2011 PE Review:IV-C: S&W Management

Michael C. Hirschi, PhD, PE, D.WREProfessor and Assistant Dean

University of Illinoismch@illinois.edu

Acknowledgements:

Chris Henry, I-C PE Review (2006-2009)Rod Huffman, PE Review coordinator

Session Topics

• Soil & Water Basics Review

• Evapotranspiration

• Subsurface Drainage

• Irrigation

• Nutrient Management

S&W Basics Review

• Soil makeup

• Infiltration & soil-water

• Soil-Water-Plant Relations

Subsurface Drainage

• Basic issues

• Design considerations

• System sizing

• System installation

Irrigation

• Plant water use

• Types of irrigation– Sprinkler– Flood– Drip

• Design considerations

Nutrient Management

• Soil loadings

• Application issues

A few comments

• Material outlined is about 3 weeks or more in a 3-semester hour class. I’m compressing at least 6 hours of lecture and 3 laboratories into 2 hours, so I will:– Review highlights and critical points– Do example problems

• You need to:– Review and tab references– Do additional example problems, or at least

thoroughly review examples in references

Basics – Soil Make Up

• Mineral

• Water

• Air

• Organic Matter

Mineral Component - Particles

• Sand

• Silt

• Clay

• Aggregates– Silt & Sand sizes– Less dense than primary particles

Particle Size Classifications

USDA Texture Triangle

Example

After soil sample dispersal to ensure only primary particles are measured, a sample is determined to be 20% clay, 30% silt and 50% sand. What is the USDA soil texture?

A: Sandy Clay LoamB: Sandy LoamC: LoamD: Clay Loam

Solution

Answer: C, Loam

20% Clay

30% Silt

Infiltration & soil-water

• Infiltration is the passage of water through the soil-air interface into pores within the soil matrix

• Movement once infiltrated can be capillary flow or macropore flow. The latter is a direct connection from the soil surface to lower portions of the soil profile because of root holes, worm burrows, or other continuous opening

• Infiltrated water can reappear as surface runoff via “interflow” and subsurface drainage

Soil, water, air

The inter-particle space (voids) is filled with either water or air. The amount of voids depends upon the soil texture and the condition (ie. tilled, compacted, etc.).

Water (moisture) content

• Special terms reflect the fraction of voids filled with water (all vary by texture and condition):– Saturation: All voids are filled with water– Field Saturation: Natural “saturated” moisture content

which is lower than full saturation due to air that is trapped.

– Field capacity: Water that can leave pores by gravity has done so (0.1 to 0.33 bars)

– Wilting point: Water that is extractable by plant roots is gone (15 bars)

– Hygroscopic point: Water that can be removed by all usual means is gone (but some remains, 30 bars)

Saturated (all pores filled)

Field Capacity (Some air, some water)

Wilting point(water too tightly held for plant use)

Plant Available Water

Soil Water Holding Capacity

(inches-water/foot-soil)Soil Texture Range Average

Sand 0.4 - 1.0 0.8 Sandy Loam 1.0 - 1.5 1.3 Loam 1.0 - 2.0 1.6 Silt Loam 1.3 – 2.6 2.0 Clay Loam 1.3 – 2.6 2.0 Clay 1.4 – 2.4 1.8

Water States by Soil Texture

0

10

20

30

40

50

60

Sand SandyLoam

Loam Silt Loam ClayLoam

Clay

Vol

umet

ric

Wat

er c

onte

nt

Gravitational

Plant Available

Unavailable

Commentary

• Later, when we discuss drainage, it is the gravitational water that is of interest, eg. saturation down to field capacity. The volume of this water, the hydraulic characteristics of the soil in question, and the wet-condition-tolerance and value of the crop being grown dictate the drainage system design and its feasibility.

• When we consider irrigation, plant available water (AW) is that held between field capacity and wilting point. It is this water that we manage via irrigation to supply water to plants. The volume of AW the soil can hold within the crop root-zone, the crop value and water use, and the crop tolerance of dry conditions dictate irrigation design and feasibility.

Moisture “release” curve

-10cm

-100cm

-1000cm

-10000cm

Any questions on general soil and water basics?

Evapotranspiration (ET)

• Evaporation

• Crop water use

• Reference Crops

• Pan Evaporation

• Crop Coefficients

Evaporation

• Transfer of water from liquid to vapor state

• Tabulated as “lake evaporation” across the US.

Generally, evaporationexceeds precipitationwest of the Mississippi River.

Example

• The mean annual lake evaporation in inches in Amarillo, TX (panhandle), is most nearly:

A. 50

B. 65

C. 75

D. 85

Evaporation

Fangmeier et al. (2006), pg 56

Evaporation

The mean annual lake evaporation in inches in Amarillo, TX (panhandle), is most nearly:

A. 50

B. 65

C. 75

D. 85

= 1900mm/25.4 mm/in = 75 in, so answer is C

Evapotranspiration (ET)

• Combined Evaporation and Transpiration

• Also called “consumptive use”

• Useful to predict soil water deficit

• Estimation methods (predict ETo, which is for Reference Crop)– Evaporation Pan– Penman-Monteith (see example in Fangmeier

et al., 2006, pages 64-66)

ET vs. Precipitation

Reference Crops

• Alfalfa (comparable to field crops)

• Grass (easy to maintain under weather station, data can be related to alfalfa data)

Crop Coefficients

• Relate crops at various stages of growth to reference crops

• ETc = Kc x ETref

Crop Coefficients

Both figures: Fangmeier et al. (2006)page 70

Crop Coefficients, by crop & stage

Fangmeier et al. (2006)page 71

Crop growth stages

Fangmeier et al. (2006) page 71

Example

Estimate ETc for corn (maize) in Sioux City, Iowa if the ETref is 8mm/day on July 1. Planting date was April 15.

A: 8mm

B: 9mm

C: 10mm

D: 11mm

Solution

ETc = Kc x ETref

Initial growth stage is 20 days, to May 5Development stage is 40 days, to June 9Mid stage is 50 days, to July 29So, on July 15, in Mid-stage, so Kc is 1.2

ETc = Kc x ETref = 1.2 x 9 = 10.8mm, or 11mm (D)

Hint: Follow Fangmeier example 4.4

Any questions on ET?

Drainage

• Removal of excess water• Benefits include

– More days to work in field– Less crop stress due to high moisture– Early germination because of warmer soil

• Liabilities include– Expense– Potential water quality issues– Outlet required, may need pump

Objective of Drainage is Financial Benefit

• Optimize crop growth– Increase yield– Reduce wetness-based disease– Reduce variability within fields and from year to year

• Improve timeliness of field work– May use smaller equipment– May increase acreage– May reduce labor costs

• Increase value of land

Drainage types

• Surface– Basic enhancement of flow patterns– Surface grading/planing– Surface ditching

• Subsurface– Irregular– Regular

• Watertable Management

Subsurface Drainage

• Removes gravitational water only

• Degree of drainage specified as depth/day

• System design dictated by crop, soil, location, topography and more…

• Can be used to manage watertable down or up

• Changes hydrologic response of field and if widely installed, the watershed

HYDROLOGIC CYCLE (with tiles)

Design Considerations

• Soil type

• Crop to be grown

• Outlet

• Topography

100

Tile Density

Pro

fita

bili

ty

Cost/Acre Crop Yield Rate of Return

Co

st o

r Y

ield

Rat

io (

%)

Spacing

Drainage system design

• Capacity to remove water is expressed as depth/day (eg. 3/8 in/day)

• Spacing, maximum and minimum depth (absolute minimum of 24” without special protection), and maximum and minimum slope are dictated by soil and topography

Depth/Spacing ChoicesDepth/Spacing ChoicesDepth/Spacing ChoicesDepth/Spacing Choices

Excellent Reference:ASABE Standards

The material that follows is directly from ASABE EP480, issued MAR1998 (R2008), “Design of Subsurface Drains in Humid Areas”

Drain Spacing

Diagram for Hooghoudt Eq.

Drain Spacing by Hooghoudt Eq

Area Drained

CPT Capacity

Example

A subsurface drainage system is to be installed on a square 160 acres (1/4 section) in East Central Illinois. The Drainage Coefficient is 3/8”/day and the Illinois Drainage Guide indicates a 120’ spacing at 4’ depth. The proposed slope is 0.1%. What diameter CPT is needed for each lateral?

A: 3”B: 4”C: 5”D: 6”

Solution

A square 160 acres is a ½ mile on each side, or 2640’. A spacing of 120’ gives an area for each lateral of 120x2640 or 316800 sq.ft. If the system removes 0.375”/day, the flow rate needs to be 316800ft2*0.375in/day/12in/ft/24hr/day/3600s/hr or 0.115 cfs.

Enter the chart

Answer: D, 6”

Irrigation

• Supplements rainfall• Need and design dictated by crop, soil, location,

topography, water availability, energy price, and more…

• Simplistic description: Use the soil as your water tank– Deplete it to some predetermined safe level– Refill it as needed– Don’t overtop and waste water (runoff)

• Plant Available Water is soil moisture held between Field Capacity and Wilting Point.

Irrigation methods

• Sprinkler (entire area is covered)• Surface (flood, furrow)• Drip (trickle, only plant root zone is watered)• Subirrigation

Information needed for design

• Soil texture and profile water storage• Soil infiltration rate• Water source• Available flow and pressure• Water quality• Water cost• Irrigated area• Elevation changes on site• Plants to be irrigated, root depth• Plant water use (inches/day)

Design decisions and specific computed data needs

• How much do we let the soil-water deplete prior to irrigation (“management allowed depletion”, MAD, % as decimal, typically 40-50%, though can vary depending upon crop and climate)?

• How much water is available to the plant within its root zone (total amount is “available water”, AW, in inches)?

• How much water will we replace with each irrigation (equal to MAD * AW or “readily available water”, RAW, in inches)?

• How much total water do we need per irrigation cycle (equal to RAW*total irrigated area/efficiency)?

• How often do we need to irrigate the same area of plants (“irrigation interval”, equal to AW/(plant water use, in/day))?

All these concepts and equations are in any basic book or chapter on irrigation, such as Fangmeier et al. (2006).

Available Water, AW

• Soils vary in their characteristics by depth• Soil surveys have information on each soil by

depth• For example, consider the AW with depth for two

Illinois soils (data from WebSoilSurvey):

Layer Drummer SiCL Plainfield Sand

0-9” 0.18 in/in 0.07 in/in

9-18” 0.17 in/in 0.06 in/in

18-27” 0.16 in/in 0.06 in/in

27-36” 0.16 in/in 0.06 in/in

AW for Corn

• If we assume a 36” rooting depth for corn on either soil, we get the following AW:

• Drummer: AW=0.18*9+0.17*9+0.16*18= 6.03”, so 6.0” in root zone

• Plainfield: AW=0.07*9+0.06*27=2.25”, so 2.3” in root zone

Irrigation Interval

• So, given those 2 soils, and corn has a 0.25 in/d water use, if no rain, how many days before all available water is depleted?

• Drummer: 6”/.25ipd = 24days• Plainfield: 2.3”/.25ipd=9days

Now you know why there are many irrigated acres of Plainfield and few irrigated acres of Drummer

Example

You are designing a sprinkler irrigation system for a pick-your-own strawberry field. Your references indicate that strawberries use 0.25 in/day. The soil profile has a field capacity value of 0.36 in/in and a wilting point value of 0.24 in/in. The rooting depth of strawberries is 9”. You don’t wish to deplete your soil moisture below 50% available water. How much will you irrigate and how often? Assume 100% efficiency

A: 1” every 7 days; B: 0.5” every 2 days; C: 0.25” every day; D: Not enough information

Solution

Plant available water (AW) in the root zone is (0.36in/in-0.24in/in)*9in = 1”. The amount of water you wish to replace is half that amount (MAD), or 0.5” (RAW), which is your irrigation depth. Given the strawberries use 0.25 ipd, you will have to irrigate 0.5” (irrigation depth) every 2 days (irrigation interval) if it doesn’t rain.

Answer: B

Lateral Size

• Assume you will use a single lateral of pipe that you are able to move across the strawberries. It is Schedule 40 PVC and you chose four Rainbird 20JA impact sprinklers. The technical specs indicate the nozzles deliver 4.5gpm at 40psi while delivering water to a radius of 40’. Your plan for the lateral is to have the 1st sprinkler at 20’, then at 40’ intervals. What size PVC do you need between each sprinkler if your planned variation in pressure from high to low is +/- 10%?

Lateral Size

Use the friction factor equation to determine how much loss/100’ of pipe is allowable and choose lateral sizes accordingly:

Ff = (Po)*(Pv)/Lc

Where:

Ff is the maximum pipe friction factor (psi/100’),

Po is the design operating pressure (psi),

Pv is the allowable pressure variation (+/-, as decimal, psi), and

Lc is the critical length (distance to furthest sprinkler, ft)

Lateral Size

Now, Po is 40psi, Pv is +/- 10% or 0.2 expressed as decimal, Lc is 20’+40’+40’+40’ = 140’, So, Ff = 40*0.2/140 = 0.057 psi or 0.05 psi

The first section of pipe has flow for all four nozzles, or 18gpm. The next section has three nozzles flow, or 13.5gpm, the last two sections have 9gpm and 4.5gpm, respectively

Standard tables are available in many texts for pressure loss in pipes due to friction. Such a standard table is on the next page (from Rainbird website).

Lateral Size

• So, if we are to keep friction factor at 0.05 psi/100’ or less, we need to begin with 3” PVC, and it needs to stay 3” after the first nozzle, but can reduce to 2” after the second nozzle. If a bit more variation is OK (eg. +/- 15%, or 0.086psi/100’), the lateral can reduce to 2-1/2” after the first nozzle, 2” after the second and 1-1/2” after the third.

Example

You have determined that you will have to supply 2” of water every 10 days to meet a corn field water demand. You will use a lateral move system to apply the water in a 16-hr period every 10 days. The field in question is 20 acres (933 feet square). Assume an 80% sprinkler efficiency. How much water will you apply each irrigation and at what flow rate?

Solution

2” every 10 days means volume is

2”/12 in/ft*933ft*933ft = 145081 ft3 or 1,085,200 gal

Prior to efficiency being considered, flow rate is 1,085,200gal/(16hr*60min/hr) or 1130 gpm

At 80% efficiency, 1085200/.8 gal need to be sprayed or 1,356,500gal for a flow rate of 1413 gpm

Reference Recommendations

• ASABE Standards

• Fangmeier et al. (2006) or Schwab et al. (1993)

• MWPS Sprinkler Irrigation Manual

Questions on irrigation?

Nutrient Management

• One feeder pig produces 10.3 lbs of manure per day. Assuming that manure has the same density as water, how much manure, in cubic feet, is most nearly produced annually from a 1000 head barn that has 3 sets (or turns) per year.

a) 40,000b) 70,000 c) 76,000d) 257,000

Nutrient Management

• One feeder pig produces 10.3 lbs of manure per day. Assuming that manure has the same density as water, how much manure, in cubic feet, is most nearly produced annually from a 1000 head barn that has 3 sets (or turns) per year.

a) 40,000b) 60,000 c) 76,000d) 257,000

Answer B, 10.3/62.4*1000*365=60,248

Nutrient Management/Facilities

• The maximum loading rate (pounds of volatile solids per 1000 cubic foot per day) for an anaerobic lagoon for animal waste in West Central Illinois is most nearly:

a) 2.0b) 3.0c) 4.0d) 5.0

Nutrient Management/Facilities

• The maximum loading rate for an anaerobic lagoon for animal waste in West Central Illinois is most nearly:

a) 2.0

b) 3.0

c) 4.0

d) 5.0

C, 4.0, EP403.3

Loading of Soils Conversions

K 1.2 K2O

N 4.43 NO3

P 2.29 P2O5

N 1.29 NH4

cu ft 7.48 gallons

Questions?