Groundwater Hydrology Tutorial

DEPARTMENT OF CIVIL ENGINEERING FACULTY OF ENGINEERING & TECHNOLOGY UNIVERSITY OF ILORIN ILORIN, NIGERIA

GROUNDWATER HYDROLOGY (Well hydraulics solved examples)

By: Dr Olayinka Okeola

2 Page

Preface This is a comprehensive solution to analytical questions on Groundwater Hydrology. It is useful for both engineering and geology students desiring a better understanding on analytical solution procedures on well hydraulics.

Question No 1.

A tube of 30 cm diameter penetrates fully in an aquifer. The strainer length is 15cm. Calculate the yield of the well under a drawdown of 3m. The aquifer consists of sand of effective size of 0.2mm having a coefficient of permeability equal to 50m/ day. Assume a radius of drawdown equal to 150m.

Solution:

Well diameter= 30cm, radius r = 15cm = 0.15m Strainer length L =15cm = 0.15m Drawdown s = 3m Coefficient of Permeability, K = 50m/ day. =5.787 x10-4m/s

K

Radius of drawdown R= 150m Yield Q=? Aquifer Confined Aquifer Strainer’s length

Q

2.72bks 2.72  0.15  5.787  10  R 150 log10 log10 r 0.15 =0.023l/s

4

log10

3

50m 1d   m/s d 24  3600 s

150  log10 1000  3 0.15

m 3 1000l l   s s 1m 3

Question No 2. A tube well penetrates fully an unconfined aquifer. Calculate the discharge from the tube well under the following conditions. Diameter of well = 30cm Drawdown = 2m Coefficient of permeability=0.05cm/s Radius of zero drawdown=300m Length of strainer = 10m

Solution:

Tube well diameter= 30cm r = 0.15m Drawdown s = 2m Strainer’s length L = 10m Radius of zero drawdown R = 300m Coefficient of Permeability K = 0.05 cm/s = 5x 10-4m/s

k  0.05

cm 1m  m s s 100cm

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s  2.72ks L   2  Q R log10 r

3

Yield Q=? Strainer’s length magnitude  Unconfined aquifer



2.72  5.0  10 4  20.1  1 300 log10 0.15

 9.0667

 9.0667m 3 / s

m3 l  1000 3  0.907l / s s m

Question No 3.

Design a tube well for the following data: Yield required = 0.08m3/s Radius of circle of influence= 300m Drawdown = 5m Aquifer thickness = 30m Permeability = 60m/day

Solution:

Yield Q required = 0.08m3/s Radius of circle of influence = 300m Aquifer thickness = 30m Drawdown s = 5m Permeability K = 60m/day



60m 1d   6.944  10 4 m / s d 24  3600s

Unconfined Aquifer  Q  2.72bks log10

log10

R r

R 2.72bks  r Q

300 2.72  30  6.944  104  5   3.541 r 0.08 log10 300  log10 r  3.541 log10 r  log10 300  3.541  1.064 log10 r  1.064 log10

r  101.064

 0.086m  9cm

m

100cm 1m

Question No 4

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(b)A well pumping an unconfined aquifer having an initial saturated thickness of 8.2m is pumped at a rate of 65 L/s until a steady state cone of depression is established. The drawdowns measured at

4

(a)Derive the expression for the coefficient of permeability of an aquifer of depth H when a well of radius r is pumping at a steady rate Qo , such that the depth of water in two observation wells located at distances r1 and r2 from pumped well are h1 and h2 respectively.

two observation wells situated 15m and 31m from the pumped well are found to be 1.8m and 1.4m respectively. Determine the transmissivity of the aquifer. Observation wells

Solution: (a)

Qo

G.L W. T

H

h1

h2

ho

r1

ro

r2 The governing Equation is Darcy law given as :

Q  VA  kiA  ki 2rH  k h dh 

dh 2rh dr

Qo dr 2rk

Q  Qo  cons tan t Integrating gives

h 2 h2 2 h1



r Q ln r 2 r1 2k

Qo r 2 ln k r 1

Page

h22  h12 

5

Substituting gives

k



Qo

 h22  h12



ln

r2 r1

Now Q  65 L / s  0.065 m 3 / s

h2  8.2  1.4  6.8m h1  8.2  1.8  6.4m r1  15m k



r2  31m 0.065 2

 6.8  6.4

2



ln

31 15

k  0.003918  0.7259 k  0.00284 m/s Transimissivity = kH  0.00284  8.2 Transimissivity = 0.0233 m3/s.m

Question No 5.

In order to determine the field permeability of a field permeability of a free aquifer, pumping out test was performed and the following observations were made. Diameter of well= 20cm Discharge from well =240m3/hr Level of water surface before pumping started =240.5m Level of water at constant pumping =235.6m Level of impervious layer =210.0 m Level of water in observation well =239.8m Radial distance of observation well from tube well =50m

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Given: Diameter of well = 20cm Discharge from well =240m3/ hr Level of water surface before pumping started = 240.5m Level of water at constant pumping = 235.6m Level of impervious layer =210.0m Level of water in observation well = 239.8m Radial distance of observation well from tube well = 50 m

6

Calculate: 1. K 2. error in K if observation are not taken in observation well and radius of influence is assumed to be 300m 3. actual radius of influence based on observation from the well.

Solution:

For test Well: diameter (d) = 20cm = 0.2m radius (r) = 0.1m Q = 240 m3/ h = 240

m3 1h .  0.0667 m3/s h 60s  60s

H elevation = 240.5m H elevation = 235.6m Impervious elevation = 210m Hence H = 240.5  210  30.5m h = 235.6  210  25.6m s = H  h  4.8m For observation Well: Level of water = 239.8m

r1  50m h  239.8  210  29.8m s  H  h  30.5  29.8  0.6m Radial distance R Q 0.1m _ _ _ _ _ _ _ _ _ _ _ _ _r1_ _50_m _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _

H= 30.4m

h1

h

Impervious layer

210 m

NB: Free aquifer = unconfined





Page

r1 r k 1.36 h12  h 2 Q log 10

7

(i)

50 0.1 k 1.36 29.8 2  25.6 2 0.0667  log 10







0.18 316.44

 5.69  10  4 m/s

k  49.16 m/day m 24  3600 s  m d 1 day s

(ii) R = 300m 300 R 0.0667 log10 0.1 r  k 1.36 H 2  h 2 1.36 30.4 2  25.6 2 Q log10











0.232 365.57

 6.346  10  4 m/s

k  54.83 m/day Error 

54.83  49.16  100  11.5%

(iii)

log10 log10

49.16





R 1.36 H 2  h 2  r Q R 1.36  5.69  10  4 30.5 2  25.6 2   0.1 0.0667

log 10 R  log 10 r

 3 . 189

log10 R  3.189  log10 0.1 log10 R  3.189  1  2.189 R  10 2.189

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8

R = 154.5m

Question No 6 A 30cm diameter well penetrates vertically through an aquifer to an impermeable stratum located 18.0 m below the static water table. After a long period of pumping at a rate of 1.2 m3/min, the drawdown in test holes 11m and 35 m from the pumped well is found to be 3.05 and 1.62m respective. What is the hydraulic conductivity of the aquifer? Express in meters per day. What is its transmissivity? Express in cubic meters per day per meter. What is the drawdown in the pumped well?

Pumped well

Observatory well

Observatory well

GS

s1

s2

B =18m

r1  11m r2  35m

Given B = 18.0m; Q = 1.2 m3/min = 0.02m3/s; r1  11m;

r2  35 m; s1  3.05 m; s 2  1.62 ; d  0.3 m

2kbh2  h1  r  2.303 log 2   r1 

Page

Q

9

Confined Aquifer

But H  h  s Substitute: h1  H  s1 and h2  H  s 2 in equation (1)

Q

2kbs1  s 2  r  2.303 log 2   r1 

0.02 

k

(ii)

2k  183.05  1.62  35  2.303 log   11 

0.02  2.303  0.503 161.73

 1.4  10  4 m/s = 12.1m/day

T  kb  12.1  18  217 .8 m3/day/m

(iii) Drawdown in the pumped well Q

2kbh1  h  r  2.303 log 1   r2 

H  hs Therefore h  H  s and h1  H  s1 Hence,

Q

Q

2kbH  s1  H  s  r  2.303 log 1  r 2kbs  s1  r  2.303 log 1  r

0.0158s  3.05  0.02  2.303  1.865

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10

s  8.49 m (drawdown)

Question No 7 (A.) With the aid of a diagram, derive the unsteady well equation;

 2 h 1 h S h defining   . 2 r r T t r

the terms involved.

Solution Initialpiezometric surface

slope 

Q

h r

slope 

h  2 h  dr   r r 2

Instantaneous drawdown

Q2

Q1

h

b

dr

r Fig. 1: Showing a fully penetrated well in a confined aquifer

Annular Ring

dr r

r

•

Q2

Section

the annular cylinder is equal to the rate of change of volume of water within the annular space.

Page

From the principle of continuity of flow, the difference of rate of inflow and the rate of outflow from

11

Fig2: Showing annular ring of well

Thus;

Q1  Q2 

V t

(1)

From Darcy’s law, Q  k  i   flow area

i  slope of hydraulic gradient line For inner surface, i 

h r  h    2 h      2  dr   r   r  

But for outer surface, i  

  2h     2  dr  2 r  dr  b  r  r    h

Inflow, Q1  k 



 h   2r  b  r 

Outflow, Q 2  k 

(2)

(3)

Also, from the definition of storage coefficient, S which is define as volume of water released per unit surface area per unit change in head normal to the surface. Mathematically;

S

V 2r  dr dh

V  S 2r  dr dh

(4)

Change in volume with time will give

V h  S ( 2r ) dr t t

(5)

Where t = time since pumping started Substituting Equations (2), (3) and (5) into Equation (1) Hence;

S 2r  dr

 h   2 h   h  h   k    2  dr  2 r  dr b  kb 2r  t  r   r  r  

(6)

After expansion Equation (6) divide through by kb2  Equation (6) becomes

h 2 t   h   h dr r  dr   h r   r r 2  kb r  

S r dr

(7)

(8)

Page

S h h 2h  2h 2 r dr  dr  r 2 dr   2 dr  T t r r r

12

But T =kb

Dividing Equation (8) through by r dr

S h 1 h  2 h  2 h dr    T t r r r 2 r 2 r But

 2h and dr are small  r 2

(9)

 2h dr  is negligible r 2

Hence Equation (9) becomes

 2 h 1 h S h   r 2 r r T t

(10)

Equation (10) is the basic unsteady flow Equation towards a well. S = storage coefficient (dimensionless) T = transmissivity (m3/day/m)

h  slope of hydraulic gradient line dim ensionless  r

2h  sec ond derivative of the hydraulic gradient line dim ensionless  r 2

(B) Discuss Theis’ solution to the equation above? Theis’ solution to the Equation (10) can be achieved base on analogy between groundwater flow towards a well and flow of heat towards a sink. The well coefficients, S and T can be evaluated using Theis’ method by conducting pumping tests. The well is pumped at a constant discharge rate, Q and the change in drawdown with time in one or more observation wells is measured.

4T r2 t

  u 

Where s,

r2

T 

t

, W u  and

Q W u  4 s

u  are points corresponding to the coordinate P.

Page

S

13

The steps involve in the Theis’ method are: 1. W (u) and (u) are plotted on log-log paper using standard values from table W (u) against (u) known as type curve. 2. r2/t is computed for different values of drawdown 3. drawdown (s) is plotted against r2/t on log-log tracing paper using the same scale as W (u) against (u), this curve is known as data curve. 4. Coordinates of the axes of the two curves are kept parallel 5. The data curve is superimposed on the type curve such that most points on the curves fall on each other. 6. An arbitrary point, P called the matching point on the overlapped curves is selected. 7. Coordinates of the point P for both curves are selected 8. The values of Storativity (S) and Transmissivity (T) are calculated from the following equations :

s = drawdown (m) r = radial distance between the wells (m) t = pumping period (day)

(C.) Discuss the concept of “safe yield” from wells, emphasizing the factors affecting well yield ? The term safe yield of groundwater basin can be defined as the amount of water which can be withdrawn from a basin annually without producing an undesirable result. It is also the amount of water withdrawn continuously without affecting the supply of other landowners. Any draft in excess of safe yield is overdraft. The determination of safe yield of groundwater basin requires analysis of undesirable results which may occur if safe yield is exceeded. There are four factors governing safe yield namely: water supply, economies of pumpage, water quality and water rights. 1. Water supply is the most important and also the most subjective to qualitative determination. The water supply to a basin can be limited either by the physical size of groundwater basin or by the rate at which water moves through the basin from the recharge area the withdrawal area. 2. Economics factor may govern water supply in basins where the cost of pumping groundwater becomes excessive. Excessive pumping costs are usually associated with lowered groundwater levels, which may necessitate deepening of wells, lowering of pump bowls and installing larger pumps. 3. Water quality:- safe yield may be exceeded if a basin produces a groundwater of inferior quality. A quality limitation on safe yield depends on the minimum acceptable standard of water quality, which in turn depends on the made of the pumped water. 4. Water rights:- legal consideration may limit safe yield of a basin if there is interference with prior water rights within a basin or in adjacent basins. Any legal restriction on pumpage would have to be established before the safe yield could be determined.

Question No 8

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Determine (a) the field permeability coefficient and (b) radius of zero draw down.

14

The following observations were recorded during a pumping out test on a tube well penetrating fully in a free aquifer: Well diameter = 25cm Discharge from the well = 300 m3/hr R.L of original water surface before pumping started = 122.0 m R.L of water in well at constant pumping = 117.1 m R.L of water in the observation well = 121.3 m R.L of impervious layer = 92.0 m Radial distance of observation well from the tube well = 50 m

Solution For test well d  25 cm  0.25 m

r

0.25  0.125 m 2

Q  300 m

3

h

 300

3 m3 1h   0.083 m s h 60  60 s

H elevation  122.0 m helevation  117.1 m Im pervious elevation  92.0 m 

H  122.0  92.0  30 m ;

h  117.1  92 .0  25.1 m

For observation well Level of water = 121.3 m

h1  121.3  92.0  29.3m

r1  50 m S = H – h = 30.0 – 29.3 = 0.7m

Q = 0.083 m3/s

Radial distance, R

r = 0.125m

r1=50m

H=30m h1=29.3m

h=25.1m

92.0m

Impervious layer

Free aquifer = unconfined aquifer









Page

r   50  Q log 1  0.083 log  r 0.125     6.95  10  4 m k  s 1.36 h12  h 2 1.36 29.3 2  25.12

15

(a) Field permeability coefficient (k) is given as:

k  6.95  10  4

m 60  60  24s   60.05 m day 1day s

(b) Radius of zero drawdown = Radius of influence (R)

Q



1.36 k H 2  h 2 R log  r

0.083 

 

1.36  6.95  10 4 30 2  25.12  R  log   0.125 





4 2 2  R  1.36  6.95  10 30  25.1 log  0.083  0.125 



 R  log10    3.075  0.125  R  103.075 0.125 R  148.6m

 R  log   3.075  0.125 

log R  log 0.125  3.075 log R  3.075  log 0.125 log R  3.066  0.9031 log R  2.172 R  10 2.172  148 .6 m Hence the radius of zero drawdown, R = 148.6 m

Question No 9 An extensive horizontal confined aquifer having a constant thickness of 40m is pumped steadily by a series of identical wells located at 300m intervals in a N-S line. Each well completely penetrates the aquifer and delivers 20 Ls-1. Field measurements a long way to the east of the well line shows a hydraulic gradient of 1 in 175 and similarly to the west of the wells a hydraulic gradient of 1 in 500. Both piezometric surfaces slope in westerly direction. Estimate the Darcy hydraulic conductivity of the aquifer.

Solution B = 40m Q = 20 Ls-1 =

3 20  0.02m s 1000

dh dr

Page

i = hydraulic gradient =

16

Distance between wells = r =300m

1  0.0057 in eastern direction 175 1  0.02 in western direction i2 = 1 in 500 = 500 i1 = 1 in175 =

N E

W

Q =0.02 m3/s r1=300m

S

i1 =0.057 i2 = 0.02

b = 40m

The basic equation for confined aquifer is given as:

Impervious stratum

dh dr dh  hydraulic gradient line Where i  dr 0.02  2    k  40  300  0.0057 Q  2 k b r

Note that i1 is used because the Darcy hydraulic conductivity is required for western direction and

i2 is already in western direction. 0.02 2    40  300  0.0057 k  4.65  10 5 ms 1 k  4.65  10 5  60  60  24 k  4.02 m day k

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17

Hence, Darcy hydraulic conductivity of the aquifer is k  4.02 md 1

Question No 10 (A) A well penetrates fully a 10m thick water bearing stratum of medium sand having a coefficient of permeability of 0.005m/s. The well radius is 10cm and is to be worked under a drawdown of 4m at the well face. Calculate the discharge from the well. What will be the % increase in discharge if the radius of the well is doubled? Take R = 300 m in each case. (B) An artesian tube well has a diameter of 20cm. The thickness of aquifer is 30m and its permeability is 36 m/day. Find its yield under a drawdown of 4 m at the well face. Use: (i) Radius of influence as recommended by Sichardt (ii) Assume a radius of influence of 300m (iii) Comment on the difference between the values of R estimated in (i) and (ii) (C) A tube well penetrates fully a 8m thick water bearing stratum (confined) of medium sand having a coefficient of permeability of 0.004m/s. The well radius is 15cm and is to be worked under a drawdown of 3m at the well face. Calculate the discharge from the well. What will be the % increase in the discharge if the radius of the well is doubled? Take radius of zero drawdown to be 400m in each case.

Solution 10A The aquifer is confined Q

2.72 b k s R log10 r

b = 10 m; Q

k = 5 x 10-3 m3/s;

2.72  10  5  10 3  4 ; 300 log10 0.1

R = 300 m; r = 10 cm = 0.1m; s = 4 m

Q = 0.156 m3/s

When the radius is doubled, r = 20cm = 0.2m 2.72  10  5  10 3  4 ; 300 log10 0. 2

0.173  0.156 = 11 % 0.156

Comment: The size of the well has very little influence on the discharge. It can be seen that discharge only increased by a mere 11 % despite doubling of radius.

18

% increase =

Q = 0.173 m3/s

Page

Q

Solution 10B Q

2.72 b k s R log10 r

Radius of well as recommended by Sichardt, R is given as R  3000 r k ; R=

s = 4 m;

k = 36 m/day;

b = 30 m

d  10 cm  0.1 m 2

R  3000  4 

36  245 m 24  3600

(i) Q

36  4 24  3600 ; 245 log10 0.1

2.72  30 

Q = 0.040 m3/s = 40 L/s

(ii) Using R = 300 m

Q

36  4 24  3600 ; 300 log10 0.1

2.72  30 

Q = 0.039 m3/s = 39 L/s

Comment: No significant difference between the values of discharge using Sichardt formula with the assumption R = 300 m.

Solution 10C The aquifer is confined 2.72 b k s ;b = 8m; k = 0.004 m/s; R log r 2.72  8  0.004  3 Q1  400 log 10 0.15 Q

r = 15 cm = 0.15 m;

R = 400m;

s = 3 m; Q = ?

Q1 = 0.0762 m3/s = 274 m3/hr If radius of well is doubled, r = 30 cm = 0.3 m

Q2 = 0.0836 m3/s = 301 m3/hr; % increase = 27  100  9.8 % 274

Increase in discharge = 301 – 274 = 27 m3/hr

19

2.72  8  0.004  3 400 log10 0.3

Page

Q2 

Page

20

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Dr Olayinka Okeola (email: [email protected]; [email protected]; mobile no: +234 (0) 703 230 7770