Quantitive Methods in Hydro
Quantitive Methods in Hydro HYD 510
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Date Created: 10/15/15
Hydrology 431 Case Study or Example Prof John L Wilson Department of Earth amp Environmental Science New Mexico Tech NEW MEXICO INSTITUTE OF MINING AND TECHNOLOGY Department of Earth and Environmental Science HYDROLOGY PROGRAM Case Study Lets apply basic math especially calculus to the evaluation of a groundwater hydrology problem Consider the following model of groundwater in a phreatic aquifer bounded by two parallel streams stream N recharge vHHHHHH water table stream h1 hm h2 39 WlA x lt L gt one elmensionalTsteady flew flowfnermal tothe stream switheanxicoOrdinate system essentially horizontal vertically integrated ow Dupuit model of phreatic aquifer If we assume a horizontal bottom the Dupuit approxi mation gives 39 ThKh 1217 39 1 where hx is water table elev above the bottom saturated thickness 1 K is homogeneous amp isotropic hydraulic conductivity lt and T Th 2t is transmissivity Even though K is Constant T is not Aquifer ux per unit width of aquifer integrated over the vertical from bottom to wa ter table is given by Q Thdhdx K2dh2dx 1227 2 Note that if you assume Tconstant you are making additional approximations to the Dupuit approximation Department of Earth amp Environmental Science Hydrology 431 Case Study or Example New Mexico Tech Prof John L Wilson Applying conservation of mass principles you should be able to show that the aquifer can be de scribed by the following mathematical statement a m d 11 dd13 rd2h2 ODE39 drThdx dxKhdx 2dxdx 2 dx2 quot N BCl 110 121 BC2 hL 112 where N represents distributed recharge Assume that the recharge is constant A Dupuit model is linear in 112 and not h We should rewrite the boundary conditions as BCl h20 hf BC2 h2L 113 The solution can be found by method of constant coef cients or separation of variables amp simple integration and broken into separate components for each in uence a the stream prescribed head boundaries and b the uniform recharge These can then be superposed for any elevation of the two streams and for any recharge rate You should be able to find the solutions which are given below I have also added a solution c for a point sink pumping well at some location xw in the aquifer This was done by breaking the aquifer into two parts one on each side of the well and solv ing for flow to the well When there are no other in uences eg no recharge the flow from the stream on the left and the stream on the right should balance the pumping rate In addition the draw down in the well should be the same for both sides that is the heads from each side should match at the well Check it out a Streams of unequal water level elevation but no recharge In the absence of recharge the solution for h is given by a straight line in h2 hgoc h2x hf LLJ kg h h3 hf 12 3 where I ve used the subscript a to differentiate this case from the two below and L is the distance between streams h 1 is the prescribed head at the stream on the left at x0 and h2 is the head in the stream on the right at xLquot The gradient and ux per unit width are given by 2 h2h2 h2 h 39 1 1L2 H Qa x rim 7 1227 4ab b Uniform recharge to aquifer discharge to streams of equal ZERO water level elevation In the presence of uniform recharge N and setting it h20 what is the distribution of 112 2 2 2 2 hbx h x K L x x K lac 96 l 5 where the subscript b is just for designation purposes The gradient and flux per unit width are given by 2 Jim L Zt H Qb x L 2x Mr 6am l Department of Earth amp Environmental Science Hydrology 431 Case Study or Example New Mexico Tech Prof John L Wilson Then since the system is linear in hz you can superpose these solutions 3 and 5 for head or 4 and 6 for ow to get the solutions for a combination of recharge and boundary conditions 0 Pumping at xxm with no recharge streams of equal ZERO water level elevation Consider a pumping gallery or 1D well at location xxw with pumping rate Qw lzt It s really more like a drain than a well since it must by de nition by in nitely long into the page one dimensional models have their limitations Setting h1h2 0 and NO what is the distribution of h2 in this aquifer due to pumping 11300 1200 915 xw m x x lt xw 12 W 622 35 14 xw x gt xw 7b The gradient and flux per unit width are given by 1 h 2x erIW xw L x lt xw Q Cx L xw L x lt xw 8017 dx 7c 2 d i x x i xw x gt xw Q cx Q2 Lw xw x gt xw Again since the system is linear in hzi you can superpose 7 with 3 and 5 to get head for a com bination of a well recharge and boundary conditions or 8 with 4 amp 6 to get gradients or uxes SAW Notice that I set h h2 0 for the solutions to pumping and recharge This is a necessary conditi on to superpose the solutions Otherwise I d be double counting We ll use superposition of 112 to answer the following questions taking the square root to get It only when we must to finish the calculation to get to the desired product In general because of superposition h2x hx hx hx Q x Q39ax Q bx Q cx Questions 1 If there is no recharge N 0 no pumping Q w0 and zero head on the right h2 0 but finite head on the left h gt O a What is the simplest symbolic equation for h2 x b What is the simplest symbolic equation for hx c What is the shape of the water table Give a functional or algebraic name eg quadratic or exponential d What is the mathematical relationship between the discharge per unit width Q x at x0 and at xL 2 Suppose that h h2 0 and Q w 0 butN 0 The recharge is increased from N N0 to a new value N N0 N where N is the change in recharge After a new equilibrium is reached how much more Department of Earth amp Environmental Science Hydrology 431 Case Study or Example New Mexico Tech Prof John L Wilson water will be in storage in the aquifer Assume an aquifer storage coef cient specific yield in this case given by S Keep superposition in mind then find the difference in head and integrate over space to find the change in volume 3 Check 1307 and Q cx the head squared and discharge per unit width flux for the well case in the absence of recharge and pumping ie check 7 and 8 In this case the pumping rate must equal the flow in from the stream on the right and the stream on the left Check sequentially for things like units sign mass volume balance and does it make sense Annotate concisely your checks and if you find a problem document it and using the same checks correct it 4 Suppose h h2 0 but N gt0 and there is a pumping or injection well at xw L4 Using superposi tion and derivatives determine the pumping or injection rate needed to move the groundwater divide from L2 for the no pumping case left to L3 in this case Note that if dhdx 0 then thdx 0 Check your result 5 Suppose that N 0 and Q w0 but h gth2 gt0 There is ew from left to right The travel time from the left to the right boundary is given by L L 1 W quot quot0 cm dquot JO Q ax dquot Integrate and nd a symbolic expression for travel time t Department of Earth amp Environmental Science Hydrology 431 Case Study or Example New Mexico Tech Prof John L VVllson Questions and Answers 1 If there is no recharge N20 no pumping Q w0 and zero head on the right h2 0 but finite head on the left h1 gt O a What is the simplest symbolic equation for h2x 2 2L x hx h1 L b What is the simplest symbolic equation for hx hx h1 fi c What is the shape of the water table Give a functional or algebraic name eg quadratic or exponential iPower Law 12 power d What is the mathematical relationship between the discharge per unit width Q x atx O and at xL 39 2 2 K hfhz 2 They are equal to each other Q x Q a const L 2 Suppose that h 1 h2 0 and Q w 0 butN 0 The recharge is increased from N No to a new value N N0 N where N f is the change in recharge After a new equilibrium is reached how much more water w ill be ins torageinetheaquiferAssume an aquifer storagecoef cientfspecificyieldrinsthisw case given by S Keep superposition in mind then find the difference in head and integrate over space to find the change in volume V VlUL L Volume removed or added to storage 5 I hf hi dx 2 0 where hf is the nal head after equilibrium and h is the initial head These are found from 3 N N N h2x f0 L x x hf2x 47 L x x But we need h not h2 to compute the volume Thus we must take the square root L H W W yp me it 39 I L L S N0l I Lx x2 dx SV O Lx xz dx 0 L S my he J r L2dx K K 0 using power laws from algebra The integral in this is nothing more than Department of Earth amp Environmental Science Hydrology 431 Case Study or Example New Mexico Tech Prof John L Wilson I VZax xzdx x a axx2 an 1 C CRC tables where aL2 or L L J Lx x2 dx 22x L tree x2 Lzsinl sz IL 0 0 l2L le LZ O Lzsin 12LL L2sin1 8 LI ILI L2 1L1L 3 L2 1143 8S quot ILI 5 quot ILI 8 2 2 quot 8 8 Now substituting into the expression for volume yields VanLz N0N39 IVo 8 K K Is this 0k Sign ifN gt0 we get more storage Units 2 are ok Mass balance can t do anything with that unless we integrate the transient over time We don t know the transient Mass balance doesn t help I checked the result by rederiving it using Mathematica Incidentally doing this I found an error I had 16 in the denominator in my hand calculation Multiplication error I f you gure out a way to check something Do it 3 Check hx and Q cx the head squared and discharge per unit Width ux for the well case intheabsenceof recharge and pumping iereheek Giand 8Tln thisease the pumpmg ratemust equal the ow in from the stream on the right and the stream on the left Check sequentially for things likennitssigmmassaolumehalanceeand doeseitmakesense52iAnnotateeoneisely your eheeks and if you find a problem document it and using the same checks correct it My annotation here is far more detailed than I believe you would have I m simply recording in writ ten detail how I thought through each issue age h2x xw L x x lt xw 12 7n 39L 1 x L x x gt xw 7b Q ex L 3 xii L x lt xw 8a 22ng xgtxw 8b Lets check Units 7 has units of 12 on the lhs left hand side and units of 1221 12112i 12 on the right side Units are ok 8 has units of lzt on the lhs and units of12t ll lzt on the right side It checks out too 39 Lets check signs 7 is negative for 0ltxltL or in other words drawdown sx hcx is positive But what about the h2 being negative Take its square root and you get a complex number What does Department of Earth amp Environmental Science Hydrology 431 Case Study or Example New Mexico Tech Prof John L Wilson that mean A well cannot pump from this Dupuit aquifer unless there is water in it That is to apply this theory N h 1 and h2 must be large enough to insure a positive saturated thickness The T constant model is not nearly so picky What about the ux It s positive for xltxw when the ow is from the left stream to the well and negative for xgtxw when the ow is from the right stream This too is ok Lets check mass balance The pumping must come from the two streams There is no other source of recharge and this is a steady state Thus taking into account signs ow is from the left and from the right I have Qlw Qcxw lt x quot QC xw gt x quot9 3 xw L xw 9 22 2L 2 WH39OOPSL Mass doesn t balance 8 should not have the 2 in the denominator Is 7 ok Well if you differentiate it you get 8 They are consistent and consequently 7 is in error too It should have a 2 in thenumerator The correct equations are hx h2x 2 xiv L x x lt xw 12 7a 2Q w KL x L Aw x gt xw 7b Q t Q V3 x L 39 x lt r 8a C L W t quot W Q W xw x gt xw 8b Do they make sense Well h2 and Q increase in magnitude with Q w That s appropriate as is the lowering of hz with larger K The heads hZ go to zero on the boundaries and are at maximum in the center That too makes sense for this symmetric system with prescribed head boundaries Flux is constant between the streams and the well as one would expect Also the flux is maximized by putt ing the well as far from the stream as possible Does that make sense 4 Suppose h h2 0 but N gt0 and there is a pumping or injection well at xw L4 Using superposi tion and derivatives determine the pumping or injection rate needed to move the groundwater divide from L2 for the no pumping case left to US in this case Note that if dhdx 0 then thdx 0 Check your result Well this just asks us to gure out what Q w will lead to tha x 0 at xL3 which is the same thing as asking when Q L3 0 In this case Q x Q bx Qcx where the Q s are given by 6 and 8 Don t forget to use the corrected version of 8 from above Also which case of 8 do we use The place where we want the divide is at L3 which is to the right of the well at L4 That is xL3gtxw L4 Thus with xgtxw and we should use the second case I are J ZV L 2x Q39cx 9518 xw or Q x aux Qcx L zx QL W xw Department of Earth amp Environmental Science Hydrology 431 Case Study or Example New Mexico Tech Prof John L Wilson Q L3 ohm3 octLs 14 QL W 1 1VL Q39n 6 4 Solving for Q w by setting Q L3 0 yields 39 Q 4NL 2NL w 6 3 That is one should inject minus pumping at xsz4 at the rate 2NL3 in order to stabilize the groundwater divide atxL3 Note we are injecting ata rate that is 23 s of the natural rechargedis charge rate 5 Suppose that N20 an39d39Q w0 but h gt712 gt0 There is ow from left to right The travel time from the left to the right boundary is given by 39 L L t n 1 dx n E dx 0 105 0 Q ax Integrate and nd a symbolic expression for travel time t The key here is to note that Q a is a constant as is clear from 4b 2h2 Qa39x I 1L 2 Qa constant thurwecanetakethis itxtermbut oftheintegrairleavingoniythe ltx7 ermrwhicirwe knowira square root of 3 where 3 is n n n I v a v A r A 1 x have h x hf L h f hf kg Izsz 3 Let s take the second form of 3 which will t better in integral formulae 11 h hg h f Substituting 3 and 4b into the travel time integral we have L L L W L L t nJO Qax dx Qa0 hxdx QaJO a bx dx where a h and b hi h L The integral is evaluated as I a bx 3 a bx3 C CRC tables or substitution Thus L n 2 3 2 32 32 t Qa 3ba bx l0 3ana bL a Since a32 h3 l y Hydrology 431 Case Study or Example Department of Earth amp Environmental Science Prof John L Wilson New Mexico Tech 32 32 3 2 a bL t rig h LL kg kg then using the these substitutions and the de nition of b 3 3 3 3 ET t2nL h2 h1 4nL2 hl hz 3 2 23K2222 Q 12 kl kl h2h1 h2 C heck Units ok Sign 0k positive Mass balance can t see how to use Make sense39 t increases with n L and JK That s ok What about h and hz Let s check by looking at limiting condition as the saturated thickness is large compared to the di erence in h and h2 That is let s check out things when we are approximating a Tconstant case In that limit we expect t5 nLZKh1 h2 Does the expression above converge to this as h gt hg st h1 h2 h1 ltlt 0 In limit we have these approxi mations h1 2 h22 h1 h2h1 h2 s h 412 23 where we de ne 3011 1122 ht 31123 h1h2h12h1h2h22 E ht 412 332 where in the limit the expression h 2h1 hg h2 2 E 332 and these can be used in the travel time ex pression to check it It checks out Our Dupuit model result converges to the expected resuit for a T constant model in the limit