=========================== Previous exams with answers =========================== :Author: Prof. dr.ir. T.N.Olsthoorn Open-book exam (1h), Feb 7, 2022 ================================ Question 1 ---------- Questions 1.1 and 1.2 A well is installed at 250 m distance from a river. The well fully penetrates the aquifer. The aquifer is unconfined, it has a conductivity of k= 30 m/d and a representative thickness of 40 m, which may be assumed constant. It further has a storage coefficient (specific yield) of S=0.20. The well extracts water at a rate of *Q*\ =1200 m\ :sup:`3`/d during exactly 7 days, after which it stops. The questions refer to the drawdown in point p at 50 m from the well on the line through the well perpendicular to the river as shown in the figure. .. container:: centering .. figure:: pictures/2022_1.png :alt: Well adjacent to a straight fully penetrating river. The aquifer is homogeneous and extends to infinity.This is clearly a well in an aquifer with constant transmissivity, for which the well-known Theis solution is applicable. To obtain values for the Theis well function, you can make use of the Theis type curve shown below. :width: 80.0% Well adjacent to a straight fully penetrating river. The aquifer is homogeneous and extends to infinity.This is clearly a well in an aquifer with constant transmissivity, for which the well-known Theis solution is applicable. To obtain values for the Theis well function, you can make use of the Theis type curve shown below. .. container:: centering .. figure:: pictures/2022_2.png :alt: Theis type curve, i.e., the Theis well function as a function of :math:`1/u` :width: 80.0% Theis type curve, i.e., the Theis well function as a function of :math:`1/u` #. What will be the drawdown at point p at :math:`t=7` d after the start of the extraction? -------------- Generalities: The transient drawdown due to a well in an aquifer of constant transmissivity extracting at a constant rate from :math:`t=0` is according to the Theis solution .. math:: s(r,t)=\frac{Q}{4\pi\text{kD}}\mathrm{W}(u)\mbox{, with }u=\frac{r^{2}S}{4kDt} But we must take care of the river, which creates a fixed-head boundary condition. This is done by placing a mirror well of opposite sign at the exact opposite side of the river and at the same distance. The formula for the well and its mirror well is .. math:: s_{r,t}=\frac{Q}{4\pi kD}\left(\mathrm{W}\left(\frac{r_{1}^{2}S}{4kDt}\right)-\mathrm{W}\left(\frac{r_{2}^{2}S}{4kDt}\right)\right) In which :math:`r_{1}=50` m and :math:`r_{2}=450` m, :math:`\text{kD}\ =\ 1200` m2/d, :math:`S=0.2` We also have :math:`\frac{Q}{4\pi\text{kD}}=\frac{1200}{4\pi1200}=0.08` Then for t=7 days :math:`u_{1}=1.49\times10^{-2}`, :math:`u_{2}=1.21`, :math:`1\text{/}u_{1}=6.72\times10^{+1}`, :math:`1\text{/}u_{2}=8.30\times10^{-1}`, Then read the values for :math:`W(1/u)` from the Theis type graph, which yields the following numbers (however yours may be a bit different because it’s difficult to read exact numbers from a graph. :math:`W\left(1.49\times10^{-2}\right)=3.6`, :math:`W(1.21)=0.16` So the drawdown after 7 days in point p is :math:`s_{t=7}=0.08\times(3.5-0.16)=0.28` m -------------- #. What will be the drawdown at point p 14 days after the start of the extraction, i.e. 7 days after the well has started pumping? -------------- For t = 14 days continuous pumping the values become :math:`u_{1}=7.44\times10^{-3}`, :math:`u_{2}=6.03\times10^{-1}`, :math:`1\text{/}u_{1}=1.43\times10^{+2}`, :math:`1\text{/}u_{2}=1.66`, read the values of :math:`\mathrm{W}(1/u)` from the Theis type curve to get the following values (which again may be a bit different from yours due to the accuracy with which you can read the numbers from a graph): :math:`\mathrm{W}\left(7.44\times10^{-3}\right)=4.3`, :math:`\mathrm{W}\left(6.03\times10^{-1}\right)=0.45` Hence the drawdown for :math:`t=14\,\mathrm{d}` of continued pumping would be :math:`s_{t=14}=0.08\times(4.30-0.45)=0.31\,\mathrm{m}`. The actual drawdown after 14 days will be that of continuous pumpoing (0.31 m) minus that of 7 days continuous pumping (0.28 m) because the pump was switched off after 7 days. We, therefore, after 14 days we only have a remaining drawdown equal to 0.31 – 0.28 = 0.03 m. -------------- Question 2 ---------- The picture shows an aquifer bounded by a fully penetrating river at *x=0*. The aquifer is unbounded to the right and has a transmissivity and a specific yield as indicated in the picture. Note that the transmissivity may be considered constant. The river water level varies continuously according to a sine-wave with a cycle time of *T= 1* d and an amplitude of *A=1.2* m. .. container:: centering .. figure:: pictures/2022_3.png :alt: Aquifer bounded by fully penetrating water body with fluctuating water level at x=0. The aquifer extents at the right to infinity. Shown is the water table (or head) at an arbitrary time. :width: 80.0% Aquifer bounded by fully penetrating water body with fluctuating water level at x=0. The aquifer extents at the right to infinity. Shown is the water table (or head) at an arbitrary time. #. What is the maximum and minimum head at *x = 25* m and at *x = 100* m? #. What will be the drawdown at point p 14 days after the start of the extraction, i.e., 7 days after the well has started pumping? #. By how much (i.e., by how large a factor) does this delay change if the storage coefficient would be 100 times as small as the given value, i.e., if it would be :math:`S=0.001` instead of :math:`S=0.1`? -------------- The formula for the head in this aquifer is .. math:: s(x,t)=Ae^{-\text{ax}}\sin\left(\omega t-ax\right) With :math:`a=\sqrt{\frac{\omega S}{2\text{kD}}}`, the damping factor and :math:`\omega T=2\pi\rightarrow\omega=\frac{2\pi}{T}` The maximum and minimum head at any x is given by :math:`s_{\max,\min}=\pm Ae^{-\text{ax}}`, the so-called envelope. :math:`\omega=\frac{2\pi}{T}=\frac{2\pi}{1}=6.28` radians/day :math:`a=\sqrt{\frac{\omega S}{2\text{kD}}}=\sqrt{\frac{6.28\times0.1}{2\times600}}=0.023` [1/m] At :math:`x=25\,\mathrm{m}`, we have :math:`s_{\min,\max}=\pm Ae^{-\text{ax}}=\pm1.2e^{-0.023\times25}=\pm0.68\,\mathrm{m}`. At :math:`x=100\,\mathrm{m}`, we have :math:`s_{\min,\max}=\pm Ae^{-\text{ax}}=\pm1.2e^{-0.023\times100}=\pm0.12\,\mathrm{m}`. What is the delay of the wave at *x = 100* m relative to the wave at *x = 0* m? The velocity of the wave follows from the argument of the sin being constant .. math:: \omega t-ax=\mathrm{const} Taking the derivative with respect to *t* yields .. math:: \omega-a\frac{dx}{dt}=0 And so :math:`v=\frac{dx}{dt}=\frac{\omega}{a}=\frac{6.28}{0.023}\approx270\,\mathrm{m/d}`. Hence, the delay at :math:`x=100\,\mathrm{m}` is :math:`100\text{/}270=0.37\,\mathrm{d}\approx9\,\mathrm{h}`. -------------- Question 3: ----------- The picture below shows an aquifer of limited lateral extent. To the left, at *x = 0*, it is bounded by a fully penetrating surface water body, such as a lake. To the right, at *x = L*, it is bounded by an impervious land mass as shown. The aquifer properties are shown in the picture, but you don’t need them to answer the questions. The water level of the lake and the groundwater table are initially flat at a level equal to *h=0* m as indicated by the horizontal blue line. At :math:`t\ =\ 0`, the water level of the lake suddenly changes upward by an amount *A* as indicated. Ignore other hydrological features like rain and evapotranspiration. Only the effect of the sudden change of the lake level is considered. .. container:: centering .. figure:: pictures/2022_4.png :alt: Picture of the aquifer with fully penetrating water body at :math:`x=0` and impervious mass at :math:`x=L` :width: 80.0% Picture of the aquifer with fully penetrating water body at :math:`x=0` and impervious mass at :math:`x=L` #. What are the boundary conditions at *x=0* and *x=L*? -------------- The boundary condtions are a) fixed head at :math:`x=0` and b) zero flow at :math:`x=L`. -------------- #. Describe how the head in the aquifer will develop over time due to the sudden change at *x=0* and *t=0*. Your description must include the situation at :math:`t=0` and at :math:`t=\infty`. -------------- At :math:`t<0`, :math:`s=0` everywhere. At :math:`t=0^{+}`, :math:`s=A` at :math:`x=0` and :math:`s=0` everywhere else. At :math:`t>0`, :math:`s=A` at :math:`x=0` and :math:`A>s>0` everywhere else with :math:`s_{x>x_{1}}L` then is .. math:: t_{2}=\frac{L}{v_{1}}+\frac{x-L}{v_{2}}\mbox{, where }v_{1}=\frac{\omega}{a_{1}}\mbox{ and }v_{2}=\frac{\omega}{a_{2}} -------------- Question 3: ----------- Consider a well in an infinite water-table (phreatic) aquifer. Drawdowns are considered small compared to the thickness of the aquifer, so that :math:`kD=900\,\mathrm{m^{2}/d}` may be considered constant. The specific yield, :math:`S_{y}=0.15`, is also constant. As there are no boundary conditions, the drawdown by the well follows the Theis equation. An approximation of which is .. math:: s\approx\frac{2.3Q}{4\pi kD}\log\left(\frac{2.25kDt}{r^{2}S}\right) #. Derive a mathematical expression for the so-called radius of influence, that is, the distance beyond which the drawdown can be neglected at a given time :math:`t` after the well was first switched on. -------------- We just have to set :math:`s=0`, that is, the argument of the logarithm is 1 .. math:: \frac{2.25kDt}{r^{2}S_{y}}=1\rightarrow r=\sqrt{\frac{2.25kDt}{S}} -------------- #. As you can see the drawdown is logarithmic in time. Derive a mathematical expression for the increase of the drawdown per log cycle, that is between for instance *:math:`t=6\,\mathrm{d}`* and *:math:`t=60\,\mathrm{d}`* days, or *:math:`t=2\,\mathrm{d}`* days and *:math:`t=20\,\mathrm{d}`*. -------------- In this case we compare the drawdown after :math:`t=10\Delta t` with :math:`t=\Delta t`: .. math:: s_{10\Delta t}-s_{\Delta t}=\frac{2.3Q}{4\pi kD}\left[\log\left(\frac{2.25kD\left(10\Delta t\right)}{r^{2}S}\right)-\log\left(\frac{2.25kD\Delta t}{r^{2}S}\right)\right]=\frac{2.3Q}{4\pi kD} -------------- #. Assume the well has been continuously pumping for time :math:`t=t_{1}`, after which the extraction was stopped. What is the drawdown at distance :math:`r_{0}` at time :math:`t=t_{1}+\Delta t` , where :math:`\Delta t` is any time passed since :math:`t_{1}`. -------------- .. math:: s_{t_{1}+\Delta t}-s_{t_{1}}=\frac{2.3Q}{4\pi kD}\log\left(\frac{t_{1}+\Delta t}{t_{1}}\right) -------------- Closed-book exam (1h), Feb 2014 =============================== Question 1: ----------- #. What types of storage or storage coefficients are associated with transient groundwater flow? -------------- - Phreactic of water-table storage,due to filling and emptying of pores. The corresponding storage coefficient is called specific yield :math:`S_{y}\,\mathrm{[-]}` - Elastic storage which is due to compression and expansion of the water and the porous matrix. The corresponding storage coefficient is called :math:`S\,\mathrm{[-]}` -------------- #. Explain how these types of storage physically work. -------------- #. Storage due to drainage of pores. #. Storage due to elasticity of water and the porous medium -------------- #. Explain the relation between capillary fringe and air entry pressure. -------------- The capillary fringe and the air-entry pressure are equivalent. It is the pressure required to flow air through the largest pores down to the water table. It is also the height over which water is sucked up into the largest pores from the water table. -------------- #. Explain the difference between the loading efficiency (:math:`LE`) and the barometer efficiency (:math:`BE`)? -------------- The loading efficiency is the increase in water pressure (or head) in a confined aquifer due to and relative to a load placed uniformly on ground surface. The barometric efficiency is same caused by an increase of the barometric pressure. However, due to the barometric pressure also working on the water surface in the piezometer, the head in the piezometer declines. The decline is such that :math:`LE+BE=1`. -------------- #. Why does the specific yield of unconfined aquifers with a shallow groundwater table depend on the depth of the water table? -------------- This is due to the moisture profile above the water table. When the water table is shallow, part of this profile is above ground surface and will no longer contribute to drainage and therefore, reduces the specific yield. -------------- #. What is halftime when considering decay of a water mound between rivers? How would you describe it? -------------- The halftime is the time in which de difference between the head or the elevation of the water table between the river is reduced by 50%. The halftime is determined by :math:`L`, :math:`kD` and :math:`S`. -------------- #. Assume the well has been continuously pumping for time :math:`t=t_{1}`, after which the extraction was stopped. What is the drawdown at distance :math:`r_{0}` at time :math:`t=t_{1}+\Delta t` , where :math:`\Delta t` is any time passed since :math:`t_{1}`. -------------- A formula for the characteristic time of a groundwater basin of width :math:`L` is *:math:`T=L^{2}S/\left(4kDt\right)`* with halftime :math:`T_{\mathrm{50\%}}=\ln\left(2\right)T`. The formula shows that *:math:`L`* and *:math:`S`* increase the halftime and *:math:`kD`* decreases the halftime. -------------- Question 2: ----------- Consider a confined aquifer in direct contact with the ocean in which the head fluctuates along with the tide of the ocean. The tide has an amplitude of :math:`a=2.5\,\mathrm{m}`. The groundwater head in the aquifer at time *t* and distance *x* from the ocean obeys to the following expression: .. math:: s=a\,e^{-ax}\cos\left(\omega t-ax\right)\mbox{, where }a=\sqrt{\frac{\omega S}{2kD}} The frequency :math:`f` of the tide is one complete cycle per 24 hours, i.e. :math:`f=1/d`, with, of course, :math:`\omega=2\pi f`. We don’t know the value of *:math:`kD`* and :math:`S`, but we have measured the amplitude of the groundwater head fluctuation at . This amplitude is 25 cm, one tenth of that of the ocean. #. Explain the parameters in the expression and give their dimension. -------------- :math:`x`: distance [L] :math:`t`: time [T] :math:`s`: drawdown [L] or difference from average :math:`A`: amplitude of head wave [L] :math:`\omega`: angle velocity [/T] or [radians/T] :math:`S`: storage coefficient [-] :math:`kD`: transmissivity [L\ :math:`^{2}`/T] -------------- #. Give an expression for the amplitude at distance x from the ocean. -------------- The amplitude is independent of the cosine. Hence, .. math:: \pm A_{x}=\pm A_{0}\,e^{-ax} -------------- #. With the given information, compute parameter *a*, and the diffusivity of the aquifer, i.e. the ratio :math:`kD/S`. -------------- The amplitude, :math:`s_{0}=2.5\,\mathrm{m}`, is given at distance :math:`x=0` at the sea and as :math:`s_{1}=0.1s_{0}` at :math:`x=x_{1}=500\,\mathrm{m}`, therefore, .. math:: ax_{1}=\ln\left(\frac{s_{0}}{s_{1}}\right)\rightarrow a=\frac{\ln10}{500}\approx\frac{1}{220} The diffusivity then follows from :math:`a^{2}=\frac{\omega S}{2kD}`, therefore, with :math:`T=1\,\mathrm{d}`, .. math:: \frac{kD}{S}=\frac{\omega}{2a^{2}}=\frac{2\pi/1}{2}220^{2}=1.5\times10^{5}\,\mathrm{m^{2}/d} -------------- #. Give an expression for the velocity of the wave of the groundwater-head in the subsurface? How much is this velocity? -------------- The velocity of the wave of the head is found from the argument under the cosine, it must be constant, or zero for convenience .. math:: \omega t-ax=0\rightarrow\frac{x}{t}=v=\frac{\omega}{a}=\frac{2\pi}{1}\times220\approx1380\,\mathrm{m/d}=58\,\mathrm{m/h} -------------- Qestion 3: ---------- Consider an unconfined aquifer with conductivity :math:`k=10\,\mathrm{m/d}`, a specific yield of :math:`S_{y}=0.1` and an initial thickness . A well is located in this aquifer on each of the four corners of a square with sides of :math:`L=200\,\mathrm{m}`. The wells start pumping at :math:`t=0`. They pump with a rate of :math:`Q=120\,\mathrm{m^{3}/d}` for 1d, after which they stop. The drawdown according to Theis is .. math:: s=\frac{Q}{4\pi kD}\mathrm{W}\left(u\right)\mbox{, where }u=\frac{r^{2}S}{4kDt} The Theis well function is graphically given in Figure 1 below. #. Compute the drawdown in the center of the square after :math:`t=2\,\mathrm{d}`. You may neglect the change of the transmissivity caused by the change of the water depth in the aquifer. Question 4 ---------- Given that the well function can be computed by the following infinite series .. math:: \mathrm{W}\left(u\right)=-\gamma-\ln u+u-\frac{u^{2}}{2\times2!}+\frac{u^{3}}{3\times3!}-\frac{u^{4}}{4\times4!}+\cdots with :math:`\gamma=0.577216\cdots` and :math:`u=r^{2}S/\left(4kDt\right)` #. What would be a good approximation of the drawdown for small values of :math:`u`? (Assume for instance that *u*\ <0.01). Notice for mathematical convenience, that :math:`\gamma=\ln\left(e^{\gamma}\right)`. -------------- Truncate the series beyond and including :math:`u`, which is allowed for small values of :math:`u`. We are then left with the following approximation .. math:: \mathrm{W}\left(u\right)\approx-\gamma-\ln u The fill in :math:`u` and :math:`\gamma` to obtain .. math:: \mathrm{W}\left(u\right)\approx\ln\left(\frac{2.25kDt}{r^{2}S}\right) -------------- #. How could you define the radius of influence of the drawdown? Use the formula for the drawdown from the previous question together with the approximation from the previous question. -------------- Set the drawdown in the approximation of the Theis well function equal to 0, i.e. :math:`\mathrm{W}\left(u\right)=0` to obtain .. math:: \frac{2.25kDt}{r^{2}S}=1\rightarrow r=\sqrt{\frac{2.25kDt}{S}} -------------- .. container:: centering .. figure:: pictures/2014_1.png :alt: Theis well function type curve, ie, :math:`\mathrm{W}\left(u\right)` versus :math:`1/u`. :width: 80.0% Theis well function type curve, ie, :math:`\mathrm{W}\left(u\right)` versus :math:`1/u`. Closed-book exam (1h), Feb 3, 2011 ================================== Question 1: Pressure in confined aquifer ---------------------------------------- Question 1: Pressure in confined aquifer A water level in a piezometer in a confined aquifer is affected if the weight on a ground surface is suddenly changed. Compare two situations a) Sudden change by a load placed on the ground, such as sand or flooding by water and b) Sudden increase of the barometric pressure. Case a: — a load is placed on ground surface #. How does the water pressure change in the piezometer? (Up? Down? Not?) -------------- Up -------------- #. How does the head change in the piezometer? (Up? Down? Not?) -------------- Up -------------- Case b: — barometer pressure increased #. How does the water pressure change in the piezometer? (Up? Down? Not? -------------- Up -------------- #. How does the head change in the piezometer? (Up? Down? Not?) -------------- Down -------------- #. If there is a difference between the two cases, then why is that? -------------- Although the water pressure in increased by the same amount in both cases, the water level in the piezometer goes down in the case of the increased barometric pressure, because the air pressure works for 100% on the water level inside the piezometer which is greater than the increase of water pressure in the aquifer due to the air pressure on ground surface. In the case of an increased load there is no increase of pressure on the water level in the piezometer, and so the change of water level in the piezometer fully reflects the change of water pressure in the aquifer, which is some fraction of the increase of pressure on ground surface. -------------- Question 2: Tidal waves ----------------------- The groundwater head variation in a confined aquifer due to a tidal wave at x=0 can be expressed mathematically as follows: .. math:: s\left(x,t\right)=\phi\left(x,t\right)-\phi_{0}+A\,\exp\left(-ax\right)\sin\left(\omega t-ax\right)\mbox{, in which }a=\sqrt{\frac{\omega S}{2kD}} and :math:`\omega=\frac{2\pi}{T}` with :math:`T` the period of the wave. #. What is the amplitude of the wave at distance :math:`x`? -------------- .. math:: A_{x}=A_{0}\,e^{-ax} -------------- #. What is the velocity of the wave? -------------- .. math:: \omega t-ax=\mathrm{const}=0\rightarrow v=\frac{x}{t}=\frac{\omega}{a} -------------- #. If the wave would be just observable in a piezometer at x=1000 m from the coast, then at what distance would the wave be just observable on another spot along the coast where the storage coefficient is be 100 times greater than at the current spot and the transmissivity is the same? -------------- So how to get the same amplitude at :math:`x=0` for both waves? .. math:: A_{1}=A_{0}\,e^{-a_{1}L_{1}}=A_{0}\,e^{-a_{2}L_{2}} so .. math:: a_{1}L_{1}=a_{2}L_{2} .. math:: \frac{\omega S_{1}}{2kD}L_{1}=\frac{\omega S_{2}}{2kD}L_{2} .. math:: S_{1}L_{1}=S_{2}L_{2} Therefore, .. math:: L_{2}=L_{1}\frac{S_{1}}{S_{2}}=1000\times\frac{1}{100}=10\,\mathrm{m} -------------- #. A tidal wave occurring daily is just observable in the aquifer at a distance of x=500 m from shore. At what distance from the shore will a 14-day wave be just observable occurring due to the monthly moon cycle? Assume the same amplitude for both waves. -------------- .. math:: \frac{\omega_{1}S}{2kD}L_{1}=\frac{\omega_{2}S}{2kD}L_{2} .. math:: \frac{\omega_{1}}{\omega_{2}}=\frac{L_{2}}{L_{1}} With :math:`\omega_{1}=14\omega_{2}` and :math:`L_{1}=500\,\mathrm{m}`, we have :math:`L_{2}=14L_{1}=7000\,\mathrm{m}`. -------------- Question 2: Characteristic time of groundwater systems ------------------------------------------------------ In class we discussed the somewhat complicated solution by series expansion of the evolution of the head after a sudden rain shower of :math:`P` [m] in a strip of land of width :math:`L` [m] between parallel fixed-head boundaries with water level :math:`\phi_{0}` [m]. We have seen that after some time :math:`t` [d], only the first term matters, which is .. math:: s\left(x,t\right)=\phi\left(x,t\right)-\phi_{0}=\frac{P}{S_{y}}\frac{4}{\pi}\cos\left(\pi\frac{x}{L}\right)\exp\left(-\pi^{2}\frac{kD}{L^{2}S_{y}}t\right) Whenever possible express your answers mathematically: #.   What is the shape of the head? -------------- A cosine -------------- What is the maximum head, take :math:`\phi_{0}=0`? .. math:: s\left(0,t\right)=\frac{P}{S_{y}}\frac{4}{\pi}\exp\left(-\pi^{2}\frac{kDt}{L^{2}S_{y}}\right) #. What would you consider the characteristic time of this system? -------------- The most suitable definition may be this: .. math:: T=\frac{L^{2}S}{\pi^{2}kD} Because it reduces the formula for the maximum head to: .. math:: s\left(0,t\right)=\frac{P}{S_{y}}\frac{4}{\pi}\exp\left(-\frac{t}{T}\right) One could also choose another definition like .. math:: T=\frac{L^{2}S}{4kD}=\frac{b^{2}S}{kD} Which may be better memorizable. It turn the formula into .. math:: s\left(0,t\right)=\frac{P}{S_{y}}\frac{4}{\pi}\exp\left(-\frac{\pi^{2}}{4}\frac{t}{T}\right) -------------- #. What would be the half-time of this groundwater system? -------------- The half-time is the time it takes for something to become half as large. For exponential declining systems this is a constant. This is the case here with the head relative to its final value. Let the half-time be , then we can formally express the half-time as the time it takes such that the ratio of the head at time is half the head at time : .. math:: \exp\left(-\xi\left(\frac{t+T_{50\%}}{T}\right)\right)=0.5\exp\left(-\xi\left(\frac{t}{T}\right)\right) .. math:: -\xi\left(t+T_{50\%}\right)=-T\ln2-\xi t .. math:: T_{50\%}=\frac{\ln2}{\xi}T where :math:`\xi` follows from the expression under the exponent, and threfore, from the definition of the characteristic time. -------------- Question 4: Wells ----------------- The solution by Theis is given by .. math:: s\left(r,t\right)=\phi_{0}-\phi\left(r,t\right)=\frac{Q_{0}}{4\pi kD}\mathrm{W}\left(u\right)\mbox{, where }u=\frac{r^{2}S}{4kDt} #. What flow conditions are described by Theis’ well solution? -------------- The Theis solution describes the drawdown due to a well in a confined (constant :math:`kD`) aquifer extracting a flow :math:`Q_{0}` from :math:`t=0`. -------------- #. What are its parameters and what are their dimensions? -------------- :math:`s`: [L] drawdown; [m] head relative to fixed datum :math:`Q_{0}`: [L\ :sup:`3`/T] is the constant extraction from the well. :math:`kD`: [L\ :sup:`2`/T] aquifer transmissivity :math:`u`: [-] argument of well function :math:`r`: [m] distance from well :math:`S`: [-] storage coefficient of aquifer :math:`S_{y}`\ or :math:`S_{S}D` :math:`t`: [T] time since the well started its extraction. -------------- As you know, the function W(:math:`u`) is the exponential integral, which may be written as a series expansion : .. math:: \mathrm{W}\left(u\right)=-0.577316-\ln u+u-\frac{u^{2}}{2\times2!}-\frac{u^{3}}{3\times3!}+\frac{u^{4}}{4\times4!}\cdots #. How can you mathematically approximate the Theis’ solution for very small values of :math:`u` given that :math:`-0.577216\approx\ln\left(0.5615\right)`? -------------- For sufficiently small :math:`u` we may truncate all terms above :math:`\ln u` in the series expression, to get .. math:: \mathrm{W}\left(u\right)\approx-0.577361-\ln u\mbox{, where }u=\frac{r^{2}S}{4kDt} Fill in :math:`u` into the formula to obtain .. math:: \mathrm{W}\left(u\right)\approx\frac{2.3Q}{4\pi kD}\log\left(\frac{2.25kDt}{r^{2}S}\right) -------------- #. With this approximation, mathematically give the difference between the drawdown obtained at time t and the drawdown at time 10t in a piezometer at some arbitrary distance r from the well. -------------- .. math:: s_{10t}-s_{t}=\frac{2.3Q}{4\pi kD}\left[\log\left(\frac{2.25kD\left(10t\right)}{r^{2}S}\right)-\log\left(\frac{2.25kDt}{r^{2}S}\right)\right]=\frac{2.3Q}{4\pi kD} -------------- #. Also give the difference between the drawdown in a piezometer at distance r from the well and in a piezometer at distance 10r from the well, both at the same time. .. math:: s_{10t}-s_{t}=\frac{2.3Q}{4\pi kD}\left[\log\left(\frac{2.25kDt}{r^{2}S}\right)-\log\left(\frac{2.25kDt}{10^{2}r^{2}S}\right)\right]=\frac{2.3Q}{4\pi kD}\log\left(10^{2}\right)=\frac{2.3Q}{2\pi kD} Closed-book exam (1h), Feb 5, 2010 ================================== Question 1: ----------- #. What is liquefaction? -------------- Loosely packed fine sand may become liquefied and turn into quicksand by a shock (earthquake) by which grains become untached, the matrix tries to resettle into a more compact configuration, but the water cannot escape immediately due to the small conductivity of the fine sand. -------------- #. What is the difference between specific yield and elastic storage? -------------- Water table storage versus storage due to the elastic properties of both the water and the matrix. -------------- #. How does the specific yield change if an already shallow water table rises? -------------- In a phreatic aquifer with a shallow water table, specific yield gets smaller the shallower the water table. This is due to the moister profile intersecting ground surface. See syllabus, -------------- #. Why does this happen (make a sketch and explain)? -------------- See syllabus page 12. -------------- #. Which of the two materials, gravel and fine sand, has the highest specific yield and why (assume both have the same porosity)? -------------- The coarser material has the highest specific yield because it has the lowest specific retention, because less water can adhere against gravity in the unsaturated zone because of the much smaller specific surface in each volume of coarse material than the same volume of fine material. -------------- Question 2: ----------- The head in an aquifer connected to the ocean fluctuates due to tide. This fluctuation is given by the following formula, in which *s* expresses the head variation caused by the tide as a function of time *t* and the inland distance from the shore *x*: .. math:: s\left(x,t\right)=A\,\exp\left(-ax\right)\sin\left(\omega t-ax\right)\mbox{, with }a=\sqrt{\frac{\omega S}{2kD}} #. What is the expression for the maximum head fluctuation as a function of *x*? This expression is just the part of the tide formula without the sine: -------------- .. math:: s\left(x,t\right)=A\,e^{-ax} -------------- #. Sketch the head change s as a function of x at time t=0 and sketch also the envelope (maximum and minimum value of s as a function of x) .. container:: centering .. figure:: pictures/2010_0.png :alt: Sketch of wave as a function of :math:`x` :width: 80.0% Sketch of wave as a function of :math:`x` #. Which parameters increase the inland penetration of the tide and which parameters decrease this inland penetration? -------------- Parameters increasing the penetration depth of the tide wave are those that reduce the value of the damping alfa, that is a lower frequency omega, a lower storage coefficient and a higher transmissivity. -------------- Question 3: ----------- Consider an extraction canal in direct contact with an aquifer of infinite extent. The aquifer has transmissivity :math:`kD=400\,\mathrm{m^{2}/d}` and specific yield :math:`S_{y}=0.1`. As long as :math:`t<0`, the head in the aquifer is everywhere 0 m (we take the initial water level as our reference level). At time :math:`t=0\,\mathrm{d}`, the water level in the canal suddenly changes to 2 m. Then, at time :math:`t=2~\mathrm{d}`, the water level in the canal suddenly changes back to its original value of 0 m and remains constant afterwards. The head change and the head-change gradient are: .. math:: s=s_{0}\mathrm{erfc}\left(\sqrt{\frac{x^{2}S}{4kDt}}\right)\mbox{ and }\frac{\partial s}{\partial x}=-s_{0}\sqrt{\frac{S}{\pi kDt}}\exp\left(-\frac{x^{2}S}{4kDt}\right) To obtain values for the **:math:`\mathrm{erfc}`** function, use the graph below. Answer the following two questions. #. Compute the head at :math:`x=100\,\mathrm{m}` at :math:`t=3\,\mathrm{d}`. Show the formula you use and include the dimension in your answer! -------------- This problem is solved by superposition in time: .. math:: s=s_{0}\mathrm{erfc\left(\sqrt{\frac{x^{2}S}{4kDt}}\right)-s_{0}\mathrm{erfc}\left(\sqrt{\frac{x_{2}S}{4kD\left(t-3\right)}}\right)} Fill in the numbers to get your answer. -------------- #. Compute the discharge at :math:`x=0` at :math:`t=3\,\mathrm{d}`. Show the formula you use and include the dimension in your answer? -------------- The discharge at x=0 as a function of time equals .. math:: q\left(0,t\right)=-kD\frac{\partial s}{\partial x}=-kDs_{0}\sqrt{\frac{S}{\pi kDt}} so, .. math:: q\left(0,3\right)=s_{0}\sqrt{\frac{kDS}{\pi}}\left(\sqrt{\frac{1}{t}}-\sqrt{\frac{1}{t-3}}\right)=2\sqrt{\frac{400\times0.1}{\pi}}\left(\sqrt{\frac{1}{3}}-\sqrt{\frac{1}{3-1}}\right)=-3.02\,\mathrm{m^{2}/d} -------------- .. container:: centering .. figure:: pictures/2010_1.png :alt: Function :math:`\mathrm{erfc}\left(u\right)` :width: 70.0% Function :math:`\mathrm{erfc}\left(u\right)` Quesiton 4: ----------- Consider a well in a system of infinite extent which starts extracting at time *t*\ =0. We know that Theis’ formula applies: .. math:: s\left(r,t\right)=\frac{Q}{4\pi kD}\mathrm{W}\left(u\right)\mbox{, where }u=\frac{r^{2}S}{4kDt} We also know that for small values of *u*, the well function, W(:math:`u`), can be approximated by a straight line on log-t scale, which is given by: .. math:: \mathrm{W}\left(u\right)\approx2.3\log\left(\frac{0.5625}{u}\right)=2.3\log\left(\frac{2.25kDt}{r^{2}S}\right) Consider a pumping test on this well, starting the constant extraction :math:`Q=800\,\mathrm{m^{3}/d}` at :math:`t=0`. The drawdown is measured over a number of days at an observation well at 25 m distance. The measured drawdowns are shown in the figure below, which clearly reveals the straight portion of the drawdown that we expect from the expression above for large-enough values of time. #. Using the straight line through the measured data, compute the transmissivity *:math:`kD`* and the storage coefficient *:math:`S`* of this aquifer. -------------- The drawdown difference per log cycle is the most convenient way to compute the transmissivity. First draw a line through the straight portion of the graph (see figure below) and then measure the drawdown increase per log cycle, which is about 0.32 m. Then with log(10)=1, compute the transmissivity from .. math:: s_{10t}-s_{t}\approx\frac{2.3Q}{4\pi kD}\rightarrow kD=\frac{2.3Q}{4\pi\left(s_{10t}-s_{t}\right)}=\frac{1.3\times800}{4\pi0.32}\approx460\,\mathrm{m/d} Next we get the storage coefficient from the point where the straight drawdown line of the above expression is zero, which is at about 0.12 days: .. math:: \frac{2.25kDt}{r^{2}S}=1\rightarrow S=\frac{2.25kDt}{r^{2}}=\frac{2.25\times460\times0.12}{25^{2}}\approx0.2 -------------- .. container:: centering .. figure:: pictures/2010_2.png :alt: Measured drawdown during pumping test. :width: 80.0% Measured drawdown during pumping test. Closed-book exam (3h), Feb 2009 =============================== Question 1: ----------- #. What is the difference between specific yield and elastic storage? -------------- Storage coefficient in respectively unconfined and (semi)-confined aquifers. -------------- #. How does the specific yield change if an already shallow water table rises further and becomes even shallower? -------------- The specific yield becomes smaller. -------------- #. Why does this happen (make a sketch and explain)? -------------- Part of the unsaturated profile that would store or yield water now extends above ground surface, and no longer exists so that it cannot contribute to storage. -------------- #. Which of the two materials, gravel and fine sand, has the highest specific yield and why (assume both have the same porosity)? -------------- Fine sand, due to its much larger surface area and number contact points between grains. -------------- Question 2: ----------- #. What do we mean by Loading Efficiency (:math:`LE`) and what do we mean by Barometric Efficiency (:math:`BE`)? -------------- :math:`LE` is the portion of the increased total pressure supported by the water and increasing the head. The :math:`BE` is the portion of the barometric pressure change causing change of head. Both :math:`LE` and :math:`BE` only work in (semi)-confined aquifers. -------------- #. What is the difference in terms of head change if we compare a loading on land surface with an equal increase of the barometer pressure? And why? -------------- In case of a load change, the head changes in the same direction ad the load, with barometric change it changes in the opposite direction. An increase of barometric pressure caused a decrease in head. The difference is due to the face that the barometric pressure also works in the piezometer, while a load increase does not. -------------- Question 3: ----------- Tidal flow in a confined aquifer may be described mathematically by .. math:: s=A\,e^{-ax}\sin\left(\omega t-ax\right)\mbox{, where }a=\sqrt{\frac{\omega S}{2kD}} #. What are the different quantities in these expressions and what are their dimensions? -------------- :math:`s`: head change due to the wave at :math:`x=0` [m] *:math:`A`:* is amplitude of tide [m] :math:`a`: is dampoing factor [1/m] :math:`\omega`: is tidal frequency[radians/day] *:math:`S`:* [-] is storage coefficient [-] *:math:`kD`:* is transmissivity. [m\ :math:`^{2}`/d] -------------- #. By what expression is the envelope given (the envelope describes the maximum amplitude as a function of :math:`x`?) -------------- .. math:: A_{x}=A_{0}\,e^{-ax} -------------- #. How does the envelope change if the frequency of the tide would double? It is reduced. Less penetration of the wave into the aquifer as follows from. .. math:: \frac{A_{x}}{A_{0}}=\exp\left(-\sqrt{\frac{\omega S}{2kD}}x\right) #. How will the envelope change if the transmissivity would be two times less and the storage coefficient 100 times less? The damping factor is .. math:: a=\sqrt{\frac{\omega S}{2kD}} The larger :math:`a` the more damping and the less penetration of the wave into the aquifer. A higher :math:`\omega` (frequency) increase the damping. :math:`\omega=0` implies the lowest possible frequency, which yields no damping, and, therefore the same :math:`s` independent of :math:`x`, which is equivalent to the steady stat situation. The damping is also increased with higher :math:`S` and with lower :math:`kD`. Question 4: ----------- The picture below shows a strip of land of width *L* bounded by two canals. Both the strip and the canals run perpendicular to the paper (so the picture is a cross section). Suddenly the water level in the left canal is raised by *A* m as is indicated in the figure. This causes the head to change in the strip. At the right hand side the water level is unchanged. There exists an expression, which mathematically describes the effect of a sudden level rise in a strip that is unbounded on one side. We want to use this expression to computed the head in the strip. We can do this by means of mirror canals. .. container:: centering .. figure:: pictures/2009_1.png :alt: Strip of land of width *L* bounded by open water. The water level at the left hand side was suddenly raised by *A* m. This causes the head in the aquifer of the strip to change dynamically. :width: 100.0% Strip of land of width *L* bounded by open water. The water level at the left hand side was suddenly raised by *A* m. This causes the head in the aquifer of the strip to change dynamically. #. Irrespective of what the mathematical looks like, where would you put the mirror canals and which are positive and which are negative? Just draw an arrow respectively up or down (see figure) at the locations where you would put the mirror canal. .. container:: centering .. figure:: pictures/2009_2.png :alt: Same as previous figure, but now with superposition scheme shows by the arrows. :width: 100.0% Same as previous figure, but now with superposition scheme shows by the arrows. Question 5: ----------- The characteristic dynamics of a groundwater systems (i.e., the time it takes for the head of a groundwater system to reach equilibrium) is related to the argument of transient groundwater flow solutions, This argument is :math:`\sqrt{\frac{x^{2}S}{4kDt}}` in solutions for one-dimensional flow and :math:`\frac{r^{2}S}{4kDt}` for radial flow such as in the well functions of Theis and Hantush. #. Explain how the characteristic dynamics relate to these arguments? -------------- In a system of given width :math:`L` the factor :math:`T=\frac{L^{2}S}{4kDt}` has dimension time and is directly related to the dynamics of a groundwater basin. It can be considered the characteristic time of the basin/system. -------------- #. Compare the characteristic dynamics of two systems. System two is twice as wide as system one and its transmissivity is 3 times as large and its storage coefficient 100 times as small as that of system one. How do the dynamics of these two systems relate to each other, that is: how many times faster or slower is system two compared to system one in reaching piezometric equilibrium? The factor :math:`T=\frac{L^{2}S}{4kDt}` with dimensions time is a measure for the characteristic time of the groundwater system. System #2 thus has a characteristic time that is :math:`2^{2}\times0.01/3=0.013\approx1/75` times larger or 75 times smaller than system #1. Question 6: ----------- Consider a well in a semi-confined aquifer with :math:`kD=900\,\mathrm{m^{2}/d}`, :math:`S=0.001` and :math:`c=400\,\mathrm{d}` that is pumped at a discharge :math:`Q=2400\,\mathrm{m^{3}/d}`. #. How long does it take before the drawdown at 60 m distance from the well becomes stationary? .. math:: \lambda=\sqrt{kDc}=\sqrt{900\times400}=600\,\mathrm{m}\rightarrow\frac{r}{\lambda}=\frac{60}{600}=0.1 See where the Hantush type curve for :math:`r/\lambda=0.1` becomes horizontal (stationary). Read the :math:`1/u` value, which is about 1000, and compute the time. .. math:: \frac{1}{u}=\frac{4kDt}{r^{2}S}\rightarrow t=\frac{r^{2}S}{4kDu}=\frac{60^{2}\times0.001}{4\times900}\times1000\approx1.0\,\mathrm{d} #. What is the final drawdown? This drawdown, because it is steady state can be computed either by the De Glee formula (with the Bessel Function) or with the Hantush formula .. math:: s=\frac{Q}{2\pi kD}\mathrm{K_{0}}\text{\ensuremath{\left(\frac{r}{\lambda}\right)}=\ensuremath{\frac{Q}{4\pi kD}\mathrm{W}\left(u,\frac{r}{\lambda}\right)}} This is true, because the flow is steady state. Using the type curves given we apply Hantush which yields with :math:`1/u=100`\ 0 and :math:`r/L=0.1`, :math:`\mathrm{W\left(\cdots\right)}=1.9`, so .. math:: s=\frac{2400}{4\pi900}1.9\approx0.4\,\mathrm{m} Quesiton 7: ----------- A pumping test has been carried out in a confined aquifer. The drawdown and the Theis type curves are given in the graphs below. Theses graphs have been drawn on the same type of double logarithmic paper. The extraction of the well during the test was 1000 m\ :sup:`3`/d. Determine the transmissivity and the storage coefficient of this groundwater system. .. container:: centering .. figure:: pictures/2009_3.png :alt: Theis well function Type curve. :width: 80.0% Theis well function Type curve. .. container:: centering .. figure:: pictures/2009_4.png :alt: Measured drawdown versus :math:`t/r^{2}.` :width: 80.0% Measured drawdown versus :math:`t/r^{2}.` .. container:: centering .. figure:: pictures/2009_5.png :alt: Theis and Hantush well function type curves. :width: 80.0% Theis and Hantush well function type curves. -------------- Fold the paper and tear off the lower graph. Shift the two graphs over each other until they match (keep axes parallel). Then choose a “match point” and read the combined value of :math:`s` and :math:`\mathrm{W}` to obtain :math:`kD` from .. math:: s=\frac{Q}{4\pi kD}\mathrm{W\rightarrow kD=\frac{Q}{4\pi s}\mathrm{W}} With numbers read from both graphs once overlaid (:math:`\mathrm{W}=1` and :math:`s=0.1\,\mathrm{m}`) .. math:: kD=\frac{Q}{4\pi S}\mathrm{W}=\frac{1000}{4\pi0.1}\times1\approx795\,\mathrm{m/d} Then read the combined values of :math:`1/u` and :math:`t/r^{2}` from the graphs and determine :math:`S/kD` from .. math:: \frac{1}{u}=\frac{4kD}{S}\frac{t}{r^{2}}\rightarrow\frac{S}{kD}=4u\frac{t}{r^{2}} In numbers with :math:`1/u\approx1.0` and :math:`t/r^{2}\approx3\times10^{-7}` we get .. math:: \frac{S}{kD}=4u\frac{t}{r^{2}}=4\times1\times3\times10^{-7}=1.2\times10^{-6}\,\mathrm{d/m^{2}} Finally compute :math:`S=1.2\times10^{-6}\times kD=1.2\times10^{-6}\times795\approx10^{-3}`. -------------- Closed-book exam (3h), Feb 2007 =============================== Question 1: Conceptual ---------------------- #. What is specific yield? -------------- Specific yield, :math:`S_{y}`, is the reversible storage caused by filling and emptying of pores. Hence, it is the storage coeffiicent for the phreatic or water-table aquifer. -------------- #. How does specific yield depend on the distance of the water table below ground level? -------------- The shallower the water table above some depth, the lower the specific yield. This is due to the moisture curve intersecting ground surface, due to which the contents below the remaining moisture curve is less than when it does not intersect ground surface. -------------- #. What happens to the water table in a piezometer in a confined aquifer when the barometer pressure goes up, why? -------------- The water table goes down by the amount of the increased (changed) barometer pressure (divided by :math:`\rho g` of course, to convert from pressure to water column) that is not supported by the aquifer grain matrix. -------------- Question 2: Diffusion equation ------------------------------ The diffusion equation for transient flow in one dimension is :math:`D\frac{\partial^{2}s}{\partial x^{2}}=\frac{\partial s}{\partial t}` #. What is the dimension of the diffusivity *D*? -------------- It’s always [L\ :math:`^{2}`/T]. (This follows immediately from the partial differential equation.) -------------- #. What is diffusivity *D* in the case of groundwater flow? -------------- Ease of flow divided by storage, hence :math:`kD/S`. -------------- #. What is diffusivity *D* in the case of heat flow? -------------- Ease of flow divided by strorage, hence, :math:`\frac{\lambda}{\rho c}`, i.e., head conductance over specific head capacity .. math:: \frac{\lambda}{\rho c}=\mathrm{[\frac{E}{TL^{2}}/(\frac{K}{L})]/[\frac{M}{L^{3}}\frac{E}{MK}]=[L^{2}/T],\,e.g.,\,m^{2}/s,\,mm^{2}/s\,etc}. -------------- Question 3: Fluctuation groundwater ----------------------------------- In the case of a tidal fluctuation in a river in direct contact with an aquifer having transmissivity the fluctuation of the head may be described by .. math:: s=s_{0}\exp\left(-ax\right)\sin\left(\omega t-ax\right)\mbox{, with }a=\sqrt{\frac{\omega S}{2kD}} #. What is *:math:`s`* and what does this function look like? Make a sketch of *s* as a function of *x*, and show its envelopes. (The envelope is the curve of the values between which the function fluctuates, as a function of :math:`x`). -------------- Lowercase *:math:`s`* [m] is the head change that is only due to the wave (driven by the given fluctuation at :math:`x=0`. The function is a damped sine wave, i.e. it fluctuatuates between its envelopes defined by :math:`\pm A\exp(-ax)`. -------------- #. In the case of a double-day tide, :math:`\omega=\frac{4\pi}{24}\,\mathrm{h^{-1}}`, what would be the speed of the wave into the aquifer if :math:`S=0.001` and :math:`kD=500\,\mathrm{m^{2}/d}`? (Notice the dimensions!) -------------- The speed of the wave follows from keeping the argument of the sine constant. This leads to the expression :math:`\omega t-ax=\mathrm{const}`. To get the velocity, differentiate with respect to *t* to get :math:`\omega-a\frac{dx}{dt}=0`, to :math:`v=\frac{dx}{dt}=\frac{\omega}{a}`. At what distance from the river is the amplitude of the head fluctuation still only half of that in the river at :math:`x=0`? -------------- #. What happens to this distance in case the transmissivity would be 9 times a big? -------------- Note that :math:`a=\sqrt{\frac{\omega S}{2kD}}`. Hence a 9 times larger :math:`\text{\ensuremath{kD}}` causes a 3 times lower value of the damping factor ‘\ :math:`a`\ ‘ and so, a 3 times higher velocity of the wave. -------------- Question 4: Flow to an extraction canal --------------------------------------- Consider an extraction canal in direct full contact with an aquifer with transmissivity :math:`kD=400\,\mathrm{m^{2}/d}` and specific yield :math:`S_{y}=0.1`. The water level in the canal suddenly changes by 2 m downward. The head and gradient are given by: .. math:: s=s_{0}\mathrm{erfc}\left(\sqrt{\frac{x^{2}S}{4kDt}}\right)\mbox{ and }\frac{\partial s}{\partial x}=-s_{0}\sqrt{\frac{S}{\pi kDt}}\exp\left(-\frac{x^{2}S}{4kDt}\right) #. Compute the discharge into the canal after 1d. Show the formula you use and include the dimension in your answer! -------------- Easy, just fill in the numbers. -------------- #. What is the head change *:math:`s`* at 100 m from the canal after 1 and after 2 d? (Use erfc-curve further down). -------------- Just fill in the numbers after reading the right values from the curve of the erfc-function. -------------- #. What is the head change at 100 m from the canal after 2 days if the head in the river would change back by 2 m at :math:`t=1\,\mathrm{d}`? -------------- This requires superposition. Let the effect of the sudden head change at t=0 continue forever and subtract the effect of the sudden head change occurring at :math:`t=2\,\mathrm{d}` forever. -------------- Question 5: Well in semi-confined aquifer ----------------------------------------- Consider a transient well in a semi0-confined aquifer so that Hantush’s solution is valid, hence, .. math:: s=\frac{Q}{4\pi kD}\mathrm{W}\left(u,\frac{r}{\lambda}\right)\mbox{, with }u=\frac{r^{2}S}{4kDt}\mbox{ and }\lambda=\sqrt{kDc} with :math:`kD=600\,\mathrm{m^{3}/d}`, :math:`c=900\,\mathrm{d}`, :math:`S=0.001` and pumping at a rate :math:`Q=2400\,\mathrm{m^{3}/d}`. #. How long does it take before steady state is reached for a point at *r*\ =300 m from the well (why)? Use Hantush type curves (see graphic at the end of this exam). -------------- Compute :math:`r\text{/}\lambda` and look in the curve below when the line for this value becomes horizontal, therefore, steady state. Use the corresponding value of :math:`1\text{/}u` which is :math:`4kDt\text{/}\left(r^{2}S\right)`, fill in the know values of :math:`\text{\ensuremath{kD}}`, :math:`r` and :math:`S` and compute the time :math:`t` -------------- Question 6: Drawdown due to a pumping station in an unconfined aquifer ---------------------------------------------------------------------- A well is situated at 100 m from an impermeable infinitely long wall. The well is pumping at a rate of 2400 m\ :sup:`3`/d. Even though the aquifer is unconfined, the transmissivity *:math:`kD`* may be taken as a constant equal to 600 m\ :sup:`2`/d, while the specific yield *:math:`S_{y}`* equals 0.2. The well bore has a radius of :math:`r_{0}=0.25\,\mathrm{m}`. #. What is the drawdown at the well bore after 10 days of pumping ? -------------- The wall is impermeable, so we must place a mirror well with the same pumping rate at 100 m beyond the wall to compute the drawdown. The drawdown is then obtained by the superposition of the well and its mirror well: .. math:: s(0.25,t)=\frac{Q}{4\pi\text{kD}}\left(W_{h}\left(\frac{0.25^{2}S}{4\text{kDt}},\frac{0.25}{\lambda}\right)+W_{h}\left(\frac{200^{2}S}{4\text{kDt}},\frac{200}{\lambda}\right)\right) Just fill in the numbers for :math:`\text{kD}`, :math:`S`, :math:`\lambda=\sqrt{\text{kDc}}` and :math:`t`\ =10 d, look up the value for :math:`W_{h}\left(u,\frac{r}{\lambda}\right)` in the graph and compute the answer. -------------- #. A well in a confined aquifer of infinite extent, with :math:`kD=1000\,\mathrm{m^{2}/d}` and :math:`S=0.001`, is pumping at a rate of :math:`Q=24000\,\mathrm{m^{3}/d}`. How far would the radius of influence of this well after 100 years? The radius of influence is the radius beyond which the drawdown is considered negligible. You may exploit the logarithmic approximation of the Theis well function for large times: .. math:: \mathrm{W}\left(u\right)\approx\ln\left(\frac{0.5625}{u}\right)\mbox{, with }u<0.1\mbox{ and }u=\frac{r^{2}S}{4kDt} by making it zero. The long-term drawdown in a confined aquifer due to a well pumping from :math:`t=0` at a constant rate is given by the Theis equation, which approaches a logarithmic function for large enough times (see given formula of :math:`\mathrm{W}(u)`). A practical approximation of the area of influence is where this approximation of the Theis solution, hence, the linear drawdown on half-log paper is zero. This is mathematically done by setting the argument of the log equal to 1. Hence, .. math:: \frac{0.5625}{u}=\frac{0.5625\times4kDt}{r^{2}S}=1\mbox{, and, therefore, }r=\sqrt{\frac{2.25kDt}{S}} .. container:: centering .. figure:: pictures/2007_1.png :alt: :math:`\mathrm{erfc}\left(u\right)`\ function. :width: 80.0% :math:`\mathrm{erfc}\left(u\right)`\ function. .. container:: centering .. figure:: pictures/2007_2.png :alt: Theis and Hantush type curves. In case this graph is copied in black and white only, note that the lowest type curve is for the highest value of *:math:`r/\lambda`.* Note that the :math:`L` in the title and left axis of this figure stands for :math:`\lambda=\sqrt{kDc}` value :width: 80.0% Theis and Hantush type curves. In case this graph is copied in black and white only, note that the lowest type curve is for the highest value of *:math:`r/\lambda`.* Note that the :math:`L` in the title and left axis of this figure stands for :math:`\lambda=\sqrt{kDc}` value Closed-book exam (3h) Feb 2006 ============================== Question 1: Conceptual ---------------------- #. What types of reversible storage do you know in aquifer systems, explain how it works -------------- Pore-water storage (by filling and emptying of pores) know as specific yield, and elastic storage by compression and expansion of the aquifer matrix and the water. Note that when the water pressure increases, the pore matrix expands while the water is compressed. -------------- #. What values may you expect for the respective storage coefficients? -------------- Specific yield values will range between a few percent to around 25% depending on the pore space and for very fine pores also on the time scale at which the head or water table varies. Elastic storage will be in the order of 10e-3 depending on the natural stress of the aquifer and its thickness, and the specific storage coefficient in the order of 10e-5/m mainly depending on the natural stress on the grains. -------------- #. What is barometric efficiency, explain how it works. -------------- The barometric efficiency, :math:`BE`, is the fraction of the barometer pressure change that will cause drawdown in a (semi) confined aquifer. The barometer pressure change is partly taken up by the aquifer matrix (the loading efficiency :math:`\text{\ensuremath{LE}}`) while the rest of supported by a change in head (the barometric efficiency :math:`\text{\ensuremath{BE}}`). The essence is :math:`BE+LE=1`. -------------- #. When the barometric pressure increases, does the head (water table in a piezometer) in the confined aquifer rise or fall? -------------- It falls, because the water pressure in the aquifer increases by the part not supported by the grains, i.e., by :math:`\text{LE}\times p_{a}` while the barometer pressure presses with full :math:`p_{a}` on the water surface in the piezometer. Hence its level must decline by :math:`p_{a}(1-LE)\text{/}(\rho g)` to keep pressure equilibrium. -------------- #. Between what values may the barometric efficiency vary? -------------- Between 960 and 1040 hPa, or about 9.6 and 10.4 m water column. -------------- #. What happens in a confined aquifer with the head if a load is suddenly placed on ground surface, such as a train stopping near a piezometer? What happens when it leaves? Sketch a graph showing the head versus time that you would expect in that case. -------------- The head increases initially with about :math:`\text{\ensuremath{LE}\ }\Delta p` where :math:`\Delta p` is the pressure increase in the aquifer due to the train arriving. The pressure then drops off as the local head increase below the train causes the groundwater to flow away so that after a while the head is the same as before. The opposite happens when the train leaves. -------------- Question 2: Characteristic time of groundwater basin ---------------------------------------------------- Characteristic time of groundwater basin, the partial differential equation of which reads .. math:: kD\frac{\partial^{2}\phi}{\partial x^{2}}=S\frac{\partial\phi}{\partial t} #. What is a characteristic time of a groundwater basin that may be considered as one-dimensional of characteristic size *L*? (hint: Make partial differential equation dimensionless by setting :math:`\xi=\frac{x}{L}`, :math:`\tau=\frac{t}{T}` and see what :math:`T` is. -------------- The partial differential equation will then be converted into .. math:: \frac{kD}{L^{2}}\frac{\partial\phi^{2}}{\partial\xi^{2}}=\frac{S}{T}\frac{\partial\phi}{\partial\tau} or .. math:: \frac{\partial\phi^{2}}{\partial\xi^{2}}=\frac{L^{2}S}{\text{kDT}}\frac{\partial\phi}{\partial\tau} So that setting :math:`\frac{L^{2}S}{kDT}=1` gives that :math:`T=\frac{L^{2}S}{kD}` can be considered a characteristic time of the groundwater system of size :math:`L` (with :math:`L` the approximate distance between opposite head boundaries (distance between local reivers for instance). -------------- #. To reach equilibrium, how many times slower is a large basin compared to a small one with the same transmissivity and storage coefficient? -------------- The size of the basin as measured by :math:`L` is squared in the characteristic time. So, a basin where the :math:`L` is 5 times as large as another basin is 25 times a slow in its drainage (natural drainage to equilibrium) -------------- #. Compute the characteristic time for the following cases: -------------- Large basin: *kD*\ =500 m\ :sup:`2`/d, system width *L*\ =100km, storage coefficient *S*\ =0.2, .. math:: T=\frac{\left(10^{5}\right)^{2}\times0.2}{100}=2\times10^{7}\,d=\text{many},\text{many}\,\text{years} -------------- -------------- Small basin: *kD*\ =100 m\ :sup:`2`/d, system width *L*\ =100m , storage coefficient *S*\ =0.1 .. math:: T=\frac{100^{2}\times0.1}{100}=10\,d -------------- Quesiton 3: Tides in groundwater -------------------------------- Given: The tidal fluctuation in an aquifer in a point at distance *x* from the sea due to the water level fluctuation at sea with amplitude *A* is described by the following formula .. math:: s\left(x,t\right)=A\,\exp\left(-ax\right)\sin\left(\omega t-ax\right) in which the damping factor is as follows :math:`a=\sqrt{\frac{\omega S}{2kD}}`, where :math:`\omega` is the angle velocity in radians/time or :math:`\omega=\frac{2\pi}{T}` where :math:`T` is the time of a complete wave cycle. Are the following expressions true or false? #. The wave in the aquifer has a different frequency than the tide itself. -------------- False, it’s physically unimaginable that the frequencies would be different. -------------- #. The amplitude of the wave at a given distance from the sea becomes greater when -------------- The frequency of the tide is reduced? The higher the frequency, so the higher :math:`\omega`, to higher the damping :math:`a` and, therefore, the smaller the distance over which the fluctuation in the aquifer can be felt. Only an infinitely slow frequency reaches to infinity, as this is equivalent to the steady-state situation. -------------- The storage coefficient is reduced? The smaller the storage coefficient the faster and the farther the fluctuation travels inland. With a (theoretically) zero storage coefficient, the wave would (only theoretically) travel with infinite speed, which would imply that only a steady state situation would be possible (zero storage) or that the head at all :math:`x` would be the same as at :math:`x=0`. -------------- Then the transmissivity is reduced? The transmissivity determines ease of flow. Hence if reduces flow will be less easy. With zero transmissivity the fluctuation at zero would not penetrate the aquifer at all, speed zero; on the other hand, with transmissivity infinite, the speed of the wave would also be (theoretically) infinite and yield the same result as storage coefficient equal to zero. -------------- Question 4: Aquifer with river ------------------------------ Consider an aquifer of infinite extent bounded by a fully penetrating river at :math:`x=0`. At :math:`t=0` the river level suddenly changes by a height :math:`A`. The change of head :math:`s(x,t)` in the aquifer equals in this case: .. math:: s\left(x,t\right)=A\,\mathrm{erfc}\left(\sqrt{\frac{x^{2}S}{4kDt}}\right) with :math:`\mathrm{erfc}(u)` as shown in the picture below .. container:: centering .. figure:: pictures/2006_1.png :alt: :math:`\mathrm{erfc\left(u\right)}`\ versus :math:`u` :width: 80.0% :math:`\mathrm{erfc\left(u\right)}`\ versus :math:`u` #. What is the final value of the head change (the value reached after infinite time, :math:`s\left(x,\infty\right)`? -------------- That’s simple: it must be A, the new steady level of the river. -------------- #. What value has the argument of :math:`\mathrm{erfc}(\cdots)`, i.e., :math:`\sqrt{\frac{x^{2}S}{4kDt}}` when the head change is half the final value? -------------- From the graph it’s seen that it’s a fraction less than 0.5. -------------- #. If :math:`kD=400\,\mathrm{m^{2}/d}`, :math:`S=0.1` and :math:`x=100\,\mathrm{m}`, after how much time is this the change of head equal to :math:`0.5A`? -------------- This implies that the argument must be about 0.5, i.e. .. math:: \frac{x^{2}S}{4\text{kDt}}=0.25 Just fill in :math:`\text{\ensuremath{kD}}` and :math:`S` and compute :math:`t`. -------------- #. What would be the formula if the head change occurred on time *:math:`t_{1}`* instead of time\ :math:`t=0`? -------------- .. math:: s(x,t)=A\ \text{erfc}\left(\frac{x^{2}S}{4\text{kD}\left(t-t_{1}\right)}\right)\mbox{, where }t>t_{1} Again, set the argument equal to 0.5 to compute t at which the drawdown at :math:`x` is equal to :math:`0.5\ A`. -------------- #. How could you compute the head change at point *x* if the there was a sudden change of the river level of *A*\ :sub:`1` at time *t*\ =\ *t*\ :sub:`1` and another of *A*\ :sub:`2` at *t*\ =\ *t*\ :sub:`2`? -------------- By superposition, in this case .. math:: s(r,t)=A_{1}\text{erfc}\left(\frac{x^{2}S}{4\text{kD}\left(t-t_{1}\right)}\right)+A_{2}\text{erfc}\left(\frac{x^{2}S}{4\text{kD}\left(t-t_{2}\right)}\right) -------------- Question 5: Well in a confined aquifer -------------------------------------- Consider a well in a confined aquifer starting an extraction of :math:`Q=1200\,\mathrm{m^{3}/d}` at :math:`t=0`. :math:`kD=1000\,\mathrm{m^{2}/d}`, and :math:`S=0.001`. For this case the Theis solution applies: .. math:: s=\frac{Q}{4\pi kD}\mathrm{W\left(u\right)\mbox{, with }}u=\frac{r^{2}S}{4kDt} (See the type curve of Theis well function on a separate page). #. Compute the head at :math:`r=20\,\mathrm{m}` after :math:`t=1\,\mathrm{d}`. -------------- .. math:: s(20,1)=\frac{1200}{4\pi1000}\mathrm{W}\left(\frac{20^{2}\times0.001}{4\times1000\times1}\right) Where the argument of :math:`\mathrm{W}` is :math:`u` for which :math:`\mathrm{W}` can be looked up in the type curve below. -------------- #. The pump is switched off after 1 day. What is the head after 1.1 days at :math:`r=20\,\mathrm{m}`? -------------- In that case we apply superposition. That is, we place another well at the same spot and switch it on with :math:`-Q` at :math:`t=1.1` days and add the two wells. In math: .. math:: s(20,1.1)=\frac{Q}{4\pi kD}\left(\mathrm{W}\left(\frac{20^{2}S}{4kD\times1.1}\right)-\mathrm{W}\left(\frac{20^{2}S}{4kD\times(1.1-1.0)}\right)\right) Where the first term requires :math:`t>0` and the second :math:`t>1.0` d. -------------- Question 6: Well in a leaky aquifer ----------------------------------- Consider a transient well in a leaky aquifer. :math:`kD=400\,\mathrm{m^{2}/d}`, :math:`c=400\,\mathrm{d}`, :math:`S=0.001`, so that the groundwater behaves according to Hantush’s transient well formula .. math:: s=\frac{Q}{4\pi kD}\mathrm{W}\left(u,\frac{r}{\lambda}\right)\mbox{, with }\lambda=\sqrt{kDc} #. How long does it take until the head at :math:`r=40\,\mathrm{m}` becomes steady state or virtually steady state? (Hint: look at the type curves to get u for which this is the case, note). -------------- Compute :math:`\frac{r}{\lambda}=\frac{40}{\sqrt{kDc}}` and look for the corresponding type curve in the graph with the type curves for the Hantush well function. Look for which value of :math:`1/u` the curve becomes horizontal, take the value of :math:`1/u` for the point, compute u and apply this value in the formula for :math:`u`, from which the time follows. -------------- Question 7: Well in an unconfined aquifer ----------------------------------------- Consider a well in an unconfined aquifer for which the Theis-solution applies (see type curve hereafter. Further given a pumping test with an extraction of :math:`Q=600\,\mathrm{m^{3}/d}` during which drawdown measurements were made (see the graph with the small circles). Interpret the test (that is: compute *kD* and *S*). (Hint: if you can’t see through the paper make the type curve thicker using a pen and hold both curves up against a light or in the direction of a window). -------------- Shift the curves on top of each other keeping the axes parallel until you obtain the best possible fit. Then for that chosen (by arbitrary point) read the corresponding values of :math:`1/u` and :math:`1\text{/}r^{2}` and of the drawdown :math:`s` and :math:`\mathrm{W}(u)`. The from the formula for the drawdown :math:`s=\frac{Q}{2\pi kD}\mathrm{W}(u)` with now a corresponding value of :math:`s` and :math:`\mathrm{W}(u)` known as well as :math:`Q` known, compute the transmissivity. Next, with a corresponding value for :math:`t\text{/}r^{2}` and :math:`1/u` known and :math:`kD` just compute we have :math:`\frac{4kD}{S}\frac{t}{r^{2}}=\frac{1}{u}` from which the storage coefficient, :math:`S`, follows. In the case of a semi-confined aquifer, the match by shifting the graphs should also provide the curve with a specific value of :math:`r\text{/}\lambda` that best fits the measurements. This implies that with r known and :math:`kD` just determined we now also get the resistance c from :math:`\lambda=\sqrt{kDc}`. -------------- .. container:: centering .. figure:: pictures/2006_2.png :alt: Theis type curve, :math:`\mathrm{W}\left(u\right)` versus :math:`1/u` :width: 80.0% Theis type curve, :math:`\mathrm{W}\left(u\right)` versus :math:`1/u` .. container:: centering .. figure:: pictures/2006_3.png :alt: Measured drawdown during pumping test versus *:math:`t/r^{2}`*) :width: 80.0% Measured drawdown during pumping test versus *:math:`t/r^{2}`*) .. container:: centering .. figure:: pictures/2006_4.png :alt: Theis and Hantush type curves combined, i.e., :math:`\mathrm{W}\left(u\right)` and :math:`\mathrm{W}\left(u,\frac{r}{\lambda}\right)` vs :math:`1/u`. :width: 80.0% Theis and Hantush type curves combined, i.e., :math:`\mathrm{W}\left(u\right)` and :math:`\mathrm{W}\left(u,\frac{r}{\lambda}\right)` vs :math:`1/u`.