The total mass flux (flow out of the box) is

  1.72, Groundwat er Hydrology Prof. Charles Harvey

  

Le ct u r e Pa ck e t # 4 : Con t inu it y a n d Flow N e t s

Equa t ion of Con t inu it y

• Our equations of hydrogeology are a com bination of _{o o}

  Conservat ion of m ass Som e em pir ical law ( Darcy’s Law , Fick’s Law )

• Develop a control volum e, rectangular parallelepiped, REV ( Representative Elem ent ary Volum e) z dz x, y, z dy dx x y m ass inflow rat e – m ass out flow rat e = change in m ass st orage q x = specific discharge in x- direct ion ( volum e flux per area)
_{3 2} at a point x,y,z L / L - T Consider m ass flow t hrough plane y- z at ( x,y,z) ₃

  ρ dy dz L/ T M/ L q x L L = M/ T Rat e of change of m ass flux in t he x- direct ion per unit t im e per cross- sect ion is

  ∂ [ ρ q ] dydz _x

  x, y, z

  ∂x

  dz dy dx x - dx/ 2 x x + dx/ 2 m ass flow int o t he ent ry plane y- z is

  ∂ dx [ q ] − [

ρ dydz ρ q ] dydz

_{x x}

  2

∂ x

  And m ass flow out of t he exit plane y- z is

  ∂ dx

[ ρ dydz q ] + [ ρ q ] dydz

_{x x}

∂ x

  2 I n t he x- direct ion, t he flow in m inus t he flow out is ∂

− [ ρ q ] dxdydz

x

  ∂ x

  Sim ilar ly, t he flow in t he y- direct ion t hrough t he plane dxdz ( figure on left ) x, y, z x, y, z dz dz dy dy dx dx

  ⎡ ∂ dy ⎤ ⎡ ∂ dy ⎤ ∂ q ] dxdz [ q ] dxdz q ] dxdz [ q ] dxdz [ q ] dxdydz

  ρ − ρ + ρ ρ = ρ y y y y y

  ⎢ [ ⎥ − ⎢ [ ⎥ −

  2

  2

∂ y ∂ y ∂ y

⎣ ⎦ ⎣ ⎦ for t he net y- m ass flux.

  Sim ilarly, we get for t he net m ass flux in t he z- direct ion:

  ∂

− [ q ] dxdydz

ρ

z

  ∂ z

  The t ot a l m a ss flu x ( flow out of t he box) is

  ρ q ]

q ] ∂ [ q ]⎤

  ⎡ ∂ [ ρ y ∂ [ ρ x z

  − −

⎢− ⎥ dxdydz ∂ x ∂ y ⎣ ∂ z ⎦ Let ’s consider t im e derivat ive = 0 ( St eady St at e Syst em )

  ρ q ] q ] ∂ [ q ] ⎤

  ⎡ ∂ [ ρ ∂ [ ρ ∂ M y x z dxdydz

• = = − ⎢

  

x y z t

∂ ∂ ⎥ ∂ ⎣ ∂ ⎦

  How does Darcy’s Law fit int o t his?

  h ∂

  ∂ h ∂ h q = − K

  = − q = − K q K z zz y yy x xx

  

∂

∂ x y ∂ z
• For anisotropy (with alignm ent of coordinate axes and tensor principal direct ions) , or q r − = K ∇ h
• Assum ing constant density for groundwater

  ] ∂ [ q ⎡ ∂ [ q ] ∂ [ ] q ⎤ y x z

  = − ρ + +

  subst it ut ing for q,

  ⎢ ⎥ x y z

  ∂ ∂ ∂ ⎣ ⎦ ⎡

  ⎤ ∂ ∂ h ∂ ⎛ ∂ h ⎞ ∂ ∂ h ⎛ ⎞ ⎛ ⎞ = (− )(− ) ρ K ⎜ K K ⎜ ⎟ + ⎜ ⎟ x y z

  ⎢ ⎥ ⎟⎟ + x x y y z z

  ∂ ∂ ∂ ∂ ∂ ⎝ ⎠ ⎝ ⎠ ∂ ⎝⎜ ⎠ ⎣

  ⎦

  St eady- st at e flow equat ion for het erogeneous, anisot ropic condit ions:

  ⎡ ⎤

h h ⎞ h

  ∂ ∂ ∂ ⎛ ∂ ∂ ⎛ ⎞ ⎛ ∂ ⎞ = K ⎜ K K ⎜ ⎟ + ⎜ ⎟ x y z

  ⎢ ⎥ ⎟⎟ + x x y y z z

  ∂ ∂ ∂ ∂ ⎝ ⎠ ⎝ ⎠ ∂ ⎝⎜ ∂ ⎠ ⎣

  ⎦

  For isot ropic, hom ogeneous condit ions ( K is not direct ional)

  2

  2

  2 ⎡ ∂ h ∂ h ∂ h ⎤

• = K ⎢ ⎥

  2

  2

  2 ∂ x ∂ y z ⎣ ∂ ⎦

  This is t he “ diffusion equat ion” or “ heat - flow equat ion”

  2

  2

  2 ⎡ h ∂ ∂ h ∂ h ⎤

  For St e a dy St a t e ( K cancels)

  2

  2

  2

∂ x ∂ y z

⎣ ∂ ⎦

  ⎥ = ⎢

  This is called t he La pla ce e qua t ion . ₂ ∇ is t he “ Laplacian” operat or. _{2 ∂ ⋅ ∂ ⋅ ∂ ⋅ 2 2 2 2} giving = = h

  ∇ = + ∇ + _{2 2 2} ∂ x ∂ y ∂ z Note that the full equation at this point can be written in summation notation as: ∂ ∂ h = K

  ( ) ij

  ∂ x ∂ x i i

  K K K K ^{11 K 12 K 13 xx xy xz} =

  K ^{21 K 22 K 23 K yx K yy K yz} K K K K zx K zy K zz _{31 32 33} The first Equat ion ( flow in t he x direct ion) is:

  ∂ ∂ h ∂ ∂ h ∂ ∂ h _{xx xy xz} + + K K K or x y z

  ∂ x ∂ ∂x ∂ ∂x ∂ ∂ ∂ h ∂ ∂ h ∂ ∂ h ₁₁ + + K K K _{12 13} ∂ x ∂ x ∂ x ∂ x ∂ x ∂ x _{1 1 1 2 1 3} What is t he head dist ribut ion in a SS hom ogeneous syst em ? Consider solut ion t o

  st eady- st at e problem ( 1- Dim ensional) 1D Confined Aquifer – Head Dist ribut ion K( x= 0) = H and h( x= L) = 0

  H _{2 2} ?

  ⎡ ⎤ ∂ h ∂ h

  1) ₂ = ₂ = ⎢ ⎥ →

  ∂ x ∂ x ⎣ ⎦

  x= 0 x= L St eady h dist ribut ion not f( K) – h is in de pe nde n t of K.

  Flow N e t s

  As w e have seen, t o w ork w it h t he groundw at er flow equat ion in any m eaningful w ay, w e have t o find som e kind of a solut ion t o t he equat ion. This solut ion is based on boundary condit ions, and in t he t ransient case, on init ial condit ions. Let us look at t he t w o- dim ensional, st eady- st at e case. I n ot her words, let t he follow ing equat ion apply:

  ⎞ ∂ h ∂ ⎛ ∂ h ⎛ ∂ ⎞ = K ⎜ K ( Map View) ⎜ ⎟ + _{x y} ⎟⎟

  ∂ x ∂ x x ∂ y ⎝ ⎠ ∂ ⎝⎜ ⎠

  A solut ion t o t his equat ion requires us t o specify boundary condit ion. For our purposes w it h flow net s, let us consider

  ∂ h

• No- flow boundaries ( =

  , where n is t he direct ion perpendicular t o t he

  ∂ n boundary) .

• Constant-head boundaries (h = constant)
• Water-table boundary (free surface, h is not a constant) A relat ively st raight forward graphical t echnique can be used t o find t he solut ion t o t he GW flow equat ion for m any such sit uat ions. This t echnique involves t he const ruct ion of a flow n e t . A flow net is t he set of equipot ent ial lines ( const ant head) and t he associat ed flow lines ( lines along w hich groundw at er m oves) for a part icular set of boundary condit ions.
• For a given GW flow equation and a given value of K, the boundary conditions com plet ely det erm ine t he solut ion, and t herefore a flow net .

  I n addit ion, let us first consider only hom ogeneous, isot ropic condit ions: _{2 2}

  ∂ h ∂ h

  = _{2 2} ∂ x ∂ z

• ( Cross- Sect ion)

  D C h E F h ₁ D a m h ²

  8

1 O

  G

  5 2 4

  6 Sa n d a n d 3 7 gr a ve l A B I m pe r m e a ble La ye r

  y

  C D

  x h ¹ G h

  M y E F A

  h ₂ B

  6 0 ’ 1 6 ’ h = h B C h = 0 A D q= 0 H q

¹

q ² h = h

  5 0 ’ G q ³ h ² q ⁴ q ₅ q= q

  E F Let ’s look at flow in t he vicinit y of each of t hese boundaries. ( I sot ropic, hom ogeneous condit ions) .

  ∂ h ∂ h ∂ h N o- Flow Bou n da r ie s: = 0 or = 0 or = 0

  ∂ x ∂ y ∂ n • Flow is parallel to the boundary.

• Equipotentials are perpendicular to the boundary

  Con st a n t - H e a d Bou nda r ie s: h = const ant • Flow is perpendicular to the boundary.

• Equipotentials are parallel to the boundary.

  W a t e r Ta ble Bou n da r ie s: h= z

  Anyw here in an aquifer, t ot al head is pressure head plus elevat ion head: ψ + z

  h =

  However, at t he wat er t able, ψ = 0. Therefore, h = z Neit her flow nor equipot ent ials are necessarily perpendicular t o t he boundary.

  Ru le s for Flow N e t s ( I sot r opic, H om oge ne ou s Syst e m ) :

  I n addit ion t o t he boundary condit ions t he follow ing rules m ust apply in a flow net : 1) Flow is perpendicular t o equipot ent ials everywhere.

  2) Flow lines never int ersect . 3) The areas bet ween flow lines and equipot ent ials are “ curvilinear squares” . I n ot her words, t he cent ral dim ensions of t he “ squares” are t he sam e ( but t he flow lines or equipot ent ials can curve) .

• I f you draw a circle inside the curvilinear square, it is tangential to all four sides at som e point .

  h= h ¹ D a m h= h ² Re se r voir Why are t hese circles? I t preserves dQ along any st ream t ube. dQ = K dm ; dh/ ds = K dh dQ ds dQ dm dQ

  I f dm = ds ( i.e. ellipse, not circle) , t hen a const ant fact or is used.

  Ot h e r point s:

  I t is not necessary t hat flow net s have finit e boundaries on all sides; r egions of flow t hat ext end t o infinit y in one or m ore direct ions are possible ( e.g., see t he figure above) . A flow net can have “ part ial” st ream t ubes along t he edge. A flow net can have part ial squares at t he edges or ends of t he flow syst em .

  Ca lcu la t ion s fr om Flow N e t s:

  I t is possible t o m ake m any good, quant it at ive predict ions from flow net s. I n fact , at one t im e flow nest were t he m aj or t ool used for solving t he GW flow equat ion.

  Probably t he m ost im port ant calculat ion is discharge from t he syst em . For a syst em wit h one recharge area and one discharge area, we can calculat e t he discharge wit h t he following expression:

  Q = n f K dH H = n d dH Gives: Q = n / n KH _{f d} Where Q is t he volum e discharge rat e per unit t hickness of sect ion perpendicular t o t he flow net ; n f is t he num ber of st ream t ubes ( or flow channels) ; n d is t he num ber of head drops; K is t he uniform hydraulic conduct ivit y; and H is t he t ot al head drop over t he region of flow.

• Note that neither n _{f nor n d is necessarily an int eger, but it is oft en helpful if} you const ruct t he flow net such t hat one of t hem is an int eger.
• I f you choose n _{f as an int eger, it is unlikely t hat n d w ill be an int eger.}
• Note that to do this calculation, you do not need to know any lengths.

  Flow N e t s in An isot r opic, H om oge n e ou s Syst e m s:

  Before const ruct ion of a flow net in an anisot ropic syst em ( K x = K y or K x = K z et c.) , w e have t o t ransform t he syst em .

  ∂ ∂ h ∂ ∂ h

⎛ ⎞ ⎛ ⎞

K K

  ⎜ ⎟ + ⎜ ⎟ = 0 _{x z} ∂ x ∂ x ∂ z ∂ z

⎝ ⎠ ⎝ ⎠

  For hom ogeneous K, _{2 2}

  ∂ h K ∂ h

_z

_{2 2} = + ∂ x K ∂ z

_x

  I nt roduce t he t ransform ed variable

  K _x

Z = z

K _z

  Applying t his variable gives: _{2 2}

  ∂ h ∂ h _{2 2} = + ∂ x ∂ Z

  Wit h t his equat ion w e can apply flow net s exact ly as we did before. We j ust have t o rem em ber how Z relat es t o t he act ual dim ension z. I n an anisot ropic m edium , perform t he following st eps in const ruct ing a flow net :

  1. Transform t he syst em ( t he area where a flow net is desired) by t he follow ing rat io:

  K _z

Z = Z '

K _x

  where z is t he original vert ical dim ension of t he syst em ( on your page, in cm , inches, et c.) and Z’ is t he t ransform ed vert ical dim ension.

  K x is t he hydraulic conduct ivit y horizont ally on your page, and K z is t he hydraulic conduct ivit y vert ically on your page. This t ransform at ion is not specific t o t he x- dim ension or t he y- dim ension.

  2. On t he t ransform ed syst em , follow t he exact sam e principles for flow net s as out lined for a hom ogeneous, isot ropic syst em .

  3. Perform t he inverse t ransform on t he syst em , i.e.

  K _z Z = Z ' K _x

  4. I f any flow calculat ions are needed, do t hese calculat ions on t he hom ogeneous ( st ep 2) sect ion. Use t he follow ing for hydraulic conduct ivit y:

  K ' = K K _{x z}

  Where K’ is t he hom ogeneous hydraulic conduct ivit y of t he t ransform ed sect ion. ( NOTE: This t ransform ed K’ is not real! I t is only used for calculat ions on t he t ransform ed sect ion.)

  Ex a m ple s:

  T

  I h = 0 h = 100

  Flow N e t s in H e t e r oge n e ou s Syst e m s:

  We w ill only deal w it h const ruct ion of flow net s in t he sim plest t ypes of het erogeneous syst em s. We w ill r est rict ourselves t o la ye r e d h e t e r oge n e it y. I n a layered syst em , t he sam e rules apply as in a hom ogeneous syst em , wit h t he follow ing im port ant except ions:

  1. Curvilinear squares can only be draw n in ONE layer. I n ot her w ords, in a t w o- layer syst em , you w ill only have curvilinear squares in one of t he layers.

  Which layer t o draw squares in is your choice: in general you should choose t he t hicker/ larger layer.

  2. At boundaries bet ween layers, flow lines are refract ed ( in a sim ilar way t o t he way light is refract ed bet ween t wo different m edia) . The relat ionship bet w een t he angles in t w o layers is given by t he “ t angent law” :

  U ¹ n θ ¹ m

  θ ₂ U ²

  ∂ h ∂ h _{1 2}

  h ^{1 = h 2 ( 1) No sudden head changes}

  = ∂ m ∂ m ∂ h ∂ h _{1 2}

  ( 2) Conservat ion of Mass

  K = K _{1 2} ∂ n ∂ n

  Layer

  1 Layer

  2

  ∂ h ∂ h _{1 2}

  u x :

  U sin θ = − K U sin θ = K − _{1 1 1 2 2 2} ∂ m ∂ m ∂ h ∂ h _{1 2}

  u y : θ θ

  U cos = − K U cos = K − _{1 1 1 2 2}

₂

∂ n ∂ n

  θ θ U sin U sin _{1 1 2 2} By ( 1) :

  = K K _{1 2} By ( 2) : U cos θ = U cos θ _{1 1 2 2} K tan θ _{1 1}

  = K tan ₂ θ ₂ You can rearrange t he t angent law in any way t o det erm ine one unknown quant it y.

  θ For exam ple, t o det erm ine t he angle ^{2 :}

  ⎛ K ⎞ −1 ²

  θ = tan θ _{2 1} ⎜⎜ tan ⎟⎟

  K ₁ ⎝ ⎠

  One im port ant consequence for a m edium wit h large cont rast s in K: high- K layers w ill oft en have alm ost horizont al flow ( in general) , w hile low - K layers w ill oft en have alm ost vert ical flow ( in general) .

  Ex a m ple :

• _{- 3 - 4}

  I n a t hree- layer syst em , K _o ^{1 = 1 x 10 m / s and K 2 = 1 x 10 m / s. K 3 = K 1 . Flow in} t he syst em is 14 below horizont al. What do flow in layers 2 and 3 look like? _o K ²

  76 K ¹ K ³

• 4

  ⎛ 1x10 ⎞ o o −1

  θ = tan tan _{2 3} 76 = 22 ^- ⎜⎜ ⎟⎟ ⎝ 1x10 ⎠ _o What is t he angle in layer 3? I f you do t he calculat ion, you w ill find it is 76 again. When draw ing flow net s w it h different layers, a very helpful quest ion t o ask is “ What layer allows wat er t o go from t he ent rance point t o t he exit point t he easiest ?” Or, in ot her words, “ What is t he easiest ( frict ionally speaking) way for wat er t o go from here t o t here?”

  K ² K ¹ K K ^{1 2} K ² K ¹ K ₁

  

= 10

K ₂