radial flow pumping test

Well HydraulicFlow to the well

• Assoc. Prof. Dr Ismail Yusoff (Geology,UM)

• [email protected]

• 03-79674141/017-2420372

mailto:[email protected]

Lecture Outcomes

• Understand the concept of Well Hydraulic-Radial Flow

• Understand the unsteady and steady radial flow system

• Outlines the analytical solution to the radial flow system

References• 1)Fetter, C.W. (2001): Applied Hydrogeology,

Prentice Hall, New Jersey; 598pp.

• 2)Todd, D.K. and Mays L.W., (2005): Groundwater Hydrology.(3rd Edition). John Wiley & Sons, New York

• 3) Freeze, R.A. and Cherry, J.A. (1979): Groundwater, Prentice Hall, New Jersey,604pp.

• Details Ref.: Krusemen G.P. and deRidder, N.A. 1983.Analysis and evaluation of pumping test data (3rd ed.)

Groundwater Hydrology• It is the study of characteristics,

movement and occurance of water found below the surface

Aquifer

• Water bearing geological formation that can store and yield

usable amount of water

GWater Flow Equation

• Equation is based on Darcy’s Law• Flow equations are partial differential

equations in which head, h, is described in terms of x, y, z, and t.

• Equation is derived from “law of conservation of mass and energy”/ continuity law

• Assume the aquifer is homogeneous and isotropic

• Assume the fluid moves in only one direction

Flow types• Steady flow and Unsteady flow

• Unsteady flow - flow in which head changes with time

• Steady Flow -flow in which head does not changes with time, equilibrium

• Unconfined and confined aquifer flow equation

• Unconfined and confined aquifer differ in hydraulic properties

Ground-water Flow to Wells

• ”Radial Flow”

• Unsteady flow and Steady flow

• Cone of depression - area around a discharging well where the hydraulic head in the aquifer is lowered by pumping

• Produce “Time Drawdown Data”

• compute aquifer properties : T, S, Ss and K

Radial Flow and Cone of Depression

Defination T, S, Ss• Transmissivity

The rate at which water of a specific density and viscosity is transmitted through a unit width of an aquifer or confining bed under a unit hydraulic gradient. (1)

• T = Kb (m2/d) ; b = saturated thickness K = Hyd.conductivity

• Storativity /(Storage Coefficient) (S)

• The volume of water that a permeable unit will adsorb or expel from the storage per unit surface area per unit change in head.

• Confined aquifer S ≤ 0.005

• Unconfined S≈Sy in the range of 0.02-0.30

Radial Flow

Assumptions

Graphical Explanation (Time-Drawdown)

Unsteady/Transient/Nonequilibrium Radial

Flow

Steady Radial Flow

Examples

Assumptions• (1.) Bottom confining layer• (2.) All geologic units are horizontal and of infinite extent.• (3.) Potentiometric surface is horizontal prior to the start of

pumping.• (4.) Potentiometric surface is not changing with time prior to the

start of pumping.• (5.) All changes in potentiometric surface position are due to the

effect of the pumping well.• (6.) Aquifer is homogeneous and isotropic.• (7.) All flow is radial toward the well.• (8.) Ground-water flow is horizontal.• (9.) Darcy’s Law is valid.• (10.) Ground-water has constant density and viscosity.• (11.) Pumping well and observation wells are fully penetrating.• (12.) Well has an infinitesimal diameter and is 100% efficient.

Test Setup• Single Well

• Pumping well and Observation well

• Field note

Time

Depth to water level/Auto measure

Compute Drawdown

• Graph construction

“Time Drawdown Graph”

• Time Drawdown data analysis

Obtaining T, K, S

Field Setup

900 V-Notch Q=2.5H5/2

Confined Aquifer-Aquifer Test

Unconfined Aquifer-Aquifer Test

Curve shape = aquifer type

Unsteady Radial Flow

• Time Drawdown data analysis• Curve Matching technique• Confined aquifer

Theis MethodJacob Approximation method

• Unconfined aquiferNeuman/Boulton/Walton

• Leaky aquifer• Partially penetrating well-geometrical case• Pumping at the boundary-geometrical case• Recovery Test (T)

Unsteady Flow to Wells in Confined Aquifers

Confined Aq.:Theis Method

C.V. Theis 1935

Additional Assumption

• (1.) Aquifer is confined top and bottom.

• (2.) There is no source of recharge to aquifer.

• (3.) The Aquifer is compressible and water is released instantaneously from the aquifer as the head is lowered.

• (4.) The well is pumping at a constant rate.

Unsteady Flow to a Well in a Confined Aquifer

• Continuity

• Drawdown

• Theis equation

• Well function

s u( ) =Q

4pTW u( )

W u( ) =e-h

hu

¥

ò dh

u =r2S

4Tt

s(r,t) = h0 - h r,t( )

Ground surface

Bedrock

Confined

aquifer

Q

h0

Confining Layer

b

r

h(r)

Q

Pumping

well

¶2h

¶r2+

1

r

¶h

¶r=

S

T

¶h

¶t


Well Function

U vs W(u) 1/u vs W(u)

d

euW

u

Tt

Sru

4

2


Example - Theis Equation

Q = 1500 m3/dayT = 600 m2/dayS = 4 x 10-4

Find: Drawdown 1 km from well after 1 year

u =r2S

4Tt=

(1000 m)2 (4x10-4 )

4(600m2 /d)(365d)= 4.6x10-4

Ground surface

Bedrock

Confined

aquiferQ

Confining Layer

b

r1

h1

Q

Pumping

well


Well Function

u = 4.6x10-4

W(u) = 7.12

Example - Theis EquationQ = 1500 m3/dayT = 600 m2/dayS = 4 x 10-4

Find: Drawdown 1 km from well after 1 year

s =Q

4pTW (u) =

1500 m3 /d

4p (600 m2 /d)* 7.12 =1.42 m

u = 4.6x10-4

Ground surface

Bedrock

Confined

aquiferQ

Confining Layer

b

r1

h1

Q

Pumping

well

W(u) = 7.12


Pump Test in Confined AquifersTheis Method

Pump Test Analysis – Theis Method

s = Q

4pT

æ

è ç

ö

ø ÷ * W u( )

• Q/4pT and 4T/S are constant

• Relationship between

– s and r2/t is similar to the relationship between

– W(u) and u

– So if we make 2 plots: W(u) vs u, and s vs r2/t

– We can estimate the constants T, and S

Tt

Sru

4

2

r2

t =

4T

S

æ

è ç

ö

ø ÷ * u

constants

s =Q

4pTW (u)

Ground surface

Bedrock

Confined

aquiferQ

Confining Layer

br1

h1

Q

Pumpin

g well

Example - Theis Method

• Pumping test in a sandy aquifer

• Original water level = 20 m above mean sea level (amsl)

• Q = 1000 m3/hr

• Observation well = 1000 m from pumping well

• Find: S and T

Ground surface

Bedrock

Confined

aquifer

h0 = 20 m

Confining Layer

b

r1 = 1000 m

h1

Q

Pumping

well

Bear, J., Hydraulics of Groundwater, Problem 11-4, pp 539-540, McGraw-Hill, 1979.


Theis Method

Time

Water level,

h(1000)

Drawdown,

s(1000)

min m m

0 20.00 0.00

3 19.92 0.08

4 19.85 0.15

5 19.78 0.22

6 19.70 0.30

7 19.64 0.36

8 19.57 0.43

10 19.45 0.55

…

60 18.00 2.00

70 17.87 2.13

…

100 17.50 2.50

…

1000 15.25 4.75

…

4000 13.80 6.20


Theis Method

Time r2/t s u W(u)

(min) (m2/min) (m)

0 0.00 1.0E-04 8.63

3 333333 0.08 2.0E-04 7.94

4 250000 0.15 3.0E-04 7.53

5 200000 0.22 4.0E-04 7.25

6 166667 0.30 5.0E-04 7.02

7 142857 0.36 6.0E-04 6.84

8 125000 0.43 7.0E-04 6.69

10 100000 0.55 8.0E-04 6.55

…

3000 333 5.85 8.0E-01 0.31

4000 250 6.20 9.0E-01 0.26

s vs r2/t

W(u) vs u


r2/t

s

u

W(u)

r2/t s W(u)u

0.01

0.1

1

10

10 100 1000 10000 100000 1000000

s

r2/t

0.01

0.1

1

10

0.0001 0.0010 0.0100 0.1000 1.0000 10.0000

W(u

)

u

Match PointW(u) = 1, u = 0.10s = 1, r2/t = 20000

Theis MethodPump Test Analysis – Theis Method

Theis Method

• Match Point

• W(u) = 1, u = 0.10

• s = 1, r2/t = 20000

T =Q

4p

Wmp

smp

æ

è ç ç

ö

ø ÷ ÷ =

1000 m3 /hr

4p

1

1 m

æ

è ç

ö

ø ÷ = 79.58 m2 /hr (=1910 m2 /d)

S = 4Tump

r2

tmp

æ

è

ç ç ç

ö

ø

÷ ÷ ÷

= 4(79.58 m2 /hr)0.1

20000 m2 /min* 60 min/hr

æ

è ç

ö

ø ÷ = 2.65x10-5


Pump Test in Confined AquifersJacob Method

Jacob’s Approximation

Two approximation;

a) For one pumped well and one observation well

• Time- Drawdown Method

b) For one pumped well and more than three observation wells

• Distance-Drawdown method

Jacob Approximation

• Drawdown, s

• Well Function, W(u)

• Series approximation of W(u)

• Approximation of s

s u( ) =Q

4pTW u( )

W u( ) =e-h

hu

¥

ò dh » -0.5772 - ln(u)+ u -u2

2!+

u =r2S

4Tt

W u( ) » -0.5772 - ln(u) for small u < 0.01

s(r,t) »Q

4pT-0.5772 - ln

r2S

4Tt

æ

è ç

ö

ø ÷

é

ë ê ê

ù

û ú ú

Pump Test Analysis – Jacob Method

Jacob Approximation

s =2.3Q

4pTlog(

2.25Tt

r2S)

0 =2.3Q

4pTlog(

2.25Tt0

r2S)

t0

1=2.25Tt0

r2S

S =2.25Tt0

r2


Jacob Approximation

t0

S =2.25Tt0

r2

t1 t2

s1

s2

Ds

logt2

t1

æ

è ç

ö

ø ÷ = log

10* t1t1

æ

è ç

ö

ø ÷ =1

1 LOG CYCLE

1 LOG CYCLE


Jacob Approximation

S =2.25Tt0

r2=

2.25(76.26 m2/hr)(8 min*1 hr /60 min)

(1000 m)2

= 2.29x10-5

t0

t1 t2

s1

s2

Ds

t0 = 8 min

s2 = 5 ms1 = 2.6 mDs = 2.4 m


Jacob Distance-Drawdown Method

• In case more than three observation wells

• Modification to Jacob-Time drawdown method

• Plot semilog graph drawdown(y normal axis) Vs. distance (x log axis)

• Extend the straight line to ZERO drawdown= ro

T and S calculation

• T = 2.3Q

• 2Π∆(h0-h) where

• T - ft2/d or m2/d

• Q - ft3/d or m3/d

• ∆ (h0 - h) - Drawdown / log cycle (ft)

• S = 2.25Tt where

• ro2

• S = storativity

• t= time of pumping

• ro = distance where straight line intersects the zero drawdown axis (days)

Unsteady Flow to Wells in Unconfined Aquifers

Unsteady Flow to a Well in an Unconfined Aquifer

• Water is produced by

– Dewatering of unconfined aquifer

– Compressibility factors as in a confined aquifer

– Lateral movement from other formations

2rw

Ground surface

Bedrock

Unconfined

aquifer

Q

h0

Prepumping

Water level

r1

r2

h2 h1

hw

Observation

wells

Water Table

Q

Pumping

well


Analyzing Drawdown in An Unconfined Aquifer

• Early– Release of water is from

compaction of aquifer and expansion of water – like confined aquifer.

– Water table doesn’t drop significantly

• Middle– Release of water is from gravity

drainage

– Decrease in slope of time-drawdown curve relative to Theiscurve

• Late– Release of water is due to drainage

of formation over large area

– Water table decline slows and flow is essentially horizontal


Early

Late

Unconfined Aquifer (Neuman Solution)

s =Q

4pTW (ua ,h)

ua =r2S

4Tt

s =Q

4pTW (uy ,h)

uy =r2Sy

4Tt

h =r

b

æ

è ç

ö

ø ÷

2 Kz

Kr

Early (a)

Late (y)


Procedure - Unconfined Aquifer (Neuman Solution)

• Get Neuman Well Function Curves

• Plot pump test data (drawdown s vs time t)

• Match early-time data with “a-type” curve. Note the value of

• Select the match point (a) on the two graphs. Note the values of s, t, 1/ua, and W(ua, )

• Solve for T and S

• Match late-time points with “y-type” curve with the same as the a-type curve

• Select the match point (y) on the two graphs. Note s, t, 1/uy, and W(uy, )

• Solve for T and Sy

T =Q

4psW (ua ,h)

S =4Ttua

r2

T =Q

4psW (uy ,h)

Sy =4Ttuy

r2


Procedure - Unconfined Aquifer (Neuman Solution)

• From the T value and the initial (pre-pumping) saturated thickness of the aquifer b, calculate Kr

• Calculate Kz

Kr =T

b

Kz =hKrb2

r2


Example – Unconfined Aquifer Pump Test

• Q = 144.4 ft3/min

• Initial aquifer thickness = 25 ft

• Observation well 73 ft away

• Find: T, S, Sy, Kr, Kz Ground surface

Bedrock

Unconfined aquiferQ

h0=25 ft

Prepumping

Water level

r1=73 ft

h1

hw

Observation

wells

Water Table

Q= 144.4 ft3/min

Pumping

well


Pump Test data


Early-Time Data

h = 0.06

t = 0.17 min; s = 0.57 ft

1/ua =1.0; W =1.0


Early-Time Analysis

T =Q

4psW (ua ,h)

=144.4 ft 3 /min

4p (0.57 ft)(1.0)

= 20.16 ft2 /min

(29,900 ft2 /day)

S =4Ttua

r2

=4(20.16 ft2 /min)(1.0)(0.17 min)

(73 ft)2

= 0.00257

h = 0.06

t = 0.17 min; s = 0.57 ft

1/ua =1.0; W (ua ,h) =1.0


Late-Time Data

h = 0.06

t =13 min; s = 0.57 ft

1/uy = 0.1; W =1.0


Late-Time Analysis

h = 0.06

t =13 min; s = 0.57 ft

1/uy = 0.1; W =1.0

T =Q

4psW (uy ,h)

= 20.16 ft2 /min

(29,900 ft2 /day)

Sy =4Ttuy

r2

=4(20.16 ft2 /min)(0.1)(13 min)

(73 ft)2

= 0.02

Kr =T

b

=20.16 ft2 /min

25 ft

= 0.806 ft /min

Kz =(0.06)(0.806 ft /min)(25 ft)2

(73 ft)2

= 5.67x10-3 ft /min


Unsteady Flow to Wells in Leaky/Semi Confined Aquifers

Radial Flow in a Leaky AquiferWalton’s method

K b

ground surface

bedrock

aquitard

confined aquifer

initial head

Well

s(r)

r

Q

R

h0

Cone of Depression leakage

h(r)

unconfined aquifer

B

ruW

T

Qs ,

4p

bK

T

r

B

r

/

dzz

e

B

ruW

u

zB

rz

2

2

4,

When there is leakage from other layers, the drawdown from a pumping test will be less than the fully confined case.

B=Leakage factor

Unsteady Flow to Wells in Leaky Aquifers

Leaky Well Function dzz

e

B

ruW

u

zB

rz

2

2

4,

r/B = 0.01

r/B = 3

cleveland1.cive.uh.edu/software/spreadsheets/ssgwhydro/MODEL6.XLS


r/B=0 ≈Theis curve

Leaky Aquifer Example• Given:

– Well pumping in a confined aquifer

– Confining layer b’ = 14 ft. thick

– Observation well r = 96 ft. form well

– Well Q = 25 gal/min

• Find:

– T, S, and K’

From: Fetter, Example, pg. 179

t (min) s (ft)5 0.76

28 3.341 3.5960 4.0875 4.39

244 5.47493 5.96669 6.11958 6.27

1129 6.41185 6.42

K b

ground surface

bedrock

aquitard

confined aquifer

initial head

Well

s(r)

r

Q

R

h0

Cone of Depression leakage

h(r)

unconfined aquifer


= 0.15

= 0.20

= 0.30

= 0.40

r/B

Match PointW(u, r/B) = 1, 1/u = 10s = 1.6 ft, t = 26 min, r/B = 0.15


Leaky Aquifer Example

• Match Point

• Wmp = 1, (1/u)mp = 10

• smp = 1.6 ft, tmp = 26 min, r/Bmp = 0.15

• Q = 25 gal/min * 1/7.48 ft3/gal*1440 min/d = 4800 ft3/d

• t = 26 min*1/1440 d/min = 0.01806 d

T =Q

4psmp

Wmp =4800 ft 3 /d

4p (1.6 ft)*1= 238.7 ft2/d

S =4Tump

r2

tæ

è ç

ö

ø ÷ mp

=4(238.7 ft2 /d)(0.1)(0.01806)

(96 ft)2=1.87 x10-4

¢ K =T ¢ b (r /B)2

r2=

(238.7 ft2 /d)(14 ft)(0.15)2

(96 ft)2= 0.0081 ft /d


Steady Flow to Wells in Confined Aquifers

Steady Flow to a Well in a Confined Aquifer

2rw

Ground surface

Bedrock

Confined

aquifer

Q

h0

Pre-pumping

head

Confining Layer

b

r1

r2

h2

h1

hw

Observation

wells

Drawdown curve

Q

Pumping

well

Q = Aq

= (2prb)Kdh

dr

rdh

dr=

Q

2pT

h2 = h1 +Q

2pTln(

r2r1

)

Theim Equation

In terms of head (we can write it in terms of drawdown also)

h = h1 at r = r1

h = h2 at r = r2

Example - Theim Equation

• Q = 400 m3/hr

• b = 40 m.

• Two observation wells,

1. r1 = 25 m; h1 = 85.3 m

2. r2 = 75 m; h2 = 89.6 m

• Find: Transmissivity (T)

T =Q

2p h2 - h1( )ln

r2r1

æ

è ç

ö

ø ÷ =

400 m3/hr

2p 89.6 m - 85.3m( )ln

75 m

25 m

æ

è ç

ö

ø ÷ =16.3 m2 /hr

h2 = h1 +Q

2pTln(

r2r1

)2rw

Ground surface

Bedrock

Confine

d

aquifer

Q

h0

Confining Layer

b

r1

r2

h2 h1

hw

Q

Pumping

well


Steady Radial Flow in a Confined Aquifer

• Head

• Drawdown

h = h0 +Q

2pTln

r

R

æ

èç

ö

ø÷

s =Q

2pTln

R

r

æ

èç

ö

ø÷


Theim Equation In terms of drawdown (we can write it in terms of head also)

s = h0 - h

= h0 - h0 +Q

2pTln

r

R

æ

èç

ö

ø÷

é

ëê

ù

ûú


• 1-m diameter well

• Q = 113 m3/hr

• b = 30 m

• h0= 40 m

• Two observation wells,

1. r1 = 15 m; h1 = 38.2 m

2. r2 = 50 m; h2 = 39.5 m

• Find: Head and drawdown in the well

2rw

Ground surface

Bedrock

Confined

aquifer

Q

h0

Confining Layer

b

r1

r2

h2 h1

hw

Q

Pumping

wellDrawdown

Adapted from Todd and Mays, Groundwater Hydrology

T =Q

2p s1 - s2( )ln

r2r1

æ

è ç

ö

ø ÷ =

113m3/hr

2p 1.8 m - 0.5 m( )ln

50 m

15 m

æ

è ç

ö

ø ÷ =16.66 m2 /hr

s r( ) =Q

2pTln

R

r

æ

è ç

ö

ø ÷



2rw

Ground surface

Bedrock

Confine

d

aquifer

Q

h0

Confining Layer

b

r1

r2

h2 h1

hw

Q

Drawdown

@ well

Adapted from Todd and Mays, Groundwater Hydrology

hw = h2 +Q

2pTln(

rwr2

) = 39.5 m +113m3 /hr

2p *16.66 m2 /hrln(

0.5 m

50 m) = 34.5 m

sw = h0 - hw = 40 m- 34.5 m = 5.5 m

h2 = h1 +Q

2pTln(

r2r1

)


Drawdown at the well

Steady Flow to Wells in Unconfined Aquifers

Steady Flow to a Well in an Unconfined Aquifer

Q = Aq = (2prh)Kdh

dr

= prKdh2

dr

2rw

Ground surface

Bedrock

Unconfined

aquifer

Q

h0

Pre-pumping

Water level

r1

r2

h2 h1

hw

Observation

wells

Water Table

Q

Pumping

well

rd h2( )

dr=

Q

pK

h02 - h2 =

Q

pKln

R

r

æ

è ç

ö

ø ÷

h2(r) = h02 +

Q

pKln

r

R

æ

è ç

ö

ø ÷

Unconfined aquifer

h = h0 at r = R

h 1 and h2 = aquifer saturated thickness; can be b1 and b2


2rw

Ground surface

Bedrock

Unconfined

aquifer

Q

h0

Prepumping

Water level

r1

r2

h2 h1

hw

Observation

wells

Water Table

Q

Pumping

well

2 observation wells: h1 m @ r1 m h2 m @ r2 m

K =Q

p h22 - h1

2( )ln

r2r1

æ

è ç

ö

ø ÷

h2(r) = h02 +

Q

pKln

r

R

æ

è ç

ö

ø ÷

h22 = h1

2 +Q

pKln

r2r1

æ

è ç

ö

ø ÷

• Given: – Q = 300 m3/hr

– Unconfined aquifer

– 2 observation wells,

• r1 = 50 m, h = 40 m

• r2 = 100 m, h = 43 m

• Find: K

K =Q

p h22 - h1

2( )ln

r2r1

æ

è ç

ö

ø ÷ =

300 m3 /hr / 3600 s /hr

p (43m)2 - (40 m)2[ ]ln

100 m

50 m

æ

è ç

ö

ø ÷ = 7.3x10-5 m /sec

Example – Two Observation Wells in an Unconfined Aquifer

2rw

Ground surface

Bedrock

Unconfined

aquifer

Q

h0

Prepumping

Water level

r1

r2

h2 h1

hw

Observation

wells

Water Table

Q

Pumping

well


More Calculation Examples

• Fetter, C.W. (2001)

• Pg 166-169

WL Recovery/ Residual

• Water level measurements taken immediately after test.

• Water levels not influenced by erratic pumping rates.

• Recovery analysis compared to constant rate test.

• Check valve on pump essential to eliminating slugging effects (after the pump is switched off)

• Example: Todd and May (2005) pg. 170-172

radial flow pumping test

Science

time steady flow flow

wellsradial flowunsteady

steady flow cone of

confined aquifer s

hydraulic head

volume of water

occurance of water

unit surface area