maintenance cost rates.
From the text, hand move sprinkler systems use 0.01 hr labor/ac-in applied (Table 4-15), and annual maintenance costs are about 5% of the capital cost (Table 4-18) of the system. $/acre * acres = $ for the irrigation system Maintenance costs = 5% * $ for system = $/yr for maintenance Application = inches * acres = ac-in (net) Gross = net ÷ (IE/100) = ac-in (gross) Labor = 0.01 hr/ac-in * ac-in (gross) * $/hr = $ of labor Total labor + maintenance = annual cost The correct answer is: $5523/yr
Vines spaced at 11 x 14 feet are drip irrigated with 14 gal/day. IE = 82%. This equals what rate of crop water use per day (net)?
GPM = gal/day ÷ (24 hr/day * 60 min/hr) in/hr = gpm * 96.3 ÷ Area (ft^2) in/hr * 24 hr/day = in/day (gross) Net (in/day) = gross (in/day) * IE/100 The correct answer is: 0.120 in/day
What is required to get a good "efficiency" with a drip irrigation system? Assume no under-irrigation and at least 15 psi is available.
Good DU and good timing
A 69 acre field requires a net irrigation amount of 4 inches. The irrigation efficiency is 77%. What is the gross volume that must be delivered by the pump?
Gross (inches) = Net (inches) ÷ (IE/100) Inches (gross) * acres ÷ 12 in/ft = ac-ft The correct answer is: 29.9 ac-ft
Which one of these irrigation systems has the lowest capital cost ($/acre)?
Hand move sprinkler
Plants respond directly to the moisture depletion in the soil.
Plants respond directly to the tension with which water is held by the soil. The correct answer is 'False'.
To which of the irrigation systems does each descriptive phrase best apply?
Hand move → Sprinkler - Fixed Grid, Wheel lines → Sprinkler - Fixed Grid, Border strip → Surface, Permanent Set → Sprinkler - Fixed Grid, Supernet → Micro, Dual line hose → Drip, Terrace flooding → Surface, Sloping furrow → Surface, Fanjet → Micro, Mini sprinklers → Micro
How does soil variability affect the DU of sprinkler or drip irrigation systems (assuming proper design)?
Has no effect
What factors under the control of the irrigator have the greatest affect on the DU of surface irrigation systems?
How long and how fast the water is applied
Choose the factor which most affects the DU of each irrigation system. Use a choice only once.
Level basin → Soil intake rate variability, Drip (on the surface) → Orifice clogging, SDI → Root intrusion, Sloping furrow → Opportunity time variations, Under-tree sprinklers → Pressure variations, Traveling gun sprinkler → Wind
Which one of the following irrigation systems is not a pressurized irrigation system?
Level basins are the only un-pressurized irrigation system. Sprinkler and Micro irrigation systems are considered pressurized. Level basin
Assume wet soil evaporation rates according to the following data. Compare wet soil evaporation for these two irrigation systems: (a) level basin irrigation, wetting 100% of the soil surface, irrigating every 15 days; and(b) microirrigation, wetting 35% of the soil surface, irrigating every 3 days. A) What is the wet soil evaporation (in) for 15 days under the level basin irrigation system? B) What is the wet soil evaporation (in) for 15 days under the microirrigation system?
Level irrigation system solution: day 1: 0.32 in/day x 1 day x 100% = 0.32 in day 2: 0.22 in/day x 1 day x 100% = 0.22 in day 3: 0.11 in/day x 1 day x 100% = 0.11 in day 4: 0.04 in/day x 1 day x 100% = 0.04 in days 5-15: no evaporation total for 15 days = 0.32 + 0.22 + 0.11 + 0.04 = 0.69 in Microirrigation system solution: day 1: 0.32 in/day x 1 day x 35% = 0.112 in day 2: 0.22 in/day x 1 day x 35% = 0.077 in day 3: 0.11 in/day x 1 day x 35% = 0.0385 in this cycle is repeated 5 times every 15 days total for 15 days = 5 x (0.112 + 0.077 + 0.0385) = 1.14 inches The correct answer is: Level Irrigation System =0.69 inch Microirrigation system=1.14 inch
Assume wet soil evaporation rates according to the following data. Compare wet soil evaporation for these two irrigation systems: (a) level basin irrigation, wetting 100% of the soil surface, irrigating every 15 days; and(b) microirrigation, wetting 35% of the soil surface, irrigating every 3 days. Day Stage Evaporation rate 1 1 0.35 in/day 2 2 0.25 in/day 3 2 0.11 in/day 4 2 0.05 in/day 5 and after - none What is the wet soil evaporation (in) for 15 days under the level basin irrigation system? What is the wet soil evaporation (in) for 15 days under the microirrigation system?
Level irrigation system solution: day 1: 0.35 in/day x 1 day x 100% = 0.35 in day 2: 0.25 in/day x 1 day x 100% = 0.25 in day 3: 0.11 in/day x 1 day x 100% = 0.11 in day 4: 0.05 in/day x 1 day x 100% = 0.05 in days 5-15: no evaporation total for 15 days = 0.35 + 0.25 + 0.11 + 0.05 = 0.76 in Microirrigation system solution: day 1: 0.35 in/day x 1 day x 35% = 0.1225 in day 2: 0.25 in/day x 1 day x 35% = 0.0875 in day 3: 0.11 in/day x 1 day x 35% = 0.0385 in this cycle is repeated 5 times every 15 days total for 15 days = 5 x (0.1225 + 0.0875 + 0.0385) = 1.24 inches The correct answer is: Level Irrigation System =0.76 inch Microirrigation system=1.24 inch
An irrigation system applies 3.3 inches, ETc = 0.32 in/day, IE = 67%. How long before irrigation is needed again?
Net = Gross * IE/100 = inches Time (days) = amount (in) ÷ rate (in/day) The correct answer is: 6.9 days
Tomatoes with a 4 foot root zone are grown on a sandy loam soil. AWHC = 1.49 in/ft. ETc = 0.27 in/day and the irrigation manager uses a MAD = 45%. What is the total AWHC for the root zone?
Root zone AWHC = AWHC (in/ft) * RZ (ft) = inches The correct answer is: 5.96 inches
What factor under control of the irrigator has the greatest affect on drip/micro irrigation system DU?
Pressure adjustments
The two major components of non-uniformity in above-ground drip irrigation systems are:
Pressure differences and plugging
Corn with a 6 foot root zone is grown on a 49.2 acre field with a AWHC of 1.5 in/ft. ETc is 0.32 in/day and the irrigation manager uses a MAD = 68%. IE = 74%. How many days between irrigations.
RZ AWHC (in) = AWHC (in/ft) * RZ (ft) Max allowable SMD (in) = MAD (%) * RZ AWHC (in) Time = amount ÷ rate = depletion allow ÷ ET rate inches ÷ in/day = days The correct answer is: 19.1 days
Tomatoes with a 3.8 foot root zone are grown in a sandy loam soil. AWHC = 1.29 in/ft. The irrigation manager uses a MAD = 53%. What is the total maximum allowable SMD?
RZ AWHC = AWHC (in/ft) * RZ (ft) = inches Max allowable SMD = MAD (%) * RZ AWHC = inches The correct answer is: 2.6 inches
A corn crop needs an irrigation of 2.1 inches (net) every 11 days. What is the rate of water use by the crop (in/day)?
Rate (in/day) = Amount (in) ÷ Time (days) The correct answer is: 0.19
Wheat with a 4 foot root zone is grown on a 96 acre field for which AWHC = 1.4 in/ft. ETc = 0.28 in/day, and the irrigation manager uses a MAD = 45%. IE = 79%. What is the gross volume of water needed at each irrigation, in acre-ft?
Root Zone AWHC = AWHC x RZ SMDmax = Root Zone AWHC x MAD Volume = Depth x Area Gross = Net / (IE/100) The correct answer is: 25.5
Tomatoes with a 3 foot root zone are grown on a sandy loam soil. AWHC = 1.39 in/ft. ETc = 0.23 in/day and the irrigation manager uses a MAD = 54%. What is the total AWHC for the root zone?
Root zone AWHC = AWHC (in/ft) * RZ (ft) = inches The correct answer is: 4.17 inches
Which linear move application rate is acceptable if the soil infiltration rate is 0.34 in/hr?
The application rate must be below the infiltration rate. The correct answer is: 0.31 in/hr
What is the typical water-air-solid proportion of a loam soil at saturation?
The correct answer is: 0% air, 50% water, 50% solids
For each descriptive phrase, select the most closely related soil moisture term.
The correct answer is: Can measure soil moisture tensions up to about 2 bars → Watermark, Soil moisture status corresponding to approx. 15 bars tension → PWP, Irrigate when this reaches threshold determined by MAD → SMD, Plants cannot withdraw moisture when soil is drier than this → PWP, Measures plant rather than soil status → Leaf Bomb, Reads soil moisture tension up to about 80 cb → Tensiometer, Soil moisture measuring device based on capacitance → Troxler, Can measure soil moisture content up to about 10 bars → Gypsum Block, Soil moisture measurement that determines SMD (in/ft) directly → "By Feel" Method, FC - PWP → AWHC
Choose the factor which most affects the DU of each irrigation system. Only use a choice once.
The correct answer is: Center pivot → Plugged nozzles, Level Basin → Land grading, Wheel lines → Wind, Sloping furrows → Wheel row vs non wheel row, Micro sprinkler → Poor filtration, Sub-surface drip → Root intrusion
To which of the irrigation systems does each descriptive phrase best apply?
The correct answer is: LEPA package → Sprinkler - Moving, In-line emitters → Drip, Supernet → Micro, Mini sprayers → Micro, Tape → Drip, Fanjet → Micro, Spaghetti tube → Micro, Dual line hose → Drip, Contour furrow → Surface, Mini sprinklers → Micro
What type of MAD do these conditions usually call for?
The correct answer is: Microirrigation with saline water → Relatively Small MAD, Any microirrigation system → Relatively Small MAD, Sloping furrows with long runs → Relatively Large MAD, Circumstances when less frequent irrigations are required or desired → Relatively Large MAD, Corn, during pollination → Relatively Small MAD, Tomato or citrus, just prior to bud formation → Relatively Large MAD, Wheel line or side roll sprinkler system → Relatively Large MAD, Surface irrigation systems with long runs → Relatively Large MAD, Some trees or vines prior to winter → Relatively Large MAD, Hand Move sprinkler systems → Relatively Large MAD
Choose the best match between the limitation and system affected. An answer may be used more than once.
The correct answer is: Often the water will flow faster down one side → Border Strip, Difficult to use on tall crops like corn → Hand Move, Can't irrigate all of square fields → Center Pivot, Requires very high pressures → Traveler (Big Gun), Entire area most be flat (no slope) → Level Basin, Can't achieve good DU without runoff → Sloping Furrow, Often requires an additional pump to ensure proper operation → Traveler (Big Gun), Often irrigates a portion of the field that was just irrigated → Linear Move, Difficult to use alternate sets to improve CCDU → Solid Set, Must be picked up and trailered to the other end of the field → Hand Move
The "n" value in the DUother equation represents:
The number of emitters per plant
Which of these are reduced when a plant undergoes moisture stress?
The response to stress is to close stomata, which reduces the transpiration. Since the CO2 can't enter the leaves, photosynthesis stops, reducing vegetative growth. The correct answer is: All are reduced
A micro sprayer irrigation system has an application rate 0.05 in/hr. What must the infiltration rate of the soil be in order to prevent runoff?
The soil infiltration rate needs to be higher than the application rate so that all of the water that is applied infiltrates and does not runoff the field. Greater than 0.05 in/hr
Given: A 40 inch root zone containing a homogeneous, sandy loam soil. Since the past irrigation, the crop has used 3.7 inches total. Approximately how much water was extracted by the plants in the soil layer between 10 and 20 inches deep?
Top 1/4 of RZ (0-10 in): doesn't contribute 2nd 1/4 of RZ (10-20 in): 30% * in total = inches 3rd 1.4 of RZ (20-30 in): doesn't contribute Bottom 1/4 of RZ (30-40 in): doesn't contribute Total use from between 10-20 inches deep = sum of the layers that contribute The correct answer is: 1.11 inches
Given: A 40 inch root zone containing a homogeneous, sandy loam soil. Since the past irrigation, the crop has used 7.6 inches total. Approximately how much water was extracted by the plants in the soil layer between 10 and 20 inches deep?
Top 1/4 of RZ (0-10 in): doesn't contribute 2nd 1/4 of RZ (10-20 in): 30% * in total = inches 3rd 1.4 of RZ (20-30 in): doesn't contribute Bottom 1/4 of RZ (30-40 in): doesn't contribute Total use from between 10-20 inches deep = sum of the layers that contribute The correct answer is: 2.28 inches
Center pivots, with their high application rates, are best suited to course textured soils.
True
During an irrigtaion event, most of the soil above the wetting front will be unsaturated.
True
It is difficult to apply small application amounts with surface irrigation systems.
True
Lower water and power costs help to offset higher costs for systems which can achieve high DUs.
True
Plant roots in 2 different soils at the same SMC will experience different soil moisture conditions.
True
Plant roots will experience the same level of stress in two different soils so long as the soil moisture tension in the two soils is the same.
True
For each descriptive phrase, select the most closely related soil moisture term
When SMD is the indicator, this sets the threshold → MAD, Soil moisture status corresponding to approx. 15 bars tension → PWP, Direct determination of SMD (in/ft) → "By Feel" Method, Displays soil moisture tension vs soil moisture content → Soil Moisture Characteristic Curve, Percentage of AWHC allowed to be depleted between irrigations → MAD, Measures "leaf water potential" → Leaf Bomb, Soil moisture measuring device based on capacitance → Troxler, The difference between field capacity and the soil moisture content → SMD, Shows how soil moisture content changes with tension → Soil Moisture Characteristic Curve, Measures external pressure needed to force water out of the stem → Leaf Bomb
What are the two basic questions that irrigation scheduling answers?
When to irrigate and how much to irrigate
Does the number of emitters per plant make a difference in the DU calculation (assume a new system)?
Yes, the DU increases with additional emitters per plant
What does the "k" stand for in the pressure/flow rate equation Q = kPx?
constant based on emitter size and units of pressure
DUΔpressure is caused by:
differences in pressure among emitters throughout a field
Which of the following methods is used to raise the pressure in a hose?
shut off adjacent hoses using ball valves
Drip irrigation is always the best irrigation system to use.
false
It is difficult to apply small application amounts with most sprinkler systems (except hand move).
false
Plants respond directly to the moisture content of the soil.
false
What does the "Q" in the DU due to differences in pressure equation stand for?
flow rate
Consider sandy loam and clay soils. The moisture tension at which crop relative transpiration begins to decline:
is the same for both soils.
________ are the basic causes of unwanted (i.e, non-beneficial) deep percolation.
poor DU and poor timing
Soil moisture depletion (SMD) is the depth of water:
required to fill the root zone to field capacity.
In order to obtain good distrubution uniformity (DU) from sloping furrows, which one of the following is required?
runoff
Soil water potential is defined as:
the sum of pressure, solute, and matrix potentials at a point.
How many DU values are averaged in the DUother equations (in other words, how many tests are performed for DUother)?
three
A "torpedo" is a weight pulled behind a tractor to compact furrows.
true