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CE 4603: Water Resources Engineering — David Wagner 2007/04/28 16:33

Homework #10 Due 4/30

Solve problems 6.22, 6.28, 6.34, and 6.36 of Water Resources Engineering by David Chin, 2nd edition, pages 760-762.

6.22

The hydraulic conductivity distribution in a 20-m thick stratified surficial aquifer is given by

Depth (m)Kxx (m/d)Kyy (m/d)Δzi
0-2 5 15 2
2-5 7 20 3
5-7 9 21 2
7-11 14 12 4
11-15 11 17 4
15-19 6 9 4
19-20 2 5 1

Determine the effective hydraulic conductivity when the water table is 2 m below the ground surface. Contrast this with the result when the water table is 3 m below the ground surface. [See Exmaple 6.8, page 639-640]

overline{K}_d={1/h}sum{i=1}{n}{K^i_d Delta z_i}; d ∈ (xx,yy)

overline{K}_xx(h=18)={1/h}sum{i=1}{n}{K^i_xx Delta z_i} = {1/18}(7*3+9*2+14*4+11*4+6*4+2*1) = {21+18+56+44+24+2}/18 = 165/18

= 9.17 m/d

overline{K}_yy(h=18)={1/h}sum{i=1}{n}{K^i_yy Delta z_i} = {1/18}(20*3+21*2+12*4+17*4+9*4+5*1) = {60+42+48+68+36+5}/18 = 259/18

= 14.4 m/d

overline{K}_xx(h=17)={1/h}sum{i=1}{n}{K^i_xx Delta z_i} = {1/17}(7*2+9*2+14*4+11*4+6*4+2*1) = {14+18+56+44+24+2}/17 = 158/17

=9.29 m/d

overline{K}_yy(h=17)={1/h}sum{i=1}{n}{K^i_yy Delta z_i} = {1/17}(20*2+21*2+12*4+17*4+9*4+5*1) = {40+42+48+68+36+5}/17 = 239/17

=14.1 m/d

6.28

Consider a two-layer stratified aquifer between two reservoirs. The water surfaces in the reservoirs are at elevations 5 and 4 m NGVD, respectively; the ground surface between the aquifers is at elevation 10 m NGVD; the top layer of the aquifer extends from the ground surface down to -10 m NGVD, and the base of the aquifer (and reservoirs) is at -20 m NGVD. The hydraulic conductivity of the top layer is 50 m/d and that of the bottom layer is 100 m/d. If the reservoirs are 2 km apart, find the equation of the phreatic surface, and the flowrate between the reservoirs. Neglect surface recharge. [See 6.3.1, page 648-650, and example 6.13, page 650.]

L = 2000 m; hL = 25 m; hR = 24 m

b1 = 20 m; K1 = 50 m/d; b2 = 10 m; K2 = 100 m/d;

C_1 = {K_1/{2L}}(h^2_R-h^2_L) + (K_2 - K_1){b_2/L}(h_R-h_L) = {50/{2*2000}}*(24^2 - h25^2) + (100 - 50)*{10/2000}*(24-25) = {50/4000}*(-49)+(50*0.005*(-1)) = -0.6125 -0.25

=-0.8625

C_2 = {K_1/2}h^2_L + (K_2 - K_1) b_2 h_L = {50/2}*25^2 + (100 - 50)*10*25 = 25*625 + 50*10*25 = 15625+12500=28125

{K_1/2}h^2 + (K_2 - K_1) b_2 h = C_1 x + C_2 = {50/2}*h^2 + (100 - 50)*10*h = C_1 x + C_2 = 28125-0.8625x =25h^2 + 500h

h^2 + 20h = -0.0345x + 1125Q=-C1=0.8625 m²/d

6.34

A well pumps at 0.4 m³/s from a confined aquifer whose thickness is 24 m. If the drawdown 50 m from the well is 1 m and the drawdown 100 m from the well is 0.5 m, then calculate the hydraulic conductivity and transmissivity of the aquifer. Do you expect the drawdowns at 50 m and 100 m from the well to approach a steady state? Explain your answer. If the radius of the pumping well is 0.5 m and the drawdown at the pumping well is measured to be 4 m, then calculate the radial distance to where the drawdown is equal to zero. Why is the steady-state drawdown equation not valid beyond this distance? [See example 6.14, page 654.]

b=24 m ; Qw = 0.4 m³/s (3600*24) s/d= 34560 m³/d ; r1 = 50 m ; s1 = 1 m; r2 = 100 m ; s2 = 0.5 m

T={Q_w/{2 pi (s_1 - s_2)}} ln(r_2/r_1)={34560/{2*pi*(1 - 0.5)}}*ln(100/50)=11000*0.69315=

T=7625 m²/d

K={T/b}=7625/24=

K=317.7 m/d

Steady state: Yes, this is an assumption.

R≈3000 s_w sqrt{K} = 3000*4*sqrt{317.7}=

R=214000 m

Heck, the assumption of homogeneity out to 214 km is extremely unlikely, but I suppose this question is looking for φ(r > R) > H implies the steady-state equation is invalid.

6.36

Repeat Problem 6.34 for an unconfined aquifer, assuming that the saturated thickness is 24 m prior to pumping. If actual drawdowns, rather than the modified drawdowns, are used to calculate the transmissivity of the aquifer, then what would be the percentage difference in the calculated transmissivity? [See example 6.15, page 657.]

H=24 m

s^prime_1=s_1-{s^2_1/{2H}}=1-{{1^2}/{2*24}}=1-{1/48}=0.979 s^prime_2=s_2-{s^2_2/{2H}}=0.5-{{0.5^2}/{2*24}}=0.5-{0.25/48}=0.495

T={Q_w/{2 pi (s^prime_1 - s^prime_2)}} ln(r_2/r_1)={34560/{2*pi*(0.979 - 0.495)}}*ln(100/50)=11360*0.69315=

T=7877 m²/d

{7625-7877}/7877=

-3.2%

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