User:Tmhoang81/Ground water mound problem

=Derivation of ground water mound problem=

Consider a stratum consisting of a gas-filled porous medium containing a ground water mound on top of an underlying horizontal impermeable bed. Under the influence of gravity, the mound spreads out and flows along the impermeable bed.

Basic assumptions: the mound is axially symmetric and the height of the mound $$\displaystyle h\left( {r,t} \right)$$ decreases with increasing radius from the very beginning of the motion. The fluid motion in porous medium is slow, so the water pressure within the mound obeys the hydrostatic law $$\displaystyle p = \rho g\left( {h - z} \right)$$ where $$\displaystyle \rho$$ is the ground-water density and $$\displaystyle g$$ is the gravitational acceleration (neglecting the gas pressure). Thus the total head $$\displaystyle p+\rho gz$$ remains constant throughout the height of the mound within the mound and is equal $$\displaystyle \rho gh$$.

The fundamental law in theory of filtration is the Darcy law, “the flux of fluid or the flow rate per unit area per unit time is proportional to the gradient of the total head”:
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$$   \displaystyle q = - \frac{k}{\mu }gradp = { - \frac{k}{\mu }{\partial _r}p} $$  (1) Where $$\displaystyle k$$ is the permeability coefficient and $$\displaystyle \mu$$ is the dynamic viscosity of the ground water.
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Thus the total flux of water through a cylindrical surface of area $$\displaystyle 2\pi rh$$ is:
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$$   \displaystyle \bar q\left( {r,t} \right) = q\left( {r,t} \right)2\pi rh = \left( { - \frac{k}{\mu }{\partial _r}p} \right)2\pi rh = \left( { - \frac{\mu }{\partial _r}h} \right)2\pi rh = - \frac{\mu }r{\partial _r}{h^2} $$  (2) Introducing the fractional volume of the stratum occupied by pores m, the porosity of the medium. When water enters an empty pore, it does not occupy its entire volume, but only some fraction $$\displaystyle \sigma$$; the remaining fraction is occupied by gas. At the same time, the ground water never completely flows out of an initially full pore; some fraction of the water $$\displaystyle {\sigma _0}$$ is held back by capillary forces.
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Thus, if the total volume of soil and water in a cylinder is $$\displaystyle d{V_T} = 2\pi rdrdh$$, then the volume of pores is $$\displaystyle md{V_T}$$. Hence, the water flowing out of wet soil is $$\displaystyle (\sigma -{\sigma _0} )md{V_T}$$ for the case $$\displaystyle {\partial _t}h < 0$$  and the water flowing into dry soil is $$\displaystyle \sigma md{V_T}$$  for the case $$\displaystyle {\partial _t}h > 0$$

Define
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$$   \displaystyle \sum = \begin{cases} \sigma & \text{ if } \,{\partial _t}h > 0 \\ \sigma - {\sigma _0} & \text{ if } \,{\partial _t}h < 0 \end{cases}$$ (3) So the rate of water flowing out of wet soil or flowing into dry soil is:
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$$   \displaystyle dV = m\sum d{V_T} = 2\pi m\sum \left( {{\partial _t}h} \right)rdrdt$$ (4) According the conservation law of mass, this quantity must equal the change of water in the cylinder of radius $$\displaystyle r$$ and wall thickness $$\displaystyle r+dr$$ during the interval time $$\displaystyle dt$$
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$$   \displaystyle dV = \left[ {\bar q\left( {r,t} \right) - \bar q\left( {r + dr,t} \right)} \right]dt = - \fracdrdt$$ (5) Or
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$$   \displaystyle 2\pi m\sum \left( {{\partial _t}h} \right)rdrdt = \frac{\mu }\frac{\partial }\left( {r{\partial _r}{h^2}} \right)drdt$$ (6) which is equivalent to
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$$   \displaystyle {\partial _t}h = \frac\frac{1}{r}\frac{\partial }\left( {r{\partial _r}{h^2}} \right) $$  (8) As a result, the following equation for the mound height is obtained:
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$$ \displaystyle {\partial}_{t}h= \begin{cases} \left( \kappa / r \right){\partial _r}\left( {r{\partial _r}{h^2}} \right) & \text{ if } {\partial}_{t}h >0 \\ \left( \kappa _1 / r \right){\partial _r}\left( {r{\partial _r}{h^2}} \right) & \text{ if } {\partial}_{t}h <0 \end{cases}$$ (9) Where $$\displaystyle {\kappa} = \frac$$ and $${\kappa _1} = \frac$$
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Now consider the simple case. Assume $$\displaystyle \sigma = 1$$, $$\displaystyle {\sigma_0} = 0$$ and $${\kappa} = {\kappa _1} = \frac$$

The equation for the mound height becomes:
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$$   \displaystyle \kappa \frac{1}{r}{{\partial }_{r}}\left( r{{\partial }_{r}}{{h}^{2}} \right)={{\partial }_{t}}h $$  (10) The total ground water mound:
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$$   \displaystyle u=\rho gh $$ (11) The total ground water mound head satisfies the Bounssinesq equation, which is derived by substituting (11) to (10):
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$$   \displaystyle \kappa \frac{1}{r}{\partial _r}\left( {r{\partial _r}{h^2}} \right) = {\partial _t}h $$  (12) Or
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$$   \displaystyle a{\partial _r}\left( {r{\partial _r}{u^2}} \right) = {\partial _t}u $$  (13) Where
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$$   \displaystyle a = \frac{k} = \frac{\kappa } $$  (14) Under the initial condition that the groundwater is concentrated initially in a well of infinitesimally small radius:
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$$   \displaystyle \left\{ \begin{array}{l} u\left( {r,0} \right) = 0\,\,\left( {r \ne 0} \right)\,\,\,\,\,\,\,\,\,u\left( {\infty ,t} \right) \equiv 0 \\ 2\pi m\int\limits_{r = 0}^{r = \infty } {u\left( {r,0} \right)rdr = \,} \rho g{V_0} \\ \end{array} \right. $$  (15)
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Where $$\displaystyle $$ is a finite volume of water supplied instantaneously by a well at an initial instant $$\displaystyle t = 0$$

Checking dimension:

We have:
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$$   \displaystyle \left[ p \right] = F{L^{ - 2}}$$ (16)
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$$   \displaystyle \left[ {gradp} \right] = F{L^{ - 2}}{L^{ - 1}} = F{L^{ - 3}} $$  (17)
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$$   \displaystyle q = flow\,rate\, = vol.\,\,of\,fluid/area/time\, = {L^3}{L^{ - 2}}{T^{ - 1}} = L{T^{ - 1}} = speed = \frac\left[ {gradp} \right] $$  (18)
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$$   \displaystyle \left[ \mu \right] = \frac = \frac = \left[ p \right]T = \left( {ML{T^{ - 2}}} \right)\left( {{L^{ - 2}}T} \right) = M{L^{ - 1}}{T^{ - 1}}$$ (19)
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From (1):
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$$   \displaystyle \left[ k \right] = \frac = \frac = \frac = \frac = {L^2}$$ (20) Hence:
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$$   \displaystyle \left[ \kappa \right] = \frac\left[ {\rho g} \right] = \fracM{L^{ - 3}}\frac{L} = \frac{L}{T}$$ (21)
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$$   \displaystyle \left[ u \right] = \left( {M{L^{ - 3}}} \right)\left( {L{T^{ - 2}}} \right)\left( L \right) = M{L^{ - 1}}{T^{ - 2}} = \underbrace {\left( {ML{T^{ - 2}}} \right)}_F\left( \right) = F{L^{ - 2}} = pressure\,head $$  (22)
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$$   \displaystyle \left[ m \right] = 1$$ (23)
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$$   \displaystyle \left[ a \right] = \frac = \frac$$ (24)
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$$   \displaystyle \left[ {\frac{a}{r}{\partial _r}\left( {r{\partial _r}{u^2}} \right)} \right] = \left[ a \right]\left[ {\frac{1}{r}} \right]\left[ \right]\left[ r \right]\left[  \right]\left[  \right] = \left[ {\frac\,} \right]\left[ {\frac{1}{L}} \right]\left[ {\frac{1}{L}} \right]\left[ L \right]\left[ {\frac{1}{L}} \right]{\left[ {\left[ p \right]} \right]^2} = \frac{T} = \left[ {{\partial _t}u} \right]$$ (25)
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=Building nondimensional quantities from the solution of Boussinesq equation=

Equation (12) under conditions (15) has a solution:


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$$   \displaystyle h\left( r,t \right)=\frac\left[ 8-\frac \right]$$ (26)
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where


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$$   \displaystyle Q = \frac $$  (27)
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Checking dimension:


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$$   \displaystyle \left[ Q \right]=\left[ {{V}_{0}} \right]={{L}^{3}} $$  (28)
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$$   \displaystyle \left[ {\kappa t} \right] = L $$ (29)
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Denote


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$$   \displaystyle {{h}_{0}}\left( t \right)=\frac$$ (30)
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$$   \displaystyle {{r}_{0}}\left( t \right)={{\left( Q\kappa t \right)}^{{1}/{4}\;}} $$  (31)
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the solution is:


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$$   \displaystyle h\left( t \right) = {h_0}\left( t \right)\left[ {8 - {{\left( {\frac{{r\left( t \right)}}{{{r_0}\left( t \right)}}} \right)}^2}} \right] $$  (32)
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It is obvious that:


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$$   \displaystyle \left[ {{h}_{0}} \right]={{\left[ \frac{Q}{\kappa t} \right]}^{{1}/{2}\;}}={{\left[ \frac{{{L}^{3}}}{L} \right]}^{{1}/{2}\;}}=L $$  (33)
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$$   \displaystyle \left[ {{r}_{0}} \right]={{\left[ Q\kappa t \right]}^{{1}/{4}\;}}={{\left[ Q \right]}^{{1}/{4}\;}}{{\left[ \kappa t \right]}^{{1}/{4}\;}}={{\left[ {{L}^{3}} \right]}^{{1}/{4}\;}}{{\left[ L \right]}^{{1}/{4}\;}}=L $$  (34)
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Thus, 2 nondimensional combinations are:


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$$   \displaystyle \pi =\frac{h\left( r,t \right)}{{{h}_{0}}\left( t \right)}=\frac{u{{\left( at \right)}^{{1}/{2}\;}}}=\frac{u{{\left( \kappa t \right)}^{{1}/{2}\;}}}{\rho g{{Q}^{{1}/{2}\;}}}=\frac{h{{\left( \kappa t \right)}^{{1}/{2}\;}}}$$ (35)
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$$   \displaystyle {{\pi }_{1}}=\frac{r}=\frac{r}$$ (36)
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( $$\displaystyle \left[ \pi \right]=\frac{\left[ h \right]}{\left[ {{h}_{0}} \right]}=1$$ and $$\displaystyle \left[ {{\pi }_{1}} \right]=\frac{\left[ r \right]}{\left[ {{r}_{0}} \right]}=1$$ )

It is not clear how to select the different options to nondimensionalize $$\displaystyle h$$ and $$\displaystyle r$$. These definitions were based on the formula of the exact solution. Alternatively, we can choose another combination, i.e.,


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$$   \displaystyle \bar{\pi }=\frac{h} $$  (37)
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$$   \displaystyle {{{\bar{\pi }}}_{1}}=\frac{r} $$  (38)
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which are dimensionless because

$$\displaystyle \left[ {\bar{\pi }} \right]=\left[ \frac{h} \right]=\frac{L}=1$$

$$\displaystyle \left[{{{\bar{\pi }}}_{1}} \right]=\left[ \frac{r} \right]=\frac{L}=1$$

So the question is “without knowing the exact solution, how to find the appropriate scaling?

The answer is to treat the dimensions of $$\displaystyle h$$ and $$\displaystyle r$$ separately: $$\displaystyle \left[ h \right]=H$$ and $$\displaystyle \left[ r \right]=L$$

Barenblatt distinguished the vertical dimension $$\displaystyle Z$$, $$ \displaystyle \left [Z \right ] = H$$ and the horizontal dimensions $$\displaystyle X,Y$$,  $$\displaystyle \left[ X \right] = \left[ Y \right] = L$$, as independent dimensions since the ratio $$\displaystyle {h}/{r}$$ (mound height over radius) does not appear explicitly among the governing parameters. This also infers that unit of total head $$\displaystyle u$$ is independent of $$\displaystyle \left[ r \right] = L $$,


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$$   \displaystyle \left[ u \right] = \left[ \rho gh \right]=\left[ p \right] $$  (39)
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Obviously, the solution $$\displaystyle h$$ depends on the following governing parameters: the independent variables $$\displaystyle r$$ and $$\displaystyle t$$, the parameters $$\displaystyle \kappa $$, $$\displaystyle Q$$


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$$   \displaystyle h=h\left( Q,\kappa ,r,t,V \right) $$  (40)
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We have:


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$$   \displaystyle \left[ Q \right]=\left[ {{V}_{0}} \right]=H{{L}^{2}} $$  (41)
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Look at equation (12):


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$$   \displaystyle \left[ \kappa \right]\left[ \frac{1}{r} \right]\left[ {{\partial }_{r}} \right]\left[ \left( r{{\partial }_{r}}{{h}^{2}} \right) \right]=\left[ {{\partial }_{t}}h \right] $$  (42)
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Or


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$$   \displaystyle \left[ \kappa \right]\left[ \frac{1}{L} \right]\left[ \frac{1}{L} \right]\left[ L\frac{1}{L}{{H}^{2}} \right]=\left[ \frac{1}{T}H \right] $$  (43)
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Hence,
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$$   \displaystyle \left[ \kappa \right]={{L}^{-2}}{{H}^{-1}}{{T}^{-1}} $$  (44)
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$$   \displaystyle \left[ {{\left( Q\kappa t \right)}^{{1}/{4}\;}} \right]={{\left( H{{L}^{2}}{{L}^{2}}{{H}^{-1}}{{T}^{-1}}T \right)}^{{1}/{4}\;}}=L $$  (45)
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$$   \displaystyle \left[ {{Q}^{{1}/{2}\;}}{{\left( \kappa t \right)}^{{-1}/{2}\;}} \right]={{\left( H{{L}^{2}} \right)}^{{1}/{2}\;}}{{\left( {{L}^{2}}{{H}^{-1}}{{T}^{-1}}T \right)}^{{-1}/{2}\;}}=H $$  (46)
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Thus it is now clear that we need to use

$$\displaystyle \pi =\frac{h}\,\,,\,\,\,\,{{\pi }_{1}}=\frac{r}$$

rather than

$$\displaystyle \bar{\pi }=\frac{h}\,\,\,,\,\,\,{{{\bar{\pi }}}_{1}}=\frac{r}$$

because $$\displaystyle \left[ \pi \right]=\left[ {{\pi }_{1}} \right]=1$$ while $$\displaystyle \left[ {\bar{\pi }} \right]=\frac{H}{L} \text{ and } \left[ {{{\bar{\pi }}}_{1}} \right]=\frac{L}{H}$$