2012 pe review: s&w management michael c. hirschi, phd, pe, d.wre senior engineer waterborne...

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  • Slide 1
  • 2012 PE Review: S&W Management Michael C. Hirschi, PhD, PE, D.WRE Senior Engineer Waterborne Environmental, Inc. [email protected] also Professor Emeritus University of Illinois
  • Slide 2
  • Acknowledgements: Chris Henry, I-C PE Review (2006-2009) Rod Huffman, PE Review coordinator (thru 2011)
  • Slide 3
  • Session Topics Soil & Water Basics Review Evapotranspiration Subsurface Drainage Irrigation Nutrient Management
  • Slide 4
  • S&W Basics Review Soil makeup Infiltration & soil-water Soil-Water-Plant Relations
  • Slide 5
  • Subsurface Drainage Basic issues Design considerations System sizing System installation
  • Slide 6
  • Irrigation Plant water use Types of irrigation Sprinkler Flood Drip Design considerations
  • Slide 7
  • Nutrient Management Soil loadings Application issues
  • Slide 8
  • A few comments Material outlined is about 3 weeks or more in a 3-semester hour class. Im 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
  • Slide 9
  • Basics Soil Make Up Mineral Water Air Organic Matter
  • Slide 10
  • Mineral Component - Particles Sand Silt Clay Aggregates Silt & Sand sizes Less dense than primary particles
  • Slide 11
  • Particle Size Classifications
  • Slide 12
  • USDA Texture Triangle
  • Slide 13
  • 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 Loam B: Sandy Loam C: Loam D: Clay Loam
  • Slide 14
  • Solution Answer: C, Loam 20% Clay 30% Silt
  • Slide 15
  • 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
  • Slide 16
  • 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.).
  • Slide 17
  • 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)
  • Slide 18
  • Slide 19
  • Soil Water Holding Capacity (inches-water/foot-soil)
  • Slide 20
  • Water States by Soil Texture Gravitational Plant Available Unavailable
  • Slide 21
  • 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.
  • Slide 22
  • Moisture release curve -10cm -100cm -1000cm -10000cm
  • Slide 23
  • Any questions on general soil and water basics?
  • Slide 24
  • Evapotranspiration (ET) Evaporation Crop water use Reference Crops Pan Evaporation Crop Coefficients
  • Slide 25
  • Evaporation Transfer of water from liquid to vapor state Tabulated as lake evaporation across the US. Generally, evaporation exceeds precipitation west of the Mississippi River.
  • Slide 26
  • Example The mean annual lake evaporation in inches in Amarillo, TX (panhandle), is most nearly: A.50 B.65 C.75 D.85
  • Slide 27
  • Evaporation Fangmeier et al. (2006), pg 56
  • Slide 28
  • 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
  • Slide 29
  • Evapotranspiration (ET) Combined Evaporation and Transpiration Also called consumptive use Useful to predict soil water deficit Estimation methods (predict ET o, which is for Reference Crop) Evaporation Pan Penman-Monteith (see example in Fangmeier et al., 2006, pages 64-66)
  • Slide 30
  • ET vs. Precipitation
  • Slide 31
  • Reference Crops Alfalfa (comparable to field crops) Grass (easy to maintain under weather station, data can be related to alfalfa data)
  • Slide 32
  • Crop Coefficients Relate crops at various stages of growth to reference crops ET c = K c x ET ref
  • Slide 33
  • Crop Coefficients Both figures: Fangmeier et al. (2006) page 70
  • Slide 34
  • Crop Coefficients, by crop & stage Fangmeier et al. (2006) page 71
  • Slide 35
  • Crop growth stages Fangmeier et al. (2006) page 71
  • Slide 36
  • Example Estimate ET c for corn (maize) in Sioux City, Iowa if the ET ref is 8mm/day on July 1. Planting date was April 15. A: 8mm B: 9mm C: 10mm D: 11mm
  • Slide 37
  • Solution ET c = K c x ET ref Initial growth stage is 20 days, to May 5 Development stage is 40 days, to June 9 Mid stage is 50 days, to July 29 So, on July 15, in Mid-stage, so K c is 1.2 ET c = K c x ET ref = 1.2 x 9 = 10.8mm, or 11mm (D) Hint: Follow Fangmeier example 4.4
  • Slide 38
  • Any questions on ET?
  • Slide 39
  • Drainage Removal of excess water Benefits include More days to work in field Less crop stress due to high moisture Earlier germination because of warmer soil Liabilities include Expense Potential water quality issues Outlet required, may need pump
  • Slide 40
  • 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
  • Slide 41
  • Drainage types Surface Basic enhancement of flow patterns Surface grading/planing Surface ditching Subsurface Irregular Regular Watertable Management
  • Slide 42
  • 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
  • Slide 43
  • HYDROLOGIC CYCLE (with tiles)
  • Slide 44
  • Design Considerations Soil type Crop to be grown (value and response to wet conditions) Outlet Topography
  • Slide 45
  • 100 Tile Density Profitability Cost/AcreCrop YieldRate of Return Cost or Yield Ratio (%) Spacing
  • Slide 46
  • 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
  • Slide 47
  • Depth/Spacing Choices
  • Slide 48
  • Excellent Reference: ASABE Standards The material that follows is directly from ASABE EP480, issued MAR1998 (R2008), Design of Subsurface Drains in Humid Areas
  • Slide 49
  • Drain Spacing
  • Slide 50
  • Diagram for Hooghoudt Eq.
  • Slide 51
  • Drain Spacing by Hooghoudt Eq
  • Slide 52
  • Area Drained
  • Slide 53
  • Slide 54
  • CPT Capacity
  • Slide 55
  • 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
  • Slide 56
  • Solution A square 160 acres is a 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 316800ft 2 *0.375in/day/12in/ft/24hr/day/3600s/hr or 0.115 cfs. Enter the chart
  • Slide 57
  • Answer: D, 6
  • Slide 58
  • 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 Dont overtop and waste water (runoff) Plant Available Water is soil moisture held between Field Capacity and Wilting Point.
  • Slide 59
  • Irrigation methods Sprinkler (entire area is covered) Surface (flood, furrow) Drip (trickle, only plant root zone is watered) Subirrigation
  • Slide 60
  • 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)
  • Slide 61
  • 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).
  • Slide 62
  • 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): LayerDrummer SiCLPlainfield Sand 0-90.18 in/in0.07 in/in 9-180.17 in/in0.06 in/in 18-270.16 in/in0.06 in/in 27-360.16 in/in0.06 in/in
  • Slide 63
  • 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
  • Slide 64
  • 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
  • Slide 65
  • 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 dont 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
  • Slide 66
  • 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 doesnt rain. Answer: B
  • Slide 67
  • 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 1 st 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%?
  • Slide 68
  • Lateral Size Use the friction factor equation to determine how much loss/100 of pipe is allowable and choose lateral sizes accordingly: F f = (P o )*(P v )/L c Where: F f is the maximum pipe friction factor (psi/100), P o is the design operating pressure (psi), P v is the allowable pressure variation (+/-, as decimal, psi), and L c is the critical length (distance to furthest sprinkler, ft)
  • Slide 69
  • Lateral Size Now, P o is 40psi, P v is +/- 10% or 0.2 expressed as decimal, L c is 20+40+40+40 = 140, So, F f = 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).
  • Slide 70
  • Slide 71
  • 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.
  • Slide 72
  • 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?
  • Slide 73
  • Solution 2 every 10 days means volume is 2/12 in/ft*933ft*933ft = 145081 ft 3 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
  • Slide 74
  • Reference Recommendations ASABE Standards Fangmeier et al. (2006) or Schwab et al. (1993) MWPS Sprinkler Irrigation Manual
  • Slide 75
  • Questions on irrigation?
  • Slide 76
  • 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)40,000 b)70,000 c)76,000 d)257,000
  • Slide 77
  • 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)40,000 b)60,000 c)76,000 d)257,000 Answer B, 10.3/62.4*1000*365=60,248
  • Slide 78
  • 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)2.0 b)3.0 c)4.0 d)5.0
  • Slide 79
  • 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
  • Slide 80
  • Questions?