SURFC&NEAR SURFC PROCESS
SURFC&NEAR SURFC PROCESS GEOL 315
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This 24 page Class Notes was uploaded by Brandon Dickinson on Monday October 26, 2015. The Class Notes belongs to GEOL 315 at University of South Carolina - Columbia taught by Staff in Fall. Since its upload, it has received 80 views. For similar materials see /class/229617/geol-315-university-of-south-carolina-columbia in Geology at University of South Carolina - Columbia.
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Date Created: 10/26/15
GEOL315 Homework 6 Due M onday April 839quot 2013 l The attached Fig 1 shows a well eld installed at a location in Rhodc Island The wells were installed in an uncon ned glacialtill aquifer This is a real well eld installed by Researchers at the University ol Rhode Island Below is a data table of hydraulic head data measured on a single day Using the triangulation method learned in class detemiine the ow field for this well eld STEPS 1 Examine the data range and decide on a set of equipotential lines lines of equal hydraulic head useful for this map One way to go about this is divide the total range of values by 10 and round up or down to a convenient value Let s say your total range of hydraulic head values was 28 m then 28 10 28 By rounding up you would have equipotential lines for every 3 m 2 Draw triangles between all the points No lines may cross 3 Assume that hydraulic head changes in a linear way and mark the equipotential points along each edge ofeach triangle If the distance between two points on the map is 3 inches and the head Change between them is 6 m then each 1m change in hydraulic head along that side of the triangle should be placed at every z inch 4 Connect cquipotenual points across each triangle for every interval what you determined in Part 1 All equipotcntial lines within a triangle should be parallel 5 Starting at the highest potential hydraulic head draw several ow lines Attempt to bend the low lines to make them move through all of ae triangles Make sure that the ow lines are perpendicular to the equipotential linesll PLEASE SEE THE ATTACHED MAP 1 pulan QUESTIONS a What is the hydraulic gradient along Exelcr Road t 1 Point Hydraulic gradient is the than e in h draulic head across a iven distance In this case the drop in hydraulic head along Exeter Road is a roximately 1 ft 150 ft or 0007 b Where on the map is the a divergence in the ow eld indicate it using a dashed line in a pen or pencil ol39a different color than you used for the rest of your work I I39uiim Bear in mind this may or may not show up signi cantly depending on how you draw the triangles Please refer to the map for the location of this divergence II A geologic formation has a permeability of 1039 m2 What is the hydraulic conductivity in ms of the formation with respect to Him There are a variety of sources ol lhe viscosin of various uids al various temperatures online chemistry textbooks etc K k2 t K is Hydraulic Conductivity k is permeabili p is the density of the uid g is the gravity 98 ms2 1 is the dynamic viscosity of the fluid a water at 15 C and 1 atm pressure it l uinll pl139 x inquot kgms p 9991026 kgm k1x1039 m39 K 10quot 9991026 x 9ai139x10quot 9 x 10 5 ms b air at 20 C and I aim pressure I l mmi quot1 82 x 10395 kgms p 1205 kgm k l x 1039quot m2 m mquot3911zos x 98I8x10 5 7 x Hr ms Ill The following eld notes were taken from three piezometers installed at the same location on the map but with screened tips at different depths Land surface elevation 450 m Driller s log 0 rn silty sand 35 in dense clay l50 m bottom ot hole PIEZOMETERS A B C Screened interval in below land surface 149150 99100 4950 Depth to water m below land surface 27 47 36 a Use the information from the driller s log to sketch the stratigraphy at this site Indicate the screened intervals for each piezometer in your sketch See the attached diagram I l umtl Compute the hydraulic head elevation head and pressure head at the bottom of the screened interval for each piezometer Use mean sea level as your datum Show your work for at least one example then compile your answers in a table where they will be easy to nd 2 i uiuls 0 Determine the direction ot groundwater flow and suggest a hydrogeologic situation to explain this ow pattern 1 Point 3 B C Elevation Head 450150 ft 300 ft 450 00 350 ft 45050 400 ft Pressure Head 45027300 123 ft 450 4735053 ft 45036 400 14 ft Elm Head 300123423ft 35053403 t 40014 414 r r39 c Groundwater is flowing up from the Tan Clay layer into the Sand layer and down from the Dense Clny layer into the sand Groundwater likes to follow the path of least resistance It is owing from low K layers and being focused into a higher K con ned aquifer s ru A t i x r 739 it 77 I 39 u Well A Well B Well C m asl Silty Sand I 4quot Dense Clay 350 7 Red lines show approximate thickness and location of screened intervals GEOL 3 l 5 Spring 20 3 Homework 4 l 6 points Duc Friday March lst As always report your answers in SI units show your work or no credit and turn in your answers on a separate sheet of paper 1 You have 1 m3 of quartz sand with a porosity ot 40 a What is the total volume of the pore space Bulk Volume 1 In3 V01 Porosity quot39 Valhutk Vulm 040 lm Va 040x1mquot 04m mm b What is the total volume ofthe sand grains Total volume of sand grains Total Volume Pore Volume 1 m3 040m306m 0 Assume the sand is completely dry How much does it weigh How much do the sand grains weigh I used 2650 kgm3 as a density for the quartz sand and 2650k306m 1590kg d Assume the sand is illy saturated with water Now how much does it weigh Add mass of water to previously calculated mass of sand Assume a fresh water density of 1000 kgmj Full saturated volume ofwatcr equals the pore volume calculated above Mass of water 1000 k 04 mquot 400 kg Tom saturated mass 1590 400 l990kg 2 Your grad student collected a soil sample with a porosity of33 Your lab tech tells you it has a volumetric water content of 28 What is its saturation S 1 033 0 028 These terms are related to saturation hy the equation S 1 ID 0 Solve for S S 0M 028033 085 or 85 3 Two soil samples were collected in lO cm long cylindrical sample tubes that have a S em diameter The samples were removed from the tubes weighed ovendried at 105 C and weighed again The data is listed below Sample 1 Wet weight 3763 g Dry weight 3094 g Sample 2 Wet weight 3073 g Dry weight 1924 g a Compute the bulk density porosity water content and saturation Assume that pg 2650 kgm Bulk volume for both samples is pi x r2 x h 314159 x 25 cm2 x 10 cm 196 em3 Bulk density dry bulk density is the dry mass of the sample divided by the bulk volume Porosig can be determined by the equation 0 103qu densitysolids density Assume the solid density of the grains is 265 gcm3 Water Content is the volume of the water in the sample divided by the bulk volume of the sample Assume the water density is 10 gem The mass of water ls determined by the difference between the Wet Weight and the Dry Weight listed above Saturation is calculated as in Number 2 above SAMPLE 1 SAMPLE 2 Bulk Density 3094 41936 cmJ 1924 g1963cmJ 5 Porosity l 58 g cm393265 g em39J 0982 g crud265 g cm39l 040 or 40 0629 or 63 Water Com 0 w 37633094 669 g 30731924 1149 g 66 gl0 g cm393 67 cm3 1149 gl0 g cmquot 115 cm1 67 cm3196 ch lt1 15 end196 cm3 0 34 or 34 0ss7 or 59 Saturation 034 040 085 or 85 05870629 0933 or 93 b The two samples are then analyzed for grain size using sieves and gravimeuic methods The following table lists the results for each sample by mass Sample 1 The mass in the size fraction gt 2 mm is excluded from this sort of analysis review lecture from about February 4m of the example worked in class Add the maining masses of sand silt and clay Then determine the individual percents of sand silt and clay based on the new total dry mass Mass Percent Sand 1238 g 1238 2321 053 or 53 Silt 1021 g 102 l232l 044 or 44 Clay 62 g 62 2321 003 or 3 TOTAL 232 g Sample 2 This sample does not have any grains larger than sand The total dry mass is given above All that is needed is to determine the individual percentages from the total mass Mass Percent Sand 192 g 192 1924 010 or 10 Silt 385 g 38511924 020 or 20 Clay 1347 g 1347 1924 070 or 70 Use this data to determine the percent content of each component plot these results using different pen or pencil colors on the attached soil texture diagram and state what sort of soil sample was collected for both Sample 1 and 2 See the ehartllll GEOL 315 Spring 2012 Homework 5 As always you may discuss this homework ad nauseum with other students but do your own work show your own work and turn in your own work for full credit 1 A very famous water sample from a perennial spring the Sierra Nevada has the following composition Garrels and Mackenzie 1967 Concentration mgkg Ca 104 Na 595 K 157 Mg 170 HCO339 546 sof 238 c139 106 5102 a 246 a What is the ionic strength of the solution You may nd webelementscom helpful for looking up molecular weights if you can t find a periodic chart in one of your textbooks Assume a temperature of 25 C Concentration MWAW MWAW Conc m Ion mgkg gmol mgmol molkg Z Z2 Ca2 104 4008 40080 259E04 2 4 Na 595 2299 22990 259E04 1 1 K 157 391 39100 402E05 1 1 Mg2 17 2431 24310 699E05 2 4 HC03 546 6102 61020 895E04 1 1 5042 238 9607 96070 248E05 1 1 CI 106 3545 35450 299E05 1 1 Si02aq 246 6009 60090 409E04 0 0 Divide concentration mgkg by Molecular Weight 0r Atomic Weight in mgmol to get concentration in mollkg 1 I E 2 ml 212 Si02aq has no charge and does not contribute to further ionic strength calculations Conc Ion molkg Z2 m x Z2 Ca2 259E04 4 10E03 Na 259E04 1 26E04 K 402E05 1 40E05 Mg2 699E05 4 28E04 HC03 895E04 1 89E04 SO42 248E05 1 25E05 CI 299E05 1 30E05 Si02aq 409504 o ooeoo SUM 26E03 Ionic Strength 13E03 The Ionic Strength of this solution is 13 x 10 3 molkg b What is the activity of K in this solution again at 25 C Explain your choice of equation for calculating the activity coefficient Constants for calculating the activity coef cient are in Table 21 at the bottom of p 2 Ionic Strength is very close to the cut off to use the simplest form of the Debye Huckel equation log 74 Azl2 J7 At 25 C the value for A is 05085 Square Log Act Activity Conc Activity Ion A Z2 Root l Coefficient Coeff mol kg mol kg Ca2 05085 4 361E02 733E02 845E01 259E04 22E04 Na 05085 1 361E02 183E02 959E01 259E04 25E04 K 05085 1 361E02 183E02 959E01 402E05 38E05 Mg2 05085 4 361E02 733E02 845E01 699E05 59E05 HC03 05085 1 361E02 183E02 959E01 895E04 86E04 5042 05085 1 361E02 183E02 959E01 248E05 24E05 CI 05085 1 361E02 183E02 959E01 299E05 29E05 a 38x10 5m01kg O In this relatively dilute solution what percent error would be introduced by just using the concentration of K instead of the activity Error 2 ExperimentalValue TrueValue x100 TrueValue Error 402 x 10 5 38 x 10 53s x 105 x 100 4 d What is the activity of SiOzaq in the solution Since Si02aq is an uncharged specie the activity and the concentration are the same 7 1 So the activity of Si02aq 409 x 10 4 molkg e The pH of the solution is 68 What is the activity of Hl Since pH 10g10H 68 H 10 PH 10 16 x 10 7 2 The stoichiometric equation for quartz in equilibrium with dissolved silica in water is SiOmZ lt gt Si02aq Keq 1039 a Is the spring water from 1 saturated undersaturated or supersaturated with respect to quartz m 2 Storm 10 SIOW 10 APUAP 409x10quot AP 409x10quot 4 K 1x10 29 The spring is supersaturated with respect to quartz Remember IAPK gt 1 Supersaturated IAPK 1 Equilibrium IAPK lt 1 Undersaturated b You know from class that quartz dissolves extremely slowly Can you infer anything about the rate of quartz precipitation from the saturation state of the spring water The water is supersaturated with quartz which indicates is should precipitate However it does so very very slowly or it would not be supersaturated 3 Figure 1 shows an activity diagram also called a stability diagram for common aluminosilicates on the potassium side of the family Kr feldspar Figure 1 Activity diagram for KzOAle3Sl02H20 at 250C Gibbs me a Plot the water sample from Question 1 on Figure 1 For the vertical axis you need the activities of K and H Activity of K 38 x 10 5 Activity of H from No 1e 16 X 10 7 Loglo Activity KActivity H 24 Loglo Activity SiOZaq 34 See the dot of the gure above b Which minerals are in equilibrium with the spring water Kaolinite is the only mineral in equilibrium with the spring water c If this water sample were put in contact with granite again which way would you expect the reaction 2 KAlSi3087Kspar H20 ltgt A12Si205OH4ykaol 2 K 4 SiOz 2 H to proceed Since the spring water is not in equilibrium with K Feldspar you would have increase the ratio of the activity of KH andor increase the activity of SiOzaq In order to do either of those things the reaction above would proceed to the right 2 KAlsgogstpar H20 ltgt Alei2050H4ml 2 K 4 Sio2 2 H Kfeldspar I I I I I I I I I I GIbbsILe I I I I I I I I I I 3 75 74 73 3909 55mm Homework 3 1 Rain Buckel Diameter 30 cm Radius 15 cm Water Depth 5 cm Area ot39bueket opening nrz Area 31415915 cm 710 cml 2 significant gures Or Area 31415905 em2 707 cm2 l ll accept this but it is not strictly correct a Water Volume in Liters 710 cm2 x 5 cm 4000 cm3 I signi cant gure Or 707 cm x 5 em 3535 cmquot Again OK but not the most correct Convert to L 4000 em3 4000 mL x l U 000 mL 4 L 6 1 signi cant ligure Or 3535 em3 3535 mL x l Ll000 1111 35 L b Let s assume that this bucket is a rain gauge Rain occurs the gauge captures the volume you divide the volume collected by the area oflhe gauge opening to get the inches or am of rain First using 4000 cmquot as the volume ot39rain captured 4000 cur710 cm2 2 5 cm 0 rain OR 3535 Cm1707 cm2 2 5 cm ofrain Yes this is essentially the same cm of rainfall measured in the bucket No because the depth of water in a bucket with angled sides would be different The only time the answers are the same is when the gauge has absolutely straight sides a at bottom and the opening has the same radius as the container in which volume is measured c How much water volume in I fell in your yard Y ensions 30 m x 5 m Area ot thc yard 1500 m 5 cm ot rain converted to m 5 cm x 1 m100 em 005 m Volume chain that fell in the yard area of yard x 39depth39 of rainfall Volume 1500 m2 x 005m 75 m3 of rain Convert to 1 75 m3 x 1000 111 m3 75 x104 L ofwatcr Ln correct signi cant gures 8 x 104 L of water 2 Contouring a Reliet acmss the entire map is 173 feet The highest point shown is 413 ft and the lowest point shown is 240 feet 413 7 240 I73139eet b Relief across my cross section is l60 fch The highest point in the cross section is 410 feet and the lowest point is 250 feet 410 250 160 feet See attached map and contour Elevatlon ft Elevation from A to A39 1000 2000 3000 4000 5000 6000 7000 8000 Total Distance from Point A HWB Na 3 Return Period Year Rainfall mm 1993 1984 3904 1992 3872 1990 35 08 2000 3463 1983 3443 1981 3273 1994 32 54 1985 32 26 1991 3153 1982 3116 1999 2809 1588 2801 1998 2797 1939 2731 996 2721 1997 2633 1987 26 25 1986 23 61 1995 21 55 3 What Is the inngesl Return Period Rank b Return Return Probability Period 400 40 years is the longest return period 1 year Is the shortest return period Whar is the probability rhat this area will receive 3935 mm of rainfall in any given year 34 39 1002n 1 2y Return Period IOUFa in any glven year 0 V 3 R u Q0133 Quaquot rquot quot K m l n3 lgva 5 ml 1 A SQEOM m m SDXFSTDHTSX imi 53 513 anwmg 5 LOO x195 mL 5 PM I z 3 WA quotO quot DSMLmij mr 0 m 0397 16 n13 1 C39 l l lt 50 LMCJ a MU L l39 o ttdor c J 541m 9 l f T li w mlt a fo w 74 o w 0 f7 N5 0 gm rd cmquot 23 M 3 939 ur onc e rgf forxmagk 39 J gur ongl gcx iis f V Hdays mm A swam m g r BBS41190 4 N 91 rnfjmgt 98000 rf 0 i HWZ No 2 Calculate the residence time of water in days for a person who drinks 64 oz of water per day 8 8 oz glasses right and who weighs approximately 150 pounds What is the density of a person Humans are mostly water so the density of your average human is the same as watersee below GIVEN INFORMATION Residence Time Equation T 50 5 Stock or reservoir size Q Flux in or out Some useful conversions 1 lb 0454 kg 1 fluid oz 29573 mL cm Density of Water 1 gcml or 1000 kgm3 First how much water is your average human I used 65 and I took any value from 57 to 70 Human weight 150 lbs Q of water 65 oz Assume steady state gt what comes in must go out USING MASS 150 lbs x 0454 kgl lb 68 kg Human Mass Assuming that 65 of that mass is from water alone 68 kg x 065 44 kg of water 9 This is the size mass ofthe reservoir or stock Flux in 64 02 da 64 ozday x 29573 cm31 02 x 1 m100 cm3 x 1000 kgma 19 kgday This is Q in mass units Residence Time 44 kg19 kgday 23 dayg USING VOLUME 44 kg Reservoir or stock size in mass units as above 44 kg x 1 m31000 kg 0044 m3 This is the stock or reservoir in volume units 64 ozday x 29573 Erna1 oz x 1 mlOO cm3 00019 mzday Residence Time 0044 m30i0019 maday 23 day u TUPafilr aPk nL FQJI iS39 395 fairg l Extv kon dusgncgag um rxm Cm own Q GE mam L Am I Am quotIN ham Hgo ISW B70 QM P Inc u ccx rour art 30 4 afar Cor mAr W uvcd r vs JUL 1 c 393G0ISCgt nampcm 2500 Q4 Charged mgefh 39 7 75 3n 5 15500 17 0 ON 43 6397 0 quot a w I 9 4 s d 373 CV 1 u I A Favk u y w 339 v
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