sm_ch06

Water Resources Engineering

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6 – 1 6.1.1. ' L = 150 m, h 1 = 55 m, h 2 = 48.5 m, t = 32 hrs, D = 24% = 0.24, T = 15 o C. Using Darcy’s law, L h h KA L h L h Q 2 1 ' ± ' ' d d K 0433 . 0 150 48.5 55 K q A Q ± Velocity of flow, 6875 . 4 32 150 t L V ' m/hr = 1.30 x 10 -3 m/s. Using the continuity equation, Q = Aq = A pore V where A pore is the area of cross-section of the pore spaces 1.125 5) 0.24(4.687 V V A A q pore D m/hr Hence, 0.0433K = 1.125 K = 25.96 m/hr = 7.21 x 10 -3 m/s = 7.21 mm/s. Table 6.1.1 shows that the aquifer is most probably clean sand and less likely gravel or silty sand. The solution is verified by checking the Reynolds number for this problem. Q VD R e At 15 o C, Q = 1.141 x 10 -6 m 2 /s (for water) The range of grain sizes for clean sand is 0.05 mm – 1 mm Hence, 53 . 0 2 1 0.05 D ² mm (if the arithmetic mean diameter is taken) D = 22 . 0 ) 1 ( 05 . 0 mm (if the geometric diameter is taken) 60 . 0 10 x 141 . 1 ) 10 x 53 . 0 ( 10 x 30 . 1 R 6 - -3 -3 e (for the arithmetic mean diameter) 25 . 0 10 x 141 . 1 ) 10 x 22 . 0 ( 10 x 30 . 1 R 6 - -3 -3 e (for the geometric mean diameter) In either case, R e < 1. Hence, Darcy’s law holds true and the solution procedure is correct. 6.2.1. For steady state condition (equation (6.2.12)), 0 r h r r r 1 ¸ ¹ · ¨ © § w w w w That is, 0 r 1 or 0 r h r r ¸ ¹ · ¨ © § w w w w Since the first statement can not be true, 0 r h r r ¸ ¹ · ¨ © § w w w w 1 C r h r w w
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6 – 2 h C r 1 h 1 w w h = C 1 ln r + C 2 6.2.2. K 1 = 30 m/day = 3.47 x 10 -4 m/s, L 1 = 500 m K 2 = 10 m/day = 1.16 x 10 -4 m/s, L 2 = 1100 m K 3 = 50 m/day = 5.79 x 10 -4 m/s, L 3 = 400 m b = 50 m (for all aquifers) i) ' h = ' h 1 + ' h 2 + ' h 3 = 66.4 – 60.6 = 5.8 m (i) Q 1 = Q 2 = Q 3 = Q 1 1 1 1 L h b K Q ' (considering 1 m width of the channel) = 3.47 x 10 -4 (50) 500 h 1 ' = 3.47 x 10 -5 ' h 1 Similarly, 1100 h ) 50 ( 10 x 16 . 1 L h b K Q 2 4 - 2 2 2 2 ' ' = 5.27 x 10 -6 ' h 2 400 h ) 50 ( 10 x 79 . 5 L h b K Q 3 4 - 3 3 3 3 ' ' = 7.24 x 10 -5 ' h 3 Thus, 3.47 x 10 -5 ' h 1 = 5.27 x 10 -6 ' h 2 = 7.24 x 10 -5 ' h 3 6.58 ' h 1 = ' h 2 = 13.74 ' h 3 That is, ' h 1 = 2.09 ' h 3 , ' h 2 = 13.74 ' h 3 Using the component head losses in the equation for ' h 1 and ' h 2 , 2.09 ' h 3 + 13.74 ' h 3 + Dh 3 = 5.8 16.83 ' h 3 = 5.8 ' h 3 = 0.34 m. Q = 7.24 x 10 -5 (0.34) = 2.46 x 10 -5 m 3 /s (per meter width) = 2.13 m 3 /day per meter width ' h 1 = 2.09(0.34) = 0.71 m ' h 2 = 5.8 – 0.71 – 0.34 = 4.75 m ii) If the head losses in each aquifer were to be equal, ' h 1 = ' h 2 = ' h 3 = h c ' 3 3 2 2 1 1 L h b K L h b K L h b K Q c ' c ' c ' 3 3 2 2 1 1 L K L K L K 3 -4 2 -4 1 -4 L 10 x 5.79 L 10 x 1.16 L 10 x 3.47 L 1 = 2.99L 2 , L 3 = 4.99L 2 Also, L 1 + L 2 + L 3 = 500 + 1100 + 400 2.99L 2 + L 2 + 4.99L 3 = 2000 L 2 = 223 m
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6 – 3 L 1 = 2.99(223) = 667 m L 3 = 2000 – 223 – 667 = 1110 m 6.2.3. K 1 = 30 m/day = 3.47 x 10 -4 m/s, K 2 = 6.94 x 10 -4 m/s Total flow out of the aquifer: Q = Q 1 + Q 2 ±² ± ² ± ² > @ ) 900 ( 2 50 75 50 82 10 x 47 . 3 2L h h K Q 2 2 -4 2 2 o 1 1 ³ ³ ³ ³ = 7.69 x 10 -5 m 3 /s (per meter width of the aquifer) = 6.65 m 3 /day ± ² 900 ) 75 82 )( 50 ( 10 x 94 . 6 L h h b K Q -4 2 1 2 2 ³ ³ = 2.70 x 10 -4 m 3 /s (per meter width of the aquifer) = 23.33 m 3 /day Q = 6.65 + 23.33 = 29.98 m 3 /day 6.3.1. K = 2.22 x 10 3 gpd/ft 2 (10 -3 m/s) (from the solution to example 6.3.1) h o = 10 ft = 3.05 m, h = 2 ft = 0.61 m, x = 30 ft = 9.14 m Using the Dupuit equation (equation (6.3.7)), 2 2 o h h 2X K Q ³ 2 2 3 2 10 2(30) 10 x 2.22 Q ³ = 3552 gpd/ft (in US customary units) >@ 2 2 -3 ) 61 . 0 ( ) 05 . 3 ( 2(9.14) 10 Q ³ = 4.89 x 10 -4 m 3 /s/m (in SI units) 6.3.2. For confined aquifers, L h h Kb Q 2 1 1 ³ (per unit width of the aquifer) For confined layers, 2L h h K Q 2 2 2 1 2 ³ (per meter width of the aquifer) Hence, a) if Q 1 = Q 2 , L h h 2 1 ³ = 2L h h K 2 2 2 1 ³ 20 2 16 24 2 h h b 2 1 ´ ´ m.
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This document was uploaded on 05/19/2011.

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sm_ch06 - 61 6.1.1. L = 150 m, h1 = 55 m, h2 = 48.5 m, t =...

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