Topic 2: Occurrence & Movement of Groundwater

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1 -1. Occurrence and Movement of Groundwater Properties of aquifers Porosity (φ) Pore spaces places where groundwater is stored and moves Porosity = ratio or percent that aquifer material is voids V Vvoid V V + V void φ = = tot solid void Sometimes, engineers use void ratio (e) instead of porosity Measurements of porosity Vvoid φ e = = V 1 φ solid a. Direct measurement ( Ww+ s Ws) ρw φ = V tot W w + s= weight of water saturated soil W = weight of air filled soil s V = total volume of sample tot b. Using bulk density ρdb φ = 1 ρ s ρ db = dried bulk density (M/L 3 ) ρ = density of soil particle (M/L 3 ) s Effective porosity (φ e ) % of rock that is interconnected pore space V iv φ e = V iv = Volume of interconnected voids Vtot Clay has HIGH porosity but very LOW effective porosity!!!

2 - Porosity depends on how particles are packed. Cubic packing gives 47.65% porosity (φ = 0.48) π (1 ) φ = = r = 1 Rhombohedral packing gives 5.95% porosity (φ = 0.6) Porosity depends on particle size distribution Well sorted (poorly graded or poor distribution) gives large porosity C u < 4; well sorted Well graded (good distribution) gives small porosity C u > 6; poorly sorted (or well graded) Porosity also depends on other factors such as a. Shape of particles [angular particles give higher porosity] b. Cementation, compaction, and dissolution c. Degree of fracturing (of rock aquifers)

3 -3 Groundwater storage Normally, we cannot drain all water from water filled pores. Thus, we need to define another property that describes how much water aquifer can yield. This parameter is called storage property which includes specific yield, specific retention, specific capacity, and storativity. Specific yield (S y ) = % of total volume that can be drained by gravity Specific retention (S r ) = % of total volume that is held against gravity By definition, φ = Sy + Sr Not all water can be drained due to gravity because of capillary tension. Specific storage (S s ) = volume of water that aquifer releases per unit volume of aquifer per unit change in head. This volume of water is stored due to the compressibility of the aquifer and the compressibility of aquifer

4 -4 S s = ρ g( α + φβ) w S s Specific storage [L 1 ] ρ w Density of water [M/L 3 ] g Acceleration due to gravity, 9.81 m/s [L/T ] φ Porosity [ ] α Compressibility of aquifer [1/(M/LT )] β Compressibility of water [1/(M/LT )] Storativity (S) = volume of water aquifer that is released per unit surface area per unit change in head. Storativity is a combination of specific yield and specific storage. S= S + bs For unconfined aquifer y S S= bs S For confined aquifer S b Storativity [ ] or storage coefficient aquifer thickness [L] Permeability of the porous media Permeability (k) [L ] or intrinsic permeability = the capacity of a porous medium to transmit fluid. It depends only on the characteristics of the porous media. Permeability is correlated to grain size (d) k = Cd k Permeability [L ] C Cnstant [ ] obtained from experiment d Average grain size [L] Unit of intrinsic permeability is L. Sometimes it is expressed as darcys where 1.0 darcy equals to cm 10 8 cm.

5 -5 Hydraulic conductivity (K) [L/T] The ability of porous medium to transmit water. Sometimes it is named coefficient of permeability. K can be determined from several methods. 1. Calculate from intrinsic permeability kρwg K = µ w K Hydraulic conductivity [L/T] k Intrinsic permeability [L ] ρ w Density of water [M/L 3 ] g Acceleration due to gravity [L/T ] µ w Dynamic viscosity of water [M/(LT)]. Calculate from grain size distribution data K = C( d ) 10 K d 10 C Hydraulic conductivity (cm/s) Effective grain size (cm), (10% finer than this diameter) Coefficient based on the following data

6 -6 Very fine sand, poorly sorted C = Find sand with appreciable fines C = Medium sand, well sorted C = Coarse sand, poorly sorted C = Coarse sand, well sorted, clean C = From experiment (permeameter) using Darcy s law Soil/Sand column x = 0 x = L Q ( hb ha)/ L Q A (A = cross sectional area) hb ha Q= KA L 0 Q dh = q= K A dx Darcy s law Q flow rate [L 3 /T] A cross sectional area [L ] q specific discharge [L/T], Darcy velocity dh/dx hydraulic gradient [ ] h hydraulic head [L] Driving force for GW to flow Permeameter

7 -7 Hydraulic Gradient, dh/dx 4. Determine K from auiferr test (we will discuss in later topics) a. Pumping test b. Slug test Darcy velocity (q ) [L/T] Groundwater discharge (L 3 /T) per unit area dh q= K (In 1 dimension domain) dx h qx Kx 0 0 x v h q= q y 0 Ky 0 = y h qz 0 0 Kz z for 3 D Average linear pore velocity (v ), [L/T] velocity through the pores v = q K dh φ = φ dx (φ = porosity) Example II 1 Soil column (length = 13 cm) experiment shown below has h 1 =10 cm and h =8 cm. If the hydraulic conductivity and cross sectional area of the column are 1.7 m/day and 3.1 cm, respectively. Calculate darcy velocity and determine how long must it take for a chemical to move from one end to another if porosity is 0.6.

8 -8 10 cm h 1 A h Soil Column B h 8 cm x Step 1: Calculate hydraulic gradient h 10 8 = = x 0 13 Step : Calculate discharge rate h m 100 cm Q= KA = [ ] = x day 1 m cm cm /day Step 3: Darcy velocity Q q = = = 6.15 cm/day A 3.1 Step 4: Calculate average linear pore velocity q 6.15 v = = = cm/day φ 0.6 Step 5: Calculate time it takes for contaminant to move from A to B s 13 cm t = = = 0.13 day v cm/day Hydraulic head, h [L] Hydraulic head is a measure of total energy or total potential that is stored in water at any point. 1. E 1 = potential energy due to elevation (gravity) E1 = mgz (z = elevation). E = kinetic energy due to flow velocity E 1 = mv (v = flow velocity)

9 -9 3. E 3 = potential energy stored in water molecules due to pressure P E = PdV = ( P P ) V (P 0 = reference pressure, V = volume) 3 0 P0 In hydrogeology, we use gage pressure, P 0 = 0 Total energy: E = E1 + E + E3 tot Energy per volume Energy per volume measured as total head 1 Etot = mgz+ mv + PV E V tot 1 gz ρv P = ρ + + Etot v P = z+ + = h ρgv g ρg Bernoulli s equation z v g P ρg Elevation head Velocity head Pressure head h z+ p p= P ρg Velocity head is generally small because groundwater flows is very slow!! P [=] M/(LT ) For example, kg/ms or N/m p [=] L z [=] L

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11 -11 Hydraulic heads are measured (in the filed) using piezometers wells piezometers Types of aquifers Water table a surface at which the pore water pressure is atmospheric (for unconfined aquifer) Potentiometric surface a hypothetical surface that shows how high the water can rise when aquifer is under pressure (for confined aquifer)

12 -1 An aquifer can be both confined and unconfined depending on where we are considering. Perched water table is a temporary water body that mounds on an impermeable layer Some terminologies: Aquifer a porous medium that permits appreciable amount of groundwater to pass under normal filed conditions Aquiclude impermeable & does not pass significant amount of water but may store water Aquifuge nether passes not contains significant amount of water Aquitard either an aquiclude or an aquifuge (opposite to aquifer)

13 -13 Two types of aquifer: 1. Unconfined or water table aquifer. Confined aquifer Ground Surface h b h Datum, z=0 Confined aquifer Unconfined aquifer Transmissivity, T [L /T] a filed measurement of hydraulic conductivity. It is equal to hydraulic conductivity multiplied by aquifer saturated thickness (b). For confined aquifer: T = Kb For unconfined aquifer: T = Kh ***hydraulic conductivity varies with space and direction** Heterogeneity describes spatial variation of K; K = K( x, y, z) Anisotropy describes direction of variation of K; KX Ky Kz

14 -14 For layered aquifers, average or equivalent hydraulic conductivities in x (parallel) and y (perpendicular) direction can be calculated using the following relationships: K x Kb i i = bi K y = bi bi K i Determination of groundwater flow direction - Equipotential line (equal head) contour lines of hydraulic heads - Flow line (perpendicular to the equipotential line)

15 -15 Hydraulic Head: revisit Total head = pressure head + elevation head h= p+ z OR h= hp + z Piezometer measures pressure head (h p Or p) at a point Water always flow from high to low hydraulic heads. hs hp

16 -16 If aquifer has waters with different densities (such as, salt vs. fresh waters), hydraulic head will be different too. o Point water head, h s o Fresh water head, h p h p h ρs = ρ f s ( ρ s is density of saline water, and ρ f is density of fresh water) Flow line in anisotropic material If aquifer is not isotropic, the direction of the flow line will not be perpendicular to the equipotential line. For example, Isotropic (K x = K y ) Anisotropic (K x K y ) How to construc a flow line in anisotropic material: 1 Step 1: Draw ellipse with each semi axis equal to AND 1 K x K y 1 K y y x 1 K x

17 -17 Step : Draw equipotential line and assumed flow line Step 3: Draw tangent line and actual flow line Flownet Flownet is a network of equipotential lines and a set of flow lines (or streamlines) Can be used to calculate steady state flow rate (Q) through aquifer under the following assumptions. Homogeneous & isotropic Flow is laminar and darcy s law is valid No change in potential field with time ( dh dt = 0 ) The total discharge rate per unit width of an aquifer (q =Q/W) can be calculated using the following formula: must be square

18 -18 Q K N q = = W N tube drops h q Discharge per unit width (L 3 /T/L) Q Total flow rate (L 3 /T) W Width of aquifer (L) N tube Number of flow tubes ( ) N drops Number of head drops ( ) h Total head loss along flow line (L) Example II If the hydraulic conductivity of the above aquifer is 3.0 ft/day, calculate the discharge rate per unit width of an aquifer N tube = 4 N drops = 8 h = 40 4 = 16 ft 1 3 (3 ft day )(4)(16 ft) ft ft q = = 184 = day ft day Total discharge per unit width of an aquifer is 184 ft 3 /day. Flownet underneath dam

19 -19 Refraction of flowlines in heterogeneous aquifer If flow lines are not perpendicular to the K 1 boundary between two aquifers ( K 1 and K ), there will be refraction of the flow lines. K The angles of refraction can be found from the following relationship: K1 tanσ 1 = K tanσ Example II 3 [Storage problem] An unconfined aquifer with storativity of 0.13 has an area of 13 mi. If the water table drops 5.3 ft during drought, How much water was lost from storage? From the relationship, Volume of Water Drained V S = = Volume of Draining Aquifer V wdrained, a, draining = V Area h = wdrained, S Vw S A h

20 -0 V w ft = mi mi ft = ft 9 3 [ ] Example II 4 [Equivalent K] From aquifer system shown below, determine the general directionof groundwater flow h = 7 m AQUIFER #1 Y h = 8 m AQUIFER # AQUIFER #3 h = 1 m Aquifer #1 # #3 Thickness (m) 3 K (m/day) X h = 10 m 0 m First method: Use equivalent K concept Equivalent K in gradient in x direction: y direction: x direction: y direction: K K x y Kb i i = = = b dh 1 8 = = 0. dx 0 0 i i bi = = = 3.58 m/day bi K dh 7 10 = = dy 8 0 m/day Use darcy s law to calculate darcy velocity in each direction x direction: dh qx = Kx = (3.571)(0.) = m/day dx y direction: dh qy = Ky = (.58)( 0.375) = m/day dy

21 -1 direction of darcy velocity can be determined from vector addition q v θ q v x q v y v q = (0.847) + ( 0.714) = m/day v 1 qx θ = tan v = tan = 40.1 q y Second method: Use groundwater model (only for illustration, will not be discussed in this class). As you can see, the flow line directions varyith space significantly depending on where in aquifer you are located. Results from modflow 000 simulation Steady flow through aquifers Steady flow = flow that does not change with time Darcy s law only describes flow under steady state condition nothing depends on time (dh/dt = 0, dq/dt = 0, and dq/dt = 0) I. Confined aquifer dh dh dh Darcy s law: q= K or Q= KA = KbW dx dx dx From darcy s law, one can calculate hydraulic head, h, at any point x. From the relationship shown above, rearrange the above equation Q dh = dx KbW x = x h= h 1 1 x = x h= h

22 - Using the boundary conditions as follows: Then, integrate with respect to x h x Q dh = dx KbW h1 x1 Q h h = x x KbW ( ) 1 1 x1 x The flowrate can be calculated from known heads at two points. h h 1 Q= KbW x x 1 Steady flow for confined aquifer Example II 5 A confined aquifer is 33 m thick and 7 km wide. Two observation wells are located 1. km apart in the direction of flow. The head in well #1 is 97.5 m and in well # is 89.0 m. The hydraulic conductivity is 1. m/day. What is the total daily flow of water through aquifer? h h 1 Q= KbW x x 1 m 1000 m = 1. ( 33 m)( 7 km) day 1 km = 000 m 3 /day

23 -3 What is the value of potentiometric head at a point located 0.3 km from well #1 and 0.9 km from well #? Because the flowrate must be the same, h3 h 1 Q= KbW x 3 x 1 m 1000 m h = 1. ( 33 m)( 7 km) day 1 km h = 95.3 m 3 II. Unconfined aquifer In an unconfined aquifer, hydraulic head changes from one location to another. Thus cross sectional area of the flow changes accordingly. We can still develop equation for steady flow in this case. x1 x Starting from darcy s law: dh q= K or dx dh Q= KWh dx dh dh Q= KA = KhW dx dx dh Q h = dx dx KW

24 -4 Integrate above expressing using the following boundary condition x = x h= h 1 1 x = x h= h h x Q hdh = KW h1 x1 dx 1 1 Q h h = x x KW ( ) 1 1 Rearrange the equation for Q Steady flow for unconfined aquifer 1 h h 1 Q= KW x x 1 Example II 6 An unconfined aquifer has a hydraulic conductivity of 0.00 cm/sec and an effective porosity of 0.7. The aquifer is an a bed of sand with a uniform thickness of 31 m, as measured from the land surface. At well #1, the water table is 1 m below the land surface. At well #, located some 175 m away, the water table is 3.5 m from surface. (A) What is the discharge per unit width? hydraulic head at well #1: h 1 = 31 1 = 10 m (x 1 = 0 m) hydraulic head at well #: h = = 7.5 m (x = 175 m) Q 1 h h 1 = K W x x1 1 cm 1 m sec m = sec 100 cm 1 d m 3 m = 0.16 day m (B) What are the average linear velocities at well #1 and well #? well #1: v well #: v 3 m = q Q Q 0.16 day m 0.08 m/day φ = Aφ = hwφ = (10 m)(0.7) = 3 m = q Q Q 0.16 day m 0.11 m/day φ = Aφ = hwφ = (7.5 m)(0.7) =

25 -5 (C) What is the water table elevation at midway between two wells. x 3 = 175/ = 87.5 m h 3 =? Q 1 h3 h 1 = K W x3 x1 3 m 1 cm 1 m sec h3 10 m 0.16 = day m sec 100 cm 1 d m h = 8.84 m 3 Water table elevation at midway is =.16 m below land surface.

26 -6 Exercise 1. A silty sandstone with an intrinsic permeability of 1.0 µm is saturated with water at 15.5 C What is the permeability of aquifer in darcys? 1.. What is the hydraulic conductivity in m/day? 1.3. What is the hydraulic conductivity in gal/day/ft? Table: Properties of water at different temperature. Temperature ( C) Density of Water (g/cm 3 ) Viscosity of Water (g cm 1 sec 1 ) A piezometer is screened 73.4 m above mean sea level. The water pressure head in the piezometer is 17.9 m..1. What is the total head in the aquifer at the point where piezometer is installed (i.e. screened)?.. What is the fluid pressure in the aquifer at the point where the piezometer is screened? [Ans: h = m, P = N/m ] 3. A piezometer in a saline water has a point water pressure head of m. If the saline water has density of 10 kg/m 3 and the field temperature of 18 C, what is the equivalent fresh water pressure head? [Ans: h = m] 4. If an aquifer has a hydraulic conductivity of 1 ft/day and an effective porosity of 17% (φ = 0.17) and is under a hydraulic gradient of Compute the specific discharge (i.e. Darcy Velocity). 4.. Compute average linear pore velocity. [Ans: q = ft/day, v =0.388 ft/day] 5. A confined aquifer is 8 ft thick. The potentiometric surface drops 1.33 ft between two wells that are 685 ft apart. The hydraulic conductivity is 51 ft/day and the effective porosity is How many cubic feet per day of water moving through a strip of the aquifer that is 10 ft wide? 5.. What is the average linear pore velocity? [Ans: Q = 39.0 ft 3 /day, v =1.80 ft/day] 6. An unconfined aquifer has a hydraulic conductivity of cm/sec. There are two observation wells locating 597 ft apart. Both penetrate the aquifer to the bottom. In one observation well, the water stands 8.9 ft above the bottom, and in the other it is 6. ft above the bottom What is the discharge per 100 ft wide strip of the aquifer in cubic feet per day (ft 3 /day)?

27 What is the water table elevation at a point midway between the two observation wells? [Ans: Q = 3100 ft 3 /d, h = 7.6 ft] 7. An earthen dam is constructed on an impermeable bedrock layer. It is 550 ft across (i.e., the distance from the water in the reservoir to the tailwaters below the dam is 550 ft). The average hydraulic conductivity of the material used in the dam construction is 0.77 ft/day. The water in the reservoir behind the dam is 35 ft deep and the tailwaters below the dam are 0 ft deep. Compute the volume of water that seeps from the reservoir through the dam and into the tailwaters per 100 ft wide strip in cubic feet per day. [Ans: Q = 58 ft 3 /day] 8. Sketch equipotential lines and flow lines (i.e. flownet) that show the distribution of hydraulic heads and direction of groundwater flow beneath the sheet piling in the diagram below. Then, calculate the amount of discharge per unit width (m 3 /day/m). 10 m 10 m K = m/day 9. Draw a flow net for seepage through the earthen dam shown below. If the hydraulic conductivity of the material used in the dam is 0. ft/day, what is the seepage rate per unit width per day? [Ans: q = 0.55 ft 3 /day per ft]

28 -8