Question

1. (15 points) Water flows through three confined aquifers in series as shown in the figure below. For piezometric heads in the observation wells of 66.4 m and 60.6 m, determine the flow rate per unit width of the aquifer and the headlosses in each component of the aquifers between the observation wells. 66.4 m 60.6 m Ky = 30 m/day K2 = 10 m/day K3 = 50 m/day 500 m + 1100 m- 1100m — + 400m+

Answer 1

Solution: (3) Determine the flow rate per unit width of the aquifer 9,- K, H2-lt, Here q, is the flow sate per unit width of the aquifer! ki is the coefficient of permeability of the aquifert, f, is the piezo metric head in the first observation well, Hais the piezo metric head in the second observation well, and to is the thickness of the aquifer 1. - Substitute coml day for ka, 66.um for Hi soom for Li and som forte 9,= F, H2-Hit 5 30 12 -66.47 soo 50 Determine 91 : 342-1992 +0 the flow rate per unit width of the aquitor 2. qa's K2 H3-H2 t. L2 here a, is the How rate per unit width of the aquifer 2 Kz 98 the coefficient of permeability of the aquifier 2 . He is the piezo metric head in the second observation well, H3 is the piero me trec head in the third observation well, and t is the thickness of the aquifier 2.

Substitute rom/ day for 12, 110om for L. and son for to 92. Iz 13-112 te :10("37 1 so Determine the flow rate for uni 92: 0.05, -0.45H2 the flow rate for unit widto of the aquifiers 90 k3 f4-43 at Here 93 18 the flow rate per unit width of the aquifter 3 kzisthe coefficient of permeability of the aquifier 3, Hy is the piezo metric head in the fourth observation well, H3 is the piezo metric head in the third observation well, and t is the thickness of the aquifier 3 substitute 50m/day for K3, 4oom for L3 and som 93=43 Hu -H3 .. om fort. .. 50bebom * * ) so the series are equal 93 = 378.95 . 6.25H3 > According to darcy's law the flow in 9,39,393 slove the equation (1) and (2)

3+2 -199.2 - 318.75 -6.25H3 3H2 + 6.25H3 : 57.95 - slove the equations 6 and ② 342-199.2 :0.45H3-0.45 H2 3.45412 -0.45H3 = 199.2 -> slove the equations and the obtained values for H2 is 65 65.689 m and Hs 12 60.942m Determine the head losses between the first and second observation . wells. DH,: Ha-H, Here, DH, is the head loss between the first and second observation wells. Substitute 65.689m for H₂ and 66.4 for hi SHisha - Hi 65.689 -66.4 -0.71104 show the negative sign to indicate lowering of the head. therefore, head loss of the first component is Grallm | Determine the head losses between the third and second Observation wells, third and SH2 = H3-t2 Here DH2 is the head loss between the Second observation wells. Substitute 65.689m for H2 and 60,942m for H2

OH, Hp - H2 60.942-65.6% -4.147 m ate lowering of the head show the negative sign to indicate lowering of the therefore, head loss of the second compontent is 4.440ml Belermine the head losses b/w the third and fourth observation wells DH3 = 44 H3 Here Allsed the head loss blw the fourth and third observation wells substitute Go.bm for Hy and 60.942 m for fl3 DH₃ = Hlu-H3 60.6–60.942 -0-342 m show the negative sign to indicate lowering of the head therefore, head loss of the third component is [0.342 m Determine the flow rate per unit width of the aquifter 9:9, ok, H2-HI substitute 3om/day for to 66.um for H, 65,689 m for H2 5oom to 50om for L, and som for t. 9K, H2-Hit • 3066.4 -65.6897 so soos so = 2.133 m3 1 day I m. 500

there for the flow sate per unit width of the aquifier 18 - 12.133 m² / day Im head losses in each aquifier blw the wells were to be equal. She : DH2 EDH3 Compute respective lengths of each component of the aquifier use the condition for total length of the aquifier 11 +19+13 - 20000 m - Take first condition as 9, = 92 -kiSHI 12 DH₂ take 2 OH, = DH, = DHz substitute 30ml day for ke and com/ day for ka. 30 319: , take second condition as 92=93 -K2 DH2 -K3 DH3 جانا - AH2 = SH₂ 2.17 Substitu e the soml day for t3 and com/day for K2. 512= 13.

2000 substitute equation and into equate 312 +62 ts la = 2000 912= 2000 12 : 2000 2 = 222, 22m therefore, the length of the first second and third apper fpers are 1666.66m, 1222.22.77 and (111llm respectively.