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# Class Note for HWR 518 at UA

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Date Created: 02/06/15

We have seen how the hydraulic properties of a medium vary as a function of the water content andor pressure head of the water phase Now let s consider a special case of steadystate unsaturated ow The case involves ow through a homogeneous vertical column The boundary conditions are such that the hydraulic head gradient through the column is 71 downward That is the hydraulic head gradient is equal to the elevation head gradient and the pressure head gradient is zero If the pressure head is constant everywhere the water content will also be constant everywhere For this case we can determine the hydraulic conductivity throughout the column at this pressure head from a plot like Figure Ll64 The ux will be equal to this hydraulic conductivity In general then we can see this will occur if the upper boundary condition is a constant ux that is equal to the hydraulic conductivity of the medium at the pressure head de ned at the lower boundary We can also see that there is an in nite number of possible hydraulic conductivities ranging from zero to the hydraulic conductivity at full saturation K x that can describe the ability of a medium to transmit uid under unsaturated ow conditions So the same hydraulic head gradient applied to the same medium will give a wide range of uxes Given that this is the simplest case of unsaturated ow we will return to it often to explain more complex transient ow A full treatment of multiphase ow is beyond the scope of this course Fortunately water air systems can be simpli ed by considering air to be in nitely mobile The pressure in the air phase is then zero gauge everywhere and spatial and temporal derivatives of the air phase pressure are zero as well As a result only those equations describing the ow of water are necessary to describe the ow system The resulting equation describing water ow is referred to as Richards equation Ll7l where kr H is the relative permeability which varies from 0 to 1 By replacing the water pressure with the water pressure head the relationship becomes kk pg 6 U rz39j aw 62 07 a ampTampmd39 am 1 J J For water ow the hydraulic conductivity can be used rather than the permeability a W amp Q AiKkH4447m4lt mm myaa a HWR518 Lecture 17 7 Steady State Unsaturated Flow 112 Assuming that the porosity is constant in time the water phase saturation can be replaced by the volumetric water content to give the mixed form of Richards equation a aw a 6 6 736 Ksnkrl 736 I 7d6 7d L174 l U J J The pressure dependent water capacity C L39l is de ned as ivu 2w 5 6V d C 07 L17 5 Substituting this into the storage term gives the pressure head form of Richards equation a W a W 7 K kr 77 C i L176 0amp1 s1 U 03c 03c 07 This form of Richards equation is dif cult to solve because it is highly nonlinear through kr w and C041 However pressure varies smoothly everywhere even in heterogeneous or layered media making this form very robust To extend this form of the equation for use in saturated media the saturated storage term SS must be added to the right hand side For completeness the storage mechanisms discussed for saturated ow could be added to all of the multiphase ow equations However their contributions are insigni cant compared to drainage lling of the pores The water content dependent water capacity C L39l is de ned as W W 6 6 1 6 6 277277 L177 035139 d9 035139 76 035139 Alternatively beginning with the mixed form of Richard s equation Equation Ll74 and substituting the water content dependent water capacity into the gradient term gives the water content form of Richards equation 0 1 6 6 52 6 6 7 K kr 777 7 L178 0amp1 s1 U C90 cj 03c 07 This form is more amenable to analytical solutions However the water content form of the unsaturated ow equation is not valid in water saturated regions because the water HWR518 Lecture 17 7 Steady State Unsaturated Flow 212 capacity becomes zero Furthermore the water content is discontinuous across material boundaries precluding the use of this form in layered media To begin examining water ow through a partially saturated medium we will consider the relatively simple case of steady state ow through a horizontal column lled with a homogeneous medium Constant heads are applied to the left and right boundaries such that ow is from left to right For this horizontal column the pressure head gradient alone drives ow If the medium were saturated the pressure head gradient would be constant because the hydraulic conductivity would be constant In contrast if the imposed boundary conditions lead to unsaturated conditions in the column the dependence of kr on w will cause kr to be lower at the lower pressure end of the column than at the higher w end To establish a constant ux along the column the gradient in w must be greater in regions of low w This leads to the nonlinear w distribution seen on Figure Ll7l The water content along the column can be calculated based on the w distribution 0 50 100 150 200 Distance from left cm Figure Ll7l A schematic representation of pressure head hydraulic conductivity and water content pro les along a homogeneous horizontal column undergoing steady state ow Fully saturated conditions are shown as solid lines and variably saturated conditions are shown as dashed lines Next we ll consider vertical ow through a homogeneous variably saturated column Now both pressure head and elevation head gradients can drive ow The simplest case is that of a constant pressure head everywhere Elevation head gradients alone drive ow the hydraulic head gradient is 71 downward As a result vertical ow under an elevation gradient only is referred to as ow under a unit gradient condition If the pressure head is constant everywhere then the water content and the hydraulic conductivity are constant everywhere Figure Ll72 shows two different cases of unit gradient ow Note that the ux will be less for the case with a more negative pressure head because the medium will have a lower 6 causing the hydraulic conductivity to decrease HWR518 Lecture 17 7 Steady State Unsaturated Flow 312 W1 W2 H2 H1 0 Head L Figure Ll72 Head pro les through a vertical homogeneous column undergoing steadystate downward ow under a unit gradient 100 80 H z 60 40 E Elevation above base cm CD 50 0 50 100 Head cm L o 4 o Figure Ll73 Head pro les through a vertical homogeneous column undergoing steadystate downward ow Now consider what happens if the pressure head at the top boundary is reduced to a value lower than the pressure head at the bottom boundary for case 1 shown in Figure Ll72 With this change the steadystate hydraulic head gradient is nearly 71 downward at the top of the column Figure Ll7 3 Given that the hydraulic conductivity is lower than before the pressure head was lowered the ux is reduced As a result near the base of the column where the hydraulic conductivity is closer to the original value the hydraulic head gradient is much less than one The hydraulic conductivity varies nonlinearly with HWR518 Lecture 17 7 Steady State Unsaturated Flow 412 pressure head between the two boundary values giving rise to a nonlinear change in hydraulic head with depth Note gure Ll7 3 was generated using a spreadsheet model that students develop in HWRSOS The analyses presented previously have only considered homogeneous media Regarding layered media there is an important difference between saturated and unsaturated conditions Speci cally while the pressure head is continuous across a layer boundary the water content is not To demonstrate this consider a layered medium under static conditions As we ve seen previously the pressure head decreases 11 with an increase in elevation So the characteristic curves of both materials can be superimposed to describe the water content distribution of the layered medium The discontinuity in water content across the boundary causes a discontinuity in hydraulic conductivity As a result hydraulic head pro les under steadystate ow can show a sharp change in gradient across layer boundaries E 200 a g 150 O 0quot gs 100 F 50 E 0 i 0 02 04 06 Volumetric Water Content cm3cm3 Figure Ll74 Static equilibrium water content pro les in two media dashed and the combined static equilibrium water content pro le in a layered medium solid HWR518 Lecture 17 7 Steady State Unsaturated Flow 512 HWR 518 Self Check Homework 7 key 1 Plot the height of rise of water as a function of capillary tube radius for water for radii ranging from 001 to 01 cm Calculate the capillary radius that would give a rise of 5 cm The equation relating height of rise and radius is hc Zaprr The interfacial tension is 72 cmsz pw is 1 gcm3 g is 981 cmsz Rearranging and solving for r gives a radius of 0029 cm for a height of rise of 5 cm 16 14 E 12 5 10 39 8 water 3 oil 5 6 4 2 0 i i i i i i i 0 002 004 006 008 01 012 Capillary tube radius cm 2 On the same axes plot the height of rise as a function of capillary tube radius for oil with a density of 098 gcm3 and an airoil interfacial tension of 002 Nm Calculate the capillary radius required for a 5 cm height of rise of oil The oil would require a tube ofradius 0 008 cm 3 Consider two cases 1 An initially dry porous medium was lowered into a pan of water so that the bottom 1 cm of the medium was in water and the upper 20 was above the water 2 The same dry medium is lowered as far into a pan of oil Would the medium wick up more oil or water Explain HWR518 Lecture 17 7 Steady State Unsaturated Flow 612 If the medium is represented as a bundle of capillary tubes then the number and size of the tubes is the same in both cases From the previous gure water would wick up higher than oil in each tube Therefore there would be more water wicked up than oil 4 Explain why a water content pressure head plot eg Figure Ll35 is identical to a plot of the water content vs elevation above the water table under static conditions Assume that the datum is located at the water table Then the hydraulic head is zero everywhere As the elevation head increases 1 1 with elevation the pressure head must decrease 1 to maintain a constant hydraulic head Therefore the elevation head and the negative pressure head are identical and one can replace the other on the y axis of the plot 5 Generate a plot of the water content vs negative pressure head wetting phase suction relationships for soils with the van Genuchten properties shown in Figure Ll35 Assume that the porosity and residual water content are 038 and 005 for all three soils Calculate the total length of water from 0 to 500 cm for each of the pro les 500 400 300 200 100 Negative pressure head cm Volumetric water content The profiles listed in descending order according to the legend have 1105 42 0 and 60 6 cm of water above the water table This can be calculated as the sum of the products of the water content and the pressure head change between calculated points That is if you calculate the water content for every 20 cm of pressure head multiply the HWR518 Lecture 17 7 Steady State Unsaturated Flow 712 water content at each depth by 02 then sum this up over the entire 500 cm The result is the area under the curve with units of length Note 20 cm is an arbitrary choice HWR518 Lecture 17 7 Steady State Unsaturated Flow 812 HWR 518 Graded Homework 2 7 part 2 This assignment is worth 15 of your nal course grade There are two parts Each part is worth half of the marks Each subquestion is worth 5 points The total value of each question is given in parenthesis beside the question number The types of questions have been chosen to give you an idea of what you would be expected to answer on a midterm exam Part 11 deals with unsaturated ow and will be given to you at a later date You may use your notes and other references However you are to work individually on the assignment Exams will be closed book Part1 is due Tuesday October 28m no later than the beginning of class Part II is due November 25m Assignments must be handed in electronically as text documents with typed equations and embedded plots No late assignments will be accepted Assignments are to be sent to me tyhwrarizonaedu and to Andrew Hinnell ahinnellhwrarizonaedu Part II 1 20 Consider a 100cm long onedimensional vertical column packed with a homogeneous medium The pore space is lled with air and water only There is no water ow The pressure head at the bottom of the column is zero There is no ow across the top boundary Consider the medium to be comprised of straight circular tubes There are equal fractions of ve pore sizes 0001 00009 00008 00007 00006 In That is each pore size comprises 20 ofthe pores not 20 ofthe pore space Calculate the height of rise of the airwater interface in each of the ve pore sizes Calculate the fraction of the pore space that is represented by each pore size Explain brie y why the height above the water table is equivalent to the negative water pressure head under hydrostatic conditions Assuming that the porosity is 040 cm3cm3 plot the volumetric water content xaxis versus the negative pressure head yaxis for the medium Show the relationship using a line to demonstrate the steplike changes in water content that occur in capillary tubes HWR518 Lecture 17 7 Steady State Unsaturated Flow 912 2 3 20 Consider the same medium described in question 1 Flow through a 721 8 03539 horizontal circular capillary tube follows Q 2 Write this in terms of the ux through the tube From this write an expression for the hydraulic conductivity of a tube of radius r Calculate the hydraulic conductivities of the five tubes described in question 1 The dynamic viscosity ofwater is 001 Nsmz Calculate and plot the hydraulic conductivity of the medium xaxis as a function of the negative pressure head yaxis for the medium as described in question 1 25 Consider a spill of gasoline an LNAPL into an uncon ned aquifer Write an expression defining the height of rise of the waterLNAPL interface above the water table Define the terms Using oilair and oilwater interfacial tensions of 0020 and 0030 Nm and densities of 1000 and 980 kgm3 for water and oil write an expression of the height of rise of the oilwater interface as a function of the pore radius and the thickness ofthe LNAPL pool Typically LNAPLs are visualized as oating on the water table This only occurs if the height of the waterLNAPL interface above the water table is zero Write an expression that relates the thickness of LNAPL pool that can be supported as a function of the pore radius What pore size is necessary to support a 5 cm thick pool of LNAPL above the water table Assuming that the pore size is approximately equal to the average particle size what type of material would this be sand silt clay HWR518 Lecture 17 7 Steady State Unsaturated Flow 10 12 Name three changes to properties of the medium or to the LNAPL that would increase the thickness of LNAPL that could be supported above the water table 4 20 Consider a homogeneous vadose zone The van Genuchten properties describing the medium are 0r 0045 cm3cm3 0S 043 cm3cm3 a 0 145 cm391n 268 mlln and Ks 000825 cms Plot the volumetric water content XaXis as a function of negative pressure head yaXis for pressure heads ranging from 0 to 7200 cm What is the volumetric water content at a pressure head of 720 cm For the same soil plot the hydraulic conductivity as a function of volumetric water content What is the hydraulic conductivity at a pressure head of 720 cm Steadystate in ltration into a homogeneous unsaturated medium well above the water table results in a constant water content with depth Explain why this also means that the water content of the medium must adjust to the value at which the hydraulic conductivity of the medium is equal to the applied ux Based on your answer to the previous questions describe a series of measurements that could be made to de ne the K0 relationship of a soil What would the upper and lower boundary conditions be for the system What would you measure 5 20 Consider transient ow into the medium described in question 4 0r 0045 cm3cm3 05 043 cm3cm3 a 0145 cm391n 268 mlln and Ks 000825 cn s HWR518 Lecture 17 7 Steady State Unsaturated Flow 1 112 What are the three mechanisms of storage that affect transient unsaturated ow Which is dominant under unsaturated conditions If the medium has an initial water content of Bi 010 cm3cm3 and if in ltration occurs at a constant rate equal to the saturated hydraulic conductivity Ks write an expression that describes the rate of advance of the wetting front Using the soil properties given what is the rate of advance of the wetting front in cms What is the velocity of the added water front After one day of infiltration how far will the wetting front and the added water front be separated The added water front can be represented by the isoconcentration line of C Co 2 Would the isoconcentration line of C C04 be closer to farther from or the same distance from the wetting front after one day Defend your answer HWR518 Lecture 17 7 Steady State Unsaturated Flow 12 12

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