part 11 aquifer pumping tests - university of...

Hydrogeology, 431/531 - University of Arizona - Fall 2000 Dr. Marek Zreda

PART 11 Aquifer pumping tests

Groundwater exploration methods

(1) local surficial geology (maps, air photographs, field mapping)

(2) subsurface geology (stratigraphic relationships, test-drilling, geological logging and sam-pling)

(3) surface geophysics

(a) seismic refraction (geologic boundaries)

(b) electrical resistivity (aquifer limits)

(4) subsurface geophysics - borehole geophysics (resistivity, radiation logs)

(5) wells and piezometers

(a) water levels

(b) samples (soil, water)

(c) tests (pumping, recovery)

Aquifer pumping tests

Aquifer tests are performed to obtain T and S, which are both needed to

(1) determine well placement and potential yield

(2) predict drawdown

(3) understand regional flow

(4) put in numerical models

Possible methods of getting T and S via inverse problem:

(1) flownet calculations

(2) other calculations using Darcy’s law

(3) using tracer tests

(4) using inverse solution in numerical models

(5) aquifer pumping tests

Aquifer pumping tests 78


Steady state pumping test

Assumptions:

(1) Homogeneous, isotropic, confined aquifer with steady flow, 3-D

(2) Horizontal flow: 3-D changes to 2-D

(3) Impervious top and bottom and constant aquifer thickness b

(4) Prescribed head boundary around the well, i.e., “well on an island” setting. Water comes from a lateral boundary only (Theis assumption).

(5) Radial 1-D flow towards the well

(6) Well fully penetrates (compare to assumption 2); if not, then we will have vertical component

(7) Well pumps at a constant rate Q; Q appears as a boundary condition

PDE:

h∇2 0 in 3-D=

∇xy2 h 0 or

x2

2

∂∂ h

y2

2

∂∂ h

+ 0==

1r---

r∂∂ r

h∂r∂

----- 0 Laplace equation for radial flow=

��yyyy��yyyy��yyyyh = Hh = H

lake lake

Q

b

R

h1h2

r1

r2 q increases close to the well. Why?

1r---

r∂∂ r

h∂r∂

----- 0=



Boundary conditions:

BC1: Prescribed head at the lake:

h = H at r = R

BC2: Prescribed flux at the well (= pumping rate Q):

Q = K(2πrb)(∂h/∂r) at r = rw (well radius)

or, after rearranging and putting T = Kb:

Solution:

Integrate once:

Use BC2: C1 = Q/(2πT)

Rearrange by grouping r terms together

Integrate

Use BC1:

Final solution for h becomes

Q2πT---------- r

h∂r∂

-----=

rh∂r∂

----- C1=

h∂ C1r∂

r-----=

h C1 rln C2+Q

2πT---------- rln C2+= =

C2 HQ

2πT---------- Rln–=

h H–Q

2πT---------- r

R---ln

Thiem equation=



For two observation wells (h1 and h2 in the figure above):

Graphically, the solution is

s = drawdown = s(r)

at r = R, s = 0 (no drawdown at the lake)

Suppose r1 = rw r2 = R

h1 = hw h2 = H

then

Knowing Q, T, R - can calculate drawdown

Having observation wells is advantageous because we then know drawdowns and can get T of the aquifer:

In practice, there is no constant R where the head is H. Instead, R depends on Q, and this formula does not work. Two observation wells are thus necessary to solve for T

h h1–Q

2πT----------

r2

r1----ln

=

H

h

h2

r2r

s2

on normal paper

log r

h

on semilog paper

straight line

H hw–Q

2πT---------- R

rw

-----ln sw= =

TQ

2πsw------------ R

rw----- Thiem equationln=

TQ

2π h2 h1–( )------------------------

r2

r1

---- Thiem equation for observation wellln=



Rewrite in terms of decimal logarithms (log)

and if r2 = 10r1, log10 = 1 and we have

This is usually used for more than two observation wells.

If aquifer is infinite (no lakes) and confined, supply of water is from storage → transient flow.

T2.3Q

2π h2 h1–( )------------------------

r2

r1

----log Thiem equation for observation well=

T2.3Q2π∆h-------------- Equilibrium (= Thiem) steady state equation=

��yyyy��yyyy��yyyyh = h0 at r = infinity for all t

Q



Transient pumping test

Consider an aquifer with following assumptions:

• homogeneous and isotropic medium

• horizontal, radial, 1-D flow

• confined horizontal aquifer with constant thickness

• no leakage (i.e., impervious top and bottom)

• water released from storage instantaneously

• small well diameter (no storage in well)

• constant pumping rate Q

• infinite lateral extent of aquifer

• Darcy’s law valid

• saturated (single-phase) flow

• homogeneous fluid

• isothermal conditions

PDE

in radial coordinates

or, in terms of drawdown (s), using the relationship ∂h/∂r = - ∂s/∂r

Boundary conditions

Initial condition

h∇2 ST--- h∂

t∂----- h=h(x, y, t)=

1r---

r∂∂

rh∂r∂

----- S

T--- h∂

t∂----- h=h(r, t)=

1r---

r∂∂

rs∂r∂

----- S

T--- h∂

t∂----- h=h(r, t) (1)=

BC1: s 0= @ r = ∞

BC1: Q

2πT---------- r

s∂r∂

-----

r 0→lim–=

IC: s 0 @ t = 0 for all r=



The solution of equation 1 is (Theis, 1935)

where u defined as

is called similarity variable, and W(u) is called Theis well function in hydrology (tabulated in hydrogeology books; for example on p. 318 of Freeze and Cherry) or exponential integral in mathematics (tabulated in mathematics books). The exponential integral is calculated as

where γ = 0.57721566..... is the Euler’s number

Plot the solution on a log-log graph (this is the Theis type curve; left panel below). If Q, T and S are known, s(r, t) can be calculated and plotted on a log-log graph (these are the pumping test data; right panel below). Note that the shapes of the two graphs, W(u) vs. u and s vs. t, are the same. We can, therefore, use them together to determine S and T. The procedure used is called the curve matching method.

sQ

4πT---------- e

u–

u------- ud

u

∞

∫ Q4πT----------W u( )= =

ur

2S

4Tt--------=

W u( ) uln γ uk

kk!------- 1–( )k

k 1=

∞

∑–––=

log 1/u

log W(u) Theis type curve

log t

log s Pumping test data



Curve matching method (Theis method)

Can get T and S if Q is known and h is measured in time.

Procedure:

(1) Plot W(u) vs. 1/u on a log-log graph.

(2) Plot test data (s vs. t) on the same size graph.

(3) Overlay the graphs so that the curves are on top of each other.

(4) Pick any point in the common area.

(5) For this point, read off the following values: 1/u and W(u) for the first graph, and t and s from the second graph.

(6) Calculate the transmissivity using

(7) Calculate the storativity using

TQ

4πs---------W u( )=

S4Tt

r2

--------u=



Jacob logarithmic approximation

For large t, u = r2S/4Tt becomes small. For u<0.1 (or u<0.01), the infinite series term in W(u) is negligible and W(u) is simplified to

W(u) = -ln u - γ

The solution of equation 1 becomes

and finally

or, in terms of decimal log

If in observation well we have s1 = s(t1) and s2 = s(t2), we can write

and we solve for the transmissivity

sQ

4πT---------- uln γ––( )=

Q4πT---------- r

2S

4Tt--------ln 1.78ln––

=

Q4πT---------- 4Tt

1.78r2S

------------------ln=

sQ

4πT---------- 2.25Tt

r2S---------------- Jacob solutionln=

s2.3Q4πT------------ 2.25Tt

r2S----------------log=

s2 s1– ∆sQ

4πT----------

2.25Tt2

r2S

------------------ln2.25Tt1

r2S

------------------ln– = =

∆sQ

4πT----------

t2

t1

---ln=

TQ

4π∆s-------------

t2

t1----ln=



Plot on semi-log graph.

Late-time data (large t) should form a straight line. Note, however, that early-time data (small t) are off the straight line because these small t values do not make u suffi-ciently small and the Jacob method cannot be applied for these data.

Change to decimal log

For one log cycle, we have log(t2/t1) = log 10 = 1, and the solution for T is

Because we know the intercept (at s = 0), we can also calculate the storativity S:

at s = 0 we have t = t0 and we have

which is true only if

true only if

so

log t

ss

Q4πT---------- 2.25Tt

r2S----------------ln=

t0

∆s2.3Q4πT------------

t2

t1

---log=

T2.3Q4π∆s-------------=

s2.3Q4πT------------ 2.25Tt

r2S

----------------log=

02.3Q4πT------------

2.25Tt0

r2S

------------------log=

2.25Tt0

r2S

------------------log 0=

2.25Tt0

r2S------------------ 1=

S2.25Tt0

r2

------------------=



When is u sufficiently small to use the Jacob logarithmic approximation?

Look at u = r2S/4Tt and note that

(1) u ∝ r2 ⇒ small r or long t needed

(2) u ∝ S ⇒ confined aquifer better; if unconfined then need long t

(3) u ∝ 1/t ⇒ long time always good

(4) u ∝ 1/T ⇒ large T good; if small T then need long t

In summary, long t is always good.

Distance-drawdown data

At two or more observation wells at the same time:

Let observation wells be: s(r1) = s1 and s(r2) = s2, and r2 > r1 (i.e., also s2 < s1).

Let

(Note that although this is not steady-state flow, Thiem equation works, by accident).

sQ

4πT---------- 2.25Tt

r2S---------------- Jacob approximationln=

∆h s1 s2–Q

4πT---------- 2.25Tt

r12S

----------------ln2.25Tt

r22S

----------------ln–= =

∆hQ

4πT----------

r2

r1

---- 2

ln=

∆hQ

2πT----------

r2

r1---- Thiem equation !ln=



Principle of superposition

Can be used for linear systems only.

Linear are:

• confined aquifer

Non-linear are:

• phreatic aquifer (because T = T(h))

• vadose zone (because of multiphase flow)

Drawdowns calculated for multiple wells located in different places or drawdowns calculated for the same well that has variable pumping rate at different times can be linearly added.

It means that we can calculate drawdown for each well separately and then add them to get total drawdown.

Superposition in space: 2 wells pumping at the same time t

Superposition in time: 1 well; start at rate Q1 at time t1; switch to rate Q2 at time t2

Q

s

t

Q1

Q2

t1 t2

total s

s2s1



Boundary effects

Case 1: Constant head boundary (river)

To do the calculations of drawdown, we replace the real well-river system with the equivalent system that consists of the same real well (left side in the figure below) and an injecting image well (right side, on the other side of the boundary, at the same distance from the boundary as the real well). The inject-ing image well simulates the boundary (its effect on head and drawdown is exactly the same as the effect of the river). The figure below shows the distribution of drawdown in this system.

Note that drawdown is zero along the river (this is expected because the river is a constant head boundary). We demonstrate this mathematically by calculating total draw-down as follows:

Total:

X Data

0 5 10 15 20 25 30 35 40

Y D

ata

0

5

10

15

20

25

30

35

40

2

2

2

2

4

0

0

0

0

-2

-2

-2-2

-4

QrQi = -Qr

image well(recharge)

real well(discharge)

M

rr risr

Q4πT---------- 2.25Tt

rr2S

----------------ln=

siQ–

4πT---------- 2.25Tt

ri2S

----------------ln=

sQ

2πT----------

ri

rr

----ln 0 because at the boundary rr ri== =



Case 2: No-flow (barrier) boundary (need zero gradient at boundary)

We replace the barrier with an image pumping well (pumping at the same rate as the real pump-ing well).

Because we have two pumping wells, they pro-duce big drawdown. Will never reach steady state.

Drawdown is calculated by superposition:

Q. How to demonstrate that this is a no-flow boundary?

A. Derive an expression for gradient and show that the gradient in the direction normal to the boundary is zero at the boundary.

QrQi = Qr

image well(discharge)

real well(discharge)

M

rr ri

srQ

4πT---------- 2.25Tt

rr2S

----------------ln=

siQ

4πT---------- 2.25Tt

ri2S

----------------ln=

sQ

2πT---------- 2.25Tt

rrriS----------------ln=



Effects of boundaries on Theis and Jacob plots

Ideal type curve

Recharge boundary

Impermeable boundary

log time

log

draw

dow

n

log time

draw

dow

n

ideal Jacob response

recharge boundary

impermeable boundary(slope is doubled)



Other variations in drawdown shapes

Phreatic aquifer

Initially, the aquifer behaves like a con-fined aquifer (Ss only). Then, the sig-nal propagates and pore drainage occurs (Sy).

Leakage

No leakage before pumping (two aquifers in equilibrium). Neglect storage in aquitard.

Actual drawdown is horizontal for large t because there is steady state solution for leaky systems (there is additional source of water, much like in the case of a constant head boundary). It is achieved sooner in observation wells closer to the pumping well.

log t

s

earlytime

Ss only

Sy only

latetime

actualdrawdown

log t

s

no leakage

actualdrawdown



Partial penetration

We have 3-D flow due to partial penetration.

If s = const, partially penetrating wells give less Q than fully pene-trating wells because more energy is required to get water up (less area?).

If Q = const, partially penetrating well causes more drawdown than does fully penetrating one.

Drawdown curve far away from the pumping well looks like fully penetrating and the slope of drawdown curve approaches slope of fully penetrating well draw-down.

Penetration is important in calculating S. Do not use the late curve (straight line); use the early one.

Large diameter well

Has storage, usually built in low-K aquifers. Specific yield of the well: Sy = 1. S is overestimated (t1). Use t0 to get S.

log t

s

fullypenetrating

actualdrawdown

wrongestimationof S

t0

use this to get S

log t

s

smalldiameterwell

largediameterwell

t0 t1



Skin and well screen effect

Mud sealing the well forms “skin” reducing K around the well bore. If well screened in gravel, K is increased around the well.

S is underestimated!

For given Q, more drawdown for skin well, but slope the same for both. Reason: greater gradient is necessary for water to flow to the well.

log t

s

no skin(ideal well)

skineffect

part 11 aquifer pumping tests - university of...

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