soil science lab 3 henna

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University of Guelph *

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2060

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Geography

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Jan 9, 2024

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7

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Lab # 3 Water in Soil Recommended Reading: Weil and Brady, 2017: The nature and properties of soils. 15" ed., Chapters 5 and 6 Digging into Canadian Soils 114-116, 118-132 Part 1. Water Content Use the data provided in table 1 to calculate the: Mass of water in each soil, volume of water, mass based water content (sometimes called gravimetric water content) (0m), Volumetric water content (0v), depth equivalent water content in the S0 cm depth of each soil (Dw), soil bulk density (BD), soil pore space ratio (PSR), and the percentage of the pore space occupied by water. For each calculation provide your rough work for the calculations on one of the soils and plovide your answers in table 2 below. Remember that the particle density will remain constant at 2.65 g cm™, a) Table 1 Data on soil cores for use in calculation in part 1 a. Core # Wet weigh of soil (g) Dry weight of soil (g) Volume of soil (cm?) 1 178 126 125 2 200 165 125 3 225 175 125 Table 2 Mass of | Volumeof | ©®m | Ov | Dw BD PSR | % of pore space water (g) | water (cm®) @) | ) | ewem | (@em) | () filled with water 1 523 52em® | Yl 7| UVE| 20um]| 1.0 gw's 627/ 66/ 2 3Sg S om® 204|264 M um {3 (jum, RILA Sb % , . 3 . . B 509 | SO om? [297]U07 [200m[l4 glT]| g7s | ' : PSR = |-(1 qum’/2.65 ) Core \: Om = My Ms Dw = 03) (Pp) =1-03 65 ger, " MH%L* —MJ'Z =529/1269 o) (5D o) =0.bZ xi00l =\7%4- 12 = 7 N 3 _ oMl xipo7 —200m = b2 { =521 - UL BP= Ms/\o " 7. pore Spoce = Bv /05% —1\2bg /1250m = a4l ) = Mw /Dy Oy =V /Ny *\03»‘ 0. /:.6 -— e O- b ’525/' z §20m/125 o bt 7 z 52 o’ =0l x 1007 ] = Mt/ Scanned with CamScanner
Part 2: Soil water potential at equilibrium a) Assuming equilibrium, determine the value of yn and its components (Wm, yp and yg) at points A, B and C in the above diagram. Express the potentials in units of cm of water in table 3. Figure | soil water potential diagram stnarac b Soil surface 30cm y 8 Groundwater table 65 cm Cc Table 3 soil water potential at equilibrium using position “B” as the gravitational reference POINT Potential A (at surface) B C Ve e 0 S em Ym ~ 30 um 0 i Ve 30 um 0 -65 o Yn Yh&\’j-l— Y,\:30+L'3D): 0 \Qh.; 0-0=0 \thwa*‘v?: ,bi*bf;b b) Currently in figure 1 the gravitational reference point is given at position “B” (the water table). The gravitational reference point could however be placed anywhere in the soil profile as it is just a measure of the relative gravitational energy. Move the gravitational reference point to position “A” and determine the value of yu and its components (Wm, Wp and yg) at points A, B and C in the above diagram. Express the potentials in units of cm of water in table 4. Table 4 soil water potential at equilibrium using position “A” as the gravitational reference POINT Potential A B C Ve 0 300m as om Ve 0 el Ve D -0 - a5 ¢ Wi 0 ¥=Wg+¥p > -30430=0 [y, =VYy+ Yp= -95 +25=0 relative values between positions stays the same. ¢) Make note that even though the absolute values of yi changed in table 3 and table 4 that the Scanned with CamScanner
Part 3 Soil water potential at non-equilibrium In non-equilibrium condition soil matric potential cannot be determined based on the position relative to the water table. As such the matric potential must be measured at each of the desired positions in a soil. Tensiometers, which measure metric potential, were installed at four different depths (0, 10, 30 and 60 cm below the soil surface) in a soil profile. On three different days (day 1, 3 and 10) readings of ym (in cm H20) are obtained. The data are presented below in Table 5. 1. a) Using position A in table S as the gravitational reference point, complete table 5 by filling in values for yg and yn and indicate in the fourth column for each day whether the water is flowing upward () or downward (}) or is at a standstill (0) between adjacent positions (ie between position “A*” and “B”, “B” and “C” etc.). e.g A water flow is from 1 point B to point A B Table 5 Non-equilibrium soil hydraulic potentials. Day 1 Day 3 Day 10 Depth Ym Ve [vao [T [VYm Ve |wn [T [Vm ve [wn |1 (cm) ! l l 0 0 0 0A 20 |70 4| s fo |- L |40 |0 s B 10B 60 |0 |7° 2|2 .10 |-3S v 30 |no|-40 . 30C g0 |0 =50 LB [oles . a0 |-0|-5© . 60 D 300 |-‘oe|-Ho0 320 |-100|-420 250 |-\bo| -350 ¢) What can you say about the most likely weather pattern during this ten-day period? Wader i mMovag in this ron egaliprivm soil. The hydrewvtic potenbal ot A ond B horizon s higher on Da 3, inditchng thot dhe s3U is more sedveetated. ™S means ¥ most \Yely rained on day \,3 of sust on doy 3. Twe \uoderis iS olso moving domnard on Day 3, thdieoting o grecder volume g SR Greater hy drouic pokential. A and B horizon Wave lower k;gdrow\ic ?o'}on"ha\ on o (0 Yhen 3,06 0 resnt dhe soll has dried o sver Kt ot Cvms o higner ydmulic potenkal becavie potor has flovied down after the vain on Pay 10 d) In this case (table 5), explain why was the pressure potential (yp) not determined? Pre ssure folrm’ria\ s not dedormined vecone -here 1S e roim*m\. “the sol wos eing measured ot @& non - equiiorivm stede, s, dhere tond not have peon o Pressert gotential. Scanned with CamScanner
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Part 4. Available Water a) Compare the moisture characteristic data for the following 3 soils (figure 2) and complete table 6 assuming the soil is uniform in texture and structure to 50 cm soil depth. To determine the volume of water per hectare in table 6 multiply the depth of available water in meters by the surface area of 1 hectare (10000 m?). Also remember that there are 1000L in 1 of water. Figure 2 water release curves of 3 soils for use in Part 4 a. 60 j— - 50 _— 40 \-—————+—+++1 ; 30 20 10 Soil Water Content 0v (% m3 m3) Matric Potential ¥m (bars) Y Table 6 determination of plant available water from figure 2 Water Water Plant available | depth of available | Volume (L/ha) of content (Ov) | content (6v) | water content | water in top 50 water plant at Field Cap. | Wilting Point | (6v) cm of soil available in top D=(6) (DV) 50 cm of soil 207 -(0.20) 50 =100m\ =, 00O o7 (030X 50> (S om V= 1, SO0 yn® Isoil 1 | 45 7 257 lsatz g 7 &7 M.S % £(0.145) (SDum) =73 e V=130 o soil3 | \ &7 35 /. " V(o) (5A) @ Vo = (D) (S A BNz WM = (0.10v) (10000 ) (0. 15 (10000) ~(0.013w) (10000r) = 1000 = (500 yn’ =730 Scanned with CamScanner
Part 5. Water Movement in Saturated Soils In class we discussed Darcy’s experiment. Figure 3 gives a representation of an experiment to determine the flux rate of water coming through a soil. Figure 3 Darcy’s experiment example to be used to complete table 7 and 8 " Water f R 10cm | | f 12cm o Ref point b Yoy 0o [} a) Determine the hydraulic potential (wh) at both point A and B represented in figure 2. Present the data in table 7 Table 7 hydraulic potential for the experiment indicated in figure 3 Position in figure 2 Potential A B Yp 0 om 0 Ym / 0 Ve 12 um 0 Yh 22 om 0 b) Use the information you determined in Table 7 and the information in figure 3 to determine for each of the soils in table 8: the flux rate of water (Jw) (equation below), the change in hydraulic potential (Awh) (from table 7), and the hydraulic conductivity (k) (equation below) and record the information in table 8. Be sure to show your work. Scanned with CamScanner
Table 8 presentation of results for Part 5 b. Volume of Flux rate of water (Jw) water collected v (Vw) =5 110 cm? 60 seconds : =25CE (0o3foms! 60 seconds 35cm’ Part 2 Water movement in unsaturated soils Watch the following video that describes water movement in soils. https://www.youtube.com/watch?v=DmTNFIEc2VA and/ have a look at the displays in the lab (note they show the same thing as the video) a) At approximately three minutes 10 seconds in the video (3:10) there is a demonstration of capillary rise use two glass plates. They are separated on one site and clamped directly together on the other to represent a range of pore size from large on the open side to very small on the other side. Examine the wedge capillary on the demonstration table and the height of rise above the free water surface at the closed edge of the wedge. If the water on the edge of the glass that was closed together was drawn up to a total hight of 20 cx+ cm, use the following equation calculate the theoretical radius of the pore at the edge of the wedge. 0.15 0.15 h= r Y= h -3 W=200m ' o Theoebcal ragivs is 7.5e “om or o015 0-0075um 200m -3 = 7.5€ un b) Once again from the video, describe what happened when water was moving through a non-saturated loam textured soil and encountered a sand layer (3:48). Explain why this effect happens using your knowledge of soil hydraulic potentials and hydraulic _xhra:\’ct‘ loan Yexture d %01l and Cncovr\"“ conductivity. in the son3 PoeeS poove “he vhon watey (was mDV\'Aj —meah the non p Are cond, adweSive omd Cohesive forees held the water k o A encble surgls W der ewrtuplly Llows -\"r\(wgy\ e send an o et e ’n‘o’&:;r -Y’ni ool Decomes mont, put dhe sand does not allop weter » . * pass thiovgy o n maC porental in s Soth [y o O wWoul reso\r in lo The & dnesion ond C hesion outd ¥ | . h'r:‘"flfi (:5 e \\;drav\vc ?\,1’«144;\\- Gmo\\\of Qores and o lowor Mdm"l" e,flfl.\\" winlL oW in | soll allow Weder o slay there unh) i becomes vory Stvated, that 1+ storts ooM S0t < Tv S flnrova\n the sand The woker flows throvgh dhe sand oKy af b 1224 i h\ah - oker moves do the ‘ower \nadrwl\(_ fo*tm"‘a\ below. ; Kol hydradi ¢ potente Nkely +o stay in lover ydravdic potenRals. This 1S dye Yo weker Scanned with CamScanner
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c¢) Similarly describe what happened when water was moving through a silt loam textured soil and encountered a clay layer (6:36). Explain why this effect happens using your knowledge of soil hydraulic potentials and hydraulic conductivity. The <\ loapn 4exdurd <ol ollows waler to Glow well. weder e plastsed by Ave cley layer abdor the soil obove s ek due do Small clay @of®S dbility to dvar wedor \ato Ahem. oter dranstercing dvogn e sol s ey slor o ot tokes dime for puoker do Flow Lhnudn the smoll potes, The itk loam deschntd Sol has @ hgher hydmudic potential Ahan the cley li\j“/ thos uoder mores iMo Ihe Clay loyor o5 it o o loner hydraddic potentia\s. e adhesi ard tonesion forles are B'N'D"jw n the seoller pores £ X daj (aver ond drows n ard fherefne, woter moves iate Ahe clay layer. d) Similarly explain what happened when the water encountered a diagonal band of sandy soil that extended to the soil surface (10:50). Explain why this effect happens using your knowledge of soil hydraulic potentials and hydraulic conductivity. Be sure to explain the difference in effect caused by this sand band compared to the one in part “b” NQ‘V{ &‘Coun*‘fcc} e d!'OSV\q\ P immcdl'cf\dj vhore i F .Sul?ca Info ‘\’N_bar\& ond flled Fhe pores. Tre QSR Eolntkal treked by Ao wall in Hre chanred Cavsed lavge pores i Sebeded Soil o carny vister easily, This send ditkee fore At Sand in @A B becowse WS © savrared ond pact V9" s non-satureted - This sotvrobed soil gy pressure potertial and ndter oves throosh large ores casily, while non-setueted 2ol hos malC potental ond odViesion and cohesion moinknined Hhe water in the small gores oppe the Sand layer. Scanned with CamScanner