{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

sm_ch06

# Water Resources Engineering

This preview shows pages 1–4. Sign up to view the full content.

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 KA L h KA 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

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
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
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 ³ ´ ³ ´ ³ ´ > @ ) 900 ( 2 50 75 50 82 10 x 47 . 3 2L h h K Q 2 2 -4 2 2 o 1 1 ± ± ± ± = 7.69 x 10 -5 m 3 /s (per meter width of the aquifer) = 6.65 m 3 /day ³ ´ 900 ) 75 82 )( 50 ( 10 x 94 . 6 L h h b K Q -4 2 1 2 2 ± ± = 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)), ³ ´ 2 2 o h h 2X K Q ± ³ ´ 2 2 3 2 10 2(30) 10 x 2.22 Q ± = 3552 gpd/ft (in US customary units) > @ 2 2 -3 ) 61 . 0 ( ) 05 . 3 ( 2(9.14) 10 Q ± = 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 ± (per unit width of the aquifer) For confined layers, 2L h h K Q 2 2 2 1 2 ± (per meter width of the aquifer) Hence, a) if Q 1 = Q 2 , L h h Kb 2 1 ± = 2L h h K 2 2 2 1 ± 20 2 16 24 2 h h b 2 1 ² ² m.

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

### Page1 / 15

sm_ch06 - 61 6.1.1 L = 150 m h1 = 55 m h2 = 48.5 m t = 32...

This preview shows document pages 1 - 4. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online