BRAE 340 Exam 2

To prevent plugging, a good recommendation is to:

Add acid to the water

A hand move sprinkler system costing $721 per acre irrigates 312 acres of sudan grass, which requires an annual net application of 71 inches. IE = 80% and labor costs $13 per hour. What is the total annual energy cost if the energy use is 32 kw-hr/ac-in and energy costs $0.18/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

Which one of the following crops or situations most likely uses the largest MAD?

Cotton

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.

Which is NOT one of the main causes of DUother?

Drainage problems

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.

The by feel/appearance method measures SMD. That is, the difference between soil moisture content and ________.

FC SMD = FC - soil moisture content

Center pivots, with their high application rates, are best suited to fine textured soils.

False

It is usually satisfactory to level severely undulating soils for surface irrigation.

False

Plant roots in 2 different soils at the same % depletion will experience the same soil moisture conditions.

False

During an irrigation event, most of the soil above the wetting front will be saturated.

False Only soil at the very surface is saturated. However, the wetted front will only continue to move down if the moisture content above the wetted front is above FC. It doesn't need to be saturated though.

Plants respond to the difference between moisture tension and osmotic tension (due to salinity).

False Total tension = moisture tension + osmotic tension

Given the following representative soil log for a field. 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) * Thickness (ft) Layer 1 (0-18 in): 2.17 in/ft * 1.5 ft = 3.255 in Layer 2 (18-24 in): 1.59 in/ft * 0.5 ft = 0.795 in Layer 3a (24-27 in) 1.76 in/ft * (3/12) ft = 0.440 in Layer 3b (27-30 in): below root zone Layer 4 (30-48 in): below root zone Root zone AWHC = 3.255 + 0.795 + 0.440 = 4.49 inches

Water is turned into the top end of a sloping furrow system for 21 hours (total irrigation time = 21 hrs). The time (hours) to reach the end of the furrows should be which of the following for the best DU?

For sloping furrows, you want the advance ratio to be < 0.5. That is, (advance time)/(total irrigation time) < (0.5). Hence, the advance time should be half of the total hours or less.

A hand move sprinkler system costing $351 per acre irrigates 178 acres of alfalfa, which requires an annual net application of 51 inches. What is the total annual cost for labor and maintenance if the IE = 74% and labor costs $13 per hour? Hint: Look in the "Selection of Irrigation Methods" chapter for labor and maintenance cost rates.

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

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

What is a possible solution for a significant pressure difference along the mainline?

Install adjustable pressure regulators at the heads of the manifolds or adjust existing pressure regulators

If a field continues to be irrigated after the root zone reaches field capacity, what happens to the water that infiltrates the soil after that point?

It deep percolates below the root zone and cannot be used by the plants

Which one of the following crops or situtations most likely takes the smallest MAD?

Potatoes or carrots

The two major components of non-uniformity in above-ground drip irrigation systems are:

Pressure differences and plugging

In order to obtain good distrubution uniformity (DU) from sloping furrows, which one of the following is required?

Runoff

Given the following representative soil log for a field: If the effective rooting depth of the crop is 51 inches, and the MAD is 60%, what is the SMD just prior to irrigation?

Solution: For each layer, AWHC (in) = AWHC (in/ft) x Thickness (ft) Layer 1 (0 - 18 in): 2.11 in/ft x 18/12 ft = 3.165 in Layer 2 (18 - 30 in): 0.86 in/ft x 12/12 ft = 0.860 in Layer 3 (30 - 42 in): 1.52 in/ft x 12/12 ft = 1.520 in Layer 4a (42 - 51 in): 1.16 in/ft x 9/12 ft = 0.870 in Layer 4b (51 - 60 in): below root zone root zone AWHC = 3.165 + 0.860 + 1.520 + 0.870 = 6.415 in SMD just prior to irrigation = max allowed SMD = AWHC for RZ x MAD/100 6.415 inches x 60/100 = 3.849 inches

Given the following representative soil log for a field: If the effective rooting depth of the crop is 39 inches, and the MAD is 55%, what is the SMD just prior to irrigation?

Solution: For each layer, AWHC (in) = AWHC (in/ft) x Thickness (ft) Layer 1 (0 - 24 in): 2.23 in/ft x 24/12 ft = 4.460 in Layer 2 (24 - 30 in): 0.78 in/ft x 6/12 ft = 0.390 in Layer 3a (30 - 39 in): 1.24 in/ft x 9/12 ft = 0.930 in Layer 3b (39 - 42 in): below root zone Layer 4 (42 - 48 in): below root zone root zone AWHC = 4.460 + 0.390 + 0.930 = 5.78 in SMD just prior to irrigation = max allowed SMD = AWHC for RZ x MAD/100 5.78 inches x 55/100 = 3.179 inches

Given the following representative soil log for a field: Find the AWHC of the entire soil profile (to 48 inches).

Solution: For each layer, AWHC (in) = AWHC (in/ft) x Thickness (ft) Layer 1 (0 - 18 in): 2.48 in/ft x 1.5 ft = 3.720 in Layer 2 (18 - 24 in): 2.13 in/ft x 0.5 ft = 1.065 in Layer 3 (24 - 30 in): 1.02 in/ft x 0.5 ft = 0.510 in Layer 4 (30 - 48 in): 1.38 in/ft x 1.5 ft = 2.070 in soil profile AWHC = 3.720 + 1.065 + 0.510 + 2.070 = 7.37 in

Which one of the following irrigation systems is not a pressurized irrigation system?

Terraces

Which linear move application rate is acceptable if the soil infiltration rate is 0.22 in/hr?

The application rate must be below the infiltration rate.

Which statement is true in regards to soil moisture tension (matrix potential)?

The dryer the soil, the harder it is for plants to use the water

Both moisure conditions and salinity affect plants. Plants respond to a combination of the two. Specifically, plants respond to the SUM of moisture tension and osmotic tension (due to salinity).

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

Pre-irrigation is the practice of watering a field before the crop is planted.

True

Plant roots in 2 different soils at the same SMC will experience different soil water potentials.

True Soil moisture tension is the key factor, not % depletion, SMD, or soil moisture content.

What needs to happen before any pressure measurements are taken?

Turn off the backflush

Which of the following methods is used to lower the pressure in a hose?

close down the valve on the test hose

What does the "k" stand for in the pressure/flow rate equation Q = kP^x?

constant based on emitter size and units of pressure

Soil evaporation:

decreases with time.

Distribution uniformity is a measure of ___________.

how evenly water is made available to plants throughout a field

Soil moisture depletion (SMD) is the depth of water:

required to fill the root zone to field capacity.

Poor irrigation management can lower soil infiltration rates. All of the following will lower infiltration rates except irrigating with:

sprinklers that have small droplet sizes. Irrigating with sprinklers that have small droplet sizes is the best method to minimize infiltration problems such as slow infiltration rates. This is because it does not affect the soil structure. Surface irrigation can cause erosion, which affects the soil structure. In addition, water quality, both very pure and high in salts, can affect the infiltration rates.

Soil water potential is defined as:

the sum of pressure, solute, and matrix potentials at a point.

Select the correct MAD for each condition.

-Almonds prior to harvest->Relatively Large MAD -Solid set or permanent set sprinkler system-> Relatively Small MAD -Any microirrigation system ->Relatively Small MAD -Circumstances when less frequent irrigations are required or desired ->Relatively Large MAD -Wine grapes prior to harvest -> Relatively Large MAD -Shallow root zone ->Relatively Small MAD -Uncertain water supply ->Relatively Small MAD -Surface irrigation systems with long runs - >Relatively Large MAD -Microirrigation with saline water ->Relatively Small MAD -Some trees or vines prior to winter -> Relatively Large MAD

Match the irrigation method with the irrigation type.

-Border strip->Surface -Level basin-> Surface -Terrace flooding-> Surface -Wheel lines-> Sprinkler -Fixed Grid -Side roll-> Sprinkler -Fixed Grid -Linear move-> Sprinkler - Moving -LEPA package-> Sprinkler - Moving -Traveler (big gun) -> Sprinkler - Moving -Tape ->Drip -Sub-surface tape -> Drip

Choose the best match between the limitation and system affected. An answer may be used more than once.

-Can't achieve good DU without runoff->Sloping Furrow -Labor is difficult to find because the job is hard work->Hand Move -Typically need sandier soils since the application rate can be quite high to achieve a good DU-> Center Pivot -Often the water will flow faster down one side-> Border Stripe -Can be easily plugged with silt or clay ->Drip -Requires border around entire area ->Level Basin -Often requires an additional pump to ensure proper operation ->Traveler (Big Gun) -The spacing between the fixed laterals can impact the DU -> Soild Set -Often difficult to keep in alignment-> Linear Move -Tend to get knocked over during harvest and its difficult to notice -> Micro Sprinkler

Choose the factor that most affects the DU of each irrigation system. Each factor is only used once.

-Micro sprinkler (non PC)->Changes in Elevation -Level Basin ->Land grading -Drip Tape ->Poor water quality -Wheel lines ->Spacing between laterals -Sloping furrows ->Wheel row vs non wheel row -Solid set sprinklers ->Minimal overlap at field edge

For each descriptive phrase, select the most closely related soil moisture term.

-Soil moisture status corresponding to approx. 10 cb tension-> FC -Percentage of AWHC allowed to be depleted between irrigations -> MAD -Can determine SMC directly, usually in % or in/ft-> TROXLER -Field capacity minus available water holding capacity equals this -> PWP -Can determine SMD indirectly, if FC is known-> TROXLER -Calibration may change due to dissolution of sensor in soil ->GYPSUM BLOCK -Can measure soil moisture tensions up to about 200 centibars ->WATERMARK -Shows how soil moisture content changes with tension-> SOIL MOISTURE CHARACTERISTIC CURVE -Acts as "artificial root" while measuring soil moisture status-> TENSIOMETER -Only soil moisture measurement to give SMD (in/ft) directly -> BY FEEL METHOD

For each descriptive phrase, select the most closely related soil moisture term.

-The difference between field capacity and the soil moisture content-> SMD -FC - PWP -> AWHC -Soil moisture condition corresponding to zero tension-> SATURATION -Plants cannot withdraw moisture when soil is drier than this-> PWP -Measures external pressure needed to force water out of the stem-> LEAF BOMB -Direct determination of SMD (in/ft)-> BY FEEL METHOD -Directly measures soil moisture tension, often in centibars -> TENSIOMETER -Displays soil moisture tension vs soil moisture content-> SOIL MOISTURE CHARACTERISTIC CURVE -Can measure soil moisture tensions up to about 2 bars-> WATERMARK -Can measure soil moisture content up to about 10 bars-> GYPSUM BLOCK

Where is the majority of all organic matter found in the soil?

0-1 feet below the surface

Vines spaced at 10 x 12 feet are drip irrigated with 13 gal/day. IE = 83%. This equals what rate of crop water use per day (net)?

0.144 in/day 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

What emitter exponent number is represented by the red curve in the graph?

0.5

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 16 days; and(b) microirrigation, wetting 45% of the soil surface, irrigating every 4 days. a. What is the wet soil evaporation (in) for 16 days under the level basin irrigation system? b. What is the wet soil evaporation (in) for 16 days under the microirrigation system?

0.69 in; 1.24 in Level irrigation system solution: day 1: 0.30 in/day x 1 day x 100% = 0.30 in day 2: 0.22 in/day x 1 day x 100% = 0.22 in day 3: 0.12 in/day x 1 day x 100% = 0.12 in day 4: 0.05 in/day x 1 day x 100% = 0.05 in days 5-16: no evaporation total for 16 days = 0.3 + 0.22 + 0.12 + 0.05 = 0.69 in Microirrigation system solution: day 1: 0.30 in/day x 1 day x 45% = 0.135 in day 2: 0.22 in/day x 1 day x 45% = 0.099 in day 3: 0.12 in/day x 1 day x 45% = 0.054 in day 4: 0.05 in/day x 1 day x 45% = 0.0225 in this cycle is repeated 4 times every 16 days total for 16 days = 4 x (0.135 + 0.099 + 0.054 + 0.0225) = 1.24 inches The correct answer is: Level Irrigation System =0.69 inch Microirrigation system=1.24 inch

Clay loam soil, saturation = 42%, FC = 30%, PWP = 17%, air dry = 13%. What is the SMD if the soil moisture content is 18%?

1.44 in/ft Moisture content = FC (%) - SMD (%) % moisture content * 12 in/ft = in/ft

How many emitters can you test at Locations A and B?

16

It has been 7 days since the last irrigation of a crop. ETc has averaged 0.38 in/day. What is the SMD (inches)?

2.66 inches Amount = rate * time SMD (in) = ET rate (in/day) * Time (days)

SMD totals 6.4 inches for a 40 inch root zone. How deep (inches) into the soil will 3.3 inches of rain water penetrate?

20.63 inches SMD (in/in) = inches depletion / inch root zone Depth (in) = Rain (in) / SMD (in/in)

Clay loam soil, saturation = 4.8 in/ft, AWHC = 1.8 in/ft, PWP = 2.1 in/ft, air dry = 1.4 in/ft. What is the moisture content of the soil (%) if SMD = 0.5 in/ft?

28.3% FC = AWHC (in/ft) + PWP (in/ft) Moisture content = FC - SMD = in/ft Moisture content (in/ft) ÷ 12 in/ft * 100 = %

An FDR probe measures the moisture content of a soil. The soil sample is a medium loam from 20 inches deep, it is quite dark and forms a hard ball. The moisture content is measured to be 21%. 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.

3.1 in/ft From the soil moisture measurement chart, SMD = 0.6 in/ft SMD = FC - soil moisture content ==> FC = SMD + soil moisture content Soil moisture content = % * 12 in/ft = in/ft FC = in/ft + 0.6 in/ft = in/ft

Clay loam soil, saturation = 4.2 in/ft, AWHC = 1.9 in/ft, PWP = 1.5 in/ft, air dry = 0.8 in/ft. What is FC (in/ft)?

3.4 in/ft FC = AWHC (in/ft) + PWP (in/ft)

Clay loam soil, saturation = 4.9 in/ft, AWHC = 1.8 in/ft, PWP = 1.7 in/ft, air dry = 0.8 in/ft. What is FC (in/ft)?

3.5 in/ft FC = AWHC (in/ft) + PWP (in/ft)

How much water (inches) will it take to penetrate 5 feet into the soil if the SMD is 0.56 in/ft at the driest location and the AE is 75%?

3.73 inches Water Depth (in) = Soil Depth (ft) * (SMD (in/ft) / AE (%))

Corn with a 5 foot root zone are grown in a 87 acre field with a AWHC = 1.37 in/ft. ETc = 0.31 in/day and the irrigation manager uses a MAD = 64%. IE = 77%. What is the maximum allowed SMD?

4.38 inches Root zone AWHC (in) = AWHC (in/ft) * RZ (ft) Max allowed SMD (in) = MAD * RZ AWHC

A soil has a water content at field capacity (% by volume) of 53%. At PWP, the water content is 16%. The root zone is 45 inches and the SMD just prior to irrigation is 6.87 inches. What MAD is the irrigation manager using?

41% 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 = %

PWP is 5.00 inches per foot. What is PWP in percent?

42% in/ft ÷ 12 in/ft = %

A soil has a water content at field capacity (% by volume) of 52%. At PWP, the water content is 15%. The root zone is 42 inches and the SMD just prior to irrigation is 6.70 inches. What MAD is the irrigation manager using?

43% 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)

A soil has a water content at field capacity (% by volume) of 53%. At PWP, the water content is 17%. The root zone is 42 inches and the SMD just prior to irrigation is 6.71 inches. What MAD is the irrigation manager using?

44% 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 = %

Wheat with a 6 foot root zone is grown on a 111 acre field for which AWHC = 1.5 in/ft. ETc = 0.28 in/day, and the irrigation manager uses a MAD = 41%. IE = 77%. What is the gross volume of water needed at each irrigation, in acre-ft?

44.3 ac-ft Solution: Root Zone AWHC = AWHC x RZ SMDmax = Root Zone AWHC x MAD Volume = Depth x Area Gross = Net / (IE/100)

An irrigation system applies 3.4 inches, ETc = 0.30 in/day, IE = 66%. How long before irrigation is needed again?

7.5 days Net = Gross * IE/100 = inches Time (days) = amount (in) ÷ rate (in/day)

A soil has a water content at field capacity (% by volume) of 35%. At PWP, the water content is 18%. The root zone is 46 inches and the SMD just prior to irrigation is 6.88 inches. What is the root zone AWHC?

7.82 inches AWHC = FC (%) - PWP (%) % * 12 in/ft = in/ft Root zone AWHC = AWHC (in/ft) * RZ depth (ft)

Grapes with a 3.2 foot root zone are grown in a clay loam soil. AWHC = 2.45 in/ft. The irrigation manager uses a MAD = 58%. What is the total AWHC for the root zone?

7.84 inches RZ AWHC = AWHC (in/ft) * RZ (ft) = inches

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)? Hint: See the tables in Chapter 4 of the book. Where a range of costs are listed, use the low end of the range for this determination.

Check Tables 4-13 and 4-16 in the textbook. Center pivot without a corner system has the lowest capital cost per acre. While these costs are a little dated, compared to the costs of the other systems, they are relative in terms of value.