Tutorial Solution Chapter 7 Week 6 PDF

Preview

Summary

Tutorial 5 solution for Well Hydraulics (2) Tutorial 7. The original well in a confined aquifer is pumped at a rate of 220 gal/min. Measurement of drawdown in two observation wells shows that after 1270 min of pumping, no further drawdown is occurring. Well 1 is 26 ft from the pumping well and has a...

Description

Tutorial 5 solution for Well Hydraulics (2) Tutorial 7.1. The original well in a confined aquifer is pumped at a rate of 220 gal/min. Measurement of drawdown in two observation wells shows that after 1270 min of pumping, no further drawdown is occurring. Well 1 is 26 ft from the pumping well and has a head of 29.34 ft above the top of the aquifer. Well 2 is 73ft from the pumping well and has ahead of 32.56 ft above the top of the aquifer. Use the Thiem equation to find the aquifer transmissivity. We must first convert the pumping rate of 220 gal/min to an equivalent rate in cubic feet per day. We make this conversion, even though steady-state conditions were reached, before a full day (1440 min) of pumping occurred. (Solution) 1 ft3=7.488 gal, 1 gal = 0.134 ft3

Tutorial 7.2. A fully penetrating well discharge 75 gpm from unconfined aquifer. The original water table was recorded as 35 ft. After a long time period, the water table was recorded as 30 ft (AMSL, above mean sea level) in an observation well located 75ft away and 34 ft AMSL at an observation well located 2000ft away. Determine the hydraulic conductivity of this aquifer in feet per second. (Solution) 1 gal = 0.134 ft3

K

 r2  75gpm(0.134ft 3 /gal)(1min/60sec) 2000ft ln ln   75ft   b22  b12   r1  π  342 ft 2 -302 ft 2  Q

 6.8  104 ft / sec

1

Tutorial 5 solution for Well Hydraulics (2)

Tutorial 7.3. A well in a confined aquifer was pumped at a rate of 220 gaI/min (42400ft3/d) for 500 min. The aquifer is 48 ft thick. Time-drawdown data from an observation well located 824 ft away are given in Table. Find Transmissivity (T), hydraulic conductivity (K), and Determine the transmissivity (T) and the storage coefficient (S)

(Solution) The field data are plotted on logarithmic paper (Figure 7.3). The field-data graph is then placed upon the type curve graph (Figure 7.4). The following match point is obtained: W(u) =1, 1/u=1,

h0  h =2.4ft, t=4.1 min

2

Tutorial 5 solution for Well Hydraulics (2)

3

Tutorial 5 solution for Well Hydraulics (2) Tutorial 7.4. Evaluate the pumping test data of Table in Tutorial 7.3 by the CooperJacob method. The field data are plotted on semilogarithmic paper (Figure 7.5). A straight line is fit to the later time data and extended back to the zero-drawdown axis. The value of to is 5.2min, and the drawdown per log cycle of time is 5.5ft. Tips: The time used in the time-drawdown plot is often in minutes and it must be converted to days before it is used in Equation 5.19. (Solution) 1 gal = 0.134 ft3 Pumping rate Q =220 gal/min x 0.134 ft3/gal x 1440 min/day = 42400 ft3/d

Discussion: In comparing the Cooper-Jacob solution (S= 1.7x10-5) with the Theis solution (S= 2.4x10-5) in the preceding problems, we can see that the resulting answers are almost the same. As these are graphical methods of solution, there will often be a slight variation in the answers, depending upon the accuracy of the graph construction and subjective judgments in matching field data to type curves. 4

Tutorial 5 solution for Well Hydraulics (2) Tutorial 7.5 A pumping test (Q = 120 gpm=23155 ft3/d) was conducted on a confined aquifer (aquifer thickness = 30.0 ft). The time-drawdown data at a distance 150 ft away from the well were collected and are shown in the following table. Use the Cooper–Jacob straight-line method to determine the hydraulic conductivity and storativity of the aquifer. (Analysis of pumping test data using Cooper-Jacob’s straight-line method) Time since pumping started (min)

Drawdown, s (ft)

7 20 80 200

0.15 0.45 0.90 1.16

(Solution) t0

(h0  h )  0.7 ft

1 gal = 0.134 ft3 Pumping rate Q =120 gal/min x 0.134 ft3/gal x 1440 min/day = 23155 ft3/d Given: Q=23155 ft3/d, From plot, to=4.5 min,   h0  h = 0.7 ft Calculate transmissivity

2.3 23155 ft / d  2.3Q 2 T   6065 ft / day 4   h0  h  4 3.14 (0.7) 3

Hydraulic conductivity K 

2.3Q 6065 ft 2 / day   202 ft / d 4  h0  h  30 ft

Calculate storage coefficient S

5

Tutorial 5 solution for Well Hydraulics (2) 2 2.25Tt0 2.25 6065 ft / d  4.5min 1440 min/ d   150 2 r2 42.6   1.89  10 3 (no unit) 1502

S

6

Tutorial 5 solution for Well Hydraulics (2) Tutorial 7.6. A well pumping at 77000ft3/d has observation wells located 10, 40, 150, 300, and 400ft away. After 0.14day of pumping, the following drawdowns were observed: Distance (ft) 10 40 150 300 400

Drawdown (ft) 15.1 9.4 4.4 1.7 0.25

Determine the transmissivity (T) and the storage coefficient (S). (Solution)

1. Plot the distance-drawdown data on semi-log paper as shown in the following plot;

2. Draw a best-fit straight line over the observed data and extend the line to the X axis. 3. Read the drawdown per log cycle. The drawdown per log cycle is 8.8ft and r0 is 460 ft

.

7

Tutorial 5 solution for Well Hydraulics (2) Tutorial 7.7. Determine the storage coefficient for an artesian aquifer from the pump test results shown in the table below. The measurements were made at an observation well 100 m away from the pumping well. The pumping rate was 0.0221m3/s.

Time since pumping started (min)

Drawdown, (m)

10 100 1440

1.35 3.65 6.30

(Solution) Given: Q= 0.021 m3/s; r=100m Calculate transmissivity

2.3  0.0210 m / s  2.3 Q 3 2 T   1.766  10 m / s 4 h0  h  4 3.14 (6.30  1.35) 3

Or Calculate transmissivity using the first and last times

T

t  Q ln  2  4  h2  h1   t1 

 0.0221m / s  3



1400 min  2 3 ln    1.766  10 m / s 4 3.14 (6.30  1.35)  10 min 

From plot, to=2.5 min Calculate storage coefficient

3 2 2.25Tt0 2.251.766 10 m / s   2.5m in  60sec/ min  S  2 r2 100 0.596 4   1.656  10 (no unit) 4 1.0 10

t0  2.5 min

  h 0  h   drawdown  (6.30 1.30) 4.95m

8

Tutorial 5 solution for Well Hydraulics (2) Tutorial 7.8 Fully penetrating well in a 28.0m thick artesian confined aquifer pumps at a rate of 0.0038m3/s for 1941days and cause a drawdown of 64.5m at an observation well 48m away from the pumping well. The original piezometric surface was 95.5m above the bottom confining aquifer. The hydraulic conductivity is 5.8 x 10-7m/s. How much drawdown will occur at an observation well 68m away? (Hint: Thiem equation for the confined aquifer) (Solution) Calculate hydraulic head h1 h1 =95.50 - 64.50 = 30 m Calculate Transmissivity T=Kb=(5.8 x 10-7 m/s) * (28 m)= 1.624 x 10-5 m2/s

T  Kb 

r  Q ln  2   2 (h2  h1 )  r1 

2 T (h 2  h1 ) 2(3.14)(1.624 10 m /s )(h 2  30m ) , 0.00380 m3 / s  r   68  ln   ln  2   48   r1  5

Q

2

Solve for hydraulic head h2

 68  (0.00380 m 3 / s) ln    48  h2   30  12.97  30  42.97m h2 5 2(3.14)(1.624 10 m 2 / s ) Compute the drawdown at observation well 68 m away: s2 = 95.50 – 42.97 = 51.08 m

9

Tutorial 5 solution for Well Hydraulics (2) Tutorial 7.9 A 0.5m diameter fully penetrates an unconfined aquifer which is 30m thick. The drawdown at the pumped well is 10.0m and the hydraulic conductivity of the gravel aquifer is 6.4 x 10-3 m/s. If the flow is steady and the discharge is 0.014m3/s, determine the drawdown at a site 100 m from the well. (Solution) Calculate hydraulic head b1 b1 =30 – 10 = 20 m r1=0.5/2= 0.25 m

K Q

r Q ln  2  : Thiem equation for an unconfined aquifer 2  b  b1   r1  2 2

 K  b22  b12  r  ln  2   r1 

0.014m / s  3

,

(3.14)(6.4  103 m / s )  b22  202   100  ln    0.25 

Solve for b2

b

2 2

 400 

(0.014 m 3 / s)(5.99) (3.14)(6.4  103 m / s)

b2  4.17  400  20.10m Compute the drawdown at observation well 100 m away: s2 = 30 - 20.10 = 9.90 m

10

Tutorial 5 solution for Well Hydraulics (2) Tutorial 7.10 An artesian confined aquifer 5m thick with piezometric surface 65m above the bottom confining layer is being pumped by a fully penetrating well. The hydraulic conductivity is 6.1 x 10-4m/s. A steady state drawdown of 7m is observed at an observation well located 10m away. If the pumping rate is 0.02m3/s, how far away is a second observation well that has an observed drawdown of 2m? (Solution) Calculate hydraulic head h1 and hydraulic head h2 h1 = 65 - 2.0 = 63 m h2 = 65 – 7 = 58 m Calculate Transmissivity T=Kb=(6.1 x 10-4 m/s) * (5 m)= 0.00305 m2/s Eq. (5.4), T  Kb 

Q

Q r  ln  2  2 ( h2  h1)  r1 

3 2 2 T (h2  h1 ) 2(3.14)(3.05 10 m / s )(63m  58m ) , 0.02m 3 / s  r  r  ln  2  ln  2   10   r1 

Solve for r2   r  2(3.14)(3.05 10 m / s )(63m  58m ) ln  2   0.02m3 / s  10  3

2

Taking exponential of both sides

 2(3.14)(3.05 103 m2 / s)(63 m  58 m)  r2  exp   10 0.02m3 / s   r2  exp  4.79  120.41 10 r2  10.0 120.41  1204.1m

11

Tutorial 5 solution for Well Hydraulics (2) Tutorial 7.11 Estimate the distance to be expected of a drawdown of 4.81m from a pumping well under the following conditions: Pumping rate = 0.028m3/s Pumping time = 1066day Drawdown in observation well = 9.52m Observation well is located 10m from the pumping well The hydraulic conductivity = 1.5 x 10-4m/s Assume that the well is fully penetrating in an unconfined aquifer. The original piezometric surface was 14.05m above the bottom aquifer.

(Solution) Calculate hydraulic head b1 and hydraulic head b2 b1 =14.05 – 9.52 = 4.53 m b2 = 14.05 – 4.81 = 9.24 m

K Q

r  ln  2  : Thiem equation for an unconfined aquifer, Eq. (7.19)   b  b   r1  Q

2 2

2 1

 K b22 b12  r  ln  2   r1 

(3.14)(1.5 10  m / s )  9.24  4.53 4

,

0.028m / s  3

2

2



r  ln  2  10 

Solve for r2 2 2 4  r  (3.14)(1.5 10 m / s)(85.38 m  20.52 m ) ln  2   0.02 m3 / s 10 

Taking exponential of both sides

 (3.14)(1.5 10 4 m / s )(85.38m 2  20.52m 2 )   r2   exp      exp 1.0916   2.979 3 0.02 m / s  10    r2  10.0  2.979  29.79m

12

Tutorial 5 solution for Well Hydraulics (2) Tutorial 7.12 An aquifer yields the flowing results from pumping a 0.61m diameter well at 0.0303m3/s: Drawdown = 0.98m in 8 minute Drawdown = 3.87m in 24 hour Determine its transmissivity.

(Solution) From eq. (5.19), drawdown   h2  h1  , T 

t  Q ln  2  4 h2  h1   t1 

 t2  Q 0.303 m 3 / s  (24 h)(60 min/ h)  T ln    ln   4  h2  h1   t1  4(3.14)(3.87 m  0.98 m)  8min  T

0.303m3 / s ln 180   0.000834  (5.1929)  4.33 10 3 m2 / s 36.3168

Tutorial 7.13 An aquifer yields a drawdown of 1.04m at an observation well 96.93 m from a well pumping at 0.0170m3/s after 80min of pumping. The virtual time is 0.6min and the transmissivity is 5.39 X 10-3m2/s. (1) Determine the storage coefficient (2) Find the drawdown at the observation well 80days after pumping begins. (Solution) 3 2 2.25Tt0 2.25 5.39 10 m / s (0.6 min)(60 s/ min) S   (93.93) 2 r2

4.69  10   4.905 10 5 (no units) 3 9.395 10 4

S From Eq. (5.19)

T

t  t  Q Q ln  2  ,   h2  h1   ln 2  4  h2  h1   t1  4 T  t 1 

h2 1.04 m 

0.0170 m3 / s  (80 d)(1440 min/ d)  ln   3 2 4(3.14)(5.39 10 m / s)  (80 min) 

3

0.0170m / s ln 1440   1.04 0.0677 drawdown h2  0.251  7.27   1.04  2.87m drawdown h2 

13...