Lecture Packet 8. Pump Test Analysis

  1.72, Groundwat er Hydrology Prof. Charles Harvey

  

Le ct u r e Pa ck e t # 8 : Pu m p Te st Ana lysis

  The idea of a pum p t est is t o st ress t he aquifer by pum ping or inj ect ing wat er and t o not e t he drawdown over space and t im e. Hist ory

• The earliest m odel for interpretation of pum ping test data was developed by

  Thiem ( 1906) ( Adolf and Gunt her) for _o

  Const ant pum ping rat e _o Equilibr ium condit ions _o Confined and unconfined condit ions

• Theis (1935) published the first analysis of transient pum p test for _o

  Const ant pum ping rat e _o Confined condit ions

• Since then, m any m ethods for analysis of transient well tests have been designed for increasingly com plex condit ions, including _o

  Aquit ard leakage ( st udy of hydrogeology has becom e m ore a st udy of aquit ards and less of aquifers) _o Aquit ard st orage _o Wellbore st orage _o

  Part ial w ell penet rat ion _o Anisot ropy _o Slug t est s _o Recirculat ing w ell t est s ( w at er is not rem oved) I t is im port ant t o not e t he assum pt ions for a given analysis. St e a dy Ra dia l Flow in a Confine d Aqu ife r Assum e:

• Aquifer is confined (top and bottom )
• Well is pum ped at a constant rate
• Equilibrium is reached (no drawdown change with tim e)
• Wells are fully screened and is only one pum ping

Consider Darcy’s law t hrough a cylinder, radius r, wit h flow t ow ard w ell.

  dh Q dr Q = K 2 and rearrange as dh = πrb dr

  2

πKb r

  I nt egrat e from r _{h r 2 2} ^{1 , h 1 t o r 2 , h 2} Q dr

  dh = ∫ ∫ _{h r 1} 2 π Kb r ₁ Q r

  2 − = h h

  2 1 or not ing t hat T = Kb π

  2 Kb r

  1 Q r ₂

  t his is t he Thiem equat ion

  T = ln 2 ( ) π h − h r _{2 1 1} Not es on t he Thiem equat ion:

• Good with any self consistent units L and t
• I f drawdown has stabilized can use any two observation wells
• Water is not com ing from storage (S doesn’t appear) cannot get S from this t est
• Com m only used in USGS units and log , T in gpd/ ft ( gallons per day per
₁₀ foot ) , Q in gpm ( gallons per m inut e) , r and h in ft .

  r 527 Q .

  7 ₂ T = log (h ) r ₂ − h _{1 1} Spe cific Ca pa cit y of a W e ll – Roughly est im at ing T

  Specific Capacit y = Discharge Rat e/ Draw dow n in t he w ell 1. A w ell is pum ped t o approxim at e equilibrium .

  2. A good well would be 50 gpm per foot of drawdown, or 20 feet of draw down at 1,000 gpm . 3. h e and r e are t he head and corresponding dist ance from a well where drawdown is effect ively zero.

  

Q T

  4. Specific capacit y =

  T = = ln r

(h − h ) e

_{e w} 527 . 7 log r _w

  5. Rule of Thum b – T ~ 1,800 x Specfic Capacit y 6. What is r e ? I t doesn’t m at t er t hat m uch. r e = 1,000 r w log r e / r w = 3 r e = 10,000 r w log r e / r w = 4

  7. Case A Æ T = Specific Capacit y [ 527 x 3] = 1,581 x SC Case B Æ T = Specific Capacit y [ 527 x 4] = 2,108 x SC

  8. I f you use T ~ 1,800 x Specific Capacit y you are not t oo far off. SC is gpm / ft and T is gpd/ ft .

• Well is pum ped at a constant rate
• Equilibrium is reached (no drawdown change with tim e)
• Wells are fully screened and
• There is only one pum ping well

  2

  π

  K − =

  2 _{1 2 2 1 2 2} ln ) ( r r h h Q

  or not ing t hat T = Kb

  1

  2 2 π K ln r

  Q r

  2 =

  1

  2 − h

  ∫ 2 πK r _{h 1 r 1} h

  St e a dy Ra dia l Flow in a n Un con fine d Aquife r

  Q ∫ dr

  jdh =

  I nt egrat e from r ^{1 , h 1 t o r 2 , h 2} _{h 2 r 2}

  2 πK r

  Q dr = dr

  and rearrange as hdh =

  K Q (2 πrh) dh

  Radial flow in t he unconfined aquifer is given by

  K r ² r ₁ h ¹ h ² Observat ion Pum ping aquifer Wat er t able before pum ping Wat er t able during pum ping

  Q aquiclude well well

  Assum e: • Aquifer is unconfined but underlain by an im perm eable horizontal unit.

  t his is t he Thiem equat ion for unconfined condit ions ( K not T, h ² not h, no 2)

• I f the greatest difference in head in the system is < 2% , then you can use the confined equat ion for an unconfined syst em .
• For r < 1.5h _{m ax ( t he full sat urat ed t hickness) t here w ill be errors using t his} equat ion because of vert ical flow near t he well.
• I f difference in head is > 2% but < 25% use the confined equation with the follow ing correct ion for draw down
• ⎢ ⎣ ⎥ ∂ ⎦
• Aquifer is horizontal, confined both top and bottom , infinite in horizontal ext ent , const ant t hickness, hom ogeneous, isot ropic.
• Potentiom etric surface is horizontal before pum ping, is not changing with tim e before pum ping, all changes due t o one pum ping w ell
• Darcy’s law is valid and groundwater has constant properties
• Well is fully screened and 100% efficient
• Constant pum ping from a well in such a situation is radial, and horizontal,
_{2 2}
• y x r = r h r t h t h T S
• ∂ ∂ = ∂ ∂
We want a solut ion t hat gives h( r,t ) aft er w e st art pum ping. To solve equat ion we need init ial condit ions ( I Cs) and boundary condit ions ( BCs)

  aquifer

h( r,t )

Depr ession Drawdown

  where only 1 space dim ension is needed

  Cone of well w ell Q aquiclude

  aquiclude T and S r b

  r

  1 ₂ ²

  ∂ ∂

  T = Kb = Transm issivit y [ L ² / T] b = aquifer t hickness [ L] h is head [ L] Assum e:

  ∂ y ² x ² S = S s b = St orat ivit y [ L ³ / L ³ ]

  ∂

  S = T ⎡

  ∂ ² h ² ∂ h ⎤ ∂ h t

  (h h “ m easured” reduced com pared t o confined aquifer case Tr a n sie n t Pu m pin g Te st s

  2 h b

  ∆ ² ∆ = −

  − ₁ h ) _new

  2

  Observat ion Pum ping

  I C: h( r,0) = h

  Q ⎛ ∂ h ⎞

  BCs: h( ,t ) = h and lim r for t > 0

  ∞ Æ ⎜ r ⎟ =

  ⎝ ∂ r ⎠ 2πT

  w hich is j ust t he applicat ion of Darcy’s law at t he w ell

  Th e is Equa t ion – t r a nsie n t r a dia l flow

  1935 – C.V. Theis solves t his equat ion ( wit h C.I . Lubin from heat conduct ion)

  ∞ − u

  exponent ial int egral in m at h t ables but for our case it is “ well funct ion”

  Q e du h − h( t r , ) = ∫

  4 π T u _u

  Å ₂

  r S

  where u =

  4Tt Q h r , ) W (u) , w ell funct ion is W( u)

  − h( t = 4 π T

  Det erm ining T and S from Pum ping Test I nverse m et hod: Use solut ion t o t he PDE t o ident ify t he param et er values by m at ching sim ulat ed and observed heads ( dependent variables) ; e.g., m easure aquifer drawdown response given a know n pum ping rat e and get T and S.

  1. I dent ify pum ping w ell and observat ion w ells and t heir condit ions ( e.g., fully screened) .

  2. Det erm ine aquifer t ype and m ake a quick est im at e t o predict w hat you t hink will happen during pum ping t est .

  3. Theis Met hod: Arrange Theis equat ion as follows: 114 Q .

  6 ⎤

  ( in USGS unit s) and

  ∆h = ⎡ ⎢ ⎥ W (u) ⎣ T ⎦ ₂

  2 1 . Sr ⎤

  1 ⎡ 87 r T

  ⎡ ⎤ t

  Æ

  = u = ⎢ ⎥

  ⎢ T u 1 . t S

  ⎣ 87 ⎦⎥ ⎣ ⎦ 114 Q .

  6 ⎤ log ∆ h = log⎡ ⎢ ⎥ + logW (u)

  ⎣ T ⎦ ₂ ⎡ 87 ⎤ 1 . Sr

  1 log t = log ⎢ ⎥ + log u

  ⎣ T ⎦

  4. Plot t he w ell funct ion W( u) versus 1/ u on log- log paper. ( t his is called a t ype curve)

  5. Plot drawdown vs. t im e on log- log paper of sam e scale. ( t his is from dat a at a single observat ion w ell)

  6. Superim pose t he field curve on t he t ype curve, keeping t he axes parallel.

  Adj ust t he curves so t hat m ost of t he dat a fall on t he t ype curve. You t rying t o get t he const ant s ( bracket ed t erm s) t hat m ake t he t ype curve axes t ranslat e int o your axes.

  7. Select a m at ch point ( any convenient point w ill do like W( u) = 1.0 and 1/ u = 1.0) , and read off t he values for W( u) and 1/ u. Then read off t he values for drawdown and t .

  8. Com put e t he values of T and S from : Using USGS Unit s Using Self- Consist ent Unit s

  114 6 . QW ( u ) QW ( u ) T = T =

  ( h − h) 4 π ( h − h) Ttu

  4 Ttu S = S = _{2 2}

  1 87r . r

  I f in USGS unit s of draw dow n ( ft ) , Q ( gpm ) , T ( gpd/ ft ) , r ( ft ) , t ( days) , S ( decim al fract ion) .

  Q h − h( t r , ) = W ( u) ₂ 4 πT 1 87 . r S u = Tt

  To predict drawdown – drawdown vs. dist ance or t im e

• Put in r, S, T, t and solve for u
• Find W(u) based on u and table
• Multiply Q/ T and factor The analyt ic solut ion describes:
• Geom etric characteristics to the cone of depression, steepening toward t he w ell
• For a given aquifer cone of depression increases in dept h and extent w it h t im e
• Drawdown at a tim e and location increases linearly with pum ping rate
• Drawdown at a tim e and location is greater for sm aller T • Drawdown at a tim e and location is greater for lower S

  M odifie d N on e qu ilibr iu m Solut ion – Ja cob M e t h od

  Recall from Theis solut ion:

  ∞ u − Q e du h − h ( , r t ) =

  ∫

  4 T u π _u C.E. Jacob not ed t hat t he w ell funct ion can be represent ed by a series. _{2 3 4}

  ⎡ ⎤

  Q u u u h − h ( , r t ) = − . 5772 − ln u + u − − + + ....

  ⎢ ⎥

  4 T π 2 ⋅ 2 ! 3 ⋅ 3 ! 4 ⋅ 4 ! ⎣ ⎦ For sm all values of r and large values of t , u becom es sm all. ( valid w hen u < 0.01) .

  Then m ost t erm s can be dropped leaving: ₂ Q ⎡ r S ⎤

  h − h ( , r t ) = − . 5772 − ln ⎢ ⎥

  4 T

  4Tt π

  ⎣ ⎦

  Not ing t hat –ln u = ln 1/ u and ln 1.78 = 0.5772

  Q 2 .

  25 Tt ⎡ ⎤ h h ( , r t ) ln

  − = ₂

  4 T r S π ⎢⎣ ⎥⎦

  And ln u = 2.3 log u 2 .

30 Q 2 .

  25 Tt ⎡ ⎤ h − h ( , r t ) = log _{10 2}

  4 T ⎢ r S π ⎣ ⎥⎦ Since Q, r, T, and S are const ant s, drawdown vs. log t should plot as a st raight line.

  I n USGS unit s:

  264Q 0 Tt .

  3 ⎡ ⎤ h − h ( , r t ) = log _{10 2}

  4 T ⎢ π r S ⎣ ⎥⎦

  Over one log cycle you get change in drawdown

  ⎡ t ⎤ 264Q ₂ From t ^{1 t o t 2 : ∆[h − h ] = log} ₁₀ ⎢ ⎥

  T t ₁ ⎣ ⎦

  264Q T = log 10 ( USGS unit s)

  [ ^{10 ]} ∆[h − h ]

  2 Q .

  3 T = ( self- consist ent unit s)

  4 π∆ [h − h] Pr oce du r e 1. Plot on sem i- log paper – t on log scale and drawdown on arit hm et ic scale.

  2. Pick off t w o values of t im e and t he corresponding values of drawdowns ( over one log cycle m ake it easy) .

  3. Solve for T ( j ust a funct ion of Q and ∆ h) 4. Consider basic solut ion when drawdown is zero.

  2 .

25 Tt 2 .

  30 Q ⎡ ⎤ = h − h = log _{10 2} 4 πT r S ⎢⎣ ⎥⎦ 2 .

  25 Tt 2 .

  25 Tt ⎡ ⎤

  Æ

  log = 1 or _{10 2 2} = ⎢ r S r S

  ⎣ ⎥⎦ 2 .

  25 Tt

  where t is t he t im e int ercept at w hich

  S = ₂ r draw down is zero in USGS unit s.

  2 .

  25 Tt

  where t is t he t im e int ercept in days and

  S = ₂

  t he st raight line int ersect s t he zero-

  r

  drawdown axis